GATE
ELECTRONICS & COMMUNICATION
Topicwise Solved Paper
Year 2013- 1996
By RK Kanodia & Ashish Murolia
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 1
UNIT 1
(C) x
(D) 1
2012
ENGINEERING MATHEMATICS
1.8
TWO MARKS
d 2 y (t)
dy (t)
Consider the differential equation
+2
+ y (t) = d (t)
2
dt
dt
dy
with y (t) t = 0 =- 2 and
=0
dt t = 0
dy
The numerical value of
is
dt t = 0
(A) - 2
(B) - 1
(C) 0
(D) 1
-
2013
1.1
1.2
1.3
ONE MARK
1.9
The minimum eigen value of the following matrix is
R3 5 2V
S
W
S5 12 7W
SS2 7 5WW
T
X
(A) 0
(B) 1
(D) 3
(C) 2
1.10
2013
1.4
Let A be an m # n matrix and B an n # m matrix. It is given that
determinant ^Im + AB h = determinant ^In + BAh, where Ik is the
k # k identity matrix. Using the above property, the determinant of
the matrix given below is
V
R
S2 1 1 1W
S1 2 1 1W
S1 1 2 1W
W
S
S1 1 1 2W
X
T
(A) 2
(B) 5
(C) 8
(D) 16
2
t
(D) x = 2
(C) x = t
2
1.6
1 - 2 .
z+1 z+3
If C is a counter clockwise path in the z -plane such that
z + 1 = 1, the value of 1
f (z) dz is
2pj C
(A) - 2
(B) - 1
(C) 1
(D) 2
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probability that the number of required tosses is odd, is
(A) 1/3
(B) 1/2
(C) 2/3
(D) 3/4
1.11
1.12
(D) 46
Given that
-5 -3
1 0
, the value of A3 is
A=>
and I = >
H
2 0
0 1H
(A) 15A + 12I
(B) 19A + 30I
(C) 17A + 15I
(D) 17A + 21I
2011
1.13
ONE MARK
Consider a closed surface S surrounding volume V . If rv is the
position vector of a point inside S , with nt the unit normal on S ,
the value of the integral ## 5rv $ nt dS is
S
(A) 3V
(C) 10V
1.14
Given f (z) =
If x = - 1, then the value of xx is
(A) e- p/2
(B) e p/2
The maximum value of f (x) = x3 - 9x2 + 24x + 5 in the interval [1, 6]
is
(A) 21
(B) 25
(C) 41
(B) 5V
(D) 15V
dy
= ky, y (0) = c is
dx
(B) x = kecy
(D) y = ce-kx
The solution of the differential equation
(A) x = ce-ky
(C) y = cekx
#
1.7
A fair coin is tossed till a head appears for the first time. The
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ONE MARK
With initial condition x (1) = 0.5 , the solution of the differential
equation
t dx + x = t , is
dt
(A) x = t - 1
(B) x = t 2 - 1
2
2
The direction of vector A is radially outward from the origin, with
A = krn . where r2 = x2 + y2 + z2 and k is a constant. The value of
n for which d:A = 0 is
(A) - 2
(B) 2
(C) 1
(D) 0
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TWO MARKS
2012
1.5
+
The maximum value of q until which the approximation sin q . q
holds to within 10% error is
(A) 10c
(B) 18c
(C) 50c
(D) 90c
A polynomial f (x) = a 4 x 4 + a 3 x3 + a2 x2 + a1 x - a 0 with all coefficients
positive has
(A) no real roots
(B) no negative real root
(C) odd number of real roots
(D) at least one positive and one negative real root
-
1.15
The value of the integral
z = 1 is given by
(A) 0
(C) 4/5
#
c
- 3z + 4 dz where c is the circle
(z 2 + 4z + 5)
(B) 1/10
(D) 1
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2011
1.16
1.17
Page 2
TWO MARKS
A numerical solution of the equation f (x) + x - 3 = 0 can be
obtained using Newton- Raphson method. If the starting value is
x = 2 for the iteration, the value of x that is to be used in the next
step is
(A) 0.306
(B) 0.739
(C) 1.694
(D) 2.306
The system of equations
x+y+z = 6
x + 4y + 6y = 20
x + 4y + l z = m
2010
1.22
1.23
1.24
has NO solution for values of l and μ given by
(A) l = 6, m = 20
(B) l = 6, m =
Y 20
(C) l =
Y 6, m = 20
1.18
2010
A fair coin is tossed independently four times. The probability of the
event “the number of time heads shown up is more than the number
of times tail shown up”
(A) 1/16
(B) 1/3
(C) 1/4
(D) 5/16
v = xyatx + x 2 aty , then
If A
is
(A) 0
(C) 1
1.25
(D) 1/2
ONE MARKS
The eigen values of a skew-symmetric matrix are
(A) always zero
(B) always pure imaginary
(C) either zero or pure imaginary (D) always real
1.26
The trigonometric Fourier series for the waveform f (t) shown below
contains
1.27
(A) only
(B) only
(C) only
(D) only
1.21
(B) minimum at x = e
(D) minimum at x = e-1
# Av $ dlv over the path shown in the figure
C
A fair dice is tossed two times. The probability that the second toss
results in a value that is higher than the first toss is
(A) 2/36
(B) 2/6
(C) 5/12
1.20
If ey = x1/x , then y has a
(A) maximum at x = e
(C) maximum at x = e-1
(D) l =
Y 6, m = 20
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1.19
TWO MARKS
cosine terms and zero values for the dc components
cosine terms and a positive value for the dc components
cosine terms and a negative value for the dc components
sine terms and a negative value for the dc components
A function n (x) satisfied the differential equation
d 2 n (x) n (x)
- 2 =0
dx 2
L
where L is a constant. The boundary conditions are : n (0) = K
and n (3) = 0 . The solution to this equation is
(B) n (x) = K exp (- x/ L )
(A) n (x) = K exp (x/L)
2
(D) n (x) = K exp (- x/L)
(C) n (x) = K exp (- x/L)
(B) 2
3
(D) 2 3
The residues of a complex function
1 - 2z
x (z) =
z (z - 1) (z - 2)
at its poles are
(A) 1 , - 1 and 1
(B) 1 , - 1 and - 1
2
2
2
2
(D) 1 , - 1 and 3
(C) 1 , 1 and - 3
2
2
2
2
dy (x)
Consider differential equation
- y (x) = x , with the initial
dx
condition y (0) = 0 . Using Euler’s first order method with a step
size of 0.1, the value of y (0.3) is
(A) 0.01
(B) 0.031
(C) 0.0631
(D) 0.1
3s + 1
Given f (t) = L-1 ; 3
. If lim f (t) = 1, then the value
t"3
s + 4s2 + (k - 3) s E
of k is
(A) 1
(B) 2
(C) 3
(D) 4
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2009
1.28
1.29
ONE MARK
The order of the differential equation
d2y
dy 3
4
-t
2 + c dt m + y = e
dt
is
(A) 1
(B) 2
(C) 3
(D) 4
A fair coin is tossed 10 times. What is the probability that only the
first two tosses will yield heads?
2
2
(B) 10C2 b 1 l
(A) c 1 m
2
2
10
10
(C) c 1 m
(D) 10C 2 b 1 l
2
2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
1.30
If f (z) = c 0 + c1 z-1 , then
(A) 2pc1
(C) 2pjc1
unit circle
1.33
Match each differential equation
curves from Group II
Group I
dy
y
A.
=
dx
x
dy
y
B.
=dx
x
dy
x
C.
=
dx
y
dy
D.
=- x
dx
y
(A) A - 2, B - 3, C - 3, D - 1
(B) A - 1, B - 3, C - 2, D - 1
(C) A - 2, B - 1, C - 3, D - 3
(D) A - 3, B - 2, C - 1, D - 2
1.34
1.35
1.38
TWO MARKS
Group II
1.37
Which of the following functions would have only odd powers of x
in its Taylor series expansion about the point x = 0 ?
(A) sin (x3)
(B) sin (x2)
(C) cos (x3)
(D) cos (x2)
Which of the following is a solution to the differential equation
dx (t)
+ 3x (t) = 0 ?
dt
(A) x (t) = 3e - t
(B) x (t) = 2e - 3t
(D) x (t) = 3t2
(C) x (t) =- 23 t2
in Group I to its family of solution
2008
1.40
1. Circles
TWO MARKS
The recursion relation to solve x = e - x using Newton - Raphson
method is
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2. Straight lines
3. Hyperbolas
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(A) xn + 1 = e-x
(B) xn + 1 = xn - e-x
x 2 - e-x (1 - xn) - 1
(D) xn + 1 = n
xn - e-x
n
n
-x
(C) xn + 1 = (1 + xn) e -x
1+e
n
n
The Eigen values of following matrix are
V
R
S- 1 3 5 W
S- 3 - 1 6 W
SS 0 0 3 WW
X
T
(A) 3, 3 + 5j, 6 - j
(B) - 6 + 5j, 3 + j, 3 - j
(C) 3 + j, 3 - j, 5 + j
(D) 3, - 1 + 3j, - 1 - 3j
ONE MARKS
p11 p12
are nonzero,
p21 p22 G
and one of its eigenvalue is zero. Which of the following statements
is true?
(B) p11 p22 - p12 p21 =- 1
(A) p11 p12 - p12 p21 = 1
(C) p11 p22 - p12 p21 = 0
(D) p11 p22 + p12 p21 = 0
1.41
1.42
For real values of x , the minimum value of the function
n
The residue of the function f (z) =
1.43
1.44
1.45
1
at z = 2 is
(z + 2) 2 (z - 2) 2
(B) - 1
16
(D) 1
32
0 1
Consider the matrix P = =
. The value of e p is
- 2 - 3G
2e-2 - 3e-1 e-1 - e-2
e-1 + e-1 2e-2 - e-1
(A) > -2
(B)
H
>2e-1 - 4e2 3e-1 + 2e-2H
2e - 2e-1 5e-2 - e-1
5e-2 - e-1 3e-1 - e-2
(C) > -2
H
2e - 6e-1 4e-2 + 6-1
The system of linear equations
The equation sin (z) = 10 has
(A) no real or complex solution
(B) exactly two distinct complex solutions
(C) a unique solution
(D) an infinite number of complex solutions
n
(A) - 1
32
(C) 1
16
All the four entries of the 2 # 2 matrix P = =
4x + 2y = 7
2x + y = 6 has
(A) a unique solution
(B) no solution
(C) an infinite number of solutions
(D) exactly two distinct solutions
1.36
1.39
(x - p) 2
(D) - 1 +
+ ...
3!
2008
(B) 1
(D) 0
(C) 0.5
(B) 2p (1 + c0)
(D) 2p (1 + c0)
The Taylor series expansion of sin x at x = p is given by
x-p
2
(x - p )
(x - p) 2
(A) 1 +
(B) - 1 + ...
+ ...
3!
3!
(x - p) 2
(C) 1 + ...
3!
1.32
f (x) = exp (x) + exp (- x) is
(A) 2
# 1 +zf (z) dz is given by
2009
1.31
Page 3
2e-1 - e-2
e-1 - e-2
(D) >
H
-1
-2
- 2e + 2e - e-1 + 2e-2
In the Taylor series expansion of exp (x) + sin (x) about the point
x = p , the coefficient of (x - p) 2 is
(A) exp (p)
(B) 0.5 exp (p)
(C) exp (p) + 1
(D) exp (p) - 1
The value of the integral of the function g (x, y) = 4x3 + 10y 4 along
the straight line segment from the point (0, 0) to the point (1, 2) in
the x - y plane is
(A) 33
(B) 35
(C) 40
(D) 56
Consider points P and Q in the x - y plane, with P = (1, 0) and
#
Q
Q = (0, 1). The line integral 2
(xdx + ydy) along the semicircle
P
with the line segment PQ as its diameter
(A) is - 1
(B) is 0
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) is 1
(D) depends on the direction (clockwise or anit-clockwise) of the
semicircle
2007
1.46
2
#1 ydx
is
(B) 2.5
(D) 5.0
For x << 1, coth (x) can be approximated as
(A) x
(B) x2
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(C) 1
x
(D) e - x
(A) jp
(C) - p
1.55
2
For the function e - x , the linear approximation around x = 2 is
(A) (3 - x) e - 2
(B) 1 - x
2 ) x @e - 2
2007
1.52
If the semi-circular control D of radius 2 is as shown in the figure,
1
then the value of the integral
ds is
2
(
s
1
)
D
#
(B) 1
(D) not defined
Which one of following functions is strictly bounded?
(B) ex
(A) 1/x2
(C) 63 + 3 2 - (1 -
1.51
1.54
sin b q l
2
is
lim
q"0
q
(C) x2
1.50
(A) f1 (t) and f2 (t) are orthogonal
(B) f1 (t) and f3 (t) are orthogonal
(C) f2 (t) and f3 (t) are orthogonal
D) f1 (t) and f2 (t) are orthonormal
(D) 12
x
(A) 0.5
(C) 2
1.49
Three functions f1 (t), f2 (t) and f3 (t) which are zero outside the interval
[0, T] are shown in the figure. Which of the following statements is
correct?
The following plot shows a function which varies linearly with x .
(A) 1.0
(C) 4.0
1.48
1.53
ONE MARK
The value of the integral I =
1.47
Page 4
(D) e - 2
d2 y
The solution of the differential equation k2 2 = y - y2 under the
dx
boundary conditions
(i) y = y1 at x = 0 and
(ii) y = y2 at x = 3 , where k, y1 and y2 are constants, is
1.56
(A) y = (y1 - y2) exp a- x2 k + y2
k
(C) y = ^y1 - y2h sinh a x k + y1
k
1.57
The equation x3 - x2 + 4x - 4 = 0 is to be solved using the Newton
- Raphson method. If x = 2 is taken as the initial approximation of
the solution, then next approximation using this method will be
(A) 2/3
(B) 4/3
(C) 1
(D) 3/2
It is given that X1, X2 ...XM at M non-zero, orthogonal vectors.
The dimension of the vector space spanned by the 2M vectors
X1, X2,... XM , - X1, - X2,... - XM is
(A) 2M
(B) M + 1
(C) M
(D) dependent on the choice of X1, X2,... XM
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TWO MARKS
(B) y = (y2 - y1) exp a- x k + y1
k
(D) y = ^y1 - y2h exp a- x k + y2
k
(B) - jp
(D) p
Consider the function f (x) = x2 - x - 2 . The maximum value of f (x)
in the closed interval [- 4, 4] is
(A) 18
(B) 10
(C) - 225
(D) indeterminate
An examination consists of two papers, Paper 1 and Paper 2. The
probability of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2.
Given that a student has failed in Paper 2, the probability of failing
in Paper 1 is 0.6. The probability of a student failing in both the
papers is
(A) 0.5
(B) 0.18
(C) 0.12
(D) 0.06
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2006
1.58
1.59
Page 5
ONE MARK
V
R
S1 1 1 W
The rank of the matrix S1 - 1 0 W is
SS1 1 1 WW
X (B) 1
T
(A) 0
(C) 2
(D) 3
1.67
1.62
## (4 # P) $ ds , where P is a vector, is equal to
(A) # P $ dl
(B) # 4#4# P $ dl
(C) # 4# P $ dl
(D) ### 4$ Pdv
60%
0.01
Y
30%
0.02
Z
10%
0.03
TWO MARKS
The eigenvalue and the corresponding eigenvector of 2 # 2 matrix
are given by
Eigenvalue
Eigenvector
1
l1 = 8
v1 = = G
1
1
l2 = 4
v2 = = G
-1
The matrix is
6 2
4 6
(A) =
(B) =
G
2 6
6 4G
2 4
4 8
(C) =
(D) =
G
4 2
8 4G
(A) 2
(C) 6
1.69
jp
2
jp
(C) 2
The integral
(A) 1
2
(C) 4
3
#0
p
1 dz is positive sense is
z2 + 4
z-j = 2
(B) - p
2
(D) p
2
#
sin3 qdq is given by
(B) 2
3
(D) 8
3
(B) 4
(D) 8
d2 y
For the differential equation 2 + k2 y = 0 the boundary conditions
dx
are
(i) y = 0 for x = 0 and
(ii) y = 0 for x = a
The form of non-zero solutions of y (where m varies over all integers) are
(B) y = Am cos mpx
(A) y = Am sin mpx
a
a
m
m
/
mp
(C) y = / Am x a
m
1.70
/
mpx
(D) y = / Am e - a
m
As x increased from - 3 to 3, the function f (x) =
ex
1 + ex
(A) monotonically increases
(B) monotonically decreases
(C) increases to a maximum value and then decreases
(D) decreases to a minimum value and then increases
For the function of a complex variable W = ln Z (where, W = u + jv
and Z = x + jy , the u = constant lines get mapped in Z -plane as
(A) set of radial straight lines
(B) set of concentric circles
(C) set of confocal hyperbolas
(D) set of confocal ellipses
The value of the constant integral
4 2
For the matrix =
the eigenvalue corresponding to the eigenvector
2 4G
101
=101G is
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(D) e u (t)
(A)
1.66
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t
2006
1.65
Probability of being
supplied defective
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A solution for the differential equation xo (t) + 2x (t) = d (t) with
initial condition x (0-) = 0 is
(A) e - 2t u (t)
(B) e2t u (t)
(C) e u (t)
1.64
1.68
A probability density function is of the form
p (x) = Ke- a x , x ! (- 3, 3)
The value of K is
(A) 0.5
(B) 1
(C) 0.5a
(D) a
-t
1.63
% of Computer Supplied
Given that a computer is defective, the probability that was supplied by Y is
(A) 0.1
(B) 0.2
(C) 0.3
(D) 0.4
(D) 4 (4$ P) - 4 2 P
1.61
Company
4#4# P , where P is a vector, is equal to
(A) P # 4# P - 4 2 P
(B) 4 2 P + 4 (4 # P)
(C) 4 2 P + 4# P
1.60
Three companies X, Y and Z supply computers to a university. The
percentage of computers supplied by them and the probability of
those being defective are tabulated below
2005
1.71
The following differential equation has
d2 y
dy 3
3 c 2 m + 4 c m + y2 + 2 = x
dt
dt
(A) degree = 2 , order = 1
(C) degree = 4 , order = 3
1.72
ONE MARK
(B) degree = 1, order = 2
(D) degree = 2 , order = 3
A fair dice is rolled twice. The probability that an odd number will
follow an even number is
(A) 1/2
(B) 1/6
(C) 1/3
(D) 1/4
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
1.73
A solution of the following differential equation is given by
d2 y
dy
-5
+ 6y = 0
2
dx
dx
(A) y = e2x + e-3x
(B) y = e2x + e3x
(C) y = e-2x + 33x
(D) y = e-2x + e-3x
2005
1.74
1.75
1.77
In what range should Re (s) remain so that the Laplace transform of
the function e(a + 2) t + 5 exits.
(A) Re (s) > a + 2
(B) Re (s) > a + 7
(C) Re (s) < 2
(D) Re (s) > a + 5
The derivative of the symmetric function drawn in given figure will
look like
Match the following and choose the correct combination:
Group I
Group 2
E. Newton-Raphson method
1. Solving nonlinear equations
F. Runge-kutta method
2. Solving linear simultaneous
equations
G. Simpson’s Rule
3. Solving ordinary differential
equations
H. Gauss elimination
4. Numerical integration
5. Interpolation
6. Calculation of Eigenvalues
(B) E - 1, F - 6, G - 4, H - 3
(A) E - 6, F - 1, G - 5, H - 3
(C) E - 1, F - 3, G - 4, H - 2
(D) E - 5, F - 3, G - 4, H - 1
-4 2
Given the matrix =
, the eigenvector is
4 3G
3
4
(A) = G
(B) = G
2
3
2
-1
(C) = G
(D) = G
-1
2
1.78
1.79
1
2 - 0.1
a
and A - 1 = = 2 G. Then (a + b) =
G
0 3
0 b
(A) 7/20
(B) 3/20
(C) 19/60
(D) 11/20
Let, A = =
The value of the integral I =
(A) 1
(C) 2
TWO MARKS
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1.76
Page 6
1.80
1
2p
Given an orthogonal matrix
R1 1 1 1 V
W
S
S1 1 - 1 - 1 W
A =S
1 - 1 0 0W
W
S
S0 0 1 1 W
X
T
T
6AA @- 1 is
V
R1
S 4 0 0 0W
S 0 14 0 0 W
W
(A) S
1
S0 0 2 0W
S 0 0 0 12 W
X
T
R1 0 0 0 V
W
S
S0 1 0 0 W
(C) S
0 0 1 0W
W
S
S0 0 0 1 W
X
T
2
exp c- x m dx is
8
(B) p
(D) 2p
#0
3
R1
S2
S0
(B) S
S0
S0
TR
1
S4
S0
(D) S
S0
S0
T
V
0 0 0W
1
W
2 0 0
W
1
0 2 0W
0 0 12 W
XV
0 0 0W
1
W
4 0 0
W
0 14 0 W
0 0 14 W
X
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 7
SOLUTIONS
1.1
Consider the given matrix be
R
S2
S1
Im + AB = S
S1
S1
T
where m = 4 so, we obtain
R
S2
S1
AB = S
S1
S1
T
R
S1
S1
=S
S1
S1
T
Hence, we get
Option (B) is correct.
Here, as we know
Lim sin q . 0
q"0
but for 10% error, we can check option (B) first,
q = 18c = 18c # p = 0.314
180c
sin q = sin 18c = 0.309
% error = 0.314 - 0.309 # 100% = 0.49%
0.309
Now, we check it for q = 50c
1
2
1
1
1
1
2
1
1
2
1
1
1
1
2
1
1
1
1
1
1
1
1
1
V
1W
1W
1WW
2W
X
V
V R
1W S1 0 0 0W
1W S0 1 0 0W
1WW SS0 0 1 0WW
2W S0 0 0 1W
X
X T
V R V
1W S1W 61 1 1 1@
1W S1W
=
1WW SS1WW
1W S1W
X T X
q = 50c = 50c # p = 0.873
180c
sin q = sin 50c = 0.77
% error = 0.77 - 0.873 =- 12.25%
0.873
so, the error is more than 10% . Hence, for error less than 10%,
q = 18c can have the approximation
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sin q . q
1.2
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Option (A) is correct.
For, a given matrix 6A@ the eigen value is calculated as
R V
S1W
S1W
A = S W, B = 81 1 1 1B
S1W
S1W
T X
R V
Therefore,
BA = 81 1 1 1B S1W = 4
S1W
S1W
S W
S1W
From the given property
T X
Det ^Im + AB h = Det ^Im + BAh
V
V
R
_
ZR
S2 1 1 1W
b
]S1 0 0 0W
b
]S0 1 0 0W
S1 2 1 1W
&
Det S
= Det [S
+ 4`
W
W
0 0 1 0W
S1 1 2 1W
b
]S
b
]
S0 0 0 1W
S1 1 1 2W
a
\T
X
X
T
A - lI = 0
where l gives the eigen values of matrix. Here, the minimum eigen
value among the given options is
l =0
We check the characteristic equation of matrix for this eigen value
A - lI = A
(for l = 0 )
3 5 2
= 5 12 7
2 7 5
= 3 ^60 - 49h - 5 ^25 - 14h + 2 ^35 - 24h
= 33 - 55 + 22
=0
Hence, it satisfied the characteristic equation and so, the minimum
eigen value is
= 1+4
=5
Note : Determinant of identity matrix is always 1.
l =0
1.3
Option (D) is correct.
Given, the polynomial
f ^x h = a 4 x 4 + a 3 x3 + a2 x2 + a1 x - a 0
Since, all the coefficients are positive so, the roots of equation is
given by
f ^x h = 0
It will have at least one pole in right hand plane as there will be
least one sign change from ^a1h to ^a 0h in the Routh matrix 1 st column. Also, there will be a corresponding pole in left hand plane
i.e.; at least one positive root
(in R.H.P)
and at least one negative root
(in L.H.P)
Rest of the roots will be either on imaginary axis or in L.H.P
1.4
Option (B) is correct.
1.5
Option (D) is correct.
t dx + x = t
dt
dx + x = 1
t
dt
dx + Px = Q (General form)
dt
Integrating factor,
Solution has the form,
IF = e # = e = e lnt = t
Pdt
1
# dt
t
# ^Q # IF hdt + C
x # t = # (1) (t) dt + C
x # IF =
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 8
dy (t)
= e-t u (t) + te-t u (t)
dt
dy
= e0 + 0 = 1
dt t = 0
2
xt = t + C
2
Taking the initial condition,
At t = 0+ ,
x (1) = 0.5
0.5 = 1 + C & C = 0
2
1.9
2
xt = t & x = t
2
2
So,
1.6
+
Option (C) is correct.
f (z) =
1
2 pj
# f (z) dz
C
1 - 2
z+1 z+3
= sum of the residues of the poles which lie
inside the given closed region.
C & z+1 = 1
Only pole z =- 1 inside the circle, so residue at z =- 1 is.
-z + 1
f (z) =
(z + 1) (z + 3)
(z + 1) (- z + 1) 2
= =1
= lim
2
z "- 1 (z + 1) (z + 3)
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1
2 pj
So
1.7
# f (z) dz
C
n+2 = 0
1.10
p
xx = ^ei 2 h & ^ei 2 h = e- 2
1.8
p i
1.12
dy (t)
dt
then,
sY (s) - y (0) =
So,
2
L
Y (s)
L
sY (s) - y (0)
max
= (2) 3 - 9 (2) 2 + 24 (2) + 5
= 8 - 36 + 48 + 5 = 25
Option (B) is correct.
Characteristic equation.
5l + l2 + 6 = 0
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l2 + 5l + 6 = 0
Since characteristic equation satisfies its own matrix, so
(- 2s - 3) s
+2
(s2 + 2s + 1)
A2 + 5A + 6 = 0 & A2 =- 5A - 6I
Multiplying with A
A3 + 5A2 + 6A = 0
A3 + 5 (- 5A - 6I) + 6A = 0
A3 = 19A + 30I
2
= - 2s - 32 s + 2s + 4s + 2
(s + 2s + 1)
1
sY (s) - y (0) = s + 2 2 = s + 1 2 +
(s + 1)
(s + 1)
(s + 1) 2
1
= 1 +
s + 1 (s + 1) 2
Taking inverse Laplace transform
x = 4, x = 2
A - lI = 0
-5 - l -3
=0
2
-l
Y (s) [s2 + 2s + 1] = 1 - 2s - 4
Y (s) = 2- 2s - 3
s + 2s + 1
y (t)
f (x) = x3 - 9x2 + 24x + 5
df (x)
= 3x2 - 18x + 24 = 0
dx
df (x)
= x2 - 6x + 8 = 0
dx
f (x)
2
6s Y (s) + 2s - 0@ + 2 6sY (s) + 2@ + Y (s) = 1
We know that, If,
=2
3
p
Option (D) is correct.
d 2 y (t) 2dy (t)
+
+ y (t) = d (t)
dt
dt 2
By taking Laplace transform with initial conditions
dy
2
;s Y (s) - sy (0) - dt
E + 2 [sy (s) - y (0)] + Y (s) = 1
t=0
&
1 - 14
d 2 f (x)
= 6x - 18
dx 2
d 2 f (x)
For x = 2,
= 12 - 18 =- 6 < 0
dx2
So at x = 2, f (x) will be maximum
- 1 = i = cos p + i sin p
2
2
p x
1
2
Option (B) is correct.
&
=1
x = ei 2
So,
H, TTH, TTTTH, ...........
3
5
P = 1 + b 1 l + b 1 l + .....
2
2
2
P =
1.11
n =- 2
&
Option (C) is correct.
Probability of appearing a head is 1/2. If the number of required
tosses is odd, we have following sequence of events.
Probability
Option (A) is correct.
x=
Option (A) is correct.
Divergence of A in spherical coordinates is given as
d:A = 12 2 (r 2 Ar ) = 12 2 (krn + 2)
r 2r
r 2r
= k2 (n + 2) rn + 1
r
= k (n + 2) rn - 1 = 0 (given)
1.13
Option (D) is correct.
From Divergence theorem, we have
v $ nt ds
v = #A
v Adv
### 4$
s
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 9
The position vector
Thus
rv = ^utx x + uty y + utz z h
v = 5rv, thus
Here, A
1.19
v
4$ A
= cutx 2 + uty 2 + utz 2 m : ^utx x + uty y + utz z h
2x
2y
2z
dy dz
5 = 3 # 5 = 15
= c dx +
+
dx dy dz m
So,
## 5rv $ nt ds = ### 15 dv = 15V
1.20
P (E) = No. of favourable outcomes = 15 = 5
36 12
No. of total outcomes
Option (C) is correct.
Eigen value of a Skew-symmetric matrix are either zero or pure
imaginary in conjugate pairs.
Option (C) is correct.
For a function x (t) trigonometric fourier series is
x (t) = Ao +
s
1.14
We have
#
Integrating
or
Since y (0) = c thus
So, we get,
or
or
1.15
dy
= ky
dx
dy
= # k dx + A
y
0
An = 2 # x (t) cos nwt dt
T0 T
0
Bn = 2 # x (t) sin nwt dt
T0 T
For an even function x (t), Bn = 0
Since given function is even function so coefficient Bn = 0 , only cosine
ln y = kx + A
ln c = A
ln y = kx + ln c
ln y = ln ekx + ln c
y = cekx
0
Option (A) is correct.
3z + 4 dz where C is circle z = 1
C R Integrals is # 2
z
4z + 5
+
C
# f (z) dz = 0 if poles are outside C.
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C
2
z + 4z + 5 = 0
(z + 2) 2 + 1 = 0
Thus
z1, 2 =- 2 ! j & z1, 2 > 1
So poles are outside the unit circle.
Now
1.16
and constant terms are present in its fourier series representation.
Constant term :
3T/4
T/4
3T/4
A0 = 1 #
- 2AdtD
x (t) dt = 1 : # Adt + #
T -T/4
T -T/4
T/4
= 1 :TA - 2AT D =- A
2
2
T 2
Constant term is negative.
Option (C) is correct.
f (x) = x + x - 3 = 0
f l (x) = 1 + 1
2 x
Substituting x 0 = 2 we get
f l (x 0) = 1.35355 and f (x 0) = 2 +
We have
1.21
2 - 3 = 0.414
Newton Raphson Method
x1 = x 0 -
f (x 0)
f l (x 0)
Substituting all values we have
x 1 = 2 - 0.414 = 1.694
1.3535
1.17
1.18
n=1
Ao = 1 # x (t) dt T0 " fundamental period
T0 T
Where,
Option (C) is correct.
3
/ [An cos nwt + Bn sin nwt]
Option (B) is correct.
Writing A: B we have
R
V
S1 1 1 : 6 W
S1 4 6 : 20W
S
W
S1 4 l : m W
T
X
Apply R 3 " R 3 - R2
R
V
6 W
S1 1 1 :
S1 4 6 : 20 W
S
W
S0 0 l - 6 : m - 20W
T
X
For equation to have solution, rank of A and A: B must be same.
Thus for no solution; l = 6, m ! 20
Option (C) is correct.
Total outcome are 36 out of which favorable outcomes are :
(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6);
(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) which are 15.
Option (D) is correct.
Given differential equation
d 2 n (x) n (x)
- 2 =0
dx 2
L
lx
Let n (x) = Ae
lx
So,
Al2 elx - Ae2 = 0
L
l2 - 12 = 0 & l = ! 1
L
L
Boundary condition, n (3) = 0 so take l =- 1
L
x
So,
1.22
n (x) = Ae- L
n (0) = Ae0 = K & A = K
n (x) = Ke- (x/L)
Option (A) is correct.
Given that
1
ey = x x
1
ln ey = ln x x
or
y = 1 ln x
x
1
dy
Now
= 1 1 + ln x ^- x- x h = 12 - ln2
xx
dx
x
x
For maxima and minima :
dy
= 12 (1 - ln x) = 0
dx
x
ln x = 1 " x = e 1
or
2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 10
From table, at x = 0.3, y (x = 0.3) = 0.031
d 2y
=- 23 - ln x b- 23 l - 12 b 1 l
dx 2
x x
x
x
=- 22 + 2 ln3 x - 13
x
x
x
2
d x
= -22 + 23 - 13 < 0
dy 2 at x = e
e
e
e
1
So, y has a maximum at x = e
Now
1.27
Option (D) is correct.
Given that
3s + 1
f (t) = L - 1 ; 3
s + 4s 2 + (K - 3) s E
lim f (t) = 1
1
1.23
t"3
By final value theorem
Option (D) is correct.
According to given condition head should comes 3 times or 4 times
4
3
P (Heads comes 3 times or 4 times) = 4C 4 b 1 l + 4C 3 b 1 l b 1 l
2
2 2
lim f (t) = lim sF (s) = 1
t"3
1.24
Option (C) is correct.
v = xyatx + x 2 aty
A
v = dxatx + dyaty
dl
v = # (xyatx + x 2 aty) : (dxatx + dyaty)
# Av : dl
C
1.28
C
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=
=
1.29
1.30
2/ 3
#1/
3
xdx +
1/ 3
#2/
3
3xdx +
#1
3
4 dy +
3
#3
1
1 dy
3
= 1 : 4 - 1 D + 3 :1 - 4 D + 4 [3 - 1] + 1 [1 - 3]
2 3 3
2 3 3
3
3
1.31
1 - 2z
z (z - 1) (z - 2)
Poles are located at z = 0, z = 1, and z = 2
At Z = 0 residues is
R 0 = z : X (z) Z = 0 = 1 - 2 # 0 = 1
2
(0 - 1) (0 - 2)
= 1 - 2 # 2 =- 3
2
2 (2 - 1)
y2 y 4
- + ...
3! 5!
Substituting x - p = y we get
(x - p) 2 (x - p) 4
+ ...
f (x) =- 1 +
3!
5!
or
h = 0.1, y (0) = 0
y
dy
= x+y
dx
0
0
0
f (x) = sin x
x-p
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R2 = (z - 2) : X (z) z = 2
x
Option (D) is correct.
Substituting x - p = y ,we get
sin (y + p)
sin y
== - 1 (sin y)
f (y + p) =
y
y
y
3
5
y
y
= - 1 cy - + - ...m
y
3! 5!
R1 = (Z - 1) : X (Z ) Z = 1
Option (B) is correct.
Taking step size
#
unit circle
We have
= 1-2#1 = 1
1 (1 - 2)
1.26
Option (C) is correct.
Number of elements in sample space is 210 . Only one element
1
"H, H, T, T, T, T, T, T, T, T , is event. Thus probability is 10
2
Option (C) is correct.
We have
= 2pjc1
X (z ) =
At z = 2 ,
Option (B) is correct.
The highest derivative terms present in DE is of 2nd order.
unit circle
Option (C) is correct.
Given function
at z = 1,
K =4
s"0
#
=1
1.25
or
lim
f (z) = c0 + c1 z - 1
z (1 + c0) + c1
1 + f (z) 1 + c0 + c1 z - 1
=
=
f1 (z) =
z
z
z2
Since f1 (z) has double pole at z = 0 , the residue at z = 0 is
z (1 + c0) + c1
Res f1 (z) z = 0 = lim z2 .f1 (z) = lim z2 . c
m = c1
z"0
z"0
z2
Hence
[1 + f (z)]
dz = 2pj [Residue at z = 0 ]
f1 (z) dz =
z
# (xydx + x 2 dy)
C
or
s : (3s + 1)
=1
s3 + 4s2 + (K - 3) s
s (3s + 1)
lim 2
=1
s " 0 s [s + 4s + (K - 3)]
1 =1
K-3
or
= 1: 1 +4:1 :1 = 5
16
8 2 16
s"0
yi + 1 = yi + h
dy
dx
y1 = 0 + 0.1 (0) = 0
0.1
0
0.1
y2 = 0 + 0.1 (0.1) = 0.01
0.2
0.01
0.21
y 3 = 0.01 + 0.21 # 0.1 = 0.031
0.3
0.031
1.32
f (y + p) =- 1 +
Option (A) is correct.
(A)
dy
y
=
dx
x
dy
= dx
y
x
or
#
#
or
log y = log x + log c
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
or
y = cx
Thus option (A) and (C) may be correct.
dy
y
(B)
=dx
x
dy
or
=- dx
y
x
#
or
1.34
f'( xn) = 1 + e - x
#
1.41
1.37
1.38
Option (B) is correct.
The given system is
4 2 x
7
=2 1G=y G = = 6 G
4 2
We have
A ==
2 1G
4 2
and
=0
A =
2 1
4 2 7
Now
C ==
G
2 1 6
1.40
=- 2 =- 1
32
64
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1.42
Option (D) is correct.
eP = L- 1 6(sI - A) - 1@
s 0
0 1 -1
-1
= L e=
0 s G =- 2 - 3Go
s - 1 -1
-1
o
= L e=
2 s + 3G
Rank of matrix r (A) < 2
Rank of matrix r (C) = 2
= L f>
-1
1.43
2
4
cos x = 1 + x + x + ...
2! 4!
Thus
f'( x) = 1 + e
-x
2
= ex + sin x
= ex + cos x
= ex - sin x
= e p - sin p = e p
f"( p)
Thus the coefficient of (x - p) 2 is
2!
Thus only (B) may be
1.44
f (x) = x - e - x
(x - a) 2
f"( a) + ...
2!
(x - p)
f"( x)...
f (x) = f (p) + x - p f'( p) +
1!
2!
Now
(D + 3) x (t) = 0
x = e-x
Hp
f (x) = f (a) + x - a f'( a) +
1!
For x = p we have
Option (C) is correct.
1
(s + 1)( s + 2)
s
(s + 1)( s + 2)
Option (B) is correct.
Taylor series is given as
3
5
sin x = x + x + x + ...
3! 5!
x (t) = Ce - 3t
s+3
(s + 1)( s + 2)
-2
(s + 1)( s + 2)
2e - 1 - e - 2
e-1 - e-2
==
G
- 2e - 1 + 2e - 2 - e - 1 + 2e - 2
Option (B) is correct.
dx (t)
We have
+ 3x (t) = 0
dt
or
1
dn - 1 6(z - a) n f (z)@
z=a
(n - 1)! dzn - 1
1
d (z - 2) 2
1
2
(2 - 1)! dz ;
(z - 2) (z + 2) 2 Ez = a
Option (A) is correct.
We have
n
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Option (A) is correct.
We have
f (x) = ex + e-x
For x > 0 ,
ex > 1 and 0 < e-x < 1
For x < 0 ,
0 < ex < 1 and e-x > 1
Thus f (x) have minimum values at x = 0 and that is e0 + e-0 = 2 .
Since m =- 3 ,
solution.
n
-2
1
= d ;
=
dz (z + 2) 2 Ez = a ; (z + 2) 3 Ez = a
Option (A) is correct.
sin z can have value between - 1 to + 1. Thus no solution.
or
(1 + xn) e - x
1 + e-x
Option (A) is correct.
Thus Res f (z) z = 2 =
Thus only sin (x3) will have odd power of x .
1.39
n
Here we have n = 2 and a = 2
p11 p22 - p12 p21 = 0
n
- xn
Option (D) is correct.
Sum of the principal diagonal element of matrix is equal to the sum
of Eigen values. Sum of the diagonal element is - 1 - 1 + 3 = 1.In
only option (D), the sum of Eigen values is 1.
Option (C) is correct.
The product of Eigen value is equal to the determinant of the matrix.
Since one of the Eigen value is zero, the product of Eigen value is
zero, thus determinant of the matrix is zero.
n
xn + 1 = xn - xn - e- x =
1+e
Thus
Since r (A) ! r (C) there is no solution.
1.36
f (xn) = xn - e - x
Now
Hyperbola
Thus
1.35
The Newton-Raphson iterative formula is
f (xn)
xn + 1 = xn f'( xn)
Res f (z) z = a =
or
1.33
Straight Line
log y =- log x + log c
log y = log 1 + log c
x
y =c
x
or
Page 11
f (x)
f'( x)
f"( x)
f"( p)
Option (A) is correct.
The equation of straight line from (0, 0) to (1, 2) is y = 2x .
Now
g (x, y) = 4x3 + 10y 4
or,
g (x, 2x) = 4x3 + 160x 4
Now
1
1
#0 g (x, 2x) = #0 (4x3 + 160x4) dx
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 12
= [x 4 + 32x5] 10 = 33
1.45
1.46
Option (B) is correct.
Q
I =2
#P
=2
#1
0
(xdx + ydy) = 2
xdx + 2
Q
#P
xdx + 2
or
y = x+1
Now
I =
2
(x +
2
x
k
y = (y1 - y2) e - + y2
2
#1 (x + 1) dx
=
9 4
E = - = 2.5
2 2
Option (C) is correct.
Thus
1.53
+3
1.54
#
1.55
x"3
lim e - x = 3
x"3
2
1.56
x"3
lim e - x = 1
Thus e - x is strictly bounded.
2
2
x"0
1.50
= (3 - x) e
1.51
Neglecting higher powers
-2
f'( x) = 2x - 1 = 0 " x = 1
2
Option (D) is correct.
We have
or
A.E.
or
Option (A) is correct.
We have
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f (x) = e - x = e - (x - 2) - 2 = e - (x - 2) e - 2
(x - 2) 2
= ;1 - (x - 2) +
...E e - 2
2!
= 61 - (x - 2)@ e - 2
Option (C) is correct.
For two orthogonal vectors, we require two dimensions to define
them and similarly for three orthogonal vector we require three
dimensions to define them. 2M vectors are basically M orthogonal
vector and we require M dimensions to define them.
f (x) = x2 - x + 2
Option (A) is correct.
We have
Option (A) is correct.
We know that
1 ds = 2pj
[sum of residues]
2
s
1
D
Singular points are at s = ! 1 but only s =+ 1 lies inside the given
contour, Thus Residue at s =+ 1 is
lim (s - 1) f (s) = lim (s - 1) 2 1 = 1
s"1
s"1
s -1 2
1 ds = 2pj 1 = pj
`2j
2
s -1
D
#
Option (D) is correct.
lim e - x = 0
f (x) g (x) dx = 0
i.e. common area between f (x) and g (x) is zero.
coth x . 1
x
lim 12 = 3
x"0 x
lim x2 = 3
Option (C) is correct.
For two orthogonal signal f (x) and g (x)
#- 3
Option (A) is correct.
q
sin ^ q2 h
sin ^ q2 h
1 lim sin ^ 2 h = 1 = 0.5
= lim
=
lim
q"0
q " 0 2^ q h
2 q " 0 ^ q2 h
q
2
2
We have,
Option (B) is correct.
We have
f (x) = x3 - x2 + 4x - 4
f'( x) = 3x2 - 2x + 4
Taking x0 = 2 in Newton-Raphosn method
23 - 22 + 4 (2) - 4
f (x0)
x1 = x0 = 2=4
f'( x0)
3
3 (2) 2 - 2 (2) + 4
1) 2 2
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x
k
From y (0) = y1 we get
C1 + C2 = y1 - y2
From y (3) = y2 we get that C1 must be zero.
Thus
C2 = y1 - y2
1
as x << 1, cosh x . 1 and sinh x . x
1.49
x
k
1.52
#1 ydx
- y22
1
c 2 m = y2
D2 - 12 k
k
y = C1 e - + C2 e + y2
#0 ydy = 0
coth x = cosh x
sinh x
1.48
Thus solution is
Q
#P ydy
Option (B) is correct.
The given plot is straight line whose equation is
x +y =1
-1 1
=;
1.47
P.I. =
f"( x) = 2
d2 y
= y - y2
dx2
y
d2 y
y
- 2 =- 22
2
k
dx
k
k2
Since f"( x) = 2 > 0 , thus x = 1 is minimum point. The maximum
2
value in closed interval 6- 4, 4@ will be at x =- 4 or x = 4
Now maximum value
= max [f (- 4), f (4)]
= max (18, 10) = 18
D2 - 12 = 0
k
D =! 1
k
1.57
x
k
C.F. = C1 e - + C2 e
x
k
Option (C) is correct.
Probability of failing in paper 1 is
Possibility of failing in Paper 2 is
P (A) = 0.3
P (B) = 0.2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Now x = eu cos v and y = eu sin v
Thus
x 2 + y 2 = e 2u
Probability of failing in paper 1, when
student has failed in paper 2 is
P ^ BA h = 0.6
We know that
(P + B)
Pb A l =
B
P (B)
or
P (A + B) = P (B) P b A l = 0.6 # 0.2 = 0.12
B
1.58
1.59
Option (C) is correct.
We have
V
V R
R
S1 1 1 W S1 1 1 W
A = S1 - 1 0 W + S1 - 1 0 W
SS1 1 1 WW SS0 0 0 WW
X
X T
T
Since one full row is zero, r (A) < 3
1 1
Now
=- 2 ! 0 , thus r (A) = 2
1 -1
1.65
Option (D) is correct.
We have
1 dz =
2
+
z
4
z-j = 2
#
C
1.66
Option (C) is correct.
I =
# p (x) dx
3
We know
-3
#
3
-3
-3
Keax dx +
1.67
=1
Ke- a x dx = 1
# Ke
3
- ax
0
dx = 1
1.68
Since x (0 -) = 0
1
s+2
Now taking inverse Laplace transform we have
- 2t
x (t) = e u (t)
1.64
Option (A) is correct.
Sum of the Eigen values must be equal to the sum of element of
principal diagonal of matrix.
6 2
Only matrix =
satisfy this condition.
2 6G
Option (B) is correct.
We have
or
or
W
u + jv
eu + jv
eu e jv
eu (cos v + j sin v)
= ln z
= ln (x + jy)
= x + jy
= x + jy
= x + jy
or
or
1.69
4 2
2 4G
6A - lI @ [X] = 0
4 - l 2 101
0
= 2 4 - l G=101G = = 0 G
or
sX (s) + 2X (s) = 1
A ==
Now
sX (s) - x (0) + 2X (s) = 1
X (s) =
Option (C) is correct.
We have
Taking Laplace transform both sides
1.63
`
P (y + d) = 0.3 # 0.02 = 0.006
P (d) = 0.6 # 0.1 + 0.3 # 0.02 + 0.1 # 0.03 = 0.015
y
P a k = 0.006 = 0.4
d
0.015
K eax 0 + k e- ax 3 = 1
@0
a 6 @- 3 (- a) 6
K +K =1
or
a
a
or
K =a
2
Option (A) is correct.
We have
xo (t) + 2x (t) = s (t)
or
p
Option (D) is correct.
Let d " defective and y " supply by Y
P (y + d)
y
pa k =
d
P (d)
or
1.62
#0
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Option (C) is correct.
0
sin3 qdq
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## (4 # F) $ ds = # A $ dl
#
p
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Option (A) is correct.
The Stokes theorem is
or
#0
3 sin q - sin 3q dq
sin 3q = 3 sin q - 4 sin3 q
j
4
p
p
= :- 3 cos qD = : ws3q D = 8 3 + 3 B - 8 1 + 1 B = 4
12 0
4
4 4
12 12
3
0
=
A # (B # C) = B (A $ C) - C (A $ B)
4#4# P = 4 (4$ P) - P (4$4)
Thus
z-j = 2
1
dz
(z + 2i) (z - 2i)
P (0, 2) lies inside the circle z - j = 2 and P (0, - 2) does not lie.
Thus By cauchy’s integral formula
2pi = p
1
=
I = 2pi lim (z - 2i)
z " 2i
2i + 2i
2
(z + 2i)( z - 2i)
R3 - R1
= 4 (4$ P) - 4 2 P
1.61
Equation of circle
#
#
Option (D) is correct.
The vector Triple Product is
Thus
1.60
Page 13
(101)( 4 - l) + 2 (101) = 0
l =6
Option (A) is correct.
d2 y
+ k2 y = 0
2
dx
or
D2 y + k2 y = 0
The AE is
m2 + k2 = 0
The solution of AE is m = ! ik
Thus y = A sin kx + B cos kx
From x = 0 , y = 0 we get B = 0 and x = a, y = 0 we get
A sin ka = 0
or
sin ka = 0
k = mpx
a
Thus
y = Am sin ` mpx j
a
m
We have
/
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
1.70
Page 14
(A - lI) Xi = 0
1 - (- 5) 2 x1
0
== G
G
G
=
=
4
8 - 4 x2
0
Option (A) is correct.
x
e
1 + ex
For x " 3, the value of f (x) monotonically increases.
We have
1.71
1.72
1.73
f (x) =
1 2 x1
0
=0 0G=x G = = 0 G
2
Option (B) is correct.
Order is the highest derivative term present in the equation and
degree is the power of highest derivative term.
Order = 2 , degree = 1
Option (D) is correct.
Probability of coming odd number is 12 and the probability of
coming even number is 12 . Both the events are independent to each
other, thus probability of coming odd number after an even number
is 12 # 12 = 14 .
x1 + 2x2 = 0
Let - x1 = 2 & x2 =- 1,
The A.E. is
1.78
1.79
1.77
Method-Solving nonlinear eq.
Solving ordinary differential eq.
Numerical Integration
Solving linear simultaneous eq.
-4 2
A ==
4 3G
A - lI = 0
4-l 2
=0
4 3-l
or
(- 4 - l)(3 - l) - 8 = 0
or
- 12 + l + l2 - 8 = 0
or
l2 + l - 20 = 0
or
l =- 5, 4
Eigen vector for l =- 5
1 2a - 0.1b
1 0
==
G
=0
3b
0 1G
Option (A) is correct.
Gaussian PDF is
f (x) = 1
2p s
#
3 - (x - m)2
2s2
-3
e
# f (x) dx
3
-3
Eigen values
dx
for - 3 # x # 3
=1
#
x2
8
or
1 2
2p 2
#
3 - x2
8
e
dx = 1
or
1
2p
#
3 - x2
8
dx = 1
0
0
e
Option (C) is correct.
From orthogonal matrix
[AAT ] = I
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Since the inverse of I is I , thus
[AAT ] -1 = I-1 = I
Characteristic equation is
or
or
Substituting m = 0 and s = 2 in above we get
3 1
e dx = 1
2p 2 - 3
1.80
Option (C) is correct.
We have
1 0
2 - 0.1 1 a
=0 3 G= 2 G = =0 1G
0 b
and
Option (C) is correct.
For x > 0 the slope of given curve is negative. Only (C) satisfy this
condition.
"
"
"
"
or
2a - 0.1 = 0 and 3b = 1
Thus solving above we have b = 1 and a = 1
3
60
Therefore
a+b = 1 + 1 = 7
3 60
20
Option (A) is correct.
Option (C) is correct.
Newton - Raphson
Runge - kutta Method
Simpson’s Rule
Gauss elimination
AA - 1 = I
or
We have
f (t) = e(a + 2) t + 5 = e5 .e(a + 2) t
Taking Laplace transform we get
1
Thus Re (s) > (a + 2)
F (s) = e5 ;
s - (a + 2) E
1.76
Eigen vector
Option (A) is correct.
We have
Now
The CF is
yc = C1 e3x + C2 e2x
Since Q = 0 , thus
y = C1 e3x + C2 e2x
Thus only (B) may be correct.
1.75
2
-1G
1
2 - 0.1
a
-1
and A = = 2 G
A ==
0 3 G
0 b
d2 y
dy
-5
+ 6y = 0
2
dx
dx
m2 - 5m + 6 = 0
m = 3, 2
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1.74
X ==
Thus
Option (B) is correct.
We have
R2 - 4R1
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 15
UNIT 2
NETWORKS
(A) 125/100 and 80/100
(C) 100/100 and 100/100
2013
2.1
2.2
ONE MARK
2.6
Consider a delta connection of resistors and its equivalent star
connection as shown below. If all elements of the delta connection
are scaled by a factor k , k > 0 , the elements of the corresponding
star equivalent will be scaled by a factor of
(A) k2
(B) k
(C) 1/k
(D)
The transfer function
k
V2 ^s h
of the circuit shown below is
V1 ^s h
(B) 100/100 and 80/100
(D) 80/100 and 80/100
Three capacitors C1 , C2 and C 3 whose values are 10 mF , 5 mF , and
2 mF respectively, have breakdown voltages of 10 V, 5 V and 2 V
respectively. For the interconnection shown below, the maximum
safe voltage in Volts that can be applied across the combination,
and the corresponding total charge in mC stored in the effective
capacitance across the terminals are respectively,
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(A) 0.5s + 1
s+1
(C) s + 2
s+1
2.3
(B) 3s + 6
s+2
(D) s + 1
s+2
A source vs ^ t h = V cos 100pt has an internal impedance of ^4 + j3h W
. If a purely resistive load connected to this source has to extract the
maximum power out of the source, its value in W should be
(A) 3
(B) 4
(C) 5
(D) 7
2013
2.4
(B) 7 and 119
(D) 7 and 80
TWO MARKS
Common Data For Q. 8 and 9:
In the circuit shown below, if the source voltage VS = 100+53.13c V
then the Thevenin’s equivalent voltage in Volts as seen by the load
resistance RL is
(A) 100+90c
(C) 800+90c
2.5
(A) 2.8 and 36
(C) 2.8 and 32
(B) 800+0c
(D) 100+60c
The following arrangement consists of an ideal transformer and
an attenuator which attenuates by a factor of 0.8. An ac voltage
VWX1 = 100 V is applied across WX to get an open circuit voltage
VYZ1 across YZ. Next, an ac voltage VYZ2 = 100 V is applied across
YZ to get an open circuit voltage VWX2 across WX. Then, VYZ1 /VWX1
, VWX2 /VYZ2 are respectively,
Consider the following figure
2.7
2.8
2.9
The current IS in Amps in the voltage source, and voltage VS in
Volts across the current source respectively, are
(B) 8, - 10
(A) 13, - 20
(C) - 8, 20
(D) - 13, 20
The current in the 1W resistor in Amps is
(A) 2
(B) 3.33
(C) 10
(D) 12
Two magnetically uncoupled inductive coils have Q factors q1 and q2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 16
at the chosen operating frequency. Their respective resistances are
R1 and R2 . When connected in series, their effective Q factor at the
same operating frequency is
(B) ^1/q1h + ^1/q2h
(A) q1 + q2
(C) ^q1 R1 + q2 R2h / ^R1 + R2h
(D) ^q1 R2 + q2 R1h / ^R1 + R2h
2012
2.10
In the following figure, C1 and C2 are ideal capacitors. C1 has been
charged to 12 V before the ideal switch S is closed at t = 0. The
current i (t) for all t is
(A) zero
(A) 0.8 W
(C) 2 W
ONE MARK
2.14
If VA - VB = 6 V then VC - VD is
(A) - 5 V
(C) 3 V
(B) a step function
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(B) 1.4 W
(D) 2.8 W
(B) 2 V
(D) 6 V
Common Data For Q. 48 and 49 :
With 10 V dc connected at port A in the linear nonreciprocal twoport network shown below, the following were observed :
(i) 1 W connected at port B draws a current of 3 A
(ii) 2.5 W connected at port B draws a current of 2 A
(C) an exponentially decaying function (D) an impulse function
2.11
2.12
The average power delivered to an impedance (4 - j3) W by a current
5 cos (100pt + 100) A is
(A) 44.2 W
(B) 50 W
(C) 62.5 W
(D) 125 W
2.15
In the circuit shown below, the current through the inductor is
2.16
With 10 V dc connected at port A, the current drawn by 7 W
connected at port B is
(A) 3/7 A
(B) 5/7 A
(C) 1 A
(D) 9/7 A
For the same network, with 6 V dc connected at port A, 1 W
connected at port B draws 7/3 A. If 8 V dc is connected to port A
, the open circuit voltage at port B is
(A) 6 V
(B) 7 V
(C) 8 V
(D) 9 V
2011
ONE MARK
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2 A
1+j
(C) 1 A
1+j
(A)
2012
2.13
(B) - 1 A
1+j
2.17
In the circuit shown below, the Norton equivalent current in amperes
with respect to the terminals P and Q is
(D) 0 A
TWO MARKS
Assuming both the voltage sources are in phase, the value of R for
which maximum power is transferred from circuit A to circuit B is
(A) 6.4 - j 4.8
(C) 10 + j 0
2.18
(B) 6.56 - j 7.87
(D) 16 + j 0
In the circuit shown below, the value of RL such that the power
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 17
(A) i (t) = 15 exp (- 2 # 103 t) A
(B) i (t) = 5 exp (- 2 # 103 t) A
(C) i (t) = 10 exp (- 2 # 103 t) A
(D) i (t) =- 5 exp (- 2 # 103 t) A
transferred to RL is maximum is
2010
(A) 5 W
(C) 15 W
2.19
2.23
(B) 10 W
(D) 20 W
For the two-port network shown below, the short-circuit admittance
parameter matrix is
The circuit shown below is driven by a sinusoidal input
vi = Vp cos (t/RC ). The steady state output vo is
4
(A) >
-2
(A) (Vp /3) cos (t/RC )
(C) (Vp /2) cos (t/RC )
(B) (Vp /3) sin (t/RC )
(D) (Vp /2) sin (t/RC )
2011
2.20
TWO MARKS
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2.24
(B) 2.0+0c A
(D) 3.2+0c A
In the circuit shown below, the network N is described by the
following Y matrix:
0.1 S - 0.01 S
. the voltage gain V2 is
Y =>
0.01 S
0.1 SH
V1
2.22
0.5
S
1H
4
(D) >
2
2
S
4H
For parallel RLC circuit, which one of the following statements is
NOT correct ?
(A) The bandwidth of the circuit decreases if R is increased
(B) The bandwidth of the circuit remains same if L is increased
(C) At resonance, input impedance is a real quantity
(D) At resonance, the magnitude of input impedance attains its
minimum value.
2010
2.25
(A) 1/90
(C) –1/99
- 0.5
S
1H
1
(B) >
- 0.5
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In the circuit shown below, the current I is equal to
(A) 1.4+0c A
(C) 2.8+0c A
-2
S
4H
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1
(C) >
0.5
2.21
ONE MARK
TWO MARKS
In the circuit shown, the switch S is open for a long time and is
closed at t = 0 . The current i (t) for t $ 0+ is
(B) –1/90
(D) –1/11
In the circuit shown below, the initial charge on the capacitor is 2.5
mC, with the voltage polarity as indicated. The switch is closed at
time t = 0 . The current i (t) at a time t after the switch is closed is
(A) i (t) = 0.5 - 0.125e-1000t A
(C) i (t) = 0.5 - 0.5e-1000t A
2.26
(B) i (t) = 1.5 - 0.125e-1000t A
(D) i (t) = 0.375e-1000t A
The current I in the circuit shown is
(A) - j1 A
(B) j1 A
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) 0 A
2.27
Page 18
(D) 20 A
battery deliver during this talk-time?
In the circuit shown, the power supplied by the voltage source is
(A) 220 J
(C) 13.2 kJ
(B) 12 kJ
(D) 14.4 J
GATE 2009
(A) 0 W
(B) 5 W
(C) 10 W
(D) 100 W
GATE 2009
2.28
2.31
An AC source of RMS voltage 20 V with internal impedance
Zs = (1 + 2j) W feeds a load of impedance ZL = (7 + 4j) W in the
figure below. The reactive power consumed by the load is
ONE MARK
In the interconnection of ideal sources shown in the figure, it is
known that the 60 V source is absorbing power.
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(A) 8 VAR
(C) 28 VAR
2.32
2.33
Which of the following can be the value of the current source I ?
(A) 10 A
(B) 13 A
(C) 15 A
(D) 18 A
If the transfer function of the following network is
Vo (s)
1
=
Vi (s)
2 + sCR
The value of the load resistance RL is
(B) R
(A) R
2
4
(C) R
2.30
(D) 2R
A fully charged mobile phone with a 12 V battery is good for a 10
minute talk-time. Assume that, during the talk-time the battery
delivers a constant current of 2 A and its voltage drops linearly from
12 V to 10 V as shown in the figure. How much energy does the
(B) 16 VAR
(D) 32 VAR
The switch in the circuit shown was on position a for a long time,
and is move to position b at time t = 0 . The current i (t) for t > 0
is given by
(A) 0.2e-125t u (t) mA
(C) 0.2e-1250t u (t) mA
2.29
TWO MARK
(B) 20e-1250t u (t) mA
(D) 20e-1000t u (t) mA
In the circuit shown, what value of RL maximizes the power delivered
to RL ?
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2.34
(A) 2.4 W
(B) 8 W
3
(C) 4 W
(D) 6 W
The time domain behavior of an RL circuit is represented by
L di + Ri = V0 (1 + Be-Rt/L sin t) u (t).
dt
For an initial current of i (0) = V0 , the steady state value of the
R
current is given by
(A) i (t) " V0
(B) i (t) " 2V0
R
R
(C) i (t) " V0 (1 + B)
(D) i (t) " 2V0 (1 + B)
R
R
GATE 2008
2.35
Page 19
The component values are
(A) L = 5 H, R = 0.5 W, C = 0.1 F
(B) L = 0.1 H, R = 0.5 W, C = 5 F
(C) L = 5 H, R = 2 W, C = 0.1 F
(D) L = 0.1 H, R = 2 W, C = 5 F
2.39
The circuit shown in the figure is used to charge the capacitor C
alternately from two current sources as indicated. The switches S1
and S2 are mechanically coupled and connected as follows:
For 2nT # t # (2n + 1) T , (n = 0, 1, 2,..) S1 to P1 and S2 to P2
For (2n + 1) T # t # (2n + 2) T, (n = 0, 1, 2,...) S1 to Q1 and S2 to
Q2
ONE MARK
In the following graph, the number of trees (P) and the number of
cut-set (Q) are
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(A) P = 2, Q = 2
(C) P = 4, Q = 6
2.36
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In the following circuit, the switch S is closed at t = 0 . The rate of
change of current di (0+) is given by
dt
(A) 0
(R + Rs) Is
(C)
L
(B) Rs Is
L
(D) 3
GATE 2008
2.37
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(B) P = 2, Q = 6
(D) P = 4, Q = 10
TWO MARKS
The Thevenin equivalent impedance Zth between the nodes P and
Q in the following circuit is
Assume that the capacitor has zero initial charge. Given that u (t)
is a unit step function , the voltage vc (t) across the capacitor is
given by
(A)
3
/ (- 1) n tu (t - nT)
n=1
3
(B) u (t) + 2 / (- 1) n u (t - nT)
n=1
3
(C) tu (t) + 2 / (- 1) n u (t - nT) (t - nT)
(A) 1
2.38
(B) 1 + s + 1
s
2
(C) 2 + s + 1
(D) s2 + s + 1
s
s + 2s + 1
The driving point impedance of the following network is given by
Z (s) = 2 0.2s
s + 0.1s + 2
n=1
(D) / 60.5 - e- (t - 2nT) + 0.5e- (t - 2nT) - T @
3
n=1
Common Data For Q. 2.23 & 2.24 :
The following series RLC circuit with zero conditions is excited by
a unit impulse functions d (t).
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2.40
For t > 0 , the output voltage vC ^ t h is
-1
(A) 2 ^e t - e t h
(B) 2 te 2 t
3
3
1
-1
(C) 2 e 2 t cos c 3 t m
(D) 2 e 2 t sin c 3 t m
2
2
3
3
For t > 0 , the voltage across the resistor is
-1
2
2.41
Page 20
(A) 1 _e
3
(B) e
(C)
3t
2
3
2
-e
-1t
2
(A) a low-pass filter
(C) a band-pass filter
i
3
1 sin 3 t
c 2 mG
=cos c 2 t m 3
2 e -21 t sin 3 t
c 2 m
3
-1 t
2
(B) a high-pass filter
(D) a band-reject filter
GATE 2007
2.46
-1
(D) 2 e 2 t cos c 3 t m
2
3
TWO MARKS
Two series resonant filters are as shown in the figure. Let the 3-dB
bandwidth of Filter 1 be B1 and that of Filter 2 be B2 . the value
B1 is
B2
Statement for linked Answers Questions 2.25 & 2.26:
A two-port network shown below is excited by external DC source.
The voltage and the current are measured with voltmeters V1, V2
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(A) 4
(C) 1/2
2.47
(B) 1
(D) 1/4
For the circuit shown in the figure, the Thevenin voltage and
resistance looking into X - Y are
and ammeters. A1, A2 (all assumed to be ideal), as indicated
(A)
(C)
Under following conditions, the readings obtained are:
(1) S1 -open, S2 - closed A1 = 0,V1 = 4.5 V,V2 = 1.5 V, A2 = 1 A
(2) S1 -open, S2 - closed A1 = 4 A,V1 = 6 V,V2 = 6 V, A2 = 0
2.42
1.5
1.5G
4.5
1.5 G
1.5
(B) =
1.5
4.5
(D) =
1.5
(B) 4 V, 23 W
(D) 4 V, 2 W
In the circuit shown, vC is 0 volts at t = 0 sec. For t > 0 , the capacitor
current iC (t), where t is in seconds is given by
(A) 0.50 exp (- 25t) mA
(B) 0.25 exp (- 25t) mA
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-3 -1
(B) =
3 0.67 G
3
1
(D) =
- 3 - 0.67 G
GATE 2007
2.45
4.5
4.5G
1.5
4.5G
The h -parameter matrix for this network is
-3 3
(A) =
- 1 0.67 G
3 3
(C) =
1 0.67 G
2.44
V, 2 W
V, 23 W
The z -parameter matrix for this network is
1.5
(A) =
4.5
1.5
(C) =
1.5
2.43
2.48
4
3
4
3
(C) 0.50 exp (- 12.5t) mA
ONE MARK
An independent voltage source in series with an impedance
Zs = Rs + jXs delivers a maximum average power to a load impedance
ZL when
(B) ZL = Rs
(A) ZL = Rs + jXs
(C) ZL = jXs
(D) ZL = Rs - jXs
2.49
(D) 0.25 exp (- 6.25t) mA
In the ac network shown in the figure, the phasor voltage VAB (in
Volts) is
The RC circuit shown in the figure is
(A) 0
(C) 12.5+30c
(B) 5+30c
(D) 17+30c
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
GATE 2006
2.50
Page 21
TWO MARKS
A two-port network is represented by ABCD parameters given by
V1
A B V2
= I G = =C D G=- I G
1
2
If port-2 is terminated by RL , the input impedance seen at port-1
is given by
(B) ARL + C
(A) A + BRL
BRL + D
C + DRL
(C) DRL + A
(D) B + ARL
BRL + C
D + CRL
2.51
In the two port network shown in the figure below, Z12 and Z21 and
respectively
(A) re and br0
(C) 0 and bro
2.52
2.53
(B) 0 and - br0
(D) re and - br0
The first and the last critical frequencies (singularities) of a driving
point impedance function of a passive network having two kinds of
elements, are a pole and a zero respectively. The above property
will be satisfied by
(A) RL network only
(B) RC network only
(C) LC network only
(D) RC as well as RL networks
(A) Rneg # Re Z1 (jw), 6w
(C) Rneg # Im Z1 (jw), 6w
GATE 2005
2.56
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A 2 mH inductor with some initial current can be represented as
shown below, where s is the Laplace Transform variable. The value
of initial current is
(A) 0.5 A
(C) 1.0 A
(B) 2.0 A
(D) 0.0 A
2.57
In the figure shown below, assume that all the capacitors are initially
uncharged. If vi (t) = 10u (t) Volts, vo (t) is given by
-t/0.004
(A) 8e
Volts
(C) 8u (t) Volts
2.55
ONE MARK
The condition on R, L and C such that the step response y (t) in the
figure has no oscillations, is
L
C
(C) R $ 2
(D) R = 1
LC
The ABCD parameters of an ideal n: 1 transformer shown in the
figure are
n 0
>0 x H
(A) R $ 1
2
2.54
(B) Rneg # Z1 (jw) , 6w
(D) Rneg # +Z1 (jw), 6w
(B) 8 (1 - e
(D) 8 Volts
The value of x will be
(A) n
-t/0.004
) Volts
A negative resistance Rneg is connected to a passive network N
having driving point impedance as shown below. For Z2 (s) to be
positive real,
L
C
L
C
(C) n2
2.58
(B) 1
n
(D) 12
n
In a series RLC circuit, R = 2 kW , L = 1 H, and C = 1 mF The
400
resonant frequency is
(B) 1 # 10 4 Hz
(A) 2 # 10 4 Hz
p
(C) 10 4 Hz
2.59
(B) R $
(D) 2p # 10 4 Hz
The maximum power that can be transferred to the load resistor RL
from the voltage source in the figure is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 22
the figure, then the reading in the ideal voltmeter connected between
a and b is
(A) 1 W
(C) 0.25 W
2.60
(B) 10 W
(D) 0.5 W
The first and the last critical frequency of an RC -driving point
impedance function must respectively be
(A) a zero and a pole
(B) a zero and a zero
(C) a pole and a pole
(D) a pole and a zero
GATE 2005
2.61
(A) 10 3 90c A
2
(B) 10 3 - 90c A
2
(C) 5 60c A
(D) 5 - 60c A
Impedance Z as shown in the given figure is
(B) j9 W
(D) j39 W
For the circuit shown in the figure, Thevenin’s voltage and Thevenin’s
equivalent resistance at terminals a - b is
(A) 5 V and 2 W
(C) 4 V and 2 W
2.64
The h parameters of the circuit shown in the figure are
For the circuit shown in the figure, the instantaneous current i1 (t) is
(A) j29 W
(C) j19 W
2.63
2.65
(B) 0.138 V
(D) 1 V
TWO MARKS
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2.62
(A) 0.238 V
(C) - 0.238 V
(B) 7.5 V and 2.5 W
(D) 3 V and 2.5 W
If R1 = R2 = R4 = R and R3 = 1.1R in the bridge circuit shown in
0.1 0.1
(A) =
- 0.1 0.3G
30 20
(C) =
20 20G
2.66
A square pulse of 3 volts amplitude is applied to C - R circuit
shown in the figure. The capacitor is initially uncharged. The output
voltage V2 at time t = 2 sec is
(A) 3 V
(C) 4 V
GATE 2004
2.67
10 - 1
(B) =
1 0.05G
10 1
(D) =
- 1 0.05G
(B) - 3 V
(D) - 4 V
ONE MARK
Consider the network graph shown in the figure. Which one of the
following is NOT a ‘tree’ of this graph ?
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) a
(C) c
2.68
Page 23
(B) b
(D) d
The equivalent inductance measured between the terminals 1 and 2
for the circuit shown in the figure is
GATE 2004
2.72
(A) L1 + L2 + M
(C) L1 + L2 + 2M
2.69
(B) L1 + L2 - M
(D)L1 + L2 - 2M
The circuit shown in the figure, with R = 1 W, L = 1 H and C = 3 F
3
4
has input voltage v (t) = sin 2t . The resulting current i (t) is
TWO MARKS
For the lattice shown in the figure, Za = j2 W and Zb = 2 W . The
z11 z12
values of the open circuit impedance parameters 6 z @ = =
are
z21 z22 G
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(A)
(B)
(C)
(D)
2.70
5 sin (2t + 53.1c)
5 sin (2t - 53.1c)
25 sin (2t + 53.1c)
25 sin (2t - 53.1c)
For the circuit shown in the figure, the time constant RC = 1 ms.
The input voltage is vi (t) = 2 sin 103 t . The output voltage vo (t) is
equal to
1-j
(A) =
1+j
1+j
(C) =
1-j
2.73
(A) sin (103 t - 45c)
(C) sin (103 t - 53c)
2.71
(B) sin (103 t + 45c)
(D) sin (103 t + 53c)
1+j
1 + jG
1+j
1 - jG
1-j 1+j
(B) =
-1 + j 1 - j G
1 + j -1 + j
(D) =
-1 + j 1 + j G
The circuit shown in the figure has initial current iL (0-) = 1 A
through the inductor and an initial voltage vC (0-) =- 1 V across
the capacitor. For input v (t) = u (t), the Laplace transform of the
current i (t) for t $ 0 is
For the R - L circuit shown in the figure, the input voltage
vi (t) = u (t). The current i (t) is
s+2
s2 + s + 1
(D) 2 1
s +s+1
V (s)
The transfer function H (s) = o
of an RLC circuit is given by
Vi (s)
s
s2 + s + 1
(C) 2 s - 2
s +s+1
(A)
2.74
(B)
106
s + 20s + 106
The Quality factor (Q-factor) of this circuit is
(A) 25
(B) 50
H (s) =
2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) 100
2.75
2.76
(D) 5000
(C) 100
For the circuit shown in the figure, the initial conditions are zero. Its
V (s)
is
transfer function H (s) = c
Vi (s)
1
(A) 2
s + 106 s + 106
103
(C) 2
s + 103 s + 106
Page 24
2.80
Consider the following statements S1 and S2
S1 : At the resonant frequency the impedance of a series RLC
circuit is zero.
S2 : In a parallel GLC circuit, increasing the conductance G results in increase in its Q factor.
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2.77
2.81
2.82
2.79
TWO MARKS
Twelve 1 W resistance are used as edges to form a cube. The resistance
between two diagonally opposite corners of the cube is
(A) 5 W
(B) 1 W
6
(D) 3 W
(C) 6 W
2
5
The current flowing through the resistance R in the circuit in the
figure has the form P cos 4t where P is
(A) (0.18 + j0.72)
(C) - (0.18 + j1.90)
(B) (0.46 + j1.90)
(D) - (0.192 + j0.144)
The circuit for Q. 2.66 & 2.67 is given below.
Assume that the switch S is in position 1 for a long time and
thrown to position 2 at t = 0 .
ONE MARK
The minimum number of equations required to analyze the circuit
shown in the figure is
(A) 3
(C) 6
2.78
2
(B) d 2i + 2 di + 2i (t) = cos t
dt
dt
2
(D) d 2i + 2 di + 2i (t) = sin t
dt
dt
GATE 2003
Which one of the following is correct?
(A) S1 is FALSE and S2 is TRUE
(B) Both S1 and S2 are TRUE
(C) S1 is TRUE and S2 is FALSE
(D) Both S1 and S2 are FALSE
GATE 2003
The differential equation for the current i (t) in the circuit of the
figure is
2
(A) 2 d 2i + 2 di + i (t) = sin t
dt
dt
2
(C) 2 d 2i + 2 di + i (t) = cos t
dt
dt
106
(B) 2
s + 103 s + 106
106
(D) 2
s + 106 s + 106
(D) 200
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(B) 4
(D) 7
A source of angular frequency 1 rad/sec has a source impedance
consisting of 1 W resistance in series with 1 H inductance. The load
that will obtain the maximum power transfer is
(A) 1 W resistance
(B) 1 W resistance in parallel with 1 H inductance
(C) 1 W resistance in series with 1 F capacitor
(D) 1 W resistance in parallel with 1 F capacitor
A series RLC circuit has a resonance frequency of 1 kHz and a
quality factor Q = 100 . If each of R, L and C is doubled from its
original value, the new Q of the circuit is
(A) 25
(B) 50
2.83
2.84
At t = 0+ , the current i1 is
(B) - V
(A) - V
R
2R
(C) - V
(D) zero
4R
I1 (s) and I2 (s) are the Laplace transforms of i1 (t) and i2 (t) respectively.
The equations for the loop currents I1 (s) and I2 (s) for the circuit
shown in the figure, after the switch is brought from position 1 to
position 2 at t = 0 , are
V
R + Ls + Cs1 - Ls I1 (s)
s
(A) >
== G
G
1 H=
- Ls
R + Cs I2 (s)
0
R + Ls + Cs1 - Ls I1 (s)
- Vs
(B) >
=
- Ls
R + Cs1 H=I2 (s)G = 0 G
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 25
(A) 25 V
(C) - 50 V
R + Ls + Cs1
- Ls
I1 (s)
- Vs
(C) >
=
G
G
=
=
H
- Ls
R + Ls + Cs1 I2 (s)
0
V
R + Ls + Cs1
- Cs
I1 (s)
s
(D) >
== G
G
1 H=
- Ls
R + Ls + Cs I2 (s)
0
2.85
2.86
GATE 2002
The driving point impedance Z (s) of a network has the pole-zero
locations as shown in the figure. If Z (0) = 3 , then Z (s) is
3 (s + 3)
2 (s + 3)
(B) 2
s + 2s + 3
s + 2s + 2
3 (s + 3)
2 (s - 3)
(C) 2
(D) 2
s + 2s + 2
s - 2s - 3
An input voltage v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c)
V is applied to a series combination of resistance R = 1 W and
an inductance L = 1 H. The resulting steady-state current i (t) in
ampere is
(A) 10 cos (t + 55c) + 10 cos (2t + 10c + tan-1 2)
(A)
(B) 10 cos (t + 55c) + 10 23 cos (2t + 55c)
(C) 10 cos (t - 35c) + 10 cos (2t + 10c - tan-1 2)
2.87
2.90
2
(D) 10 cos (t - 35c) +
3
2
2.91
TWO MARKS
In the network of the fig, the maximum power is delivered to RL if
its value is
(A) 16 W
(B) 40 W
3
(C) 60 W
(D) 20 W
If the 3-phase balanced source in the figure delivers 1500 W at
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cos (2t - 35c)
The impedance parameters z11 and z12 of the two-port network in
the figure are
(A)
(B)
(C)
(D)
(B) 50 V
(D) 0 V
a leading power factor 0.844 then the value of ZL (in ohm) is
approximately
z11 = 2.75 W and z12 = 0.25 W
z11 = 3 W and z12 = 0.5 W
z11 = 3 W and z12 = 0.25 W
z11 = 2.25 W and z12 = 0.5 W
(A) 90+32.44c
(C) 80+ - 32.44c
(B) 80+32.44c
(D) 90+ - 32.44c
GATE 2001
GATE 2002
2.88
2.92
The Voltage e0 in the figure is
The dependent current source shown in the figure
(A) delivers 80 W
(C) delivers 40 W
2.89
ONE MARK
ONE MARK
(A) 2 V
(C) 4 V
(B) absorbs 80 W
(D) absorbs 40 W
In the figure, the switch was closed for a long time before opening
at t = 0 . The voltage vx at t = 0+ is
2.93
2.94
(B) 4/3 V
(D) 8 V
If each branch of Delta circuit has impedance 3 Z , then each
branch of the equivalent Wye circuit has impedance
(B) 3Z
(A) Z
3
(D) Z
(C) 3 3 Z
3
The admittance parameter Y12 in the 2-port network in Figure is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 26
2.98
(A) - 0.02 mho
(C) - 0.05 mho
(B) 0.1 mho
(D) 0.05 mho
GATE 2001
2.95
The z parameters z11 and z21 for the 2-port network in the figure are
(A) z11 = 6 W; z21 = 16 W
11
11
(C) z11 = 6 W; z21 =- 16 W
11
11
TWO MARKS
The voltage e0 in the figure is
(B) z11 = 6 W; z21 = 4 W
11
11
(D) z11 = 4 W; z21 = 4 W
11
11
GATE 2000
2.99
(A) 48 V
(C) 36 V
The circuit of the figure represents a
(B) 24 V
(D) 28 V
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2.96
ONE MARK
(A) Low pass filter
(C) band pass filter
2.100
(B) High pass filter
(D) band reject filter
In the circuit of the figure, the voltage v (t) is
When the angular frequency w in the figure is varied 0 to 3, the
locus of the current phasor I2 is given by
(A) eat - ebt
(C) aeat - bebt
2.101
(B) eat + ebt
(D) aeat + bebt
In the circuit of the figure, the value of the voltage source E is
(A) - 16 V
(B) 4 V
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(C) - 6 V
2.97
In the figure, the value of the load resistor RL which maximizes the
power delivered to it is
(A) 14.14 W
(C) 200 W
(B) 10 W
(D) 28.28 W
GATE 2000
2.102
(D) 16 V
TWO MARKS
Use the data of the figure (a). The current i in the circuit of the
figure (b)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 27
2.107
(A) - 2 A
(C) - 4 A
A Delta-connected network with its Wye-equivalent is shown in the
given figure. The resistance R1, R2 and R3 (in ohms) are respectively
(A) 1.5, 3 and 9
(C) 9, 3 and 1.5
(B) 2 A
(D) 4 A
(B) 3, 9 and 1.5
(D) 3, 1.5 and 9
GATE 1998
GATE 1999
2.103
ONE MARK
2.108
Identify which of the following is NOT a tree of the graph shown in
the given figure is
2.109
(A) begh
(C) abfg
2.104
ONE MARK
A network has 7 nodes and 5 independent loops. The number of
branches in the network is
(A) 13
(B) 12
(C) 11
(D) 10
The nodal method of circuit analysis is based on
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(D) aegh
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A 2-port network is shown in the given figure. The parameter h21 for
this network can be given by
(A) KVL and Ohm’s law
(C) KCL and KVL
2.110
(A) - 1/2
(C) - 3/2
(B) + 1/2
(D) + 3/2
GATE 1999
2.105
2.111
The parallel RLC circuit shown in the figure is in resonance. In this
circuit
(A) IR < 1 mA
(C) IR + IC < 1 mA
2.112
2.106
Superposition theorem is NOT applicable to networks containing
(A) nonlinear elements
(B) dependent voltage sources
(C) dependent current sources
(D) transformers
TWO MARK
The Thevenin equivalent voltage VTH appearing between the
terminals A and B of the network shown in the given figure is given
by
(A) j16 (3 - j4)
(C) 16 (3 + j4)
(B) KCL and Ohm’s law
(D) KCL, KVL and Ohm’s law
(B) j16 (3 + j4)
(D) 16 (3 - j4)
2.113
(B) IR + IL > 1 mA
(D) IR + IC > 1 mA
0 - 1/2
The short-circuit admittance matrix a two-port network is >
1/2 0 H
The two-port network is
(A) non-reciprocal and passive
(B) non-reciprocal and active
(C) reciprocal and passive
(D) reciprocal and active
The voltage across the terminals a and b in the figure is
The value of R (in ohms) required for maximum power transfer in
the network shown in the given figure is
(A) 2
(C) 8
(B) 4
(D) 16
(A) 0.5 V
(C) 3.5 V
2.114
(B) 3.0 V
(D) 4.0 V
A high-Q quartz crystal exhibits series resonance at the frequency
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
ws and parallel resonance at the frequency wp . Then
(A) ws is very close to, but less than wp
(B) ws << wp
(C) ws is very close to, but greater than wp
(D) ws >> wp
GATE 1997
2.115
Page 28
(A) 10 V
(C) 5 V
2.119
In the circuit of the figure is the energy absorbed by the 4 W resistor
in the time interval (0, 3) is
ONE MARK
The current i4 in the circuit of the figure is equal to
(A) 36 Joules
(C) 256 Joules
2.120
(A) 12 A
(C) 4 A
2.116
(B) 15 V
(D) None of the above
(B) 16 Joules
(D) None of the above
In the circuit of the figure the equivalent impedance seen across
terminals a, b, is
(B) - 12 A
(D) None or these
The voltage V in the figure equal to
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(A) b 16 l W
3
(C) b 8 + 12j l W
3
(B) b 8 l W
3
(D) None of the above
GATE 1996
2.121
(A) 3 V
(C) 5 V
2.117
ONE MARK
In the given figure, A1, A2 and A3 are ideal ammeters. If A2 and A3
read 3 A and 4 A respectively, then A1 should read
(B) - 3 V
(D) None of these
(A) 1 A
The voltage V in the figure is always equal to
(B) 5 A
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(C) 7 A
(A) 9 V
(C) 1 V
2.118
(B) 5 V
(D) None of the above
2.122
The voltage V in the figure is
The number of independent loops for a network with n nodes and
b branches is
(A) n - 1
(B) b - n
(C) b - n + 1
(D) independent of the number of nodes
GATE 1996
2.123
(D) None of these
TWO MARKS
The voltages VC1, VC2, and VC3 across the capacitors in the circuit in
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 29
the given figure, under steady state, are respectively.
(A) 80 V, 32 V, 48 V
(C) 20 V, 8 V, 12 V
(B) 80 V, 48 V, 32 V
(D) 20 V, 12 V, 8 V
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 30
SOLUTIONS
2.1
Option (B) is correct.
In the equivalent star connection, the resistance can be given as
Rb Ra
RC =
Ra + Rb + Rc
Ra Rc
RB =
Ra + Rb + Rc
Rb Rc
RA =
Ra + Rb + Rc
As the circuit open across RL so
I2 = 0
or,
j40I2 = 0
i.e., the dependent source in loop 1 is short circuited. Therefore,
^ j4h Vs
VL1 =
j4 + 3
j40
100 53.13c
j4 + 3
40 90c
100 53.13c
=
5 53.13c
= 800 90c
VTh = 10 VL1 =
2.5
So, if the delta connection components Ra , Rb and Rc are scaled
by a factor k then
^k Rb h^k Rc h
RAl =
kRa + kRb + kRc
Option (C) is correct.
For the given transformer, we have
V = 1.25
1
VWX
2
Rb Rc
=k
k Ra + Rb + Rc
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Since,
So,
or,
at
= k RA
hence, it is also scaled by a factor k
2.2
2.3
2.4
Option (D) is correct.
For the given capacitance, C = 100mF in the circuit, we have the
reactance.
1
10 4
=
XC = 1 =
sc
s
s # 100 # 10-6
So,
10 4 + 10 4
V2 ^s h
= 4 s
V1 ^s h
10 + 10 4 + 10 4
s
s
= s+1
s+2
Option (C) is correct.
For the purely resistive load, maximum average power is transferred
when
2
2
+ XTh
RL = RTh
where RTh + jXTh is the equivalent thevinin (input) impedance of
the circuit. Hence, we obtain
RL = 42 + 32
5W
VYZ = 0.8 (attenuation factor)
V
VYZ = 0.8 1.25 = 1
^ h^
h
VWX
VYZ = VWX
V
VWX = 100 V ; YZ = 100
100
VWX
1
1
1
at
VWZ = 100 V ;
2
VWX
= 100
100
VYZ
2
2
2.6
Option (C) is correct.
The quality factor of the inductances are given by
q1 = wL1
R1
and
q2 = wL2
R2
So, in series circuit, the effective quality factor is given by
XLeq
= wL 1 + wL 2
Q =
Req
R1 + R 2
q1
q
wL 1 + wL 2
+ 2
q R + q2 R2
R
R
R
R
R
R
1 2
2
= 1 2
= 1 1
= 2
1 + 1
1 + 1
R1 + R 2
R 2 R1
R 2 R1
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2.7
Option (C) is correct.
Option (C) is correct.
For evaluating the equivalent thevenin voltage seen by the load RL
, we open the circuit across it (also if it consist dependent source).
The equivalent circuit is shown below
Consider that the voltage across the three capacitors C1 , C2 and C 3
are V1 , V2 and V3 respectively. So, we can write
V2 = C 3
V3 C 2
....(1)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2.8
Since, Voltage is inversely proportional to capacitance
Now, given that
C1 = 10 mF ; ^V1hmax = 10V
C2 = 5 mF ; ^V2hmax = 5 V
C 3 = 2 mF ; ^V3hmax = 2V
So, from Eq (1) we have
V2 = 2
5
V3
for
^V3hmax = 2
We obtain,
V2 = 2 # 2 = 0.8 volt < 5
5
i.e.,
V2 < ^V2hmax
Hence, this is the voltage at C2 . Therefore,
V3 = 2 volt
V2 = 0.8 volt
and
V1 = V2 + V3 = 2.8 volt
Now, equivalent capacitance across the terminal is
Ceq = C 2 C 3 + C1 = 5 # 2 + 10 = 80 mF
5+2
7
C2 + C3
Equivalent voltage is (max. value)
Vmax = V1 = 2.8
So, charge stored in the effective capacitance is
Q = Ceq Vmax = b 80 l # ^2.8h = 32 mC
7
Option (D) is correct.
Page 31
I (s) =
vc (0) /s
v (0)
= c
1 + 1
1 + 1
C1 C 2
C1 s C 2 s
vC (0) = 12 V
I (s) = b C1 C2 l (12 V) = 12Ceq
C1 + C 2
Taking inverse Laplace transform for the current in time domain,
i (t) = 12Ceq d (t)
2.11
Option (B) is correct.
In phasor form,
Z = 4 - j 3 = 5 - 36.86cW
I = 5 100c A
Average power delivered.
Pavg. = 1 I 2 Z cos q = 1 # 25 # 5 cos 36.86c = 50 W
2
2
Alternate method:
Z = (4 - j3) W ,
I = 5 cos (100pt + 100) A
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2
Pavg = 1 Re $ I Z . = 1 # Re "(5) 2 # (4 - j3),
2
2
= 1 # 100 = 50 W
2
2.12
At the node 1, voltage is given as
V1 = 10 volt
Applying KCL at node 1
IS + V1 + V1 - 2 = 0
2
1
10
10
IS + + - 2 = 0
2
1
IS =- 13 A
Also, from the circuit,
VS - 5 # 2 = V1
VS = 10 + V1
= 20 volt
2.9
2.10
(Impulse)
Option (C) is correct
Applying nodal analysis at top node.
V1 + 1 0c V1 + 1 0c
+
= 1 0c
1
j1
Option (C) is correct.
Again from the shown circuit, the current in 1 W resistor is
I = V1 = 10 = 10 A
1
1
Option (D) is correct.
The s -domain equivalent circuit is shown as below.
2.13
V1 (j 1 + 1) + j 1 + 1 0c = j1
V1 = - 1
1 + j1
1
V1 + 1 0c - 1 + j + 1
Current
=
I1 =
j1
j1
j
=
= 1 A
(1 + j) j 1 + j
Option (A) is correct.
We obtain Thevenin equivalent of circuit B .
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 32
So current in the branch will be
IAB = 6 = 3 A
2
We can see, that the circuit is a one port circuit looking from
terminal BD as shown below
Thevenin Impedance :
ZTh = R
Thevenin Voltage :
For a one port network current entering one terminal, equals the
current leaving the second terminal. Thus the outgoing current from
A to B will be equal to the incoming current from D to C as shown
i.e.
IDC = IAB = 3 A
VTh = 3 0c V
Now, circuit becomes as
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The total current in the resistor 1 W will be
(By writing KCL at node D )
I1 = 2 + IDC
= 2+3 = 5A
So,
VCD = 1 # (- I1) =- 5 V
2.15
I1 = 10 - 3
2+R
Power transfer from circuit A to B
P = (I 12) 2 R + 3I1
2
P = :10 - 3D R + 3 :10 - 3D
2+R
2+R
P = 49R 2 + 21
(2 + R)
(2 + R)
49R + 21 (2 + R)
P =
(2 + R) 2
P = 42 + 70R2
(2 + R)
2
dP = (2 + R) 70 - (42 + 70R) 2 (2 + R) = 0
dR
(2 + R) 4
(2 + R) [(2 + R) 70 - (42 + 70R) 2] = 0
140 + 70R - 84 - 140R = 0
56 = 70R
R = 0.8 W
Option (C) is correct.
When 10 V is connected at port A the network is
Current in the circuit,
2.14
Option (A) is correct.
In the given circuit
VA - VB = 6 V
Now, we obtain Thevenin equivalent for the circuit seen at load
terminal, let Thevenin voltage is VTh, 10 V with 10 V applied at port A
and Thevenin resistance is RTh .
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IL =
VTh,10 V
RTh + RL
For RL = 1 W , IL = 3 A
V
3 = Th,10 V
RTh + 1
For RL = 2.5 W , IL = 2 A
V
= Th,10 V
RTh + 2.5
...(i)
...(ii)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Dividing above two
3 = RTh + 2.5
2
RTh + 1
Page 33
ISC =
2.18
3RTh + 3 = 2RTh + 5
RTh = 2 W
Substituting RTh into equation (i)
Option (C) is correct.
Power transferred to RL will be maximum when RL is equal to the
Thevenin resistance. We determine Thevenin resistance by killing
all source as follows :
VTh,10 V = 3 (2 + 1) = 9 V
Note that it is a non reciprocal two port network. Thevenin voltage
seen at port B depends on the voltage connected at port A. Therefore
we took subscript VTh,10 V . This is Thevenin voltage only when 10 V
source is connected at input port A. If the voltage connected to port
A is different, then Thevenin voltage will be different. However,
Thevenin’s resistance remains same.
Now, the circuit is as shown below :
RTH = 10 # 10 + 10 = 15 W
10 + 10
2.19
For RL = 7 W ,
2.16
IL =
VTh,10 V
= 9 = 1A
2 + RL 2 + 7
Option (A) is correct.
The given circuit is shown below
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Option (B) is correct.
Now, when 6 V connected at port A let Thevenin voltage seen at
port B is VTh,6 V . Here RL = 1 W and IL = 7 A
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VTh, 6 V = RTh # 7 + 1 # 7 = 2 # 7 + 7 = 7 V
3
3
3 3
This is a linear network, so VTh at port B can be written as
VTh = V1 a + b
where V1 is the input applied at port A.
We have V1 = 10 V , VTh,10 V = 9 V
...(i)
`
9 = 10a + b
When V1 = 6 V , VTh, 6 V = 9 V
...(ii)
`
7 = 6a + b
Solving (i) and (ii)
a = 0.5 , b = 4
Thus, with any voltage V1 applied at port A, Thevenin voltage or
open circuit voltage at port B will be
So,
VTh, V = 0.5V1 + 4
For
V1 = 8 V
VTh,8 V = 0.5 # 8 + 4 = 8 = Voc (open circuit voltage)
For parallel combination of R and C equivalent impedance is
R$ 1
jwC
R
=
Zp =
1
j
1
+
wRC
R+
jwC
Transfer function can be written as
R
1 + jwRC
Vout = Z p
=
Vin
Zs + Zp
R
R+ 1 +
jwC 1 + jwRC
Option (A) is correct.
Replacing P - Q by short circuit as shown below we have
Using current divider rule the current Isc is
jwRC
jwRC + (1 + jwRC) 2
j
Here w = 1
=
2
RC
j + (1 + j)
j
=
=1
3
(1 + j) 2 + j
V
= b p l cos (t/RC)
3
=
Vout
Vin
1
2.17
25
(16 0 ) = (6.4 - j4.8) A
25 + 15 + j30
Thus
2.20
v out
Option (B) is correct.
From star delta conversion we have
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Thus
R1 =
Page 34
Ra Rb
6.6
=
= 2W
Ra + Rb + Rc 6 + 6 + 6
Given circuit is as shown below
Here
R1 = R 2 = R 3 = 2 W
Replacing in circuit we have the circuit shown below :
By writing node equation at input port
I1 = V1 + V1 - V2 = 4V1 - 2V2
0.5
0.5
By writing node equation at output port
I2 = V2 + V2 - V1 =- 2V1 + 4V2
0.5
0.5
From (1) and (2), we have admittance matrix
4 -2
Y =>
- 2 4H
Now the total impedance of circuit is
(2 + j4) (2 - j4)
+2 = 7W
Z =
(2 + j4) (2 - j4)
Current
I = 14+0c = 2+0c
7
2.21
2.24
...(2)
Option (D) is correct.
A parallel RLC circuit is shown below :
Option (D) is correct.
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From given admittance matrix we get
I1 = 0.1V1 - 0.01V2 and
...(1)
...(2)
I2 = 0.01V1 + 0.1V2
Now, applying KVL in outer loop;
V2 =- 100I2
or
I2 =- 0.01V2
From eq (2) and eq (3) we have
Now,
Z in =
At resonance
1 = wC
wL
Z in = 1 = R
1/R
(maximum at resonance)
Thus (D) is not true.
Furthermore bandwidth is wB i.e wB \ 1 and is independent of L
R
,
Hence statements A, B, C, are true.
2.25
Option (A) is correct.
Here we take the current flow direction as positive.
At t = 0- voltage across capacitor is
-3
Q
VC (0-) =- =- 2.5 # 10-6 =- 50 V
C
50 # 10
+
Thus
VC (0 ) =- 50 V
In steady state capacitor behave as open circuit thus
1
1 + 1 + jw C
R jwL
Input impedance
So,
...(3)
- 0.01V2 = 0.01V1 + 0.1V2
- 0.11V2 = 0.01V1
V2 = - 1
11
V1
2.22
...(1)
Option (A) is correct.
Let the current
i (t) = A + Be-t/t
t " Time constant
When the switch S is open for a long time before t < 0 , the circuit is
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V (3) = 100 V
VC (t) = VC (3) + (VC (0+) - VC (3)) e-t/RC
= 100 + (- 50 - 100) e
-t
10 # 50 # 10-6
3
Now
= 100 - 150e- (2 # 10 t)
ic (t) = C dV
dt
3
= 50 # 10-6 # 150 # 2 # 103 e-2 # 10 t A
3
= 15e-2 # 10 t
ic (t) = 15 exp (- 2 # 103 t) A
2.23
Option (A) is correct.
At t = 0 , inductor current does not change simultaneously, So the
circuit is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 35
Applying nodal analysis
VA - 10 + 1 + VA - 0 = 0
2
2
Current,
Current is resistor (AB)
i (0) = 0.75 = 0.375 A
2
2VA - 10 + 2 = 0 = V4 = 4 V
I1 = 10 - 4 = 3 A
2
Current from voltage source is
I 2 = I1 - 3 = 0
Since current through voltage source is zero, therefore power
delivered is zero.
Similarly for steady state the circuit is as shown below
2.28
Option (A) is correct.
Circuit is as shown below
i (3) = 15 = 0.5 A
3
-3
= 10-3 sec
t = L = 15 # 10
Req
10 + (10 || 10)
Now
and
So,
Hence
2.26
i (t)
i (0)
i (3)
B
i (t)
t
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= A + Be- 1 # 10 = A + Be-100t
= A + B = 0.375
= A = 0.5
= 0.375 - 0.5 =- 0.125
= 0.5 - 0.125e-1000 t A
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Option (A) is correct.
Circuit is redrawn as shown below
Where,
Z1 = jwL = j # 103 # 20 # 10-3 = 20j
Z2 = R || XC
1
=- 20j
XC = 1 =
jwC
j # 103 # 50 # 10-6
1 (- 20j)
Z2 =
R = 1W
1 - 20j
Since 60 V source is absorbing power. So, in 60 V source current
flows from + to - ve direction
So,
I + I1 = 12
Voltage across Z2
VZ =
2
Z2 : 20 0 =
Z1 + Z 2
=c
- 20j
c 1 - 20j m
20j
c 20j - 1 - 20j m
(- 20j)
: 20 =- j
20j + 400 - 20j m
I = 12 - I1
I is always less then 12 A So, only option (A) satisfies this conditions.
: 20
2.29
Current in resistor R is
j
V
I = Z =- =- j A
1
R
2
2.27
Option (A) is correct.
The circuit can be redrawn as
Option (C) is correct.
For given network we have
(RL XC ) Vi
V0 =
R + (RL XC )
RL
V0 (s)
RL
= 1 + sRL C
=
Vi (s)
R
RR
R
+
L
L sC + RL
R+
1 + sRL C
=
RL
1
=
R + RRL sC + RL
R
1+
+ RsC
RL
But we have been given
V (s)
1
=
T . F. = 0
2 + sCR
Vi (s)
Comparing, we get
1 + R = 2 & RL = R
RL
2.30
Option (C) is correct.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 36
The energy delivered in 10 minutes is
E =
t
t
#0 VIdt = I #0Vdt
= I # Area
= 2 # 1 (10 + 12) # 600 = 13.2 kJ
2
2.31
Option (B) is correct.
From given circuit the load current is
20+0c
=
= 20+0c
IL = V
Zs + ZL
(1 + 2j) + (7 + 4j) 8 + 6j
= 1 (8 - 6j) = 20+0c = 2+ - f where f = tan - 1 3
10+f
5
4
The open circuit voltage is
From fig
The voltage across load is
VL = IL ZL
The reactive power consumed by load is
Vx =- 4 # 12.5 =- 50 V
I2 = 100 + Vx = 100 - 50 = 12.5 A
4
4
Pr = VL IL* = IL ZL # IL* = ZL IL 2
2
= (7 # 4j) 20+0c = (7 + 4j) = 28 + 16j
8 + 6j
Isc = I1 + I2 = 25 A
Rth = Voc = 100 = 4 W
Isc
25
Thus average power is 28 and reactive power is 16.
2.32
Thus for maximum power transfer RL = Req = 4 W
Option (B) is correct.
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Voc = 100 V
I1 = 100 = 12.5 A
8
2.34
2.35
-
At t = 0 , the circuit is as shown in fig below :
Option (A) is correct.
Steady state all transient effect die out and inductor act as short
circuits and forced response acts only. It doesn’t depend on initial
current state. From the given time domain behavior we get that
circuit has only R and L in series with V0 . Thus at steady state
i (t) " i (3) = V0
R
Option (C) is correct.
The given graph is
There can be four possible tree of this graph which are as follows:
V (0-) = 100 V
Thus
V (0+) = 100 V
At t = 0+ , the circuit is as shown below
There can be 6 different possible cut-set.
I (0+) = 100 = 20 mA
5k
At steady state i.e. at t = 3 is I (3)= 0
Now
i (t) = I (0+) e-
t
RCeq
u (t)
(0.5m + 0.3m) 0.2m
= 0.16 m F
0.5m + 0.3m + 0.2m
1
1 =
= 1250
RCeq
5 # 103 # 0.16 # 10-6
i (t) = 20e-1250t u (t) mA
Ceq =
2.33
Option (C) is correct.
For Pmax the load resistance RL must be equal to thevenin resistance
Req i.e. RL = Req . The open circuit and short circuit is as shown
below
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2.36
Option (B) is correct.
Initially i (0-) = 0 therefore due to inductor i (0+) = 0 . Thus all
current Is will flow in resistor R and voltage across resistor will be
Is Rs . The voltage across inductor will be equal to voltage across Rs
as no current flow through R.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Thus
but
Thus
2.37
Page 37
vL (0+) = Is Rs
di (0+)
vL (0+) = L
dt
For 4T < t < 5T , capacitor will be charged from 0 V
di (0+)
vL (0+) Is Rs
=
=
L
L
dt
At t = 5T, Vc = T Volts
Thus the output waveform is
Vc =
t
#4Tdt = t - 4T
Option (A) is correct.
Killing all current source and voltage sources we have,
Only option C satisfy this waveform.
2.40
Zth =
=
(1 + s) ( s1 + 1)
(1 + s)( s1 + 1)
(1 + s) + ( s1 + 1)
=
[ s1 + 1 + 1 + s]
s + s1 + 1 + 1
or
Zth = 1
Alternative :
Here at DC source capacitor act as open circuit and inductor act
as short circuit. Thus we can directly calculate thevenin Impedance as 1 W
2.38
Option (D) is correct.
Since Vs (t) = d (t) " Vs (s) = 1 and
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Z (s) = R 1 sL = 2
sC
s +
We have been given
Z (s) = 2 0.2s
s + 0.1s + 2
Comparing with given we get
1 = 0.2 or C = 5 F
C
1 = 0.1 or R = 2 W
RC
1 = 2 or L = 0.1 H
LC
2.39
s
C
s
RC
+
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1
LC
Option (C) is correct.
Voltage across capacitor is
t
Vc = 1 idt
C 0
Vc (s) =
Vt = 2 e- sin c 3 t m
2
3
t
2
2.41
t
#0 dt
Option (B) is correct.
Let voltage across resistor be vR
VR (s)
s
= 1 1
=
VS (s)
( s + s + 1) (s2 + s + 1)
Since vs = d (t) " Vs (s) = 1 we get
s
=
VR (s) = 2 s
1 2
(s + s + 1) (s + 2 ) + 43
For 0 < t < T , capacitor will be charged from 0 V
#0
=
t
dt = t
or
At t = T, Vc = T Volts
For T < t < 2T , capacitor will be discharged from T volts as
Vc = T -
Vc =
t
At t = 3T, Vc = T Volts
For 3T < t < 4T , capacitor will be discharged from T Volts
Vc = T At t = 4T, Vc = 0 Volts
t
#3Tdt = 4T - t
vR (t) = e- cos
1
2
t
#T dt = 2T - t
#2Tdt = t - 2T
(s + 12 )
(s + 12 ) 2 +
= e- 2 =cos
t
At t = 2T, Vc = 0 volts
For 2T < t < 3T , capacitor will be charged from 0 V
3
2
Vc (s) = 2 =
G
3 (s + 12 ) 2 + 43
Taking inverse Laplace transform we have
Here C = 1 F and i = 1 A. Therefore
Vc =
1
(s + s + 1)
2
or
#
Vc =
Option (D) is correct.
Writing in transform domain we have
1
Vc (s)
= 1 s
= 2 1
Vs (s)
^ s + s + 1h
(s + s + 1)
2.42
3
4
-
(s +
1
2
3
1 2
2) + 4
3 t-1
2 e- sin 3 t
2
2# 3
2
1
2
3 t - 1 sin 3 t
2
2 G
3
Option (C) is correct.
From the problem statement we have
= 6 = 1.5W
z11 = v1
i1 i = 0 4
v
= 4.5 = 4.5W
z12 = 1
i2 i = 0
1
= 6 = 1.5W
z21 = v2
i1 i = 0 4
= 1.5 = 1.5W
z22 = v2
i2 i = 0
1
2
1
2
2
Thus z -parameter matrix is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
z11 z12
1.5 4.5
=z z G = =1.5 1.5 G
21 22
2.43
Option (A) is correct.
From the problem statement we
=
h12 = v1
v2 i = 0
=
h22 = i2
v2 i = 0
1
1
Page 38
2.46
have
4.5 = 3
1.5
1 = 0.67
1.5
From z matrix, we have
v1 = z11 i1 + z12 i2
v2 = z21 i1 + z22 i2
If v2 = 0
i2 = - z21 = - 1.5 =- 1 = h
Then
21
i1
z22
1.5
We know that bandwidth of series RLC circuit is R . Therefore
L
Bandwidth of filter 1 is B1 = R
L1
Bandwidth of filter 2 is B2 = R = R = 4R
L2
L1
L1 /4
B
1
Dividing above equation 1 =
B2
4
2.47
or
i2 =- i1
Putting in equation for v1, we get
v1
i1
Option (D) is correct.
Here Vth is voltage across node also. Applying nodal analysis we get
Vth + Vth + Vth - 2i = 2
2
1
1
But from circuit
i = Vth = Vth
1
v1 = (z11 - z12) i1
= h11 = z11 - z12 = 1.5 - 4.5 =- 3
v2 = 0
Therefore
Vth + Vth + Vth - 2Vth = 2
2
1
1
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or
Vth = 4 volt
From the figure shown below it may be easily seen that the short
circuit current at terminal XY is isc = 2 A because i = 0 due to
short circuit of 1 W resistor and all current will pass through short
circuit.
Hence h -parameter will be
h11 h12
-3 3
=h h G = =- 1 0.67 G
21 22
2.44
Option (D) is correct.
Option (D) is correct.
According to maximum Power Transform Theorem
ZL = Zs* = (Rs - jXs)
2.45
Option (C) is correct.
At w " 3 , capacitor acts as short circuited and circuit acts as shown
in fig below
Here we get V0 = 0
Vi
At w " 0 , capacitor acts as open circuited and circuit look like as
shown in fig below
Here we get also V0 = 0
Vi
So frequency response of the circuit is as shown in fig and circuit is
a Band pass filter.
Therefore
2.48
Rth = Vth = 4 = 2 W
isc
2
Option (A) is correct.
The voltage across capacitor is
At t = 0+ ,
Vc (0+) = 0
At t = 3 ,
VC (3) = 5 V
The equivalent resistance seen by capacitor as shown in fig is
Req = 20 20 = 10kW
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Time constant of the circuit is
t = Req C = 10k # 4m = 0.04 s
Using direct formula
or
Vc (t) = VC (3) - [Vc (3) - Vc (0)] e-t/t
= VC (3) (1 - e-t/t) + VC (0) e-t/t = 5 (1 - e-t/0.04)
Vc (t) = 5 (1 - e-25t)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Now
IC (t) = C
Page 39
dVC (t)
dt
Taking Laplace transform we get
V (s) = sLI (s) - Li (0+)
As per given in question
- Li (0+) =- 1 mV
Thus
i (0+) = 1 mV = 0.5 A
2 mH
= 4 # 10-6 # (- 5 # 25e-25t) = 0.5e-25t mA
2.49
Option (D) is correct.
Impedance
= (5 - 3j) (5 + 3j) =
=
(5 - 3j) # (5 + 3j)
5 - 3j + 5 + 3 j
(5) 2 - (3j) 2
= 25 + 9 = 3.4
10
10
2.54
VAB = Current # Impedance
= 5+30c # 34 = 17+30c
2.50
Option (D) is correct.
The network is shown in figure below.
Now
and
also
From (1) and (2)
Thus
V1 = AV2 - BI2
I1 = CV2 - DI2
V2 =- I2 RL
we get
V1 = AV2 - BI2
CV2 - DI2
I1
...(1)
...(2)
...(3)
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Substituting value of V2 from (3) we get
Input Impedance
Zin = - A # I2 RL - BI2
- C # I2 RL - DI2
or
Zin = ARL + B
CRL + D
2.51
Option (B) is correct.
At initial all voltage are zero. So output is also zero.
Thus
v0 (0+) = 0
At steady state capacitor act as open circuit.
Thus,
The equivalent resistance and capacitance can be calculate after
killing all source
Option (B) is correct.
The circuit is as shown below.
At input port
V1 = re I1
At output port
V2 = r0 (I2 - bI1) =- r0 bI1 + r0 I2
Comparing standard equation
= 1 4 = 0.8 kW
= 4 1 = 5 mF
= Req Ceq = 0.8kW # 5mF = 4 ms
= v 0 (3) - [v 0 (3) - v 0 (0+)] e-t/t
= 8 - (8 - 0) e-t/0.004
v0 (t) = 8 (1 - e-t/0.004) Volts
Req
Ceq
t
v0 (t)
V1 = z11 I1 + z12 I2
V2 = z21 I1 + z22 I2
z12 = 0 and z21 =- r0 b
2.52
Option (B) is correct.
For series RC network input impedance is
Zins = 1 + R = 1 + sRC
sC
sC
Thus pole is at origin and zero is at - 1
RC
2.55
Thus pole is at - 1 and zero is at infinity.
RC
Option (A) is correct.
We know
v = Ldi
dt
Option (A) is correct.
Here
Z2 (s) = Rneg + Z1 (s)
or
Z2 (s) = Rneg + Re Z1 (s) + j Im Z1 (s)
For Z2 (s) to be positive real, Re Z2 (s) $ 0
Thus
Rneg + Re Z1 (s) $ 0
or
Re Z1 (s) $- Rneg
But Rneg is negative quantity and - Rneg is positive quantity.
Therefore
For parallel RC network input impedance is
1 R
sC
sC
=
Zin =
1 +R
1 + sRC
sC
2.53
v0 (3) = 4 # vi = 4 # 10 = 8
5
5
or
2.56
Re Z1 (s) $ Rneg
Rneg # Re Z1 (jw)
Option (C) is correct.
Transfer function is
For all w.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 40
1
Y (s)
1
sC
=
= 2
1
U (s)
s LC + scR + 1
R + sL +
sC
For RC parallel network the driving point impedance is
R 1
R
Cs =
Zinp =
1
1
+
sRC
R+
Cs
1
LC
=
s2 + R s + 1
L
LC
2
Comparing with s + 2xwn s + = 0 we have
Here
2xwn = R ,
L
and
wn = 1
LC
Thus
x = R LC = R C
2L
2
L
For no oscillations, x $ 1
R C $1
Thus
2
L
or
2.57
Here pole is s =- 1/RC and zero is at 3, therefore first critical
frequency is a pole and last critical frequency is a zero.
wn2
R $2
2.61
i1 (t) + 5+0c = 10+60c
or
or
2.62
L
C
Option (B) is correct.
For given transformer
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2.58
2.63
V1 = AV2 + BI2
I1 = CV2 + DI2
n 0
A B
=C D G = = 0 1 G
n
x = 1
n
Option (B) is correct.
Open circuit at terminal ab is shown below
or
Vab = 7.5 = Vth
Short circuit at terminal ab is shown below
Option (B) is correct.
Short circuit current from terminal ab is
Isc = 1 + 10 = 3 A
5
Resonant frequency
f0 =
1
==
2p LC
2p
3
4
= 10 # 20 = 10 Hz
p
2p
1
1 # 1 # 10 - 6
400
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Option (C) is correct.
Maximum power will be transferred when RL = Rs = 100W
In this case voltage across RL is 5 V, therefore
Thus
Option (C) is correct.
For stability poles and zero interlace on real axis. In RC series
network the driving point impedance is
Zins = R + 1 = 1 + sRC
sC
Cs
Here pole is at origin and zero is at s =- 1/RC , therefore first
critical frequency is a pole and last critical frequency is a zero.
Rth = Vth = 7.5 = 2.5 W
Isc
3
Here current source being in series with dependent voltage source
make it ineffective.
2
Pmax = V = 5 # 5 = 0.25 W
R
100
2.60
Option (B) is correct.
If L1 = j5W and L3 = j2W the mutual induction is subtractive because
current enters from dotted terminal of j2W coil and exit from dotted
terminal of j5W. If L2 = j2W and L3 = j2W the mutual induction is
additive because current enters from dotted terminal of both coil.
Thus
Z = L1 - M13 + L2 + M23 + L3 - M31 + M32
Applying KCL at node we get
Vab + Vab - 10 = 1
5
5
We have L = 1H and C = 1 # 10-6
400
2.59
i1 (t) = 10+60c - 5+0c = 5 + 5 3j - 5
i1 (t) = 5 3 +90c = 10 3 +90c
2
= j5 + j10 + j2 + j10 + j2 - j10 + j10 = j9
I2 = V1 = n
V2
1
I1
or
I1 = I2 and V1 = nV2
n
Comparing with standard equation
Thus
Option (A) is correct.
Applying KCL we get
2.64
Option (C) is correct.
Here Va = 5 V because R1 = R2 and total voltage drop is 10 V.
Now
Vb = R3 # 10 = 1.1 # 10 = 5.238 V
R3 + R4
2.1
V = Va - Vb = 5 - 5.238 =- 0.238 V
2.65
Option (D) is correct.
For h parameters we have to write V1 and I2 in terms of I1 and V2 .
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 41
V1 = h11 I1 + h12 V2
I2 = h21 I1 + h22 V2 Applying KVL at input port
V1 = 10I1 + V2
Applying KCL at output port
V2 = I + I
1
2
20
or
I2 =- I1 + V2
20
or
Taking inverse Laplace transform
i (t) = 1 (1 - e-2t) u (t)
2
At t = 0 ,
i (t) = 0
At t = 12 ,
i (t) = 0.31
At t = 3 ,
i (t) = 0.5
Graph (C) satisfies all these conditions.
Thus from above equation we get
h11 h12
10 1
=h h G = =- 1 0.05G
12 22
2.66
Option (B) is correct.
RC = 0.1 # 10 - 6 # 103 = 10 - 4 sec
Since time constant RC is very small, so steady state will be
reached in 2 sec. At t = 2 sec the circuit is as shown in fig.
Time constant
Z (s) = s + 2
V (s)
1
=
I (s) = i
s + 2 s (s + 2)
I (s) = 1 ; 1 - 1 E
2 s s+2
Impedance
2.72
Option (D) is correct.
We know that
where
V1 = z11 I1 + z12 I2
V2 = z11 I1 + z22 I2
z11 = V1
I1 I = 0
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2
Vc = 3 V
V2 =- Vc =- 3 V
2.67
2.68
2.69
Option (B) is correct.
For a tree there must not be any loop. So a, c, and d don’t have any
loop. Only b has loop.
Option (D) is correct.
The sign of M is as per sign of L If current enters or exit the dotted
terminals of both coil. The sign of M is opposite of L If current
enters in dotted terminal of a coil and exit from the dotted terminal
of other coil.
Thus
Leq = L1 + L2 - 2M
Option (A) is correct.
Here w = 2 and V = 1+0c
Y = 1 + jwC + 1
R
jwL
= 3 + j2 # 3 + 1 1 = 3 + j4
j2 # 4
= 5+ tan - 1 4 = 5+53.11c
3
I = V * Y = (1+0c)( 5+53.1c) = 5+53.1c
i (t) = 5 sin (2t + 53.1c)
Thus
2.70
Option (A) is correct.
vi (t) =
Here w = 10 rad and Vi =
Now
3
2 sin 10 t
3
2 +0c
1
jwC
1
V
V0 =
.Vt =
+
wCR i
1
1
j
R+
jwC
1
1 + j # 103 # 10 - 3
= 1 - 45c
v0 (t) = sin (103 t - 45c)
=
2.71
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Option (C) is correct.
Input voltage
Taking Laplace transform
2 + 0c
vi (t) = u (t)
Vi (s) = 1
s
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z21 = V2
I1
I1 = 0
Consider the given lattice network, when I2 = 0 . There is two similar path in the circuit for the current I1. So I = 1 I1
2
For z11 applying KVL at input port we get
V1 = I (Za + Zb)
Thus
V1 = 1 I1 (Za + Zb)
2
z11 = 1 (Za + Zb)
2
For Z21 applying KVL at output port we get
V2 = Za I1 - Zb I1
2
2
Thus
V2 = 1 I1 (Za - Zb)
2
z21 = 1 (Za - Zb)
2
For this circuit z11 = z22 and z12 = z21. Thus
V
R
S Za + Zb Za - Zb W
z11 z12
2
2
=z z G = SS Za - Zb Za + Zb WW
21 22
S 2
2 W
X
T
Here Za = 2j and Zb = 2W
1+j j-1
z11 z12
Thus
=z z G = = j - 1 1 + j G
21 22
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2.73
Page 42
Option (B) is correct.
Applying KVL,
= minimum number of equation
Ldi (t) 1
+
v (t) = Ri (t) +
dt
C
#0
3
Number of branches = b = 8
Number of nodes = n = 5
Minimum number of equation
i (t) dt
Taking L.T. on both sides,
V (s)
v (t)
Hence
1
s
2 +1
s
or
2.74
I (s)
I (s) vc (0+)
= RI (s) + LsI (s) - Li (0 ) +
+
sC
sC
= u (t) thus V (s) = 1
s
I (s) 1
= I (s) + sI (s) - 1 +
s
s
I (s) 2
=
6s + s + 1@
s
= 2s+2
s +s+1
+
= 8-5+1 = 4
2.78
2.79
Thus
Now
2.75
2.80
Option (C) is correct.
Applying KVL we get,
di (t) 1
+
i (t) dt
dt
C
di (t)
+ i (t) dt
sin t = 2i (t) + 2
dt
#
#
or
Differentiating with respect to t , we get
2di (t) 2d2 i (t)
cos t =
+ i (t)
+
dt
dt2
2.81
Option (A) is correct.
For current i there is 3 similar path. So current will be divide in
three path
Option (D) is correct.
V0 (s)
Vi (s)
1
1
sC
=
= 2
1
s LC + sCR + 1
R + sL +
sC
1
= 2 -2
-4
s (10 # 10 ) + s (10-4 # 10 4) + 1
106
= -6 2 1
= 2
10 s + s + 1 s + 106 s + 106
Option (D) is correct.
Impedance of series RLC circuit at resonant frequency is minimum,
not zero. Actually imaginary part is zero.
Z = R + j ` wL - 1 j
wC
At resonance wL - 1 = 0 and Z = R that is purely resistive.
wC
Thus S1 is false
Now quality factor
Q =R C
L
Since G = 1 ,
Q = 1 C
G
L
R
If G - then Q . provided C and L are constant. Thus S2 is also
false.
2.77
L =Q
C
2
sin t = Ri (t) + L
2xw = 20
2x = 203 = 0.02
10
Q = 1 = 1 = 50
2x
0.02
H (s) =
2.76
L
C
When R, L and C are doubled,
2L = 1
Q' = 1
2R 2C
2R
Thus
Q' = 100 = 50
2
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106 = 103
Option (B) is correct.
Q = 1
R
s2 + 20s + 106 = 0
wn =
ZL = ZS* = Rs - jXs
ZL = 1 - 1j
Thus
Option (B) is correct.
Characteristics equation is
Comparing with s2 + 2xwn s + wn2 = 0 we have
Option (C) is correct.
For maximum power transfer
Option (B) is correct.
Number of loops = b - n + 1
so, we get
Vab - b i # 1l - b i # 1l - b 1 # 1l = 0
3
6
3
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Vab = R = 1 + 1 + 1 = 5 W
eq
6
i
3 6 3
2.82
2.83
Option ( ) is correct.
Data are missing in question as L1 &L2 are not given.
Option (A) is correct.
At t = 0 - circuit is in steady state. So inductor act as short circuit
and capacitor act as open circuit.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 43
Now
At t = 0 - ,
i1 (0 -) = i2 (0 -) = 0
vc (0 -) = V
At t = 0+ the circuit is as shown in fig. The voltage across capacitor and current in inductor can’t be changed instantaneously. Thus
i1 = i2 =- V
2R
2.84
Option (C) is correct.
When switch is in position 2, as shown in fig in question, applying
KVL in loop (1),
RI1 (s) + V + 1 I1 (s) + sL [I1 (s) - I2 (s)] = 0
s
sC
or
I1 (s) 8R + 1 + sL B - I2 (s) sL = - V
s
sc
=
10 2 cos (t + 10c) 10 5 cos (2t + 10c)
+
1+j
1 + j2
=
10 2 cos (t + 10c) 10 5 cos (2t + 10c)
+
12 + 22 + tan-1 1
12 + 22 tan-1 2
10 2 cos (t + 10c) 10 5 cos (2t + 10c)
+
2 + tan-1 45c
5 tan-1 2
i (t) = 10 cos (t - 35c) + 10 cos (2t + 10c - tan-1 2)
=
2.87
At t = 0+ ,
Z1 = R + jw1 L = 1 + j1
Z2 = R + jw2 L = 1 + j2
v (t) v (t)
i (t) = 1 + 2
Z1
Z2
Option (A) is correct.
Using 3- Y conversion
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z11 I1 + z12 I2 = V1
Applying KVL in loop 2,
sL [I2 (s) - I1 (s)] + RI2 (s) + 1 I2 (s) = 0
sC
2.85
2.86
Z12 I1 + Z22 I2 = V2
or
- sLI1 (s) + 8R + sL + 1 BI2 (s) = 0
sc
Now comparing with
Z11 Z12 I1
V1
=Z Z G=I G = =V G
21
22
2
2
we get
R
V
- sL
SR + sL + 1
W I1 (s)
-V
sC
S
W
sH
=
G
=
>
SS
- sL
R + sL + 1 WW I2 (s)
0
sC
T
X
Option (B) is correct.
Zeros =- 3
Pole1 =- 1 + j
Pole 2 =- 1 - j
K (s + 3)
Z (s) =
(s + 1 + j)( s + 1 - j)
K (s + 3)
K (s + 3)
=
=
2
2
(s + 1) - j
(s + 1) 2 + 1
From problem statement Z (0) w = 0 = 3
Thus 3K = 3 and we get K = 2
2
2 (s + 3)
Z (s) = 2
s + 2s + 2
2 # 1 = 2 = 0.5
2+1+1
4
R2 = 1 # 1 = 1 = 0.25
2+1+1
4
R3 = 2 # 1 = 0.5
2+1+1
R1 =
Now the circuit is as shown in figure below.
Now
z11 = V1
I1
2.88
Option (A) is correct.
Applying KCL at for node 2,
v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c)
1 4444 2 4444 3 1 4444
4 2 4444
43
v1
= 2 + 0.5 + 0.25 = 2.75
z12 = R3 = 0.25
Option (C) is correct.
Thus we get w1 = 1 and w2 = 2
I2 = 0
v2
V2 + V2 - V1 = V1
5
5
5
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
or
V2 = V1 = 20 V
Voltage across dependent current source is 20 thus power delivered by it is
PV2 # V1 = 20 # 20 = 80 W
5
5
Page 44
2.91
It deliver power because current flows from its +ive terminals.
2.89
Option (C) is correct.
When switch was closed, in steady state, iL (0 -) = 2.5 A
Option (D) is correct.
IP , VP " Phase current and Phase voltage
IL, VL " Line current and line voltage
Now
VP = c VL m and IP = IL
3
So,
Power = 3VP IL cos q
1500 = 3 c VL m (IL) cos q
3
also
IL = c VL m
3 ZL
1500 = 3 c VL mc VL m cos q
3
3 ZL
ZL =
As power factor is leading
So,
cos q = 0.844 " q = 32.44
As phase current leads phase voltage
At t = 0+ , iL (0+) = iL (0 -) = 2.5 A and all this current of will pass
through 2 W resistor. Thus
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(400) 2 (.844)
= 90 W
1500
ZL = 90+ - q = 90+ - 32.44c
2.92
Option (C) is correct.
Applying KCL, we get
e0 - 12 + e0 + e0 = 0
4
4 2+2
or
2.93
e0 = 4 V
Option (A) is correct.
The star delta circuit is shown as below
Vx =- 2.5 # 20 =- 50 V
2.90
Option (A) is correct.
For maximum power delivered, RL must be equal to Rth across same
terminal.
Here
Applying KCL at Node, we get
0.5I1 = Vth + I1
20
or
but
Thus
and
Vth + 10I1 = 0
I1 = Vth - 50
40
Vth + Vth - 50 = 0
4
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or
Vth = 10 V
For Isc the circuit is shown in figure below.
Now
2.94
but
ZA = ZB = ZC =
3Z 3Z
= Z
3Z+ 3Z+ 3Z
3
Option (C) is correct.
y11 y12
y1 + y3 - y3
=y y G = = - y y + y G
21 22
3
2
3
y12 =- y3
y12 =- 1 =- 0.05 mho
20
Isc = 0.5I1 - I1 =- 0.5I1
I1 =- 50 =- 1.25 A
40
Isc =- 0.5 # - 12.5 = 0.625 A
Rth = Vth = 10 = 16 W
Isc
0.625
ZAB = ZBC = ZCA = 3 Z
ZAB ZCA
ZA =
ZAB + ZBC + ZCA
ZAB ZBC
ZB =
ZAB + ZBC + ZCA
ZBC ZCA
ZC =
ZAB + ZBC + ZCA
2.95
Option (D) is correct.
We apply source conversion the circuit as shown in fig below.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 45
Now applying nodal analysis we have
e0 - 80 + e0 + e0 - 16 = 0
6
10 + 2 12
4e0 = 112
e0 = 112 = 28 V
4
or
2.96
V0 = RL
Vs
RL + Rs
At resonant frequency w =
V0 = 0 .
(finite value)
1 circuit acts as shown in fig and
LC
Option (A) is correct.
jw C
I2 = Em +01c = Em +0c
1 + jwCR2
R2 + jwC
+90c
+ tan-1 wCR2
E m wC
+ (90c - tan-1 wCR2)
I2 =
2 2
2
1 + w C R2
At w = 0
I2 = 0
and at w = 3,
I2 = Em
R2
Only fig. given in option (A) satisfies both conditions.
+I2 =
2.97
Thus it is a band reject filter.
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Option (A) is correct.
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Xs = wL = 10 W
For maximum power transfer
RL =
2.98
Rs2 + Xs2 =
102 + 102 = 14.14 W
2.100
Option (C) is correct.
Applying KVL in LHS loop
or
E1 = 2I1 + 4 (I1 + I2) - 10E1
E1 = 6I1 + 4I2
11
11
Option (D) is correct.
Applying KCL we get
Now
2.101
Thus z11 = 6
11
iL = eat + ebt
V (t) = vL = L diL = L d [eat + ebt] = aeat + bebt
dt
dt
Option (A) is correct.
Going from 10 V to 0 V
Applying KVL in RHS loop
E2 = 4 (I1 + I2) - 10E1
= 4 (I1 + I2) - 10 c 6I1 + 4I2 m
11
11
=- 16I1 + 4I2
11
11
Thus z21 =- 16
11
2.99
Option (D) is correct.
At w = 0 , circuit act as shown in figure below.
or
2.102
V0 = RL
Vs
RL + Rs
(finite value)
At w = 3 , circuit act as shown in figure below:
2.103
10 + 5 + E + 1 = 0
E =- 16 V
Option (C) is correct.
This is a reciprocal and linear network. So we can apply reciprocity
theorem which states “Two loops A & B of a network N and if an
ideal voltage source E in loop A produces a current I in loop B ,
then interchanging positions an identical source in loop B produces
the same current in loop A. Since network is linear, principle of
homogeneity may be applied and when volt source is doubled,
current also doubles.
Now applying reciprocity theorem
i = 2 A for 10 V
V = 10 V, i = 2 A
V =- 20 V, i =- 4 A
Option (C) is correct.
Tree is the set of those branch which does not make any loop and
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 46
Req = 5W 20W + 4W
connects all the nodes.
abfg is not a tree because it contains a loop l node (4) is not connected
Req = 5.20 + 4 = 4 + 4 = 8 W
5 + 20
2.107
2.104
Option (A) is correct.
For a 2-port network the parameter h21 is defined as
h21 = I2
I1 V = 0 (short circuit)
2
2.108
2.109
Applying node equation at node a we get
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Now
and
Thus
Va - V1 + Va - 0 + Va - 0 = 0
R
R
R
& Va = V1
3Va = V1
3
V1 - V1
V
3 = 2V1
1 - Va
=
I1 =
R
R
3R
0 - V1 - V
0
V
3 =
a
1
=
I2 =
R
R
3R
- V1 /3R - 1
I2
= h21 =
=
2
I1 V = 0
2V1 /3R
2.110
2.111
2.112
2.113
Option (D) is correct.
Delta to star conversion
Rab Rac
= 5 # 30 = 150 = 3 W
R1 =
50
Rab + Rac + Rbc 5 + 30 + 15
Rab Rbc
= 5 # 15 = 1.5 W
R2 =
Rab + Rac + Rbc 5 + 30 + 15
Rac Rbc
= 15 # 30 = 9 W
R3 =
Rab + Rac + Rbc 5 + 30 + 15
Option (C) is correct.
No. of branches = n + l - 1 = 7 + 5 - 1 = 11
Option (B) is correct.
In nodal method we sum up all the currents coming & going at
the node So it is based on KCL. Furthermore we use ohms law to
determine current in individual branch. Thus it is also based on
ohms law.
Option (A) is correct.
Superposition theorem is applicable to only linear circuits.
Option (B) is correct.
Option (B) is correct.
For reciprocal network y12 = y21 but here y12 =- 12 ! y21 = 12 . Thus
circuit is non reciprocal. Furthermore only reciprocal circuit are
passive circuit.
Option (C) is correct.
Taking b as reference node and applying KCL at a we get
Vab - 1 + Vab = 3
2
2
or
or
2.114
2.115
Vab - 1 + Vab = 6
Vab = 6 + 1 = 3.5 V
2
Option (A) is correct.
Option (B) is correct.
The given figure is shown below.
2
2.105
Option (A) is correct.
Applying node equation at node A
Vth - 100 (1 + j0) Vth - 0
+
=0
3
4j
or
or
By simplifying
4jVth - 4j100 + 3Vth = 0
Vth (3 + 4j) = 4j100
4j100
Vth =
3 + 4j
Vth =
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4j100
3 - 4j
3 + 4j # 3 - 4j
Vth = 16j (3 - j4)
2.106
Option (C) is correct.
For maximum power transfer RL should be equal to RTh at same
terminal.
so, equivalent Resistor of the circuit is
Applying KCL at node a we have
I = i 0 + i1 = 7 + 5 = 12 A
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
In the steady state condition all capacitors behaves as open circuit
& Inductors behaves as short circuits as shown below :
Applying KCL at node f
so
2.116
Page 47
I =- i 4
i 4 =- 12 amp
Option (A) is correct.
Thus voltage across capacitor C1 is
VC = 100 # 40 = 80 V
10 + 40
1
so
2.117
2.118
2.119
Now the circuit faced by capacitor C2 and C 3 can be drawn as
below :
V = 3 - 0 = 3 volt
Option (D) is correct.
Can not determined V without knowing the elements in box.
Option (A) is correct.
The voltage V is the voltage across voltage source and that is 10 V.
Option (B) is correct.
Voltage across capacitor
-t
VC (t) = VC (3) + (VC (0) - VC (3)) e RC
Here VC (3) = 10 V and (VC (0) = 6 V. Thus
-t
Now
= 10 - 10 + 4e
Energy absorbed by resistor
-t
RC
-t
= 4e
#0 3V RR(t) = #0 3 164e 4
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-t
RC
-t
2
E
2.120
-t
VC (t) = 10 + (6 - 10) e RC = 10 - 4e RC = 10 - 4e 8
VR (t) = 10 - VC (t)
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=
-t
#0 3 4e 4
= 16 J
Option (B) is correct.
It is a balanced whetstone bridge
R1 R 3
b R2 = R 4 l
so equivalent circuit is
Voltage across capacitor C2 and C 3 are
VC = 80 C 3 = 80 # 3 = 48 volt
5
C2 + C3
VC = 80 C2 = 80 # 2 = 32 volt
5
C2 + C3
2
3
Zeq = (4W 8W) = 4 # 8 = 8
3
4+8
2.121
Option (B) is correct.
Current in A2 ,
Inductor current can be
Current in A 3 ,
Total current
I2 = 3 amp
defined as I2 =- 3j
I3 = 4
I1 = I 2 + I 3
I1 = 4 - 3j
I =
2.122
2.123
(4) 2 + (3) 2 = 5 amp
Option (C) is correct.
For a tree we have (n - 1) branches. Links are the branches which
from a loop, when connect two nodes of tree.
so if total no. of branches = b
No. of links = b - (n - 1) = b - n + 1
Total no. of links in equal to total no. of independent loops.
Option (B) is correct.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
UNIT 3
Page 47
3.6
In the CMOS circuit shown, electron and hole mobilities are equal,
and M1 and M2 are equally sized. The device M1 is in the linear
region if
ELECTRONICS DEVICES
2013
3.1
3.2
3.3
ONE MARK
In a forward biased pn junction diode, the sequence of events that
best describes the mechanism of current flow is
(A) injection, and subsequent diffusion and recombination of minority carriers
(B) injection, and subsequent drift and generation of minority carriers
(C) extraction, and subsequent diffusion and generation of minority carriers
(D) extraction, and subsequent drift and recombination of minority
carriers
In IC technology, dry oxidation (using dry oxygen) as compared to
wet oxidation (using steam or water vapor) produces
(A) superior quality oxide with a higher growth rate
(B) inferior quality oxide with a higher growth rate
(C) inferior quality oxide with a lower growth rate
(D) superior quality oxide with a lower growth rate
In the three dimensional view of a silicon n -channel MOS transistor shown below, d = 20 nm . The transistor is of width 1 mm . The
depletion width formed at every p-n junction is 10 nm. The rela-
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tive permittivity of Si and SiO 2 , respectively, are 11.7 and 3.9, and
e0 = 8.9 # 10-12 F/m .
TWO MARKS
The small-signal resistance (i.e., dVB /dID ) in kW offered by the
n-channel MOSFET M shown in the figure below, at a bias point of
VB = 2 V is (device data for M: device transconductance parameter
kN = mn C 0' x ^W/L h = 40 mA/V2 , threshold voltage VTN = 1 V , and
neglect body effect and channel length modulation effects)
3.7
3.8
(A) 12.5
(C) 50
2012
3.5
(B) 1.875 V < Vin < 3.125 V
(D) 0 < Vin < 5 V
Common Data For Q. 2 and 3 :
In a MOSFET operating in the saturation region, the channel length
modulation effect causes
(A) an increase in the gate-source capacitance
(B) a decrease in the transconductance
(C) a decrease in the unity-gain cutoff frequency
(D) a decrease in the output resistance
2013
3.4
(A) Vin < 1.875 V
(C) Vin > 3.125 V
(B) 25
(D) 100
The source-body junction capacitance is approximately
(A) 2 fF
(B) 7 fF
(C) 2 pF
(D) 7 pF
2011
TWO MARKS
The source of a silicon (ni = 1010 per cm3) n -channel MOS transistor
has an area of 1 sq mm and a depth of 1 mm . If the dopant density in
the source is 1019 /cm3 , the number of holes in the source region with
the above volume is approximately
(A) 107
(B) 100
(C) 10
(D) 0
The gate source overlap capacitance is approximately
(A) 0.7 fF
(B) 0.7 pF
(C) 0.35 fF
(D) 0.24 pF
3.9
3.10
ONE MARK
Drift current in the semiconductors depends upon
(A) only the electric field
(B) only the carrier concentration gradient
(C) both the electric field and the carrier concentration
(D) both the electric field and the carrier concentration gradient
A Zener diode, when used in voltage stabilization circuits, is biased
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
in
(A) reverse bias region below the breakdown voltage
(B) reverse breakdown region
(C) forward bias region
(D) forward bias constant current mode
3.11
(C) epitaxial oxidation
3.16
3.17
TWO MARKS
Common Data For Q. 3.12 & 3.13 :
The channel resistance of an N-channel JFET shown in the fig-
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In a uniformly doped BJT, assume that NE , NB and NC are the
emitter, base and collector doping in atoms/cm3 , respectively. If the
emitter injection efficiency of the BJT is close unity, which one of
the following condition is TRUE
(B) NE >> NB and NB > NC
(A) NE = NB = NC
(C) NE = NB and NB < NC
(D) NE < NB < NC
Compared to a p-n junction with NA = ND = 1014 /cm3 , which
one of the following statements is TRUE for a p-n junction with
NA = ND = 1020 /cm3 ?
(A) Reverse breakdown voltage is lower and depletion capacitance
is lower
(B) Reverse breakdown voltage is higher and depletion capacitance
is lower
(C) Reverse breakdown voltage is lower and depletion capacitance
is higher
(D) Reverse breakdown voltage is higher and depletion capacitance
is higher
The silicon sample with unit cross-sectional area shown below is in
thermal equilibrium. The following information is given: T = 300 K
electronic charge = 1.6 # 10-19 C , thermal voltage = 26 mV and
electron mobility = 1350 cm2 / V-s
3.18
3.19
3.13
The channel resistance when VGS =- 3 V is
(A) 360 W
(B) 917 W
(C) 1000 W
(D) 3000 W
The magnitude of the electric field at x = 0.5 mm is
(A) 1 kV/cm
(B) 5 kV/cm
(C) 10 kV/cm
(D) 26 kV/cm
The magnitude of the electron of the electron drift current density
at x = 0.5 mm is
(B) 1.08 # 10 4 A/m2
(A) 2.16 # 10 4 A/cm2
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The channel resistance when VGS = 0 V is
(A) 480 W
(B) 600 W
(C) 750 W
(D) 1000 W
(C) 4.32 # 103 A/cm2
2009
2010
3.14
3.15
TWO MARKS
Statements for Linked Answer Question : 3.10 & 3.11 :
ure below is 600 W when the full channel thickness (tch ) of 10 μm
is available for conduction. The built-in voltage of the gate P+ N
junction (Vbi ) is - 1 V . When the gate to source voltage (VGS ) is 0
V, the channel is depleted by 1 μm on each side due to the built in
voltage and hence the thickness available for conduction is only 8
μm
3.12
(D) ion implantation
2010
A silicon PN junction is forward biased with a constant current at
room temperature. When the temperature is increased by 10ºC, the
forward bias voltage across the PN junction
(A) increases by 60 mV
(B) decreases by 60 mV
(C) increases by 25 mV
(D) decreases by 25 mV
2011
Page 48
ONE MARK
At room temperature, a possible value for the mobility of electrons
in the inversion layer of a silicon n -channel MOSFET is
(A) 450 cm2 / V-s
(B) 1350 cm2 / V-s
(C) 1800 cm2 / V-s
(D) 3600 cm2 / V-s
Thin gate oxide in a CMOS process in preferably grown using
(A) wet oxidation
(B) dry oxidation
3.20
3.21
(D) 6.48 # 102 A/cm2
ONE MARK
In an n-type silicon crystal at room temperature, which of the
following can have a concentration of 4 # 1019 cm - 3 ?
(A) Silicon atoms
(B) Holes
(C) Dopant atoms
(D) Valence electrons
The ratio of the mobility to the diffusion coefficient in a semiconductor
has the units
(A) V - 1
(B) cm.V1
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) V.cm - 1
(D) V.s
2009
3.22
Page 49
3.28
TWO MARKS
Consider the following two statements about the internal conditions
in a n -channel MOSFET operating in the active region.
S1 : The inversion charge decreases from source to drain
S2 : The channel potential increases from source to drain.
Which of the following is correct?
(A) Only S2 is true
(B) Both S1 and S2 are false
(C) Both S1 and S2 are true, but S2 is not a reason for S1
(D) Both S1 and S2 are true, and S2 is a reason for S1
The drain current of MOSFET in saturation is given by
ID = K (VGS - VT ) 2 where K is a constant.
The magnitude of the transconductance gm is
(A)
K (VGS - VT ) 2
VDS
(B) 2K (VGS - VT )
(C)
Id
VGS - VDS
(D)
K (VGS - VT ) 2
VGS
2008
3.29
TWO MARKS
The measured trans conductance gm of an NMOS transistor
operating in the linear region is plotted against the gate voltage
VG at a constant drain voltage VD . Which of the following figures
represents the expected dependence of gm on VG ?
Common Data For Q. 3.13 and 3.14
Consider a silicon p - n junction at room temperature having the
following parameters:
Doping on the n -side = 1 # 1017 cm - 3
Depletion width on the n -side = 0.1mm
Depletion width on the p -side = 1.0mm
Intrinsic carrier concentration = 1.4 # 1010 cm - 3
Thermal voltage = 26 mV
Permittivity of free space = 8.85 # 10
Dielectric constant of silicon = 12
3.23
3.24
- 14
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F.cm - 1
The built-in potential of the junction
(A) is 0.70 V
(B) is 0.76 V
(C) is 0.82 V
(D) Cannot be estimated from the data given
The peak electric field in the device is
(A) 0.15 MV . cm - 1, directed from p -region to n -region
(B) 0.15 MV . cm - 1, directed from n -region to p -region
(C) 1.80 MV . cm - 1, directed from p-region to n -region
(D) 1.80 MV . cm - 1, directed from n -region to p -region
2008
3.25
3.26
3.27
ONE MARK
Which of the following is NOT associated with a p - n junction ?
(A) Junction Capacitance
(B) Charge Storage Capacitance
(C) Depletion Capacitance
(D) Channel Length Modulations
Which of the following is true?
(A) A silicon wafer heavily doped with boron is a p+ substrate
(B) A silicon wafer lightly doped with boron is a p+ substrate
(C) A silicon wafer heavily doped with arsenic is a p+ substrate
(D) A silicon wafer lightly doped with arsenic is a p+ substrate
A silicon wafer has 100 nm of oxide on it and is furnace at a
temperature above 1000c C for further oxidation in dry oxygen. The
oxidation rate
(A) is independent of current oxide thickness and temperature
(B) is independent of current oxide thickness but depends on temperature
(C) slows down as the oxide grows
(D) is zero as the existing oxide prevents further oxidation
3.30
3.31
Silicon is doped with boron to a concentration of 4 # 1017 atoms cm3
. Assume the intrinsic carrier concentration of silicon to be 1.5 # 1010
/ cm 3 and the value of kT/q to be 25 mV at 300 K. Compared to
undopped silicon, the fermi level of doped silicon
(A) goes down by 0.31 eV
(B) goes up by 0.13 eV
(C) goes down by 0.427 eV
(D) goes up by 0.427 eV
The cross section of a JFET is shown in the following figure. Let
Vc be - 2 V and let VP be the initial pinch -off voltage. If the width
W is doubled (with other geometrical parameters and doping levels
remaining the same), then the ratio between the mutual trans
conductances of the initial and the modified JFET is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 50
(C) P - 2, Q - 2, R - 1, S- -2
(D) P - 2, Q - 1, R - 2, S - 2
3.36
(A) 4
(C) e
3.32
1 - 2/Vp
o
1 - 1/2Vp
1 - 2/Vp
(B) 1 e
2 1 - 1/2Vp o
1 - (2 - Vp )
(D)
1 - [1 (2 Vp )]
Consider the following assertions.
S1 : For Zener effect to occur, a very abrupt junction is required.
S2 : For quantum tunneling to occur, a very narrow energy barrier
is required.
Which of the following is correct ?
(A) Only S2 is true
(B) S1 and S2 are both true but S2 is not a reason for S1
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(A) P - 3, Q - 1, R - 4, S - 2
(B) P - 1, Q - 4, R - 3, S - 2
(C) P - 3, Q - 4, R - 1, S - 2
(D) P - 3, Q - 2, R - 1, S - 4
3.37
3.38
3.33
3.34
2007
3.35
TWO MARKS
Group I lists four types of p - n junction diodes. Match each device
in Group I with one of the option in Group II to indicate the bias
condition of the device in its normal mode of operation.
Group - I
Group-II
(P) Zener Diode
(1) Forward bias
(Q) Solar cell
(2) Reverse bias
(R) LASER diode
(S) Avalanche Photodiode
(A) P - 1, Q - 2, R - 1, S - 2
(B) P - 2, Q - 1, R - 1, S - 2
The DC current gain (b) of a BJT is 50. Assuming that the emitter
injection efficiency is 0.995, the base transport factor is
(A) 0.980
(B) 0.985
(C) 0.990
(D) 0.995
The figure shows the high-frequency capacitance - voltage characteristics of Metal/Sio 2 /silicon (MOS) capacitor having an area of
1 # 10 - 4 cm 2 . Assume that the permittivities (e0 er ) of silicon and
Sio2 are 1 # 10 - 12 F/cm and 3.5 # 10 - 13 F/cm respectively.
ONE MARK
The electron and hole concentrations in an intrinsic semiconductor
are ni per cm3 at 300 K. Now, if acceptor impurities are introduced
with a concentration of NA per cm3 (where NA >> ni , the electron
concentration per cm3 at 300 K will be)
(A) ni
(B) ni + NA
2
(D) ni
(C) NA - ni
NA
+
In a p n junction diode under reverse biased the magnitude of
electric field is maximum at
(A) the edge of the depletion region on the p-side
(B) the edge of the depletion region on the n -side
(C) the p+ n junction
(D) the centre of the depletion region on the n -side
A p+ n junction has a built-in potential of 0.8 V. The depletion layer
width a reverse bias of 1.2 V is 2 mm. For a reverse bias of 7.2 V, the
depletion layer width will be
(A) 4 mm
(B) 4.9 mm
(C) 8 mm
(D) 12 mm
Common Data For Q. 2.29, 2.30 and 2.31 :
(C) S1 and S2 and are both true but S2 is not a reason for S1
(D) Both S1 and S2 are false
2007
Group I lists four different semiconductor devices. match each device
in Group I with its charactecteristic property in Group II
Group-I
Group-II
(P) BJT
(1) Population iniversion
(Q) MOS capacitor
(2) Pinch-off voltage
(R) LASER diode
(3) Early effect
(S) JFET
(4) Flat-band voltage
3.39
The gate oxide thickness in the MOS capacitor is
(A) 50 nm
(B) 143 nm
(C) 350 nm
(D) 1 mm
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3.40
3.41
The maximum depletion layer width in silicon is
(A) 0.143 mm
(B) 0.857 mm
(C) 1 mm
(D) 1.143 mm
Consider the following statements about the C - V characteristics
plot :
S1 : The MOS capacitor has as n -type substrate
S2 : If positive charges are introduced in the oxide, the C - V polt
will shift to the left.
Then which of the following is true?
(A) Both S1 and S2 are true
(B) S1 is true and S2 is false
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 51
(C) S1 is false and S2 is true
(D) Both S1 and S2 are false
(B) E - 3, F - 4, G - 1, H - 3
(C) E - 2, F - 4, G - 1, H - 2
(D) E - 1, F - 3, G - 2, H - 4
2006
3.42
3.43
3.44
3.45
ONE MARK
The values of voltage (VD) across a tunnel-diode corresponding to
peak and valley currents are Vp, VD respectively. The range of tunneldiode voltage for VD which the slope of its I - VD characteristics is
negative would be
(A) VD < 0
(B) 0 # VD < Vp
(C) Vp # VD < Vv
(D) VD $ Vv
The concentration of minority carriers in an extrinsic semiconductor
under equilibrium is
(A) Directly proportional to doping concentration
(B) Inversely proportional to the doping concentration
(C) Directly proportional to the intrinsic concentration
(D) Inversely proportional to the intrinsic concentration
Under low level injection assumption, the injected minority carrier
current for an extrinsic semiconductor is essentially the
(A) Diffusion current
(B) Drift current
(C) Recombination current
(D) Induced current
The phenomenon known as “Early Effect” in a bipolar transistor
refers to a reduction of the effective base-width caused by
(A) Electron - hole recombination at the base
(B) The reverse biasing of the base - collector junction
(C) The forward biasing of emitter-base junction
(D) The early removal of stored base charge during saturation-tocut off switching
2006
3.46
3.48
(B) vR =+ 5
(D) - 5 # vR < 0
The majority carriers in an n-type semiconductor have an average
drift velocity v in a direction perpendicular to a uniform magnetic
field B . The electric field E induced due to Hall effect acts in the
direction
(A) v # B
(B) B # v
(C) along v
(D) opposite to v
Find the correct match between Group 1 and Group 2
Group 1
Group 2
E - Varactor diode
1. Voltage reference
F - PIN diode
2. High frequency switch
G - Zener diode
3. Tuned circuits
H - Schottky diode
4. Current controlled attenuator
(A) E - 4, F - 2, G - 1, H - 3
A heavily doped n - type semiconductor has the following data:
Hole-electron ratio : 0.4
Doping concentration
: 4.2 # 108 atoms/m3
Intrinsic concentration
: 1.5 # 10 4 atoms/m 3
The ratio of conductance of the n -type semiconductor to that of
the intrinsic semiconductor of same material and ate same temperature is given by
(A) 0.00005
(B) 2000
(C) 10000
(D) 20000
2005
3.50
ONE MARK
The bandgap of Silicon at room temperature is
(A) 1.3 eV
(B) 0.7 eV
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(C) 1.1 eV
3.51
TWO MARKS
In the circuit shown below, the switch was connected to position 1
at t < 0 and at t = 0 , it is changed to position 2. Assume that the
diode has zero voltage drop and a storage time ts . For 0 < t # ts, vR
is given by (all in Volts)
(A) vR =- 5
(C) 0 # vR < 5
3.47
3.49
3.52
A Silicon PN junction at a temperature of 20c C has a reverse
saturation current of 10 pico - Ameres (pA). The reserve saturation
current at 40cC for the same bias is approximately
(A) 30 pA
(B) 40 pA
(C) 50 pA
(D) 60 pA
The primary reason for the widespread use of Silicon in semiconductor
device technology is
(A) abundance of Silicon on the surface of the Earth.
(B) larger bandgap of Silicon in comparison to Germanium.
(C) favorable properties of Silicon - dioxide (SiO2)
(D) lower melting point
2005
3.53
3.54
3.55
(D) 1.4 eV
TWO MARKS
A Silicon sample A is doped with 1018 atoms/cm 3 of boron. Another
sample B of identical dimension is doped with 1018 atoms/cm 3
phosphorus. The ratio of electron to hole mobility is 3. The ratio of
conductivity of the sample A to B is
(A) 3
(B) 1
3
(C) 2
(D) 3
3
2
A Silicon PN junction diode under reverse bias has depletion region
of width 10 mm. The relative permittivity of Silicon, er = 11.7 and
the permittivity of free space e0 = 8.85 # 10 - 12 F/m. The depletion
capacitance of the diode per square meter is
(A) 100 mF
(B) 10 mF
(C) 1 mF
(D) 20 mF
A MOS capacitor made using p type substrate is in the accumulation
mode. The dominant charge in the channel is due to the presence of
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) holes
(C) positively charged icons
3.56
Page 52
(B) electrons
(D) negatively charged ions
For an n -channel MOSFET and its transfer curve shown in the
figure, the threshold voltage is
(C) a CMOS inverter
(D) a BJT inverter
3.61
(A) 1 V and the device is in active region
(B) - 1 V and the device is in saturation region
(C) 1 V and the device is in saturation region
(D) - 1 V and the device is an active region
2004
(A) 56 mA
(C) 60 mA
2004
ONE MARK
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3.57
3.58
The impurity commonly used for realizing the base region of a silicon
n - p - n transistor is
(A) Gallium
(B) Indium
(C) Boron
(D) Phosphorus
3.62
3.63
3.64
If for a silicon npn transistor, the base-to-emitter voltage (VBE ) is
0.7 V and the collector-to-base voltage (VCB) is 0.2 V, then the
transistor is operating in the
(A) normal active mode
(B) saturation mode
(C) inverse active mode
(D) cutoff mode
3.65
3.59
3.60
Assuming VCEsat = 0.2 V and b = 50 , the minimum base current (IB)
required to drive the transistor in the figure to saturation is
Consider the following statements S1 and S2.
S1 : The b of a bipolar transistor reduces if the base width is increased.
S2 : The b of a bipolar transistor increases if the dopoing concentration in the base is increased.
Which remarks of the following is correct ?
(A) S1 is FALSE and S2 is TRUE
(B) Both S1 and S2 are TRUE
(C) Both S1 and S2 are FALSE
(D) S1 is TRUE and S2 is FALSE
TWO MARKS
In an abrupt p - n junction, the doping concentrations on the p side and n -side are NA = 9 # 1016 /cm 3 respectively. The p - n
junction is reverse biased and the total depletion width is 3 mm.
The depletion width on the p -side is
(A) 2.7 mm
(B) 0.3 mm
(C) 2.25 mm
(D) 0.75 mm
The resistivity of a uniformly doped n -type silicon sample is 0.5W mc. If the electron mobility (mn) is 1250 cm 2 /V-sec and the charge of
an electron is 1.6 # 10 - 19 Coulomb, the donor impurity concentration
(ND) in the sample is
(A) 2 # 1016 /cm 3
(B) 1 # 1016 /cm 3
(C) 2.5 # 1015 /cm 3
(D) 5 # 1015 /cm 3
Consider an abrupt p - n junction. Let Vbi be the built-in potential
of this junction and VR be the applied reverse bias. If the junction
capacitance (Cj ) is 1 pF for Vbi + VR = 1 V, then for Vbi + VR = 4 V,
Cj will be
(A) 4 pF
(B) 2 pF
(C) 0.25 pF
(D) 0.5 pF
Consider the following statements Sq and S2.
S1 : The threshold voltage (VT ) of MOS capacitor decreases with
increase in gate oxide thickness.
S2 : The threshold voltage (VT ) of a MOS capacitor decreases with
increase in substrate doping concentration.
Which Marks of the following is correct ?
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(A) S1 is FALSE and S2 is TRUE
(B) Both S1 and S2 are TRUE
(C) Both S1 and S2 are FALSE
(D) S1 is TRUE and S2 is FALSE
Given figure is the voltage transfer characteristic of
3.66
(A) an NOMS inverter with enhancement mode transistor as load
(B) an NMOS inverter with depletion mode transistor as load
(B) 140 mA
(D) 3 mA
3.67
The drain of an n-channel MOSFET is shorted to the gate so that
VGS = VDS . The threshold voltage (VT ) of the MOSFET is 1 V. If the
drain current (ID) is 1 mA for VGS = 2 V, then for VGS = 3 V, ID is
(A) 2 mA
(B) 3 mA
(C) 9 mA
(D) 4 mA
The longest wavelength that can be absorbed by silicon, which has
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 53
the bandgap of 1.12 eV, is 1.1 mm. If the longest wavelength that
can be absorbed by another material is 0.87 mm, then bandgap of
this material is
(A) 1.416 A/cm 2
(B) 0.886 eV
(C) 0.854 eV
3.68
(D) 0.706 eV
The neutral base width of a bipolar transistor, biased in the active
region, is 0.5 mm. The maximum electron concentration and the
diffusion constant in the base are 1014 / cm 3 and Dn = 25 cm 2 /
sec respectively. Assuming negligible recombination in the base,
the collector current density is (the electron charge is 1.6 # 10 - 19
Coulomb)
(A) 800 A/cm 2
(B) 8 A/cm 2
(C) 200 A/cm 2
(D) 2 A/cm 2
2003
3.69
3.70
3.71
n -type silicon is obtained by doping silicon with
(A) Germanium
(B) Aluminium
(C) Boron
(D) Phosphorus
3.73
3.75
3.76
A particular green LED emits light of wavelength 5490 Ac. The
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The intrinsic carrier concentration of silicon sample at 300 K is
1.5 # 1016 /m 3 . If after doping, the number of majority carriers is
5 # 1020 /m 3 , the minority carrier density is
(A) 4.50 # 1011/m 3
(B) 3.333 # 10 4 /m 3
energy bandgap of the semiconductor material used there is
(Plank’s constant = 6.626 # 10 - 34 J - s )
(A) 2.26 eV
(B) 1.98 eV
(C) 1.17 eV
(D) 0.74 eV
-5
(D) 3.00 # 10 /m 3
Choose proper substitutes for X and Y to make the following
statement correct Tunnel diode and Avalanche photo diode are
operated in X bias ad Y bias respectively
(A) X: reverse, Y: reverse
(B) X: reverse, Y: forward
(C) X: forward, Y: reverse
(D) X: forward, Y: forward
For an n - channel enhancement type MOSFET, if the source is
connected at a higher potential than that of the bulk (i.e. VSB > 0 ),
the threshold voltage VT of the MOSFET will
(A) remain unchanged
(B) decrease
(C) change polarity
(D) increase
3.79
3.80
TWO MARKS
An n -type silicon bar 0.1 cm long and 100 mm2 i cross-sectional
area has a majority carrier concentration of 5 # 1020 /m 2 and the
carrier mobility is 0.13 m2 /V-s at 300 K. If the charge of an electron
is 1.5 # 10 - 19 coulomb, then the resistance of the bar is
(A) 106 Ohm
(B) 10 4 Ohm
(C) 10 - 1 Ohm
At 300 K, for a diode current of 1 mA, a certain germanium diode
requires a forward bias of 0.1435 V, whereas a certain silicon diode
requires a forward bias of 0.718 V. Under the conditions state above,
the closest approximation of the ratio of reverse saturation current
in germanium diode to that in silicon diode is
(A) 1
(B) 5
(D) 8 # 103
(C) 4 # 103
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2003
3.74
3.77
3.78
The Bandgap of silicon at 300 K is
(A) 1.36 eV
(B) 1.10 eV
(C) 0.80 eV
(D) 0.67 eV
(C) 5.00 # 10 /m 3
Group 2
1. Heavy doping
2. Coherent radiation
3. Spontaneous emission
4. Current gain
(A) P - 1, Q - 2, R - 4, S - 3
(B) P - 2, Q - 3, R - 1, S - 4
(C) P - 3 Q - 4, R - 1, S - 2
(D) P - 2, Q - 1, R - 4, S - 3
ONE MARK
20
3.72
Group 1
P. LED
Q. Avalanche photo diode
R. Tunnel diode
S. LASER
3.81
(D) 10 - 4 Ohm
The electron concentration in a sample of uniformly doped n -type
silicon at 300 K varies linearly from 1017 /cm 3 at x = 0 to 6 # 1016 /
cm 3 at x = 2mm . Assume a situation that electrons are supplied to
keep this concentration gradient constant with time. If electronic
charge is 1.6 # 10 - 19 coulomb and the diffusion constant Dn = 35 cm
2 /s, the current density in the silicon, if no electric field is present, is
(A) zero
(B) -112 A/cm 2
(C) +1120 A/cm 2
(D) -1120 A/cm 2
Match items in Group 1 with items in Group 2, most suitably.
When the gate-to-source voltage (VGs) of a MOSFET with threshold
voltage of 400 mV, working in saturation is 900 mV, the drain current
is observed to be 1 mA. Neglecting the channel width modulation
effect and assuming that the MOSFET is operating at saturation,
the drain current for an applied VGS of 1400 mV is
(A) 0.5 mA
(B) 2.0 mA
(C) 3.5 mA
(D) 4.0 mA
If P is Passivation, Q is n -well implant, R is metallization and S is
source/drain diffusion, then the order in which they are carried out
in a standard n -well CMOS fabrication process, is
(B) Q - S - R - P
(A) P - Q - R - S
(C) R - P - S - Q
(D) S - R - Q - P
The action of JFET in its equivalent circuit can best be represented
as a
(A) Current controlled current source
(B) Current controlled voltage source
(C) Voltage controlled voltage source
(D) Voltage controlled current source
2002
3.82
ONE MARK
In the figure, silicon diode is carrying a constant current of 1 mA.
When the temperature of the diode is 20cC, VD is found to be 700
mV. If the temperature rises to 40cC, VD becomes approximately
equal to
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 54
1998
3.89
(A) 740 mV
(C) 680 mV
3.83
(B) 660 mV
(D) 700 mV
3.90
If the transistor in the figure is in saturation, then
The electron and hole concentrations in a intrinsic semiconductor
are ni and pi respectively. When doped with a p-type material,
these change to n and p, respectively, Then
(B) n + ni = p + pi
(A) n + p = ni + pi
(C) npi = ni p
(D) np = ni pi
The fT of a BJT is related to its gm, C p and C m as follows
Cp + Cm
gm
gm
(C) fT =
Cp + Cm
(A) fT =
3.91
(A)
(B)
(C)
(D)
IC
IC
IC
IC
is always equal to bdc IB
is always equal to - bde IB
is greater than or equal to bdc IB
is less than or equal to bdc IB
3.92
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2001
3.84
3.85
MOSFET can be used as a
(A) current controlled capacitor
(C) current controlled inductor
3.93
ONE MARK
(B) voltage controlled capacitor
(D) voltage controlled inductor
3.94
The effective channel length of MOSFET in saturation decreases
with increase in
(A) gate voltage
(B) drain voltage
(C) source voltage
(D) body voltage
1999
3.86
The early effect in a bipolar junction transistor is caused by
(A) fast turn-on
(B) fast turn-off
(C) large collector-base reverse bias
(D) large emitter-base forward bias
1999
3.87
3.88
ONE MARK
An npn transistor (with C = 0.3 pF ) has a unity-gain cutoff
frequency fT of 400 MHz at a dc bias current Ic = 1 mA . The value
of its Cm (in pF) is approximately (VT = 26 mV)
(A) 15
(B) 30
(C) 50
(D) 96
A long specimen of p-type semiconductor material
(A) is positively charged
(B) is electrically neutral
(C) has an electric field directed along its length
(D) acts as a dipole
Two identical FETs, each characterized by the parameters gm and rd
are connected in parallel. The composite FET is then characterized
by the parameters
g
g
(A) m and 2rd
(B) m and rd
2
2
2
(C) 2gm and rd
(D) 2gm and 2rd
2
q
The units of
are
kT
(A) V
(B) V-1
(C) J
(D) J/K
ONE MARK
For a MOS capacitor fabricated on a p-type semiconductor, strong
inversion occurs when
(A) surface potential is equal to Fermi potential
(B) surface potential is zero
(C)
surface potential is negative and
equal to Fermi potential in magnitude
(D) surface potential is positive and equal to twice the Fermi po-
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TWO MARKS
An n -channel JEFT has IDSS = 2 mA and Vp =- 4 V . Its
transconductance gm (in milliohm) for an applied gate-to-source
voltage VGS of - 2 V is
(A) 0.25
(B) 0.5
(C) 0.75
(D) 1.0
2p (C p + C m)
gm
gm
(D) fT =
2p (C p + C m)
(B) fT =
The static characteristic of an adequately forward biased p-n
junction is a straight line, if the plot is of
(A) log I vs log V
(B) log I vs V
(C) I vs log V
(D) I vs V
1997
3.95
ONE MARK
tential
3.96
The intrinsic carrier density at 300 K is 1.5 # 1010 /cm3 , in silicon.
For n -type silicon doped to 2.25 # 1015 atoms/cm3 , the equilibrium
electron and hole densities are
(A) n = 1.5 # 1015 /cm3, p = 1.5 # 1010 /cm3
(B) n = 1.5 # 1010 /cm3, p = 2.25 # 1015 /cm3
(C) n = 2.25 # 1015 /cm3, p = 1.0 # 1015 /cm3
(D) n = 1.5 # 1010 /cm3, p = 1.5 # 1010 /cm3
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
1996
3.97
3.98
3.99
3.100
3.102
3.103
ONE MARK
The p-type substrate in a conventional pn -junction isolated
integrated circuit should be connected to
(A) nowhere, i.e. left floating
(B) a DC ground potential
(C) the most positive potential available in the circuit
(D) the most negative potential available in the circuit
If a transistor is operating with both of its junctions forward biased,
but with the collector base forward bias greater than the emitter
base forward bias, then it is operating in the
(A) forward active mode
(B) reverse saturation mode
(C) reverse active mode
(D) forward saturation mode
The common-emitter short-circuit current gain b of a transistor
(A) is a monotonically increasing function of the collector current
IC
(B) is a monotonically decreasing function of IC
(C) increase with IC , for low IC , reaches a maximum and then decreases with further increase in IC
(D) is not a function of IC
A n -channel silicon (Eg = 1.1 eV) MOSFET was fabricated using
n +poly-silicon gate and the threshold voltage was found to be 1 V.
Now, if the gate is changed to p+ poly-silicon, other things remaining
the same, the new threshold voltage should be
(A) - 0.1 V
(B) 0 V
(C) 1.0 V
(D) 2.1 V
1996
3.101
Page 55
TWO MARKS
In a bipolar transistor at room temperature, if the emitter current
is doubled the voltage across its base-emitter junction
(A) doubles
(B) halves
(C) increases by about 20 mV
(D) decreases by about 20 mV
An npn transistor has a beta cut-off frequency fb of 1 MHz and
common emitter short circuit low-frequency current gain bo of
200 it unity gain frequency fT and the alpha cut-off frequency fa
respectively are
(B) 200 MHz, 199 MHz
(A) 200 MHz, 201 MHz
(C) 199 MHz, 200 MHz
(D) 201 MHz, 200 MHz
A silicon n MOSFET has a threshold voltage of 1 V and oxide
thickness of Ao.
[er (SiO 2) = 3.9, e0 = 8.854 # 10-14 F/cm, q = 1.6 # 10-19 C]
The region under the gate is ion implanted for threshold voltage
tailoring. The dose and type of the implant (assumed to be a sheet
charge at the interface) required to shift the threshold voltage to
- 1 V are
(B) 1.08 # 1012 /cm2 , n-type
(A) 1.08 # 1012 /cm2 , p-type
(C) 5.4 # 1011 /cm2 , p-type
(D) 5.4 # 1011 /cm2 , n-type
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 56
SOLUTIONS
= 10-8 cm2 # 10-4 cm
= 10-12 cm3
So total no. of holes is,
p = p 0 # V = 10 # 10-12 = 10-11
Which is approximately equal to zero.
3.1
3.2
3.3
Option (A) is correct.
The potential barrier of the pn junction is lowered when a forward
bias voltage is applied, allowing electrons and holes to flow across
the space charge region (Injection) when holes flow from the p
region across the space charge region into the n region, they become
excess minority carrier holes and are subject to diffuse, drift and
recombination processes.
Option (D) is correct.
In a MOSFET operating in the saturation region, the channel length
modulation effect causes a decrease in output resistance.
Option (A) is correct.
Given,
VB = 2V
VTN = 1V
So, we have
Drain voltage
2
or,
ID = kN ^VB - 1h
Differentiating both side with respect to ID
1 = kN 2 ^VB - 1hdVB
dID
Since,
Hence, we obtain
VBQ = 2 volt (at D.C. Voltage)
dVB =
1
dID
2kN ^VB - 1h
1
=
2 # 40 # 10-6 # ^2 - 1h
= 12.5 # 103 W
= 12.5 kW
Option (D) is correct.
For the semiconductor,
Volume of given device,
= 1 mm 2 # 1 mm
Since all the parameters of PMOS and NMOS are equal.
So,
mn = mp
COX bW l = COX bW l = COX bW l
L M2
L
L M1
Given that M1 is in linear region. So, we assume that M2 is either in
cutoff or saturation.
Case 1 : M2 is in cut off
So,
I 2 = I1 = 0
Where I1 is drain current in M1 and I2 is drain current in M2 .
m C
2
Since,
I1 = p OX bW l82VSD ^VSG - VTp h - V SD
B
2
L
m C
2
0 = p OX bW l [2VSD ^VSG - VTp h - V SD
&
]
2
L
Solving it we get,
2 ^VSG - VTp h = VSD
2 ^5 - Vin - 1h = 5 - VD
Vin = VD + 3
2
&
VD = 2 volt
VG = 2 volt
VS = 0 (Ground)
Therefore,
VGS = 2 > VTN
and
VDS = 2 > VGS - VTN
So, the MOSFET is in the saturation region. Therefore, drain current is
ID = kN ^VGS - VTN h2
3.5
Option (A) is correct.
Given the circuit as below :
Option (D) is correct.
In IC technology, dry oxidation as compared to wet oxidation
produces superior quality oxide with a lower growth rate
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3.4
3.6
n 0 p 0 = n i2
2
20
p 0 = n i = 1019 = 10 per cm3
n 0 10
V = Area # depth
&
For
So,
So for the NMOS
I1 = 0 , VD = 5 V
Vin = 5 + 3 = 4 V
2
VGS = Vin - 0 = 4 - 0 = 4 V and VGS > VTn
So it can’t be in cutoff region.
Case 2 : M2 must be in saturation region.
So,
I1 = I 2
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mp COX W
mn COX W
2
2
2 (VSG - VTp) VSD - V SD
@ = 2 L (VGS - VTn)
2 L6
2
2 (VSG - VTp) VSD - V SD
= (VGS - VTn) 2
&
2 (5 - Vin - 1) (5 - VD) - (5 - VD) 2 = (Vin - 0 - 1) 2
&
2 (4 - Vin) (5 - VD) - (5 - VD) 2 = (Vin - 1) 2
Substituting VD = VDS = VGS - VTn and for N -MOS & VD = Vin - 1
&
2 (4 - Vin) (6 - Vin) - (6 - Vin) 2 = (Vin - 1) 2
&
48 - 36 - 8Vin =- 2Vin + 1
&
6Vin = 11
&
Vin = 11 = 1.833 V
6
&
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 57
So for M2 to be in saturation Vin < 1.833 V or Vin < 1.875 V
3.7
Option (B) is correct.
Gate source overlap capacitance.
Co = dWeox e0 (medium Sio 2 )
tox
= 0.69 # 10-15 F
3.8
rl =
3.13
-9
-6
-12
= 20 # 10 # 1 # 10 #-93.9 # 8.9 # 10
1 # 10
3.9
3.10
Option (C) is correct.
At VGS = 0 V ,
Thus
3.14
Option (B) is correct.
Source body junction capacitance.
Cs = Aer e0
d
A
= (0.2 mm + 0.2 mm + 0.2 mm) # 1 mm + 2 (0.2 mm # 0.2 mm)
= 0.68 mm2
d = 10 nm (depletion width of all junction)
-12
8.9 # 10-12
Cs = 0.68 # 10 # 11.7 -#
9
10 # 10
= 7 # 10-15 F
3.15
3.16
Option (B) is correct.
Dry oxidation is used to achieve high quality oxide growth.
Option (B) is correct.
Emitter injection efficiency is given as
1
g =
1 + NB
NE
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Option (C) is correct.
Drift current
Id = qnmn E
It depends upon Electric field E and carrier concentration n
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Option (B) is correct.
Zener diode operates in reverse breakdown region.
3.17
3.12
Option (D) is correct.
For every 1c C increase in temperature, forward bias voltage across
diode decreases by 2.5 mV. Thus for 10c C increase, there us 25 mV
decreases.
Option (B) is correct.
Full channel resistance is
r L
r # = 600 W
W#a
If VGS is applied, Channel resistance is
r L
where b = a c1 rl = #
W#b
Pinch off voltage,
qN
Vp = D a2
2e
If depletion on each side is d = 1 μm at VGS = 0 .
qN
Vj = D d2
2e
qND
qN
or
1 = D (1 # 10-6) 2 &
= 1012
2e
2e
Now from equation (2), we have
12
...(1)
g = 1, NE >> NB
VGS
Vp m
Option (C) is correct.
Reverse bias breakdown or Zener effect occurs in highly doped
PN junction through tunneling mechanism. In a highly doped PN
junction, the conduction and valence bands on opposite sides of
the junction are sufficiently close during reverse bias that electron
may tunnel directly from the valence band on the p-side into the
conduction band on n -side.
Breakdown voltage VB \ 1
NA ND
So, breakdown voltage decreases as concentration increases
Depletion capacitance
1/2
ees NA ND
C ='
1
2 (Vbi + VR) (NA + ND)
Thus
C \ NA ND
Depletion capacitance increases as concentration increases
3.18
...(2)
Option (C) is correct.
Sample is in thermal equilibrium so, electric field
1 = 10 kV/cm
1 mm
Option (A) is correct.
Electron drift current density
E =
3.19
Jd = ND mn eE = 1016 # 1350 # 1.6 # 10-19 # 10 # 1013
= 2.16 # 10 4 A/cm2
3.20
-6 2
Vp = 10 # (5 # 10 )
or
Vp =- 25 V
At VGS =- 3 V ;
- 3 mm = 3.26 mm
b = 5 b1 - 25 l
since 2b = 8 mm
b = 4 mm
rL
a = 600 5 = 750 W
rl =
#4
Wa # b
Option (A) is correct.
At room temperature mobility of electrons for Si sample is given
mn = 1350 cm2 /Vs . For an n -channel MOSFET to create an inversion
layer of electrons, a large positive gate voltage is to be applied.
Therefore, induced electric field increases and mobility decreases.
So, Mobility mn < 1350 cm2 /Vs for n -channel MOSFET
To achieve
3.11
rL
rL
a = 600
5
=
# 3.26 = 917 W
W # b Wa # b
3.21
Option (C) is correct.
Only dopant atoms can have concentration of 4 # 1019 cm - 3 in n type silicon at room temperature.
Option (A) is correct.
2
Unit of mobility mn is = cm
V. sec
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 58
2
Unit of diffusion current Dn is = cm
sec
2
2
m
Thus unit of n is
= cm / cm = 1 = V-1
V $ sec sec V
Dn
3.22
3.23
Option (D) is correct.
Both S1 and S2 are true and S2 is a reason for S1.
17
E2 - E1 = 25 # 10-3 e ln 4 # 10 10 = 0.427 eV
1.5 # 10
Hence fermi level goes down by 0.427 eV as silicon is doped with
boron.
3.31
VP = VP1
VP1 = W12 = W2
Now
VP2
W22 (2W) 2
or
4VP1 = VP2
Initial transconductance
gm = Kn ;1 - Vbi - VGS E
Vp
Let
NA WP = ND WN
17
-6
NA = ND WN = 1 # 10 # 0.1-6# 10 = 1 # 1016
WP
1 # 10
The built-in potential is
D
Vbi = VT 1n c NA N
n i2 m
17
16
# 10 = 0.760
= 26 # 10-3 ln e 1 # 10 # 1 10
o
(1.4 # 10 ) 2
3.24
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3.25
3.26
3.27
3.28
3.29
3.30
For first condition
gm1
0 - (- 2)
= Kn =1 = Kn ;1 VP1 G
Option (B) is correct.
The peak electric field in device is directed from p to n and is
MV/cm
Option (D) is correct.
Channel length modulation is not associated with a p - n junction.
It is being associated with MOSFET in which effective channel
length decreases, producing the phenomenon called channel length
modulation.
gm2 = Kn =1 -
Hence
3.32
3.33
NA = 4 # 1017
ni = 1.5 # 1010
Option (A) is correct.
Option (D) is correct.
As per mass action law
Thus
or
3.34
3.35
p = NA
nNA = ni2
2
n = ni
NA
Option (C) is correct.
The electric field has the maximum value at the junction of p+ n .
Option (B) is correct.
Zener diode and Avalanche diode works in the reverse bias and laser
diode works in forward bias.
In solar cell diode works in forward bias but photo current is in
reverse direction. Thus
Zener diode : Reverse Bias
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Option (B) is correct.
gm = 2ID = 2 K (VGS - VT ) 2 = 2K (VGS - VT )
2VGS
2VGS
E2 - E1 = kT ln NA
ni
2
4VP1 E
np = ni2
If acceptor impurities are introduces
Option (D) is correct.
Oxidation rate is zero because the existing oxide prevent the further
oxidation.
Option (C) is correct.
0 - (- 2)
= K2 ;1 VP2 G
1 - 2/VP1
gm1
=f
p
gm2
1 - 1/ (2VP1)
VP = VP1
Dividing
Option (A) is correct.
Trivalent impurities are used for making p - type semiconductors.
So, Silicon wafer heavily doped with boron is a p+ substrate.
Solar Cell : Forward Bias
Laser Diode : Forward Bias
Avalanche Photo diode : Reverse Bias
3.36
Which is straight line.
2
VP1 E
For second condition
from p to n
E =- eND xn
es
from n to p
= eND xn
es
-19
17
-5
# 1 # 10 = 0.15
= 1.6 # 10 # 1 #-10
14
8.85 # 10 # 12
Option (C) is correct.
As
VD = constant
Thus
gm \ (VGS - VT )
2
VP = eW ND
es
Pinch off voltage
Option (B) is correct.
We know that
or
Option (C) is correct.
Option (C) is correct.
In BJT as the B-C reverse bias voltage increases, the B-C space
charge region width increases which xB (i.e. neutral base width) > A
change in neutral base width will change the collector current. A
reduction in base width will causes the gradient in minority carrier
concentration to increase, which in turn causes an increased in the
diffusion current. This effect si known as base modulation as early
effect.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
In JFET the gate to source voltage that must be applied to achieve
pinch off voltage is described as pinch off voltage and is also called
as turn voltage or threshold voltage.
In LASER population inversion occurs on the condition when
concentration of electrons in one energy state is greater than that in
lower energy state, i.e. a non equilibrium condition.
In MOS capacitor, flat band voltage is the gate voltage that must be
applied to create flat ban condition in which there is no space charge
region in semiconductor under oxide.
Therefore
BJT : Early effect
MOS capacitor : Flat-band voltage
LASER diode : Population inversion
JFET : Pinch-off voltage
3.37
Page 59
Now
= 0.857 mm
3.41
3.42
C2 = 7 pF
6
- 12
-4
D2 = e0 er2 A = 1 # 710 #- 1210 = 6 # 10 - 4 cm
C2
7
6 # 10
Option (C) is correct.
Depletion region will not be formed if the MOS capacitor has n type
substrate but from C-V characteristics, C reduces if V is increased.
Thus depletion region must be formed. Hence S1 is false
If positive charges is introduced in the oxide layer, then to equalize
the effect the applied voltage V must be reduced. Thus the C - V
plot moves to the left. Hence S2 is true.
Option (C) is correct.
For the case of negative slope it is the negative resistance region
Option (A) is correct.
W = K V + VR
Now
2m = K 0.8 + 1.2
From above two equation we get
0.8 + 7.2 =
0.8 + 1.2
W2 = 4 m m
W =
2m
or
3.38
8 =2
2
Option (B) is correct.
a=
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b
= 50 = 50
b + 1 50 + 1
51
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Current Gain = Base Transport Factor # Emitter injection Efficiency
or
3.39
3.40
a = b1 # b2
50
b1 = a =
= 0.985
51 # 0.995
b2
Option (A) is correct.
At low voltage when there is no depletion region and capacitance is
decide by SiO2 thickness only,
C = e0 er1 A
D
-13
10-4 = 50 nm
or
D = e0 er1 A = 3.5 # 10 -#
12
C
7 # 10
3.43
Option (B) is correct.
The construction of given capacitor is shown in fig below
Option (A) is correct.
For n -type p is minority carrier concentration
np = ni2
np = Constant
p \ 1
n
Since ni is constant
Thus p is inversely proportional to n .
3.44
3.45
When applied voltage is 0 volts, there will be no depletion region
and we get
C1 = 7 pF
When applied voltage is V , a depletion region will be formed as
shown in fig an total capacitance is 1 pF. Thus
CT = 1 pF
or
CT = C1 C2 = 1 pF
C1 + C2
1 = 1 + 1
or
C1 C2
CT
Substituting values of CT and C1 we get
3.46
Option (A) is correct.
Diffusion current, since the drift current is negligible for minority
carrier.
Option (B) is correct.
In BJT as the B-C reverse bias voltage increases, the B-C space
charge region width increases which xB (i.e. neutral base width) > A
change in neutral base width will change the collector current. A
reduction in base width will causes the gradient in minority carrier
concentration to increases, which in turn causes an increases in the
diffusion current. This effect si known as base modulation as early
effect.
Option (A) is correct.
For t < 0 diode forward biased and VR = 5 . At t = 0 diode abruptly
changes to reverse biased and current across resistor must be 0.
But in storage time 0 < t < ts diode retain its resistance of forward
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 60
sp = pqmp
sp
m
= p =1
3
sn
mn
biased. Thus for 0 < t < ts it will be ON and
VR =- 5 V
3.47
Option (B) is correct.
According to Hall effect the direction of electric field is same as that
of direction of force exerted.
E =- v # B
E = B#v
or
3.48
3.54
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3.55
3.56
Option (D) is correct.
mP
We have
= 0.4
mn
3.57
3.58
3.59
mp
mn h
Option (C) is correct.
For silicon at 0 K,
3.60
3.61
At T = 300 K,
Eg300 = 1.21 - 3.6 # 10 - 4 # 300 = 1.1 eV
This is standard value, that must be remembered.
Option (B) is correct.
The reverse saturation current doubles for every 10cC rise in
temperature as follows :
(T - T1)/10
I0 (T) = I 01 # 2
Thus at 40c C, I0 = 40 pA
sn = nqmn
b =
a
1-a
Option (C) is correct.
Option (A) is correct.
Applying KVL we get
or
Now
3.62
IC = VCC - VCE = 3 - 0.2 = 2.8 mA
RC
1k
IB = IC = 2.8m = 56 mA
50
b
Option (B) is correct.
We know that
Wp NA = Wn ND
Option (A) is correct.
Silicon is abundant on the surface of earth in the from of SiO2 .
Option (B) is correct.
Option (D) is correct.
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EgT = Eg0 - 3.6 # 10 T
3.53
Option (A) is correct.
Here emitter base junction is forward biased and base collector
junction is reversed biased. Thus transistor is operating in normal
active region.
VCC - IC RC - VCE = 0
-4
3.52
Option (C) is correct.
Trivalent impurities are used for making p type semiconductor.
Boron is trivalent.
Thus
Eg0 = 1.21 eV
At any temperature
3.51
Option (C) is correct.
From the graph it can be easily seen that Vth = 1 V
Now
VGS = 3 - 1 = 2 V
a -" b a ." b .
If the base width increases, recombination of carrier in base region
increases and a decreases & hence b decreases. If doping in base
region increases, recombination of carrier in base increases and a
decreases thereby decreasing b . Thus S1 is true and S2 is false.
Conductance of n type semiconductor
sn = nqmn
Conductance of intrinsic semiconductor
3.50
Option (B) is correct.
In accumulation mode for NMOS having p -substrate, when positive
voltage is applied at the gate, this will induce negative charge near
p - type surface beneath the gate. When VGS is made sufficiently
large, an inversion of electrons is formed and this in effect forms and
n - channel.
We have
si = ni q (mn + mp)
nmn
sn =
n
Ratio is
=
n
(
)
n
1
s
m
+
m
i
i
n
p
i^ +
8
4
.
4
2
10
#
=
= 2 # 10
1.5 # 10 4 (1 + 0.4)
C = e0 er A
d
C = e0 er = 8.85 # 10-12 # 11.7 = 10.35 m F
d
A
10 # 10-6
and
VDS = 5 - 1 = 4 V
Since
VDS > VGS $ VDS > VGS - Vth
Thus MOSFET is in saturation region.
Schottky diode : High frequency switch
3.49
Option (B) is correct.
or
Option (B) is correct.
The varacter diode is used in tuned circuit as it can provide frequently
stability.
PIN diode is used as a current controlled attenuator.
Zener diode is used in regulated voltage supply or fixed voltage
reference.
Schottkey diode has metal-semiconductor function so it has fast
switching action so it is used as high frequency switch
Varactor diode : Tuned circuits
PIN Diode : Current controlled attenuator
Zener diode : Voltage reference
(n = p)
3 m # 1016
= 0.3 m m
Wp = Wn # ND =
NA
9 # 1016
Option (B) is correct.
Conductivity
s = nqun
or
3.63
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
r = 1 = 1
nqmn
s
n = 1
qrmn
16
= 10 /cm 3
or resistivity
Thus
At T = 300 K,
Eg300 = 1.21 - 3.6 # 10 - 4 # 300 = 1.1 eV
This is standard value, that must be remembered.
1
1.6 # 10 # 0.5 # 1250
For n type semiconductor n = ND
=
3.64
- 19
Page 61
3.71
np = ni2
2
16
.5 # 1016 = 4.5 # 1011
p = ni = 1.5 # 10 # 120
n
5 # 10
Option (D) is correct.
We know that
eeS NA ND
Cj = ;
2 (Vbi + VR)( NA + ND) E
1
Cj \
(Vbi + VR)
C j2
(Vbi + VR) 1
1 =1
=
=
C j1
(Vbi + VR) 2
4
2
C
Cj2 = j1 = 1 = 0.5 pF
2
2
1
2
Thus
Now
or
3.65
3.66
Option (C) is correct.
Increase in gate oxide thickness makes difficult to induce charges
in channel. Thus VT increases if we increases gate oxide thickness.
Hence S1 is false.
Increase in substrate doping concentration require more gate voltage
because initially induce charges will get combine in substrate. Thus
VT increases if we increase substrate doping concentration. Hence S2
is false.
3.72
Option (D) is correct.
3.74
Option (A) is correct.
2
ID2 = (3 - 1) = 4
ID1
(2 - 1) 2
ID2 = 4IDI = 4 mA
or
Eg \ 1
l
Eg2
= l1 = 1.1
Eg1
0.87
l2
Eg2 = 1.1 # 1.12 = 1.416 eV
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3.70
From above relation we have
R =
= 106 W
3.75
EgT = Eg0 - 3.6 # 10 - 4 T
dn = 6 # 1016 - 1017 =- 2 # 1020
dx
2 # 10 - 4 - 0
Jn = nqme E + Dn q dn
dx
Since no electric field is present, E = 0 and we get
So,
Jn = qDn dn
dx
= 1.6 # 10 - 19 # 35 # (- 2 # 1020) =- 1120 A/cm 2
3.76
3.77
Option (C) is correct.
LED works on the principal of spontaneous emission.
In the avalanche photo diode due to the avalanche effect there is
large current gain.
Tunnel diode has very large doping.
LASER diode are used for coherent radiation.
Option (C) is correct.
We know that
I = Io `e h V - 1j
where h = 1 for germanium and h = 2 silicon. As per question
VD1
si
T
Io `e e - 1j = Io `e hV - 1j
VDsi
VDGe
hVT
n
Option (D) is correct.
Pentavalent make n -type semiconductor and phosphorous is
pentavalent.
Option (C) is correct.
For silicon at 0 K Eg0 = 1.21 eV
At any temperature
1
0.1 # 10 - 2
=
20
nqmn A
5 # 10 # 1.6 # 10 - 19 # 0.13 # 100 # 10 - 12
Option (D) is correct.
Now
Option (B) is correct.
Concentration gradient
dn =
1014
= 2 # 1018
-4
dx
0.5 # 10
q = 1.6 # 10 - 19 C
Dn = 25
1014
dn =
dx
0.5 # 10 - 4
JC = qDn dn = 1.6 # 10 - 19 # 25 # 2 # 1018 = 8 A/cm 2
dx
rl
, r = 1 and a = nqun
A
s
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Option (A) is correct.
Thus
3.69
R =
We that
Option (D) is correct.
We know that
or
3.68
Option (C) is correct.
Tunnel diode shows the negative characteristics in forward bias. It
is used in forward bias.
Avalanche photo diode is used in reverse bias.
3.73
ID = K (VGS - VT ) 2
(VGS2 - VT ) 2
I
DS
Thus
=
IDI
(VGS1 - VT ) 2
Substituting the values we have
3.67
Option (A) is correct.
By Mass action law
T
Ge
VDsi
or
si
T
DGe
si
3.78
0.718
Io
# 10 - 1
= 4 # 103
= eVhV - 1 = e 2 #026.1435
Io
e 26 # 10 - 1
e hV - 1
T
-3
-3
Option (A) is correct.
In eV
-34
8
Eg = hc = 6.626 # 10 # -310# 10 = 3.62 J
l
54900 # 10
-19
E (J)
Eg (eV) = g
= 3.62 # 10-19 = 2.26 eV
e
1.6 # 10
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 62
Alternatively
At a given value of vBE , increasing the reverse-bias voltage on the
collector-base junction and thus increases the width of the depletion
region of this junction. This in turn results in a decrease in the
effective base width W . Since IS is inversely proportional to W , IS
increases and that iC increases proportionally. This is early effect.
1.24
Eg = 1.24 eV =
= 2.26 eV
l (mm)
5490 # 10-4 mm
3.79
Option (D) is correct.
We know that
ID = K (VGS - VT ) 2
2
ID2 = (VGS2 - VT )
Thus
ID1
(VGS1 - VT ) 2
Substituting the values we have
or
3.80
2
ID2 = (1.4 - 0.4) = 4
ID1
(0.9 - 0.4) 2
ID2 = 4IDI = 4 mA
3.88
Option (D) is correct.
For a JFET in active region we have
2
IDS = IDSS c1 - VGS m
VP
Option (B) is correct.
At constant current the rate of change of voltage with respect to
temperature is
dV =- 2.5 mV per degree centigrade
dT
Here
Thus
Therefore,
3.83
3.84
3 T = T2 - T1 = 40 - 20 = 20cC
3 VD =- 2.5 # 20 = 50 mV
VD = 700 - 50 = 650 mV
or
or
3.85
3.89
3.86
Option (C) is correct.
C m 15.3 # 10-12 - 0.3 # 10-12 = 15 # 10-12 15 pF
Option (D) is correct.
For any semiconductor (Intrinsic or extrinsic) the product n p
remains constant at a given temperature so here
np = ni pi
3.90
Option (D) is correct.
fT =
3.91
gm
2p (C p + C m)
Option (B) is correct.
For a Forward Bias p-n junction, current equation
I = I 0 (eV/kT - 1)
or
or
I + 1 = eV/kT
I0
kT log b I + 1l = V
I0
So if we plot log I vs V we get a straight line.
3.92
3.93
3.94
Option (B) is correct.
A specimen of p - type or n - type is always electrical neutral.
Option (C) is correct.
Option (B) is correct.
The unit of q is e and unit of kT is eV. Thus unit of e/kT is
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Option (B) is correct.
The metal area of the gate in conjunction with the insulating
dielectric oxide layer and semiconductor channel, form a parallel
plate capacitor. It is voltage controlled capacitor because in active
region the current voltage relationship is given by
Option (D) is correct.
In MOSFET the body (substrate) is connected to power supply in
such a way to maintain the body (substrate) to channel junction in
cutoff condition. The resulting reverse bias voltage between source
and body will have an effect on device function. The reverse bias
will widen the depletion region resulting the reduction in channel
length.
gm = IC = 1
26
VT
gm
fT =
2p (C p + C m)
1/26
400 =
2p (0.3 # 10-12 + C m)
1
= 15.3 # 10-12
(0.3 # 10-12 + C m) =
2p # 26 # 400
or
Option (D) is correct.
Condition for saturation is IC < bIB
IDS = K (VGS - VT ) 2
Option (A) is correct.
Now
From above equation it is clear that the action of a JFET is voltage controlled current source.
3.82
Option (B) is correct.
For an n -channel JEFT trans-conductance is
-3
(- 2)
gm = - 2IDSS b1 - VGS l = - 2 # 2 # 10 =1 -4
VP
VP
(- 4)G
= 10-3 # 1 = 0.5 mho
2
We have
Option (B) is correct.
In n -well CMOS fabrication following are the steps :
(A) n - well implant
(B) Source drain diffusion
(C) Metalization
(D) Passivation
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3.81
3.87
e/eV = V-1 .
3.95
3.96
Option (D) is correct.
Option (C) is correct.
We have
ni = 1.5 # 1010 /cm3
Nd = 2.25 # 1015 atoms/cm3
For n type doping we have electron concentration
n - Nd = 2.25 # 1015 atom/cm3
For a given temperature
np = n i2
Hole concentration
2
(1.5 # 1010) 2
p = ni =
= 1.0 # 105 /cm3
n
2.25 # 1015
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
3.97
3.98
Page 63
Option (D) is correct.
In p n -junction isolated circuit we should have high impedance, so
that p n junction should be kept in reverse bias. (So connect p to
negative potential in the circuit)
Option (B) is correct.
If both junction are forward biased and collector base junction is
more forward biased then IC will be flowing out wards (opposite
direction to normal mode) the collector and it will be in reverse
saturation mode.
3.99
Option (C) is correct.
For normal active mode we have
b = IC
IB
For small values of IC , if we increases IC , b also increases until we
reach (IC ) saturation. Further increases in IC (since transistor is in
saturation mode know) will increases IB and b decreases.
3.100
Option (C) is correct.
For a n -channel mosfet thresholds voltage is given by
VTN = VGS - VDS (sat)
for p-channel [p+ polysilicon used in gate]
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VTP = VSD (sat) - VGS
so
VTP =- VDS (sat) + VGS
so threshold voltage will be same.
3.101
Option (C) is correct.
Emitter current is given by
IE = I 0 (eV /kT - 1)
IE = I 0 eV /kT
eV
VBE = kT ln b IE l
I0
(VBE ) 1 = kT ln b IE 1 l
I0
(VBE ) 2 = kT ln b IE 2 l
I0
(VBE ) 2 - (VBE ) 1 = kT ;ln b IE 2 lE = kT ln b 2IE 1 l
IE 1
IE 1
BE
or
or
Now
or
BE
BE
/kT
>> 1
Now if emitter current is double i.e. IE 2 = 2IE1
(VBE ) 2 = (VBE ) 1 + (25 # 0.60) m volt
= (VBE ) 1 + 15 m volt
Thus if emitter current is doubled the base emitter junction voltage is increased by 15 mV.
3.102
Option (A) is correct.
Unity gain frequency is given by
fT = fB # b = 106 # 200 = 200 MHz
a-cutoff frequency is given by
f
fb
fa = b =
= fb (b + 1)
1-a
b
1b+1
= 106 # (200 + 1) = 201 MHz
3.103
Option (A) is correct.
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 64
UNIT 4
ANALOG CIRCUITS
2013
4.1
ONE MARK
In the circuit shown below what is the output voltage ^Vouth if a
silicon transistor Q and an ideal op-amp are used?
(A) 125 and 125
(C) 250 and 125
4.4
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(A) - 15 V
(C) + 0.7 V
4.2
(B) - 0.7 V
(D) + 15 V
In a voltage-voltage feedback as shown below, which one of the
following statements is TRUE if the gain k is increased?
(A) The input
es
(B) The input
increases
(C) The input
decreases
(D) The input
es
impedance increases and output impedance decreas-
impedance decreases and output impedance also
4.3
In the circuit shown below, the knee current of the ideal Zener
dioide is 10 mA. To maintain 5 V across RL , the minimum value of
RL in W and the minimum power rating of the Zener diode in mW
, respectively, are
In the circuit shown below the op-amps are ideal. Then, Vout in Volts
is
(A) 4
(C) 8
impedance decreases and output impedance increas-
TWO MARKS
(B) 32
(D) 200
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impedance increases and output impedance also
4.6
2013
The ac schematic of an NMOS common-source state is shown in the
figure below, where part of the biasing circuits has been omitted for
simplicity. For the n -channel MOSFET M, the transconductance
gm = 1 mA/V , and body effect and channel length modulation effect
are to be neglected. The lower cutoff frequency in HZ of the circuit
is approximately at
(A) 8
(C) 50
4.5
(B) 125 and 250
(D) 250 and 250
(B) 6
(D) 10
In the circuit shown below, Q1 has negligible collector-to-emitter
saturation voltage and the diode drops negligible voltage across it
under forward bias. If Vcc is + 5 V , X and Y are digital signals with
0 V as logic 0 and Vcc as logic 1, then the Boolean expression for Z is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 65
The current in the circuit is
(A) 10 mA
(C) 6.67 mA
(A) XY
(C) XY
4.7
(B) XY
(D) XY
4.11
The diodes and capacitors in the circuit shown are ideal. The voltage
v (t) across the diode D1 is
A voltage 1000 sin wt Volts is applied across YZ . Assuming ideal
diodes, the voltage measured across WX in Volts, is
(A) cos (wt) - 1
(A) sin wt
(C) ^sin wt - sin wt h /2
4.8
(B) _sin wt + sin wt i /2
(D) 0 for all t
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In the circuit shown below, the silicon npn transistor Q has a
very high value of b . The required value of R2 in kW to produce
IC = 1 mA is
(A) 20
(C) 40
(C) 1 - cos (wt)
(B) 100 W
(D) 10.1 kW
2012
ONE MARK
The current ib through the base of a silicon npn transistor is
1 + 0.1 cos (10000pt) mA At 300 K, the rp in the small signal model
of the transistor is
(D) 1 - sin (wt)
The impedance looking into nodes 1 and 2 in the given circuit is
(A) 50 W
(C) 5 kW
(B) 30
(D) 50
2012
(B) sin (wt)
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4.12
4.9
(B) 9.3 mA
(D) 6.2 mA
4.13
The circuit shown is a
1
rad/s
(R1 + R2) C
(B) high pass filter with f3dB = 1 rad/s
R1 C
(C) low pass filter with f3dB = 1 rad/s
R1 C
1
(D) high pass filter with f3dB =
rad/s
(R1 + R2) C
(A) low pass filter with f3dB =
(A) 250 W
(C) 25 W
4.10
(B) 27.5 W
(D) 22.5 W
The i -v characteristics of the diode in the circuit given below are
v - 0.7 A, v $ 0.7 V
i = * 500
0A
v < 0. 7 V
4.14
The voltage gain Av of the circuit shown below is
TWO MARKS
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 66
and the threshold voltage VT = 1 V . The voltage Vx at the source of
the upper transistor is
(A) Av . 200
(C) Av . 20
(B) Av . 100
(D) Av . 10
(A) 1 V
(C) 3 V
2011
4.15
ONE MARK
4.18
In the circuit shown below, capacitors C1 and C2 are very large and
are shorts at the input frequency. vi is a small signal input. The gain
magnitude vo at 10 M rad/s is
vi
(B) 2 V
(D) 3.67 V
For the BJT,
in the circuit shown below,
Q1
b = 3, VBEon = 0.7 V, VCEsat = 0.7 V . The switch is initially closed.
At time t = 0 , the switch is opened. The time t at which Q1 leaves
the active region is
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(A) 10 ms
(C) 50 ms
4.19
(B) 25 ms
(D) 100 ms
For a BJT, the common base current gain a = 0.98 and the collector
base junction reverse bias saturation current ICO = 0.6 mA . This
BJT is connected in the common emitter mode and operated in the
active region with a base drive current IB = 20 mA . The collector
current IC for this mode of operation is
(A) 0.98 mA
(B) 0.99 mA
(C) 1.0 mA
(D) 1.01 mA
Statement for Linked Answer Questions: 4.6 & 4.7
(A) maximum
(C) unity
4.16
The circuit below implements a filter between the input current ii
and the output voltage vo . Assume that the op-amp is ideal. The
filter implemented is a
(A) low pass filter
(C) band stop filter
2011
4.17
(B) minimum
(D) zero
(B) band pass filter
(D) high pass filter
TWO MARKS
In the circuit shown below, for the MOS transistors, mn Cox = 100 mA/V 2
In the circuit shown below, assume that the voltage drop
across a forward biased diode is 0.7 V. The thermal voltage
Vt = kT/q = 25 mV . The small signal input vi = Vp cos ^wt h where
Vp = 100 mV.
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
4.20
4.21
The bias current IDC through the diodes is
(A) 1 mA
(B) 1.28 mA
(C) 1.5 mA
(D) 2 mA
2010
The ac output voltage vac is
(A) 0.25 cos ^wt h mV
(C) 2 cos (wt) mV
Consider the common emitter amplifier shown below with the following circuit parameters:
b = 100, gm = 0.3861 A/V, r0 = 259 W, RS = 1 kW, RB = 93 kW,
RC = 250 kW, RL = 1 kW, C1 = 3 and C2 = 4.7 mF
TWO MARKS
Common Data For Q. 4.11 & 4.12 :
(B) 1 cos (wt) mV
(D) 22 cos (wt) mV
2010
4.22
Page 67
ONE MARK
The amplifier circuit shown below uses a silicon transistor. The
capacitors CC and CE can be assumed to be short at signal frequency
and effect of output resistance r0 can be ignored. If CE is disconnected
from the circuit, which one of the following statements is true
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(A) The input resistance Ri increases and magnitude of voltage
gainAV decreases
(B) The input resistance Ri decreases and magnitude of voltage
gain AV increases
(C) Both input resistance Ri and magnitude of voltage gain AV
decreases
(D) Both input resistance Ri and the magnitude of voltage gain
AV increases
4.23
In the silicon BJT circuit shown below, assume that the emitter
area of transistor Q1 is half that of transistor Q2
The value of current Io is approximately
(A) 0.5 mA
(B) 2 mA
(C) 9.3 mA
(D) 15 mA
4.24
Assuming the OP-AMP to be ideal, the voltage gain of the amplifier
shown below is
(A) - R2
R1
R || R 3
(C) - 2
R1
(B) - R 3
R1
(D) -b R2 + R 3 l
R1
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4.25
4.26
4.27
The resistance seen by the source vS is
(A) 258 W
(B) 1258 W
(C) 93 kW
(D) 3
The lower cut-off frequency due to C2 is
(A) 33.9 Hz
(B) 27.1 Hz
(C) 13.6 Hz
(D) 16.9 Hz
The transfer characteristic for the precision rectifier circuit shown
below is (assume ideal OP-AMP and practical diodes)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2009
4.28
4.30
In the circuit below, the diode is ideal. The voltage V is given by
(A) min (Vi, 1)
(C) min (- Vi, 1)
4.29
TWO MARKS
Page 68
(B) max (Vi, 1)
(D) max (- Vi, 1)
4.31
In the following a stable multivibrator circuit, which properties of
v0 (t) depend on R2 ?
4.32
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For small increase in VG beyond 1V, which of the following gives the
correct description of the region of operation of each MOSFET
(A) Both the MOSFETs are in saturation region
(B) Both the MOSFETs are in triode region
(C) n-MOSFETs is in triode and p -MOSFET is in saturation
region
(D) n- MOSFET is in saturation and p -MOSFET is in triode
region
Estimate the output voltage V0 for VG = 1.5 V. [Hints : Use the
appropriate current-voltage equation for each MOSFET, based on
the answer to Q.4.16]
(B) 4 + 1
(A) 4 - 1
2
2
(C) 4 - 3
(D) 4 + 3
2
2
In the circuit shown below, the op-amp is ideal, the transistor has
VBE = 0.6 V and b = 150 . Decide whether the feedback in the circuit
is positive or negative and determine the voltage V at the output of
the op-amp.
(A) Positive feedback, V = 10 V
(B) Positive feedback, V = 0 V
(C) Negative feedback, V = 5 V
(D) Negative feedback, V = 2 V
4.33
(A) Only the frequency
(B) Only the amplitude
(C) Both the amplitude and the frequency
(D) Neither the amplitude nor the frequency
Statement for Linked Answer Question 4.16 and 4.17
Consider for CMOS circuit shown, where the gate voltage v0 of
the n-MOSFET is increased from zero, while the gate voltage of
the p -MOSFET is kept constant at 3 V. Assume, that, for both
transistors, the magnitude of the threshold voltage is 1 V and the
product of the trans-conductance parameter is 1mA. V - 2
A small signal source Vi (t) = A cos 20t + B sin 106 t is applied to a
transistor amplifier as shown below. The transistor has b = 150 and
hie = 3W . Which expression best approximate V0 (t)
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 69
(A) 0 V
(C) 0.7 V
4.37
(B) 0.1 V
(D) 1.1 V
The OPAMP circuit shown above represents a
(A) V0 (t) =- 1500 (A cos 20t + B sin 106 t)
(B) V0 (t) = - 1500( A cos 20t + B sin 106 t)
(C) V0 (t) =- 1500B sin 106 t
(D) V0 (t) =- 150B sin 106 t
2008
4.34
ONE MARK
In the following limiter circuit, an input voltage Vi = 10 sin 100pt
is applied. Assume that the diode drop is 0.7 V when it is forward
biased. When it is forward biased. The zener breakdown voltage is
6.8 V
The maximum and minimum values of the output voltage respectively are
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(A) high pass filter
(C) band pass filter
(A) 6.1 V, - 0.7 V
(C) 7.5 V, - 0.7 V
2008
4.35
(B) 0.7 V, - 7.5 V
(D) 7.5 V, - 7.5 V
4.38
TWO MARSK
Two identical NMOS transistors M1 and M2 are connected as shown
below. Vbias is chosen so that both transistors are in saturation. The
equivalent gm of the pair is defied to be 2Iout at constant Vout
2Vi
The equivalent gm of the pair is
For the circuit shown in the following figure, transistor M1 and M2
are identical NMOS transistors. Assume the M2 is in saturation and
the output is unloaded.
4.39
4.36
(B) low pass filter
(D) band reject filter
The current Ix is related to Ibias as
(B) Ix = Ibias
(A) Ix = Ibias + Is
(D) Ix = Ibias - Is
(C) Ix = Ibias - cVDD - Vout m
RE
Consider the following circuit using an ideal OPAMP. The I-V
V
characteristic of the diode is described by the relation I = I 0 _eV - 1i
where VT = 25 mV, I0 = 1m A and V is the voltage across the diode
(taken as positive for forward bias). For an input voltage Vi =- 1 V
, the output voltage V0 is
t
(A) the sum of individual gm ' s of the transistors
(B) the product of individual gm ’s of the transistors
(C) nearly equal to the gm of M1
g
(D) nearly equal to m of M2
g0
Consider the Schmidt trigger circuit shown below
A triangular wave which goes from -12 to 12 V is applied to the
inverting input of OPMAP. Assume that the output of the OPAMP swings from +15 V to -15 V. The voltage at the non-inverting
input switches between
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) - 12V to +12 V
(C) -5 V to +5 V
Page 70
(B) -7.5 V to 7.5 V
(D) 0 V and 5 V
Statement for Linked Answer Question 3.26 and 3.27:
In the following transistor circuit, VBE = 0.7 V, r3 = 25 mV/IE , and
b and all the capacitances are very large
(A) -2 V
(C) -0.5 V
4.45
4.40
The value of DC current IE is
(A) 1 mA
(C) 5 mA
(B) 2 mA
(D) 10 mA
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4.41
4.43
In the Op-Amp circuit shown, assume that the diode current
follows the equation I = Is exp (V/VT ). For Vi = 2V, V0 = V01, and for
Vi = 4V, V0 = V02 .
The relationship between V01 and V02 is
(A) V02 = 2 Vo1
(C) Vo2 = Vo1 1n2
ONE MARK
The correct full wave rectifier circuit is
4.47
(B) Vo2 = e2 Vo1
(D) Vo1 - Vo2 = VT 1n2
In the CMOS inverter circuit shown, if the trans conductance
parameters of the NMOS and PMOS transistors are
W
kn = kp = mn Cox Wn = mCox p = 40mA/V2
Ln
Lp
and their threshold voltages ae VTHn = VTHp = 1 V the current I is
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In a transconductance amplifier, it is desirable to have
(A) a large input resistance and a large output resistance
(B) a large input resistance and a small output resistance
(C) a small input resistance and a large output resistance
(D) a small input resistance and a small output resistance
2007
4.44
(B) saturation
(D) reverse active
The mid-band voltage gain of the amplifier is approximately
(A) -180
(B) -120
(C) -90
(D) -60
2007
4.42
For the BJT circuit shown, assume that the b of the transistor is
very large and VBE = 0.7 V. The mode of operation of the BJT is
(A) cut-off
(C) normal active
4.46
(B) -1 V
(D) 0.5 V
TWO MARKS
For the Op-Amp circuit shown in the figure, V0 is
(A) 0 A
(C) 45 mA
4.48
(B) 25 mA
(D) 90 mA
For the Zener diode shown in the figure, the Zener voltage at knee is
7 V, the knee current is negligible and the Zener dynamic resistance
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 71
is 10 W. If the input voltage (Vi) range is from 10 to 16 V, the output
voltage (V0) ranges from
(A) 0 Volt
(C) 9.45 Volts
4.54
(A) 7.00 to 7.29 V
(C) 7.14 to 7.43 V
(B) 6.3 Volt
(D) 10 Volts
For the circuit shown below, assume that the zener diode is ideal
with a breakdown voltage of 6 volts. The waveform observed across
R is
(B) 7.14 to 7.29 V
(D) 7.29 to 7.43 V
Statement for Linked Answer Questions 4.35 & 4.36:
Consider the Op-Amp circuit shown in the figure.
4.49
4.50
The transfer function V0 (s)/ Vi (s) is
(A) 1 - sRC
(B) 1 + sRC
1 + sRC
1 - sRC
1
1
(C)
(D)
1 - sRC
1 + sRC
4.52
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ONE MARK
The input impedance (Zi) and the output impedance (Z0) of an ideal
trans-conductance (voltage controlled current source) amplifier are
(B) Zi = 0, Z0 = 3
(A) Zi = 0, Z0 = 0
(C) Zi = 3, Z0 = 0
(D) Zi = 3, Z0 = 3
An n-channel depletion MOSFET has following two points on its
ID - VGs curve:
(i) VGS = 0 at ID = 12 mA and
(ii) VGS =- 6 Volts at ID = 0 mA
Which of the following Q point will given the highest trans conductance gain for small signals?
(B) VGS =- 3 Volts
(A) VGS =- 6 Volts
(C) VGS = 0 Volts
(D) VGS = 3 Volts
2006
4.53
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If Vi = V1 sin (wt) and V0 = V2 sin (wt + f), then the minimum and
maximum values of f (in radians) are respectively
(B) 0 and p
(A) - p and p
2
2
2
(C) - p and 0
(D) - p and 0
2
2006
4.51
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TWO MARKS
For the circuit shown in the following figure, the capacitor C is
initially uncharged. At t = 0 the switch S is closed. The Vc across
the capacitor at t = 1 millisecond is
In the figure shown above, the OP-AMP is supplied with ! 15V .
Common Data For Q. 4.41, 4.42 and 4.43 :
In the transistor amplifier circuit shown in the figure below, the
transistor has the following parameters:
bDC = 60 , VBE = 0.7V, hie " 3
The capacitance CC can be assumed to be infinite.
In the figure above, the ground has been shown by the symbol 4
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
4.55
4.56
4.57
Under the DC conditions, the collector-or-emitter voltage drop is
(A) 4.8 Volts
(B) 5.3 Volts
(C) 6.0 Volts
(D) 6.6 Volts
Page 72
4.61
If bDC is increased by 10%, the collector-to-emitter voltage drop
(A) increases by less than or equal to 10%
(B) decreases by less than or equal to 10%
(C) increase by more than 10%
(D) decreases by more than 10%
The small-signal gain of the amplifier vc is
vs
4.62
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(A) -10
(C) 5.3
(B) 10 kW
(C) 40 kW
(D) infinite
The effect of current shunt feedback in an amplifier is to
(A) increase the input resistance and decrease the output resistance
(B) increases both input and output resistance
(C) decrease both input and output resistance
(D) decrease the input resistance and increase the output resistance
The cascade amplifier is a multistage configuration of
(A) CC - CB
(B) CE - CB
(C) CB - CC
(D) CE - CC
2005
4.63
(B) -5.3
(D) 10
4.64
Common Data For Q. 4.44 & 4.45:
(A) 30 kW
4
A regulated power supply, shown in figure below, has an unregulated input (UR) of 15 Volts and generates a regulated output Vout .
Use the component values shown in the figure.
In an ideal differential amplifier shown in the figure, a large value
of (RE ).
(A) increase both the differential and common - mode gains.
(B) increases the common mode gain only.
(C) decreases the differential mode gain only.
(D) decreases the common mode gain only.
For an npn transistor connected as shown in figure VBE = 0.7 volts.
Given that reverse saturation current of the junction at room
temperature 300 K is 10 - 13 A, the emitter current is
(A) 30 mA
(C) 49 mA
4.65
4.58
4.59
The power dissipation across the transistor Q1 shown in the figure is
(A) 4.8 Watts
(B) 5.0 Watts
(C) 5.4 Watts
(D) 6.0 Watts
(B) 39 mA
(D) 20 mA
The voltage e0 is indicated in the figure has been measured by an
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ideal voltmeter. Which of the following can be calculated ?
If the unregulated voltage increases by 20%, the power dissipation
across the transistor Q1
(A) increases by 20%
(B) increases by 50%
(C) remains unchanged
(D) decreases by 20%
2005
4.60
TWO MARKS
ONE MARK
The input resistance Ri of the amplifier shown in the figure is
(A) Bias current of the inverting input only
(B) Bias current of the inverting and non-inverting inputs only
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 73
(C) Input offset current only
(D) Both the bias currents and the input offset current
4.66
The Op-amp circuit shown in the figure is filter. The type of filter
and its cut. Off frequency are respectively
4.70
(A) high pass, 1000 rad/sec.
(C) high pass, 1000 rad/sec
4.67
(B) Low pass, 1000 rad/sec
(D) low pass, 10000 rad/sec
The circuit using a BJT with b = 50 and VBE = 0.7V is shown in
the figure. The base current IB and collector voltage by VC and
respectively
4.71
Zi and Z0 of the circuit are respectively
(B) 2 MW and 20 kW
(A) 2 MW and 2 kW
11
(C) infinity and 2 MW
(D) infinity and 20 kW
11
ID and VDS under DC conditions are respectively
(A) 5.625 mA and 8.75 V
(B) 1.875 mA and 5.00 V
(C) 4.500 mA and 11.00 V
(D) 6.250 mA and 7.50 V
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(A) 43 mA and 11.4 Volts
(C) 45 mA and 11 Volts
4.68
The Zener diode in the regulator circuit shown in the figure has a
Zener voltage of 5.8 volts and a zener knee current of 0.5 mA. The
maximum load current drawn from this current ensuring proper
functioning over the input voltage range between 20 and 30 volts, is
(A) 23.7 mA
(C) 13.7 mA
4.69
(B) 40 mA and 16 Volts
(D) 50 mA and 10 Volts
(B) 14.2 mA
(D) 24.2 mA
Both transistors T1 and T2 show in the figure, have a b = 100 ,
threshold voltage of 1 Volts. The device parameters K1 and K2 of
T1 and T2 are, respectively, 36 mA/V2 and 9 mA/V 2 . The output
voltage Vo i s
(A) 1 V
(C) 3 V
(B) 2 V
(D) 4 V
Common Data For Q. 4.58, 4.59 and 4.60 :
Given, rd = 20kW , IDSS = 10 mA, Vp =- 8 V
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4.72
4.73
Transconductance in milli-Siemens (mS) and voltage gain of the
amplifier are respectively
(A) 1.875 mS and 3.41
(B) 1.875 ms and -3.41
(C) 3.3 mS and -6
(D) 3.3 mS and 6
Given the ideal operational amplifier circuit shown in the figure
indicate the correct transfer characteristics assuming ideal diodes
with zero cut-in voltage.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2004
4.74
4.75
4.76
Page 74
ONE MARK
An ideal op-amp is an ideal
(A) voltage controlled current source
(B) voltage controlled voltage source
(C) current controlled current source
(D) current controlled voltage source
Voltage series feedback (also called series-shunt feedback) results in
(A) increase in both input and output impedances
(B) decrease in both input and output impedances
(C) increase in input impedance and decrease in output impedance
(D) decrease in input impedance and increase in output impedance
The circuit in the figure is a
(A) 1 mF
2p
1
mF
(C)
2p 6
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(A) low-pass filter
(C) band-pass filter
2004
4.77
4.78
(B) high-pass filter
(D) band-reject filter
4.81
TWO MARKS
A bipolar transistor is operating in the active region with a collector
current of 1 mA. Assuming that the b of the transistor is 100 and
the thermal voltage (VT ) is 25 mV, the transconductance (gm) and
the input resistance (rp) of the transistor in the common emitter
configuration, are
(A) gm = 25 mA/V and rp = 15.625 kW
(B) gm = 40 mA/V and rp = 4.0 kW
(C) gm = 25 mA/V and rp = 2.5 k W
(D) gm = 40 mA/V and rp = 2.5 kW
The value of C required for sinusoidal oscillations of frequency 1
kHz in the circuit of the figure is
(D) 2p 6 mF
In the op-amp circuit given in the figure, the load current iL is
(A) - Vs
R2
(C) - Vs
RL
4.80
(B) 2p mF
(B) Vs
R2
(D) Vs
R1
In the voltage regulator shown in the figure, the load current can
vary from 100 mA to 500 mA. Assuming that the Zener diode is ideal
(i.e., the Zener knee current is negligibly small and Zener resistance
is zero in the breakdown region), the value of R is
(A) 7 W
(B) 70 W
(D) 14 W
(C) 70 W
3
In a full-wave rectifier using two ideal diodes, Vdc and Vm are the dc
and peak values of the voltage respectively across a resistive load. If
PIV is the peak inverse voltage of the diode, then the appropriate
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4.82
relationships for this rectifier are
(A) Vdc = Vm , PIV = 2Vm
(B) Idc = 2 Vm , PIV = 2Vm
p
p
(C) Vdc = 2 Vm , PIV = Vm
(D) Vdc Vm , PIV = Vm
p
p
Assume that the b of transistor is extremely large and VBE = 0.7V, IC
and VCE in the circuit shown in the figure
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 75
resistance of the current-shunt negative feedback amplifier using the
above amplifier with a feedback factor of 0.2, is
(B) 1 kW
(A) 1 kW
5
11
(C) 5 kW
4.89
(A) IC = 1 mA, VCE = 4.7 V
(C) IC = 1 mA, VCE = 2.5 V
(B) IC = 0.5 mA, VCE = 3.75 V
(D) IC = 0.5 mA, VCE = 3.9 V
2003
4.83
4.84
In the amplifier circuit shown in the figure, the values of R1 and R2
are such that the transistor is operating at VCE = 3 V and IC = 1.5
mA when its b is 150. For a transistor with b of 200, the operating
point (VCE , IC ) is
ONE MARK
Choose the correct match for input resistance of various amplifier
configurations shown below :
Configuration
Input resistance
CB : Common Base
LO : Low
CC : Common Collector
MO : Moderate
CE : Common Emitter
HI : High
(A) CB - LO, CC - MO, CE - HI
(B) CB - LO, CC - HI, CE - MO
(C) CB - MO, CC - HI, CE - LO
(D) CB - HI, CC - LO, CE - MO
(A) (2 V, 2 mA)
(C) (4 V, 2 mA)
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The circuit shown in the figure is best described as a
(A) bridge rectifier
(C) frequency discriminator
The oscillator circuit shown in the figure has an ideal inverting
amplifier. Its frequency of oscillation (in Hz) is
(B) ring modulator
(D) voltage double
If the input to the ideal comparators shown in the figure is a
sinusoidal signal of 8 V (peak to peak) without any DC component,
then the output of the comparators has a duty cycle of
1
(2p 6 RC)
1
(C)
( 6 RC)
(A)
(A) 1/2
(C) 1/6
4.86
4.87
(B) 1/3
(D) 1/2
4.91
If the differential voltage gain and the common mode voltage gain
of a differential amplifier are 48 dB and 2 dB respectively, then
common mode rejection ratio is
(A) 23 dB
(B) 25 dB
(C) 46 dB
(D) 50 dB
Generally, the gain of a transistor amplifier falls at high frequencies
due to the
(A) internal capacitances of the device
(B) coupling capacitor at the input
(C) skin effect
(D) coupling capacitor at the output
2003
TWO MARKS
An amplifier without feedback has a voltage gain of 50, input
resistance of 1 k W and output resistance of 2.5 kW. The input
(B)
1
(2pRC)
(D)
6
(2pRC)
The output voltage of the regulated power supply shown in the
figure is
(A) 3 V
(C) 9 V
4.92
4.88
(B) (3 V, 2 mA)
(D) (4 V, 1 mA)
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4.90
4.85
(D) 11 kW
(B) 6 V
(D) 12 V
If the op-amp in the figure is ideal, the output voltage Vout will be
equal to
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) 1 V
(C) 14 V
4.93
4.94
Page 76
(B) 6 V
(D) 17 V
Three identical amplifiers with each one having a voltage gain of 50,
input resistance of 1 kW and output resistance of 250 W are cascaded.
The opened circuit voltages gain of the combined amplifier is
(A) 49 dB
(B) 51 dB
(C) 98 dB
(D) 102 dB
An ideal sawtooth voltages waveform of frequency of 500 Hz and
amplitude 3 V is generated by charging a capacitor of 2 mF in every
cycle. The charging requires
(A) Constant voltage source of 3 V for 1 ms
2002
4.98
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(B) Constant voltage source of 3 V for 2 ms
(C) Constant voltage source of 1 mA for 1 ms
(D) Constant voltage source of 3 mA for 2 ms
2002
4.95
4.96
4.97
In a negative feedback amplifier using voltage-series (i.e. voltagesampling, series mixing) feedback.
(A) Ri decreases and R0 decreases
(B) Ri decreases and R0 increases
(C) Ri increases and R0 decreases
(D) Ri increases and R0 increases
(Ri and R0 denote the input and output resistance respectively)
The circuit in the figure employs positive feedback and is
intended to generate sinusoidal oscillation. If at a frequency
V (f) 1
= +0c, then to sustain oscillation at this frequency
f0, B (f) = 3 f
V0 (f)
6
(A) R2 = 5R1
(C) R2 = R1
6
ONE MARK
4.99
4.100
A 741-type opamp has a gain-bandwidth product of 1 MHz. A noninverting amplifier suing this opamp and having a voltage gain of 20
dB will exhibit a -3 dB bandwidth of
(A) 50 kHz
(B) 100 kHz
(D) 1000 kHz
(C) 1000 kHz
7.07
17
Three identical RC-coupled transistor amplifiers are cascaded. If
each of the amplifiers has a frequency response as shown in the
figure, the overall frequency response is as given in
TWO MARKS
(B) R2 = 6R1
(D) R2 = R1
5
An amplifier using an opamp with a slew-rate SR = 1 V/m sec has
a gain of 40 dB. If this amplifier has to faithfully amplify sinusoidal
signals from dc to 20 kHz without introducing any slew-rate induced
distortion, then the input signal level must not exceed.
(A) 795 mV
(B) 395 mV
(C) 79.5 mV
(D) 39.5 mV
A zener diode regulator in the figure is to be designed to meet the
specifications: IL = 10 mA V0 = 10 V and Vin varies from 30 V to 50
V. The zener diode has Vz = 10 V and Izk (knee current) =1 mA. For
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satisfactory operation
(A) R # 1800W
(C) 3700W # R # 4000W
4.101
(B) 2000W # R # 2200W
(D) R $ 4000W
The voltage gain Av = v0 of the JFET amplifier shown in the figure
vt
is IDSS = 10 mA Vp =- 5 V(Assume C1, C2 and Cs to be very large
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 77
(D) Pz = 115 mW, PT = 11.9 W
4.107
(A) +16
(C) +8
(B) -16
(D) -6
4
2001
4.102
The current gain of a BJT is
(A) gm r0
(C) gm rp
4.103
4.104
4.106
(A) Hartely oscillator with foscillation = 79.6 MHz
(B) Colpitts oscillator with foscillation = 50.3 MHz
(C) Hartley oscillator with foscillation = 159.2 MHz
(D) Colpitts oscillator with foscillation = 159.3 MHz
ONE MARK
gm
r
g
(D) m
rp
(B)
4.108
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Consider the following two statements :
Statement 1 :
A stable multi vibrator can be used for generating square wave.
Statement 2:
Bistable multi vibrator can be used for storing binary information.
(A) Only statement 1 is correct
(B) Only statement 2 is correct
(C) Both the statements 1 and 2 are correct
(D) Both the statements 1 and 2 are incorrect
of 100.
TWO MARKS
-14
-13
An npn BJT has gm = 38 mA/V, C m = 10
F, C p = 4 # 10
and DC current gain b0 = 90 . For this transistor fT and fb are
(A) fT = 1.64 # 108 Hz and fb = 1.47 # 1010 Hz
(B) fT = 1.47 # 1010 Hz and fb = 1.64 # 108 Hz
(C) fT = 1.33 # 1012 Hz and fb = 1.47 # 1010 Hz
(D) fT = 1.47 # 1010 Hz and fb = 1.33 # 1012 Hz
The inverting OP-AMP shown in the figure has an open-loop gain
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Thee ideal OP-AMP has the following characteristics.
(B) Ri = 0, A = 3, R0 = 0
(A) Ri = 3, A = 3, R0 = 0
(C) Ri = 3, A = 3, R0 = 3
(D) Ri = 0, A = 3, R0 = 3
2001
4.105
The oscillator circuit shown in the figure is
The closed-loop gain V0 is
Vs
(A) - 8
(C) - 10
F,
4.109
The transistor shunt regulator shown in the figure has a regulated
output voltage of 10 V, when the input varies from 20 V to 30 V.
The relevant parameters for the zener diode and the transistor are
: Vz = 9.5 , VBE = 0.3 V, b = 99 , Neglect the current through RB .
Then the maximum power dissipated in the zener diode (Pz ) and the
transistor (PT ) are
In the figure assume the OP-AMPs to be ideal. The output v0 of
the circuit is
(A) 10 cos (100t)
(C) 10 - 4
2000
4.110
(A) Pz = 75 mW, PT = 7.9 W
(B) Pz = 85 mW, PT = 8.9 W
(C) Pz = 95 mW, PT = 9.9 W
(B) - 9
(D) - 11
t
#0 cos (100t) dt
(B) 10
t
#0 cos (100t) dt
(D) 10 - 4 d cos (100t)
dt
ONE MARK
In the differential amplifier of the figure, if the source resistance of
the current source IEE is infinite, then the common-mode gain is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) zero
(C) indeterminate
4.111
4.116
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4.112
4.113
4.114
(A) precision integrator
(B) Hartely oscillator
(C) Butterworth high pass filter (D) Wien-bridge oscillator
(B) infinite
(D) Vin1 + Vin2
2VT
In the circuit of the figure, V0 is
(A) -1 V
(C) +1 V
Page 78
(A) square wave
(C) parabolic wave
4.117
(B) 2 V
(D) +15 V
Introducing a resistor in the emitter of a common amplifier stabilizes
the dc operating point against variations in
(A) only the temperature
(B) only the b of the transistor
(C) both temperature and b
(D) none of the above
The current gain of a bipolar transistor drops at high frequencies
because of
(A) transistor capacitances
(B) high current effects in the base
(C) parasitic inductive elements
(D) the Early effect
Assume that the op-amp of the figure is ideal. If vi is a triangular
wave, then v0 will be
The most commonly used amplifier is sample and hold circuits is
(A) a unity gain inverting amplifier
(B) a unity gain non-inverting amplifier
(C) an inverting amplifier with a gain of 10
(D) an inverting amplifier with a gain of 100
2000
4.118
(B) triangular wave
(D) sine wave
TWO MARKS
In the circuit of figure, assume that the transistor is in the active
region. It has a large b and its base-emitter voltage is 0.7 V. The
value of Ic is
If the op-amp in the figure, is ideal, then v0 is
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(A) zero
(C) - (V1 + V2) sin wt
4.115
(B) (V1 - V2) sin wt
(D) (V1 + V2) sin wt
The configuration of the figure is a
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 79
4.125
A dc power supply has a no-load voltage of 30 V, and a full-load
voltage of 25 V at a full-load current of 1 A. Its output resistance
and load regulation, respectively, are
(A) 5 W and 20%
(B) 25 W and 20%
(C) 5 W and 16.7%
(D) 25 W and 16.7%
1998
4.126
(A) Indeterminate since Rc is not given (B) 1 mA
(C) 5 mA
(D) 10 mA
4.119
ONE MARK
The circuit of the figure is an example of feedback of the following
type
If the op-amp in the figure has an input offset voltage of 5 mV and
an open-loop voltage gain of 10000, then v0 will be
(A) current series
(C) voltage series
(A) 0 V
(C) + 15 V or -15 V
4.120
4.121
4.122
4.124
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ONE MARK
The first dominant pole encountered in the frequency response of a
compensated op-amp is approximately at
(A) 5 Hz
(B) 10 kHz
(C) 1 MHz
(D) 100 MHz
Negative feedback in an amplifier
(A) reduces gain
(B) increases frequency and phase distortions
(C) reduces bandwidth
(D) increases noise
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4.127
4.128
In the cascade amplifier shown in the given figure, if the commonemitter stage (Q1) has a transconductance gm1 , and the common
base stage (Q2) has a transconductance gm2 , then the overall
transconductance g (= i 0 /vi) of the cascade amplifier is
(A) gm1
g
(C) m1
2
4.123
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(B) 5 mV
(D) +50 V or -50 V
1999
(B) current shunt
(D) voltage shunt
(B) gm2
g
(D) m2
2
4.129
In a differential amplifier, CMRR can be improved by using an
increased
(A) emitter resistance
(B) collector resistance
(C) power supply voltages
(D) source resistance
From a measurement of the rise time of the output pulse of an
amplifier whose is a small amplitude square wave, one can estimate
the following parameter of the amplifier
(A) gain-bandwidth product
(B) slow rate
(C) upper 3–dB frequency
(D) lower 3–dB frequency
The emitter coupled pair of BJT’s given a linear transfer relation
between the differential output voltage and the differential output
voltage and the differential input voltage Vid is less a times the
thermal voltage, where a is
(A) 4
(B) 3
(C) 2
(D) 1
Crossover distortion behavior is characteristic of
(A) Class A output stage
(B) Class B output stage
(C) Class AB output stage
(D) Common-base output stage
In a shunt-shunt negative feedback amplifier, as compared to the
basic amplifier
(A) both, input and output impedances,decrease
(B) input impedance decreases but output impedance increases
(C) input impedance increase but output
(D) both input and output impedances increases.
1999
1998
4.130
TWO MARK
An amplifier has an open-loop gain of 100, an input impedance of
1 kW,and an output impedance of 100 W. A feedback network with
a feedback factor of 0.99 is connected to the amplifier in a voltage
series feedback mode. The new input and output impedances,
respectively, are
(A) 10 W and 1W
(B) 10 W and 10 kW
(C) 100 kW and 1 W
(D) 100 kW and 1 kW
4.131
TWO MARKS
A multistage amplifier has a low-pass response with three real poles
at s =- w1 - w2 and w3 . The approximate overall bandwidth B of the
amplifier will be given by
(B) 1 = 1 + 1 + 1
(A) B = w1 + w2 + w3
w1 w2 w3
B
(C) B = (w1 + w2 + w3) 1/3
(D) B =
w12 + w22 + w23
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
4.132
4.133
4.134
One input terminal of high gain
ground and a sinusoidal voltage
output of comparator will be
(A) a sinusoid
(C) a half rectified sinusoid
Page 80
comparator circuit is connected to
is applied to the other input. The
(B) a full rectified sinusoid
(D) a square wave
In a series regulated power supply circuit, the voltage gain Av of the
‘pass’ transistor satisfies the condition
(B) 1 << Av < 3
(A) Av " 3
(C) Av . 1
(D) Av << 1
For full wave rectification, a four diode bridge rectifier is claimed to
have the following advantages over a two diode circuit :
(A) less expensive transformer,
(B) smaller size transformer, and
(C) suitability for higher voltage application.
Of these,
(A) only (1) and (2) are true
(B) only (1) and (3) are true
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(A) decrease the voltage gain and decrease the input impedance
(B) increase the voltage gain and decrease the input impedance
(C) decrease the voltage gain and increase the input impedance
(D) increase the voltage gain and increase the input impedance
4.138
4.139
(C) only (2) and (3) are true
(D) (1), (2) as well as (3) are true
4.135
4.136
4.137
In a common emitter BJT amplifier, the maximum usable supply
voltage is limited by
(A) Avalanche breakdown of Base-Emitter junction
(B) Collector-Base breakdown voltage with emitter open (BVCBO)
(C) Collector-Emitter breakdown voltage with base open (BVCBO)
(D) Zener breakdown voltage of the Emitter-Base junction
1997
In the MOSFET amplifier of the figure is the signal output V1 and
V2 obey the relationship
(A) V1 = V2
2
(B) V1 =-V2
2
(C) V1 = 2V2
(D) V1 =- 2V2
4.140
4.141
ONE MARK
In the BJT amplifier shown in the figure is the transistor is based in
the forward active region. Putting a capacitor across RE will
TWO MARKS
In the circuit of in the figure is the current iD through the ideal
diode (zero cut in voltage and forward resistance) equals
(A) 0 A
(C) 1 A
For small signal ac operation, a practical forward biased diode can
be modelled as
(A) a resistance and a capacitance in series
(B) an ideal diode and resistance in parallel
(C) a resistance and an ideal diode in series
(D) a resistance
1997
A cascade amplifier stags is equivalent to
(A) a common emitter stage followed by a common base stage
(B) a common base stage followed by an emitter follower
(C) an emitter follower stage followed by a common base stage
(D) a common base stage followed by a common emitter stage
(B) 4 A
(D) None of the above
The output voltage V0 of the circuit shown in the figure is
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) - 4 V
(C) 5 V
4.142
A half wave rectifier uses a diode
voltage is Vm sin wt and the load
is given by
(A) Vm
2 RL
2
V
m
(C)
p
(B) 6 V
(D) - 5.5 V
with a forward resistance Rf . The
resistance is RL . The DC current
Vm
p (R f + RL)
(D) Vm
RL
(B)
1996
4.143
ONE MARK
4.147
(B) 0.5 gm1
(D) 0.5 gm2
Value of R in the oscillator circuit shown in the given figure, so
chosen that it just oscillates at an angular frequency of w. The value
of w and the required value of R will respectively be
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(D) 0.4 mamp
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The circuit shown in the figure is that of
(A) a non-inverting amplifier
(C) an oscillator
(B) an inverting amplifier
(D) a Schmitt trigger
1996
4.145
(A) gm1
(C) gm2
In the circuit of the given figure, assume that the diodes are ideal
and the meter is an average indicating ammeter. The ammeter will
read
(A) 0.4 2 A
(C) 0.8 A
p
4.144
Page 81
TWO MARKS
In the circuit shown in the given figure N is a finite gain amplifier
with a gain of k , a very large input impedance, and a very low
output impedance. The input impedance of the feedback amplifier
with the feedback impedance Z connected as shown will be
(A) 105 rad/ sec, 2 # 10 4 W
(C) 2 # 10 4 rad/ sec, 105 W
4.148
(A) Z b1 - 1 l
k
Z
(C)
(k - 1)
4.146
(B) Z (1 - k)
(D) Z
(1 - k)
(B) 2 # 10 4 rad/ sec, 2 # 10 4 W
(D) 105 rad/ sec, 105 W
A zener diode in the circuit shown in the figure is has a knee current
of 5 mA, and a maximum allowed power dissipation of 300 mW
. What are the minimum and maximum load currents that can
be drawn safely from the circuit, keeping the output voltage V0
constant at 6 V?
A Darlington stage is shown in the figure. If the transconductance of
c
Q1 is gm1 and Q2 is gm2 , then the overall transconductance gmc ;T i cc E
vbe
is given by
(A) 0 mA, 180 mA
(C) 10 mA, 55 mA
(B) 5 mA, 110 mA
(D) 60 mA, 180 mA
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 82
***********
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Page 83
SOLUTIONS
4.1
(1)
Since, voltage across zener diode is 5 V so, current through 100 W
resistor is obtained as
Is = 10 - 5 = 0.05 A
100
Therefore, the load current is given by
IL = 5
RL
Since, for proper operation, we must
have
IZ $ Iknes
So, from Eq. (1), we write
0.05 A - 5 $ 10 mA
RL
50 mA - 5 $ 10 mA
RL
40 mA $ 5
RL
-3
40 # 10 $ 5
RL
Option (B) is correct.
For the given ideal op-amp, negative terminal will be also ground
(at zero voltage) and so, the collector terminal of the BJT will be
at zero voltage.
i.e.,
VC = 0 volt
The current in 1 kW resistor is given by
I = 5 - 0 = 5 mA
1 kW
This current will flow completely through the BJT since, no current will flow into the ideal op-amp ( I/P resistance of ideal opamp is infinity). So, for BJT we have
VC = 0
VB = 0
IC = 5 mA
i.e.,the base collector junction is reverse biased (zero voltage)
therefore, the collector current (IC ) can have a value only if baseemitter is forward biased. Hence,
VBE = 0.7 volts
&
VB - VE = 0.7
&
0 - Vout = 0.7
or,
Vout =- 0.7 volt
4.2
4.3
Option (A) is correct.
The i/p voltage of the system is given as
Vin = V1 + Vf
= V1 + k Vout
= V1 + k A 0 V1
^Vout = A 0 V1h
= V1 ^1 + k A 0h
Therefore, if k is increased then input voltage is also increased so,
the input impedance increases. Now, we have
Vout = A 0 V1
Vin
= A0
^1 + k A 0h
A
0 Vin
=
^1 + k A 0h
Since, Vin is independent of k when seen from output mode, the
output voltage decreases with increase in k that leads to the decrease
of output impedance. Thus, input impedance increases and output
impedance decreases.
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RL
1
-3 # 5
40 # 10
5
# RL
40 # 10-3
or,
125 W # RL
Therefore, minimum value of RL = 125 W
Now, we know that power rating of Zener diode is given by
PR = VZ IZ^maxh
IZ^maxh is maximum current through zener diode in reverse bias.
Maximum currrent through zener diode flows when load current is
zero. i.e.,
IZ^maxh = Is = 10 - 5 = 0.05
100
Therefore,
PR = 5 # 0.05 W
= 250 mW
4.4
Option (A) is correct.
For the given circuit, we obtain the small signal model as shown in
figure below :
Option (B) is correct.
We obtain the node voltage at V1 as
V1 +
V1
+ gm Vi = 0
RD R + 1
L
sC
&
From the circuit, we have
or,
Is = IZ + I L
IZ = Is - I L
V1 =
- gm Vi
1 +
1
RD R + 1
L
sC
Therefore, the output voltage V0 is obtained as
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
V0 = V1 RL
RL + 1
sC
=
RL
RL + 1
sC
4.6
- gm Vi
J
K 1
1
K RD +
K
RL + 1
sC
L
so, the transfer function is
V0 = - RD RL sCgm
Vi
1 + sC ^RD + RL h
1
Then, we have the pole at w =
C ^RD + RL h
It gives the lower cutoff frequency of transfer function.
1
i.e.,
w0 =
C ^RD + RL h
1
or,
f0 =
2pC ^RD + RL h
1
=
2p # 10-6 # 20 # 103
= 7.97
. 8 Hz
4.5
Page 84
N
O
O
O
P
Logic 0 means voltage is v = 0 volt and logic 1 means voltage is
5 volt
For x = 0 , y = 0 , Transistor is at cut off mode and diode is forward
biased. Since, there is no drop across forward biased diode.
So,
Z =Y=0
For x = 0 , y = 1, Again Transistor is in cutoff mode, and diode is
forward biased. with no current flowing through resistor.
So,
Z =Y=1
For x = 1, y = 0 , Transistor is in saturation mode and so, z directly
connected to ground irrespective of any value of Y .
i.e.,
Z = 0 (ground)
Similarly for X = Y = 1
Z = 0 (ground)
Hence, from the obtained truth table, we get
Z =XY
Option (C) is correct.
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Option (B) is correct.
For the given circuit, we can make the truth table as below
X
Y
Z
0
0
0
0
1
1
1
0
0
1
1
0
4.7
Option (D) is correct.
Given, the input voltage
VYZ = 100 sin wt
For + ve half cycle
VYZ > 0
i.e., VY is a higher voltage than VZ
So, the diode will be in cutoff region. Therefore, there will no voltage difference between X and W node.
i.e.,
VWX = 0
Now, for - ve half cycle all the four diodes will active and so, X
and W terminal is short circuited
For the given ideal op-Amps we can assume
V 2- = V 2+ = V2 (ideal)
V 1+ = V 1- = V1 (ideal)
So, by voltage division
V1 = Vout # 1
2
Vout = 2V1
and, as the I/P current in Op-amp is always zero therefore, there
will be no voltage drop across 1 KW in II op-amp
i.e.,
V2 = 1 V
Therefore,
V1 - V2 = V2 - ^- 2h
1
1
&
V1 - 1 = 1 + 2
or,
V1 = 4
Hence,
Vout = 2V1 = 8 volt
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i.e.,
Hence,
4.8
VWX = 0
VWX = 0 for all t
Option (C) is correct.
The equivalent circuit can be shown as
VTh = VCC
R2
R1 + R 2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 85
3R2
R1 + R 2
and
RTh = R2 R1
R 2 + R1
Since, IC = bIB has b . 3 (very high) so, IB is negative in
comparison to IC . Therefore, we can write the base voltage
VB = VTh
So,
VTh - 0.7 - IC RE = 0
=
or,
or,
or,
or,
Hence,
4.9
4.10
3R2 - 0.7 - 10-3 500 = 0
^
h^ h
R1 + R 2
3R2
= 0.7 + 0.5
60 kW + R2
The peak rectifier adds + 1 V to peak voltage, so overall peak voltage
lowers down by - 1 volt.
So,
vo = cos wt - 1
4.12
3R2 = ^60 kWh^1.2h + 1.2R2
Option (A) is correct.
We put a test source between terminal 1, 2 to obtain equivalent
impedance
1.8R2 = ^60 kWh # ^1.2h
R2 = 60 # 1.2 = 40 kW
1.8
Option (C) is correct.
Given
ib = 1 + 0.1 cos (1000pt) mA
So,
IB = DC component of ib
= 1 mA
In small signal model of the transistor
bVT
VT " Thermal voltage
rp =
IC
IC = I
= VT = VT
B
I
b
IC /b
B
= VT
IB
So,
rp = 25 mV = 25 W
VT = 25 mV, IB = 1 mA
1 mA
Option (D) is correct.
Let v > 0.7 V and diode is forward biased. By applying Kirchoff’s
voltage law
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10 - i # 1k - v = 0
10 - :v - 0.7 D (1000) - v = 0
500
So,
4.11
10 - (v - 0.7) # 2 - v = 0
10 - 3v + 1.4 = 0
v = 11.4 = 3.8 V > 0.7
3
i = v - 0.7 = 3.8 - 0.7 = 6.2 mA
500
500
ZTh = Vtest
Itest
Applying KCL at top right node
Vtest + Vtest - 99I = I
test
b
9 k + 1k 100
Vtest + Vtest - 99I = I
test
b
10 k 100
...(i)
(Assumption is true)
Option (A) is correct.
The circuit composed of a clamper and a peak rectifier as shown.
But
Clamper clamps the voltage to zero voltage, as shown
4.13
Ib =- Vtest =-Vtest
9k + 1k
10k
Substituting Ib into equation (i), we have
Vtest + Vtest + 99Vtest = I
test
10 k 100
10 k
100Vtest + Vtest = I
test
10 # 103 100
2Vtest = I
test
100
ZTh = Vtest = 50 W
Itest
Option (B) is correct.
First we obtain the transfer function.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 86
IC - IE = 13.7 - VC = (b + 1) IB
12k
13.7 - VC = 100I
...(ii)
B
12 # 103
Solving equation (i) and (ii),
IB = 0.01 mA
Small Signal Analysis :
Transforming given input voltage source into equivalent current
source.
0 - Vi (jw) 0 - Vo (jw)
+
=0
1 +R
R2
1
jwC
Vo (jw)
- Vi (jw)
=
1 +R
R2
1
j wC
Vi (jw) R2
R1 - j 1
wC
1 " 3, so V = 0
o
wC
Vo (jw) =At w " 0 (Low frequencies),
This is a shunt-shunt feedback amplifier.
Given parameters,
rp = VT = 25 mV = 2.5 kW
IB
0.01 mA
b
100
= 0.04 s
gm = =
rp 2.5 # 1000
Writing KCL at output node
v0 + g v + v0 - vp = 0
m p
RC
RF
v 0 : 1 + 1 D + v p :gm - 1 D = 0
RC RF
RF
Substituting RC = 12 kW, RF = 100 kW, gm = 0.04 s
v 0 (9.33 # 10-5) + v p (0.04) = 0
v 0 =- 428.72Vp
...(i)
Writing KCL at input node
vi = v p + v p + v p - vo
Rs
Rs rp
RF
vi = v 1 + 1 + 1 - v 0
p:
Rs
Rs rp RF D RF
vi = v (5.1 10-4) - v 0
#
p
Rs
RF
Substituting Vp from equation (i)
At w " 3 (higher frequencies)
1 " 0, so V (jw) =- R2 V (jw)
o
R1 i
wC
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The filter passes high frequencies so it is a high pass filter.
H (jw) = Vo = - R2
Vi
R1 - j 1
wC
R
R
2
2
=
H (3) =
R1
R1
At 3 dB frequency, gain will be
6H (3)@
2 times of maximum gain
H ^ jw0h = 1 H (3)
2
R2
R2
1
=
b
l
2 R1
R 12 + 21 2
w0 C
So,
2R 12 = R 12 +
vi = - 5.1 # 10-4 v - v 0
0
428.72
Rs
RF
1
w C2
2
0
vi
=- 1.16 # 10-6 v 0 - 1 # 10-5 v 0
10 # 103
R = 21 2
w C
w0 = 1
R1 C
2
1
4.14
Option (D) is correct.
DC Analysis :
Rs = 10 kW
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(source resistance)
vi
=- 1.116 # 10-5
10 # 103
1
- 8.96
Av = v 0 =
3
vi
10 # 10 # 1.116 # 10-5
4.15
Using KVL in input loop,
VC - 100IB - 0.7 = 0
VC = 100IB + 0.7
...(i)
Option (A) is correct.
For the parallel RLC circuit resonance frequency is,
1
wr = 1 =
= 10 M rad/s
-6
LC
10 # 10 # 1 # 10-9
Thus given frequency is resonance frequency and parallel RLC
circuit has maximum impedance at resonance frequency
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 87
Gain of the amplifier is gm # (ZC RL) where ZC is impedance of
parallel RLC circuit.
At w = wr , ZC = R = 2 kW = ZC max .
Hence at this frequency (wr ), gain is
Gain w = w = gm (ZC RL) = gm (2k 2k) = gm # 103 which is
maximum. Therefore gain is maximum at wr = 10 M rad/ sec .
r
4.16
In active region, for common emitter amplifier,
Substituting ICO
4.19
Option (D) is correct.
The given circuit is shown below :
...(1)
IC = bIB + (1 + b) ICO
= 0.6 mA and IB = 20 mA in above eq we have,
IC = 1.01 mA
Option (C) is correct.
In active region
VBEon = 0.7 V
Emitter voltage
VE = VB - VBEon =- 5.7 V
V - (- 10) - 5.7 - (- 10)
Emitter Current
=
= 1 mA
IE = E
4.3k
4.3k
Now
IC . IE = 1 mA
Applying KCL at collector
i1 = 0.5 mA
i1 = C dVC
dt
VC = 1 # i1 dt = i1 t
C
C
Since
or
From diagram we can write
Ii = Vo + Vo
R1 sL1
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Transfer function
or
H (s) = Vo = sR1 L1
I1 R1 + sL1
jwR1 L1
H (jw) =
R 1 + j wL 1
At w = 0
At w = 3
4.17
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H (jw) = R1 = constant .
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Hence HPF.
Option (C) is correct.
Given circuit is shown below.
with time, the capacitor charges and voltage across collector changes
from 0 towards negative.
When saturation starts,
VCE = 0.7 & VC =+ 5 V (across
capacitor)
Thus from (1) we get,
+ 5 = 0.5 mA T
5 mA
For transistor M2 ,
VGS = VG - VS = Vx - 0 = Vx
VDS = VD - VS = Vx - 0 = Vx
Since VGS - VT = Vx - 1 < VDS , thus M2 is in saturation.
By assuming M1 to be in saturation we have
IDS (M ) = IDS (M )
mn C 0x
m C
(4) (5 - Vx - 1) 2 = n 0x 1 (Vx - 1) 2
2
2
1
or
Taking positive root,
At Vx = 3 V for M1,VGS
is true and Vx = 3 V .
4.18
...(1)
Option (D) is correct.
We have
Now
2
or
4.20
4 (4 - Vx ) 2 = (Vx - 1) 2
2 (4 - Vx ) = ! (Vx - 1)
4.21
8 - 2Vx = Vx - 1
Vx = 3 V
= 5 - 3 = 2 V < VDS . Thus our assumption
a = 0.98
b = a = 4.9
1-a
-6
T = 5 # 5 # 10
= 50 m sec
-3
0.5 # 10
Option (A) is correct.
The current flows in the circuit if all the diodes are forward biased.
In forward biased there will be 0.7 V drop across each diode.
12.7 - 4 (0.7)
Thus
= 1 mA
IDC =
9900
Option (B) is correct.
The forward resistance of each diode is
r = VT = 25 mV = 25 W
IC
1 mA
4 (r)
Thus
Vac = Vi # e
4 (r) + 9900 o
= 100 mV cos (wt) 0.01
= 1 cos (wt) mV
4.22
Option (A) is correct.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
The equivalent circuit of given amplifier circuit (when CE is
connected, RE is short-circuited)
Page 88
4.24
Option (A) is correct.
The circuit is as shown below :
Input impedance
Ri = RB || r p
Voltage gain
AV = gm RC
Now, if CE is disconnected, resistance RE appears in the circuit
So,
or
4.25
Input impedance
Input resistance seen by source vs
R in = vs = Rs + Rs || rs
is
= (1000 W) + (93 kW || 259 W) = 1258 W
4.26
Input impedance increases
4.23
Option (B) is correct.
By small signal equivalent circuit analysis
R in = RB || [rp + (b + 1)] RE
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AV =
Voltage gain
0 - Vi + 0 - Vo = 0
R1
R2
Vo =- R2
R1
Vi
gm RC
1 + gm R E
fo =
Voltage gain decreases.
2
1
1
= 271 Hz
2 # 3.14 # 1250 # 4.7 # 10-6
Lower cut-off frequency
f
fL . o = 271 = 27.1 Hz
10
10
=
2
The circuit is as shown below :
4.27
VB =- 10 - (- 0.7) =- 9.3 V
Collector current
I1 =
0 - (- 9.3)
= 1 mA
(9.3 kW)
b 1 = 700 (high), So IC . IE
Applying KCL at base we have
1 - IE = IB + IB
1 - (b 1 + 1) IB = IB + IB
1
1
2
1
2
1 = (700 + 1 + 1)
IB
+ IB
2
1
2
2
IB . 2
702
2
I 0 = IC = b 2 : IB = 715 # 2 . 2 mA
702
2
2
1
2p (RC + RL) C2
fo
Option (B) is correct.
Since, emitter area of transistor Q1 is half of transistor Q2 , so current
IE = 1 IE and IB = 1 IB
2
2
1
Option (B) is correct.
Cut-off frequency due to C2
Option (B) is correct.
The circuit is as shown below
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 89
I = 20 - 0 + Vi - 0 = 5 + Vi
4R
R
R
If I > 0, diode D2 conducts
So, for 5 + VI > 0 & VI > - 5, D2 conducts
2
Equivalent circuit is shown below
IC = 10 - 5 = 1 mA
5k
Thus
Current
IE = IC
VE = IE RE = 1m # 1.4k = 1.4V
= 0.6 + 1.4 = 2V
Thus the feedback is negative and output voltage is V = 2V .
4.33
Option (D) is correct.
The output voltage is
V0 = Ar Vi .-
hfe RC
Vi
hie
Here RC = 3 W and hie = 3 kW
V0 . - 150 # 3k Vi
3k
Thus
Output is Vo = 0 . If I < 0 , diode D2 will be off
5 + VI < 0 & V < - 5, D is off
I
2
R
The circuit is shown below
.- 150 (A cos 20t + B sin 106 t)
Since coupling capacitor is large so low frequency signal will be
filtered out, and best approximation is
V0 .- 150B sin 106 t
4.34
0 - Vi + 0 - 20 + 0 - Vo = 0
R
4R
R
4.28
4.29
4.30
4.31
4.32
or
Vo =- Vi - 5
At Vi =- 5 V,
At Vi =- 10 V,
Vo = 0
Vo = 5 V
Option (A) is correct.
Let diode be OFF. In this case 1 A current will flow in resistor and
voltage across resistor will be V = 1.V
Diode is off, it must be in reverse biased, therefore
Vi - 1 > 0 " Vi > 1
Thus for Vi > 1 diode is off and V = 1V
Option (B) and (C) doesn’t satisfy this condition.
Let Vi < 1. In this case diode will be on and voltage across diode will
be zero and V = Vi
Thus
V = min (Vi, 1)
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For the positive half of Vi , the diode D1 is forward bias, D2 is reverse
bias and the zener diode is in breakdown state because Vi > 6.8 .
Thus output voltage is
V0 = 0.7 + 6.8 = 7.5 V
For the negative half of Vi, D2 is forward bias thus
Then
V0 =- 0.7 V
4.35
Option (C) is correct.
Option (B) is correct.
By Current mirror,
^Lh
Ix = W 2 Ibias
^ L h1
Since MOSFETs are identical,
W
W
Thus
b L l =b L l
2
2
W
Option (A) is correct.
The R2 decide only the frequency.
Option (D) is correct.
For small increase in VG beyond 1 V the n - channel MOSFET goes
into saturation as VGS "+ ive and p - MOSFET is always in active
region or triode region.
Option (C) is correct.
Hence
4.36
Ix = Ibias
Option (B) is correct.
The circuit is using ideal OPAMP. The non inverting terminal of
OPAMP is at ground, thus inverting terminal is also at virtual
ground.
Option (D) is correct.
The circuit is shown in fig below
Thus current will flow from -ive terminal (0 Volt) to -1 Volt source.
Thus the current I is
0 - (- 1)
I =
= 1
100k
100k
The voltage at non inverting terminal is 5 V because OP AMP is
ideal and inverting terminal is at 5 V.
The current through diode is
I = I 0 _eV - 1i
V
t
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 90
Now VT = 25 mV and I0 = 1 mA
Thus
or
Now
V
4.37
V
I = 10-6 8e 25 # 10 - 1B = 1 5
10
V = 0.06 V
V0 = I # 4k + V = 1 # 4k + 0.06 = 0.1
100k
-3
The Thevenin resistance and voltage are
VTH = 10 # 9 = 3 V
10 + 20
and total
RTH = 10k # 20k = 6.67 kW
10k + 20k
Option (B) is correct.
The circuit is using ideal OPAMP. The non inverting terminal of
OPAMP is at ground, thus inverting terminal is also at virtual
ground.
Since b is very large, therefore IB is small and can be ignored
Thus
IE = VTH - VBE = 3 - 0.7 = 1 mA
RE
2.3k
4.41
Option (D) is correct.
The small signal model is shown in fig below
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gm =
Vo =- gm Vp # (3k 3k )
=- 1 Vin (1.5k)
25
Thus we can write
vi
=
R1 + sL
or
-v
R2
sR2 C2 + 1
v0 =R2
vi
(R1 + sL)( sR2 C2 + 1)
or
4.42
and from this equation it may be easily seen that this is the standard form of T.F. of low pass filter
K
H (s) =
(R1 + sL)( sR2 C2 + 1)
and form this equation it may be easily seen that this is the standard form of T.F. of low pass filter
H (s) = 2 K
as + bs + b
4.38
4.39
4.43
Option ( ) is correct.
The current in both transistor are equal. Thus gm is decide by M1.
Hence (C) is correct option.
Option (C) is correct.
Let the voltage at non inverting terminal be V1, then after applying
KCL at non inverting terminal side we have
15 - V1 + V0 - V1 = V1 - (- 15)
10
10
10
or
V1 = V0
3
IC
= 1m = 1 A/V
VT
25m
25
Vp = Vin
=- 60Vin
Am = Vo =- 60
Vin
Option (C) is correct.
The circuit shown in (C) is correct full wave rectifier circuit.
Option (A) is correct.
In the transconductance amplifier it is desirable to have large input
resistance and large output resistance.
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4.44
Option (C) is correct.
We redraw the circuit as shown in fig.
If V0 swings from -15 to +15 V then V1 swings between -5 V to +5
V.
4.40
IC . IE
Option (A) is correct.
For the given DC values the Thevenin equivalent circuit is as follows
Applying voltage division rule
v+ = 0.5 V
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
v+ = vv- = 0.5 V
i = 1 - 0.5 = 0.5 mA
1k
i = 0.5 - v0 = 0.5 mA
2k
We know that
Thus
Now
and
v0 = 0.5 - 1 =- 0.5 V
or
4.45
Page 91
Option (B) is correct.
If we assume b very large, then IB = 0 and IE = IC ; VBE = 0.7 V. We
assume that BJT is in active, so applying KVL in Base-emitter loop
IE = 2 - VBE = 2 - 0.7 = 1.3 mA
RE
1k
The range of current through 200 kW is
3 = 15 mA to 9 = 45 mA
200k
200k
The range of variation in output voltage
15m # RZ = 0.15 V to 45m # RZ = 0.45
Thus the range of output voltage is 7.15 Volt to 7.45 Volt
4.49
V+ =
4.46
Option (D) is correct.
Here the inverting terminal is at virtual ground and the current in
resistor and diode current is equal i.e.
or
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V0 = 1 - sRC
Vi
1 + sRC
4.50
Option (C) is correct.
V0 = H (s) = 1 - sRC
Vi
1 + sRC
1 - jwRC
H (jw) =
1 + jwRC
For the first condition
VD = 0 - Vo1 = VT 1n 4
Is R
+H (jw) = f =- tan - 1 wRC - tan - 1 wRC
Subtracting above equation
Vo1 - Vo2 = VT 1n 4 - VT 1n 2
Is R
Is R
Vo1 - Vo2 = VT 1n 4 = VT 1n2
2
Option (D) is correct.
We have
Vthp = Vthp = 1 V
WP
W
and
= N = 40mA/V2
LP
LN
From figure it may be easily seen that Vas for each NMOS and
PMOS is 2.5 V
mA
Thus
ID = K (Vas - VT ) 2 = 40 2 (2.5 - 1) 2 = 90 m A
V
4.48
1
V
1 + sCR i
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VD = 0 - Vo1 = VT 1n 2
Is R
4.47
1
V
1 + sCR i
T
For the first condition
or
Vi =
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Vi = I eV /V
s
R
VD = VT 1n Vi
Is R
D
R + sC1
Applying voltage division rule
(V + Vi)
V+ = R1 (V0 + Vi) = o
R1 + R1
2
(Vo + Vi)
1
or
V =
1 + sCR i
2
Vo =- 1 +
2
or
Vi
1 + sRC
IR = ID
or
1
sC
V- = V+ =
Now
Since b is very large, we have IE = IC , thus
IC = 1.3 mA
Now applying KVL in collector-emitter loop
10 - 10IC - VCE - IC = 0
or
VCE =- 4.3 V
Now
VBC = VBE - VCE
= 0.7 - (- 4.3) = 5 V
Since VBC > 0.7 V, thus transistor in saturation.
Option (A) is correct.
The voltage at non-inverting terminal is
Minimum value,
Maximum value,
4.51
4.52
4.53
fmin
fmax
=- 2 tan - 2 wRC
= - p (at w " 3)
= 0( at w = 0)
Option (D) is correct.
In the transconductance amplifier it is desirable to have large input
impedance and large output impedance.
Option (C) is correct.
Option (D) is correct.
The voltage at inverting terminal is
V- = V+ = 10 V
Here note that current through the capacitor is constant and that
is
I = V- = 10 = 10 mA
1k 1k
Option (C) is correct.
We have VZ = 7 volt, VK = 0, RZ = 10W
Circuit can be modeled as shown in fig below
Thus the voltage across capacitor at t = 1 msec is
1m
1m
VC = 1 Idt = 1 10mdt
C 0
1m 0
Im
4
= 10 dt = 10 V
#
#
#0
4.54
Since Vi is lies between 10 to 16 V, the range of voltage across 200
kW
V200 = Vi - VZ = 3 to 9 volt
Option (A) is correct.
In forward bias Zener diode works as normal diode.
Thus for negative cycle of input Zener diode is forward biased and
it conducts giving VR = Vin .
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
For positive cycle of input Zener diode is reversed biased
when 0 < Vin < 6 , Diode is OFF and VR = 0
when Vin > 6 Diode conducts and voltage across diode is 6 V. Thus
voltage across is resistor is
VR = Vin - 6
Only option (B) satisfy this condition.
4.55
Page 92
Now
VCE = 15 - 9 = 6 V
The power dissipated in transistor is
P = VCE IC = 6 # 0.9 = 5.4 W
4.59
Option (C) is correct.
The circuit under DC condition is shown in fig below
Option (B) is correct.
If the unregulated voltage increase by 20%, them the unregulated
voltage is 18 V, but the VZ = Vin = 6 remain same and hence Vout
and IC remain same. There will be change in VCE
Thus,
VCE - 18 - 9 = 9 V
IC = 0.9 A
Power dissipation
P = VCE IC = 9 # 0.9 = 8.1 W
Thus % increase in power is
8.1 - 5.4 # 100 = 50%
5.4
4.60
Applying KVL we have
VCC - RC (IC + IB) - VCE = 0
...(1)
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4.61
Option (B) is correct.
Since the inverting terminal is at virtual ground, the current flowing
through the voltage source is
Is = Vs
10k
Vs = 10 kW = R
or
in
Is
Option (D) is correct.
The effect of current shunt feedback in an amplifier is to decrease
the input resistance and increase the output resistance as :
Rif = Ri
1 + Ab
Rof = R0 (1 + Ab)
Ri " Input resistance without feedback
Rif " Input resistance with feedback.
where
and
VCC - RB IB - VBE = 0
Substituting IC = bIB in (1) we have
VCC - RC (bIB + IB) - VCE = 0
Solving (2) and (3) we get
VCE = VCC - VCC - VBE
RB
1+
RC (1 + b)
Now substituting values we get
12 - 0.7
= 5.95 V
VCE = 12 53
1+
1 + (1 + 60)
4.56
...(2)
...(3)
4.62
...(4)
4.63
ACM =- RC
2RE
b' = 110 # 60 = 66
100
And differential mode gain
Substituting b' = 66 with other values in (iv) in previous solutions
12 - 0.7
= 5.29 V
VCE = 12 53
1+
1 + (1 + 66)
Thus change is
4.58
Option (D) is correct.
Common mode gain
Option (B) is correct.
We have
4.57
Option (B) is correct.
The CE configuration has high voltage gain as well as high
current gain. It performs basic function of amplifications. The CB
configuration has lowest Ri and highest Ro . It is used as last step to
match a very low impedance source and to drain a high impedance
load
Thus cascade amplifier is a multistage configuration of CE-CB
ADM =- gm RC
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= 5.29 - 59.5 # 100 =- 4.3%
5.95
Thus only common mode gain depends on RE and for large value
of RE it decreases.
Option (A) is correct.
Option (C) is correct.
The Zener diode is in breakdown region, thus
V+ = VZ = 6 V = Vin
R
We know that
Vo = Vin c1 + f m
R1
or
Vout = Vo = 6`1 + 12k j = 9 V
24k
The current in 12 kW branch is negligible as comparison to 10 W.
Thus Current
IC . IE . = Vout = 9 = 0.9 A
RL
10
4.64
Option (C) is correct.
IE = Is `e nV - 1j
VBE
T
= 10
4.65
- 13
0.7
c e1 # 26 # 10 - 1m = 49 mA
-3
Option (C) is correct.
The circuit is as shown below
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 93
IL = 24.2 - 0.5 = 23.7 mA
or
4.69
4.70
Option (D) is correct.
Option (B) is correct.
The small signal model is as shown below
Writing equation for I- have
e 0 - V- = I
1M
or
e0 = I- (1M) + VWriting equation for I+ we have
0 - V+
= I+
1M
...(1)
or
V+ = - I+ (1M)
Since for ideal OPAMP V+ = V- , from (1) and (2) we have
...(2)
From the figure we have
Zin = 2 MW
e0 = I- (1M) - I + (1M)
= (I- - I+) (1M) = IOS (1M)
Thus if e0 has been measured, we can calculate input offset current
IOS only.
4.66
4.67
Option (C) is correct.
At low frequency capacitor is open circuit and voltage acr s noninverting terminal is zero. At high frequency capacitor act as short
circuit and all input voltage appear at non-inverting terminal. Thus,
this is high pass circuit.
The frequency is given by
1
= 1000
w = 1 =
3
RC
1 # 10 # 1 # 10 - 6
rad/sec
Z0 = rd RD = 20k 2k = 20 kW
11
and
4.71
Option (A) is correct.
The circuit in DC condition is shown below
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Option (B) is correct.
The circuit under DC condition is shown in fig below
Since the FET has high input resistance, gate current can be neglect
and we get VGS =- 2 V
Since VP < VGS < 0 , FET is operating in active region
2
(- 2) 2
Now
ID = IDSS c1 - VGS m = 10 c1 (- 8) m
VP
= 5.625 mA
Now
Applying KVL we have
VCC - RB IB - VBE - RE IE = 0
or
VCC - RB IB - VBE - RE (b + 1) IB = 0
Since IE = IB + bIB
or
IB = VCC - VBE
RB + (b + 1) RE
20 - 0.7
=
= 40m A
430k + (50 + 1)1 k
Now
4.68
= 8.75 V
4.72
Option (B) is correct.
The transconductance is
gm =
or,
The gain is
IC = bIB = 50 # 40m = 2 mA
VC = VCC - RC IC = 20 - 2m # 2k = 16 V
Option (A) is correct.
The maximum load current will be at maximum input voltage i.e.
Vmax = 30 V i.e.
Vmax - VZ = I + I
L
Z
1k
30 - 5.8 = I = 0.5 m
or
L
1k
VDS = VDD - ID RD = 20 - 5.625 m # 2 k
So,
4.73
VP
2
ID IDSS
= 2 5.625mA # 10mA = 1.875 mS
8
A =- gm (rd RD)
= 1.875ms # 20 K =- 3.41
11
Option (B) is correct.
Only one diode will be in ON conditions
When lower diode is in ON condition, then
Vu = 2k Vsat = 2 10 = 8 V
2.5k
2.5
when upper diode is in ON condition
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 94
Vu = 2k Vsat = 2 (- 10) =- 5 V
2.5k
4
4.74
4.75
or
2V+ - Vo + IL R2 = 0
Since V- = V+ , from (1) and (2) we have
Option (B) is correct.
An ideal OPAMP is an ideal voltage controlled voltage source.
Option (C) is correct.
In voltage series feed back amplifier, input impedance increases
by factor (1 + Ab) and output impedance decreases by the factor
(1 + Ab).
Vs + IL R2 = 0
4.77
IL =- Vs
R2
or
4.80
Rif = Ri (1 + Ab)
Ro
Rof =
(1 + Ab)
4.76
...(2)
Option (D) is correct.
If IZ is negligible the load current is
12 - Vz = I
L
R
as per given condition
100 mA # 12 - VZ # 500 mA
R
At IL = 100 mA 12 - 5 = 100 mA
R
Option (A) is correct.
This is a Low pass filter, because
V0 = 0
At w = 3
Vin
V0 = 1
and at w = 0
Vin
VZ = 5 V
R = 70W
or
At IL = 500 mA 12 - 5 = 500 mA
R
VZ = 5 V
or
R = 14 W
Thus taking minimum we get
Option (D) is correct.
When IC >> ICO
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R = 14 W
4.81
4.82
Option (B) is correct.
Option (C) is correct.
The Thevenin equivalent is shown below
IC
= 1mA = 0.04 = 40 mA/V
VT
25mV
b
rp =
= 100 - 3 = 2.5 kW
gm
40 # 10
gm =
4.78
4.79
Option (A) is correct.
The given circuit is wein bridge oscillator. The frequency of oscillation
is
2pf = 1
RC
1
or
= 1 m
C = 1 =
3
3
2pRf
2p
2p # 10 # 10
VT =
R1 V = 1
#5 = 1 V
R1 + R2 C
4+1
Since b is large is large, IC . IE , IB . 0 and
IE = VT - VBE = 1 - 0.7 = 3 mA
RE
300
VCE = 5 - 2.2kIC - 300IE
= 5 - 2.2k # 1m - 300 # 1m
= 2.5 V
Now
Option (A) is correct.
The circuit is as shown below
4.83
Option (B) is correct.
For the different combinations the table is as follows
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CE
CE
CC
CB
We know that for ideal OPAMP
Ai
High
High
Unity
V- = V+
Applying KCL at inverting terminal
V- - Vs + V- - V0 = 0
R1
R1
Av
High
Unity
High
Ri
Medium
High
Low
Ro
Medium
Low
High
or
2V- - Vo = Vs
Applying KCL at non-inverting terminal
V+
V - Vo
+ IL + +
=0
R2
R2
...(1)
4.84
4.85
Option (D) is correct.
This circuit having two diode and capacitor pair in parallel, works
as voltage doubler.
Option (B) is correct.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 95
If the input is sinusoidal signal of 8 V (peak to peak) then
or
Vi = 4 sin wt
The output of comparator will be high when input is higher than
Vref = 2 V and will be low when input is lower than Vref = 2 V.
Thus the waveform for input is shown below
In second case IB2
Thus
4.90
4.91
From fig, first crossover is at wt1 and second crossover is at wt2
where
4 sin wt1 = 2V
wt1 = sin - 1 1 = p
2
6
wt2 = p - p = 5p
6
6
5p
p
-6
Duty Cycle = 6
=1
2p
3
Thus
4.86
Thus the output of comparators has a duty cycle of 1 .
3
Option (C) is correct.
CMMR = Ad
Ac
R2 = 2kW
IB1 = IC1 = 1.5m = 0.01 mA
150
b1
will we equal to IB1 as there is no in R1.
IC2 = b2 IB2 = 200 # 0.01 = 2 mA
VCE2 = VCC - IC2 R2 = 6 - 2m # 2 kW = 2 V
Option (A) is correct.
The given circuit is a R - C phase shift oscillator and frequency of
its oscillation is
1
f =
2p 6 RC
Option (C) is correct.
If we see th figure we find that the voltage at non-inverting terminal
is 3 V by the zener diode and voltage at inverting terminal will be 3
V. Thus Vo can be get by applying voltage division rule, i.e.
20 V = 3
20 + 40 o
V0 = 9 V
or
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4.92
Option (B) is correct.
The circuit is as shown below
20 log CMMR = 20 log Ad - 20 log Ac
= 48 - 2 = 46 dB
Where Ad "Differential Voltage Gain
and AC " Common Mode Voltage Gain
or
4.87
Option (B) is correct.
The gain of amplifier is
Ai =
8 (3) = 8 kW
1+8
3
V+ = V- = 8 V
3
- gm
gb + jwC
V+ =
Thus the gain of a transistor amplifier falls at high frequencies
due to the internal capacitance that are diffusion capacitance and
transition capacitance.
4.88
4.89
Now applying KCL at inverting terminal we get
V- - 2 + V- - Vo = 0
1
5
Option (A) is correct.
We have Ri = 1kW, b = 0.2, A = 50
Ri
Thus,
Rif =
= 1 kW
(1 + Ab)
11
Option (A) is correct.
The DC equivalent circuit is shown as below. This is fixed bias
circuit operating in active region.
4.93
Option (C) is correct.
The equivalent circuit of 3 cascade stage is as shown in fig.
or
1k
50V1 = 40V1
1k + 0.25k
1k
V3 =
50V2 = 40V2
1k + 0.25k
V2 =
In first case
VCC - IC1 R2 - VCE1 = 0
6 - 1.5mR2 - 3 = 0
Vo = 6V- - 10
= 6 # 8 - 10 = 6 V
3
or
Similarly
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
or
or
or
4.94
Page 96
V3 = 40 # 40V1
Vo = 50V3 = 50 # 40 # 40V1
AV = Vo = 50 # 40 # 40 = 8000
V1
Thus from above equation for sustained oscillation
6 = 1 + R2
R1
20 log AV = 20 log 8000 = 98 dB
Option (D) is correct.
If a constant current is made to flow in a capacitor, the output
voltage is integration of input current and that is sawtooth waveform
as below :
t
VC = 1 idt
C 0
The time period of wave form is
T = 1 = 1 = 2 m sec
f
500
4.99
20 # 10
1
idt
6
2 # 10 0
or
i (2 # 10 - 3 - 0) = 6 # 10 - 6
or
i = 3 mA
Thus the charging require 3 mA current source for 2 msec.
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4.95
Slew Rate
or
4.100
-6
Option (A) is correct.
The circuit is shown as below
Option (C) is correct.
In voltage-amplifier or voltage-series amplifier, the Ri increase and
Ro decrease because
I
For satisfactory operations
Vin - V0
R
When Vin = 30 V,
30 - 10
R
20
or
R
Option (B) is correct.
Let x be the gain and it is 20 db, therefore
6
Hz
1
2n
-1
1
23
or
Thus R # 1818W
$ 11 mA
R # 3636W
Option (D) is correct.
We have
1
2 2 - 1 = 0.5 kHz
Option (A) is correct.
As per Barkhousen criterion for sustained oscillations Ab $ 1 and
phase shift must be or 2pn .
$ (10 + 1) mA
40 $ 11 # 10 - 3
R
4.101
fHn = fH
[IZ + IL = I]
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-1
The higher cutoff frequency is
> IZ + IL
50 - 10 $ (10 + 1) mA
R
Option (A) is correct.
In multistage amplifier bandwidth decrease and overall gain increase.
From bandwidth point of view only options (A) may be correct
because lower cutoff frequency must be increases and higher must
be decreases. From following calculation we have
We have fL = 20 Hz and fH = 1 kHz
For n stage amplifier the lower cutoff frequency is
fL
20
f =
=
= 39.2 . 40
Ln
= IZ + IL
R # 1818 W
or
when Vin = 50 V
6
BW = 10 = 10 = 105 Hz = 100 kHz
Gain
10
4.98
dVO
= AV Vm w = AV Vm 2pf
c dt m
max
Vm = SR
AV V2pf
1
10 # 100 # 2p # 20 # 103
or
VM = 79.5 mV
=
20 log x = 20
or
x = 10
Since Gain band width product is 106 Hz, thus
So, bandwidth is
4.97
VO = VV Vi = Vm sin wt
dVO = A V w cos wt
V m
dt
Now
Rif = Ri (1 + Ab)
Ro
Rof =
(1 + Ab)
4.96
Option (C) is correct.
Let the gain of OPAMP be AV then we have
20 log AV = 40 dB
or
AV = 100
Let input be Vi = Vm sin wt then we have
-3
#
3=
R2 = 5R1
or
#
Thus
VO (f)
= 1 + R2
Vf (f)
R1
V (f)
b (f) = 1 +0 = f
6
VO (f)
A=
Now from circuit
Now
and
Thus
IDSS
VG
VS
VGS
= 10 mA and VP =- 5 V
=0
= ID RS = 1 # 2.5W = 2.5 V
= VG - VS = 0 - 2.5 =- 2.5 V
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
gm = 2IDSS 81 - ` - 2.5 jB = 2 mS
VP
-5
AV = V0 =- gm RD
Vi
Now
9
= 1 # 10 = 50.3 MHz
2p 10
Option (C) is correct.
The current gain of a BJT is
4.108
hfe = gm rp
4.103
4.105
4.106
Option (D) is correct.
The circuit is as shown below
Option (A) is correct.
The ideal op-amp has following characteristic :
and
4.104
1
2p LCeq
Ceq = C1 C2 = 2 # 2 = 1 pF
C1 + C2
4
1
f =
2p 10 # 10 - 6 # 10 - 12
f =
=- 2ms # 3k =- 6
So,
4.102
Page 97
Ri " 3
R0 " 0
A"3
Option (C) is correct.
Both statements are correct because
(1) A stable multivibrator can be used for generating square wave,
because of its characteristic
(2) Bi-stable multivibrator can store binary information, and this
multivibrator also give help in all digital kind of storing.
Option (B) is correct.
If fT is the frequency at which the short circuit common emitter gain
attains unity magnitude then
gm
38 # 10 - 3
=
fT =
2p (Cm + Cp)
2p # (10 - 14 + 4 # 10 - 13)
or
= 1.47 # 1010 Hz
If fB is bandwidth then we have
10
f
fB = T = 1.47 # 10 = 1.64 # 108 Hz
90
b
Let V- be the voltage of inverting terminal, since non inverting
terminal a at ground, the output voltage is
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Vo = AOL VNow applying KCL at inverting terminal we have
V- - Vs + V- - V0 = 0
R1
R2
Option (C) is correct.
If we neglect current through RB then it can be open circuit as
shown in fig.
4.109
Maximum power will dissipate in Zener diode when current through
it is maximum and it will occur at Vin = 30 V
I = Vin - Vo = 30 - 10 = 1 A
20
20
or
I IC + IZ = bIB + IZ
= bIZ + IZ = (b + 1) IZ
IZ = I = 1 = 0.01 A
99 + 1
b+1
...(1)
...(2)
From (1) and (2) we have
VO = A =
- R2
CL
Vs
R - R2 + R1
ROL
Substituting the values we have
- 10k
=- 1000 . - 11
ACL =
10
89
k
1
k
+
1k 100k
Option (A) is correct.
The first OPAMP stage is the differentiator and second OPAMP
stage is integrator. Thus if input is cosine term, output will be also
cosine term. Only option (A) is cosine term. Other are sine term.
However we can calculate as follows. The circuit is shown in fig
Since IC = bIB
since IB = IZ
Power dissipated in zener diode is
PZ = VZ IZ = 9.5 # 0.01 = 95 mW
IC = bIZ = 99 # 0.1 = 0.99 A
VCE = Vo = 10 V
Power dissipated in transistor is
PT = VC IC = 10 # 0.99 = 9.9 W
4.107
Option (B) is correct.
From the it may be easily seen that the tank circuit is having
2-capacitors and one-inductor, so it is colpits oscillator and frequency
is
Applying KCL at inverting terminal of first OP AMP we have
V1 = - wjL = - 100 # 10 # 10 - 3 = - 1
R
10
10
VS
- jVS
or
V1 =
= j cos 100t
10
Applying KCL at inverting terminal of second OP AMP we have
VO = - 1/jwC
100
V1
1
== j10
j100 # 10 # 10 - 6 # 100
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
or
4.110
Page 98
V0 = j10V2 = j10 (- j cos 100t)
V0 = 10 cos 100t
Option (A) is correct.
Common mode gain is
AC = aRC
REE
Since source resistance of the current source is infinite REE = 3 ,
common mode gain AC = 0
4.111
4.112
Option (D) is correct.
In positive feed back it is working as OP-AMP in saturation region,
and the input applied voltage is +ve.
So,
V0 =+ Vsat = 15 V
Option (C) is correct.
With the addition of RE the DC abis currents and voltages remain
closer to the point where they were set by the circuit when the
outside condition such as temperature and transistor parameter b
change.
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4.113
R1 V = 5
# 15 = 5 V
R1 + R2 C
10 + 5
Since b is large is large, IC . IE , IB . 0 and
IE = VT - VBE
RE
4.3
= 5 - 0.7 =
= 10 mA
0.430KW
0.430kW
VT =
4.119
4.120
4.121
4.122
Option (A) is correct.
At high frequency
and
Option (A) is correct.
Option (A) is correct.
Negative feedback in amplifier reduces the gain of the system.
Option (A) is correct.
By drawing small signal equivalent circuit
gm
+ jw (C)
1
Ai \
Capacitance
1
Ai a
frequency
Ai =-
or,
Option (C) is correct.
The output voltage will be input offset voltage multiplied by open
by open loop gain. Thus
So
V0 = 5mV # 10, 000 = 50 V
But
V0 = ! 15 V in saturation condition
So, it can never be exceeds ! 15 V
So,
V0 = ! Vset = ! 15V
'
gbc
Thus due to the transistor capacitance current gain of a bipolar
transistor drops.
4.114
4.115
4.116
4.117
4.118
by applying KCL at E2
Option (C) is correct.
As OP-AMP is ideal, the inverting terminal at virtual ground due
to ground at non-inverting terminal. Applying KCL at inverting
terminal
sC (v1 sin wt - 0) + sC (V2 sin wt - 0) + sC (Vo - 0) = 0
or
Vo =- (V1 + V2) sin wt
gm1 Vp 1
Vp
= gm2 Vp
rp
2
2
2
at C2
from eq (1) and (2)
i 0 =- gm2 Vp
2
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Option (D) is correct.
There is R - C , series connection in parallel with parallel R - C
combination. So, it is a wein bridge oscillator because two resistors
R1 and R2 is also in parallel with them.
gm1 Vp +
Option (A) is correct.
The given circuit is a differentiator, so the output of triangular wave
will be square wave.
1
i 0 =- i
0
gm2 rp
2
gm1 Vp =- i 0 :1 + 1 D
gm2 rp
1
2
Option (B) is correct.
In sampling and hold circuit the unity gain non-inverting amplifier
is used.
gm2 rp = b >> 1
so
gm1 Vp =- i 0
i 0 =- g
m1
Vp
i0 = g
a Vp = Vi
m1
Vi
Option (B) is correct.
Crossover behavior is characteristic of calss B output stage. Here 2
2
1
Option (D) is correct.
The Thevenin equivalent is shown below
1
1
4.123
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
transistor are operated one for amplifying +ve going portion and
other for -ve going portion.
4.124
Option (C) is correct.
In Voltage series feedback mode input impedance is given by
R in = Ri (1 + bv Av)
where
bv = feedback factor ,
Av = openloop gain
and
Ri = Input impedance
So,
R in = 1 # 103 (1 + 0.99 # 100) = 100 kW
Similarly output impedance is given by
R0
ROUT =
R 0 = output impedance
(1 + bv Av)
100
Thus
= 1W
ROUT =
(1 + 0.99 # 100)
4.125
4.127
4.128
4.133
4.134
4.135
Option (D) is correct.
This is a voltage shunt feedback as the feedback samples a portion
of output voltage and convert it to current (shunt).
Option (A) is correct.
In a differential amplifier CMRR is given by
(1 + b) IQ R 0
CMRR = 1 ;1 +
E
2
VT b
So where R 0 is the emitter resistance. So CMRR can be improved
by increasing emitter resistance.
Option (C) is correct.
In series voltage regulator the pass transistor is in common collector configuration having voltage gain close to unity.
Option (D) is correct.
In bridge rectifier we do not need central tap transformer, so its less
expensive and smaller in size and its PIV (Peak inverse voltage)
is also greater than the two diode circuit, so it is also suitable for
higher voltage application.
Option (C) is correct.
In the circuit we have
V2 = IS # RD
2
and
Option (B) is correct.
Regulation = Vno - load - Vfuel - load
Vfull - load
= 30 - 25 # 100 = 20%
25
Output resistance = 25 = 25 W
1
4.126
Page 99
V1 = IS # RD
V2 = 1
2
V1
V1 = 2V2
4.136
Option (C) is correct.
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4.137
Option (C) is correct.
The equivalent circuit of given amplifier circuit (when CE is
connected, RE is short-circuited)
Option (C) is correct.
We know that rise time (tr ) is
tr = 0.35
fH
where fH is upper 3 dB frequency. Thus we can obtain upper 3 dB
frequency it rise time is known.
4.129
4.130
Input impedance
Ri = RB || r p
Voltage gain
AV = gm RC
Now, if CE is disconnected, resistance RE appears in the circuit
Option (D) is correct.
In a BJT differential amplifier for a linear response Vid < VT .
Option (D) is correct.
In a shunt negative feedback amplifier.
Input impedance
Ri
R in =
(1 + bA)
where
Ri = input impedance of basic amplifier
b = feedback factor
A = open loop gain
Input impedance
R in = RB || [rp + (b + 1)] RE
Input impedance increases
gm RC
Voltage gain
Voltage gain decreases.
AV =
1 + gm R E
So, R in < Ri
Similarly
ROUT =
R0
(1 + bA)
ROUT < R 0
Thus input & output impedances decreases.
4.131
4.132
Option (A) is correct.
Option (D) is correct.
Comparator will give an output either equal to + Vsupply or - Vsupply .
So output is a square wave.
4.138
4.139
Option (A) is correct.
In common emitter stage input impedance is high, so in cascaded
amplifier common emitter stage is followed by common base stage.
Option (C) is correct.
We know that collect-emitter break down voltage is less than
compare to collector base breakdown voltage.
BVCEO < BVCBO
both avalanche and zener break down. Voltage are higher than
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 100
2Va - 4 + Va - V0 = 0
V0 = 3Va - 4
Va - V0 + Va - 0 = 0
100
10
BVCEO .So BVCEO limits the power supply.
4.140
Option (C) is correct.
If we assume consider the diode in reverse bias then Vn should be
greater than VP .
So
VP < Vn
by calculating
VP = 10 # 4 = 5 Volt
4+4
V0 =- 5.5 Volts
4.142
Vn = 2 # 1 = 2 Volt
here VP > Vn (so diode cannot be in reverse bias mode).
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apply node equation at node a
Va - 10 + Va + Va = 2
1
4
4
so current
4.141
6Va - 10 = 8
Va = 3 Volt
Ib = 0 - 3 + 10 - 3
4
4
6
10
= 1 amp
Ib =
4
Option (D) is correct.
By applying node equation at terminal (2) and (3) of OP -amp
Va - V0 + 10Va = 0
11Va = V0
Va = V0
11
V0 = 3V0 - 4
11
8V0 =- 4
11
Option (B) is correct.
Circuit with diode forward resistance looks
So the DC current will
IDC =
4.143
Vm
p (R f + RL)
Option (D) is correct.
For the positive half cycle of input diode D1 will conduct & D2 will
be off. In negative half cycle of input D1 will be off & D2 conduct so
output voltage wave from across resistor (10 kW) is –
Ammeter will read rms value of current
so
I rms = Vm (half wave rectifier)
pR
4
=
= 0.4 mA
p
(10 kW) p
4.144
4.145
Option (D) is correct.
In given circuit positive feedback is applied in the op-amp., so it
works as a Schmitt trigger.
Option (D) is correct.
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Gain with out feedback factor is given by
V0 = kVi
after connecting feedback impedance Z
Va - Q Va - V0
=0
+
5
10
given input impedance is very large, so after connecting Z we have
Ii = Vi - V0
V0 = kVi
Z
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Ii = Vi - kVi
Z
Zin = Vi = Z
Ii
(1 - k)
input impedance
4.146
4.147
Page 101
Option (A) is correct.
Option (A) is correct.
For the circuit, In balanced condition It will oscillated at a frequency
w
1 =
1
= 105 rad/ sec
-3
-6
LC
10 # 10 # .01 # 10
In this condition
R1 = R 3
R2
R4
5 =R
100
1
=
R = 20 kW = 2 # 10 4 W
4.148
Option (C) is correct.
V0 kept constant at
so current in 50 W resistor
V0 = 6 volt
I = 9-6
50 W
I = 60 m amp
Maximum allowed power dissipation in zener
PZ = 300 mW
Maximum current allowed in zener
PZ = VZ (IZ ) max = 300 # 10-3
&
= 6 (IZ ) max = 300 # 10-3
&
= (IZ ) max = 50 m amp
Given knee current or minimum current in zener
In given circuit
(IZ ) min = 5 m amp
I = IZ + I L
I L = I - IZ
(IL) min = I - (IZ ) max
= (60 - 50) m amp = 10 m amp
(IL) max = I - (IZ ) min
= (60 - 5) = 55 m amp
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
UNIT 5
Page 101
2012
5.4
ONE MARK
Consider the given circuit
DIGITAL CIRCUITS
2013
5.1
5.2
A bulb in a staircase has two switches, one switch being at the
ground floor and the other one at the first floor. The bulb can be
turned ON and also can be turned OFF by any one of the switches
irrespective of the state of the other switch. The logic of switching
of the bulb resembles
(A) and AND gate
(B) an OR gate
(C) an XOR gate
(D) a NAND gate
For 8085 microprocessor, the following program is executed.
MVI A, 05H;
MVI B, 05H;
PTR: ADD B;
DCR B;
JNZ PTR;
ADI 03H;
HLT;
At the end of program, accumulator contains
(A) 17H
(B) 20H
(C) 23H
(D) 05H
2013
5.3
ONE MARK
TWO MARKS
In this circuit, the race around
(A) does not occur
(B) occur when CLK = 0
(C) occur when CLK = 1 and A = B = 1
(D) occur when CLK = 1 and A = B = 0
5.5
The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input
A is greater than the 2-bit input B . The number of combinations for
which the output is logic 1, is
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(A) 4
(C) 8
5.6
There are four chips each of 1024 bytes connected to a 16 bit address
bus as shown in the figure below, RAMs 1, 2, 3 and 4 respectively
are mappped to addresses
In the circuit shown
(A) Y = A B + C
(C) Y = (A + B ) C
5.7
5.8
(B) Y = (A + B) C
(D) Y = AB + C
/
In the sum of products function f (X, Y, Z) = (2, 3, 4, 5), the prime
implicants are
(A) XY, XY
(B) XY, X Y Z , XY Z
(C) XY Z , XYZ, XY
(D) XY Z , XYZ, XY Z , XY Z
2012
(A) 0C00H-0FFFH, 1C00H-1FFFH, 2C00H-2FFFH, 3C00H3FFFH
(B) 1800H-1FFFH, 2800H-2FFFH, 3800H-3FFFH, 4800H-4FFFH
(C) 0500H-08FFH, 1500H-18FFH, 3500H-38FFH, 5500H-58FFH
(D) 0800H-0BFFH, 1800H-1BFFH, 2800H-2BFFH, 3800H-3BFFH
(B) 6
(D) 10
TWO MARKS
In the CMOS circuit shown, electron and hole mobilities are equal,
and M1 and M2 are equally sized. The device M1 is in the linear
region if
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) Vin < 1.875 V
(C) Vin > 3.125 V
5.9
(A) changed from “0” to “1”
(C) changed in either direction
(B) 1.875 V < Vin < 3.125 V
(D) 0 < Vin < 5 V
The state transition diagram for the logic circuit shown is
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2011
5.10
Page 102
5.12
The logic function implemented by the circuit below is (ground
implies a logic “0”)
(A) F = AND ^P, Q h
(C) F = XNOR ^P, Q h
2011
5.13
(B) changed from “1” to “0”
(D) not changed
(B) F = OR ^P, Q h
(D) F = XOR ^P, Q h
TWO MARKS
The output of a 3-stage Johnson (twisted ring) counter is fed to
a digital-to analog (D/A) converter as shown in the figure below.
Assume all states of the counter to be unset initially. The waveform
which represents the D/A converter output Vo is
ONE MARK
The output Y in the circuit below is always ‘1’ when
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(A) two or more of the inputs P, Q, R are ‘0’
(B) two or more of the inputs P, Q, R are ‘1’
(C) any odd number of the inputs P, Q, R is ‘0’
(D) any odd number of the inputs P, Q, R is ‘1’
5.11
When the output Y in the circuit below is “1”, it implies that data
has
5.14
Two D flip-flops are connected as a synchronous counter that goes
through the following QB QA sequence 00 " 11 " 01 " 10 " 00 " ....
The connections to the inputs DA and DB are
(A) DA = QB, DB = QA
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
5.15
(B) DA = Q A, DB = Q B
(C) DA = (QA Q B + Q A QB), DB = QA
(D) DA = (QA QB + Q A Q B), DB = Q B
(A) A = 1, B = 1, C = 0
(C) A = 0, B = 1, C = 0
An 8085 assembly language program is given below. Assume that
the carry flag is initially unset. The content of the accumulator after
the execution of the program is
2010
(A) 8CH
(C) 23H
5.19
(B) 64H
(D) 15H
2010
5.16
Page 103
(B) A = 1, B = 0, C = 0
(D) A = 0, B = 0, C = 1
TWO MARKS
Assuming that the flip-flop are in reset condition initially, the count
sequence observed at QA , in the circuit shown is
(A) 0010111...
(C) 0101111...
(B) 0001011...
(D) 0110100....
ONE MARK
Match the logic gates in Column A with their equivalents in Column
B
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5.20
(A) P-2, Q-4, R-1, S-3
(C) P-2, Q-4, R-3, S-1
5.17
The Boolean function realized by the logic circuit shown is
(B) P-4, Q-2, R-1, S-3
(D) P-4, Q-2, R-3, S-1
(A) F = Sm (0, 1, 3, 5, 9, 10, 14)
(C) F = Sm (1, 2, 4, 5, 11, 14, 15)
In the circuit shown, the device connected Y5 can have address in
the range
5.21
For the 8085 assembly language program given below, the content of
the accumulator after the execution of the program is
(A) 00H
(C) 67H
(A) 2000 - 20FF
(C) 2E00 - 2EFF
(B) 2D00 - 2DFF
(D) FD00 - FDFF
2009
5.22
5.18
For the output F to be 1 in the logic circuit shown, the input
combination should be
(B) F = Sm (2, 3, 5, 7, 8, 12, 13)
(D) F = Sm (2, 3, 5, 7, 8, 9, 12)
(B) 45H
(D) E7H
ONE MARK
The full form of the abbreviations TTL and CMOS in reference to
logic families are
(A) Triple Transistor Logic and Chip Metal Oxide Semiconductor
(B) Tristate Transistor Logic and Chip Metal Oxide Semiconductor
(C) Transistor Transistor Logic and Complementary Metal Oxide
Semiconductor
(D) Tristate Transistor Logic and Complementary Metal Oxide
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 104
Silicon
5.23
In a microprocessor, the service routine for a certain interrupt starts
from a fixed location of memory which cannot be externally set, but
the interrupt can be delayed or rejected Such an interrupt is
(A) non-maskable and non-vectored
(B) maskable and non-vectored
(C) non-maskable and vectored
(D) maskable and vectored
(1) push buttons pressed/not pressed in equivalent to logic 1/0
respectively.
(2) a segment glowing/not glowing in the display is equivalent to
logic 1/0 respectively.
5.27
5.28
2009
5.24
5.25
TWO MARKS
If X = 1 in logic equation 6X + Z {Y + (Z + XY )}@ {X + X (X + Y)} = 1
, then
(A) Y = Z
(B) Y = Z
(C) Z = 1
(D) Z = 0
What are the minimum number of 2- to -1 multiplexers required to
generate a 2- input AND gate and a 2- input Ex-OR gate
(A) 1 and 2
(B) 1 and 3
5.29
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(C) 1 and 1
5.26
If segments a to g are considered as functions of P1 and P2 , then
which of the following is correct
(A) g = P 1 + P2, d = c + e
(B) g = P1 + P2, d = c + e
(C) g = P1 + P2, e = b + c
(D) g = P1 + P2, e = b + c
What are the minimum numbers of NOT gates and 2 - input OR
gates required to design the logic of the driver for this 7 - Segment
display
(A) 3 NOT and 4 OR
(B) 2 NOT and 4 OR
(C) 1 NOT and 3 OR
(D) 2 NOT and 3 OR
Refer to the NAND and NOR latches shown in the figure. The
inputs (P1, P2) for both latches are first made (0, 1) and then, after a
few seconds, made (1, 1). The corresponding stable outputs (Q1, Q2)
are
(A) NAND: first (0, 1) then (0, 1) NOR: first (1, 0) then (0, 0)
(B) NAND : first (1, 0) then (1, 0) NOR : first (1, 0) then (1, 0)
(C) NAND : first (1, 0) then (1, 0) NOR : first (1, 0) then (0, 0)
(D) NAND : first (1, 0) then (1, 1) NOR : first (0, 1) then (0, 1)
(D) 2 and 2
What are the counting states (Q1, Q2) for the counter shown in the
figure below
2008
5.30
(A) 11, 10, 00, 11, 10,...
(C) 00, 11, 01, 10, 00...
TWO MARKS
The logic function implemented by the following circuit at the
terminal OUT is
(B) 01, 10, 11, 00, 01...
(D) 01, 10, 00, 01, 10...
Statement for Linked Answer Question 5.18 & 5.19 :
Two products are sold from a vending machine, which has two
push buttons P1 and P2 .
When a buttons is pressed, the price of the corresponding product
is displayed in a 7 - segment display. If no buttons are pressed, '0'
is displayed signifying ‘Rs 0’.
If only P1 is pressed, ‘2’ is displayed, signifying ‘Rs. 2’
If only P2 is pressed ‘5’ is displayed, signifying ‘Rs. 5’
If both P1 and P2 are pressed, 'E' is displayed, signifying ‘Error’
The names of the segments in the 7 - segment display, and the
glow of the display for ‘0’, ‘2’, ‘5’ and ‘E’ are shown below.
Consider
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(A) P NOR Q
(C) P OR Q
5.31
5.32
(B) P NAND Q
(D) P AND Q
The two numbers represented in signed 2’s complement form are
P + 11101101 and Q = 11100110 . If Q is subtracted from P , the
value obtained in signed 2’s complement is
(A) 1000001111
(B) 00000111
(C) 11111001
(D) 111111001
Which of the following Boolean Expressions correctly represents the
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 105
3
relation between P, Q, R and M1
per the relation VDAC =
2n - 1 bn Volts, where b3 (MSB), b1, b2 and
0
The counter starts from the
b0 (LSB) are the countern =outputs.
clear state.
/
(A) M1 = (P OR Q) XOR R
(B) M1 = (P AND Q) X OR R
(C) M1 = (P NOR Q) X OR R
(D) M1 = (P XOR Q) XOR R
5.33
For the circuit shown in the figure, D has a transition from 0 to 1
after CLK changes from 1 to 0. Assume gate delays to be negligible
Which of the following statements is true
5.35
The stable reading of the LED displays is
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(A) Q goes to 1 at the CLK transition and stays at 1
(B) Q goes to 0 at the CLK transition and stays 0
(C) Q goes to 1 at the CLK tradition and goes to 0 when D goes
to 1
(D) Q goes to 0 at the CLK transition and goes to 1 when D goes
to 1
5.34
For each of the positive edge-triggered J - K flip flop used in the
following figure, the propagation delay is 3 t .
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(A) 06
(C) 12
5.36
5.37
(B) 07
(D) 13
The magnitude of the error between VDAC and Vin at steady state in
volts is
(A) 0.2
(B) 0.3
(C) 0.5
(D) 1.0
For the circuit shown in the following, I0 - I3 are inputs to the 4:1
multiplexers, R(MSB) and S are control bits.
The output Z can be represented by
Which of the following wave forms correctly represents the output
at Q1 ?
(A) PQ + PQS + QRS
(B) PQ + PQR + PQS
(C) PQR + PQR + PARS + QRS
(D) PQR + PQRS + PQRS + QRS
5.38
Statement For Linked Answer Question 5.26 & 5.27 :
In the following circuit, the comparators output is logic “1” if
V1 > V2 and is logic "0" otherwise. The D/A conversion is done as
An 8085 executes the following instructions
2710 LXI H, 30A0 H
2713 DAD H
2414 PCHL
All address and constants are in Hex. Let PC be the contents of
the program counter and HL be the contents of the HL register
pair just after executing PCHL. Which of the following statements
is correct ?
PC = 30A0H
PC = 2715H
(B)
(A)
HL = 2715H
HL = 30A0H
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
PC = 6140H
(C)
HL = 6140H
Page 106
(C) Q1: normal active; Q2: cut-off; Q3: cut-off; Q4: saturation
(D) Q1: saturation; Q2: saturation; Q3: saturation; Q4: normal active
PC = 6140H
(D)
HL = 2715H
5.44
2007
5.39
5.40
X = 01110 and Y = 11001 are two 5-bit binary numbers represented
in two’s complement format. The sum of X and Y represented in
two’s complement format using 6 bits is
(A) 100111
(B) 0010000
(C) 000111
(D) 101001
The Boolean function Y = AB + CD is to be realized using only 2 input NAND gates. The minimum number of gates required is
(A) 2
(B) 3
(C) 4
(D) 5
2007
5.41
ONE MARK
(A)
(B)
(C)
(D)
TWO MARKS
The Boolean expression Y = ABC D + ABCD + ABC D + ABC D
can be minimized to
5.45
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(A) Y = ABC D + ABC + AC D
(B) Y = ABC D + BCD + ABC D
(C) Y = ABCD + BC D + ABC D
(D) Y = ABCD + BC D + ABC D
5.42
5.43
P = 1, Q = 0;
P = 1, Q = 0;
P = 1, Q = 0;
P = 1, Q = 0;
P = 1, Q = 0; P = 1, Q = 0 or P = 0, Q = 1
P = 0, Q = 1; or P = 0, Q = 1; P = 0, Q = 1
P = 1, Q = 1; P = 1, Q = 0 or P = 0, Q = 1
P = 1, Q = 1; P = 1, Q = 1
An 8255 chip is interfaced to an 8085 microprocessor system as an
I/O mapped I/O as show in the figure. The address lines A0 and A1
of the 8085 are used by the 8255 chip to decode internally its thee
ports and the Control register. The address lines A3 to A7 as well
as the IO/M signal are used for address decoding. The range of
addresses for which the 8255 chip would get selected is
(A) F8H - FBH
(C) F8H - FFH
(B) F8GH - FCH
(D) F0H - F7H
Statement for Linked Answer Question 5.37 and 5.38 :
In the following circuit, X is given by
(A)
(B)
(C)
(D)
The following binary values were applied to the X and Y inputs of
NAND latch shown in the figure in the sequence indicated below :
X = 0,Y = 1; X = 0, Y = 0; X = 1; Y = 1
The corresponding stable P, Q output will be.
In the Digital-to-Analog converter circuit shown in the figure below,
VR = 10V and R = 10kW
X = ABC + ABC + ABC + ABC
X = ABC + ABC + ABC + ABC
X = AB + BC + AC
X = AB + BC + AC
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The circuit diagram of a standard TTL NOT gate is shown in the
figure. Vi = 25 V, the modes of operation of the transistors will be
5.46
5.47
The current is
(A) 31.25mA
(C) 125mA
(B) 62.5mA
(D) 250mA
The voltage V0 is
(A) - 0.781 V
(C) - 3.125 V
(B) - 1.562 V
(D) - 6.250 V
Statement for Linked Answer Questions 5.39 & 5.40 :
(A) Q1: revere active; Q2: normal active; Q3: saturation; Q4: cut-off
(B) Q1: revere active; Q2: saturation; Q3: saturation; Q4: cut-off
An 8085 assembly language program is given below.
Line 1:
MVI A, B5H
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2:
3:
4:
5:
6:
7:
8:
5.48
5.49
5.50
Page 107
MVI B, OEH
XRI 69H
ADD B
ANI 9BH
CPI 9FH
STA 3010H
HLT
- out shift registers loaded with the data shown are used to feed the
data to a full adder. Initially, all the flip - flops are in clear state.
After applying two clock pulse, the output of the full-adder should
be
The contents of the accumulator just execution of the ADD
instruction in line 4 will be
(A) C3H
(B) EAH
(C) DCH
(D) 69H
After execution of line 7 of the program, the status of the CY and
Z flags will be
(B) CY = 0, Z = 1
(A) CY = 0, Z = 0
(C) CY = 1, Z = 0
(D) CY = 1, Z = 1
(A) S = 0, C0 = 0
(C) S = 1, C0 = 0
5.54
For the circuit shown, the counter state (Q1 Q0) follows the sequence
(B) S = 0, C0 = 1
(D) S = 1, C0 = 1
A new Binary Coded Pentary (BCP) number system is proposed
in which every digit of a base-5 number is represented by its
corresponding 3-bit binary code. For example, the base-5 number
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(A) 00, 01, 10, 11, 00
(C) 00, 01, 11, 00, 01
2006
5.51
24 will be represented by its BCP code 010100. In this numbering
system, the BCP code 10001001101 corresponds of the following
number is base-5 system
(A) 423
(B) 1324
(C) 2201
(D) 4231
ONE MARK
The number of product terms in the minimized sum-of-product
expression obtained through the following K - map is (where, "d"
denotes don’t care states)
(A) 2
(C) 4
5.55
TWO MARKS
In the figure shown above, the ground has been shown by the symbol 4
An I/O peripheral device shown in Fig. (b) below is to be interfaced
to an 8085 microprocessor. To select the I/O device in the I/O
address range D4 H - D7 H, its chip-select (CS ) should be connected
to the output of the decoder shown in as below :
5.56
(A) output 7
(C) output 2
5.53
A 4 - bit D/A converter is connected to a free - running 3 - big UP
counter, as shown in the following figure. Which of the following
waveforms will be observed at V0 ?
(B) 3
(D) 5
2006
5.52
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(B) 00, 01, 10, 00, 01
(D) 00, 10, 11, 00, 10
(B) output 5
(D) output 0
For the circuit shown in figures below, two 4 - bit parallel - in serial
Following is the segment of a 8085 assembly language program
LXI SP, EFFF H
CALL 3000 H
:
:
:
3000 H LXI H, 3CF4
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
PUSH PSW
SPHL
POP PSW
RET
On completion of RET execution, the contents of SP is
(A) 3CF0 H
(B) 3CF8 H
(C) EFFD H
(D) EFFF H
5.57
Page 108
2005
5.61
TWO MARKS
The transistors used in a portion of the TTL gate show in the figure
have b = 100 . The base emitter voltage of is 0.7 V for a transistor
in active region and 0.75 V for a transistor in saturation. If the sink
current I = 1 A and the output is at logic 0, then the current IR will
be equal to
Two D - flip - flops, as shown below, are to be connected as a
synchronous counter that goes through the following sequence
00 " 01 " 11 " 10 " 00 " ...
The inputs D0 and D1 respectively should be connected as,
(A) Q 1 and Q0
(C) Q1 Q0 and Q 1 Q0
(A) 0.65 mA
(C) 0.75 mA
(B) Q 0 and Q1
(D) Q 1 Q 0 and Q1 Q0
5.62
(B) 0.70 mA
(D) 1.00 mA
The Boolean expression for the truth table shown is
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5.58
The point P in the following figure is stuck at 1. The output f will
be
(A) B (A + C)( A + C )
(C) B (A + C )( A + C)
5.63
(A) ABC
(C) ABC
(B) A
(D) A
2005
5.59
5.60
5.64
The present output Qn of an edge triggered JK flip-flop is logic 0.
If J = 1, then Qn + 1
(A) Cannot be determined
(B) Will be logic 0
(C) will be logic 1
(D) will rave around
The given figure shows a ripple counter using positive edge triggered
flip-flops. If the present state of the counter is Q2 Q1 Q0 = 001 then is
next state Q2 Q1 Q will be
ONE MARK
Decimal 43 in Hexadecimal and BCD number system is respectively
(A) B2, 0100 011
(B) 2B, 0100 0011
(C) 2B, 0011 0100
(D) B2, 0100 0100
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The Boolean function f implemented in the figure using two input
multiplexes is
(A) ABC + ABC
(C) ABC + ABC
(B) B (A + C )( A + C)
(D) B (A + C)( A + C )
(A) 010
(C) 100
(B) ABC + ABC
(D) ABC + ABC
5.65
(B) 111
(D) 101
What memory address range is NOT represents by chip # 1 and
chip # 2 in the figure A0 to A15 in this figure are the address lines
and CS means chip select.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 109
(C) 5
5.71
5.72
(A) 0100 - 02FF
(C) F900 - FAFF
(B) 1500 - 16FF
(D) F800 - F9FF
Consider an 8085 microprocessor system.
5.67
The following program starts at location 0100H.
LXI SP, OOFF
LXI H, 0701
MVI A, 20H
SUB M
The content of accumulator when the program counter reaches
0109 H is
(A) 20 H
(B) 02 H
(C) 00 H
(D) FF H
If in addition following code exists from 019H onwards,
ORI 40 H
ADD M
What will be the result in the accumulator after the last instruction is executed ?
(A) 40 H
(B) 20 H
(C) 60 H
(D) 42 H
2004
5.68
5.69
5.70
Choose the correct one from among the alternatives A, B, C, D after
matching an item from Group 1 most appropriate item in Group 2.
Group 1
Group 2
P. Shift register
1. Frequency division
Q. Counter
2. Addressing in memory chips
R. Decoder
3. Serial to parallel data conversion
(B) P - 3, Q - 1, R - 2
(A) P - 3, Q - 2, R - 1
(C) P - 2, Q - 1, R - 3
(D) P - 1, Q - 2, R - 2
The figure the internal schematic of a TTL AND-OR-OR-Invert
(AOI) gate. For the inputs shown in the figure, the output Y is
(A) 0
(C) AB
Statement For Linked Answer Questions 5.57 & 5.58 :
5.66
(D) 7
(B) 1
(D) AB
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2004
5.73
5.74
TWO MARKS
11001, 1001, 111001 correspond to the 2’s complement representation
of which one of the following sets of number
(A) 25,9, and 57 respectively
(B) -6, -6, and -6 respectively
(C) -7, -7 and -7 respectively
(D) -25, -9 and -57 respectively
In the modulo-6 ripple counter shown in figure, the output of the
2- input gate is used to clear the J-K flip-flop
The 2-input gate is
ONE MARK
A master - slave flip flop has the characteristic that
(A) change in the output immediately reflected in the output
(B) change in the output occurs when the state of the master is
affected
(C) change in the output occurs when the state of the slave is affected
(D) both the master and the slave states are affected at the same
time
The range of signed decimal numbers that can be represented by
6-bits 1’s complement number is
(A) -31 to +31
(B) -63 to +63
(C) -64 to +63
(D) -32 to +31
A digital system is required to amplify a binary-encoded audio
signal. The user should be able to control the gain of the amplifier
from minimum to a maximum in 100 increments. The minimum
number of bits required to encode, in straight binary, is
(A) 8
(B) 6
(A) a NAND gate
(C) an OR gate
5.75
5.76
5.77
(B) a NOR gate
(D) a AND gare
The minimum number of 2- to -1 multiplexers required to realize a
4- to -1 multiplexers is
(A) 1
(B) 2
(C) 3
(D) 4
The Boolean expression AC + BC is equivalent to
(A) AC + BC + AC
(B) BC + AC + BC + ACB
(C) AC + BC + BC + ABC
(D) ABC + ABC + ABC + ABC
A Boolean function f of two variables x and y is defined as follows :
f (0, 0) = f (0, 1) = f (1, 1) = 1; f (1, 0) = 0
Assuming complements of x and y are not available, a minimum
cost solution for realizing f using only 2-input NOR gates and 2-
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 110
input OR gates (each having unit cost) would have a total cost of
(A) 1 unit
(B) 4 unit
(C) 3 unit
(D) 2 unit
5.78
The 8255 Programmable Peripheral Interface is used as described
below.
(i) An A/D converter is interface to a microprocessor through an
8255.
The conversion is initiated by a signal from the 8255 on Port C. A
signal on Port C causes data to be stobed into Port A.
(ii) Two computers exchange data using a pair of 8255s. Port A
works as a bidirectional data port supported by appropriate handshaking signals.
The appropriate modes of operation of the 8255 for (i) and (ii)
would be
(A) Mode 0 for (i) and Mode 1 for (ii)
(B) Mode 1 for (i) and Mode 2 for (ii)
(C) Mode for (i) and Mode 0 for (ii)
(D) Mode 2 for (i) and Mode 1 for (ii)
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5.79
5.80
5.81
The number of memory cycles required to execute the following
8085 instructions
(i) LDA 3000 H
(ii) LXI D, FOF1H
would be
(A) 2 for (i) and 2 for (ii)
(B) 4 for (i) and 3 for (ii)
(C) 3 for (i) and 3 for (ii)
(D) 3 for (i) and 4 for (ii)
Consider the sequence of 8085 instructions given below
LXI H, 9258
MOV A, M
CMA
MOV M, A
Which one of the following is performed by this sequence ?
(A) Contents of location 9258 are moved to the accumulator
(B) Contents of location 9258 are compared with the contents of
the accumulator
(C) Contents of location 8529 are complemented and stored in
location 8529
(D) Contents of location 5892 are complemented and stored in
location 5892
It is desired to multiply the numbers 0AH by 0BH and store the
result in the accumulator. The numbers are available in registers B
and C respectively. A part of the 8085 program for this purpose is
given below :
MVI A, 00H
LOOP
--------------HLT
END
The sequence of instructions to complete the program would be
(A) JNX LOOP, ADD B, DCR C
(B) ADD B, JNZ LOOP, DCR C
(C) DCR C, JNZ LOOP, ADD B
(D) ADD B, DCR C, JNZ LOOP
2003
5.82
5.83
5.84
5.85
5.86
The number of distinct Boolean expressions of 4 variables is
(A) 16
(B) 256
(C) 1023
(D) 65536
The minimum number of comparators required to build an 8-bits
flash ADC is
(A) 8
(B) 63
(C) 255
(D) 256
The output of the 74 series of GATE of TTL gates is taken from a
BJT in
(A) totem pole and common collector configuration
(B) either totem pole or open collector configuration
(C) common base configuration
(D) common collector configuration
Without any additional circuitry, an 8:1 MUX can be used to obtain
(A) some but not all Boolean functions of 3 variables
(B) all functions of 3 variables but non of 4 variables
(C) all functions of 3 variables and some but not all of 4 variables
(D) all functions of 4 variables
A 0 to 6 counter consists of 3 flip flops and a combination circuit of
2 input gate (s). The common circuit consists of
(A) one AND gate
(B) one OR gate
(C) one AND gate and one OR gate
(D) two AND gates
2003
5.87
ONE MARK
TWO MARKS
The circuit in the figure has 4 boxes each described by inputs P, Q, R
and outputs Y, Z with Y = P 5 Q 5 R and Z = RQ + PR + QP
The circuit acts as a
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(A) 4 bit adder giving P + Q
(B) 4 bit subtractor giving P - Q
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 111
(C) 4 bit subtractor giving Q-P
(D) 4 bit adder giving P + Q + R
5.88
5.89
5.90
Propagation delay is
minimum
If the function W, X, Y and Z are as follows
W = R + PQ + RS
X = PQRS + PQRS + PQRS
Y = RS + PR + PQ + P .Q
Z = R + S + PQ + P .Q .R + PQ .S
Then,
(A) W = Z, X = Z
(B) W = Z, X = Y
(C) W = Y
(D) W = Y = Z
The correct column is
(A) P
(C) R
5.92
ECL
TTL
TTL
(B) Q
(D) S
The circuit shown in figure converts
A 4 bit ripple counter and a bit synchronous counter are made using
flip flops having a propagation delay of 10 ns each. If the worst case
delay in the ripple counter and the synchronous counter be R and
S respectively, then
(A) R = 10 ns, S = 40 ns
(B) R = 40 ns, S = 10 ns
(C) R = 10 ns S = 30 ns
(D) R = 30 ns, S = 10 ns
In the circuit shown in the figure, A is parallel-in, parallel-out 4 bit
register, which loads at the rising edge of the clock C . The input
lines are connected to a 4 bit bus, W . Its output acts at input to a
16 # 4 ROM whose output is floating when the input to a partial
table of the contents of the ROM is as follows
Data
0011
1111
0100
1010
1011
1000
0010
1000
Address
0
2
4
6
8
10
11
14
The clock to the register is shown, and the data on the W bus at
time t1 is 0110. The data on the bus at time t2 is
(A) BCD to binary code
(C) Excess -3 to gray code
(A) 1111
(C) 1000
(B) Binary to excess - 3 code
(D) Gray to Binary code
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5.93
5.94
5.91
CMOS
In an 8085 microprocessor, the instruction CMP B has been executed
while the content of the accumulator is less than that of register B
. As a result
(A) Carry flag will be set but Zero flag will be reset
(B) Carry flag will be rest but Zero flag will be set
(C) Both Carry flag and Zero flag will be rest
(D) Both Carry flag and Zero flag will be set
The circuit shown in the figure is a 4 bit DAC
The input bits 0 and 1 are represented by 0 and 5 V respectively.
The OP AMP is ideal, but all the resistance and the 5 v inputs
have a tolerance of ! 10%. The specification (rounded to nearest
multiple of 5%) for the tolerance of the DAC is
(A) ! 35%
(B) ! 20%
(C) ! 10%
(D) ! 5%
(B) 1011
(D) 0010
The DTL, TTL, ECL and CMOS famil GATE of digital ICs are
compared in the following 4 columns
(P)
(Q)
(R)
(S)
Fanout is minimum
DTL
DTL
TTL
CMOS
Power consumption is
minimum
TTL
CMOS
ECL
DTL
2002
5.95
5.96
ONE MARK
4 - bit 2’s complement representation of a decimal number is 1000.
The number is
(A) +8
(B) 0
(C) -7
(D) -8
If the input to the digital circuit (in the figure) consisting of a
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 112
cascade of 20 XOR - gates is X , then the output Y is equal to
(A) 0
(C) X
5.97
The number of comparators required in a 3-bit comparators type
ADC
(A) 2
(B) 3
(C) 7
(D) 8
2002
5.98
(B) 1
(D) X
TWO MARKS
(A) gray code numbers
(C) excess - 3 code numbers
5.101
The circuit in the figure has two CMOS NOR gates. This circuit
functions as a:
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(B) 2 4 2 1 BCD numbers
(D) none of the above
Consider the following assembly language program
MVI B, 87H
MOV A, B
START :
JMP NEXT
MVI B, 00H
XRA B
OUT PORT1
HLT
NEXT :
XRA B
JP START
OUT PORT2
HTL
The execution of above program in an 8085 microprocessor will
result in
(A) an output of 87H at PORT1
(B) an output of 87H at PORT2
(C) infinite looping of the program execution with accumulator
data remaining at 00H
(D) infinite looping of the program execution with accumulator
data alternating between 00H and 87H
2001
(A) flip-flop
(C) Monostable multivibrator
5.99
(B) Schmitt trigger
(D) astable multivibrator
The gates G1 and G2 in the figure have propagation delays of 10 ns
and 20 ns respectively. If the input V1, makes an output change from
logic 0 to 1 at time t = t0 , then the output waveform V0 is
5.102
5.103
ONE MARKS
The 2’s complement representation of -17 is
(A) 101110
(B) 101111
(C) 111110
(D) 110001
For the ring oscillator shown in the figure, the propagation delay of
each inverter is 100 pico sec. What is the fundamental frequency of
the oscillator output
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5.100
If the input X3, X2, X1, X0 to the ROM in the figure are 8 4 2 1 BCD
numbers, then the outputs Y3, Y2, Y1, Y0 are
(A) 10 MHz
(C) 1 GHz
5.104
(B) 100 MHz
(D) 2 GHz
Ab 8085 microprocessor based system uses a 4K # 8 bit RAM whose
starting address is AA00H. The address of the last byte in this RAM
is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) OFFFH
(C) B9FFH
2001
5.105
Page 113
(B) 1000H
(D) BA00H
TWO MARKS
In the TTL circuit in the figure, S2 and S0 are select lines and X7
and X0 are input lines. S0 and X0 are LSBs. The output Y is
(A) 5 V; 3 V; 7 V
(C) 5 V; 5 V; 5 V
(B) 4 V; 3 V; 4 V
(D) 4 V; 4 V; 4 V
2000
5.109
(A) indeterminate
(C) A 5 B
5.106
(B) A 5 B
(D) C (A 5 B ) + C (A 5 B)
5.110
In the figure, the LED
ONE MARKS
An 8 bit successive approximation analog to digital communication
has full scale reading of 2.55 V and its conversion time for an analog
input of 1 V is 20 ms. The conversion time for a 2 V input will be
(A) 10 ms
(B) 20 ms
(C) 40 ms
(D) 50 ms
The number of comparator in a 4-bit flash ADC is
(A) 4
(B) 5
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(A) emits light when both S1 and S2 are closed
(B) emits light when both S1 and S2 are open
(C) emits light when only of S1 and S2 is closed
(D) does not emit light, irrespective of the switch positions.
5.107
(C) 15
5.111
For the logic circuit shown in the figure, the required input condition
(A, B, C) to make the output (X) = 1 is
The digital block in the figure is realized using two positive edge
triggered D-flip-flop. Assume that for t < t0, Q1 = Q2 = 0 . The circuit
in the digital block is given by
(A) 1,0,1
(C) 1,1,1
5.112
5.113
5.114
In the DRAM cell in the figure, the Vt of the NMOSFET is 1 V. For
the following three combinations of WL and BL voltages.
5.115
(B) 0,0,1
(D) 0,1,1
The number of hardware interrupts (which require an external signal
to interrupt) present in an 8085 microprocessor are
(A) 1
(B) 4
(C) 5
(D) 13
In the microprocessor, the RST6 instruction transfer the program
execution to the following location :
(A)30 H
(B) 24 H
(C) 48 H
(D) 60 H
2000
5.108
(D) 16
TWO MARKS
The contents of register (B) and accumulator (A) of 8085
microprocessor are 49J are 3AH respectively. The contents of A and
status of carry (CY) and sign (S) after execution SUB B instructions
are
(A) A = F1, CY = 1, S = 1
(B) A = 0F, CY = 1, S = 1
(C) A = F0, CY = 0, S = 0
(D) A = 1F, CY = 1, S = 1
For the logic circuit shown in the figure, the simplified Boolean
expression for the output Y is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 114
(A) y = AB
(C) y = A + B
5.120
(A) A + B + C
(C) B
5.116
(B) A
(D) C
For the 4 bit DAC shown in the figure, the output voltage V0 is
5.121
5.122
(A) 10 V
(C) 4 V
(B) 5 V
(D) 8 V
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5.117
A sequential circuit using D flip-flop and logic gates is shown in
the figure, where X and Y are the inputs and Z is the inputs. The
circuit is
5.124
5.125
(A) S - R
(B) S - R
(C) J - K
(D) J - K
5.118
A Darlington emitter follower circuit is sometimes used in the output
stage of a TTL gate in order to
(A) increase its IOL
(B) reduce its IOH
(C) increase its speed of operation
(D) reduce power dissipation
Commercially available ECL gears use two ground lines and one
negative supply in order to
(A) reduce power dissipation
(B) increase fan-out
(C) reduce loading effect
(D) eliminate the effect of power line glitches or the biasing circuit
The resolution of a 4-bit counting ADC is 0.5 volts. For an analog
input of 6.6 volts, the digital output of the ADC will be
(A) 1011
(B) 1101
(C) 1100
(D) 1110
1999
5.123
Flip-Flop with inputs X = R and Y = S
Flip-Flop with inputs X = S and Y = R
Flip-Flop with inputs X = J and Y = K
Flip-Flop with input X = K and Y = J
The minimized form of the logical expression
(ABC + ABC + ABC + ABC ) is
(A) AC + BC + AB
(B) AC + BC + AB
(C) AC + BC + AB
(D) AC + BC + AB
For a binary half-subtractor having two inputs A and B, the correct
set of logical expressions for the outputs D (= A minus B) and
X (= borrow) are
(A) D = AB + AB, X = AB
(B)
D = AB + AB + AB , X = AB
(C) D = AB + AB , X = AB
(D) D = AB + AB , X = AB
The ripple counter shown in the given figure is works as a
(A) mod-3 up counter
(C) mod-3 down counter
5.126
(B) 1.0 kHz
(D) 0.77 kHz
1999
5.119
The logical expression y = A + AB is equivalent to
ONE MARK
TWO MARKS
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In the figure, the J and K inputs of all the four Flip-Flips are made
high. The frequency of the signal at output Y is
(A) 0.833 kHz
(C) 0.91 kHz
(B) y = AB
(D) y = A + B
(B) mod-5 up counter
(D) mod-5 down counter
If CS = A15 A14 A13 is used as the chip select logic of a 4 K RAM in
an 8085 system, then its memory range will be
(A) 3000 H - 3 FFF H
(B) 7000 H - 7 FFF H
(C) 5000 H - 5 FFF H and 6000 H - 6 FFF H
(D) 6000 H - 6 FFF H and 7000 H - 7 FFF H
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
1998
5.127
ONE MARK
Page 115
5.135
The minimum number of 2-input NAND gates required to implement
of Boolean function Z = ABC , assuming that A, B and C are
available, is
(A) two
(B) three
(C) five
(D) six
5.136
5.128
5.129
The noise margin of a TTL gate is about
(A) 0.2 V
(B) 0.4 V
(C) 0.6 V
(D) 0.8 V
In the figure is A = 1 and B = 1, the input B is now replaced by a
sequence 101010....., the output x and y will be
(A)
(B)
(C)
(D)
5.130
5.131
5.133
5.134
An instruction used to set the carry Flag in a computer can be
classified as
(A) data transfer
(B) arithmetic
(C) logical
(D) program control
The figure is shows a mod-K counter, here K is equal to
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fixed at 0 and 1, respectively
x = 1010.....while y = 0101......
x = 1010.....and y = 1010......
fixed at 1 and 0, respectively
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An equivalent 2’s complement representation of the 2’s complement
number 1101 is
(A) 110100
(B) 01101
(C) 110111
(D) 111101
The threshold voltage for each transistor in the figure is 2 V. For
this circuit to work as an inverter, Vi must take the values
(A) - 5 V and 0 V
(C) - 0 V and 3 V
5.132
5.137
For the identity AB + AC + BC = AB + AC , the dual form is
(A) (A + B) (A + C) (B + C) = (A + B) (A + C)
(B) (A + B ) (A + C ) (B + C ) = (A + B ) (A + C )
(C) (A + B) (A + C) (B + C) = (A + B ) (A + C )
(D) AB + AC + BC = AB + AC
(A) 1
(C) 3
5.138
(B) - 5 V and 5 V
(D) 3 V and 5 V
An I/O processor control the flow of information between
(A) cache memory and I/O devices
(B) main memory and I/O devices
(C) two I/O devices
(D) cache and main memories
The current I through resistance r in the circuit shown in the figure
is
(A) - V
12R
(C) V
6R
5.139
(B) 2
(D) 4
(B) V
12R
(D) V
3T
The K -map for a Boolean function is shown in the figure is the
number of essential prime implicates for this function is
Two 2’s complement number having sign bits x and y are added
and the sign bit of the result is z . Then, the occurrence of overflow
is indicated by the Boolean function
(A) xyz
(B) x y z
(C) x yz + xyz
(D) xy + yz + zx
The advantage of using a dual slope ADC in a digital voltmeter is
that
(A) its conversion time is small
(B) its accuracy is high
(C) it gives output in BCD format
(D) it does not require a
(A) 4
(C) 6
(B) 5
(D) 8
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
1997
5.140
5.141
5.142
Page 116
ONE MARK
Each cell of a static Random Access Memory contains
(A) 6 MOS transistors
(B) 4 MOS transistors and 2 capacitors
(C) 2 MOS transistors and 4 capacitors
(D) 1 MOS transistors and 1 capacitors
(C) 30 H to 33 H
5.147
A 2 bit binary multiplier can be implemented using
(A) 2 inputs ANSs only
(B) 2 input XORs and 4 input AND gates only
(C) Two 2 inputs NORs and one XNO gate
(D) XOR gates and shift registers
In standard TTL, the ‘totem pole’ stage refers to
(A) the multi-emitter input stage
(B) the phase splitter
(C) the output buffer
(D) open collector output stage
5.144
The inverter 74 ALSO4 has the following specifications
IOH max =- 0.4 A, IOL max = 8 mA, IIH max = 20 mA, IIL max =- 0.1 mA
The fan out based on the above will be
(A) 10
(B) 20
(C) 60
(D) 100
5.146
ADDRESS (HEX)
INSTRUCTION
6010
LXI H, 8 A 79 H
6013
MOV A, L
6015
ADDH
6016
DAA
6017
MOV H, A
6018
PCHL
5.148
A signed integer has been stored in a byte using the 2’s complement
format. We wish to store the same integer in a 16 bit word. We
should
(A) copy the original byte to the less significant byte of the word
and fill the more significant with zeros
(B) copy the original byte to the more significant byte of the word
and fill the less significant byte with zeros
(C) copy the original byte to the less significant byte of the word
and make each fit of the more significant byte equal to the
most significant bit of the original byte
(D) copy the original byte to the less significant byte as well as the
more significant byte of the word
1997
5.149
TWO MARKS
For the NMOS logic gate shown in the figure is the logic function
implemented is
The output of the logic gate in the figure is
(A) 0
(C) A
5.145
The following instructions have been executed by an 8085 mP
From which address will the next instruction be fetched ?
(A) 6019
(B) 6379
(C) 6979
(D) None of the above
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5.143
(D) 70 H to 73 H
(B) 1
(D) F
In an 8085 mP system, the RST instruction will cause an interrupt
(A) only if an interrupt service routine is not being executed
(B) only if a bit in the interrupt mask is made 0
(C) only if interrupts have been enabled by an EI instruction
(D) None of the above
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The decoding circuit shown in the figure is has been used to generate
the active low chip select signal for a microprocessor peripheral.
(The address lines are designated as AO to A7 for I/O address)
(A) ABCDE
(C) A : (B + C) + D : E
5.150
The peripheral will correspond to I/O address in the range
(A) 60 H to 63 H
(B) A4 to A 7H
(B) (AB + C ) : (D + E )
(D) (A + B ) : C + D : E
In a J–K flip-flop we have J = Q and K = 1. Assuming the flip flop
was initially cleared and then clocked for 6 pulses, the sequence at
the Q output will be
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) 010000
(C) 010010
5.151
5.152
5.154
5.155
5.156
5.158
5.159
5.160
(D) 13
The following sequence of instructions are executed by an 8085
microprocessor:
1000
LXI SP, 27 FF
1003
CALL 1006
1006
POP H
The contents of the stack pointer (SP) and the HL, register pair
on completion of execution of these instruction are
(A) SP = 27 FF, HL = 1003
(B) SP = 27 FD, HL = 1003
(C) SP = 27 FF, HL = 1006
(D) SP = 27 FD, HL = 1006
The boolean function A + BC is a reduced form of
(A) AB + BC
(B) (A + B) : (A + C)
(C) AB + ABC
(D) (A + C) : B
ONE MARK
Schottky clamping is resorted in TTl gates
(A) to reduce propagation delay (B) to increase noise margins
(C) to increase packing density (D) to increase fan-out
A pulse train can be delayed by a finite number of clock periods
using
(A) a serial-in serial-out shift register
(B) a serial-in parallel-out shift register
(C) a parallel-in serial-out shift register
(D) a parallel-in parallel-out shift register
A 12-bit ADC is operating with a 1 m sec clock period and the total
conversion time is seen to be 14 m sec . The ADC must be of the
(A) flash type
(B) counting type
(C) intergrating type
(D) successive approximation
type
The total number of memory accesses involved (inclusive of the opcode fetch) when an 8085 processor executes the instruction LDA
2003 is
(A) 1
(B) 2
(C) 3
(D) 4
1996
5.157
(C) 8
The gate delay of an NMOS inverter is dominated by charge time
rather than discharge time because
(A) the driver transistor has larger threshold voltage than the load
transistor
(B) the driver transistor has larger leakage currents compared to
the load transistor
(C) the load transistor has a smaller W/L ratio compared to the
driver transistor
(D) none of the above
1996
5.153
(B) 011001
(D) 010101
Page 117
TWO MARKS
A dynamic RAM cell which hold 5 V has to be refreshed every 20 m
sec, so that the stored voltage does not fall by more than 0.5 V
. If the cell has a constant discharge current of 1 pA, the storage
capacitance of the cell is
(A) 4 # 10-6 F
(B) 4 # 10-9 F
(C) 4 # 10-12 F
(D) 4 # 10-15 F
A 10-bit ADC with a full scale output voltage of 10.24 V is designed
to have a ! LSB/2 accuracy. If the ADC is calibrated at 25c C
and the operating temperature ranges from 0c C to 25c C , then the
maximum net temperature coefficient of the ADC should not exceed
(A) ! 200 mV/cC
(B) ! 400 mV/cC
(C) ! 600 mV/cC
(D) ! 800 mV/cC
A memory system of size 26 K bytes is required to be designed using
memory chips which have 12 address lines and 4 data lines each.
The number of such chips required to design the memory system is
(A) 2
(B) 4
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 118
SOLUTIONS
5.1
For S1 S 0 = 0 0
We have
A13 = A12 = 0
and for I/p = 1we obtain
A10 = 1 or A10 = 0
A11 = 1
A14 = 1 or A14 = 0
A15 = 1 or A15 = 0
Since, A 0 - A 9 can have any value 0 or 1
Therefore, we have the address range as
Option (C) is correct.
Let A denotes the position of switch at ground floor and B denotes
the position of switch at upper floor. The switch can be either in up
position or down position. Following are the truth table given for
different combinations of A and B
A15 A14 A13 A12 A11 A10 A 9
A6
A5
A4
A3
A2
A1
A0
B
Y(Bulb)
From
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
up(1)
up(1)
OFF(0)
to
0
0
0
0
1
0
1
1
1
1
1
1
1
1
1
1
Down(0)
Down(0)
OFF(0)
up(1)
Down(0)
ON(1)
In Hexadecimal & 0800 H to 0BFFH
Similarly, for chip 2, we obtain the range as follows
E = 1 for S1 S 0 = 0 1
so,
A13 = 0 and A12 = 1
and also the I/P = 1 for
A10 = 0 , A11 = 1, A14 = 0 , A15 = 0
so, the fixed I/ps are
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Output of ADD B (Stored value at
A)
05
05 + 05
04
05 + 05 + 04
03
05 + 05 + 04 + 03
02
05 + 05 + 04 + 03 + 02
01
05 + 05 + 04 + 03 + 02 + 01
00
System is out of loop
i.e.,
A = 05 + 05 + 04 + 03 + 02 + 01 = 144
At this stage, the 8085 microprocessor exits from the loop and
reads the next instruction. i.e., the accumulator is being added to
03 H. Hence, we obtain
A = A + 03 H = 14 + 03 = 17 H
Option (D) is correct.
For chip-1, we have the following conclusions:
it is enable when (i)
S1 S 0 = 0 0
and (ii)
Input = 1
A13
A12
A11
A10
0
0
0
1
1
0
A8
A7
A6
A5
A4
A3
A2
A1
A0
From
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
to
0
0
0
1
1
0
1
1
1
1
1
1
1
1
1
1
In hexadecimal it is from 1800 H to 1BFFH . There is no need to
obtain rest of address ranged as only (D) is matching to two results.
i.e., the XOR gate
Content in B
A14
A15 A14 A13 A12 A11 A10 A 9
Y = A5B
Option (A) is correct.
The program is being executed as follows
MVI A, 0.5H;
A = 05H
MVI B, 0.5H;
B = 05H
At the next instruction, a loop is being introduced in which for the
instruction “DCR B” if the result is zero then it exits from loop so,
the loop is executed five times as follows :
A15
Therefore, the address range is
position leads to the ON state of bulb. Hence, from the truth table,
we get
5.3
A7
A
Down(0)
up(1)
ON(1)
When the switches A and B are both up or both down, output
will be zero (i.e. Bulb will be OFF). Any of the switch changes its
5.2
A8
5.4
Option (A) is correct.
The given circuit is
Condition for the race-around
It occurs when the output of the circuit (Y1, Y2) oscillates between
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‘0’ and ‘1’ checking it from the options.
1. Option (A): When CLK = 0
Output of the NAND gate will be A1 = B1 = 0 = 1. Due to these
input to the next NAND gate, Y2 = Y1 : 1 = Y1 and Y1 = Y2 : 1 = Y2 .
If Y1 = 0 ,
Y2 = Y1 = 1 and it will remain the same and doesn’t
oscillate.
If Y2 = 0 ,
Y1 = Y2 = 1 and it will also remain the same for the
clock period. So, it won’t oscillate for CLK = 0 .
So, here race around doesn’t occur for the condition CLK = 0 .
2. Option (C): When CLK = 1, A = B = 1
A1 = B1 = 0 and so Y1 = Y2 = 1
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 119
And it will remain same for the clock period. So race around
doesn’t occur for the condition.
3. Option (D): When CLK = 1, A = B = 0
So,
A1 = B1 = 1
And again as described for Option (B) race around doesn’t occur
for the condition.
5.5
mp COX W
2
2V V - VTp h - V SD
B
2 b L l8 SD ^ SG
m C
2
0 = p OX bW l [2VSD ^VSG - VTp h - V SD
]
2
L
I1 =
Since,
&
Solving it we get,
2 ^VSG - VTp h = VSD
2 ^5 - Vin - 1h = 5 - VD
Vin = VD + 3
2
&
Option ( ) is correct.
&
I1 = 0 , VD = 5 V
Vin = 5 + 3 = 4 V
2
For
So,
So for the NMOS
Y = 1, when A > B
VGS = Vin - 0 = 4 - 0 = 4 V and VGS > VTn
So it can’t be in cutoff region.
Case 2 : M2 must be in saturation region.
So,
I1 = I 2
A = a1 a 0, B = b1 b 0
a1
a0
b1
b0
Y
0
1
0
0
1
1
0
0
0
1
1
0
0
1
1
1
1
0
0
1
1
1
0
1
1
1
1
1
0
1
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Total combination = 6
5.6
5.7
Option (A) is correct.
Parallel connection of MOS & OR operation
Series connection of MOS & AND operation
The pull-up network acts as an inverter. From pull down network
we write
Y = (A + B) C = (A + B) + C = A B + C
mp COX W
mn COX W
2
2
2 (VSG - VTp) VSD - V SD
@ = 2 L (VGS - VTn)
2 L6
&
Option (A) is correct.
Prime implicants are the terms that we get by solving K-map
2 (5 - Vin - 1) (5 - VD) - (5 - VD) 2 = (Vin - 0 - 1) 2
&
2 (4 - Vin) (5 - VD) - (5 - VD) 2 = (Vin - 1) 2
Substituting VD = VDS = VGS - VTn and for N -MOS & VD = Vin - 1
&
&
&
&
Option (A) is correct.
Given the circuit as below :
2 (4 - Vin) (6 - Vin) - (6 - Vin) 2 = (Vin - 1) 2
48 - 36 - 8Vin =- 2Vin + 1
6Vin = 11
&
Vin = 11 = 1.833 V
6
So for M2 to be in saturation Vin < 1.833 V or Vin < 1.875 V
F = XY + XY
1prime
44 2
44 3
implicants
5.8
2
2 (VSG - VTp) VSD - V SD
= (VGS - VTn) 2
5.9
Option (D) is correct.
Let Qn + 1 is next state and Qn is the present state. From the given
below figure.
If A = 0,
If A = 1,
So state diagram is
Since all the parameters of PMOS and NMOS are equal.
So,
mn = mp
COX bW l = COX bW l = COX bW l
L M2
L
L M1
Given that M1 is in linear region. So, we assume that M2 is either in
cutoff or saturation.
Case 1 : M2 is in cut off
So,
I 2 = I1 = 0
Where I1 is drain current in M1 and I2 is drain current in M2 .
5.10
D = Y = AX 0 + AX1
Qn + 1 = D = AX 0 + AX1
Qn + 1 = A Qn + AQn X 0 = Q , X1 = Q
Qn + 1 = Qn (toggle of previous state)
Qn + 1 = Qn
Option (B) is correct.
The given circuit is shown below:
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 120
0
1
1
0
1
QB ^t + 1h
0
0
0
(PQ QR ) PR = (PQ + QR PR )
= PQ + QR + PR
= PQ + QR + PR
If any two or more inputs are ‘1’ then output y will be 1.
5.11
QB ^t + 1h = Q A
Option (A) is correct.
For the output to be high, both inputs to AND gate should be high.
The D-Flip Flop output is the same, after a delay.
Let initial input be 0;
(Consider Option A)
st
then Q = 1 (For 1 D-Flip Flop). This is given as input to 2nd FF.
Let the second input be 1. Now, considering after 1 time interval;
The output of 1st Flip Flop is 1 and 2nd FF is also 1. Thus Output
DA = Q A Q B + QA QB
5.15
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RRC
Thus A = 23 H
= 1.
5.12
5.16
Option (D) is correct.
Option (C) is correct.
Initially Carry Flag, C = 0
MVI A, 07 H
; A = 0000 0111
RLC
; Rotate left without carry. A = 0000 1110
MVO B, A
; B = A = 0000 1110
RLC
; A = 0001 1100
RLC
; A = 0011 1000
ADD B
; A = 0011 1000
+ 0000 1110
;
0100 0110
;
; Rotate Right with out carry, A = 0010 0011
Option ( ) is correct.
F = S 1 S 0 I 0 + S 1 S 0 I 1 + S 1 S 0 I 2 + S1 S 0 I 3
I0 = I3 = 0
( S1 = P, S 0 = Q )
F = PQ + PQ = XOR (P, Q)
5.13
Option (A) is correct.
All the states of the counter are initially unset.
5.17
State Initially are shown below in table :
5.14
Q2
Q1
Q0
0
0
0
0
1
0
0
4
1
1
0
6
1
1
1
7
0
1
1
3
0
0
1
1
0
0
0
0
Option (B) is correct.
Since G2 is active low input, output of NAND gate must be 0
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G2 = A15 : A14 A13 A12 A11 = 0
So,
A15 A14 A13 A12 A11 = 00101
To select Y5 Decoder input
ABC = A 8 A9 A10 = 101
Address range
A15 A14 A13 A12 A11 A10 A 9 A 8 ...............A 0
0011101........A 0
S
S
2
D
^2D00 - 2DFF h
Option (D) is correct.
The sequence is QB QA
00 " 11 " 01 " 10 " 00 " ...
QB
QA
QB (t + 1)
QA (t + 1)
0
0
1
1
1
1
0
1
5.18
Option (A) (B) (C) are correct.
In the circuit
F = (A 5 B) 9 (A 9 B) 9 C
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
For two variables
A5B = A9B
So,
(A 5 B) 9 (A 9 B) = 0 (always)
F = 09C = 0$C+1$C = C
So, F = 1 when C = 1 or C = 0
5.19
Option (D) is correct.
Let QA (n), QB (n), QC (n) are present states and QA (n + 1), QB (n + 1),
QC (n + 1) are next states of flop-flops.
In the circuit
QA (n + 1) = QB (n) 9 QC (n)
QB (n + 1) QA (n)
QC (n + 1) QB (n)
Initially all flip-flops are reset
1st clock pulse
Page 121
A = A 5 B = 00100010 5 01000101 = 01100111 = 674
5.22
5.23
5.24
QA = 0 9 0 = 1
QB = 0
QC = 0
5.25
2 nd clock pulse
Option (C) is correct.
TTL " Transistor - Transistor logic
CMOS " Complementary Metal Oxide Semi-conductor
Option (D) is correct.
Vectored interrupts : Vectored interrupts are those interrupts in
which program control transferred to a fixed memory location.
Maskable interrupts : Maskable interrupts are those interrupts which
can be rejected or delayed by microprocessor if it is performing some
critical task.
Option (D) is correct.
We have 6X + Z {Y + (Z + XY )}@[X + Z (X + Y)] = 1
Substituting X = 1 and X = 0 we get
[1 + Z {Y + (Z + 1Y )}][ 0 + Z (1 + Y)] = 1
or
[1][ Z (1)] = 1
1 + A = 1 and 0 + A = A
or
Z =1)Z=0
Option (A) is correct.
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QA = 0 9 0 = 1
QB = 1
QC = 0
3 rd clock pulse
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QA = 1 9 0 = 0
QB = 1
QC = 1
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4 th clock pulse
The AND gate implementation by 2:1 mux is as follows
QA = 1 9 1 = 1
QB = 0
QC = 1
So, sequence
5.20
QA = 01101.......
Y = AI 0 + AI1 = AB
Option (D) is correct.
Output of the MUX can be written as
The EX - OR gate implementation by 2:1 mux is as follows
F = I 0 S 0 S1 + I1 S 0 S1 + I 2 S 0 S 1 + I 3 S 0 S 1
Here, I 0 = C, I1 = D, I2 = C , I 3 = CD
and S 0 = A, S1 = B
So,
F = C A B + D A B + C A B + C DA B
Writing all SOP terms
F = A B C D + A B C D + A BCD + A B C D
1 44 2
44 3 1 44 2
44 3 S
1 44 2
4
43
m
m
m
m
3
7
2
5
+A B C D + A B C D + ABC D
1 44 2
4
4 3 1 44 2
44 3 S
m
m
m
9
F = / m (2, 3, 5, 7, 8, 9, 12)
5.21
8
12
Y = BI0 + BI1 = AB + BA
5.26
Option (A) is correct.
The given circuit is as follows.
Option (C) is correct.
By executing instruction one by one
MVI A, 45 H & MOV 45 H into accumulator, A = 45 H
STC & Set carry, C = 1
CMC & Complement carry flag, C = 0
RAR & Rotate accumulator right through carry
The truth table is as shown below. Sequence is 00, 11, 10, 00 ...
A = 00100010
XRA B & XOR A and B
CLK
J1
K1
Q1
J2
K2
Q2
1
1
1
0
1
1
0
2
1
1
1
1
1
1
3
0
0
1
0
1
0
4
1
1
0
1
1
0
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
5.27
Page 122
From the figure shown below it may be easily seen upper MOSFET
are shorted and connected to Vdd thus OUT is 1 only when the node
S is 0,
Option (B) is correct.
The given situation is as follows
The truth table is as shown below
P1
P2
a
b
c
d
e
f
g
0
0
1
1
1
1
1
1
0
0
1
1
0
1
1
0
1
1
1
0
1
1
0
1
1
0
1
1
1
1
0
0
1
1
1
1
Since the lower MOSFETs are shorted to ground, node S is 0 only
when input P and Q are 1. This is the function of AND gate.
5.31
From truth table we can write
a =1
b = P 1 P 2 + P1 P 2 = P 2
and
P - Q = (- 19) - (- 26) = 7
Thus 7 signed two’s complements form is
1 NOT Gate
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c = P1 P2 + P1 P2 = P1
d = 1 = c+e
1 NOT Gate
c = P1 P2 = P1 + P2
1 OR GATE
1 OR GATE
f = P1 P2 = P1 + P2
1 OR GATE
g = P1 P2 = P1 + P2
Thus we have g = P1 + P2 and d = 1 = c + e . It may be observed
easily from figure that
Led g does not glow only when both P1 and P2 are 0. Thus
Option (B) is correct.
MSB of both number are 1, thus both are negative number. Now
we get
11101101 = (- 19) 10
and
11100110 = (- 26) 10
(7) 10 = 00000111
5.32
Option (D) is correct.
The circuit is as shown below
So
and
5.33
X = PQ
Y = (P + Q)
Z = PQ (P + Q)
= (P + Q )( P + Q) = PQ + PQ = P 5 Q
M1 = Z 5 R = (P 5 Q) 5 R
Option (A) is correct.
The circuit is as shown below
g = P1 + P2
LED d is 1 all condition and also it depends on
d = c+e
5.28
5.29
Option (D) is correct.
As shown in previous solution 2 NOT gates and 3-OR gates are
required.
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Option (C) is correct.
For the NAND latche the stable states are as follows
The truth table is shown below. When CLK make transition Q goes
to 1 and when D goes to 1, Q goes to 0
5.34
For the NOR latche the stable states are as follows
5.30
Option (D) is correct.
Option (B) is correct.
Since the input to both JK flip-flop is 11, the output will change
every time with clock pulse. The input to clock is
The output Q0 of first FF occurs after time 3 T and it is as shown
below
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 123
Thus after execution above instruction contests of PC and HL are
same and that is 6140H
5.39
The output Q1 of second FF occurs after time 3 T when it gets
input (i.e. after 3 T from t1) and it is as shown below
5.40
5.35
Option (D) is correct.
VDAC =
We have
/ 2n - 1bn = 2- 1b0 + 20 b1 + 21b2 + 22 b3
or
VDAC = 0.5b0 + b1 + 2b2 + 4b3
The counter outputs will increase by 1 from 0000 till Vth > VDAC .
Option (B) is correct.
Y = AB + CD = AB .CD
This is SOP form and we require only 3 NAND gate
3
n=0
Option (C) is correct.
MSB of Y is 1, thus it is negative number and X is positive number
Now we have
X = 01110 = (14) 10
and
Y = 11001 = (- 7) 10
X + Y = (14) + (- 7) = 7
In signed two’s complements from 7 is (7) 10 = 000111
5.41
Option (A) is correct.
The circuit is as shown below
The output of counter and VDAC is as shown below
Clock
b3 b3 b2 b0
VDAC
1
0001
0
2
0010
0.5
3
0011
1
4
0100
1.5
5
0101
2
6
0110
2.5
7
0111
3
8
1000
3.5
9
1001
4
10
1010
4.5
11
1011
5
12
1100
5.5
13
1101
6
14
1110
6.5
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Y = AB + AB
and when VADC = 6.5 V (at 1101), the output of AND is zero and
the counter stops. The stable output of LED display is 13.
5.36
5.37
Option (B) is correct.
The VADC - Vin at steady state is
= 6.5 - 6.2 = 0.3V
and
5.42
5.43
5.44
Z = PQ + PQS + QRS
5.38
Option (C) is correct.
2710H LXI H, 30A0H
2713H DAD H
2714H PCHL
; Load 16 bit data 30A0 in HL pair
; 6140H " HL
; Copy the contents 6140H of HL in PC
Option (D) is correct.
Y = ABCD + ABCD + ABC D + ABC D
= ABCD + ABC D + ABC D + ABC D
= ABCD + ABC D + BC D (A + A)
= ABCD + ABC D + BC D
Option (A) is correct.
Z = I0 RS + I1 RS + I2 RS + I3 RS
= (P + Q ) RS + PRS + PQRS + PRS
= PRS + QRS + PRS + PQRS + PRS
The k - Map is as shown below
X = YC + YC = (AB + AB ) C + (AB + AB ) C
= (AB + AB) C + (AB + AB ) C
= ABC + ABC + ABC + ABC
5.45
A+A = 1
Option (B) is correct.
In given TTL NOT gate when Vi = 2.5 (HIGH), then
Q1 " Reverse active
Q2 " Saturation
Q3 " Saturation
Q4 " cut - off region
Option (C) is correct.
For X = 0, Y = 1
For X = 0, Y = 0
For X = 1, Y = 1
P = 1, Q = 0
P = 1, Q = 1
P = 1, Q = 0 or P = 0, Q = 1
Option (C) is correct.
Chip 8255 will be selected if bits A3 to A7 are 1. Bit A0 to A2 can
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
be 0 or.
1. Thus address range is
11111000
11111111
5.46
Page 124
; Since 8 AH < 9BH, CY = 1
; Store the contents of A to location 3010
7 : STA 3010 H
H
F8H
FFH
Option (B) is correct.
Since the inverting terminal is at virtual ground the resistor network
can be reduced as follows
8 : HLT
; Stop
Thus the contents of accumulator after execution of ADD instruction
is EAH.
5.49
5.50
Option (C) is correct.
The CY = 1 and Z = 0
Option (A) is correct.
For this circuit the counter state (Q1, Q0) follows the sequence 00, 01,
10, 00 ... as shown below
Clock
D1 D0
Q1 Q0
Q1 NOR Q0
00
1
1st
01
10
0
2nd
10
01
0
3rd
00
00
0
The current from voltage source is
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5.51
I = VR = 10 = 1 mA
R
10k
This current will be divide as shown below
5.52
5.53
-3
i = I = 1 # 10 = 62.5 m A
16
16
Now
5.47
5.48
Option (C) is correct.
The net current in inverting terminal of OP - amp is
I- = 1 + 1 = 5I
4 16
16
So that
V0 =- R # 5I =- 3.125
16
Option (B) is correct.
Line
1 : MVI A, B5H
2 : MVI B, 0EH
3 : XRI 69H
4 : ADDB
5 : ANI 9BH
6 : CPI 9FH
; Move B5H to A
; Move 0EH to B
; [A] XOR 69H and store in A
; Contents of A is CDH
; Add the contents of A to contents of B and
; store in A, contents of A is EAH
; [a] AND 9BH, and store in A,
; Contents of A is 8 AH
; Compare 9FH with the contents of A
5.54
Option (A) is correct.
As shown below there are 2 terms in the minimized sum of product
expression.
1
0
0
1
0
d
0
0
0
0
d
1
1
0
0
1
Option (B) is correct.
The output is taken from the 5th line.
Option (D) is correct.
After applying two clock poles, the outputs of the full adder is S = 1
, C0 = 1
A
B
Ci
S
Co
1st
1
0
0
0
1
2nd 1
1
1
1
1
Option (D) is correct.
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100010011001
S
S
S
S
4
2
3
1
5.55
Option (B) is correct.
In this the diode D2 is connected to the ground. The following table
shows the state of counter and D/A converter
Q2 Q1 Q0
D3 = Q2
D2 = 0
D1 = Q1
D0 = Q0
Vo
000
0
0
0
0
0
001
0
0
0
1
1
010
0
0
1
0
2
011
0
0
1
1
3
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
100
1
0
0
0
8
101
1
0
0
1
9
110
1
0
1
0
10
111
1
0
1
1
11
000
0
0
0
0
0
001
0
0
0
1
1
Page 125
The circuit is as shown below
Thus option (B) is correct
5.56
5.57
5.58
Option (B) is correct.
LXI, EFFF H
; Load SP with data EFFH
CALL 3000 H
; Jump to location 3000 H
:
:
:
3000H LXI H, 3CF4
; Load HL with data 3CF4H
PUSH PSW
; Store contnets of PSW to Stack
POP PSW
; Restore contents of PSW from stack
PRE
; stop
Before instruction SPHL the contents of SP is 3CF4H.
After execution of POP PSW, SP + 2 " SP
After execution of RET, SP + 2 " SP
Thus the contents of SP will be 3CF4H + 4 = 3CF8H
If output is at logic 0, the we have V0 = 0 which signifies BJT Q3 is
in saturation and applying KVL we have
or
or
5.62
Option (A) is correct.
We have
f = ABC + ABC
= B (AC + AC ) = B (A + C)( A + C )
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Option (A) is correct.
The inputs D0 and D1 respectively should be connected as Q1 and Q0
where Q0 " D1 and Q1 " D0
Option (D) is correct.
If the point P is stuck at 1, then output f is equal to A
VBE3 = IR # 1k
0.75 = IR # 1k
IR = 0.75 mA
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5.63
Option (C) is correct.
Characteristic equation for a jk flip-flop is written as
Where
So,
5.59
Option (B) is correct.
Dividing 43 by 16 we get
2
16 43
32
g
11
11 in decimal is equivalent is B in hexamal.
Thus
4310 * 2B16
Now
410 * 01002
310 * 00112
Thus
4310 * 01000011BCD
5.60
5.64
5.65
5.66
5.67
5.61
Option (C) is correct.
= JQ n + K Qn
is the present output
is next output
= 10 + K : 0
Qn = 0
=1
Option (C) is correct.
Since T2 T1 T0 is at 111, at every clock Q2 Q1 Q0 will be changes. Ir
present state is 011, the next state will be 100.
Option (D) is correct.
Option (C) is correct.
0100H LXI SP, 00FF
0103H LXI H, 0701
0106H MVI A, 20H
0108 H SUB M
; Load SP with 00FFG
; Load HL with 0107H
; Move A with 20 H
; Subtract the contents of memory
; location whose address is stored in HL
; from the A and store in A
0109H ORI 40H
; 40H OR [A] and store in A
010BH ADD M
; Add the contents of memeory location
; whose address is stored in HL to A
; and store in A
HL contains 0107H and contents of 0107H is 20H
Thus after execution of SUB the data of A is 20H - 20H = 00
Option (A) is correct.
The diagram is as shown in fig
f' = BC + BC
f = f'A + f'0 = f'A = ABC + ABC
Qn + 1
Qn
Qn + 1
Qn + 1
Qn + 1
Option (C) is correct.
Before ORI instruction the contents of A is 00H. On execution the
ORI 40H the contents of A will be 40H
00H = 00000000
40H = 01000000
ORI 01000000
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
After ADD instruction the contents of memory location whose address is stored in HL will be added to and will be stored in A
40H + 20 H = 60 H
5.68
Page 126
multiplexers is required.
5.76
AC + BC = AC1 + BC 1
= AC (B + B ) + BC (A + A)
= ACB + ACB + BC A + BC A
Option (C) is correct.
A master slave D-flip flop is shown in the figure.
5.77
In the circuit we can see that output of flip-flop call be triggered
only by transition of clock from 1 to 0 or when state of slave latch
is affected.
5.69
5.78
Option (A) is correct.
The range of signed decimal numbers that can be represented by
n - bits 1’s complement number is - (2n - 1 - 1) to + (2n - 1 - 1).
Thus for n = 6 we have
Range =- (26 - 1 - 1) to + (26 - 1 - 1)
=- 31 to + 31
5.71
5.72
5.73
5.74
5.75
5.79
2n $ 100
n $7
Option (B) is correct.
Shift Register " Serial to parallel data conversion
Counter " Frequency division
Decoder " Addressing in memory chips.
Option (A) is correct.
For the TTL family if terminal is floating, then it is at logic 1.
Thus
Y = (AB + 1) = AB .0 = 0
5.80
Option (C) is correct.
11001
1001
111001
00110
0110
000110
+1
+1
+1
00111
0111
000111
7
7
7
Thus 2’s complement of 11001, 1001 and 111001 is 7. So the number
given in the question are 2’s complement correspond to -7.
Option (C) is correct.
In the modulo - 6 ripple counter at the end of sixth pulse (i.e. after
101 or at 110) all states must be cleared. Thus when CB is 11 the
all states must be cleared. The input to 2-input gate is C and B
and the desired output should be low since the CLEAR is active low
Thus when C and B are 0, 0, then output must be 0. In all other
case the output must be 1. OR gate can implement this functions.
Option (C) is correct.
Number of MUX is 4 = 2 and 2 = 1. Thus the total number 3
2
3
Option (D) is correct.
For 8255, various modes are described as following.
Mode 1 : Input or output with hand shake
In this mode following actions are executed
1. Two port (A & B) function as 8 - bit input output ports.
Each port uses three lines from C as a hand shake signal
Input & output data are latched.
Form (ii) the mode is 1.
Mode 2 : Bi-directional data transfer
This mode is used to transfer data between two computer. In this
mode port A can be configured as bidirectional port. Port A uses
five signal from port C as hand shake signal.
For (1), mode is 2
Option (D) is correct.
The minimum number of bit require to encode 100 increment is
or
Option (D) is correct.
We have
f (x, y) = xy + xy + xy = x (y + y) + xy
= x + xy
or
f (x, y) = x + y
Here compliments are not available, so to get x we use NOR gate.
Thus desired circuit require 1 unit OR and 1 unit NOR gate giving
total cost 2 unit.
2.
3.
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5.70
Option (D) is correct.
Option (B) is correct.
LDA 16 bit & Load accumulator directly this instruction copies
data byte from memory location (specified within the instruction)
the accumulator.
It takes 4 memory cycle-as following.
1. in instruction fetch
2. in reading 16 bit address
1. in copying data from memory to accumulator
LXI D, (F0F1) 4 & It copies 16 bit data into register pair D and E.
It takes 3 memory cycles.
Option (A) is correct.
LXI H, 9258H
MOV A, M
CMa
; 9258H " HL
; (9258H) " A
; A"A
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MOV M, A
; A"M
This program complement the data of memory location 9258H.
5.81
Option (D) is correct.
MVI A, 00H
; Clear accumulator
LOOP ADD B
; Add the contents of B to A
DCR C
; Decrement C
JNZ LOOP
; If C is not zero jump to loop
HLT
END
This instruction set add the contents of B to accumulator to contents
of C times.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
5.82
16
2 = 2 = 65536
5.85
5.86
5.87
Thus W = Z and X = Z
5.89
Option (C) is correct.
In the flash analog to digital converter, the no. of comparators is
equal to 2n - 1, where n is no. of bit.s
So,
5.84
= R + S + PQ + PR + PQS + QRS
Option (D) is correct.
The number of distinct boolean expression of n variable is 22n . Thus
24
5.83
Page 127
2n - 1 = 28 - 1 = 255
Option (B) is correct.
When output of the 74 series gate of TTL gates is taken from
BJT then the configuration is either totem pole or open collector
configuration .
Option (D) is correct.
A 2n: 1 MUX can implement all logic functions of (n + 1) variable
without andy additional circuitry. Here n = 3 . Thus a 8 : 1 MUX
can implement all logic functions of 4 variable.
Option (D) is correct.
Counter must be reset when it count 111. This can be implemented
by following circuitry
5.90
Option (C) is correct.
After t = t1, at first rising edge of clock, the output of shift register
is 0110, which in input to address line of ROM. At 0110 is applied
to register. So at this time data stroed in ROM at 1010 (10), 1000
will be on bus.
When W has the data 0110 and it is 6 in decimal, and it’s data
value at that add is 1010
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Option (B) is correct.
We have
Y = P5Q5R
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Z = RQ + PR + QP
Here every block is a full subtractor giving P - Q - R where R is
borrow. Thus circuit acts as a 4 bit subtractor giving P - Q .
5.88
Option (B) is correct.
Propagation delay of flip flop is
tpd = 10 nsec
Propagation delay of 4 bit ripple counter
R = 4tpd = 40 ns
and in synchronous counter all flip-flop are given clock simultaneously, so
S = tpd = 10 ns
then 1010 i.e. 10 is acting as odd, at time t2 and data at that
movement is 1000.
Option (A) is correct.
5.91
W = R + PQ + RS
X = PQRS + PQRS + PQRS
Y = RS + PR + PQ + PQ
= RS + PR $ PQ $ PQ
= RS + (P + R )( P + Q)( P + Q)
= RS + (P + PQ + PR + QR )( P + Q)
= RS + PQ + QR (P + P ) + QR
= RS + PQ + QR
5.92
5.93
Z = R + S + PQ + PQR + PQS
= R + S + PQ $ PQR $ PQS
5.94
Option (B) is correct.
The DTL has minimum fan out and CMOS has minimum power
consumption. Propagation delay is minimum in ECL.
Option (D) is correct.
Let input be 1010;
output will be 1101
Let input be 0110;
output will be 0100
Thus it convert gray to Binary code.
Option (A) is correct.
CMP B
& Compare the accumulator content with context of
Register B
If A < R
CY is set and zero flag will be reset.
Option (A) is correct.
Vo =- V1 :R bo + R b1 + R b2 + R b 3D
4R
R
2R
4R
= R + S + (P + Q )( P + Q + R)( P + Q + S)
= R + S + PQ + PQ + PQS + PR + PQR
+ PRS + PQ + PQS + PQR + QRS
Exact value when V1 = 5 , for maximum output
VoExact =- 5 :1 + 1 + 1 + 1 D =- 9.375
2 4 8
= R + S + PQ + PQS + PR + PQR + PRS
Maximum Vout due to tolerance
Vo max =- 5.5 :110 + 110 + 110 + 110 D
90
2 # 90 4 # 90 8 # 90
+ PQS + PQR + QRS
Tolerance
= R + S + PQ (1 + S) + PR (1 + P ) + PRS
+ PQS + PQR + QRS
= R + S + PQ + PR + PRS + PQS
+ PQR + QRS
5.95
5.96
= R + S + PQ + PR (1 + Q ) + PQS + QRS
=- 12.604
= 34.44% = 35%
Option (D) is correct.
If the 4- bit 2’s complement representation of a decimal number is
1000, then the number is -8
Option (B) is correct.
Output of 1 st XOR = = X $ 1 + X $ 1 = X
Output of 2 nd XOR = X X + XX = 1
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 128
Its 1’s complement is 101110
So 2’s compliment is
So after 4,6,8,...20 XOR output will be 1.
5.97
5.98
Option (C) is correct.
In the comparator type ADC, the no. of comparators is equal to 2n - 1
, where n is no. of bit.s
So,
23 - 1 = 7
Option (C) is correct.
The circuit is as shown below
+
101111
5.103
5.104
The circuit shown is monostable multivibrator as it requires an
external triggering and it has one stable and one quasistable state.
5.99
Option (B) is correct.
They have prorogation delay as respectively,
G1 " 10 nsec
G2 " 20 nsec
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For abrupt change in Vi from 0 to 1 at time t = t0 we have to assume the output of NOR then we can say that option (B) is correct waveform.
5.100
5.101
Option (B) is correct.
Let X3 X2 X1 X0 be 1001 then Y3 Y2 Y1 Y0 will be 1111.
Let X3 X2 X1 X0 be 1000 then Y3 Y2 Y1 Y0 will be 1110
Let X3 X2 X1 X0 be 0110 then Y3 Y2 Y1 Y0 will be 1100
So this converts 2-4-2-1 BCD numbers.
Option (B) is correct.
MVI B, 87H
MOV A, B
START :
JMP NEXT
XRA B
JP START
NEXT :
JMP NEXT
XRA
JP
START
OUT PORT2
5.102
; B = 87
; A = B = 87
; Jump to next
; A 5 B " A,
; A = 00, B = 87
; Since A = 00 is positive
; so jump to START
;Jump to NEXT ; unconditionally
; B ; A 5 B " A, A = 87 ,
; B = 87 H
; will not jump as D7 , of A is 1
; A = 87 " PORT2
Option (B) is correct.
The two’s compliment representation of 17 is
17 = 010001
101110
1
Option (C) is correct.
The propagation delay of each inverter is tpd then The fundamental
frequency of oscillator output is
1
= 1 GHz
f = 1 =
2ntpd
2 # 5 # 100 # 10 - 12
Option (C) is correct.
4K # 8 bit means 102410 location of byte are present
Now
102410 * 1000H
It starting address is AA00H then address of last byte is
AA00H + 1000H - 0001H = B9FFH
5.105
Option (D) is correct.
or
5.106
5.107
Y = I0 + I3 + I5 + I6
= C BA + C AB + CBA + CBA
Y = C (A 5 B ) + C (A 5 B)
Option (D) is correct.
For the LED to glow it must be forward biased. Thus output of
NAND must be LOW for LED to emit light. So both input to
NAND must be HIGH. If any one or both switch are closed, output
of AND will be LOW. If both switch are open, output of XOR will
be LOW. So there can’t be both input HIGH to NAND. So LED
doesn’t emit light.
Option (C) is correct.
The output of options (C) satisfy the given conditions
5.108
Option (B) is correct.
5.109
Option (B) is correct.
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Conversion time of successive approximate analog to digital
converters is independent of input voltage. It depends upon the
number of bits only. Thus it remains unchanged.
5.110
5.111
Option (C) is correct.
In the flash analog to digital converter, the no. of comparators is
equal to 2n - 1, where n is no. of bits.
So,
2 4 - 1 = 15
Option (D) is correct.
As the output of AND is X = 1, the all input of this AND must be
1. Thus
...(1)
AB + AB = 1
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 129
...(2)
BC + BC = 1
...(3)
C =1
From (2) and (3), if C = 1, then B = 1
If B = 1, then from (1) A = 0 . Thus A = 0, B = 1 and C = 1
5.112
5.113
5.114
Option (C) is correct.
Interrupt is a process of data transfer by which an external device
can inform the processor that it is ready for communication. 8085
microprocessor have five interrupts namely TRAP, INTR, RST 7.5,
RST 6.5 and RST 5.5
Option (A) is correct.
Accumulator contains A = 49 H
Register B = 3 AH
SUB B = A minus B
Carry
so here outputA
Carry CY
Sign flag S
Vo = 8 # 5 = 5V
8
5.117
Option (D) is correct.
The truth table is shown below
Z = XQ + YQ
Comparing from the truth table of J - K FF
Y = J,
X =K
X
Y
Z
0
0
Q
0
1
0
1
0
1
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1
=1
=0F
=1
=1
5.118
Option (C) is correct.
The circuit is as shown below :
5.119
5.120
Y = B + (B + C ) = B (B + C ) = B
Option (B) is correct.
The circuit is as shown below
5.121
5.122
1
Q1
Option (B) is correct.
In the figure the given counter is mod-10 counter, so frequency of
output is 10k = 1k
10
Option (D) is correct.
We have
y = A + AB
we know from Distributive property
Thus
5.116
...(2)
From (1) and (2) we have
Option (A) is correct.
For any RST instruction, location of program transfer is obtained
in following way.
RST x
& (x ) 8) 10 " convert in hexadecimal
So for RST 6
& (6 ) 8) 10 = (48) 10 = (30) H
A = 49 H = 01001001
B = 3 AH = 00111010
2’s complement of (- B) = 11000110
A - B = A + (- B)
010 010 01
& +1 1 0 0 0 1 1 0
0 0 0 0 1111
5.115
Now applying voltage divider rule
V- = 1k V% = 1 Vo
1k + 7k
8
x + yz = (x + y) (x + z)
y = (A + A) (A + B) = A + B
Option (C) is correct.
Darligton emitter follower provides a low output impedance in both
logical state (1 or 0). Due to this low output impedance, any stray
capacitance is rapidly charged and discharged, so the output state
changes quickly. It improves speed of operation.
Option (D) is correct.
Option (B) is correct.
For ADC we can write
Analog input = (decimal eq of digital output) # resol
6.6 = (decimal eq. of digital output) # 0.5
6.6 = decimal eq of digital. output
0.5
13.2 = decimal equivalent of digital output so output of ADC is = 1101.
The voltage at non-inverting terminal is
V+ = 1 + 1 = 5
8 2
8
V- = V+ = 5
8
5.123
...(1)
Option (A) is correct.
We use the K -map as below.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 130
address range
A15 A14 A13 A12
initial
1 1 1
address
final
1 1 1
address
so address range is (7 0
So given expression equal to
= AC + BC + AB
5.124
Option (C) is correct.
For a binary half-subtractor truth table si given below.
5.127
(111)
A11 A10 A 9 A 8 A7 A6 A5 A 4 A 3 A2 A1 A 0
0 0 0 0 0 0 0 0 0 0 0 0 0
&7000H
1 1 1 1 1 1 1 1 1 1 1 11
&7FFFH
0 0 H – 7 F F F H)
Option (C) is correct.
Given boolean function is
Z = ABC
Z = ABC = ACB = AC + B
Now
Thus
Z = AC + B
we have
Z = X + Y (1 NOR gate)
where
X = AC (1 NAND gate)
To implement a NOR gate we required 4 NAND gates as shown
below in figure.
from truth table we can find expressions of D & X
D = A 5 B = AB + AB
X = AB
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5.125
here total no. of NAND gates required
= 4+1 = 5
5.128
Option (D) is correct.
From the given figure we can write the output
Option (B) is correct.
For TTL worst cases low voltages are
VOL (max) = 0.4 V
VIL (max) = 0.8 V
Worst case high voltages are
VOH (min) = 2.4 V
VIH (min) = 2 V
The difference between maximum input low voltage and maximum
output low voltage is called noise margin. It is 0.4 V in case of TTL.
5.129
For the state 010 all preset = 1 and output QA QB QC = 111 so here
total no. of states = 5 (down counter)
5.126
Option (B) is correct.
We have 4 K RAM (12 address lines)
Option (D) is correct.
From the figure we can see
If
A =1
then
y =1
If
A =1
then also
y =1
B=0
x=0
B=1
x=0
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so for sequence B = 101010....output x and y will be fixed at 0 and
1 respectively.
5.130
5.131
5.132
so here chip select logic CS = A15 A14 A13
5.133
Option (D) is correct.
Given 2’s complement no. 1101; the no. is 0011
for 6 digit output we can write the no. is – 000011
2’s complement representation of above no. is 111101
Option (A) is correct.
Option (B) is correct.
An I/O Microprocessor controls data flow between main memory
and the I/O device which wants to communicate.
Option (D) is correct.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
5.134
5.135
Option (B) is correct.
Dual slope ADC is more accurate.
The given gate is ex-OR so output
Option (A) is correct.
Dual form of any identity can be find by replacing all AND function
to OR and vice-versa. so here dual form will be
Here input
(A + B) (A + C) (B + C) = (A + B) (A + C)
5.136
5.137
Option (B) is correct.
Carry flag will be affected by arithmetic instructions only.
Option (C) is correct.
This is a synchronous counter. we can find output as
QA QB
0 0
1 0
0 1
0 0
h
So It counts only three states. It is a mod-3 counter.
K =3
5.138
5.139
Page 131
Option (B) is correct.
Option (A) is correct.
Essential prime implicates for a function is no. of terms that we get
by solving K -map. Here we get 4 terms when solve the K -map.
F = AB + AB
B = 0 so,
F = A1 + A0 = A
5.145
Option (C) is correct.
EI = Enabled Interput flag ,RST will cause an Interrupt
only it we enable EI .
5.146
5.147
Option (A) is correct.
Here only for the range 60 to 63 H chipselect will be 0, so peripheral
will correspond in this range only chipselect = 1 for rest of the given
address ranges.
Option (B) is correct.
By executing instructions one by one
LXI H, 8A79 H (Load HL pair by value 8A79)
H = 8AH L = 79 H
MOV A, L (copy contain of L to accumulator)
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A = 79 H
ADDH (add contain of H to accumulator)
A = 79 H =
0 1111 0 0 1
H = 8AH = add 1 0 0 0 1 0 1 0
=A=
0 0 0 0 0 0 11
Carry = 1
DAA (Carry Flag is set, so DAA adds 6 to high order four bits)
y = B D + A C D + C AB + CA B
so no of prime implicates is 4
5.140
5.141
Option (A) is correct.
Option (B) is correct.
For a 2 bit multiplier
B1
# A1
A 0 B1
B0
A0
A0 B0
# A1 B1 A1 B 0
C3
C2
C1
C0
This multiplication is identical to AND operation and then addition.
5.142
5.143
H = 63 H
PCHL (Load program counter by HL pair)
PC = 6379 H
5.148
5.149
Option (B) is correct.
Consider high output state
fan out = IOH max = 400 mA = 20
IIH max
20 mA
Consider low output state
fan out = IOL max = 8 mA = 80
IIL max
0.1 mA
Option (A) is correct.
Option (C) is correct.
Option (C) is correct.
NMOS In parallel makes OR Gate & in series makes AND so here
we can have
F = A (B + C) + DE
we took complement because there is another NMOS given above
(works as an inverter)
Option (C) is correct.
In totem pole stage output resistance will be small so it acts like a
output buffer.
Thus fan out is 20
5.144
0 1111 0 0 1
DAA add 1 0 0 0 1 0 1 0
A = 0 0 0 0 0 0 1 1 = 63 H
MOV H, A (copy contain of A to H)
5.150
Option (D) is correct.
For a J -K flip flop we have characteristic equation as
Q (t + 1) = JQ (t) + KQ (t)
Q (t) & Q (t + 1) are present & next states.
In given figure
J = Q (t), K = 1 so
Q (t + 1) = Q (t) Q (t) + 0Q (t)
Q (t + 1) = Q (t)[complement of previous state]
we have initial input Q (t) = 0
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 132
so for 6 clock pulses sequence at output Q will be 010101
5.151
5.152
Option (B) is correct.
By distributive property in boolean algebra we have
(A + BC) = (A + B) (A + C)
(A + B) (A + C) = AA + AC + AB + BC
= A (1 + C) + AB + BC
= A + AB + BC
= A (1 + B) + BC = A + BC
5.153
5.154
Option (A) is correct.
The current in a p n junction diode is controlled by diffusion of
majority carriers while current in schottky diode dominated by the
flow of majority carrier over the potential barrier at metallurgical
junction. So there is no minority carrier storage in schottky diode,
so switching time from forward bias to reverse bias is very short
compared to p n junction diode. Hence the propagation delay will
reduces.
= 200 mV/cC
5.159
10.24
2 # 1024 # (50 - 25) cC
Option (D) is correct.
210 # 8 = 13
No. of chips = 26 #
12
2 #4
5.160
Option (C) is correct.
Given instruction set
1000 LXI SP 27FF
1003 CALL
1006
1006 POP H
First Instruction will initialize the SP by a value
27FF
SP ! 27FF
CALL 1006 will “Push PC” and Load PC by value 1006
PUSH PC will store value of PC in stack
PC = 1006
Option (B) is correct.
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5.155
Tcoff =
or
Option (C) is correct.
Option (D) is correct.
The total conversion time for different type of ADC are given as–
t is clock period
For flash type & 1t
Counter type & (2n - t) = 4095 m sec
now POP H will be executed
which load HL pair by stack values
HL = 1006
and
SP = SPl + 2
SP = SPl + 2 = SP - 2 + 2 = SP
SP = 27FF
n = no.of bits
Integrating type conver time > 4095 m sec
successive approximation type nt = 12 m sec
here n = 12 so
nt = 12
12t = 12
so this is succ. app. type ADC.
5.156
5.157
Option (D) is correct.
LDA 2003 (Load accumulator by a value 2003 H) so here total no.
of memory access will be 4.
1 = Fetching instruction
2 = Read the value from memory
1 = write value to accumulator
Option (D) is correct.
Storage capacitance
-12
C = i = 1 # 10
5 - 0.5
dv
b dt l
b 20 10-3 l
#
= 1 # 10
5.158
-12
# 20 # 10
4.5
-3
= 4.4 # 10-15 F
Option (A) is correct.
or
Accuracy ! 1 LSB = Tcoff # DT
2
1 10.24 = T
coff # DT
2 # 210
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
UNIT 6
Page 133
2013
6.8
SIGNALS & SYSTEMS
6.9
2013
6.1
6.2
ONE MARK
Two systems with impulse responses h1 ^ t h and h2 ^ t h are connected
in cascade. Then the overall impulse response of the cascaded system
is given by
(A) product of h1 ^ t h and h2 ^ t h
(B) sum of h1 ^ t h and h2 ^ t h
(C) convolution of h1 ^ t h and h2 ^ t h
(D) subtraction of h2 ^ t h from h1 ^ t h
The impulse response of a system is h ^ t h = tu ^ t h. For an input
u ^t - 1h, the output is
2
t ^t - 1h
(A) t u ^ t h
(B)
u ^t - 1h
2
2
^t - 1h2
(C)
u ^t - 1h
2
6.3
6.4
6.5
6.6
6.7
2
(D) t - 1 u ^t - 1h
2
6.10
The impulse response of a continuous time system is given by
h ^ t h = d ^t - 1h + d ^t - 3h. The value of the step response at t = 2 is
(A) 0
(B) 1
(C) 2
(D) 3
A
system
described
by
the
differential
equation
2
dy
dy
+ 5 + 6y ^ t h = x ^ t h. Let x ^ t h be a rectangular pulse given by
dt
dt2
1
0<t<2
x^t h = *
0
otherwise
dy
Assuming that y ^0 h = 0 and
= 0 at t = 0 , the Laplace transdt
form of y ^ t h is
1 - e-2s
e-2s
(B)
(A)
s ^s + 2h^s + 3h
s ^s + 2h^s + 3h
-2s
e
1 - e-2s
(C)
(D)
^s + 2h^s + 3h
^s + 2h^s + 3h
A system described by a linear, constant coefficient, ordinary, first
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For
a
periodic
signal
v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p/4h, the fundamental
frequency in rad/s
(A) 100
(B) 300
(C) 500
(D) 1500
order differential equation has an exact solution given by y ^ t h for
t > 0 , when the forcing function is x ^ t h and the initial condition
is y ^0 h. If one wishes to modify the system so that the solution
becomes - 2y ^ t h for t > 0 , we need to
(A) change the initial condition to - y ^0 h and the forcing function
to 2x ^ t h
(B) change the initial condition to 2y ^0 h and the forcing function
to - x ^ t h
(C) change the initial condition to j 2 y ^0 h and the forcing function to j 2 x ^ t h
(D) change the initial condition to - 2y ^0 h and the forcing function
to - 2x ^ t h
A band-limited signal with a maximum frequency of 5 kHz is to
be sampled. According to the sampling theorem, the sampling
frequency which is not valid is
(A) 5 kHz
(B) 12 kHz
(C) 15 kHz
(D) 20 kHz
Which one of the following statements is NOT TRUE for a continuous
time causal and stable LTI system?
(A) All the poles of the system must lie on the left side of the jw
axis
(B) Zeros of the system can lie anywhere in the s-plane
(C) All the poles must lie within s = 1
(D) All the roots of the characteristic equation must be located on
the left side of the jw axis.
TWO MARKS
Assuming zero initial condition, the response y ^ t h of the system
given below to a unit step input u ^ t h is
The DFT of a vector 8a b c dB is the vector 8a b g dB . Consider
the product
V
R
Sa b c d W
Sd a b c W
8p q r sB = 8a b c dBSc d a b W
W
S
Sb c d aW
X
T
The DFT of the vector 8p q r sB is a scaled version of
(B) 9 a b
(A) 9a2 b2 g2 d2C
g
dC
(C) 8a + b b + d d + g g + aB
(D) 8a b g dB
(A) u ^ t h
2
(C) t u ^ t h
2
2012
(B) tu ^ t h
(D) e-t u ^ t h
6.12
Let g ^ t h = e- pt , and h ^ t h is a filter matched to g ^ t h. If g ^ t h is
applied as input to h ^ t h, then the Fourier transform of the output is
(A) e- pf
(B) e- pf /2
2
2
(C) e- p f
6.11
2
(D) e-2pf
2
ONE MARK
The unilateral Laplace transform of f (t) is 2 1
. The unilateral
s +s+1
Laplace transform of tf (t) is
(A) - 2 s
(B) - 2 2s + 1 2
(s + s + 1) 2
(s + s + 1)
(C) 2 s
(D) 2 2s + 1 2
2
(s + s + 1)
(s + s + 1)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
6.13
If x [n] = (1/3) n - (1/2) n u [n], then the region of convergence (ROC)
of its z -transform in the z -plane will be
(A) 1 < z < 3
(B) 1 < z < 1
3
3
2
(C) 1 < z < 3
(D) 1 < z
2
3
2012
6.14
6.19
6.20
TWO MARKS
The input x (t) and output y (t) of a system are related as
y (t) =
If the unit step response of a network is (1 - e- at), then its unit
impulse response is
(A) ae- at
(B) a-1 e- at
(C) (1 - a-1)e- at
(D) (1 - a) e- at
t
-3
The Fourier transform of a signal h (t) is H (jw) = (2 cos w) (sin 2w) /w
. The value of h (0) is
(A) 1/4
(B) 1/2
(C) 1
(D) 2
2011
6.21
6.22
GATE Electronics & Communication
by RK Kanodia
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6.16
A system is defined by its impulse response h (n) = 2n u (n - 2). The
system is
(A) stable and causal
(B) causal but not stable
(C) stable but not causal
(D) unstable and non-causal
# x (t) cos (3t) dt . The system is
(A) time-invariant and stable
(B) stable and not time-invariant
(C) time-invariant and not stable (D) not time-invariant and not
stable
6.15
Page 134
Let y [n] denote the convolution of h [n] and g [n], where
h [n] = (1/2) n u [n] and g [n] is a causal sequence. If y [0] = 1 and
y [1] = 1/2, then g [1] equals
(A) 0
(B) 1/2
(C) 1
Two systems H1 (Z ) and H2 (Z ) are connected in cascade as shown
below. The overall output y (n) is the same as the input x (n) with a
one unit delay. The transfer function of the second system H2 (Z ) is
1 - 0.6z-1
z-1 (1 - 0.4z-1)
z-1 (1 - 0.4z-1)
(C)
(1 - 0.6z-1)
(D) 3/2
2011
An input x (t) = exp (- 2t) u (t) + d (t - 6) is applied to an LTI system
with impulse response h (t) = u (t) . The output is
(A) [1 - exp (- 2t)] u (t) + u (t + 6)
(B) [1 - exp (- 2t)] u (t) + u (t - 6)
(C) 0.5 [1 - exp (- 2t)] u (t) + u (t + 6)
(D) 0.5 [1 - exp (- 2t)] u (t) + u (t - 6)
(A)
6.23
ONE MARK
TWO MARKS
(B)
z-1 (1 - 0.6z-1)
(1 - 0.4z-1)
(D)
1 - 0.4 z-1
z-1 (1 - 0.6z-1)
The first six points of the 8-point DFT of a real valued sequence are
5, 1 - j 3, 0, 3 - j 4, 0 and 3 + j 4 . The last two points of the DFT
are respectively
(A) 0, 1 - j 3
(B) 0, 1 + j 3
(C) 1 + j3, 5
(D) 1 - j 3, 5
2
6.17
The differential equation 100 ddty - 20 dy
dt + y = x (t) describes a system
with an input x (t) and an output y (t). The system, which is initially
relaxed, is excited by a unit step input. The output y ^ t h can be
represented by the waveform
2
2010
6.24
ONE MARK
The trigonometric Fourier series for the waveform f (t) shown below
contains
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6.18
The trigonometric Fourier series of an even function does not have
the
(A) dc term
(B) cosine terms
(C) sine terms
(D) odd harmonic terms
(A) only cosine terms and zero values for the dc components
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 135
(B) only cosine terms and a positive value for the dc components
(C) only cosine terms and a negative value for the dc components
(D) only sine terms and a negative value for the dc components
6.25
6.26
6.27
Consider the z -transform x (z) = 5z2 + 4z-1 + 3; 0 < z < 3. The
inverse z - transform x [n] is
(A) 5d [n + 2] + 3d [n] + 4d [n - 1]
(B) 5d [n - 2] + 3d [n] + 4d [n + 1]
(C) 5u [n + 2] + 3u [n] + 4u [n - 1]
(D) 5u [n - 2] + 3u [n] + 4u [n + 1]
Two discrete time system with impulse response h1 [n] = d [n - 1]
and h2 [n] = d [n - 2] are connected in cascade. The overall impulse
response of the cascaded system is
(B) d [n - 4]
(A) d [n - 1] + d [n - 2]
(C) d [n - 3]
(D) d [n - 1] d [n - 2]
For a N -point FET algorithm N = 2m which one of the following
statements is TRUE ?
(A) It is not possible to construct a signal flow graph with both
input and output in normal order
(B) The number of butterflies in the m th stage in N/m
(C) In-place computation requires storage of only 2N data
(D) Computation of a butterfly requires only one complex multiplication.
2010
6.28
6.29
6.30
6.32
6.33
3s + 1
Given f (t) = L ; 3
. If lim f (t) = 1, then the value
t"3
s + 4s2 + (k - 3) s E
of k is
(A) 1
(B) 2
(C) 3
(D) 4
The transfer function of a discrete time LTI system is given by
2 - 3 z-1
4
H (z) =
3
-1
1 - z + 1 z-2
8
4
Consider the following statements:
S1: The system is stable and causal for ROC: z > 1/2
S2: The system is stable but not causal for ROC: z < 1/4
S3: The system is neither stable nor causal for ROC:
1/4 < z < 1/2
Which one of the following statements is valid ?
(A) Both S1 and S2 are true
(B) Both S2 and S3 are true
(C) Both S1 and S3 are true
(D) S1, S2 and S3 are all true
The Fourier series of a real periodic function has only
(P) cosine terms if it is even
(Q) sine terms if it is even
(R) cosine terms if it is odd
(S) sine terms if it is odd
The ROC of z -transform of the discrete time sequence
n
n
x (n) = b 1 l u (n) - b 1 l u (- n - 1) is
3
2
(B) z > 1
(A) 1 < z < 1
2
3
2
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(C) z < 1
3
-1
A continuous time LTI system is described by
d 2 y (t)
dy (t)
dx (t)
+ 4x (t)
+4
+ 3y (t) = 2
2
dt
dt
dt
Assuming zero initial conditions, the response y (t) of the above
system for the input x (t) = e-2t u (t) is given by
(B) (e-t - e-3t) u (t)
(A) (et - e3t) u (t)
(C) (e-t + e-3t) u (t)
(D) (et + e3t) u (t)
A function is given by f (t) = sin2 t + cos 2t . Which of the following
is true ?
(A) f has frequency components at 0 and 1 Hz
2p
(B) f has frequency components at 0 and 1 Hz
p
(C) f has frequency components at 1 and 1 Hz
p
2p
(D) f has frequency components at 0.1 and 1 Hz
p
2p
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TWO MARKS
2009
6.31
Which of the above statements are correct ?
(A) P and S
(B) P and R
(C) Q and S
(D) Q and R
(D) 2 < z < 3
2009
6.34
TWO MARKS
Given that F (s) is the one-side Laplace transform of f (t), the Laplace
transform of
t
#0 f (t) dt is
(A) sF (s) - f (0)
(C)
6.35
6.36
#0
s
F (t) dt
(B) 1 F (s)
s
(D) 1 [F (s) - f (0)]
s
A system with transfer function H (z) has impulse response h (.)
defined as h (2) = 1, h (3) =- 1 and h (k) = 0 otherwise. Consider the
following statements.
S1 : H (z) is a low-pass filter.
S2 : H (z) is an FIR filter.
Which of the following is correct?
(A) Only S2 is true
(B) Both S1 and S2 are false
(C) Both S1 and S2 are true, and S2 is a reason for S1
(D) Both S1 and S2 are true, but S2 is not a reason for S1
Consider a system whose input x and output y are related by the
equation y (t) =
# x (t - t) g (2t) dt where h (t) is shown in the graph.
3
-3
ONE MARK
Which of the following four properties are possessed by the system
?
BIBO : Bounded input gives a bounded output.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Causal : The system is causal,
LP : The system is low pass.
LTI : The system is linear and time-invariant.
(A) Causal, LP
(B) BIBO, LTI
(C) BIBO, Causal, LTI
(D) LP, LTI
6.37
6.38
comes zero are
(A) p, 2p
(C) 0, p
6.43
The 4-point Discrete Fourier Transform (DFT) of a discrete time
sequence {1,0,2,3} is
(A) [0, - 2 + 2j , 2, - 2 - 2j ]
(B) [2, 2 + 2j , 6, 2 - 2j ]
(C) [6, 1 - 3j , 2, 1 + 3j ]
(D) [6, - 1 + 3j , 0, - 1 - 3j ]
An LTI system having transfer function s +s 2+s 1+ 1 and input
x (t) = sin (t + 1) is in steady state. The output is sampled at a rate
ws rad/s to obtain the final output {x (k)}. Which of the following
is true ?
(A) y (.) is zero for all sampling frequencies ws
(B) y (.) is nonzero for all sampling frequencies ws
(C) y (.) is nonzero for ws > 2 , but zero for ws < 2
(D) y (.) is zero for ws > 2 , but nonzero for w2 < 2
2
2008
6.40
6.42
6.44
6.45
ONE MARK
The input and output of a continuous time system are respectively
denoted by x (t) and y (t). Which of the following descriptions
corresponds to a causal system ?
(B) y (t) = (t - 4) x (t + 1)
(A) y (t) = x (t - 2) + x (t + 4)
(C) y (t) = (t + 4) x (t - 1)
(D) y (t) = (t + 5) x (t + 5)
A discrete time linear shift - invariant system has an impulse
response h [n] with h [0] = 1, h [1] =- 1, h [2] = 2, and zero otherwise
The system is given an input sequence x [n] with x [0] = x [2] = 1,
and zero otherwise. The number of nonzero samples in the output
sequence y [n], and the value of y [2] are respectively
(A) 5, 2
(B) 6, 2
(C) 6, 1
(D) 5, 3
The signal x (t) is described by
1 for - 1 # t # + 1
x (t) = )
0 otherwise
Two of the angular frequencies at which its Fourier transform be-
{x (n)} is a real - valued periodic sequence with a period N . x (n)
and X (k) form N-point Discrete Fourier Transform (DFT) pairs.
N-1
The DFT Y (k) of the sequence y (n) = 1 / x (r) x (n + r) is
N r=0
N-1
(B) 1 / X (r) X (k + r)
(A) X (k) 2
N r=0
(D) 0
Statement for Linked Answer Question 6.31 and 6.32:
In the following network, the switch is closed at t = 0- and the
sampling starts from t = 0 . The sampling frequency is 10 Hz.
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TWO MARKS
A linear, time - invariant, causal continuous time system has a
rational transfer function with simple poles at s =- 2 and s =- 4
and one simple zero at s =- 1. A unit step u (t) is applied at the
input of the system. At steady state, the output has constant value
of 1. The impulse response of this system is
(A) [exp (- 2t) + exp (- 4t)] u (t)
(B) [- 4 exp (- 2t) - 12 exp (- 4t) - exp (- t)] u (t)
(C) [- 4 exp (- 2t) + 12 exp (- 4t)] u (t)
(D) [- 0.5 exp (- 2t) + 1.5 exp (- 4t)] u (t)
Let x (t) be the input and y (t) be the output of a continuous time
system. Match the system properties P1, P2 and P3 with system
relations R1, R2, R3, R4
Properties
Relations
P1 : Linear but NOT time - invariant R1 : y (t) = t2 x (t)
P2 : Time - invariant but NOT linear
R2 : y (t) = t x (t)
P3 : Linear and time - invariant
R3 : y (t) = x (t)
R4 : y (t) = x (t - 5)
(A) (P1, R1), (P2, R3), (P3, R4)
(B) (P1, R2), (P2, R3), (P3, R4)
(C) (P1, R3), (P2, R1), (P3, R2)
(D) (P1, R1), (P2, R2), (P3, R3)
N-1
(C) 1 / X (r) X (k + r)
N r=0
The impulse response h (t) of a linear time invariant continuous
time system is described by h (t) = exp (at) u (t) + exp (bt) u (- t)
where u (- t) denotes the unit step function, and a and b are real
constants. This system is stable if
(A) a is positive and b is positive
(B) a is negative and b is negative
(C) a is negative and b is negative
(D) a is negative and b is positive
2008
6.41
(B) 0.5p, 1.5p
(D) 2p, 2.5p
2
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6.39
Page 136
6.46
6.47
The samples x (n), n = (0, 1, 2, ...) are given by
(A) 5 (1 - e-0.05n)
(B) 5e-0.05n
(C) 5 (1 - e-5n)
(D) 5e-5n
The expression and the region of convergence of the z -transform of
the sampled signal are
5z , z < e-0.05
(A) 5z 5 , z < e-5
(B)
z
e-0.05
z-e
5z , z > e-0.05
(C)
(D) 5z -5 , z > e-5
-0.05
z-e
z-e
Statement for Linked Answer Question 6.33 & 6.34:
The impulse response h (t) of linear time - invariant continuous
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 137
time system is given by h (t) = exp (- 2t) u (t), where u (t) denotes
the unit step function.
6.48
6.49
The frequency response H (w) of this system in terms of angular
frequency w, is given by H (w)
1
(B) sin w
(A)
w
1 + j2w
jw
1
(C)
(D)
2 + jw
2 + jw
#
#
6.58
The output of this system, to the sinusoidal input x (t) = 2 cos 2t for
all time t , is
(A) 0
(B) 2-0.25 cos (2t - 0.125p)
(C) 2-0.5 cos (2t - 0.125p)
(D) 2-0.5 cos (2t - 0.25p)
6.59
2007
6.50
ONE MARK
1
If the Laplace transform of a signal Y (s) =
, then its final
s (s - 1)
value is
(A) - 1
(B) 0
(C) 1
(D) Unbounded
2007
6.51
TWO MARKS
The 3-dB bandwidth of the low-pass signal e-t u (t), where u (t) is the
unit step function, is given by
(B) 1
(A) 1 Hz
2 - 1 Hz
2p
2p
(C) 3
6.52
(A) 5
(C) 16p
6.53
6.54
6.55
p
-p
6.56
6.57
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1 sin t - p
`
4j
2
(C) 1 e-t sin t
2
(A)
(B) 10p
(D) 5 + j10p
(B) non-causal system
(D) low-pass system
The frequency response of a linear, time-invariant system is given by
H (f) = 1 + j510pf . The step response of the system is
t
(A) 5 (1 - e-5t) u (t)
(B) 5 61 - e- 5@ u (t)
t
(C) 1 (1 - e-5t) u (t)
(D) 1 ^1 - e- 5 h u (t)
2
5
6.60
6.61
6.62
ONE MARK
Let x (t) * X (jw) be Fourier Transform pair. The Fourier Transform
of the signal x (5t - 3) in terms of X (jw) is given as
j3w
j3w
jw
jw
(A) 1 e- 5 X b l
(B) 1 e 5 X b l
5
5
5
5
jw
jw
(C) 1 e-j3w X b l
(D) 1 e j3w X b l
5
5
5
5
The Dirac delta function d (t) is defined as
1
t=0
(A) d (t) = )
0
otherwise
(B)
1 sin t + p
`
4j
2
(D) sin t - cos t
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2006
<2
3
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X (e jw) dw is
The z -transform X (z) of a sequence x [n] is given by X [z] = 1 -0.25z .
It is given that the region of convergence of X (z) includes the unit
circle. The value of x [0] is
(A) - 0.5
(B) 0
(C) 0.25
(D) 05
A Hilbert transformer is a
(A) non-linear system
(C) time-varying system
<3
In the system shown below, x (t) = (sin t) u (t) In steady-state, the
response y (t) will be
(D) 1 Hz
#
1 < z < 2 then the
3
3
SPECIAL EDITION ( STUDY MATERIAL FORM )
At market Book is available in 3 volume i.e. in 3 book binding
form. But at NODIA Online Store book is available in 10 book
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A 5-point sequence x [n] is given as x [- 3] = 1, x [- 2] = 1, x [- 1] = 0,
x [0] = 5 and x [1] = 1. Let X (eiw) denoted the discrete-time Fourier
transform of x [n]. The value of
3 t=0
(B) d (t) = )
0
otherwise
3
1 t=0
(C) d (t) = )
and d (t) dt = 1
-3
0 otherwise
3
3 t=0
(D) d (t) = )
and d (t) dt = 1
-3
0 otherwise
If the region of convergence of x1 [n] + x2 [n] is
region of convergence of x1 [n] - x2 [n] includes
(A) 1 < z < 3
(B) 2 < z
3
3
(C) 3 < z < 3
(D) 1 < z
2
3
6.63
TWO MARKS
Consider the function f (t) having Laplace transform
F (s) = 2 w0 2 Re [s] > 0
s + w0
The final value of f (t) would be
(A) 0
(B) 1
(C) - 1 # f (3) # 1
(D) 3
A system with input x [n] and output y [n] is given as y [n] = (sin 56 pn) x [n]
. The system is
(A) linear, stable and invertible
(B) non-linear, stable and non-invertible
(C) linear, stable and non-invertible
(D) linear, unstable and invertible
The unit step response of a system starting from rest is given by
c (t) = 1 - e-2t for t $ 0 . The transfer function of the system is
1
(B) 2
(A)
2+s
1 + 2s
(C) 1
(D) 2s
2+s
1 + 2s
The unit impulse response of a system is f (t) = e-t, t $ 0 . For this
system the steady-state value of the output for unit step input is
equal to
(A) - 1
(B) 0
(C) 1
(D) 3
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2005
6.64
6.65
Page 138
ONE MARK
Choose the function f (t); - 3 < t < 3 for which a Fourier series
cannot be defined.
(A) 3 sin (25t)
(B) 4 cos (20t + 3) + 2 sin (710t)
(C) exp (- t ) sin (25t)
(D) 1
(B) 41
(C) 42
(D) 82
2005
6.70
The function x (t) is shown in the figure. Even and odd parts of a
unit step function u (t) are respectively,
TWO MARKS
The output y (t) of a linear time invariant system is related to its
input x (t) by the following equations
y (t)= 0.5x (t - td + T) + x (t - td ) + 0.5x (t - td + T)
The filter transfer function H (w) of such a system is given by
(A) (1 + cos wT) e-jwt
d
(B) (1 + 0.5 cos wT) e-jwt
(C) (1 - cos wT) e-jwt
(D) (1 - 0.5 cos wT) e-jwt
d
d
d
(A) 1 , 1 x (t)
2 2
(C) 1 , - 1 x (t)
2
2
(B) - 1 , 1 x (t)
2 2
(D) - 1 , - 1 x (t)
2
2
6.71
GATE Electronics & Communication
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6.66
6.67
The region of convergence of z - transform of the sequence
5 n
6 n
b 6 l u (n) - b 5 l u (- n - 1) must be
(B) z > 5
(A) z < 5
6
6
(C) 5 < z < 6
(D) 6 < z < 3
6
5
5
Match the following and choose the correct combination.
Group 1
E. Continuous and aperiodic signal
F. Continuous and periodic signal
G. Discrete and aperiodic signal
H. Discrete and periodic signal
Group 2
1. Fourier representation is continuous and aperiodic
2. Fourier representation is discrete and aperiodic
3. Fourier representation is continuous and periodic
4. Fourier representation is discrete and periodic
(A)
(B)
(C)
(D)
6.72
Which of the following can be impulse response of a causal system ?
E - 3,
E - 1,
E - 1,
E - 2,
F - 2,
F - 3,
F - 2,
F - 1,
G - 4,
G - 2,
G - 3,
G - 4,
H-1
H-4
H-4
H-3
A signal x (n) = sin (w0 n + f) is the input to a linear time- invariant
system having a frequency response H (e jw). If the output of the
system Ax (n - n0) then the most general form of +H (e jw) will be
(A) - n0 w0 + b for any arbitrary real
(B) - n0 w0 + 2pk for any arbitrary integer k
(C) n0 w0 + 2pk for any arbitrary integer k
(D) - n0 w0 f
Statement of linked answer question 6.59 and 6.60 :
A sequence x (n) has non-zero values as shown in the figure.
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6.68
Let x (n) = ( 12 ) n u (n), y (n) = x2 (n) and Y (e jw) be the Fourier
transform of y (n) then Y (e j0)
(A) 1
4
(B) 2
(C) 4
(D) 4
3
6.69
The power in the signal s (t) = 8 cos (20p - p2 ) + 4 sin (15pt) is
(A) 40
x ( n2 - 1),
6.73
The sequence y (n) = *
0,
For n even
will be
For n odd
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 139
The system is stable only if
(A) a = 2 , b < 2
(C) a < 2 , any value of b
6.81
(B) a > 2, b > 2
(D) b < 2 , any value of a
The impulse response h [n] of a linear time invariant system is given
as
-2 2
n = 1, - 1
h [ n] = * 4 2
n = 2, - 2
0
otherwise
If the input to the above system is the sequence e jpn/4 , then the
output is
(A) 4 2 e jpn/4
(B) 4 2 e-jpn/4
(C) 4e jpn/4
(D) - 4e jpn/4
6.82
Let x (t) and y (t) with Fourier transforms F (f) and Y (f) respectively
be related as shown in Fig. Then Y (f) is
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6.74
6.75
The Fourier transform of y (2n) will be
(A) e-j2w [cos 4w + 2 cos 2w + 2]
(B) cos 2w + 2 cos w + 2
-jw
(C) e [cos 2w + 2 cos w + 2]
(D) e-j2w [cos 2w + 2 cos + 2]
For a signal x (t) the Fourier transform is X (f). Then the inverse
Fourier transform of X (3f + 2) is given by
j4pt
(A) 1 x` t j e j3pt
(B) 1 x` t j e - 3
2 2
3 3
(C) 3x (3t) e-j4pt
(D) x (3t + 2)
2004
6.76
6.77
6.78
ONE MARK
The impulse response h [n] of a linear time-invariant system is given
by h [n] = u [n + 3] + u [n - 2) - 2n [n - 7] where u [n] is the unit step
sequence. The above system is
(A) stable but not causal
(B) stable and causal
(C) causal but unstable
(D) unstable and not causal
The Fourier transform of a conjugate symmetric function is always
(A) imaginary
(B) conjugate anti-symmetric
(C) real
(D) conjugate symmetric
6.80
6.83
A causal LTI system is described by the difference equation
2y [n] = ay [n - 2] - 2x [n] + bx [n - 1]
(B) - 1 X (f/2) e j2pf
2
(C) - X (f/2) e j2pf
(D) - X (f/2) e-j2pf
ONE MARK
The Laplace transform of i (t) is given by
2
I (s) =
s (1 + s)
At t " 3, The value of i (t) tends to
(A) 0
(B) 1
(C) 2
(D) 3
6.84
The Fourier series expansion of a real periodic signal with
fundamental frequency f0 is given by gp (t) =
cn e j2pf t . It is given
n =- 3
that c3 = 3 + j5 . Then c-3 is
(A) 5 + j3
(B) - 3 - j5
(C) - 5 + j3
(D) 3 - j5
/
TWO MARKS
Consider the sequence x [n] = [- 4 - j51 + j25]. The conjugate antisymmetric part of the sequence is
(B) [- j2.5, 1, j2.5]
(A) [- 4 - j2.5, j2, 4 - j2.5]
(C) [- j2.5, j2, 0]
(D) [- 4, 1, 4]
(A) - 1 X (f/2) e-jpf
2
2003
The z -transform of a system is H (z) = z -z0.2 . If the ROC is z < 0.2
, then the impulse response of the system is
(A) (0.2) n u [n]
(B) (0.2) n u [- n - 1]
(C) - (0.2) n u [n]
(D) - (0.2) n u [- n - 1]
2004
6.79
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6.85
6.86
0
Let x (t) be the input to a linear, time-invariant system. The required
output is 4p (t - 2). The transfer function of the system should be
(A) 4e j4pf
(B) 2e-j8pf
(C) 4e-j4pf
(D) 2e j8pf
A sequence x (n) with the z -transform X (z) = z 4 + z2 - 2z + 2 - 3z-4
is applied as an input to a linear, time-invariant system with the
impulse response h (n) = 2d (n - 3) where
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
1, n = 0
d (n) = )
0, otherwise
The output at n = 4 is
(A) - 6
(B) zero
(C) 2
(D) - 4
2003
6.87
Page 140
(A) e f u (f)
(C) e f u (- f)
6.93
TWO MARKS
Let P be linearity, Q be time-invariance, R be causality and S be
stability. A discrete time system has the input-output relationship,
x (n)
n$1
y (n) = *0,
n=0
x (n + 1) n # - 1
where x (n) is the input and y (n) is the output. The above system
has the properties
(A) P, S but not Q, R
(B) P, Q, S but not R
(C) P, Q, R, S
6.94
6.95
(D) Q, R, S but not P
6.96
The system under consideration is an RC low-pass filter (RC-LPF)
with R = 1 kW and C = 1.0 mF.
Let H (f) denote the frequency response of the RC-LPF. Let f1 be
H (f1)
the highest frequency such that 0 # f # f1
$ 0.95 . Then f1
H (0)
(in Hz) is
(A) 324.8
(B) 163.9
(C) 52.2
(D) 104.4
6.97
6.98
Let tg (f) be the group delay function of the given RC-LPF and
f2 = 100 Hz. Then tg (f2) in ms, is
(A) 0.717
(B) 7.17
(C) 71.7
(D) 4.505
2002
6.90
6.91
6.92
TWO MARKS
The Laplace transform of continuous - time signal x (t) is X (s) =
. If the Fourier transform of this signal exists, the x (t) is
(A) e2t u (t) - 2e-t u (t)
(B) - e2t u (- t) + 2e-t u (t)
(C) - e2t u (- t) - 2e-t u (t)
(D) e2t u (- t) - 2e-t u (t)
5-s
s2 - s - 2
If the impulse response of discrete - time system is
h [n] =- 5n u [- n - 1],
then the system function H (z) is equal to
(A) - z and the system is stable
z-5
(B) z and the system is stable
z-5
(C) - z and the system is unstable
z-5
(D) z and the system is unstable
z-5
2001
Common Data For Q. 6.73 & 6.74 :
6.89
A linear phase channel with phase delay Tp and group delay Tg must
have
(A) Tp = Tg = constant
(B) Tp \ f and Tg \ f
(C) Tp = constant and Tg \ f ( f denote frequency)
(D) Tp \ f and Tp = constant
2002
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6.88
(B) e-f u (f)
(D) e-f u (- f)
ONE MARK
The transfer function of a system is given by H (s) = 2 1
. The
s (s - 2)
impulse response of the system is
(B) (t * e2t) u (t)
(A) (t2 * e-2t) u (t)
(C) (te-2 t) u (t)
(D) (te-2t) u (t)
The region of convergence of the z - transform of a unit step function
is
(A) z > 1
(B) z < 1
(C) (Real part of z ) > 0
(D) (Real part of z ) < 0
Let d (t) denote the delta function. The value of the integral
3
d (t) cos b 3t l dt is
2
-3
#
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ONE MARK
Convolution of x (t + 5) with impulse function d (t - 7) is equal to
(A) x (t - 12)
(B) x (t + 12)
(C) x (t - 2)
(D) x (t + 2)
(A) 1
(C) 0
Which of the following cannot be the Fourier series expansion of a
periodic signal?
(A) x (t) = 2 cos t + 3 cos 3t
(B) x (t) = 2 cos pt + 7 cos t
(C) x (t) = cos t + 0.5
(D) x (t) = 2 cos 1.5pt + sin 3.5pt
6.99
1
The Fourier transform F {e-1 u (t)} is equal to
. Therefore,
1
+
j2pf
1
is
F'
1 + j2pt 1
6.100
(B) - 1
(D) p2
If a signal f (t) has energy E , the energy of the signal f (2t) is equal
to
(A) 1
(B) E/2
(C) 2E
(D) 4E
2001
TWO MARKS
The impulse response functions of four linear systems S1, S2, S3, S4
are given respectively by
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 141
h1 (t) = 1, h2 (t) = u (t),
u (t)
and
h3 (t) =
t+1
h 4 (t) = e-3t u (t)
where u (t) is the unit step function. Which of these systems is
time invariant, causal, and stable?
(A) S1
(B) S2
(C) S3
(D) S4
2000
6.101
ONE MARK
L [f (t)] = s2+ 2 ,
s +1
t
h (t) = f (t) g (t - t) dt .
0
L [h (t)] is
Given
that
L [g (t)] =
#
2
(A) s + 1
s+3
(C)
6.102
(B)
s2 + 1
+ s+2
(s + 3)( s + 2) s2 + 1
s2 + 1
(s + 3) (s + 2)
and
1
s+3
6.107
(D) None of the above
2
The Fourier Transform of the signal x (t) = e-3t is of the following
form, where A and B are constants :
(A) Ae-B f
(B) Ae-Bf
(C) A + B f 2
(D) Ae-Bf
2
6.103
6.104
2000
6.105
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A system with an input x (t) and output y (t) is described by the
relations : y (t) = tx (t). This system is
(A) linear and time - invariant
(B) linear and time varying
(C) non - linear and time - invariant
(D) non - linear and time - varying
A linear time invariant system has an impulse response e2t, t > 0 . If
the initial conditions are zero and the input is e3t , the output for
t > 0 is
(A) e3t - e2t
(B) e5t
(C) e3t + e2t
(D) None of these
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df (w)
f (w0)
, dw w = w
w0
df (w)
(C) wo , d (w) w = w
f (wo)
(A) -
6.106
and
and
and
and
n-2
n-3
n-2
n-2
(D) wo f (wo),
o
d2 f (w0)
dw2
#
wo
-3
6.108
6.109
ONE MARK
The z -transform F (z) of the function f (nT) = anT is
(A) z T
(B) z T
z-a
z+a
z
z
(C)
(D)
z - a-T
z + a-T
If [f (t)] = F (s), then [f (t - T)] is equal to
(A) esT F (s)
(B) e-sT F (s)
F (s)
F (s)
(C)
(D)
sT
1-e
1 - e-sT
A signal x (t) has a Fourier transform X (w). If x (t) is a real and odd
function of t , then X (w) is
(A) a real and even function of w
(B) a imaginary and odd function of w
(C) an imaginary and even function of w
(D) a real and odd function of w
1999
6.111
TWO MARKS
The Fourier series representation of an impulse train denoted by
s (t) =
3
/ d (t - nT0) is given by
n =- 3
Let u (t) be the step function. Which of the waveforms in the figure
corresponds to the convolution of u (t) - u (t - 1) with u (t) - u (t - 2)
?
3
j2pnt
(A) 1 / exp T0 n =- 3
T0
3
jpnt
(C) 1 / exp
T0 n =- 3
T0
6.112
w = wo
f (l)
1999
6.110
n-3
n-2
n-1
n-4
(B) f (wo), -
0
One period (0, T) each of two periodic waveforms W1 and W2 are shown
in the figure. The magnitudes of the nth Fourier series coefficients of
W1 and W2 , for n $ 1, n odd, are respectively proportional to
(A)
(B)
(C)
(D)
A system has a phase response given by f (w), where w is the
angular frequency. The phase delay and group delay at w = w0 are
respectively given by
3
jpnt
(B) 1 / exp T0 n =- 3
T0
3
j2pnt
(D) 1 / exp
T0 n =- 3
T0
The z -transform of a signal is given by
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
C (z) =
1997
1z-1 (1 - z-4)
4 (1 - z-1) 2
Its final value is
(A) 1/4
(C) 1.0
6.120
(B) zero
(D) infinity
1998
6.113
6.114
6.115
Page 142
ONE MARK
The function f (t) has the Fourier Transform g (w). The Fourier
Transform
ff (t) g (t) e =
(A) 1 f (w)
2p
ONE MARK
If F (s) = 2 w 2 , then the value of Limf (t)
t"3
s +w
(A) cannot be determined
(B) is zero
(C) is unity
(D) is infinite
The trigonometric Fourier series of a even time function can have
only
(A) cosine terms
(B) sine terms
(C) cosine and sine terms
(D) d.c and cosine terms
(C) 2pf (- w)
6.121
A periodic signal x (t) of period T0 is given by
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6.123
-3
(B) 1 f (- w)
2p
(D) None of the above
The Laplace Transform of eat cos (at) is equal to
(s - a)
(s + a)
(A)
(B)
2
2
(s - a) + a
(s - a) 2 + a2
1
(C)
(D) None of the above
(s - a) 2
1996
6.122
3
# g (t) e-jwt dt o is
ONE MARK
The trigonometric Fourier series of an even function of time does
not have the
(A) dc term
(B) cosine terms
(C) sine terms
(D) odd harmonic terms
The Fourier transform of a real valued time signal has
(A) odd symmetry
(B) even symmetry
(C) conjugate symmetry
(D) no symmetry
1, t < T1
x (t) = *
0, T1 < t < T0
2
The dc component of x (t) is
(A) T1
T0
(C) 2T1
T0
6.116
6.117
The unit impulse response of a linear time invariant system is the
unit step function u (t). For t > 0 , the response of the system to an
excitation e-at u (t), a > 0 will be
(A) ae-at
(B) (1/a) (1 - e-at)
(C) a (1 - e-at)
(D) 1 - e-at
The z-transform of the time function
(A) z - 1
z
k=0
(B)
z
z-1
(z - 1) 2
(D)
z
A 22 + A 32 + .....
A 22 + A 32 + .....
(D)
c
m
A1
A 12 + A 22 + A 32 + ....
The Fourier transform of a function x (t) is X (f). The Fourier
dX (t)
transform of
will be
df
dX (f)
(B) j2pfX (f)
(A)
df
X (f)
(D)
(C) jfX (f)
jf
(C)
6.119
3
/ d (n - k) is
z
(z - 1) 2
A distorted sinusoid has the amplitudes A1, A2, A 3, .... of the
fundamental, second harmonic, third harmonic,..... respectively. The
total harmonic distortion is
2
2
(B) A 2 + A 3 + .....
(A) A2 + A 3 + ....
A1
A1
(C)
6.118
(B) T1
2T0
(D) T0
T1
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 143
SOLUTIONS
6.1
Given, the maximum frequency of the band-limited signal
fm = 5 kHz
According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist frequency which is given as
fN = 2fm = 2 # 5 = 10 kHz
So, the sampling frequency fs must satisfy
Option (C) is correct.
If the two systems with impulse response h1 ^ t h and h2 ^ t h are
connected in cascaded configuration as shown in figure, then the
overall response of the system is the convolution of the individual
impulse responses.
fs $ fN
fs $ 10 kHz
only the option (A) doesn’t satisfy the condition therefore, 5 kHz
is not a valid sampling frequency.
6.5
6.2
Option (C) is correct.
Given, the input
x ^ t h = u ^t - 1h
It’s Laplace transform is
-s
X ^s h = e
s
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The impulse response of system is given
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h^t h = t u^t h
Its Laplace transform is
H ^s h = 12
s
Hence, the overall response at the output is
Y ^s h = X ^s h H ^s h
-s
=e3
s
its inverse Laplace transform is
^t - 1h2
y^t h =
u ^t - 1h
2
6.3
system stability and causality both.
But, Option (C) is not true for the stable system as, S = 1
have one pole in right hand plane also.
6.6
Option (A) is correct.
Given, the signal
v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p4 h
So we have
w1 = 100 rad/s
w2 = 300 rad/s
w3 = 500 rad/s
Therefore, the respective time periods are
T1 = 2p = 2p sec
w1 100
T2 = 2p = 2p sec
w2
300
T3 = 2p sec
500
So, the fundamental time period of the signal is
LCM ^2p, 2p, 2ph
L.C.M. ^T1, T2 T3h =
HCF ^100, 300, 500h
or,
T0 = 2p
100
Hence, the fundamental frequency in rad/sec is
w0 = 2p = 100 rad/s
10
6.4
Option (A) is correct.
Option (C) is correct.
For a system to be casual, the R.O.C of system transfer function
H ^s h which is rational should be in the right half plane and to the
right of the right most pole.
For the stability of LTI system. All poles of the system
should lie in the left half of S -plane and no repeated pole should
be on imaginary axis. Hence, options (A), (B), (D) satisfies an LTI
Option (B) is correct.
The Laplace transform of unit step fun n is
U ^s h = 1
s
So, the O/P of the system is given as
Y ^s h = b 1 lb 1 l
s s
= 12
s
For zero initial condition, we check
dy ^ t h
u^t h =
dt
&
U ^s h = SY ^s h - y ^0 h
&
U ^s h = s c 12 m - y ^0 h
s
1
or,
U ^s h =
s
^y ^0 h = 0h
Hence, the O/P is correct which is
Y ^s h = 12
s
its inverse Laplace transform is given by
y ^ t h = tu ^ t h
6.7
No Option is correct.
The matched filter is characterized by a frequency response that is
given as
H ^ f h = G * ^ f h exp ^- j2pfT h
f
where
g^t h
G^f h
Now, consider a filter matched to a known signal g ^ t h. The fourier
transform of the resulting matched filter output g 0 ^ t h will be
G0 ^ f h = H^ f hG^ f h
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 144
ordinary, first order differential equation with forcing function x ^ t h
is y ^ t h so, we can define a function relating x ^ t h and y ^ t h as below
dy
P + Qy + K = x ^ t h
dt
= G * ^ f h G ^ f h exp ^- j2pfT h
= G ^ f h 2 exp ^- j2pfT h
T is duration of g ^ t h
Assume exp ^- j2pfT h = 1
So,
G0 ^ f h = G_ f i 2
Since, the given Gaussian function is
g ^ t h = e- pt
Fourier transform of this signal will be
f
g ^ t h = e- pt
e- pf = G ^ f h
Therefore, output of the matched filter is
2
G 0 ^ f h = e- pf
where P , Q , K are constant. Taking the Laplace transform both
the sides, we get
....(1)
P sY ^s h - Py ^0 h + Q Y ^s h = X ^s h
Now, the solutions becomes
y1 ^ t h =- 2y ^ t h
or,
Y1 ^s h =- 2Y ^s h
So, Eq. (1) changes to
2
2
2
2
6.8
Option (B) is correct.
Given, the impulse response of continuous time system
h ^ t h = d ^t - 1h + d ^t - 3h
From the convolution property, we know
x ^ t h * d ^t - t 0h = x ^t - t 0h
So, for the input
x ^ t h = u ^ t h (Unit step fun n )
The output of the system is obtained as
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y^t h = u^t h * h^t h
= u ^ t h * 6d ^t - 1h + d ^t - 3h@
= u ^t - 1h + u ^t - 3h
at t = 2
6.9
y ^2 h = u ^2 - 1h + u ^2 - 3h
=1
Option (B) is correct.
Given, the differential equation
d2y
dy
2 + 5 dt + 6y ^ t h = x ^ t h
dt
Taking its Laplace transform with zero initial conditions, we have
s2 Y ^s h + 5sY ^s h + 6Y ^s h = X ^s h
....(1)
Now, the input signal is
1 0<t<2
0 otherwise
i.e.,
x ^ t h = u ^ t h - u ^t - 2h
Taking its Laplace transform, we obtain
-2s
X ^s h = 1 - e
s
s
-2s
= 1-e
s
Substituting it in equation (1), we get
2s
X ^s h
Y ^s h = 2
= 21 - e
s + 5s + 6
s ^s + 5s + 6h
x^t h = *
=
6.10
1 - e-2s
s ^s + 2h^s + 3h
Option (D) is correct.
The solution of a system described by a linear, constant coefficient,
P sY1 ^s h - P y1 ^0 h + Q Y1 ^s h = X1 ^s h
or,
....(2)
- 2PSY ^s h - P y1 ^0 h - 2QY1 ^s h = X1 ^s h
Comparing Eq. (1) and (2), we conclude that
X1 ^s h =- 2X ^s h
y1 ^0 h =- 2y ^0 h
Which makes the two equations to be same. Hence, we require to
change the initial condition to - 2y ^0 h and the forcing equation to
- 2x ^ t h
6.11
Option (A) is correct.
Given, the DFT of vector 8a b c dB as
D.F.T. %8a b c dB/ = 8a b g dB
Also, we have
V
R
Sa b c d W
Sd a b c W
...(1)
8p q r sB = 8a b c dBSc d a b W
W
S
Sb c d aW
X
T
For matrix circular convolution, we know
Rh h h VRx V
S 0 2 1WS 0W
x 6n@ * h 6n@ = Sh1 h 0 h2WSx1W
SSh h h WWSSx WW
2
1
0
1
T
XT X
where "x 0, x1, x2, are three point signals for x 6n@ and similarly for
h 6n@, h 0 , h1 and h2 are three point signals. Comparing this transformation to Eq(1), we get
VT
R
Sa d c W
Sb a d W
6p q r s@ = Sc b aW 8a b c dB
W
S
Sd c b W
X
T
= 6a b c d @T * 6a b c d @T
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Now, we know that
So,
R V
R V
Sa W
Sa W
Sb W
Sb W
=S W * S W
Sc W
Sc W
Sd W
Sd W
T X
T X
x1 6n@ * x2 6n@ = X1DFT 6k @ X2, DFT 6k @
R V
Sa W
Sb W
Sc W
S W
Sd W
T X
*
R V R V
Sa W SaW
Sb W SbW
Sc W = Sg W
S W S W
Sd W Sd W
T X T X
*
R V
SaW
SbW
Sg W
S W
Sd W
T X
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 145
= 9a2 b 2 g2 d 2C
6.12
Option (D) is correct.
Using s -domain differentiation property of Laplace transform.
If
f (t)
F (s)
dF (s)
L
tf (t)
ds
2s + 1
=
L [tf (t)] = - d ; 2 1
ds s + s + 1E (s2 + s + 1) 2
So,
6.13
L
h (0) = h1 (0) + h2 (0) = 1 + 1 = 1
2 2
Option (C) is correct.
n
n
x [ n] = b 1 l - b 1 l u [ n ]
3
2
n
-n
n
x [n] = b 1 l u [n] + b 1 l u [- n - 1] - b 1 l u (n)
3
3
2
Taking z -transform
X 6z @
3
3
So, inverse Fourier transform of H (jw)
h (t) = h1 (t) + h2 (t)
6.16
y [n] = h [n] * g [n] =
For causal sequence,
3
1 n -n
1 -n -n
1 n -n
=
b 3 l z u [n ] +
b 3 l z u [- n - 1] b 2 l z u [ n]
n =- 3
n =- 3
n =- 3
/
/
/
-1
3
=
3
/
3
/
/ b 31z l + / b 13 z l
n
n=0
m
-
m=1
/ b 21z l
n
Taking m =- n
n=0
1 44 2
44 3
II
14
42
4
43
I
3
14
42
4
43
III
1 < 1 or z > 1
3
3z
Series II converges if 1 z < 1 or z < 3
3
Series III converges if 1 < 1 or z > 1
2z
2
Region of convergence of X (z) will be intersection of above three
So,
ROC : 1 < z < 3
2
Option (D) is correct.
y (t) =
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For n = 0 ,
For n = 1,
# x (t) cos (3t) dt
-3
# d (t) cos (3t) dt
-3
y (t, t 0) =
Delayed output,
From equation (i),
# d (t - t ) cos (3t) dt
0
-3
y (t - t 0) = u (t - t 0)
y (t, t 0) ! y (t - t 0)
= u (t) cos (3t 0)
#
As t " 3, y (t) " 3 (unbounded)
6.15
So,
6.17
System is not time invariant.
Stability :
Consider a bounded input x (t) = cos 3t
t
t
1 - cos 6t = 1
y (t) =
cos2 3t =
2
2
-3
-3
#
y [1]
y [1]
1
2
= u (t) cos (0) = u (t)
For a delayed input (t - t 0) output is
t
t
-3
-3
y [0] = h [0] g [0] + h [1] g [- 1] + ...........
y [0] = h [0] g [0] g [- 1] = g [- 2] = ....0
...(i)
y [0] = h [0] g [0]
= h [1] g [1] + h [1] g [0] + h [1] g [- 1] + ....
= h [1] g [1] + h [1] g [0]
1
= 1 g [1] + 1 g [0] h [1] = b 1 l = 1
2
2
2
2
1 = g [1] + g [0]
g [1] = 1 - g [0]
y [0] 1
= =1
g [0] =
h [0] 1
g [1] = 1 - 1 = 0
Option (A) is correct.
d2 y
dy
We have
100 2 - 20 + y = x (t)
dt
dt
Applying Laplace transform we get
100s2 Y (s) - 20sY (s) + Y (s) = X (s)
Y (s)
1
or
=
H (s) =
X (s) 100s2 - 20s + 1
1/100
A
= 2
=
s - (1/5) s + 1/100 s2 + 2xwn s + w2
Here wn = 1/10 and 2xwn =- 1/5 giving x =- 1
Roots are s = 1/10, 1/10 which lie on Right side of s plane thus
unstable.
# 1dt - 12 # cos 6t dt
t
/ h [n] g [n - k]
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Time Invariance :
Let,
x (t) = d (t)
y (t) =
3
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t
t
/ h [n] g [n - k]
k =- 3
y [n] = h [n] g [n] + h [n] g [n - 1] + h [n] g [n - 2] + .....
Series I converges if
6.14
y [n] =
3
k=0
3
1 n -n
1 -n -n
1 n -n
=
b3l z +
b3l z b2l z
n=0
n =- 3
n=0
/
Option (A) is correct.
Convolution sum is defined as
System is not stable.
Option (C) is correct.
(2 cos w) (sin 2w)
H (jw) =
= sin 3w + sin w
w
w
w
We know that inverse Fourier transform of sin c function is a
rectangular function.
6.18
6.19
Option (C) is correct.
For an even function Fourier series contains dc term and cosine term
(even and odd harmonics).
Option (B) is correct.
Function h (n) = an u (n) stable if a < 1 and Unstable if a H 1
We We have
h (n)
= 2n u (n - 2);
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 146
Here a = 2 therefore h (n) is unstable and since h (n) = 0 for n < 0
Therefore h (n) will be causal. So h (n) is causal and not stable.
6.20
Constant term is negative.
6.25
Option (A) is correct.
6.21
6.26
Option (D) is correct.
We have
x (t) = exp (- 2t) m (t) + s (t - 6) and h (t) = u (t)
Taking Laplace Transform we get
X (s) = b 1 + e-6s l and H (s) = 1
s+2
s
Now
or
Thus
Y (s) = H (s) X (s)
-6s
1
= 1 : 1 + e-6sD =
+e
s s+2
s
s (s + 2)
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or
Now
6.27
6.28
6.23
6.24
x (t) = Ao +
Where,
3
/ [An cos nwt + Bn sin nwt]
n=1
Ao 1 # x (t) dt
T0 T
T0 "fundamental period
h [n] = d [n - 3]
Option (D) is correct.
For an N-point FET algorithm butterfly operates on one pair of
samples and involves two complex addition and one complex
multiplication.
Option (D) is correct.
3s + 1
f (t) = L - 1 ; 3
s + 4s 2 + (k - 3) s E
lim f (t) = 1
t"3
By final value theorem
lim f (t) = lim sF (s) = 1
t"3
-1
Option (C) is correct.
For a function x (t) trigonometric fourier series is
X (z) = 5z + 4z + 3
x [n] = 5d [n + 2] + 4d [n - 1] + 3d [n]
=z
and
Y (z)
= H (z) = z-1
X (z)
Option (B) is correct.
For 8 point DFT, x* [1] = x [7]; x* [2] = x [6]; x* [3] = x [5] and it is
conjugate symmetric about x [4], x [6] = 0 ; x [7] = 1 + j3
ad [n ! a]
-1
-3
We have
y (n) = x (n - 1)
Y (z) = z-1 X (z)
H1 (z) H2 (z) = z
1 - 0.4z-1
-1
c 1 - 0.6z-1 m H2 (z) = z
z-1 (1 - 0.6z-1)
H2 (z) =
(1 - 0.4z-1)
-2
= z :z
or,
6.29
s"0
or
s. (3s + 1)
=1
s + 4s2 + (k - 3) s
s (3s + 1)
lim 2
=1
s " 0 s [s + 4s + (k - 3)]
1 =1
k-3
or
k =4
Option (B) is correct.
or
2
H (z ) = H1 (z ) : H2 (z )
or
6.22
Inverse Z - transform
Option (C) is correct.
We have
h1 [n] = d [n - 1] or H1 [Z ] = Z - 1
and
h 2 [n] = d [n - 2] or H2 (Z ) = Z - 2
Response of cascaded system
-1
-6s
1
+e
Y (s) = 1 2s 2 (s + 2)
s
y (t) = 0.5 [1 - exp (- 2t)] u (t) + u (t - 6)
aZ ! a
We know that
Given that
Inverse z-transform
Impulse response = d (step response)
dt
= d (1 - e- at)
dt
= 0 + ae- at = ae- at
Option (A) is correct.
lim
s"0
3
Option (B) is correct.
System is described as
d 2 y (t)
dt (t)
dx (t)
+ 4x (t)
+4
+ 3y (t) = 2
2
dt
dt
dt
Taking Laplace transform on both side of given equation
s 2 Y (s) + 4sY (s) + 3Y (s) = 2sX (s) + 4X (s)
(s 2 + 4s + 3) Y (s) = 2 (s + 2) X (s) s
Transfer function of the system
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0
An = 2 # x (t) cos nwt dt
T0 T
Bn = 2 # x (t) sin nwt dt
T0 T
For an even function x (t), Bn = 0
Since given function is even function so coefficient Bn = 0 , only cosine
and constant terms are present in its fourier series representation
3T/4
Constant term
A0 = 1 #
x (t) dt
T -T/4
T/4
3T/4
= 1 : # Adt + #
- 2AdtD
T -T/4
T/4
= 1 :TA - 2AT D =- A
2
2
T 2
and
H (s) =
0
0
Input
or,
2 (s + 2)
Y (s)
2 (s + 2)
=
=
(s + 3) (s + 1)
X (s) s 2 + 4s + 3
x (t) = e-2t u (t)
X (s) = 1
(s + 2)
Y (s) = H (s) : X (s)
2 (s + 2)
Y (s) =
: 1
(s + 3) (s + 1) (s + 2)
By Partial fraction
Y (s) = 1 - 1
s+1 s+3
Taking inverse Laplace transform
Output
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 147
y (t) = (e-t - e-3t) u (t)
6.30
Option (C) is correct.
We have
2 - 34 z - 1
1 - 34 z - 1 + 18 z - 2
By partial fraction H (z ) can be written as
1
1
H (z ) =
-1
1 -1 +
^1 - 2 z h ^1 - 14 z h
For ROC : z > 1/2
H (z) =
It has the finite magnitude values. So it is a finite impulse response
filter. Thus S2 is true but it is not a low pass filter. So S1 is false.
6.36
n
n
h [n] = b 1 l u [n] + b 1 l u [n], n > 0
2
4
1
= an u [n], z > a
1 - z -1
Thus system is causal. Since ROC of H (z ) includes unit circle, so
it is stable also. Hence S1 is True
For ROC : z < 1
4
n
n
h [n] =-b 1 l u [- n - 1] + b 1 l u (n), z > 1 , z < 1
2
2
4
4
System is not causal. ROC of H (z ) does not include unity circle,
so it is not stable and S 3 is True
6.31
6.32
Option (A) is correct.
The Fourier series of a real periodic function has only cosine terms
if it is even and sine terms if it is odd.
6.37
Option (B) is correct.
Here h (t) ! 0 for t < 0 . Thus system is non causal. Again any
bounded input x (t) gives bounded output y (t). Thus it is BIBO
stable.
Here we can conclude that option (B) is correct.
Option (D) is correct.
We have
x [n] = {1, 0, 2, 3) and N = 4
X [k ] =
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For N = 4 ,
n =- 1
n
n=0
n=3
Now
n =- 3
n =- 1
1 z-1 < 1 "
3
1 z-1 > 1 "
2
First term gives
Second term gives
#
Here function is defined for 0 < t < t , Thus
t
F (s)
L
f (t)
s
0
#
6.35
Option (A) is correct.
We have h (2) = 1, h (3) =- 1 otherwise h (k) = 0 . The diagram of
response is as follows :
3
/ x [ n]
3
/ x [n] e
-jpn/2
n=0
= x [0] + x [1] e-jp/2 + x [2] e-jp + x [3] e-jp3/2
= 1 + 0 - 2 + j3 =- 1 + j3
X [2] =
3
/ x [n] e
-jpn
n=0
= x [0] + x [1] e-jp + x [2] e-j2p + x [3] e-jp3
= 1+0+2-3 = 0
1< z
3
1> z
2
#
X [0] =
x [1] =
n
-
k = 0, 1,... 3
= 1+0+2+3 = 6
1 n -n
b2l z
Option (B) is correct.
By property of unilateral Laplace transform
t
F (s) 1 0
L
f (t) dt
f (t) dt
+
s
s -3
-3
-j2pnk/4
n=0
X [3] =
3
/ x [n] e
-j3pn/2
n=0
Thus its ROC is the common ROC of both terms. that is
1< z <1
3
2
6.34
3
/ x [n] e
= x [0] + x [1] + x [2] + x [3]
= / b 1 z-1 l - / b 1 z-1 l
3
2
n=0
n =- 3
n
X [k ] =
n=0
Taking z transform we have
/ b 13 l z-n - /
k = 0, 1...N - 1
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The function has a DC term and a cosine function. The frequency
of cosine terms is
w = 2 = 2pf " f = 1 Hz
p
The given function has frequency component at 0 and 1 Hz.
p
Option (A) is correct.
n
n
x [n] = b 1 l u (n) - b 1 l u (- n - 1)
3
2
n=3
-j2pnk/N
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Option (B) is correct.
Given function is
X (z) =
/ x [n] e
n=0
f (t) = sin2 t + cos 2t = 1 - cos 2t + cos 2t = 1 + 1 cos 2t
2
2 2
6.33
N-1
= x [0] + x [1] e-j3p/2 + x [2] e-j3p + x [3] e-j9p/2
= 1 + 0 - 2 - j3 =- 1 - j3
Thus [6, - 1 + j3, 0, - 1 - j3]
6.38
6.39
Option (A) is correct.
Option (C) is correct.
The output of causal system depends only on present and past states
only.
In option (A) y (0) depends on x (- 2) and x (4).
In option (B) y (0) depends on x (1).
In option (C) y (0) depends on x (- 1).
In option (D) y (0) depends on x (5).
Thus only in option (C) the value of y (t) at t = 0 depends on x (- 1)
past value. In all other option present value depends on future value.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
6.40
6.41
Page 148
The y (t) = t x (t) is not linear, thus option (B) is wrong and (a) is
correct. We can see that
R1: y (t) = t2 x (t) Linear and time variant.
R2: y (t) = t x (t) Non linear and time variant.
R3: y (t) = x (t) Non linear and time invariant
R4: y (t) = x (t - 5) Linear and time invariant
Option (D) is correct.
We have
h (t) = eat u (t) + e bt u (- t)
This system is stable only when bounded input has bounded output For stability at < 0 for t > 0 that implies a < 0 and bt > 0
for t > 0 that implies b > 0 . Thus, a is negative and b is positive.
Option (C) is correct.
K (s + 1)
, and R (s) = 1
G (s) =
(s + 2)( s + 4)
s
K (s + 1)
C (s) = G (s) R (s) =
s (s + 2)( s + 4)
K
= K +
- 3K
8s 4 (s + 2) 8 (s + 4)
c (t) = K :1 + 1 e-2t - 3 e-4tD u (t)
8 4
8
Thus
6.45
Given :
Hence
6.46
1
x (t) = )
0
Fourier transform is
#- 33e-jwt x (t) dt
=
for - 1 # t # + 1
otherwise
x (n) = 5e
6.47
6.48
Option (C) is correct.
=
6.49
#- 33 h (t) e-jwt dt
#0 3e-2t e-jwt dt = #0 3e-(2 + jw)t dt
=
1
(2 + jw)
Option (D) is correct.
H (jw) =
1
(2 + jw)
The phase response at w = 2 rad/sec is
+H (jw) =- tan-1 w =- tan-1 2 =- p =- 0.25p
2
2
4
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Magnitude response at w = 2 rad/sec is
1
= 1
H (jw) =
22 + w2
2 2
Input is
x (t) = 2 cos (2t)
Output is
= 1 # 2 cos (2t - 0.25p)
2 2
= 1 cos [2t - 0.25p]
2
3
/ x (k) h (n - k)
3
/ x (k) h (2 - k)
k =- 3
Option (B) is correct.
Mode function are not linear. Thus y (t) = x (t) is not linear but
this functions is time invariant. Option (A) and (B) may be correct.
= 5e-0.05n For t > 0
Option (C) is correct.
Since
x (n) = 5e-0.05n u (n) is a causal signal
Its z transform is
1
5z
=
X (z) = 5 :
1 - e-0.05 z-1 D z - e-0.05
Its ROC is e-0.05 z-1 > 1 " z > e-0.05
H (jw) =
= x (0) h (2 - 0) + x (1) h (2 - 1) + x (2) h (2 - 2)
= h (2) + 0 + h (0) = 1 + 2 = 3
There are 5 non zero sample in output sequence and the value of
y [2] is 3.
-n
2 # 10
h (t) = e-2t u (t)
k =- 3
6.44
I = I (0+) e-t/RC
I (0+) = V = 5 = 25mA
R
200k
t
2
Option (D) is correct.
Given
h (n) = [1, - 1, 2]
x (n) = [1, 0, 1]
y (n) = x (n)* h (n)
The length of y [n] is = L1 + L2 - 1 = 3 + 3 - 1 = 5
y (2) =
X (k) 2
= VR # R = 5e- V
Here the voltages across the resistor is input to sampler at frequency of 10 Hz. Thus
1
y (n) = x (n) * h (n) =
DFT
t
2
This is zero at w = p and w = 2p
6.43
y (n) = rxx (n)
I = 25e- m A
#-1 e-jwt 1dt
= 1 [e-jwt]-11
- jw
= 1 (e-jw - e jw) = 1 (- 2j sin w)
- jw
- jw
= 2 sin w
w
r=0
RC = 200k # 10m = 2 sec
Option (A) is correct.
We have
N-1
/x (r) x (n + r)
Option (B) is correct.
Current through resistor (i.e. capacitor) is
Here,
h (t) = L-1 G (s) = (- 4e-2t + 12e-4t) u (t)
6.42
y (n) = 1
N
It is Auto correlation.
At steady-state , c (3) = 1
K = 1 or K = 8
Thus
8
8 (s + 1)
Then,
= 12 - 4
G (s) =
(s + 4) (s + 2)
(s + 2)( s + 4)
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Option (A) is correct.
6.50
Option (D) is correct.
Y (s) =
1
s (s - 1)
Final value theorem is applicable only when all poles of system lies
in left half of S -plane. Here s = 1 is right s -plane pole. Thus it is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 149
a t = 0 . The area of dirac delta function is unity.
unbounded.
6.51
Option (A) is correct.
6.58
-t
x (t) = e u (t)
Taking Fourier transform
X (jw) = 1
1 + jw
X (jw) = 1 2
1+w
Magnitude at 3dB frequency is
1
2
1 =
1
2
1 + w2
w = 1 rad
f = 1 Hz
2p
Thus
or
or
6.52
6.59
Option (B) is correct.
For discrete time Fourier transform (DTFT) when N " 3
p
x [ n] = 1
X (e jw) e jwn dw
2p - p
6.60
Option (D) is correct.
The ROC of addition or subtraction of two functions x1 (n) and x2 (n)
is R1 + R2 . We have been given ROC of addition of two function and
has been asked ROC of subtraction of two function. It will be same.
Option (A) is correct.
As we have
x (t) = sin t ,
Now
H (s) = 1
s+1
or
H (jw) = 1 = 1
jw + 1
j+1
or
H (jw) = 1 + - 45c
2
Thus
y (t) = 1 sin (t - p4 )
2
Option (C) is correct.
F (s) =
#
Putting n = 0 we get
= 1
2p
#
p
-p
x [0] = 1
2p
X (e jw) dw
or
6.53
#
p
-p
#
p
-p
X (e jw) e jw0 dw
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X (e ) dw = 2px [0] = 2p # 5 = 10p
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Option (B) is correct.
0.5
1 - 2z-1
Since ROC includes unit circle, it is left handed system
x (n) =- (0.5) (2) -n u (- n - 1)
x (0) = 0
If we apply initial value theorem
x (0) = lim X (z) = lim 0.5 -1 = 0.5
z"3
z " 31 - 2z
That is wrong because here initial value theorem is not applicable
because signal x (n) is defined for n < 0 .
6.55
L-1 F (s) = sin wo t
f (t) = sin wo t
Thus the final value is - 1 # f (3) # 1
6.61
Let
Now
6.62
Option (A) is correct.
A Hilbert transformer is a non-linear system.
H (s)
6.57
Step response
Y (s)
or
Y (s)
or
y (t)
x (n) = d (n)
y (n) = sin 0 = 0 (bounded)
BIBO stable
Option (B) is correct.
c (t) = 1 - e-2t
Taking Laplace transform
C (s)
2
=
C (s) =
#s = 2
s (s + 2)
s+2
U (s)
Option (B) is correct.
5
1 + j10pf
5
= 5 =
= 1
1 + 5s
5^s + 15 h s +
=1 a 1
s ^s + 5 h
=1 11 =5- 51
s ^s + 5 h
s s+ 5
-t/5
= 5 (1 - e ) u (t)
Option (C) is correct.
y (n) = b sin 5 pn l x (n)
6
H (f) =
6.56
w0
s + w2
2
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X (z) =
6.54
thus w = 1
6.63
1
s+1
L
x (t) = u (t)
X (s) = 1
s
Y (s) = H (s) X (s) = 1 # 1 = 1 - 1
s+1 s
s s+1
h (t) = e-t
1
5
Option (A) is correct.
F
x (t)
X (jw)
Using scaling we have
F
1 X jw
x (5t)
5 c 5 m
Using shifting property we get
F
1 X jw e- j35w
x ;5 bt - 3 lE
5
5 b5l
Option (D) is correct.
Dirac delta function d (t) is defined at t = 0 and it has infinite value
Option (C) is correct.
L
H (s) =
y (t) = u (t) - e-t
In steady state i.e. t " 3, y (3) = 1
6.64
6.65
Option (C) is correct.
Fourier series is defined for periodic function and constant.
3 sin (25t) is a periodic function.
4 cos (20t + 3) + 2 sin (710t) is sum of two periodic function and also
a periodic function.
e- t sin (25t) is not a periodic function, so FS can’t be defined for it.
1 is constant
Option (A) is correct.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
g (t) + g (- t)
2
g (t) - g (- t)
odd{g (t)} =
2
6.70
Ev{g (t)} =
Here
Thus
Page 150
g (t) = u (t)
u (t) + u (- t)
=
ue (t) =
2
u (t) - u (- t)
=
uo (t) =
2
Option (A) is correct.
y (t)
= 0.5x (t - td + T) + x (t - td ) + 0.5x (t - td - T)
Taking Fourier transform we have
Y (w)
1
2
x (t)
2
= 0.5e
-jw (- td + T)
or
X (w) + e-jwt X (w) + 0.5e-jw (- t - T) X (w)
Y (w)
= e-jwt [0.5e jwT + 1 + 0.5e-jwT ]
X (w)
d
d
d
= e-jwt [0.5 (e jwT + e-jwT ) + 1]
d
6.66
Option (C) is correct.
Here
x1 (n) = ` 5 jn u (n)
6
1
X1 (z) =
1 - ^ 65 z-1h
= e-jwt [cos wT + 1]
d
6.71
x2 (n) =-` 6 jn u (- n - 1)
5
1
X1 (z) = 1 1 - ^ 65 z-1h
ROC : R2 " z < 6
5
Thus ROC of x1 (n) + x2 (n) is R1 + R2 which is 5 < z < 6
6
5
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6.67
6.68
6.72
Option (C) is correct.
For continuous and aperiodic signal Fourier representation is
continuous and aperiodic.
For continuous and periodic signal Fourier representation is discrete
and aperiodic.
For discrete and aperiodic signal Fourier representation is continuous
and periodic.
For discrete and periodic signal Fourier representation is discrete
and periodic.
Option (B) is correct.
Y (e jw) = Ae-jw n X (e jw)
Y (e jw)
or
= Ae-jw n
H (e jw) =
X (e jw)
Thus
+H (e jw) =- wo no
For LTI discrete time system phase and frequency of H (e jw) are
periodic with period 2p. So in general form
o
n
x (n) = b 1 l u (n)
2
2n
Y (e j0) =
n
n=3
/ ` 14 j
n=0
or
Y (e j0) =
1
1-
1
4
=
n=3
...(1)
/ b 14 l e-jwn
n
n=0
n =- 3
or
6.73
2 n
n
y (n) = ;b 1 l E u (n) = b 1 l u (n)
2
4
/ y (n) e-jwn
1
3
4
= 1 +b1l +b1l+b1l +b1l
4
4
4
4
=4
3
n=2
n = 3,
n=4
n = 5,
Option (A) is correct.
s (t) = 8 cos ` p - 20pt j + 4 sin 15pt
2
= 8 sin 20pt + 4 sin 15pt
Here A1 = 8 and A2 = 4 . Thus power is
2
2
2
2
P = A1 + A2 = 8 + 4 = 40
2
2
2
2
Option (A) is correct.
From
x (n) = [ 12 , 1, 2, 1, 1, 12 ]
y (n) = x ^ n2 - 1h, n even
= 0 , for n odd
n =- 2 ,
y (- 2) = x ( -22 - 1) = x (- 2) = 12
n =- 1,
y (- 1) = 0
n = 0,
y (0) = x ( 20 - 1) = x (- 1) = 1
n = 1,
y (1) = 0
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Alternative :
Taking z transform of (1) we get
1
Y (z) =
1 - 14 z-1
Substituting z = e jw we have
1
Y (e jw) =
1 - 14 e-jw
Y (e j0) = 1 1 = 4
3
1- 4
6.69
o
q (w) =- no wo + 2pk
y (n) = x2 (n) = b 1 l u2 (n)
2
n=3
o
o
Option (D) is correct.
Y (e jw) =
d
y (n) = Ax (n - no)
Taking Fourier transform
Option (D) is correct.
For causal system h (t) = 0 for t # 0 . Only (D) satisfy this condition.
or
Y (w)
= e-jwt (cos wT + 1)
X (w)
H (w) =
or
ROC : R1 " z > 5
6
6.74
y (2)
y (3)
y (4)
y (5)
y (6)
= x ( 22 - 1) = x (0) = 2
=0
= x ( 24 - 1) = x (1) = 1
=0
= x ( 26 - 1) = x (2) = 12
n=6
Hence
y (n)
= 1 d (n + 2) + d (n) + 2d (n - 2) + d (n - 4) + 1 d (n - 6)
2
2
Option (C) is correct.
Here y (n) is scaled and shifted version of x (n) and again y (2n) is
scaled version of y (n) giving
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 151
z (n) = y (2n) = x (n - 1)
= 1 d (n + 1) + d (n) + 2d (n - 1) + d (n - 2) + 1 d (n - 3)
2
2
Taking Fourier transform.
Z (e jw) = 1 e jw + 1 + 2e-jw + e-2jw + 1 e-3jw
2
2
= e-jw b 1 e2jw + e jw + 2 + e-jw + 1 e-2jw l
2
2
=e
6.80
Option (B) is correct.
F
x (at)
F
xb 1 f l
3
F
X (f)
Using scaling we have
Thus
1 X f
a ca m
3X (3f)
1e
3
Thus
1 e-jp
3
6.76
x (t) = X (f + f0)
F
X (3f + 2)
xb 1 t l
3
F
t
xb 1 t l
3X (3 (f + 23 ))
3
F
t
xb 1 t l
X [3 (f + 23 )]
3
-j 4 pt
3
2
3
4
3
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Option (A) is correct.
3
/
A system is stable if
a <2
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-j2pf0 t
e-j2p
-
Option (C) is correct.
We have
2y (n) = ay (n - 2) - 2x (n) + bx (n - 1)
Taking z transform we get
or
Using shifting property we get
e
4 - j 25 ]
2j
2Y (z) = aY (z) z-2 - 2X (z) + bX (z) z-1
Y (z)
bz-1 - 2
or
...(i)
=c
m
X (z)
2 - az-2
z ( b - z)
or
H (z) = 22 a
(z - 2 )
It has poles at ! a/2 and zero at 0 and b/2 . For a stable system
poles must lie inside the unit circle of z plane. Thus
a <1
2
e2jw + e-2jw + e jw + 2 + e-jw
b
l
2
x (t)
x (n) - x* (- n)
2
= [- 4 - j 25 ,
Z (e jw) = e-jw [cos 2w + 2 cos w + 2]
or
6.75
-jw
xcas (n) =
h (n) < 3. The plot of given h (n) is
n =- 3
But zero can lie anywhere in plane. Thus, b can be of any value.
6.81
Option (D) is correct.
x (n) = e jpn/4
h (n)
We have
and
= 4 2 d (n + 2) - 2 2 d (n + 1) - 2 2 d (n - 1)
+ 4 2 d (n - 2)
y (n) = x (n)* h (n)
Now
3
/
Thus
6
/
h (n) =
=
h (n)
z
H (z) =
z - 0.2
z < 0.2
6.79
p
4
1
1 - az-1
h [n] =- (0.2) n u [- n - 1]
Option (A) is correct.
We have
x (n) = [- 4 - j5, 1 + 2j,
4]
1 - 2j, - 4 + j5]
-
+ ej
p
4
(n + 1)
- 2 2 ej
@ - 2 2 6e j
(n - 2)
p
4
p
4
(n - 1)
(n + 1)
+ 4 2 ej
+ ej
p
4
p
4
(n - 2)
@
(n - 1)
p
2
p
2
p
2
p
4
p
4
r
4
6.82
Option (B) is correct.
From given graph the relation in x (t) and y (t) is
y (t) =- x [2 (t + 1)]
x (t)
F
x (at)
F
x (2t)
F
X (f)
Using scaling we have
x *( n) = [- 4 + j5, 1 - 2j, 4]
x *( - n) = [4,
(n + 2)
p
4
p
4
-
- 2 2 ej
= 4 2 e j n 6e j + e-j @ - 2 2 e j n 6e j + e-j @
= 4 2 e j n [0] - 2 2 e j n [2 cos p4 ]
or
y (n) =- 4e j n
z <a
Option (C) is correct.
The Fourier transform of a conjugate symmetrical function is always
real.
(n + 2)
p
4
p
4
- an u [- n - 1] *
Thus
y (n) = x (n + 2) h (- 2) + x (n + 1) h (- 1)
+ x (n - 1) h (1) + x (n - 2) h (2)
= 4 2 6e j
Option (D) is correct.
We know that
6.78
= 4 2 ej
/ x (n - k) h (k)
k =- 2
or
= 1+1+1+1+2+2+2+2+2
= 15 < 3
Hence system is stable but h (n) ! 0 for n < 0 . Thus it is not
causal.
6.77
/
2
x (n - k) h (k) =
k =- 3
n =- 3
n =- 3
3
Thus
Using shifting property we get
1 X f
a ca m
1X f
2 c2m
p
4
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 152
x (t - t0) = e-j2pft X (f)
0
x [2 (t + 1)]
F
- x [2 (t + 1)]
F
Thus
6.83
or
or
or
Option (C) is correct.
From the Final value theorem we have
lim i (t) = lim sI (s) = lim s
t"3
6.84
Now
j2pf
f
f
e-j2pf (- 1) 1 X b l = e X b l
2 2
2
2
j2p f
f
-e
Xc m
2
2
s"0
s"0
2
= lim 2
=2
s (1 + s) s " 0 (1 + s)
or
Option (D) is correct.
Here
C3 = 3 + j5
For real periodic signal
or
or
Ck*
Thus
6.85
C-k =
C-3 = Ck = 3 - j5
or
Thus
Option (C) is correct.
6.89
y (t) = 4x (t - 2)
Taking Fourier transform we get
Y (e j2pf ) = 4e-j2pf2 X (e j2pf )
Thus
6.86
Y (e j2pf )
= 4e-4jpf
X (e j2pf )
H (e j2pf ) = 4e-4jpf
Option (B) is correct.
We have
h (n) = 3d (n - 3)
or
Taking z transform
H (z) = 2z-3
4
2
-4
X (z) = z + z - 2z + 2 - 3z
Now
Y (z) = H (z) X (z)
= 2z-3 (z 4 + z2 - 2z + 2 - 3z-4)
= 2 (z + z-1 - 2z-2 + 2z-3 - 3z-7)
Taking inverse z transform we have
6.88
Option (C) is correct.
The frequency response of RC-LPF is
1
H (f) =
1 + j2pfRC
1
1 + jwRC
q (w) =- tan-1 wRC
dq (w)
RC
tg ==
dw
1 + w2 R2 C2
10-3
=
= 0.717 ms
2
1 + 4p # 10 4 # 10-6
6.90
6.91
Option (C) is correct.
If
x (t)* h (t) = g (t)
Then
x (t - t1)* h (t - t2) = y (t - t1 - t2)
Thus
x (t + 5)* d (t - 7) = x (t + 5 - 7) = x (t - 2)
Option (B) is correct.
In option (B) the given function is not periodic and does not satisfy
Dirichlet condition. So it cant be expansion in Fourier series.
x (t) = 2 cos pt + 7 cos t
T1 = 2p = 2
w
T2 = 2p = 2p
1
T1 = 1 = irrational
T2
p
6.92
Option (C) is correct.
From the duality property of fourier transform we have
If
Then
x (t)
FT
X (f)
X (t)
FT
x (- f)
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Option (A) is correct.
System is non causal because output depends on future value
For n # 1
y (- 1) = x (- 1 + 1) = x (0)
Time varying
y (n - n0) = x (n - n0 + 1)
Depends on Future
y (n) = x (n + 1)
i.e.
None causal
y (1) = x (2)
For bounded input, system has bounded output. So it is stable.
y (n) = x (n) for n $ 1
= 0 for n = 0
= x (x + 1) for n #- 1
So system is linear.
f1 max
f1 # 52.2 Hz
= 52.2 Hz
Option (A) is correct.
Time Shifting property
y (n) = 2[ d (n + 1) + d (n - 1) - 2d (n - 2)
+ 2d (n - 3) - 3d (n - 7)]
At n = 4 ,
y (4) = 0
6.87
1 + 4p2 f12 R2 C2 # 1.108
4p2 f12 R2 C2 # 0.108
2pf1 RC # 0.329
f1 # 0.329
2pRC
f1 # 0.329
2pRC
f1 # 0.329
2p1k # 1m
H (w) =
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or
H (0) = 1
H (f1)
1
=
$ 0.95
H (0)
1 + 4p2 f12 R2 C2
Therefore if
Then
6.93
e-t u (t)
1
1 + j2pt
FT
1
1 + j2pf
FT
e f u (- f)
Option (A) is correct.
and
Thus
q (w) =- wt0
- q (w)
= t0
tp =
w
dq (w)
= t0
tg =dw
tp = tg = t0 = constant
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
6.94
Option (*) is correct.
5-s
= -2 + 1
X (s) = 2 5 - s =
s+1 s-2
(
s
1
)(
s
2
)
+
s -s-2
Here three ROC may be possible.
Re (s) < - 1
Re (s) > 2
- 1 < Re (s) < 2
Since its Fourier transform exits, only - 1 < Re (s) < 2 include
imaginary axis. so this ROC is possible. For this ROC the inverse
Laplace transform is
Page 153
and
Option (B) is correct.
For left sided sequence we have
z
Thus
- 5n u (- n - 1)
z
or
- 5n u (- n - 1)
z
1
1 - az-1
1
1 - 5z-1
z
z-5
6.100
where z < 5
Since ROC is z < 5 and it include unit circle, system is stable.
Alternative :
h (n) =- 5n u (- n - 1)
3
/ h (n) z
H (z) =
-n
-1
/-5 z
=
n =- 3
n -n
-1
/ (5z
=-
n =- 3
-3
/ (5z
-1 -m
)
= 1-
n =- 1
3
/ (5
-1
6.96
)
5-1 z < 1 or z < 5
Thus
6.102
6.103
x1 (t)* x2 (t)
3
/ x (n) .z
-n
=
3
/ 1.z
-n
n=0
n =- 3
=
3
/ (z
-1 n
)
n=0
n=0
Option (A) is correct.
6.104
# d (t) = 1
3
6.99
# d (t) x (t) = # x (0) d (t) dt
3
-3
-3
Thus system is linear
Option (A) is correct.
We have
h (t) = e2t
LS
and
x (t) = e3t
LS
-3
Let x (t) = cos ( 23 t), then x (0) = 1
3
Option (B) is correct.
Let
x (t) = ax1 (t) + bx2 (t)
ay1 (t) = atx1 (t)
by2 (t) = btx2 (t)
Adding above both equation we have
y (t - t0) = (t - t0) x (t - t0)
which is not equal. Thus system is time varying.
and this is possible when z-1 < 1. Thus ROC is z-1 < 1 or
z >1
We know that d (t) x (t) = x (0) d (t) and
p2 f2
a
FT
yd (d) = tx (t - t0)
If output is delayed then we have
3
/ (z-1) n < 3
Now
Option (B) is correct.
Since normalized Gaussion function have Gaussion FT
p eThus
e-at
a
ay1 (t) + by2 (t) = atx1 (t) + btx2 (t)
= t [ax1 (t) + bx2 (t)]
= tx (t)
or
ay1 (t) + by2 (t) = y (t)
If input is delayed then we have
H (z) is convergent if
6.98
f (t)* g (t) = h (t)
H (s) = F (s) # G (s)
s2 + 1
= 1
H (s) = s2+ 2 #
+
+
(
s
2
)(
s
3
)
s+3
s +1
2
Option (A) is correct.
We have
h (n) = u (n)
H (z) =
3
L
L
1 # 1
(t * e2t) u (t)
2
s
2
s
Here we have used property that convolution in time domain is
multiplication in s - domain
6.97
[f (p)] 2 dp = E
2
-3
#
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m=0
LT
-3
dp 1
=
2
2
[f (p)] 2
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-1 n
Option (B) is correct.
1
= 12 # 1
2
s-2
s
s (s - 2)
X1 (s) X2 (s)
3
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z) -m
1
,
1 - 5-1 z
= 1- 5 = z
z-5
5-z
#
Option (B) is correct.
n =- 3
= 1-
[f (2t)] 2 dt
Option (B) is correct.
Since h1 (t) ! 0 for t < 0 , thus h1 (t) is not causal
h2 (t) = u (t) which is always time invariant, causal and stable.
u (t)
is time variant.
h3 (t) =
1+t
h 4 (t) = e-3t u (t) is time variant.
Let n =- m, then
H (z) =-
3
[f (t)] 2 dt
h (t) = f (t)* g (t)
We know that convolution in time domain is multiplication in s domain.
where z < a
where z < 5
-3
-3
E1 =
6.101
- an u (- n - 1)
3
Substituting 2t = p we get
x (t) = [- 2e-t u (t) - 2e2t u (- t)]
6.95
#
E1 = #
E =
=
# d (t) dt = 1
3
Now output is
-3
Option (B) is correct.
Let E be the energy of f (t) and E1 be the energy of f (2t), then
Thus
1
s-2
X (s) = 1
s-3
H (s) =
Y (s) = H (s) X (s)
= 1 # 1 = 1 - 1
s-2 s-3 s-3 s-2
y (t) = e3t - e2t
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
6.105
z-1 (1 - z-4) (z - 1)
z"1
4 (1 - z-1) 2
z-1 z-4 (z 4 - 1) (z - 1)
= lim
z"1
4z-2 (z - 1) 2
-3 (z - 1) (z + 1) (z 2 + 1) (z - 1)
= lim z
z"1 4
(z - 1) 2
-3
= lim z (z + 1) (z2 + 1) = 1
z"1 4
Option (C) is correct.
We know that for a square wave the Fourier series coefficient
sin nw t
Cnsq = At nw t2
T
2
1
Cnsq \
n
0
Thus
= lim
...(i)
0
If we integrate square wave, triangular wave will be obtained,
Hence
Cntri \ 12
n
6.106
Page 154
6.113
Option (B) is correct.
u (t) - u (t - 1) = f (t)
L
u (t) - u (t - 2) = g (t)
L
L
f (t)* g (t)
w
s2 + w2
lim f (t) final value theorem states that:
t"3
lim f (t) = lim sF (s)
t"3
t"3
6.114
So
x (t) = A 0 +
3
/ An cos nwt
n=1
6.115
Option (C) is correct.
Given periodic signal
1,
t < T1
x (t) = *
0, T1 < t < T0
2
The figure is as shown below.
Option (A) is correct.
Option (A) is correct.
We have
f (nT) = anT
Taking z -transform we get
3
/ anT z-n
3
/ (aT ) n z-n
=
n =- 3
= z T
z-a
=
3
n=0
n =- 3
T n
/ b az
l
For x (t) fourier series expression can be written as
x (t) = A 0 +
Option (B) is correct.
If
L [f (t)] = F (s)
Applying time shifting property we can write
6.110
Option (A) is correct.
6.111
Option (A) is correct.
n=1
where dc term
T /2
A 0 = 1 # x (t) dt = 1 # x (t) dt
T0 T
T0 -T /2
T
T
= 1 : # x (t) dt + # x (t) dt +
T0 -T /2
-T
= 1 60 + 2T1 + 0@
T0
A 0 = 2T1
T0
0
0
Option (C) is correct.
Given z transform
0
1
0
C (z) =
z-1 (1 - z-4)
4 (1 - z-1) 2
Applying final value theorem
lim f (n) = lim (z - 1) f (z)
n"3
z"1
lim (z - 1) F (z) = lim (z - 1)
z"1
3
/ [An cos nwt + Bn sin nwt]
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L [f (t - T)] = e-sT F (s)
6.112
n=1
Series will contain only DC & cosine terms.
L
F (z) =
6.109
3
/ [An cos nwt + Bn sin nwt]
For even function x (t), Bn = 0
f (t)* g (t)
= t - (t - 2) u (t - 2) - (t - 1) u (t - 1) + (t - 3) u (t - 3)
The graph of option (B) satisfy this equation.
6.108
Option (D) is correct.
Trigonometric Fourier series of a function x (t) is expressed as :
x (t) = A 0 +
-2s
-s
-3s
= 12 - e 2 - e 2 + e 2
s
s
s
s
Taking inverse Laplace transform we have
6.107
s"0
It must be noted that final value theorem can be applied only if
poles lies in –ve half of s -plane.
Here poles are on imaginary axis (s1, s2 = ! jw) so can not apply
final value theorem. so lim f (t) cannot be determined.
F (s) G (s)
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f (t)* g (t)
F (s) =
We have
F (s) = 1 [1 - e-s]
s
G (s) = 1 [1 - e-2s]
s
= 12 [1 - e-s] [1 - e-2s]
s
= 12 [1 - e-2s - e-s + e-3s]
s
or
Option (A) is correct.
z"1
-1
6.116
-4
z (1 - z )
4 (1 - z-1) 2
1
1
#T
T0 /2
x (t) dtD
1
Option (B) is correct.
The unit impulse response of a LTI system is u (t)
Let
h (t) = u (t)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Taking LT we have
H (s) = 1
s
If the system excited with an input x (t) = e-at u (t), a > 0 , the
response
Y (s) = X (s) H (s)
1
(s + a)
Y (s) = 1 1 = 1 :1 - 1 D
a s s+a
(s + a) s
Page 155
6.123
Option (C) is correct.
The conjugation property allows us to show if x (t) is real, then
X (jw) has conjugate symmetry, that is
[ x (t) real]
X (- jw) = X)(jw)
Proof :
X (s) = L [x (t)] =
so
replace w by - w then
Taking inverse Laplace, the response will be
y (t) = 1 61 - e-at@
a
6.117
x [ n] =
X (z) =
X)(jw) = =
if x (t) real x)(t) = x (t)
/ d (n - k)
/ x [n] z-n
=
k=0
then
3
3
/ ; / d (n - k) z-nE
Option (B) is correct.
6.119
Option (B) is correct.
F
x (t)
X (f)
by differentiation property;
dx (t)
F;
= jwX (w)
dt E
dx (t)
or
F;
= j2pfX (f)
dt E
6.120
F
so
2pf (- w)
3
# g (t) e-jwt dt = 2pf (- w)
F [g (t)] =
-3
6.121
Option (B) is correct.
Given function
Now
x (t) = eat cos (at)
L
s
cos (at)
s2 + a2
x (t)
L
X (s)
s0 t
e x (t)
L
eat cos (at)
L
X (s - s 0)
(s - a)
(s - a) 2 + a2
If
then
so
6.122
3
# x)(t) e jwt dt
-3
3
# x (t) e jwt dt = X (- jw)
-3
shifting in s-domain
Option (C) is correct.
For a function x (t), trigonometric fourier series is :
x (t) = A 0 +
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We have
f (t)
g (w)
by duality property of fourier transform we can write
F
-3
=
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Option (C) is correct.
g (t)
X)(jw) =
)
# x (t) e-jwt dtG
n =- 3 k = 0
Since d (n - k) defined only for n = k so
3
1
= z
X (z) = / z-k =
(z - 1)
(1 - 1/z)
k=0
6.118
3
# x (t) e jwt dt
-3
3
3
k=0
3
-3
X (- jw) =
Option (B) is correct.
We have
3
# x (t) e-jwt dt
X (jw) =
3
/ [An cos nwt + Bn sin nwt]
n=1
A 0 = 1 # x (t) dt
T0 =Fundamental period
T0 T
An = 2 # x (t) cos nwtdt
T0 T
Bn = 2 # x (t) sin nwtdt
T0 T
For an even function x (t), coefficient Bn = 0
for an odd function x (t),
A0 = 0
where
0
0
0
An = 0
so if x (t) is even function its fourier series will not contain sine
terms.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 156
UNIT 7
approximate value of Ka that will reduce the time constant of the
closed loop system by one hundred times as compared to that of the
open-loop system is
CONTROL SYSTEMS
2013
7.1
(A) 1
(C) 10
ONE MARK
The Bode plot of a transfer function G ^s h is shown in the figure
below.
7.3
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s+1
5s2 + 6s + 2
(C) 2 s + 1
s + 4s + 2
The state diagram of a system is shown below. A system is
o = AX + Bu ;
described by the state-variable equations X
y = CX + Du
7.2
and - 8 dB at 1 rad/s and
is negative for all w. Then G ^s h is
(B) 392.8
s
(D) 322
s
TWO MARKS
The open-loop transfer function of a dc motor is given as
w ^s h
= 10 . When connected in feedback as shown below, the
Va ^s h 1 + 10s
s+1
s 2 + 6s + 2
(D) 2 1
5s + 6s + 2
(B)
Statement for Linked Answer Questions 4 and 5:
7.5
2013
The signal flow graph for a system is given below. The transfer
Y ^s h
function
for this system is
U ^s h
(A)
7.4
The gain _20 log G ^s h i is 32 dB
10 rad/s respectively. The phase
(A) 39.8
s
(C) 32
s
(B) 5
(D) 100
The state-variable equations of the system shown in the figure above
are
o = >- 1 0 H X + >- 1H u
o = >- 1 0 H X + >- 1H u
X
X
(A)
(B)
1 -1
1
-1 -1
1
y = 61 - 1@X + u
y = 6- 1 - 1@X + u
o = >- 1 0 H X + >- 1H u
o = >- 1 - 1H X + >- 1H u
X
X
(C)
(D)
-1 -1
1
0 -1
1
y = 6- 1 - 1@X - u
y = 61 - 1@ X - u
The state transition matrix eAt of the system shown in the figure
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above is
e-t 0
(A) > -t -tH
te e
e-t
0
(B) >
H
-t
- te e-t
e-t 0
(C) > -t -tH
e e
e-t - te-t
(D) >
H
0 e-t
2012
7.6
ONE MARK
(s2 + 9) (s + 2)
(s + 1) (s + 3) (s + 4)
is excited by sin (wt). The steady-state output of the system is zero
at
A system with transfer function G (s) =
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) w = 1 rad/s
(C) w = 3 rad/s
(B) w = 2 rad/s
(D) w = 4 rad/s
2012
7.7
Page 157
TWO MARKS
The feedback system shown below oscillates at 2 rad/s when
2011
7.11
(A) K = 2 and a = 0.75
(C) K = 4 and a = 0.5
7.8
(B) K = 3 and a = 0.75
(D) K = 2 and a = 0.5
The state variable description of an LTI system is given by
Jxo1N J 0 a1 0NJx1N J0N
K O K
OK O K O
Kxo2O = K 0 0 a2OKx2O + K0O u
Kxo O Ka
0 0OKx 3O K 1O
3
3
L P L
PL P L P
Jx1N
K O
y = _1 0 0iKx2O
Kx 3O
L P
where y is the output and u is the input. The system is controllable for
(B) a1 = 0, a2 ! 0, a 3 ! 0
(A) a1 ! 0, a2 = 0, a 3 ! 0
(C) a1 = 0, a 3 ! 0, a 3 = 0
(D) a1 ! 0, a2 ! 0, a 3 = 0
2011
7.9
The block diagram of a system with one input u and two outputs y1
and y2 is given below.
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ONE MARK
The root locus plot for a system is given below. The open loop
transfer function corresponding to this plot is given by
s (s + 1)
(s + 2) (s + 3)
(s + 1)
(B) G ^s h H ^s h = k
s (s + 2) (s + 3) 2
1
(C) G ^s h H ^s h = k
s (s - 1) (s + 2) (s + 3)
(s + 1)
(D) G ^s h H ^s h = k
s (s + 2) (s + 3)
(A) G ^s h H ^s h = k
7.10
TWO MARKS
For the transfer function G (jw) = 5 + jw , the corresponding Nyquist
plot for positive frequency has the form
A state space model of the above system in terms of the state vector x and the output vector y = [y1 y2]T is
(A) xo = [2] x + [1] u ; y = [1 2] x
1
(B) xo = [- 2] x + [1] u; y = > H x
2
-2 0
1
(C) xo = >
x + > H u ; y = 81 2B x
H
0 -2
1
2 0
1
1
(D) xo = > H x + > H u ; y = > H x
0 2
1
2
Common Data For Q. 7.4 & 7.5
100
.
s (s + 10) 2
The plant is placed in a unity negative feedback configuration as
shown in the figure below.
The input-output transfer function of a plant H (s) =
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
7.12
7.13
Page 158
The gain margin of the system under closed loop unity negative
feedback is
(A) 0 dB
(B) 20 dB
(C) 26 dB
(D) 46 dB
(A) 10s + 1
0.1s + 1
(C) 100s
10s + 1
The signal flow graph that DOES NOT model the plant transfer
function H (s) is
2010
7.17
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7.14
1
s+1
(C) 2
(D) 2
s+1
s+3
Y (s)
A system with transfer function
has an output
= s
X (s) s + p
p
y (t) = cos a2t - k
3
for the input signal x (t) = p cos a2t - p k. Then, the system param2
eter p is
(A) 3
(B) 2/ 3
(A) 0
7.15
(C) 1
7.16
7.19
The state variable representation of the system can be
0
-1 1
1 1
0
o= >
xo = >
x +> Hu
H
x
x
u
+
H
>
H
(A)
(B)
2
-1 0
-1 0
2
yo = 80 0.5B x
yo = [0 0.5] x
1 1
0
xo = >
x +> Hu
H
(C)
(D)
2
-1 0
yo = 80.5 0.5B x
The transfer function of the system is
(A) s2+ 1
(B)
s +1
(C) 2 s + 1
(D)
s +s+1
-1
xo = >
-1
yo = 80.5
1
0
x +> Hu
H
0
2
0.5B x
s-1
s2+1
s-1
s2+s+1
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(B)
(D)
A unity negative feedback closed loop system has a plant with the
transfer function G (s) = 2 1
and a controller Gc (s) in the
s + 2s + 2
feed forward path. For a unit set input, the transfer function of the
controller that gives minimum steady state error is
(A) Gc (s) = s + 1
(B) Gc (s) = s + 2
s+2
s+1
(s + 1) (s + 4)
(C) Gc (s) =
(D) Gc (s) = 1 + 2 + 3s
s
(s + 2) (s + 3)
The signal flow graph of a system is shown below:
ONE MARK
The transfer function Y (s) /R (s) of the system shown is
TWO MARKS
Common Data For Q. 7.10 & 7.11 :
7.18
2010
(B) 100s + 1
0.1s + 1
(D) 0.1s + 1
10s + 1
2009
7.20
ONE MARK
The magnitude plot of a rational transfer function G (s) with real
coefficients is shown below. Which of the following compensators
has such a magnitude plot ?
3 /2
For the asymptotic Bode magnitude plot shown below, the system
transfer function can be
(A) Lead compensator
(B) Lag compensator
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) PID compensator
7.21
Page 159
(D) Lead-lag compensator
(C) 6 dB and 90c
(D) 3 dB and 90c
Consider the system
dx = Ax + Bu with A = =1 0G and B = = p G
0 1
q
dt
where p and q are arbitrary real numbers. Which of the following
statements about the controllability of the system is true ?
(A) The system is completely state controllable for any nonzero
values of p and q
(B) Only p = 0 and q = 0 result in controllability
(C) The system is uncontrollable for all values of p and q
(D) We cannot conclude about controllability from the given data
2009
7.22
2008
7.26
ONE MARKS
Step responses of a set of three second-order underdamped systems
all have the same percentage overshoot. Which of the following
diagrams represents the poles of the three systems ?
TWO MARKS
The feedback configuration and the pole-zero locations of
2
G (s) = s2 - 2s + 2
s + 2s + 2
are shown below. The root locus for negative values of k , i.e. for
- 3 < k < 0 , has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to
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7.23
(A) ! 2 and 0c
(B) ! 2 and 45c
(C) ! 3 and 0c
(D) ! 3 and 45c
The unit step response of an under-damped second order system
has steady state value of -2. Which one of the following transfer
functions has theses properties ?
(B) 2 - 3.82
(A) 2 - 2.24
s + 1.91s + 1.91
s + 2.59s + 1.12
- 382
(C) 2 - 2.24
(D) 2
s - 2.59s + 1.12
s - 1.91s + 1.91
Common Data For Q. 7.16 and 7.17 :
The Nyquist plot of a stable transfer function G (s) is shown in the
figure are interested in the stability of the closed loop system in
the feedback configuration shown.
7.27
7.24
7.25
The pole-zero given below correspond to a
Which of the following statements is true ?
(A) G (s) is an all-pass filter
(B) G (s) has a zero in the right-half plane
(C) G (s) is the impedance of a passive network
(D) G (s) is marginally stable
The gain and phase margins of G (s) for closed loop stability are
(A) 6 dB and 180c
(B) 3 dB and 180c
(A) Law pass filter
(C) Band filter
(B) High pass filter
(D) Notch filter
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2008
7.28
Page 160
TWO MARKS
Group I lists a set of four transfer functions. Group II gives a list
of possible step response y (t). Match the step responses with the
corresponding transfer functions.
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(A) P - 3, Q - 1, R - 4, S - 2
(C) P - 2, Q - 1, R - 4, S - 2
7.29
(B) P - 3, Q - 2, R - 4, S - 1
(D) P - 3, Q - 4, R - 1, S - 2
A signal flow graph of a system is given below
The set of equalities that corresponds to this signal flow graph is
Jx1N R b - g 0 VJx1N R0 0 V
W u1
WK O S
K O S
(A) d K x2O = S g a 0 WK x2O+ S0 1 We o
u2
dt K O S
x3
S- a b 0 WWK x3O SS1 0 WW
L P
L P
Jx1N RT0 a g XVJx1N TR1 0 XV
W u1
W
S
K O
K O S
(B) d K x2O = S0 - a - g WK x2O+ S0 1 We o
u2
dt K O S
x3
S0 b - b WWK x3O SS0 0 WW
L P TR
L P
Jx1N - a b 0 VXJx1N RT1 0 VX
W u1
WK O S
K O S
(C) d K x2O = S- b - g 0 WK x2O+ S0 1 We o
u2
dt K O S
x3
S a g 0 WWK x3O SS0 0 WW
L P TR
L P
Jx1N - a 0 b XVJx1N TR1 0XV
W u1
WK O S
K O S
(D) d K x2O = S g 0 a WK x2O+ S0 1 We o
u2
dt K O S
x3
S- b 0 - a WWK x3O SS0 0 WW
L P T
X
XL P T
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7.30
A certain system has transfer function
G (s) = 2 s + 8
s + as - 4
where a is a parameter. Consider the standard negative unity
feedback configuration as shown below
Which of the following statements is true?
(A) The closed loop systems is never stable for any value of a
(B) For some positive value of a, the closed loop system is stable,
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 161
but not for all positive values.
(C) For all positive values of a, the closed loop system is stable.
(D) The closed loop system stable for all values of a, both positive
and negative.
7.31
7.32
7.33
The number of open right half plane of
10
is
G (s) = 5
4
3
s + 2s + 3s + 6s2 + 5s + 3
(A) 0
(B) 1
(C) 2
(D) 3
7.37
The magnitude of frequency responses of an underdamped second
order system is 5 at 0 rad/sec and peaks to 10 at 5 2 rad/sec.
3
The transfer function of the system is
500
(B) 2 375
(A) 2
s + 5s + 75
s + 10s + 100
720
(C) 2
(D) 2 1125
s + 12s + 144
s + 25s + 225
Group I gives two possible choices for the impedance Z in the diagram.
The circuit elements in Z satisfy the conditions R2 C2 > R1 C1. The
transfer functions V0 represents a kind of controller.
Vi
5
(s + 5)( s2 + s + 1)
The second-order approximation of T (s) using dominant pole concept is
5
1
(B)
(A)
(s + 5)( s + 1)
(s + 5)( s + 1)
(C) 2 5
(D) 2 1
s +s+1
s +s+1
The open-loop transfer function of a plant is given as G (s) = s 1- 1 .
If the plant is operated in a unity feedback configuration, then the
lead compensator that an stabilize this control system is
10 (s + 4)
10 (s - 1)
(B)
(A)
s+2
s+2
10 (s + 2)
2 (s + 2)
(C)
(D)
s + 10
s + 10
T (s) =
7.38
2
A unity feedback control system has an open-loop transfer function
K
G (s) =
2
s (s + 7s + 12)
The gain K for which s = 1 + j1 will lie on the root locus of this
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Match the impedances in Group I with the type of controllers in
Group II
system is
(A) 4
(C) 6.5
7.39
(A) Q - 1, R - 2
(C) Q - 2, R - 3
2007
7.34
7.35
The asymptotic Bode plot of a transfer function is as shown in the
figure. The transfer function G (s) corresponding to this Bode plot is
(B) Q - 1, R - 3
(D) Q - 3, R - 2
ONE MARK
If the closed-loop transfer function of a control system is given as
s-5
, then It is
T (s)
(s + 2)( s + 3)
(A) an unstable system
(B) an uncontrollable system
(C) a minimum phase system
(D) a non-minimum phase system
2007
1
1
(B)
(s + 1)( s + 20)
s (s + 1)( s + 20)
100
100
(C)
(D)
s (s + 1)( s + 20)
s (s + 1)( 1 + 0.05s)
The state space representation of a separately excited DC servo
motor dynamics is given as
(A)
7.40
dw
dt
dio
dt
> H = =- 1 - 10G=ia G + =10Gu
TWO MARKS
A control system with PD controller is shown in the figure. If the
velocity error constant KV = 1000 and the damping ratio z = 0.5 ,
then the value of KP and KD are
7.36
The transfer function of a plant is
(B) KP = 100, KD = 0.9
(D) KP = 10, KD = 0.9
-1
1 w
0
where w is the speed of the motor, ia is the armature current and
w (s)
of the motor
u is the armature voltage. The transfer function
U (s)
is
10
s2 + 11s + 11
(C) 2 10s + 10
s + 11s + 11
(A)
(A) KP = 100, KD = 0.09
(C) KP = 10, KD = 0.09
(B) 5.5
(D) 10
1
s2 + 11s + 11
(D) 2 1
s + s + 11
(B)
Statement for linked Answer Question 8.33 & 8.34 :
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Consider a linear system whose state space representation is
1
x (t) = Ax (t). If the initial state vector of the system is x (0) = = G,
2
e-2x
then the system response is x (t) = >
H. If the itial state vector
- 2e-2t
1
of the system changes to x (0) = = G, then the system response
-2
e-t
becomes x (t) = > -tH
-e
Page 162
7.47
7.48
7.41
7.42
The eigenvalue and eigenvector pairs (li vi) for the system are
1
1
1
1
(A) e- 1 = Go and e- 2 = Go
(B) e- 1, = Go and e2, = Go
-1
-2
-1
-2
1
1
1
1
(C) e- 1, = Go and e- 2, = Go
(D) e- 2 = Go and e1, = Go
-1
-2
-1
-2
The system matrix A is
0 1
(A) =
- 1 1G
2 1
(C) =
- 1 - 1G
Statement for Linked Answer Questions 7.41 & 7.42 :
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7.43
a2 + 4b , a2 + 4b
The Nyquist plot of G (jw) H (jw)for a closed loop control system,
passes through (- 1, j0) point in the GH plane. The gain margin of
the system in dB is equal to
(A) infinite
(B) greater than zero
(C) less than zero
7.46
(D)
(D) zero
The positive values of K and a so that the system shown in the
figures below oscillates at a frequency of 2 rad/sec respectively are
(A) 1, 0.75
(C) 1, 1
(B) 2, 0.75
(D) 2, 2
With the value of a set for a phase - margin of p , the value of unit
4
- impulse response of the open - loop system at t = 1 second is equal
to
(A) 3.40
(B) 2.40
(C) 1.84
(D) 1.74
ONE MARK
A linear system is equivalently represented by two sets of state
equations :
Xo = AX + BU and Wo = CW + DU
The eigenvalues of the representations are also computed as [l]
and [m]. Which one of the following statements is true ?
(A) [l] = [m] and X = W
(B) [l] = [m] and X ! W
(C) [l] ! [m] and X = W
(D) [l] = [m] and X ! W
TWO MARKS
a2 - 4b , a2 - 4b
Consider a unity - gain feedback control system whose open - loop
1
transfer function is : G (s) = as +
s2
The value of a so that the system has a phase - margin equal to p
4
is approximately equal to
(A) 2.40
(B) 1.40
(C) 0.84
(D) 0.74
2005
7.51
Consider two transfer functions G1 (s) = 2 1
and
s + as + b
s
.
G2 (s) = 2
s + as + b
The 3-dB bandwidths of their frequency responses are, respectively
(A) a2 - 4b , a2 + 4b
(B) a2 + 4b , a2 - 4b
(C)
7.45
7.50
The open-loop function of a unity-gain feedback control system is
given by
K
G (s) =
(s + 1)( s + 2)
The gain margin of the system in dB is given by
(A) 0
(B) 1
(C) 20
(D) 3
7.44
7.49
ONE MARK
2006
A linear system is described by the following state equation
0 1
Xo (t) = AX (t) + BU (t), A = =
- 1 0G
The state transition matrix of the system is
cos t sin t
- cos t sin t
(A) =
(B) =
G
- sin t cos t
- sin t - cos t G
- cos t - sin t
cos t - sin t
(C) =
(D) =
G
- sin t cos t
cos t sin t G
1 1
(B) =
- 1 - 2G
0 1
(D) =
- 2 - 3G
2006
The transfer function of a phase lead compensator is given by
Gc (s) = 1 + 3Ts where T > 0 The maximum phase shift provide by
1 + Ts
such a compensator is
(B) p
(A) p
3
2
(C) p
(D) p
4
6
7.52
Which one of the following polar diagrams corresponds to a lag
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network ?
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
7.53
Despite the presence of negative feedback, control systems still have
problems of instability because the
(A) Components used have non- linearities
(B) Dynamic equations of the subsystem are not known exactly.
(C) Mathematical analysis involves approximations.
(D) System has large negative phase angle at high frequencies.
2005
7.54
Page 163
TWO MARKS
The polar diagram of a conditionally stable system for open loop
gain K = 1 is shown in the figure. The open loop transfer function
of the system is known to be stable. The closed loop system is stable
for
Statement for Linked Answer Question 40 and 41 :
7.59
The open loop transfer function of a unity feedback system is given
by
-2s
G (s) = 3e
s (s + 2)
The gain and phase crossover frequencies in rad/sec are, respectively
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7.55
7.56
7.57
7.58
(A) K < 5 and 1 < K < 1
(B) K < 1 and 1 < K < 5
2
8
8
2
(C) K < 1 and 5 < K
(D) K > 1 and 5 > K
8
8
In the derivation of expression for peak percent overshoot
- px
Mp = exp e
o # 100%
1 - x2
Which one of the following conditions is NOT required ?
(A) System is linear and time invariant
(B) The system transfer function has a pair of complex conjugate
poles and no zeroes.
(C) There is no transportation delay in the system.
(D) The system has zero initial conditions.
A ramp input applied to an unity feedback system results in 5%
steady state error. The type number and zero frequency gain of the
system are respectively
(A) 1 and 20
(B) 0 and 20
(D) 1 and 1
(C) 0 and 1
20
20
A double integrator plant G (s) = K/s2, H (s) = 1 is to be compensated
to achieve the damping ratio z = 0.5 and an undamped natural
frequency, wn = 5 rad/sec which one of the following compensator
Ge (s) will be suitable ?
(A) s + 3
(B) s + 99
s + 99
s+3
(C) s - 6
(D) s - 6
s + 8.33
s
K (1 - s)
An unity feedback system is given as G (s) =
.
s (s + 3)
Indicate the correct root locus diagram.
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(A) 0.632 and 1.26
(C) 0.485 and 0.632
7.60
(B) 0.632 and 0.485
(D) 1.26 and 0.632
Based on the above results, the gain and phase margins of the
system will be
(B) 7.09 dB and 87.5c
(A) -7.09 dB and 87.5c
(C) 7.09 dB and - 87.5c
(D) - 7.09 and - 87.5c
2004
7.61
7.62
ONE MARK
The gain margin for the system with open-loop transfer function
2 (1 + s)
, is
G (s) H (s) =
s2
(A) 3
(B) 0
(C) 1
(D) - 3
K
Given G (s) H (s) =
.The point of intersection of the
s (s + 1)( s + 3)
asymptotes of the root loci with the real axis is
(A) - 4
(B) 1.33
(C) - 1.33
(D) 4
2004
7.63
TWO MARKS
Consider the Bode magnitude plot shown in the fig. The transfer
function H (s) is
(A)
(s + 10)
(s + 1)( s + 100)
(B)
10 (s + 1)
(s + 10)( s + 100)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
7.64
7.65
7.66
102 (s + 1)
103 (s + 100)
(C)
(D)
(s + 10)( s + 100)
(s + 1)( s + 10)
A causal system having the transfer function H (s) = 1/ (s + 2) is
excited with 10u (t). The time at which the output reaches 99% of
its steady state value is
(A) 2.7 sec
(B) 2.5 sec
(C) 2.3 sec
(D) 2.1 sec
A system has poles at 0.1 Hz, 1 Hz and 80 Hz; zeros at 5 Hz, 100
Hz and 200 Hz. The approximate phase of the system response at
20 Hz is
(A) - 90c
(B) 0c
(C) 90c
(D) - 180c
(D) controllable and observable
7.71
(A)
1 - (be + cf + dg)
abcd
abcd
1 - (be + cf + dg) + bedg
-2 2
If A = =
, then sin At is
1 - 3G
(C)
7.69
7.70
1 0
, the state transition matrix eAt is given by
0 1G
0 e-t
(A) > -t
H
e
0
et 0
(B) =
G
0 et
e-t 0
(C) >
H
0 e-t
0 et
(D) = t G
e 0
2003
7.72
ONE MARK
Fig. shows the Nyquist plot of the open-loop transfer function
G (s) H (s) of a system. If G (s) H (s) has one right-hand pole, the
closed-loop system is
bedg
1 - (be + cf + dg)
1 - (be + cf + dg) + bedg
(D)
abcd
(A) always stable
(B) unstable with one closed-loop right hand pole
(C) unstable with two closed-loop right hand poles
(D) unstable with three closed-loop right hand poles
7.73
A PD controller is used to compensate a system. Compared to the
uncompensated system, the compensated system has
(A) a higher type number
(B) reduced damping
(C) higher noise amplification
(D) larger transient overshoot
(B)
2003
7.74
sin (- 4t) + 2 sin (- t) - 2 sin (- 4t) + 2 sin (- t)
(A) 1 =
G
3 - sin (- 4t) + sin (- t) 2 sin (- 4t) + sin (- t)
sin (- 2t) sin (2t)
(B) =
sin (t) sin (- 3t)G
sin (4t) + 2 sin (t) 2 sin (- 4t) - 2 sin (- t)
(C) 1 =
2 sin (4t) + sin (t) G
3 sin (- 4t) + sin (t)
cos (- t) + 2 cos (t) 2 cos (- 4t) + 2 cos (- t)
(D) 1 =
G
3 - cos (- 4t) + cos (- t) - 2 cos (- 4t) + cos (t)
7.68
Given A = =
Consider the signal flow graph shown in Fig. The gain x5 is
x1
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7.67
Page 164
The open-loop transfer function of a unity feedback system is
K
G (s) =
2
s (s + s + 2)( s + 3)
The range of K for which the system is stable is
(B) 13 > K > 0
(A) 21 > K > 0
4
(C) 21 < K < 3
(D) - 6 < K < 3
4
For the polynomial P (s) = s2 + s 4 + 2s3 + 2s2 + 3s + 15 the number
of roots which lie in the right half of the s -plane is
(A) 4
(B) 2
(C) 3
(D) 1
The state variable equations of a system are : xo1 =- 3x1 - x2 = u, xo2 = 2x1
and y = x1 + u . The system is
(A) controllable but not observable
(B) observable but not controllable
(C) neither controllable nor observable
The signal flow graph of a system is shown in Fig. below. The
transfer function C (s)/ R (s) of the system is
(A)
6
s2 + 29s + 6
(B)
s (s + 2)
s + 29s + 6
(D)
6s
s2 + 29s + 6
s (s + 27)
s + 29s + 6
K
The root locus of system G (s) H (s) =
has the breaks (s + 2)( s + 3)
(C)
7.75
TWO MARKS
2
2
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away point located at
(A) (- 0.5, 0)
(C) (- 4, 0)
7.76
(B) (- 2.548, 0)
(D) (- 0.784, 0)
The approximate Bode magnitude plot of a minimum phase system
is shown in Fig. below. The transfer function of the system is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 165
7.82
7.83
(s + 0.1) 3
(s + 0.1) 3
7
(B)
10
(s + 10)( s + 100)
(s + 10) 2 (s + 100)
2
(s + 0.1)
(s + 0.1) 3
(C)
(D)
(s + 10) 2 (s + 100)
(s + 10)( s + 100) 2
A second-order system has the transfer function
C (s)
= 2 4
R (s)
s + 4s + 4
With r (t) as the unit-step function, the response c (t) of the system
is represented by
(A) 108
7.77
The phase margin of a system with the open - loop transfer function
(1 - s)
G (s) H (s) =
(1 + s)( 2 + s)
(A) 0c
(B) 63.4c
(C) 90c
(D) 3
The transfer function Y (s)/ U (s) of system described by the state
equation xo (t) =- 2x (t) + 2u (t) and y (t) = 0.5x (t) is
1
(A) 0.5
(B)
(s - 2)
(s - 2)
1
(C) 0.5
(D)
(s + 2)
(s + 2)
2002
7.84
7.85
TWO MARKS
The system shown in the figure remains stable when
(A) k < - 1
(B) - 1 < k < 3
(C) 1 < k < 3
(D) k > 3
The transfer function of a system is G (s) =
100
. For a
(s + 1)( s + 100)
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unit - step input to the system the approximate settling time for 2%
criterion is
7.78
7.79
The gain margin and the phase margin of feedback system with
8
are
G (s) H (s) =
(s + 100) 3
(A) dB, 0c
(B) 3, 3
(C) 3, 0c
(D) 88.5 dB, 3
2002
7.81
7.86
(C) unstable
ONE MARK
Consider a system with transfer function G (s) = 2s + 6 . Its
ks + s + 6
damping ratio will be 0.5 when the value of k is
(B) 3
(A) 2
6
(D) 6
(C) 1
6
Which of the following points is NOT on the root locus of a system
k
with the open-loop transfer function G (s) H (s) =
s (s + 1)( s + 3)
(A) s =- j 3
(B) s =- 1.5
(C) s =- 3
(D) s =- 3
The characteristic polynomial of a system is
The system is
(A) stable
7.87
et
(B) = G
t
t
(D) = t G
te
(B) 4 sec
(D) 0.01 sec
q (s) = 2s5 + s 4 + 4s3 + 2s2 + 2s + 1
The zero-input response of a system given by the state-space equation
x1 (0)
1
1 0 x1
xo1
=xo G = =1 1G=x G and =x (0)G = = 0 G is
2
2
2
tet
(A) = G
t
et
(C) = t G
te
7.80
(A)100 sec
(C) 1 sec
The
system
with
the
open
loop
transfer
1
has a gain margin of
G (s) H (s) =
2
s
(
s
)
+
s
+
1
(A) - 6 db
(B) 0 db
(C) 35 db
(D) 6 db
2001
7.88
7.89
(B) marginally stable
(D) oscillatory
function
ONE MARK
The Nyquist plot for the open-loop transfer function G (s) of a unity
negative feedback system is shown in the figure, if G (s) has no pole
in the right-half of s -plane, the number of roots of the system
characteristic equation in the right-half of s -plane is
(A) 0
(B) 1
(C) 2
(D) 3
The equivalent of the block diagram in the figure is given is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
7.90
7.91
Page 166
If the characteristic equation of a closed - loop system is s2 + 2s + 2 = 0
, then the system is
(A) overdamped
(B) critically damped
(C) underdamped
(D) undamped
The root-locus diagram for a closed-loop feedback system is shown
in the figure. The system is overdamped.
7.93
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Z3 (s)
- Z3 (s)
,
Z1 (s) + Z3 (s) + Z4 (s) Z1 (s) + Z3 (s)
- Z3 (s)
- Z3 (s)
(B)
,
Z2 (s) - Z3 (s) + Z4 (s) Z1 (s) + Z3 (s)
Z3 (s)
Z3 (s)
(C)
,
Z2 (s) + Z3 (s) + Z4 (s) Z1 (s) + Z3 (s)
- Z3 (s)
Z3 (s)
(D)
,
Z2 (s) - Z3 (s) + Z4 (s) Z1 (s) + Z3 (s)
The open-loop DC gain of a unity negative feedback system with
closed-loop transfer function 2 s + 4
is
s + 7s + 13
(B) 4
(A) 4
9
13
(A)
(C) 4
7.94
The feedback control system in the figure is stable
(A) for all K $ 0
(C) only if 0 # K < 1
2000
7.95
2001
7.92
ONE MARK
TWO MARKS
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(B) only if 1 < k < 5
(D) if 0 # k < 1 or k > 5
TWO MARK
(B) only if K $ 0
(D) only if 0 # K # 1
An amplifier with resistive negative feedback has tow left half plane
poles in its open-loop transfer function. The amplifier
(A) will always be unstable at high frequency
(B) will be stable for all frequency
(C) may be unstable, depending on the feedback factor
(D) will oscillate at low frequency.
2000
(A) only if 0 # k # 1
(C) only if k > 5
(D) 13
7.96
An electrical system and its signal-flow graph representations are
shown the figure (A) and (B) respectively. The values of G2 and H
, respectively are
1
A system described by the transfer function H (s) = 3
2
+
a
+ ks + 3
s
s
is stable. The constraints on a and k are.
(A) a > 0, ak < 3
(B) a > 0, ak > 3
(C) a < 0, ak > 3
(D) a > 0, ak < 3
1999
7.97
ONE MARK
For a second order system with the closed-loop transfer function
T (s) = 2 9
s + 4s + 9
the settling time for 2-percent band, in seconds, is
(A) 1.5
(B) 2.0
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) 3.0
7.98
(D) 4.0
the eigen values of the closed-loop system will be
(A) 0, - 1, - 2
(B) 0, - 1, - 3
(C) - 1, - 1, - 2
(D) 0, - 1, - 1
The gain margin (in dB) of a system a having the loop transfer
function
G (s) H (s) =
(A) 0
(C) 6
7.99
Page 167
2
is
s (s + 1)
(B) 3
(D) 3
1998
The system modeled described by the state equations is
0 1
0
X =>
x + > Hu
H
2 -3
1
7.105
Y = 81 1B x
7.100
7.101
7.102
7.103
7.104
7.106
The phase margin (in degrees) of a system having the loop transfer
function G (s) H (s) = 2 3 is
s (s + 1)
(A) 45c
(B) - 30c
(C) 60c
(D) 30c
1999
The number of roots of s3 + 5s2 + 7s + 3 = 0 in the left half of the s
-plane is
(A) zero
(B) one
(C) two
(A) controllable and observable
(B) controllable, but not observable
(C) observable, but not controllable
(D) neither controllable nor observable
TWO MARKS
An amplifier is assumed to have a single-pole high-frequency transfer
function. The rise time of its output response to a step function input
is 35 n sec . The upper 3 dB frequency (in MHz) for the amplifier to
as sinusoidal input is approximately at
(A) 4.55
(B) 10
(C) 20
(D) 28.6
If the closed - loop transfer function T (s) of a unity negative feedback
system is given by
an - 1 s + an
T (s) = n
n-1
s + a1 s + .... + an - 1 s + an
then the steady state error for a unit ramp input is
(A) an
(B) an
an - 1
an - 2
(C) an - 2
(D) zero
an - 2
Consider the points s1 =- 3 + j4 and s2 =- 3 - j2 in the s-plane.
Then, for a system with the open-loop transfer function
G (s) H (s) = K 4
(s + 1)
(A) s1 is on the root locus, but not s2
(B) s2 is on the root locus, but not s1
(C) both s1 and s2 are on the root locus
(D) neither s1 nor s2 is on the root locus
For the system described by the state equation
R 0 1 0V
R0V
S
W
S W
xo = S 0 0 1W x + S0W u
SS0.5 1 2WW
SS1WW
T
X
T X
If the control signal u is given by u = [- 0.5 - 3 - 5] x + v , then
ONE MARK
(D) three
The transfer function of a tachometer is of the form
(A) Ks
(B) K
s
K
(C) K
(D)
(s + 1)
s (s + 1)
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7.107
7.108
7.109
7.110
7.111
Consider a unity feedback control system with open-loop transfer
K .
function G (s) =
s (s + 1)
The steady state error of the system due to unit step input is
(A) zero
(B) K
(C) 1/K
(D) infinite
The transfer function of a zero-order-hold system is
(A) (1/s) (1 + e-sT )
(B) (1/s) (1 - e-sT )
(C) 1 - (1/s) e-sT
(D) 1 + (1/s) e-sT
In the Bode-plot of a unity feedback control system, the value of
phase of G (jw) at the gain cross over frequency is - 125c. The phase
margin of the system is
(A) - 125c
(B) - 55c
(C) 55c
(D) 125c
Consider a feedback control system with loop transfer function
K (1 + 0.5s)
G (s) H (s) =
s (1 + s) (1 + 2s)
The type of the closed loop system is
(A) zero
(B) one
(C) two
(D) three
The transfer function of a phase lead controller is 1 + 3Ts . The
1 + Ts
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
maximum value of phase provided by this controller is
(A) 90c
(B) 60c
(C) 45c
(D) 30c
7.112
7.113
Page 168
(C) 0
(D) None of the above
***********
The Nyquist plot of a phase transfer function g (jw) H (jw) of a system
encloses the (–1, 0) point. The gain margin of the system is
(A) less than zero
(B) zero
(C) greater than zero
(D) infinity
2s2 + 6s + 5
(s + 1) 2 (s + 2)
The characteristic equation of the system is
(A) 2s2 + 6s + 5 = 0
(B) (s + 1) 2 (s + 2) = 0
(C) 2s2 + 6s + 5 + (s + 1) 2 (s + 2) = 0
The transfer function of a system is
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(D) 2s2 + 6s + 5 - (s + 1) 2 (s + 2) = 0
7.114
In a synchro error detector, the output voltage is proportional to
[w (t)] n, where w (t) is the rotor velocity and n equals
(A) –2
(B) –1
(C) 1
(D) 2
1997
7.115
ONE MARK
In the signal flow graph of the figure is y/x equals
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(A) 3
(B) 5
2
(C) 2
(D) None of the above
7.116
A certain linear time invariant system has the state and the output
equations given below
1 - 1 X1
0
Xo1
> o H = >0 1 H>X H + >1H u
2
X2
y = 81 1B: X1 D
X2
If X1 (0) = 1, X2 (0) =- 1, u (0) = 0, then
(A) 1
(B) –1
dy
dt
is
t=0
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 169
SOLUTIONS
7.1
Pk2 = ^1 h^s-1h^1 h^1 h = s-1
since, all the loops are touching to the paths Pk1 and Pk2 so,
D k 1 = Dk 2 = 1
Now, we have
D = 1 - (sum of individual loops)
+ (sum of product of nontouching
loops)
Here, the loops are
Option (B) is correct.
From the given plot, we obtain the slope as
20 log G2 - 20 log G1
Slope =
log w2 - log w1
L1 = ^- 4h^1 h =- 4
L2 = ^- 4h^s-1h = 4s-1
L 3 = ^- 2h^s-1h^s-1h =- 2s-2
L 4 = ^- 2h^s-1h^1 h =- 2s-1
As all the loop L1, L2, L 3 and L 4 are touching to each other so,
From the figure
and
20 log G2
20 log G1
w1
w2
=- 8 dB
= 32 dB
= 1 rad/s
= 10 rad/s
D = 1 - ^L1 + L2 + L 3 + L 4h
= 1 - ^- 4 - 4s-1 - 2s-2 - 2s-1h
= 5 + 6s1 + 2s2
So, the slope is
Slope = - 8 - 32
log 10 - log 1
=- 40 dB/decade
Therefore, the transfer function can be given as
G ^s h = k2
S
at w = 1
G ^ jwh = k 2 = k
w
In decibel,
20 log G ^ jwh = 20 log k = 32
32
7.2
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or,
k = 10 = 39.8
Hence, the Transfer function is
G ^s h = k2 = 392.8
s
s
Option (C) is correct.
Given, open loop transfer function
G ^s h = 10Ka = Ka 1
1 + 10s s + 10
By taking inverse Laplace transform, we have
g ^ t h = e- t
20
From Mason’s gain formulae
Y ^s h
= SPk Dk
D
U ^s h
s-2 + s-1
=
5 + 6s-1 + 2s-2
= 2s+1
5s + 6s + 2
1
10
Comparing with standard form of transfer function, Ae-t/t , we get
the open loop time constant,
tol = 10
Now, we obtain the closed loop transfer function for the given
system as
G ^s h
10Ka
H ^s h =
=
1 + G ^s h 1 + 10s + 10Ka
Ka
=
s + ^Ka + 101 h
By taking inverse Laplace transform, we get
h ^ t h = ka .e-^k + ht
So, the time constant of closed loop system is obtained as
tcl = 1 1
ka + 10
or,
tcl = 1
ka
(approximately)
a
1
10
Now, given that ka reduces open loop time constant by a factor of
100. i.e.,
tcl = tol
100
1
or,
= 10
100
ka
Hence,
ka = 10
7.3
Option (A) is correct.
For the given SFG, we have two forward paths
Pk1 = ^1 h^s-1h^s-1h^1 h = s-2
7.4
Option (A) is correct.
For the shown state diagram we can denote the states x1 , x2 as below
So, from the state diagram, we obtain
xo1 =- x1 - u
xo2 =- x2 + ^1 h^- 1h^1 h^- 1h u + ^- 1h^1 h^- 1h x1
xo2 =- x2 + x1 + u
and
y
= ^- 1h^1 h x2 + ^- 1h^1 h^- 1h x1 + ^1 h^- 1h^1 h^- 1h^1 h u
= x1 - x 2 + u
Hence, in matrix form we can write the state variable equations
- 1 0 x1
-1
xo1
> o H = > 1 - 1H >x H + > 1 H u
x2
2
x1
and
y = 81 - 1B > H + u
x2
which can be written in more general form as
-1 0
-1
Xo = >
X +> H
1 - 1H
1
y = 81 - 1B X + u
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
7.5
Page 170
Option (A) is correct.
From the obtained state-variable equations
We have
-1 0
A =>
1 - 1H
s = j k+2
jw = j k + 2
S+1 0
SI - A = >
- 1 S + 1H
So,
1 >S + 1 0 H
^S + 1h2 1 S + 1
R 1
V
S
0 W
S+1
W
=S 1
1 W
S
S^S + 1h2 S + 1W
T
X
Hence, the state transition matrix is obtained as
eAt = L-1 ^SI - Ah-1
V_
ZR 1
]]S
0 Wbb
S+1
W
= L-1 [S 1
1 W`
S
]S^S + 1h2 S + 1Wb
\T
Xa
^SI - Ah-1 =
and
7.8
e-1 0
= > -t -tH
te e
Option (C) is correct.
(s2 + 9) (s + 2)
(s + 1) (s + 3) (s + 4)
(- w2 + 9) (jw + 2)
=
(jw + 1) (jw + 3) (jw + 4)
The steady state output will be zero if
G (s) =
G (jw) = 0
-w 2 + 9 = 0
7.7
&
w = 3 rad/s
7.9
7.10
Option (A) is correct.
K (s + 1)
[R (s) - Y (s)]
s + as2 + 2s + 1
K (s + 1)
K (s + 1)
R (s)
= 3
Y (s) ;1 + 3
E
2
2
1
s
+ as2 + 2s + 1
s + as + s +
Y (s) [s3 + as2 + s (2 + k) + (1 + k)] = K (s + 1) R (s)
Y (s)
Transfer Function,
H (s) =
R (s)
K (s + 1)
= 3
s + as2 + s (2 + k) + (1 + k)
Routh Table :
Y (s) =
3
7.11
Option (D) is correct.
General form of state equations are given as
xo = Ax + Bu
yo = Cx + Du
For the given problem
R 0 a 0V
R0V
1
S
W
S W
A = S 0 0 a2W, B = S0W
SSa
SS1WW
0 0WW
3
RT 0 a 0VXR0V R 0VT X
1
S
WS W S W
AB = S 0 0 a2WS0W = Sa2W
SSa
0 0WWSS1WW SS 0WW
3
RT 0
0XT aX1 a2VWTRS0XVW RSa1 a2VW
S
0
0WS0W = S 0W
A2 B = Sa2 a 3
SS 0 a a
0WWSS1WW SS 0WW
3 1
T
XT X T
X
For controllability it is necessary that following matrix has a tank
of n = 3 .
R0 0 a a V
1 2W
S
0W
U = 6B : AB : A2 B@ = S0 a2
SS1 0
0WW
So,
a2 ! 0
T
X
a 3 may be zero or not.
a1 a 2 ! 0 & a1 ! 0
Option (B) is correct.
For given plot root locus exists from - 3 to 3, So there must be odd
number of poles and zeros. There is a double pole at s =- 3
Now
poles = 0, - 2, - 3, - 3
zeros =- 1
k (s + 1)
Thus transfer function
G (s) H (s) =
s (s + 2) (s + 3) 2
Option (A) is correct.
We have
G (jw) = 5 + jw
Here s = 5 . Thus G (jw) is a straight line parallel to jw axis.
Option (B) is correct.
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Now
For oscillation,
a (2 + K) - (1 + K)
=0
a
a = K+1
K+2
w = k + 2 = 2 (Oscillation frequency)
k =2
a = 2 + 1 = 3 = 0.75
2+2 4
and
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7.6
A (s) = as2 + (k + 1) = 0
s2 =- k + 1 = - k + 1 (k + 2) =- (k + 2)
a
(k + 1)
Auxiliary equation
dy
x = y1 and xo = 1
dx
y1
x
1
y = > H = > H = > Hx
y2
2x
2
y1 = 1 u
s+2
y1 (s + 2)
yo1 + 2y1
xo + 2x
xo
xo
=u
=u
=u
=- 2x + u
= [- 2] x + [1] u
Drawing SFG as shown below
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 171
Y (s)
= 1
R (s) s + 1
7.15
Y (s)
= s
X (s) s + p
jw
H (jw) =
jw + p
H (s) =
xo1 = [- 2] x1 + [1] u
y1 = x1 ; y2 = 2x1
Thus
y1
1
y = > H = > H x1
y2
2
Amplitude Response
H (jw) =
x1 = x
Here
7.12
Option (C) is correct.
Phase Response
100
s (s + 10) 2
100
Now
G (jw) H (jw) =
jw (jw + 10) 2
If wp is phase cross over frequency +G (jw) H (jw) = 180c
G (s) H (s) =
We have
- 180c = 100 tan-1 0 - tan-1 3 - 2 tan-1 a
Thus
or
or
or
or
Now
Option (B) is correct.
Transfer function is given as
Input
Output
wp
10 k
- 180c =- 90 - 2 tan-1 (0.1wp)
45c = tan-1 (0.1wp)
tan 45c 0.1wp = 1
wp = 10 rad/se
100
G (jw) H (jw) =
w (w2 + 100)
7.14
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100
= 1
10 (100 + 100) 20
1 =
p
or
or
Alternative :
Option (D) is correct.
From option (D)
TF = H (s)
100
100
=
!
s (s + 10) 2
s (s2 + 100)
So,
Option (B) is correct.
From the given block diagram
or
7.16
H (s) = Y (s) - E (s) $
1
s+1
E (s) = R (s) - H (s)
= R (s) - Y (s) +
E (s)
(s + 1)
1
= R (s) - Y (s)
s + 1D
sE (s)
= R (s) - Y (s)
(s + 1)
E (s)
Y (s) =
s+1
E (s) :1 -
From (1) and (2)
Transfer function
sY (s) = R (s) - Y (s)
(s + 1) Y (s) = R (s)
y (t) = H (jw) x (t - qh) = cos a2t - p k
3
w
H (jw) = p =
w2+ p 2
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Gain Margin =- 20 log 10 G (jw) H (jw)
=- 20 log 10 b 1 l
20
= 26 dB
7.13
qh (w) = 90c - tan-1 a w k
p
x (t) = p cos a2t - p k
2
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At w = wp
G (jw) H (jw) =
w
w +p2
2
...(1)
2
,
(w = 2 rad/ sec)
4+p2
4p 2 = 4 + p 2 & 3p 2 = 4
p = 2/ 3
qh = 9- p - a- p kC = p
3
2
6
p = p - tan-1 w
apk
6
2
tan-1 a w k = p - p = p
p
2 6
3
w = tan p = 3
a3k
p
2 = 3 , (w = 2 rad/ sec)
p
p = 2/ 3
Option (A) is correct.
Initial slope is zero, so K = 1
At corner frequency w 1 = 0.5 rad/ sec , slope increases by + 20 dB/
decade, so there is a zero in the transfer function at w 1
At corner frequency w 2 = 10 rad/ sec , slope decreases by - 20 dB/
decade and becomes zero, so there is a pole in transfer function at
w2
K a1 + s k
w1
Transfer function
H (s) =
s
a1 + w 2 k
...(2)
7.17
Option (D) is correct.
Steady state error is given as
eSS = lim
s"0
1 a1 + s k (1 + 10s)
0.1
=
=
s
(1 + 0.1s)
+
1
a
0.1 k
sR (s)
1 + G (s) GC (s)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
R (s) = 1
s
Page 172
D = 1 - [L1 + L2] = 1 - :- 1 - 12 D = 1 + 1 + 12
s s
s s
(unit step unit)
1
1 + G (s) GC (s)
1
= lim
s"0
G (s)
1+ 2 C
s + 2s + 2
eSS will be minimum if lim GC (s) is maximum
s"0
In option (D)
lim GC (s) = lim 1 + 2 + 3s = 3
s
s"0
s"0
So,
eSS = lim 1 = 0 (minimum)
s"0 3
eSS = lim
s"0
7.18
Option (D) is correct.
Assign output of each integrator by a state variable
So,
7.20
7.21
Option (C) is correct.
This compensator is roughly equivalent to combining lead and lad
compensators in the same design and it is referred also as PID
compensator.
Option (C) is correct.
Here
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xo1 =- x1 + x2
xo2 =- x1 + 2u
y = 0.5x1 + 0.5x2
State variable representation
-1 1
0
xo = >
x + > Hu
H
-1 0
2
yo = [0.5 0.5] x
7.19
Option (C) is correct.
By masson’s gain formula
Transfer function
Y (s)
=
H (s) =
U (s)
Forward path given
D1 = 1, D2 = 2
Y (s)
= P1 D 1 + P2 D 2
H (s) =
D
U (s)
1 :1+1:1
2
(1 + s)
s
=s
= 2
1
1
(s + s + 1)
1+ + 2
s s
0
p
and B = = G
G
1
q
0 p
p
== G
G
G
=
1 q
q
p q
S = 8B AB B = =
q pG
1
0
1
AB = =
0
A ==
S = pq - pq = 0
Since S is singular, system is completely uncontrollable for all values of p and q .
7.22
Option (B) is correct.
The characteristic equation is
1 + G (s) H (s) = 0
K (s2 - 2s + 2)
or
1+
=0
s 2 + 2s + 2
or
s2 + 2s + 2 + K (s2 - 2s + 2) = 0
2
or
K =- s2 + 2s + 2
s - 2s + 2
For break away & break in point differentiating above w.r.t. s we
have
2
2
dK =- (s - 2s + 2)( 2s + 2) - (s + 2s + 2)( 2s - 2) = 0
ds
(s2 - 2s + 2) 2
Thus
(s2 - 2s + 2)( 2s + 2) - (s2 + 2s + 2)( 2s - 2) = 0
s =! 2
Let qd be the angle of departure at pole P , then
or
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/ PK DK
D
P1 (abcdef ) = 2 # 1 # 1 # 0.5 = 12
s
s
s
P2 (abcdef ) = 2 # 1 # 1 # 0.5
3
Loop gain
L1 (cdc) =- 1
s
L2 (bcdb) = 1 # 1 # - 1 = -21
s
s
s
- qd - qp1 + qz1 + qz2 = 180c
- qd = 180c - (- qp1 + qz1 + q2)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 173
= 180c - (90c + 180 - 45c) =- 45c
7.23
1 + G (s) H (s) = 0
1+ 2 s+8
=0
s + as - 4
or
s 2 + as - 4 + s + 8 = 0
or
s2 + (a + 1) s + 4 = 0
This will be stable if (a + 1) > 0 " a > - 1. Thus system is stable
for all positive value of a.
Option (B) is correct.
For under-damped second order response
kwn2
where x < 1
s2 + 2xwn s + wn2
Thus (A) or (B) may be correct
For option (A)
wn = 1.12 and 2xwn = 2.59 " x = 1.12
For option (B)
wn = 1.91 and 2xwn = 1.51 " x = 0.69
T (s) =
7.24
7.25
7.26
7.27
7.31
Option (B) is correct.
The plot has one encirclement of origin in clockwise direction. Thus
G (s) has a zero is in RHP.
1 + G (s) = 0
or
s5 + 2s 4 + 3s3 + 6s2 + 5s + 3 = 0
Substituting s = z1 we have
Option (C) is correct.
The Nyzuist plot intersect the real axis ate - 0.5. Thus
G. M. =- 20 log x =- 20 log 0.5 = 6.020 dB
And its phase margin is 90c.
Option (C) is correct.
Transfer function for the given pole zero plot is:
(s + Z1)( s + Z2)
(s + P1)( s + P2)
From the plot Re (P1 and P2 )>(Z1 and Z2 )
So, these are two lead compensator.
Hence both high pass filters and the system is high pass filter.
Option (C) is correct.
The characteristic equation is
3z5 + 5z 4 + 6z3 + 3z2 + 2z + 1 = 0
The routh table is shown below. As there are tow sign change in
first column, there are two RHS poles.
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Option (C) is correct.
Percent overshoot depends only on damping ratio, x .
2
Mp = e- xp 1 - x
If Mp is same then x is also same and we get
x = cos q
Thus
q = constant
The option (C) only have same angle.
7.28
Option (D) is correct.
P = 2 25
2xwn = 0, x = 0 " Undamped
s + 25
2
6
Q= 2
s + 20s + 62
R=
S=
7.29
2
2xwn = 20, x > 1 " Overdamped
2xwn = 12, x = 1 " Critically
72
s + 7s + 72
2xwn = 7, x < 1 " underdamped
2
Graph 4
7.32
6
s + 12s + 62
2
Graph 3
Graph 1
z5
3
6
2
z4
5
3
1
z3
21
5
7
5
z2
4
3
3
z1
- 74
z0
1
Option (C) is correct.
For underdamped second order system the transfer function is
Kwn2
s2 + 2xwn s + wn2
It peaks at resonant frequency. Therefore
T (s) =
Graph 2
wr = wn 1 - 2x2
Resonant frequency
Option (C) is correct.
We labeled the given SFG as below :
and peak at this frequency
mr =
7.30
From this SFG we have
xo1 =- gx1 + bx3 + m1
xo2 = gx1 + ax3
xo3 =- bx1 - ax3 + u2
R V R
VR V R V
Sx1 W S- g 0 b WSx1 W S0 1 W u1
Sx2 W = S g 0 a WSx2 W+ S0 0 We o
Thus
SSx WW SS- b 0 - a WWSSx WW SS1 0 WW u2
3
3
X
T X T
XT X T
Option (C) is correct.
The characteristic equation of closed lop transfer function is
5
2x 1 - x2
We have wr = 5 2 , and mr = 10 . Only options (A) satisfy these
3
values.
wn = 10, x = 1
2
where
wr = 10 1 - 2` 1 j = 5 2
4
and
Hence satisfied
mr = 1 5 1 = 10
22 1- 4
3
7.33
Option (B) is correct.
The given circuit is a inverting amplifier and transfer function is
Vo = - Z = - Z (sC1 R1 + 1)
R
R1
Vi
sC R + 1
1
1
1
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(sC2 R2 + 1)
sC2
(sC2 R2 + 1) (sC1 R1 + 1)
PID Controller
=#
sC2
R1
R2
=
(sC2 R2 + 1)
(sC1 R1 + 1)
R2
=#
(sC2 R2 + 1)
R1
Z =
For Q ,
Vo
Vi
For R,
Z
Vo
Vi
Page 174
7.38
1 + G (s) = 0
K
or
1+
=0
2
s (s + 7s + 12)
or
s (s2 + 7s + 12) + K = 0
Point s =- 1 + j lie on root locus if it satisfy above equation i.e
Since R2 C2 > R1 C1, it is lag compensator.
7.34
7.35
Option (D) is correct.
In a minimum phase system, all the poles as well as zeros are on the
left half of the s -plane. In given system as there is right half zero
(s = 5), the system is a non-minimum phase system.
(- 1 + j)[( - 1 + j) 2 + 7 (- 1 + j) + 12) + K] = 0
or
K =+ 10
7.39
Option (B) is correct.
We have
Kv = lim sG (s) H (s)
s"0
1000 = lim s
or
s"0
(Kp + KD s) 100
= Kp
s (s + 100)
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Hence
7.40
or
and
Kp = 100
or
2xwn = 10 + 100KD
KD = 0.9
or
or
Option (D) is correct.
We have
or
7.41
2
In given transfer function denominator is (s + 5)[( s + 0.5) +
3
4]
. We can see easily that pole at s =- 0.5 ! j 23 is dominant then
pole at s =- 5 . Thus we have approximated it.
7.37
Option (A) is correct.
1 =
1
2
(
s
1
)(
s - 1)
+
s -1
The lead compensator C (s) should first stabilize the plant i.e.
1
term. From only options (A), C (s) can remove this
remove
(s - 1)
term
G (s) =
Thus
satisfies.
dw
dt
dia
dt
> H = =- 1 - 10G=in G + =10Gu
-1
1 w
0
dw =- w + i
n
dt
dia =- w - 10i + 10u
a
dt
sw (s) =- w (s) = Ia (s)
or
(s + 1) w (s) = Ia (s)
Taking Laplace transform (ii) we get
or
s2 + (10 + 100KD) s + 10 4 = 0
Comparing with s2 + 2xwn + wn2 = 0 we get
5
T (s) =
(s + 5)( s2 + s + 1)
5
=
= 2 1
5`1 + s j (s2 + s + 1)
s +s+1
5
100
s (s + 1)( 1 + .05s)
...(1)
...(2)
Taking Laplace transform (i) we get
Now characteristics equation is
7.36
G (s) =
Option (A) is correct.
We have
(K + KD s) 100
= Kp
1000 = lims " 0 s p
s (s + 100)
or
K =5
Thus
1 + G (s) H (s) = 0
or
Option (D) is correct.
At every corner frequency there is change of -20 db/decade in slope
which indicate pole at every corner frequency. Thus
K
G (s) =
s (1 + s)`1 + s j
20
Bode plot is in (1 + sT) form
= 60 dB = 1000
20 log K
w w = 0. 1
Now characteristics equations is
1 + G (s) H (s) = 0
(100 + KD s) 100
1+
=0
s (s + 10)
Option (D) is correct.
For ufb system the characteristics equation is
10 (s - 1)
1
G (s) C (s) =
#
(s + 1)( s - 1)
(s + 2)
10
Only option (A)
=
(s + 1)( s + 2)
sIa (s) =- w (s) - 10Ia (s) + 10U (s)
w (s) = (- 10 - s) Ia (s) + 10U (s)
= (- 10 - s)( s + 1) w (s) + 10U (s)
w (s) =- [s2 + 11s + 10] w (s) + 10U (s)
(s2 + 11s + 11) w (s) = 10U (s)
w (s)
= 2 10
U (s)
(s + 11s + 11)
...(3)
From (3)
Option (A) is correct.
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We have
Let
xo (t) = Ax (t)
p q
A ==
r sG
1
For initial state vector x (0) = = G the system response is
-2
e-2t
x (t) = >
H
- 2e-2t
Thus
e-2t
> d (- 2e-2t)H
dt
d
dt
t=0
or
- 2e-2 (0)
p q
==
p q 1
r s G=- 2G
1
> 4e-2 (0) H = =r s G=- 2G
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
-2
p - 2q
= 4 G = = r - 2s G
p - 2q =- 2 and r - 2s = 4
1
For initial state vector x (0) = = G the system response is
-1
e-t
x (t) = > -tH
-e
We get
Thus
e-t
> d (- e-t)H
dt
d
dt
t=0
==
Page 175
Given system is 2nd order and for 2nd order system G.M. is infinite.
7.44
...(i)
7.45
7.46
p q 1
- e- (0)
> e- (0) H = =r s G=- 1G
-1
p-q
= 1G = = r - s G
The characteristic equation
lI - A = 0
l -1
=0
2 l+3
Option (D) is correct.
If the Nyquist polt of G (jw) H (jw) for a closed loop system pass
through (- 1, j0) point, the gain margin is 1 and in dB
GM =- 20 log 1
= 0 dB
p q 1
r s G=- 1G
We get
p - q =- 1 and r - s = 1
Solving (1) and (2) set of equations we get
p q
0 1
=r s G = =- 2 - 3G
Option (D) is correct.
Option (B) is correct.
The characteristics equation is
1 + G (s) H (s) = 0
1+
K (s + 1)
=0
s + as2 + 2s + 1
3
s3 + as2 + (2 + K) s + K + 1 = 0
The Routh Table is shown below. For system to be oscillatory
stable
...(2)
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or
l (l + 3) + 2 = 0
or
l =- 1, - 2
Thus Eigen values are - 1 and - 2
Eigen vectors for l1 =- 1
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(l1 I - A) X1 = 0
or
l1 - 1 x11
= 2 l + 3G=x G = 0
1
21
- 1 - 1 x11
= 2 2 G=x G = 0
21
or
as2 + K + 1 = 0
At 2 rad/sec we have
s = jw " s2 =- w2 =- 4 ,
Thus
- 4a + K + 1 = 0
Solving (i) and (ii) we get K = 2 and a = 0.75 .
7.47
7.43
Option (D) is correct.
2+K
s2
a
1+K
s1
(1 + K) a - (1 + K)
a
s0
1+K
Option (D) is correct.
The transfer function of given compensator is
Gc (s) = 1 + 3Ts
1 + Ts
Comparing with
Gc (s) = 1 + aTs we get a = 3
1 + Ts
The maximum phase sift is
Option (D) is correct.
As shown in previous solution the system matrix is
0 1
A ==
- 2 - 3G
or
7.48
...(2)
1
s3
or
- x11 - x21 = 0
or
x11 + x21 = 0
We have only one independent equation x11 =- x21 .
Let x11 = K, then x21 =- K , the Eigen vector will be
x12
K
1
=x G = =- 2K G = K =- 2G
22
7.42
...(1)
Then we have
or
- x11 - x21 = 0
or
x11 + x21 = 0
We have only one independent equation x11 =- x21 .
Let x11 = K , then x21 =- K , the Eigen vector will be
1
x11
K
=x G = =- K G = K =- 1G
21
Now Eigen vector for l2 =- 2
(l2 I - A) X2 = 0
l2 - 1 x12
or
= 2 l + 3G=x G = 0
2
22
- 2 - 1 x11
or
= 2 1 G=x G = 0
21
a (2 + K) - (K + 1)
=0
a
a = K+1
K+2
fmax = tan-1 a - 1
2 a
= tan-1 3 - 1 = tan-1 1
2 3
3
fmax = p
6
Option (A) is correct.
T>0
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(sI - A) = =
(sI - A)
-1
s 0
0 1
s -1
-=
==
G
G
0 s
-1 0
1 sG
s
s -1
s +1
1
= 2
G = > -1
=
s +1 1 s
s +1
2
2
f (t) = eAt = L-1 [(sI - A)] -1
7.49
1
s2 + 1
s
s2 + 1
T (s) = 1 + sT
1 + sbT
H
or
and
At w = 0 ,
At w = 0 ,
cos t sin t
==
- sin t cos t G
p = p + tan-1 (w a) - p
g
4
p = tan-1 (w a)
g
4
At w = 3 ,
7.54
7.55
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+T (jw) = tan-1 (wT) - tan-1 (wbT)
T (jw) = 1
+T (jw) =- tan-1 0 = 0
T (jw) = 1
b
At w = 3 ,
7.53
b > 1; T > 0
1 + w2 T2
1 + w2 b2 T2
T (jw) =
Option (C) is correct.
1
G (s) = as +
We have
s2
+G (jw) = tan-1 (wa) - p
Since PM is p i.e. 45c, thus
4
p = p + +G (jw ) w " Gain cross over Frequeng
g
4
cy
or
Page 176
+T (jw) = 0
Option (A) is correct.
Despite the presence of negative feedback, control systems still have
problems of instability because components used have nonlinearity.
There are always some variation as compared to ideal characteristics.
Option (B) is correct.
Option (C) is correct.
The peak percent overshoot is determined for LTI second order
closed loop system with zero initial condition. It’s transfer function
is
wn2
s2 + 2xwn s + wn2
Transfer function has a pair of complex conjugate poles and zeroes.
T (s) =
7.56
Option (A) is correct.
For ramp input we have R (s) = 12
s
Now
ess = lim sE (s)
s"0
or
awg = 1
At gain crossover frequency G (jwg) = 1
Thus
or
or
7.50
s"0
1 + a2 wg2
=1
wg2
1 + 1 = wg2
wg = (2)
or
C (s) = G (s) R (s) = G (s)
= 0.84s2 + 1 = 12 + 0.84
s
s
s
Taking inverse Laplace transform
7.52
c (t) = (t + 0.84) u (t)
c (1 sec) = 1 + 0.84 = 1.84
Option (C) is correct.
We have
where l is set of Eigen values
Xo = AX + BU
and
where m is set of Eigen values
Wo = CW + DU
If a liner system is equivalently represented by two sets of state
equations, then for both sets, states will be same but their sets of
Eigne values will not be same i.e.
X = W but l ! m
Option (D) is correct.
The transfer function of a lag network is
Finite
kv is finite for type 1 system having ramp input.
1
4
G (s) = 0.84s2 + 1
s
Due to ufb system H (s) = 1 and due to unit impulse response
R (s) = 1, thus
7.51
But
(as awg = 1)
Option (C) is correct.
For a = 0.84 we have
At t = 1,
R (s)
1
= lim
1 + G (s) s " 0 s + sG (s)
ess = lim 1 = 5% = 1
s " 0 sG (s)
20
kv = 1 = lim sG (s) = 20
s"0
ess
= lim s
7.57
7.58
Option (A) is correct.
Option (C) is correct.
Any point on real axis of s - is part of root locus if number of OL
poles and zeros to right of that point is even. Thus (B) and (C) are
possible option.
The characteristics equation is
1 + G (s) H (s) = 0
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or
or
1+
K (1 - s)
=0
s (s + 3)
2
K = s + 3s
1-s
For break away & break in point
dK = (1 - s)( 2s + 3) + s2 + 3s = 0
ds
or
- s2 + 2s + 3 = 0
which gives s = 3 , - 1
Here - 1 must be the break away point and 3 must be the break in
point.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
7.59
Page 177
Option (D) is correct.
-2s
G (s) = 3e
s (s + 2)
-2jw
or
G (jw) = 3e
jw (jw + 2)
3
G (jw) =
w w2 + 4
Let at frequency wg the gain is 1. Thus
3
=1
wg (wg2 + 4)
7.62
7.63
or
or
or
Now
wg4 +
4wg2
-9 = 0
wg2 = 1.606
wg = 1.26 rad/sec
+G (jw) =- 2w - p - tan-1 w
2
2
Let at frequency wf we have +GH =- 180c
w
- p =- 2wf - p - tan-1 f
2
2
w
or
2wf + tan-1 f = p
2
2
w
w 3
or
2wf + c f - 1 ` f j m = p
2
2
3 2
or
or
7.60
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At w = 1 change in slope is +20 dB " 1 zero at w = 1
At w = 10 change in slope is - 20 dB " 1 poles at w = 10
At w = 100 change in slope is - 20 dB " 1 poles at w = 100
K (s + 1)
Thus
T (s) = s
s
+ 1)
( 10 + 1)( 100
Now
20 log10 K =- 20 " K = 0.1
0.1 (s + 1)
100 (s + 1)
Thus
=
T (s) = s
s
( 10 + 1)( 100 + 1) (s + 10)( s + 100)
wf = 0.63 rad
1
2
7.64
The phase at gain cross over frequency is
w
+G (jwg) =- 2wg - p - tan-1 g
2
2
=- 2 # 1.26 - p - tan-1 1.26
2
2
or
7.61
Option (C) is correct.
The given bode plot is shown below
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5wf wf3
=p
2
24
2
5wf
.p
2
2
Option (D) is correct.
The gain at phase crossover frequency wf is
3
3
=
G (jwg) =
wf (wf2 + 4)
0.63 (0.632 + 4)
or
G (jwg) = 2.27
G.M. =- 20 log G (jwg)
- 20 log 2.26 =- 7.08 dB
Since G.M. is negative system is unstable.
w2 = 3
G (jw) H (jw) = 2 1 +
w2
Thus gain margin is = 1 = 0 and in dB this is - 3 .
3
Option (C) is correct.
Centroid is the point where all asymptotes intersects.
SReal of Open Loop Pole - SReal Part of Open Loop Pole
s =
SNo.of Open Loop Pole - SNo.of Open Loop zero
= - 1 - 3 =- 1.33
3
C (s) = H (s) $ R (s) =
=- 4.65 rad or - 266.5c
PM = 180c + +G (jwg) = 180c - 266.5c =- 86.5c
Option (D) is correct.
The open loop transfer function is
2 (1 + s)
G (s) H (s) =
s2
Substituting s = jw we have
2 (1 + jw)
...(1)
G (jw) H (jw) =
- w2
+G (jw) H (jw) =- 180c + tan-1 w
The frequency at which phase becomes - 180c, is called phase
crossover frequency.
Thus
- 180 =- 180c + tan-1 wf
or
tan-1 wf = 0
or
wf = 0
The gain at wf = 0 is
Option (C) is correct.
We have
r (t) = 10u (t)
or
R (s) = 10
s
Now
H (s) = 1
s+2
or
1 $ 10 10
s + 2 s s (s + 2)
C (s) = 5 - 5
s s+2
c (t) = 5 [1 - e-2t]
The steady state value of c (t) is 5. It will reach 99% of steady
state value reaches at t , where
or
or
or
7.65
5 [1 - e-2t] = 0.99 # 5
1 - e-2t = 0.99
e-2t = 0.1
- 2t = ln 0.1
t = 2.3 sec
Option (A) is correct.
Approximate (comparable to 90c) phase shift are
Due to pole at 0.01 Hz " - 90c
Due to pole at 80 Hz " - 90c
Due to pole at 80 Hz " 0
Due to zero at 5 Hz " 90c
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 178
Due to zero at 100 Hz " 0
Due to zero at 200 Hz " 0
Thus approximate total - 90c phase shift is provided.
7.66
We have only one independent equation x12 = 2x22
Let x22 = K , then x12 = 2K . Thus Eigen vector will be
x12
2K
2
=x G = = K G = K = 1 G
22
Option (C) is correct.
Mason Gain Formula
T (s) =
Spk 3 k
3
Digonalizing matrix
In given SFG there is only one forward path and 3 possible loop.
p1 = abcd
31 = 1
3= 1 - (sum of indivudual loops) - (Sum of two non touching
loops)
Now Diagonal matrix of sin At is D where
sin (l1 t)
0
sin (- 4t)
0
D ==
==
G
0
sin (l2 t)
0
sin (l2 t)G
L1 L2 = bedg
Now matrix
C (s)
p1 3 1
=
1 - (be + cf + dg) + bedg
R (s)
=
abcd
1 - (be + cf + dg) + bedg
- sin (- 4t) - 2 sin (- t) 2 sin (- 4t) - 2 sin (- t)
=-` 1 j=
sin (- 4t) + 2 sin (t) - 2 sin (- 4t) - sin (- t)G
3
- sin (- 4t) - 2 sin (- t) 2 sin (- 4t) - 2 sin (- t)
=-` 1 j=
sin (- 4t) - sin (- t) - 2 sin (- 4t) + 2 sin (- t)G
3
sin (- 4t) + 2 sin (- t) - 2 sin (- 4t) + 2 sin (- t)
= ` 1 j=
Gs
3 - sin (- 4t + sin (- t) 2 sin (- 4t) + sin (- t)
7.68
Option (A) is correct.
We have
-2 2
A ==
1 - 3G
l + 2 -2
=0
-1 l + 3
or
(l + 2)( l + 3) - 2 = 0
or
l2 + 5l + 4 = 0
Thus
l1 =- 4 and l2 =- 1
Eigen values are - 4 and - 1.
Eigen vectors for l1 =- 4
s4
Now Eigen vector for l2 =- 1
(l2 I - A) X2 = 0
l2 + 2 - 2 x12
or
= - 1 l + 3G=x G = 0
2
22
1
5
K
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(l1 I - A) X1 = 0
l1 + 2 - 2 x11
= 1 l + 3G=x G = 0
1
21
- 2 - 2 x11
=- 1 - 1G=x G = 0
21
or
- 2x11 - 2x21 = 0
or
x11 + x21 = 0
We have only one independent equation x11 =- x21 .
Let x21 = K , then x11 =- K , the Eigen vector will be
-K
-1
x11
=x G = = K G = K = 1 G
21
1 + G (s)
K
=0
s (s2 + 2s + 2)( s + 3)
s 4 + 4s3 + 5s2 + 6s + K = 0
The routh table is shown below. For system to be stable,
(21 - 4K)
0 < K and 0 <
2/7
This gives
0 < K < 21
4
1+
[lI - A] = 0
or
Option (A) is correct.
For ufb system the characteristic equation is
1 + G (s) = 0
Characteristic equation is
or
B = sin At = MDM-1
- 1 2 sin (- 4t)
0
1 -2
=-` 1 j=
G
G
=
=
1 1
0
sin (- t) - 1 - 1G
3
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7.67
-1 2
x11 x12
=
M ==
x21 x22 G = 1 1G
1 -2
M-1 = ` - 1 j=
G
3 -1 -1
Now
= 1 - (L1 + L2 + L3) + (L1 L3)
Non touching loop are L1 and L3 where
Thus
1 - 2 x12
=- 1 2 G=x G = 0
22
or
7.69
s3
4
6
s2
7
2
K
s1
21 - 4K
7/2
0
s0
K
0
Option (B) is correct.
We have
P (s) = s5 + s 4 + 2s3 + 3s + 15
The routh table is shown below.
If e " 0+ then 2e +e 12 is positive and -15e2-e +2412e - 144 is negative. Thus
there are two sign change in first column. Hence system has 2 root
2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 179
on RHS of plane.
7.70
s5
1
2
3
s4
1
2
15
s3
e
- 12
0
s2
2e + 12
e
15
0
s1
2
-15e - 24e - 144
2e + 12
s0
0
7.75
Option (D) is correct.
1
- 3 - 1 x1
x1
We have
=x G = = 2 0 G=x G + = 0 Gu
2
2
1
x1
and
Y = [1 0]= G + = G u
2
x2
-3 -1
1
Here
, B = = G and C = [1 0]
A ==
G
2 0
0
The controllability matrix is
1 -3
QC = [B AB ] = =
0 2G
det QC ! 0
The observability matrix is
Thus controllable
Q0 = [CT AT CT ]
1 -3
==
!0
0 - 1G
det Q0 ! 0
7.71
where L1 and L3 are non-touching
C (s)
This
R (s)
p1 3 1
=
1 - (loop gain) + pair of non - touching loops
^ s +s27 h
=
=
1 - ^ -s3 - 24s - s2 h + -s2 . -s3
s (s + 27)
= 2
s + 29s + 6
which gives
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Thus observable
The location of poles on s - plane is
1
0
(s - 1)
s-1
1
=
= 0
(s - 1)G > 0
(s - 1) 2
0
1
s-1
H
Option (A) is correct.
Since breakpoint must lie on root locus so s =- 0.748 is possible.
7.76
Option (D) is correct.
Mason Gain Formula
Spk 3 k
3
In given SFG there is only forward path and 3 possible loop.
p1 = 1
31 = 1 + 3 + 24 = s + 27
s
s
s
L1 = - 2 , L2 = - 24 and L3 = - 3
s
s
s
Option (A) is correct.
The given bode plot is shown below
At w = 0.1 change in slope is + 60 dB " 3 zeroes at w = 0.1
At w = 10 change in slope is - 40 dB " 2 poles at w = 10
At w = 100 change in slope is - 20 dB " 1 poles at w = 100
K ( 0s.1 + 1) 3
Thus
T (s) = s
s
+ 1)
( 10 + 1) 2 ( 100
Now
20 log10 K = 20
or
K = 10
10 ( 0s.1 + 1) 3
108 (s + 0.1) 3
Thus
=
T (s) = s
s
+ 1)
(s + 10) 2 (s + 100)
( 10 + 1) 2 ( 100
Option (C) is correct.
PD Controller may accentuate noise at higher frequency. It does not
effect the type of system and it increases the damping. It also reduce
the maximum overshoot.
T (s) =
s = - 10 ! 100 - 72 =- 0.784, - 2.548
6
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Z = P-N
N " Net encirclement of (- 1 + j0) by Nyquist plot,
P " Number of open loop poles in right hand side of s - plane
Z " Number of closed loop poles in right hand side of s - plane
Here N = 1 and P = 1
Thus
Z =0
Hence there are no roots on RH of s -plane and system is always
stable.
7.74
K =- s (s2 + 5s2 + 6s)
dK =- (3s2 + 10s + 6) = 0
ds
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et 0
eAt = L-1 [(sI - A)] -1 = =
G
0 et
7.73
1 + G (s) H (s) = 0
K
=0
1+
s (s + 2)( s + 3)
or
s 0
1 0
s-1 0
-=
==
(sI - A) = =
G
G
0 s
0 1
0 s - 1G
7.72
s + 27
Option (D) is correct.
We have
or
Option (B) is correct.
(sI - A) -1 =
^ s h
1 + 29s + s62
7.77
Option (B) is correct.
The characteristics equation is
s2 + 4s + 4 = 0
Comparing with
s2 + 2xwn + wn2 = 0
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
we get
Thus
7.78
7.79
2xwn = 4 and wn2 = 4
x =1
ts = 4 = 4 = 2
xwn
1#2
Critically damped
or
or
Option (C) is correct.
We have
1
xo1
=xo G = =1
2
1
A ==
1
s
(sI - A) = =
0
7.84
0 x1
x1 (0)
1
and =
=
1G=x2 G
x2 (0)G = 0 G
0
1G
0
1 0
s-1 0
-= G = =
G
s
1 1
- 1 s - 1G
7.85
1
(s - 1)
0
s-1
1
(sI - A) =
>
H = > +1
(s - 1) 2 + 1 (s - 1)
(s - 1)
t
e 0
L-1 [(sI - A) -1] = eAt = = t t G
te e
et 0 1
et
x (t) = eAt # [x (t0)] = = t t G= G = = t G
te e 0
te
0
2
1
s-1
H
7.86
2
ks + s + 6 = 0
s2 + 1 s + 6 = 0
K
K
Comparing with s2 + 2xwn s + wn2 = 0 we have
we get
2xwn = 1 and wn2 = 6
K
K
or
2 # 0.5 # 6 Kw = 1
K
6 = 1 & K =1
or
6
K
K2
7.82
7.83
or
or
Now
y (t) = 0.5x (t)
Option (B) is correct.
Routh table is shown below. Here all element in 3rd row are zero, so
system is marginal stable.
s5
2
4
2
s4
1
2
1
s3
0
0
0
s1
s0
Given x = 0.5
Option (D) is correct.
From the expression of OLTF it may be easily see that the maximum
magnitude is 0.5 and does not become 1 at any frequency. Thus gain
cross over frequency does not exist. When gain cross over frequency
does not exist, the phase margin is infinite.
sX (s) =- 2X (s) + 2U (s)
(s + 2) X (s) = 2U (s)
2U (s)
X (s) =
(s + 2)
Overdamped
s2
Option (B) is correct.
Any point on real axis lies on the root locus if total number of poles
and zeros to the right of that point is odd. Here s =- 1.5 does not
lie on real axis because there are total two poles and zeros (0 and
- 1) to the right of s =- 1.5 .
Option (D) is correct.
We have
xo (t) =- 2x (t) + 2u (t)
Taking Laplace transform we get
Option (B) is correct.
The characteristics equation is
For overdamped system settling time can be determined by the
dominant pole of the closed loop system. In given system dominant
pole consideration is at s =- 1. Thus
1 =1
and Ts = 4 = 4 sec
T
T
Option (C) is correct.
The characteristics equation is
or
Option (D) is correct.
From Mason gain formula we can write transfer function as
K
Y (s)
K
s
=
=
3
R (s)
1 - ( s + -sK ) s - 3 (3 - K)
For system to be stable (3 - K) < 0 i.e. K > 3
(s + 1)( s + 100) = 0
s2 + 101s + 100 = 0
Comparing with s2 + 2xwn + wn2 = 0 we get
2xwn = 101 and wn2 = 100
Thus
x = 101
20
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7.81
Y (s) = 0.5X (s)
0.5 # 2U (s)
Y (s) =
s+2
Y (s)
1
=
U (s)
(s + 2)
Option (B) is correct.
-1
7.80
Page 180
...(i)
7.87
Option (B) is correct.
The open loop transfer function is
1
G (s) H (s) =
2
s (s + s + 1)
Substituting s = jw we have
1
G (jw) H (jw) =
jw (- w2 + jw + 1)
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+G (jw) H (jw) =- p - tan-1 w 2
2
(1 - w )
The frequency at which phase becomes - 180c, is called phase
crossover frequency.
wf
Thus
- 180 =- 90 - tan-1
1 - wf2
wf
or
- 90 =- tan-1
1 - wf2
or
1 - w2f = 0
wf = 1 rad/sec
The gain margin at this frequency wf = 1 is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 181
GM =- 20 log10 G (jwf) H (jwf)
Comparing (1) and (6) we have
Z3 (s)
H =
Z1 (s) + Z3 (s)
= 20 log10 (wf (1 - w2f) 2 + w2f
=- 20 log 1 = 0
7.88
Option (A) is correct.
Z = P-N
N " Net encirclement of (- 1 + j0) by Nyquist plot,
P " Number of open loop poles in right had side of s - plane
Z " Number of closed loop poles in right hand side of s - plane
Here N = 0 (1 encirclement in CW direction and other in CCW)
and P = 0
Thus Z = 0
Hence there are no roots on RH of s - plane.
7.89
7.90
7.93
or
or
Option (D) is correct.
Take off point is moved after G2 as shown below
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Option (C) is correct.
The characteristics equation is
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7.94
2
and
x = 1
2
Since x < 1 thus system is under damped
Option (D) is correct.
If roots of characteristics equation lie on negative axis at different
positions (i.e. unequal), then system response is over damped.
From the root locus diagram we see that for 0 < K < 1, the roots
are on imaginary axis and for 1 < K < 5 roots are on complex plain.
For K > 5 roots are again on imaginary axis.
Thus system is over damped for 0 # K < 1 and K > 5 .
= G1 Vi (s) + HI2 (s)
= G2 I1 (s)
= G3 I2 (s)
KVL in given block diagram we have
Vi (s) = I1 (s) Z1 (s) + [I1 (s) - I2 (s)] Z3 (s)
0 = [I2 (s) - I1 (s)] Z3 (s) + I2 (s) Z2 (s) + I2 (s) Z4 (s)
From (4) we have
or
Vi (s) = I1 (s)[ Z1 (s) + Z3 (S)] - I2 (s) Z3 (S)
Z3 (s)
1
or
+ I2
I1 (s) = Vi
Z1 (s) + Z3 (s)
Z1 (s) + Z3 (s)
1 + G (s) H (s) = 0
K (s - 2)
1+
(s - 2) = 0
(s + 2) 2
or
(s + 2) 2 + K (s - 2) 2 = 0
or
(1 + K) s2 + 4 (1 - K) s + 4K + 4 = 0
Routh Table is shown below. For System to be stable 1 + k > 0 ,
and 4 + 4k > 0 and 4 - 4k > 0 . This gives - 1 < K < 1
As per question for 0 # K < 1
...(1)
...(2)
...(3)
...(4)
...(5)
7.95
...(6)
From (5) we have
or
I1 (s) Z3 (S) = I2 (s)[ Z2 (s) + Z3 (s) + Z4 (s)]
I1 (s) Z3 (s)
Is (s) =
Z3 (s) + Z2 (s) + Z4 (s)
Comparing (2) and (7) we have
Z3 (s)
G2 =
Z3 (s) + Z2 (s) + Z4 (s)
7.96
...(7)
Option (C) is correct.
From the Block diagram transfer function is
G (s)
T (s) =
1 + G (s) H (s)
K (s - 2)
Where
G (s) =
(s + 2)
and
H (s) = (s - 2)
The Characteristic equation is
Option (C) is correct.
From SFG we have
I1 (s)
I2 (s)
V0 (s)
Now applying
G (0) = 4
9
Thus
wn =
7.92
1 = s2 + 7s + 13 - 1 = s2 + 6s + 9
G (s)
s+4
s+4
G (s) = 2 s + 4
s + 6s + 9
For DC gain s = 0 , thus
s2 + 2s + 2 = 0
Comparing with s2 + 2xwn + wn2 = 0 we get
2xwn = 2 and wn2 = 2
7.91
Option (B) is correct.
For unity negative feedback system the closed loop transfer function
is
G (s)
s+4
,
CLTF =
=
G (s) " OL Gain
1 + G (s) s2 + 7s + 13
2
1 + G (s)
or
= s + 7s + 13
G (s)
s+4
s2
1+k
4 + 4k
s1
4 - 4k
0
s0
4 + 4k
Option (B) is correct.
It is stable at all frequencies because for resistive network feedback
factor is always less than unity. Hence overall gain decreases.
Option (B) is correct.
The characteristics equation is s2 + as2 + ks + 3 = 0
The Routh Table is shown below
For system to be stable a > 0 and aK - 3 > 0
a
Thus a > 0 and aK > 3
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
s3
1
K
s2
a
3
s1
aK - 3
a
0
s0
3
Page 182
The observability matrix is
1 2
!0
Q0 = [CT AT CT ] = =
1 - 2G
7.100
Option (D) is correct.
G (s) H (s) = 2 3
s (s + 1)
we have
7.97
Option (B) is correct.
Closed loop transfer function is given as
T (s) = 2 9
s + 4s + 9
by comparing with standard form we get natural freq.
G (jw) H (jw) at w = w = 1
g
x =
wg =
g
g
= 180 - 150 = 30c
7.101
Option (D) is correct.
Given loop transfer function is
2
s (s + 1)
2
jw (jw + 1)
Phase cross over frequency can be calculated as
G (jw) H (jw) =
1
G (jw) H (jwp) o
A ==
Here input
0
0
1
, B = = G and C = [1 1]
G
1
2 -3
The controllability matrix is
QC = [B AB ] = =
det QC ! 0
7.103
0 1
1 - 3G
G (s) =
R (s) = 12 (unit Ramp)
s
1
so
E (s) = lim 12
s " 0 s 1 + G (s)
n
a1 sn - 1 + .... + an - 2 s2
= lim 12 s +
s"0 s
sn + a1 sn - 1 + .... + an
= an - 2
an
2
=0
wp w2p + 1
G.M. = 20 log 10 b 1 l = 3
0
Option (A) is correct.
Here
Option (B) is correct.
7.104
Thus controllable
an - 1 s + an
sn + a1 sn - 1 + ....an - 2 s2
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1
at w = wp
G (jw) H (jw) G
G (jwp) H (jwp) =
7.99
G (s) H (s) =
an - 1 s + an
sn + a1 sn - 1 + ....an - 2 s2
Steady state error is given by
1
E (s) = lim R (s)
s"0
1 + G (s) H (s)
for unity feed back H (s) = 1
Thus
Gain margin
so
Option (C) is correct.
Closed-loop transfer function is given by
an - 1 s + an
T (s) = n
s + a1 sn - 1 + ... + an - 1 s + an
an - 1 s + an
n
n-1
2
= s + a1 s + ...an - 2 s
an - 1 s + an
1+ n
s + a1 sn - 1 + ...an - 2 s2
For unity feed back H (s) = 1
p
G.M. = 20 log 10 e
Option (B) is correct.
Thus
f (w) at w = w =- 180c
f (w) =- 90c - tan-1 (w)
- 90c - tan-1 (wp) =- 180c
tan-1 (wp) = 90c
wp = 3
20 log 10 =
3
f (w) at w = w =- 90 - tan-1 (wg)
=- 90 - tan-1 3 =- 90 - 60 =- 150
Phase margin = 180 + f (w) at w = w
7.102
G (s) H (s) =
w1, w2 = ! 3
which gives
4
=4 =2
ts = 4 =
xwn
3 # 2/3 2
So here
2 3
=1
w w2 + 1
12 = w2 (w2 + 1)
w4 + w2 - 12 = 0
(w2 + 4) (w2 - 3) = 0
w2 = 3 and w2 =- 4
or
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2 3
jw (jw + 1)
Gain cross over frequency
4 = 2/3
2#3
for second order system the setting time for 2-percent band is given
by
7.98
G (jw) H (jw) =
or
wA2 = 9
wn = 3
2xwn = 4
damping factor
Thus observable
det Q0 ! 0
7.105
Option (A) is correct.
7.106
Option (A) is correct.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 183
By applying Routh’s criteria
3
Phase is
f (w) = tan-1 (3Tw) - tan-1 (Tw)
w
f (w) = tan-1 ; 3Tw - T
1 + 3T 2 w2 E
f (w) = tan-1 ; 2Tw2 2 E
1 + 3T w
For maximum value of phase
df (w)
=0
dw
2
s + 5s + 7s + 3 = 0
s3
1
s2
5
s1
7#5-3
5
s0
3
7
3
=
32
5
0
There is no sign change in the first column. Thus there is no root
lying in the left-half plane.
7.107
7.108
Option (A) is correct.
Techometer acts like a differentiator so its transfer function is of the
form ks .
Option (A) is correct.
Open loop transfer function is
K
G (s) =
s (s + 1)
Steady state error
sR (s)
E (s) = lim
s " 0 1 + G (s) H (s)
R (s) = input
Where
H (s) = 1 (unity feedback)
R (s) = 1
s
7.109
1 = 3T 2 w2
Tw = 1
3
So maximum phase is
or
s1
s (s + 1)
s
so
=0
= lim 2
E (s) = lim
s"0
s"0 s + s + K
K
1+
s (s + 1)
Option (B) is correct.
Fig given below shows a unit impulse input given to a zero-order hold
circuit which holds the input signal for a duration T & therefore,
the output is a unit step function till duration T .
fmax = tan-1 ; 2Tw2 2 E at Tw = 1
1 + 3T w
3
R
V
1
S 2
W
-1
3 W = tan-1 1 = 30c
= tan S
; 3E
SS1 + 3 # 1 WW
3
T
X
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7.113
7.114
Option (A) is correct.
G (jw) H (jw) enclose the (- 1, 0) point so here G (jwp) H (jwp) > 1
wp = Phase cross over frequency
1
Gain Margin = 20 log 10
G (jwp) H (jwp)
so gain margin will be less than zero.
Option (B) is correct.
The denominator of Transfer function is called the characteristic
equation of the system. so here characteristic equation is
(s + 1) 2 (s + 2) = 0
h (t) = u (t) - u (t - T)
Taking Laplace transform we have
H (s) = 1 - 1 e-sT = 1 61 - e-sT @
s s
s
7.110
7.111
7.112
7.115
7.116
Option (C) is correct.
Phase margin = 180c + qg where qg = value of phase at gain
crossover frequency.
Here
qg =- 125c
so
P.M = 180c - 125c = 55c
Option (B) is correct.
Open loop transfer function is given by
K (1 + 0.5s)
G (s) H (s) =
s (1 + s) (1 + 2s)
Close looped system is of type 1.
It must be noted that type of the system is defined as no. of poles
lies at origin in OLTF.
lying
Option (D) is correct.
Transfer function of the phase lead controller is
1 + (3Tw) j
T.F = 1 + 3Ts =
1+s
1 + (Tw) j
Option (C) is correct.
In synchro error detector, output voltage is proportional to [w (t)],
where w (t) is the rotor velocity so here n = 1
Option (C) is correct.
By masson’s gain formulae
/ Dk Pk
y
=
x
D
Forward path gain
so gain
7.117
P1 = 5 # 2 # 1 = 10
D = 1 - (2 # - 2) = 1 + 4 = 5
D1 = 1
y
= 10 # 1 = 2
5
x
Option (C) is correct.
By given matrix equations we can have
Xo1 = dx1 = x1 - x2 + 0
dt
Xo2 = dx2 = 0 + x2 + m
dt
x1
y = [1 1] > H = x1 + x2
x2
dy
= dx1 + dx2
dt
dt
dt
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 184
dy
= x1 + m
dt
dy
dt
= x1 (0) + m (0)
t=0
= 1+0 = 0
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
UNIT 8
Page 184
8.6
COMMUNICATION SYSTEMS
The optimum threshold to achieve minimum bit error rate (BER) is
(A) 1
(B) 4
2
5
(C) 1
(D) 3
2
2012
2013
8.1
ONE MARK
The bit rate of a digital communication system is R kbits/s . The
modulation used is 32-QAM. The minimum bandwidth required for
ISI free transmission is
(A) R/10 Hz
(B) R/10 kHz
(C) R/5 Hz
(D) R/5 kHz
2013
8.2
8.7
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8.8
P ^3V F 2U h is
8.3
8.4
(B) 1/2
(D) 5/9
Consider two identically distributed zero-mean random variables U
and V . Let the cumulative distribution functions of U and 2V be
F ^x h and G ^x h respectively. Then, for all values of x
(A) F ^x h - G ^x h # 0
(B) F ^x h - G ^x h $ 0
(C) ^F (x) - G (x)h .x # 0
(D) ^F (x) - G (x)h .x $ 0
8.9
8.10
Let U and V be two independent and identically distributed random
variables such that P ^U =+ 1h = P ^U =- 1h = 1 . The entropy
2
H ^U + V h in bits is
(A) 3/4
(B) 1
(C) 3/2
(D) log 2 3
Bits 1 and 0 are transmitted with equal probability. At the receiver, the pdf of the respective received signals for both bits are
as shown below.
If the detection threshold is 1, the BER will be
(A) 1
(B) 1
2
4
(C) 1
(D) 1
8
16
(A) 6000/p, 0
(B) 6400/p, 0
(C) 6400/p, 20/ (p 2 )
(D) 6000/p, 20/ (p 2 )
In a baseband communications link, frequencies upto 3500 Hz are
used for signaling. Using a raised cosine pulse with 75% excess
bandwidth and for no inter-symbol interference, the maxi mum
possible signaling rate in symbols per second is
(A) 1750
(B) 2625
(C) 4000
(D) 5250
A source alphabet consists of N symbols with the probability of the
first two symbols being the same. A source encoder increases the
probability of the first symbol by a small amount e and decreases
that of the second by e. After encoding, the entropy of the source
(A) increases
(B) remains the same
(C) increases only if N = 2
(D) decreases
Two independent random variables X and Y are uniformly
distributed in the interval 6- 1, 1@. The probability that max 6X, Y @
is less than 1/2 is
(A) 3/4
(B) 9/16
(C) 1/4
(D) 2/3
2012
TWO MARKS
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Common Data for Questions 5 and 6:
8.5
The power spectral density of a real process X (t) for positive
frequencies is shown below. The values of E [X 2 (t)] and E [X (t)] ,
respectively, are
TWO MARKS
Let U and V be two independent zero mean Gaussain random
variables of variances 1 and 1 respectively. The probability
9
4
(A) 4/9
(C) 2/3
ONE MARK
8.11
8.12
A BPSK scheme operating over an AWGN channel with noise
power spectral density of N 0 /2, uses equiprobable signals
s1 (t) = 2E sin (wc t) and s2 (t) =- 2E sin (wc t) over the symbol
T
T
interval (0, T). If the local oscillator in a coherent receiver is ahead
in phase by 45c with respect to the received signal, the probability
of error in the resulting system is
(A) Q c 2E m
(B) Q c E m
N0
N0
E
E
(C) Q c
(D) Q c
2N 0 m
4N 0 m
A binary symmetric channel (BSC) has a transition probability of
1/8. If the binary symbol X is such that P (X = 0) = 9/10, then the
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 185
probability of error for an optimum receiver will be
(B) 63/80
(A) 7/80
(C) 9/10
(D) 1/10
8.13
(D) P-2, Q-4, R-3, S-1
2011
The signal m (t) as shown is applied to both a phase modulator
(with k p as the phase constant) and a frequency modulator (with k f
as the frequency constant) having the same carrier frequency.
The ratio k p /k f (in rad/Hz) for the same maximum phase deviation is
(A) 8p
(B) 4p
(C) 2p
(D) p
Statement for Linked Answer Question 14 and 15 :
8.14
8.15
The transfer function of a compensator is given as
Gc (s) = s + a
s+b
Gc (s) is a lead compensator if
(A) a = 1, b = 2
(B) a = 3, b = 2
(D) a = 3, b = 1
(C) a =- 3, b =- 1
8.18
6 rad/s
X (t) is a stationary random process with auto-correlation function
RX (t) = exp (- pt 2). This process is passed through the system
shown below. The power spectral density of the output process Y (t)
is
(A) (4p 2 f 2 + 1) exp (- pf 2)
(C) (4p 2 f 2 + 1) exp (- pf )
8.19
(B) (4p 2 f 2 - 1) exp (- pf 2)
(D) (4p 2 f 2 - 1) exp (- pf )
A message signal m (t) = cos 2000pt + 4 cos 4000pt modulates the
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carrier c (t) = cos 2pfc t where fc = 1 MHz to produce an AM signal.
For demodulating the generated AM signal using an envelope
detector, the time constant RC of the detector circuit should satisfy
(A) 0.5 ms < RC < 1 ms
(B) 1 μs << RC < 0.5 ms
(C) RC << 1 μs
(D) RC >> 0.5 ms
The phase of the above lead compensator is maximum at
(A) 2 rad/s
(B) 3 rad/s
(C)
TWO MARKS
(D) 1/ 3 rad/s
Common Data For Q. 8.5 & 8.6
2011
8.16
8.17
ONE MARK
A four-phase and an eight-phase signal constellation are shown in
the figure below.
An analog signal is band-limited to 4 kHz, sampled at the Nyquist
rate and the samples are quantized into 4 levels. The quantized
levels are assumed to be independent and equally probable. If we
transmit two quantized samples per second, the information rate is
(A) 1 bit/sec
(B) 2 bits/sec
(C) 3 bits/sec
(D) 4 bits/sec
The Column -1 lists the attributes and the Column -2 lists the
modulation systems. Match the attribute to the modulation system
that best meets it.
Column -1
Column -2
P.
Power efficient transmission of 1.
signals
Conventional
AM
Q.
Most
bandwidth
efficient 2.
transmission of voice signals
FM
R.
Simplest receiver structure
S.
Bandwidth efficient transmission 4.
of signals with significant dc
component
(A) P-4, Q-2, R-1, S-3
(B) P-2, Q-4, R-1, S-3
(C) P-3, Q-2, R-1, S-4
3.
8.20
8.21
VSB
SSB-SC
For the constraint that the minimum distance between pairs of
signal points be d for both constellations, the radii r 1 , and r 2 of the
circles are
(B) r 1 = 0.707d, r 2 = 1.932d
(A) r 1 = 0.707d, r2 = 2.782d
(C) r 1 = 0.707d, r 2 = 1.545d
(D) r 1 = 0.707d, r 2 = 1.307d
Assuming high SNR and that all signals are equally probable, the
additional average transmitted signal energy required by the 8-PSK
signal to achieve the same error probability as the 4-PSK signal is
(A) 11.90 dB
(B) 8.73 dB
(C) 6.79 dB
(D) 5.33 dB
2010
8.22
ONE MARK
Suppose that the modulating signal is m (t) = 2 cos (2pfm t) and the
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 186
carrier signal is xC (t) = AC cos (2pfC t), which one of the following is
a conventional AM signal without over-modulation
(A) x (t) = AC m (t) cos (2pfC t)
(B) x (t) = AC [1 + m (t)] cos (2pfC t)
(C) x (t) = AC cos (2pfC t) + AC m (t) cos (2pfC t)
4
gain and cut-off frequency 1 MHz. Let Yk represent the random
variable y (tk ).
Yk = Nk , if transmitted bit bk = 0
Yk = a + Nk if transmitted bit bk = 1
Where Nk represents the noise sample value. The noise sample has
a probability density function, PNk (n) = 0.5ae- a n (This has mean
zero and variance 2/a 2 ). Assume transmitted bits to be equiprobable and threshold z is set to a/2 = 10-6 V .
(D) x (t) = AC cos (2pfm t) cos (2pfC t) + AC sin (2pfm t) sin (2pfC t)
8.23
Consider an angle modulated signal
x (t) = 6 cos [2p # 106 t + 2 sin (800pt)] + 4 cos (800pt)
The average power of x (t) is
(A) 10 W
(B) 18 W
(C) 20 W
(D) 28 W
8.24
Consider the pulse shape s (t) as shown below. The impulse response
h (t) of the filter matched to this pulse is
8.25
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8.26
8.27
8.28
The value of the parameter a (in V - 1 ) is
(A) 1010
(B) 107
(C) 1.414 # 10-10
(D) 2 # 10-20
The probability of bit error is
(A) 0.5 # e-3.5
(C) 0.5 # e-7
(B) 0.5 # e-5
(D) 0.5 # e-10
The Nyquist sampling rate for the signal
sin (500pt) sin (700) pt
is given by
s (t) =
#
pt
pt
(A) 400 Hz
(B) 600 Hz
(C) 1200 Hz
(D) 1400 Hz
X (t) is a stationary process with the power spectral density Sx (f ) > 0
, for all f . The process is passed through a system shown below
Let Sy (f ) be the power spectral density of Y (t). Which one of the
following statements is correct
(A) Sy (f ) > 0 for all f
(B) Sy (f ) = 0 for f > 1 kHz
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(C) Sy (f ) = 0 for f = nf0, f0 = 2 kHz kHz, n any integer
(D) Sy (f ) = 0 for f = (2n + 1) f0 = 1 kHz , n any integer
2009
2010
TWO MARKS
Statement for linked Answer Question : 8.10 & 8.11 :
Consider a baseband binary PAM receiver shown below. The
additive channel noise n (t) is with power spectral density
Sn (f ) = N 0 /2 = 10-20 W/Hz . The low-pass filter is ideal with unity
8.29
ONE MARK
For a message siganl m (t) = cos (2pfm t) and carrier of frequency fc ,
which of the following represents a single side-band (SSB) signal ?
(B) cos (2pfc t)
(A) cos (2pfm t) cos (2pfc t)
(C) cos [2p (fc + fm) t]
(D) [1 + cos (2pfm t) cos (2pfc t)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2009
8.30
TWO MARKS
Consider two independent random variables X and Y with identical
distributions. The variables X and Y take values 0, 1 and 2 with
probabilities 12 , 14 and 1 respectively. What is the conditional
4
probability P (X + Y = 2 X - Y = 0) ?
(A) 0
(C) 1/6
8.31
8.33
2008
8.37
TWO MARKS
The probability density function (pdf) of random variable is as
shown below
(B) 1/16
(D) 1
The corresponding commutative distribution function CDF has the
form
A discrete random variable X takes values from 1 to 5 with
probabilities as shown in the table. A student calculates the mean
X as 3.5 and her teacher calculates the variance of X as 1.5. Which
of the following statements is true ?
k
1
2
3
4
5
P (X = k)
0.1
0.2
0.3
0.4
0.5
(A) Both the student and the teacher are right
(B) Both the student and the teacher are wrong
(C) The student is wrong but the teacher is right
(D) The student is right but the teacher is wrong
8.32
Page 187
A message signal given by m (t) = ( 12 ) cos w1 t - ( 12 ) sin w2 t
amplitude - modulated with a carrier of frequency wC to generator
s (t)[ 1 + m (t)] cos wc t . What is the power efficiency achieved by this
modulation scheme ?
(B) 11.11%
(A) 8.33%
(C) 20%
(D) 25%
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A communication channel with AWGN operating at a signal to noise
ration SNR >> 1 and bandwidth B has capacity C1. If the SNR is
doubled keeping constant, the resulting capacity C2 is given by
(B) C2 . C1 + B
(A) C2 . 2C1
(C) C2 . C1 + 2B
(D) C2 . C1 + 0.3B
Common Data For Q. 8.19 & 8.20 :
The amplitude of a random signal is uniformly distributed between
-5 V and 5 V.
8.34
If the signal to quantization noise ratio required in uniformly
quantizing the signal is 43.5 dB, the step of the quantization is
approximately
(A) 0.033 V
(B) 0.05 V
(C) 0.0667 V
(D) 0.10 V
8.38
8.35
If the positive values of the signal are uniformly quantized with a
step size of 0.05 V, and the negative values are uniformly quantized
with a step size of 0.1 V, the resulting signal to quantization noise
ration is approximately
(A) 46 dB
(B) 43.8 dB
(C) 42 dB
(D) 40 dB
2008
8.36
8.39
ONE MARK
Consider
the
amplitude
modulated
(AM)
signal
Ac cos wc t + 2 cos wm t cos wc t . For demodulating the signal using
envelope detector, the minimum value of Ac should be
(A) 2
(B) 1
(C) 0.5
(D) 0
8.40
A memory less source emits n symbols each with a probability p.
The entropy of the source as a function of n
(A) increases as log n
(B) decreases as log ( n1 )
(C) increases as n
(D) increases as n log n
Noise with double-sided power spectral density on K over all
frequencies is passed through a RC low pass filter with 3 dB cut-off
frequency of fc . The noise power at the filter output is
(A) K
(B) Kfc
(D) 3
(C) kpfc
Consider a Binary Symmetric Channel (BSC) with probability of
error being p. To transmit a bit, say 1, we transmit a sequence of
three 1s. The receiver will interpret the received sequence to represent
1 if at least two bits are 1. The probability that the transmitted bit
will be received in error is
(B) p3
(A) p3 + 3p2 (1 - p)
(D) p3 + p2 (1 - p)
(C) (1 - p3)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
8.41
8.42
8.43
Page 188
(A) E [X2] - E2 [X]
(C) E [X2]
Four messages band limited to W, W, 2W and 3W respectively are
to be multiplexed using Time Division Multiplexing (TDM). The
minimum bandwidth required for transmission of this TDM signal is
(A) W
(B) 3W
(C) 6W
(D) 7W
Consider the frequency modulated signal
10 cos [2p # 105 t + 5 sin (2p # 1500t) + 7.5 sin (2p # 1000t)]
with carrier frequency of 105 Hz. The modulation index is
(A) 12.5
(B) 10
(C) 7.5
(D) 5
The signal cos wc t + 0.5 cos wm t sin wc t is
(A) FM only
(B) AM only
(C) both AM and FM
(D) neither AM nor FM
2007
8.50
8.51
8.52
Common Data For Q. 8.29, 8.30 and 8.31 :
A speed signal, band limited to 4 kHz and peak voltage varying
between +5 V and - 5 V, is sampled at the Nyquist rate. Each
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(B) E [X2] + E2 [X]
(D) E2 [X]
A Hilbert transformer is a
(A) non-linear system
(C) time-varying system
(B) non-causal system
(D) low-pass system
In delta modulation, the slope overload distortion can be reduced by
(A) decreasing the step size
(B) decreasing the granular
noise
(C) decreasing the sampling rate (D) increasing the step size
The raised cosine pulse p (t) is used for zero ISI in digital
communications. The expression for p (t) with unity roll-off factor
is given by
sin 4pWt
p (t) =
4pWt (1 - 16W2 t2)
The value of p (t) at t = 1 is
4W
(A) - 0.5
(C) 0.5
8.53
TWO MARKS
(B) 0
(D) 3
In the following scheme, if the spectrum M (f) of m (t) is as shown,
then the spectrum Y (f) of y (t) will be
sample is quantized and represented by 8 bits.
8.44
8.45
8.46
If the bits 0 and 1 are transmitted using bipolar pulses, the minimum
bandwidth required for distortion free transmission is
(A) 64 kHz
(B) 32 kHz
(C) 8 kHz
(D) 4 kHz
Assuming the signal to be uniformly distributed between its peak to
peak value, the signal to noise ratio at the quantizer output is
(A) 16 dB
(B) 32 dB
(C) 48 dB
(D) 4 kHz
Assuming the signal to be uniformly distributed between its peak to
peak value, the signal to noise ratio at the quantizer output is
(A) 1024
(B) 512
(C) 256
(D) 64
2007
8.47
8.48
If S (f) is the power spectral density of a real, wide-sense stationary
random process, then which of the following is ALWAYS true?
(B) S (f) $ 0
(A) S (0) # S (f)
(D)
3
#- 3 S (f) df = 0
If E denotes expectation, the variance of a random variable X is
given by
During transmission over a certain binary communication channel,
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ONE MARK
If R (t) is the auto correlation function of a real, wide-sense stationary
random process, then which of the following is NOT true
(A) R (t) = R (- t)
(B) R (t) # R (0)
(C) R (t) =- R (- t)
(D) The mean square value of the process is R (0)
(C) S (- f) =- S (f)
8.49
8.54
bit errors occur independently with probability p. The probability
of AT MOST one bit in error in a block of n bits is given by
(A) pn
(B) 1 - pn
(C) np (1 - p) n - 1 + (1 + p) n
8.55
(D) 1 - (1 - p) n
In a GSM system, 8 channels can co-exist in 200 kHz bandwidth
using TDMA. A GSM based cellular operator is allocated 5 MHz
bandwidth. Assuming a frequency reuse factor of 1 , i.e. a five-cell
5
repeat pattern, the maximum number of simultaneous channels that
can exist in one cell is
(A) 200
(B) 40
(C) 25
(D) 5
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
8.56
In a Direct Sequence CDMA system the chip rate is 1.2288 # 106
chips per second. If the processing gain is desired to be AT LEAST
100, the data rate
(A) must be less than or equal to 12.288 # 103 bits per sec
(B) must be greater than 12.288 # 103 bits per sec
(C) must be exactly equal to 12.288 # 103 bits per sec
(D) can take any value less than 122.88 # 103 bits per sec
Common Data For Q. 8.41 & 8.42 :
Page 189
2006
8.61
8.58
If these constellations are used for digital communications over an
AWGN channel, then which of the following statements is true ?
(A) Probability of symbol error for Constellation 1 is lower
(B) Probability of symbol error for Constellation 1 is higher
(C) Probability of symbol error is equal for both the constellations
(D) The value of N0 will determine which of the constellations has
a lower probability of symbol error
8.62
8.60
The values of a and b are
(B) a = 1 and b = 3
(A) a = 1 and b = 1
12
5
40
6
(C) a = 1 and b = 1
(D) a = 1 and b = 1
4
16
3
24
Assuming that the reconstruction levels of the quantizer are the
mid-points of the decision boundaries, the ratio of signal power to
quantization noise power is
(B) 64
(A) 152
3
9
(C) 76
(D) 28
3
A signal with bandwidth 500 Hz is first multiplied by a signal g (t)
where
3
/(- 1)k d (t - 0.5 # 10- 4 k)
R =- 3
The resulting signal is then passed through an ideal lowpass filter
with bandwidth 1 kHz. The output of the lowpass filter would be
(A) d (t)
(B) m (t)
(C) 0
(D) m (t) d (t)
The minimum sampling frequency (in samples/sec) required to
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reconstruct the following signal from its samples without distortion
3
2
x (t) = 5` sin 2p100t j + 7` sin 2p100t j would be
pt
pt
3
3
(B) 4 # 10
(A) 2 # 10
3
(C) 6 # 10
(D) 8 # 103
8.64
The minimum step-size required for a Delta-Modulator operating
at 32k samples/sec to track the signal (here u (t) is the unit-step
function)
x (t) = 125[ u (t) - u (t - 1) + (250t)[ u (t - 1) - u (t - 2)]
so that slope-overload is avoided, would be
(A) 2 - 10
(B) 2 - 8
(C) 2 - 6
8.65
8.66
8.59
TWO MARKS
g (t) =
Statement for Linked Answer Question 8.44 & 8.45 :
An input to a 6-level quantizer has the probability density function
f (x) as shown in the figure. Decision boundaries of the quantizer
are chosen so as to maximize the entropy of the quantizer output.
It is given that 3 consecutive decision boundaries are’ - 1'.'0' and
' 1' .
(D) A (w) = C, f (w) = kw- 1
2006
8.63
The if ratio or the average energy of Constellation 1 to the average
energy of Constellation 2 is
(A) 4a2
(B) 4
(C) 2
(D) 8
A low-pass filter having a frequency response H (jw) = A (w) e jf (w)
does not produce any phase distortions if
(B) A (w) = Cw2, f (w) = kw
(A) A (w) = Cw3, f (w) = kw3
(C) A (w) = Cw, f (w) = kw2
Two 4-array signal constellations are shown. It is given that f1
and f2 constitute an orthonormal basis for the two constellation. Assume that the four symbols in both the constellations are
equiprobable. Let N0 denote the power spectral density of white
2
Gaussian noise.
8.57
ONE MARK
8.67
(D) 2 - 4
A zero-mean white Gaussian noise is passes through an ideal lowpass
filter of bandwidth 10 kHz. The output is then uniformly sampled
with sampling period ts = 0.03 msec. The samples so obtained would
be
(A) correlated
(B) statistically independent
(C) uncorrelated
(D) orthogonal
A source generates three symbols with probabilities 0.25, 0.25,
0.50 at a rate of 3000 symbols per second. Assuming independent
generation of symbols, the most efficient source encoder would have
average bit rate is
(A) 6000 bits/sec
(B) 4500 bits/sec
(C) 3000 bits/sec
(D) 1500 bits/sec
The diagonal clipping in Amplitude Demodulation (using envelop
detector) can be avoided it RC time-constant of the envelope detector
satisfies the following condition, (here W is message bandwidth and
w is carrier frequency both in rad/sec)
(B) RC > 1
(A) RC < 1
W
W
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
8.68
Page 190
(C) RC < 1
(D) RC > 1
w
w
A uniformly distributed random variable X with probability density
function
fx (x) = 1 pu (x + 5) - u (x - 5)]
10
signal and modulation index would be
(A) t, 0.5
(B) t, 1.0
(C) t, 2.0
(D) t2, 0.5
where u (.) is the unit step function is passed through a transformation given in the figure below. The probability density function
of the transformed random variable Y would be
The following two question refer to wide sense stationary stochastic process
Common Data For Q. 8.58 & 8.59 :
8.73
(A) fy (y) = 1 [u (y + 2.5) - u (y - 2.25)]
5
(B) fy (y) = 0.5d (y) + 0.5d (y - 1)
(C) fy (y) = 0.25d (y + 2.5) + 0.25d (y - 2.5) + 5d (y)
(D) fy (y) = 0.25d (y + 2.5) + 0.25d (y - 2.5) + 1 [u (y + 2.5) - u (y - 2.5)]
10
8.74
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8.69
Consider the following Amplitude Modulated (AM) signal, where
fm < B
XAM (t) = 10 (1 + 0.5 sin 2pfm t) cos 2pfc t
8.75
8.76
(A) 3
8.70
2
(C) 3
(D) 3
12
L
A message signal with bandwidth 10 kHz is Lower-Side Band SSB
modulated with carrier frequency fc1 = 106 Hz. The resulting signal
is then passed through a Narrow-Band Frequency Modulator with
carrier frequency fc2 = 109 Hz.
The bandwidth of the output would be
(B) 2 # 106 Hz
(A) 4 # 10 4 Hz
(D) 2 # 1010 Hz
(C) 2 # 109 Hz
The parameters of the system obtained in previous Q would be
(A) first order R-L lowpass filter would have R = 4W L = 1H
(B) first order R-C highpass filter would have R = 4W C = 0.25F
(C) tuned L-C filter would have L = 4H C = 4F
(D) series R-L-C lowpass filter would have R = 1W , L = 4H ,
C = 4F
Common Data For Q. 8.60 & 8.61 :
In the following figure the minimum value of the constant "C" , which
is to be added to y1 (t) such that y1 (t) and y2 (t) are different , is
(B) 3
2
It is desired to generate a stochastic process (as voltage process)
with power spectral density S (w) = 16/ (16 + w2) by driving a
Linear-Time-Invariant system by zero mean white noise (As voltage
process) with power spectral density being constant equal to 1. The
system which can perform the desired task could be
(A) first order lowpass R-L filter
(B) first order highpass R-C filter
(C) tuned L-C filter
(D) series R-L-C filter
The average side-band power for the AM signal given above is
(A) 25
(B) 12.5
(C) 6.25
(D) 3.125
The AM signal gets added to a noise with Power Spectral Density
Sn (f) given in the figure below. The ratio of average sideband power
to mean noise power would be :
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Common Data For Q. 8.56 & 8.57 :
Let g (t) = p (t)*( pt), where * denotes convolution &
p (t) = u (t) - u (t - 1) lim with u (t) being the unit step function
z"3
8.71
8.72
The impulse response of filter matched
s (t) = g (t) - d (1 - 2)* g (t) is given as :
(A) s (1 - t)
(B) - s (1 - t)
(C) - s (t)
(D) s (t)
to
the
signal
An Amplitude Modulated signal is given as
xAM (t) = 100 [p (t) + 0.5g (t)] cos wc t
in the interval 0 # t # 1. One set of possible values of modulating
(A)
25
8N0 B
(B)
25
4N0 B
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C)
25
2N0 B
(D) 25
N0 B
2005
8.77
Page 191
the probability density function as shown in the figure. The mean
square value of v is
ONE MARK
Find the correct match between group 1 and group 2.
Group 1
Group 2
W. Phase modulation
P. {1 + km (t) A sin (wc t)}
Q. km (t) A sin (wc t)
X. Frequency modulation
R. A sin {wc t + km (t)}
Y. Amplitude modulation
t
S. A sin ; wc t + k m (t) dt E
Z. DSB-SC modulation
(A) 4
(C) 8
#- 3
(A)
(B)
(C)
(D)
8.78
P - Z, Q - Y, R - X, S - W
P - W, Q - X, R - Y, S - Z
P - X, Q - W, R - Z, S - Y
P - Y, Q - Z, R - W, S - X
Which of the following analog
minimum transmitted power and
(A) VSB
(C) SSB
2005
8.79
8.80
8.83
modulation scheme requires the
minimum channel bandwidth ?
(B) DSB-SC
(D) AM
TWO MARKS
A device with input X (t) and output y (t) is characterized by:
Y (t) = x2 (t). An FM signal with frequency deviation of 90 kHz and
modulating signal bandwidth of 5 kHz is applied to this device. The
bandwidth of the output signal is
(A) 370 kHz
(B) 190 kHz
(C) 380 kHz
(D) 95 kHz
(B) 6
(D) 9
A carrier is phase modulated (PM) with frequency deviation of 10
kHz by a single tone frequency of 1 kHz. If the single tone frequency
is increased to 2 kHz, assuming that phase deviation remains
unchanged, the bandwidth of the PM signal is
(A) 21 kHz
(B) 22 kHz
(C) 42 kHz
(D) 44 kHz
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Common Data For Q. 8.69 and 8.70 :
Asymmetric three-level midtread quantizer is to be designed assuming equiprobable occurrence of all quantization levels.
A signal as shown in the figure is applied to a matched filter. Which
of the following does represent the output of this matched filter ?
8.84
8.85
If the probability density function is divide into three regions as
shown in the figure, the value of a in the figure is
(B) 2
(A) 1
3
3
(C) 1
(D) 1
2
4
The quantization noise power for the quantization region between
- a and + a in the figure is
(B) 1
(A) 4
9
81
(C) 5
(D) 2
81
81
2004
8.86
8.81
8.82
Noise with uniform power spectral density of N0 W/Hz is passed
though a filter H (w) = 2 exp (- jwtd ) followed by an ideal pass filter
of bandwidth B Hz. The output noise power in Watts is
(A) 2N0 B
(B) 4N0 B
(C) 8N0 B
(D) 16N0 B
An output of a communication channel is a random variable v with
In a PCM system, if the code word length is increased from 6 to 8
bits, the signal to quantization noise ratio improves by the factor
(B) 12
(A) 8
6
(C) 16
8.87
ONE MARK
(D) 8
An AM signal is detected using an envelop detector. The carrier
frequency and modulating signal frequency are 1 MHz and 2 kHz
respectively. An appropriate value for the time constant of the
envelop detector is
(A) 500m sec
(B) 20m sec
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) 0.2m sec
8.88
8.89
8.90
Page 192
symbols are represented as shown in the figure.
The source output is transmitted using two modulation schemes,
namely Binary PSK (BPSK) and Quadrature PSK (QPSK). Let
B1 and B2 be the bandwidth requirements of the above rectangular
pulses is 10 kHz, B1 and B2 are
(D) 1m sec
An AM signal and a narrow-band FM signal with identical carriers,
modulating signals and modulation indices of 0.1 are added together.
The resultant signal can be closely approximated by
(A) broadband FM
(B) SSB with carrier
(C) DSB-SC
(D) SSB without carrier
In the output of a DM speech encoder, the consecutive pulses are of
opposite polarity during time interval t1 # t # t2 . This indicates that
during this interval
(A) the input to the modulator is essentially constant
(B) the modulator is going through slope overload
(C) the accumulator is in saturation
(D) the speech signal is being sampled at the Nyquist rate
(A) B1 = 20 kHz, B2 = 20 kHz
(C) B1 = 20 khz, B2 = 10 kHz
8.94
The distribution function Fx (x) of a random variable x is shown in
the figure. The probability that X = 1 is
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2004
8.91
8.92
8.93
(B) 0.25
(D) 0.30
TWO MARKS
A 1 mW video signal having a bandwidth of 100 MHz is transmitted
to a receiver through cable that has 40 dB loss. If the effective oneside noise spectral density at the receiver is 10 - 20 Watt/Hz, then the
signal-to-noise ratio at the receiver is
(A) 50 dB
(B) 30 dB
(C) 40 dB
(D) 60 dB
(A) 1 sec - 1
(B) - 1 sec - 1
2
(C) - 1 sec - 1
(D) 1 sec - 1
2
A source produces binary data at the rate of 10 kbps. The binary
(B)
1 + sin (2p # 106 t)
5 - sin (2p - 106 t)
5 + cos (2p # 106 t)
(D)
4
4
Two sinusoidal signals of same amplitude and frequencies 10 kHz
and 10.1 kHz are added together. The combined signal is given to an
ideal frequency detector. The output of the detector is
(A) 0.1 kHz sinusoid
(B) 20.1 kHz sinusoid
(C) a linear function of time
(D) a constant
(C)
8.96
Consider the signal x (t) shown in Fig. Let h (t) denote the impulse
response of the filter matched to x (t), with h (t) being non-zero only
in the interval 0 to 4 sec. The slope of h (t) in the interval 3 < t < 4
sec is
A 100 MHz carrier of 1 V amplitude and a 1 MHz modulating signal
of 1 V amplitude are fed to a balanced modulator. The ourput of
the modulator is passed through an ideal high-pass filter with cutoff frequency of 100 MHz. The output of the filter is added with 100
MHz signal of 1 V amplitude and 90c phase shift as shown in the
figure. The envelope of the resultant signal is
(A) constant
8.95
(A) zero
(C) 0.55
(B) B1 = 10 kHz, B2 = 20 kHz
(D) B1 = 10 kHz, B2 = 10 kHz
Consider a binary digital communication system with equally likely
0’s and 1’s. When binary 0 is transmitted the detector input can
lie between the levels - 0.25 V and + 0.25 V with equl probability :
when binary 1 is transmitted, the voltage at the detector can have
any value between 0 and 1 V with equal probability. If the detector
has a threshold of 0.2 V (i.e., if the received signal is greater than
0.2 V, the bit is taken as 1), the average bit error probability is
(A) 0.15
(B) 0.2
(C) 0.05
(D) 0.5
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8.97
8.98
A random variable X with uniform density in the interval 0 to 1 is
quantized as follows :
If 0 # X # 0.3 ,
xq = 0
If 0.3 < X # 1,
xq = 0.7
where xq is the quantized value of X.
The root-mean square value of the quantization noise is
(A) 0.573
(B) 0.198
(D) 0.266
(C) 2.205
Choose the current one from among the alternative A, B, C, D after
matching an item from Group 1 with the most appropriate item in
Group 2.
Group 1
Group 2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
1.
2.
3.
4.
P. Slope overload
Q. m-law
R. Envelope detector
S. Hilbert transform
T. Hilbert transform
U. Matched filter
(A) 1 - T, 2 - P, 3 - U, 4 - S
(B) 1 - S, 2 - U, 3 - P, 4 - T
(C) 1 - S, 2 - P, 3 - U, 4 - Q
(D) 1 - U, 2 - R, 3 - S, 4 - Q
8.99
8.100
Page 193
Common Data For Q. 8.90 & 8.91 :
FM
DM
PSK
PCM
Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600
Hz, are sampled at their respective Nyquist rates, encoded with
12 bit words, and time division multiplexed. The bit rate for the
multiplexed. The bit rate for the multiplexed signal is
(A) 115.2 kbps
(B) 28.8 kbps
(C) 57.6 kbps
(D) 38.4 kbps
Consider a system shown in the figure. Let X (f) and Y (f) and denote
the Fourier transforms of x (t) and y (t) respectively. The ideal HPF
has the cutoff frequency 10 kHz.
X (t) is a random process with a constant mean value of 2 and the
auto correlation function Rxx (t) = 4 (e - 0.2 t + 1).
8.105
Let X be the Gaussian random variable obtained by sampling the
process at t = ti and let
3
Q (a) =
- 1 e dy
a
2p
The probability that 6x # 1@ is
(B) Q (0.5)
(A) 1 - Q (0.5)
1
(C) Q c
(D) 1 - Q c 1 m
m
2 2
2 2
Let Y and Z be the random variable obtained by sampling X (t) at
t = 2 and t = 4 respectively. Let W = Y - Z . The variance of W is
(A) 13.36
(B) 9.36
(C) 2.64
(D) 8.00
#
8.106
8.107
x2
2
A sinusoidal signal with peak-to-peak amplitude of 1.536 V is
quantized into 128 levels using a mid-rise uniform quantizer. The
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The positive frequencies where Y (f) has spectral peaks are
(A) 1 kHz and 24 kHz
(B) 2 kHz and 244 kHz
(C) 1 kHz and 14 kHz
(D) 2 kHz and 14 kHz
2003
8.101
8.102
8.103
ONE MARK
(B) 48 # 10 - 6 V2
(B) 12 # 10 - 6 V2
(D) 3.072 V
Let x (t) = 2 cos (800p) + cos (1400pt). x (t) is sampled with the
rectangular pulse train shown in the figure. The only spectral
components (in kHz) present in the sampled signal in the frequency
range 2.5 kHz to 3.5 kHz are
The input to a coherent detector is DSB-SC signal plus noise. The
noise at the detector output is
(A) the in-phase component
(B) the quadrature - component
(C) zero
(D) the envelope
The noise at the input to an ideal frequency detector is white. The
detector is operating above threshold. The power spectral density of
the noise at the output is
(A) raised - cosine
(B) flat
(C) parabolic
(D) Gaussian
(A) 2.7, 3.4
(C) 2.6, 2.7, 3.3, 3.4, 3.6
8.109
At a given probability of error, binary coherent FSK is inferior to
binary coherent PSK by.
(A) 6 dB
(B) 3 dB
(C) 2 dB
(D) 0 dB
2003
8.104
8.108
quantization-noise power is
(A) 0.768 V
(B) 3.3, 3.6
(D) 2.7, 3.3
A DSB-SC signal is to be generated with a carrier frequency fc = 1
MHz using a non-linear device with the input-output characteristic
V0 = a0 vi + a1 vi3 where a0 and a1 are constants. The output of the
non-linear device can be filtered by an appropriate band-pass filter.
Let Vi = Aci cos (2pfi ct) + m (t) is the message signal. Then the value
of fci (in MHz) is
(A) 1.0
(B) 0.333
(B) 0.5
(D) 3.0
TWO MARKS
Let X and Y be two statistically independent random variables
uniformly distributed in the ranges (- 1, 1) and (- 2, 1) respectively.
Let Z = X + Y . Then the probability that (z #- 1) is
(A) zero
(B) 1
6
(C) 1
(D) 1
3
12
Common Data For Q. 8.95 & 8.96 :
Let m (t) = cos [(4p # 103) t] be the message signal &
c (t) = 5 cos [(2p # 106 t)] be the carrier.
8.110
c (t) and m (t) are used to generate an AM signal. The modulation
index of the generated AM signal is 0.5. Then the quantity
Total sideband power
is
Carrier power
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) 1
2
(C) 1
3
8.111
8.112
(B) 1
4
(D) 1
8
c (t) and m (t) are used to generated an FM signal. If the peak
frequency deviation of the generated FM signal is three times the
transmission bandwidth of the AM signal, then the coefficient of the
term cos [2p (1008 # 103 t)] in the FM signal (in terms of the Bessel
coefficients) is
(B) 5 J8 (3)
(A) 5J4 (3)
2
(C) 5 J8 (4)
(D) 5J4 (6)
2
Choose the correct one from among the alternative A, B, C, D after
matching an item in Group 1 with most appropriate item in Group
2.
Group 1
Group 2
P. Ring modulator
1. Clock recovery
Q. VCO
2. Demodulation of FM
R. Foster-Seely discriminator
3. Frequency conversion
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S. Mixer
(A) P - 1; Q - 3; R - 2; S - 4
(C) P - 6; Q - 1; R - 3; S - 2
8.113
8.114
8.115
8.116
4. Summing the two inputs
5. Generation of FM
6. Generation of DSB-Sc
(B) P - 6; Q = 5; R - 2; S - 3
(D) P - 5; Q - 6; R - 1; S - 3
Page 194
(A) r = 0.5, S is required
(C) r = 0.5, S is not required
8.117
The input to a linear delta modulator having a step-size 3= 0.628
is a sine wave with frequency fm and peak amplitude Em . If the
sampling frequency fx = 40 kHz, the combination of the sine-wave
frequency and the peak amplitude, where slope overload will take
place is
Em
fm
(A) 0.3 V
8 kHz
(B) 1.5 V
4 kHz
(C) 1.5 V
2 kHz
(D) 3.0 V
1 kHz
If S represents the carrier synchronization at the receiver and r
represents the bandwidth efficiency, then the correct statement for
the coherent binary PSK is
A signal is sampled at 8 kHz and is quantized using 8 - bit uniform
quantizer. Assuming SNRq for a sinusoidal signal, the correct
statement for PCM signal with a bit rate of R is
(A) R = 32 kbps, SNRq = 25.8 dB
(B) R = 64 kbps, SNRq = 49.8 dB
(C) R = 64 kbps, SNRq = 55.8 dB
(D) R = 32 kbps, SNRq = 49.8 dB
2002
8.118
8.119
ONE MARK
A 2 MHz sinusoidal carrier amplitude modulated by symmetrical
square wave of period 100 m sec . Which of the following frequencies
will NOT be present in the modulated signal ?
(A) 990 kHz
(B) 1010 kHz
(C) 1020 kHz
(D) 1030 kHz
Consider a sample signal y (t) = 5 # 10 - 6 # (t)
+3
/ d (t - nTs)
n =- 3
where x (t) = 10 cos (8p # 103) t and Ts = 100m sec.
When y (t) is passed through an ideal lowpass filter with a cutoff
frequency of 5 KHz, the output of the filter is
(A) 5 # 10 - 6 cos (8p # 103) t
(b) 5 # 10 - 5 cos (8p # 103) t
(C) 5 # 10 - 1 cos (8p # 103) t
8.120
8.121
A superheterodyne receiver is to operate in the frequency range 550
kHz - 1650 kHz, with the intermediate frequency of 450 kHz. Let
R = Cmax /Cmin denote the required capacitance ratio of the local
oscillator and I denote the image frequency (in kHz) of the incoming
signal. If the receiver is tuned to 700 kHz, then
(B) R = 2.10, I - 1150
(A) R = 4.41, I = 1600
(C) R = 3.0, I = 600
(D) R = 9.0, I = 1150
If Eb , the energy per bit of a binary digital signal, is 10 - 5 wattsec and the one-sided power spectral density of the white noise,
N0 = 10 - 6 W/Hz, then the output SNR of the matched filter is
(A) 26 dB
(B) 10 dB
(C) 20 dB
(D) 13 dB
(B) r = 1.0, S is required
(D) r = 1.0, S is not required
(D) 10 cos (8p # 103) t
For a bit-rate of 8 Kbps, the best possible values of the transmitted
frequencies in a coherent binary FSK system are
(A) 16 kHz and 20 kHz
(C) 20 kHz and 32 kHz
(C) 20 kHz and 40 kHz
(D) 32 kHz and 40 kHz
The line-of-sight communication requires the transmit and receive
antennas to face each other. If the transmit antenna is vertically
polarized, for best reception the receiver antenna should be
(A) horizontally polarized
(B) vertically polarized
(C) at 45c with respect to horizontal polarization
(D) at 45c with respect to vertical polarization
2002
8.122
TWO MARKS
An angle-modulated signal is given by
s (t) = cos 2p (2 # 106 t + 30 sin 150t + 40 cos 150t).
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The maximum frequency and phase deviations of s (t) are
(A) 10.5 kHz, 140p rad
(B) 6 kHz, 80p rad
(C) 10.5 kHz, 100p rad
(D) 7.5 kHz, 100p rad
8.123
In the figure m (t) = 2 sin 2pt , s (t) = cos 200pt and n (t) = sin 199pt
t
t
.
The output y (t) will be
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
8.124
8.125
(A) sin 2pt
(B) sin 2pt + sin pt cos 3pt
t
t
t
(C) sin 2pt + sin 0.5pt cos 1.5pt (D) sin 2pt + sin pt cos 0.75pt
t
t
t
t
3
A signal x (t) = 100 cos (24p # 10 ) t is ideally sampled with a sampling
period of 50m sec ana then passed through an ideal lowpass filter
with cutoff frequency of 15 kHz. Which of the following frequencies
is/are present at the filter output ?
(A) 12 kHz only
(B) 8 kHz only
(C) 12 kHz and 9 kHz
(D) 12 kHz and 8 kHz
If the variance ax2 of d (n) = x (n) - x (n - 1) is one-tenth the
variance ax2 of stationary zero-mean discrete-time signal x (n), then
R (k)
the normalized autocorrelation function xx 2 at k = 1 is
(A) 0.95
(B) 0.90 ax
(C) 0.10
8.127
of K0 m (t) is greater than 1. Which of the following could be the
detector output ?
(A) Ac m (t)
(B) Ac2 [1 + Ka m (t)] 2
(C) [Ac (1 + Ka m (t)]
(D) Ac [1 + Ka m (t)] 2
8.133
ONE MARK
A bandlimited signal is sampled at the Nyquist rate. The signal can
be recovered by passing the samples through
(A) an RC filter
(B) an envelope detector
(C) a PLL
(D) an ideal low-pass filter with the appropriate bandwidth
The frequency range for satellite communication is
(A) 1 KHz to 100 KHz
(B) 100 KHz to 10 KHz
(C) 10 MHz to 30 MHz
(D) 1 GHz to 30 GHz
2000
8.134
(D) 0.05
2001
8.126
Page 195
8.135
TWO MARKS
In a digital communication system employing Frequency Shift
Keying (FSK), the 0 and 1 bit are represented by sine waves of 10
KHz and 25 KHz respectively. These waveforms will be orthogonal
for a bit interval of
(A) 45m sec
(B) 200m sec
(C) 50m sec
(D) 250m sec
A message m (t) bandlimited to the frequency fm has a power of Pm
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The PDF of a Gaussian random variable X is given by
(x - 4)
px (x) = 1 e - 18 . The probability of the event {X = 4} is
3 2p
1
1
(B)
(A)
2
3 2p
(C) 0
(D) 1
4
2
2001
8.128
8.129
8.130
8.131
The Nyquist sampling interval, for the signal sin c (700t) + sin c (500t)
is
(A) 1 sec
(B) p sec
350
350
(C) 1 sec
(D) p sec
700
175
During transmission over a communication channel, bit errors occur
independently with probability p. If a block of n bits is transmitted,
the probability of at most one bit error is equal to
(B) p + (n - 1)( 1 - p)
(A) 1 - (1 - p) n
n-1
n
(D) (1 - p) + np (1 - p)
n-1
8.136
8.137
8.138
In a FM system, a carrier of 100 MHz modulated by a sinusoidal
signal of 5 KHz. The bandwidth by Carson’s approximation is 1
MHz. If y (t) = (modulated waveform) 3 , than by using Carson’s
approximation, the bandwidth of y (t) around 300 MHz and the and
the spacing of spectral components are, respectively.
(A) 3 MHz, 5 KHz
(B) 1 MHz, 15 KHz
(C) 3 MHz, 15 KHz
(D) 1 MHz, 5 KHz
ONE MARK
The input to a channel is a bandpass signal. It is obtained by linearly
modulating a sinusoidal carrier with a single-tone signal. The output
of the channel due to this input is given by
y (t) = (1/100) cos (100t - 10-6) cos (106 t - 1.56)
The group delay (tg) and the phase delay (t p) in seconds, of the
channel are
(A) tg = 10-6, t p = 1.56
(B) tg = 1.56, t p = 10-6
(C) tg = 108, t p = 1.56 # 10-6
(D) tg = 108, t p = 1.56
ONE MARK
The amplitude modulated waveform s (t) = Ac [1 + Ka m (t)] cos wc t
is fed to an ideal envelope detector. The maximum magnitude
(B) Pm
4
2
2
(C) Pm sin q
(D) Pm cos q
4
4
The Hilbert transform of cos w1 t + sin w2 t is
(A) sin w1 t - cos w2 t
(B) sin w1 t + cos w2 t
(C) cos w1 t - sin w2 t
(D) sin w1 t + sin w2 t
1999
The PSD and the power of a signal g (t) are, respectively, Sg (w) and
Pg . The PSD and the power of the signal ag (t) are, respectively,
(A) a2 Sg (w) and a2 Pg
(B) a2 Sg (w) and aPg
(C) aSg (w) and a2 Pg
(D) aSg (w) and aPs
2000
8.132
(A) Pm cos q
2
TWO MARKS
A video transmission system transmits 625 picture frames per second.
Each frame consists of a 400 # 400 pixel grid with 64 intensity levels
per pixel. The data rate of the system is
(A) 16 Mbps
(B) 100 Mbps
(C) 600 Mbps
(D) 6.4 Gbps
(C) np (1 - p)
. The power of the output signal in the figure is
8.139
A modulated signal is given by s (t) = m1 (t) cos (2pfc t) + m2 (t) sin (2pfc t)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
where the baseband signal m1 (t) and m2 (t) have bandwidths of
10 kHz, and 15 kHz, respectively. The bandwidth of the modulated
signal, in kHz, is
(A) 10
(B) 15
(C) 25
(D) 30
8.140
A modulated signal is given by s (t) = e-at cos [(wc + Dw) t] u (t),
where a wc and Dw are positive constants, and wc >> Dw . The
complex envelope of s (t) is given by
(A) exp (- at) exp [j (wc + Dw) t] u (t)
(B) exp (- at) exp (jDwt) u (t)
(C) exp (jDwt) u (t)
(D) exp [jwc + Dw) t]
1999
8.141
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8.142
8.143
8.144
(D) 1000
8.146
8.148
8.149
The input to a matched filter is given by
6
-4
sec
"10
0 sin (2p # 10 t) 0 < 1 < 10
s (t) =
otherwise
The peak amplitude of the filter output is
(A) 10 volts
(B) 5 volts
(C) 10 millivolts
(D) 5 millivolts
8.151
8.153
8.154
The image channel selectivity of superheterodyne receiver depends
upon
(A) IF amplifiers only
(B) RF and IF amplifiers only
(C) Preselector, RF and IF amplifiers
(D) Preselector, and RF amplifiers only
In a PCM system with uniform quantisation, increasing the number
of bits from 8 to 9 will reduce the quantisation noise power by a
factor of
(A) 9
(B) 8
(C) 4
(D) 2
Flat top sampling of low pass signals
(A) gives rise to aperture effect (B) implies oversampling
(C) leads to aliasing
(D) introduces delay distortion
A DSB-SC signal is generated using the carrier cos (we t + q) and
modulating signal x (t). The envelope of the DSB-SC signal is
(A) x (t)
(B) x (t)
(C) only positive portion of x (t) (D) x (t) cos q
Quadrature multiplexing is
(A) the same as FDM
(B) the same as TDM
(C) a combination of FDM and TDM
(D) quite different from FDM and TDM
The Fourier transform of a voltage signal x (t) is X (f). The unit of
X (f) is
(A) volt
(B) volt-sec
(C) volt/sec
(D) volt 2
Compression in PCM refers to relative compression of
(A) higher signal amplitudes
(B) lower signal amplitudes
(C) lower signal frequencies
(D) higher signal frequencies
For a give data rate, the bandwidth B p of a BPSK signal and the
bandwidth B 0 of the OOK signal are related as
(A) B p = B 0
(B) B p = B 0
2
4
(C) B p = B 0
Four independent messages have bandwidths of 100 Hz,
200 Hz and 400 Hz , respectively. Each is sampled at the Nyquist
rate, and the samples are time division multiplexed (TDM) and
transmitted. The transmitted sample rate (in Hz) is
(A) 1600
(B) 800
(C) 400
(D) 200
8.155
(D) B p = 2B 0
The spectral density of a real valued random process has
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(A) an even symmetry
(C) a conjugate symmetry
ONE MARK
The amplitude spectrum of a Gaussian pulse is
(A) uniform
(B) a sine function
(C) Gaussian
(D) an impulse function
The ACF of a rectangular pulse of duration T is
(A) a rectangular pulse of duration T
(B) a rectangular pulse of duration 2T
(C) a triangular pulse of duration T
(D) a triangular pulse of duration 2T
8.150
8.152
The peak-to-peak input to an 8-bit PCM coder is 2 volts. The signal
power-to-quantization noise power ratio (in dB) for an input of
0.5 cos (wm t) is
(A) 47.8
(B) 49.8
(C) 95.6
(D) 99.6
1998
8.145
8.147
TWO MARKS
The Nyquist sampling frequency (in Hz) of a signal given by
6 # 10 4 sin c2 (400t) * 106 sin c3 (100t) is
(A) 200
(B) 300
(C) 500
Page 196
8.156
The probability density function of the envelope of narrow band
Gaussian noise is
(A) Poisson
(B) Gaussian
(C) Rayleigh
(D) Rician
1997
8.157
(B) an odd symmetry
(D) no symmetry
ONE MARK
The line code that has zero dc component for pulse transmission of
random binary data is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 197
(A) Non-return to zero (NRZ)
(B) Return to zero (RZ)
(C) Alternate Mark Inversion (AM)
(D) None of the above
8.158
8.159
from
(A) IF stages only
(B) RF stages only
(C) detector and RF stages only (D) detector RF and IF stages
1996
2
A probability density function is given by p (x) = Ke-x /2 - 3 < x < 3
. The value of K should be
2
(B)
(A) 1
p
2p
(C) 1
(D) 1
2 p
p 2
A deterministic signal has the power spectrum given in the figure
is, The minimum sampling rate needed to completely represent this
signal is
8.164
The number of bits in a binary PCM system is increased from n to
n + 1. As a result, the signal to quantization noise ratio will improve
by a factor
(A) n + 1
(B) 2(n + 1)/n
n
(C) 22 (n + 1)/n
8.165
8.166
TWO MARKS
(D) which is independent of n
The auto correlation function of an energy signal has
(A) no symmetry
(B) conjugate symmetry
(C) odd symmetry
(D) even symmetry
An FM signal with a modulation index 9 is applied to a frequency
tripler. The modulation index in the output signal will be
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form. But at NODIA Online Store book is available in 10 book
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(A) 1 kHz
(C) 3 kHz
8.160
8.161
8.163
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A communication channel has first order low pass transfer function.
The channel is used to transmit pulses at a symbol rate greater than
the half-power frequency of the low pass function. Which of the
network shown in the figure is can be used to equalise the received
pulses?
The power spectral density of a deterministic signal is given by
[sin (f) /f 2] where f is frequency. The auto correlation function of
this signal in the time domain is
(A) a rectangular pulse
(B) a delta function
(C) a sine pulse
(D) a triangular pulse
1996
8.162
(B) 2 kHz
(D) None of these
ONE MARK
A rectangular pulse of duration T is applied to a filter matched to
this input. The out put of the filter is a
(A) rectangular pulse of duration T
(B) rectangular pulse of duration 2T
(C) triangular pulse
(D) sine function
The image channel rejection in a superheterodyne receiver comes
(A) 0
(C) 9
(B) 3
(D) 27
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 198
SOLUTIONS
8.1
Option (B) is correct.
In ideal Nyquist Channel, bandwidth required for ISI (Inter Symbol
reference) free transmission is
W = Rb
2
Here, the used modulation is 32 - QAM (Quantum Amplitude
modulation
i.e.,
q = 32
or
2v = 32
v = 5 bits
So, the signaling rate (sampling rate) is
Rb = R
5
(R " given bit rate)
for which we have
X = 2U = 0
Y = 2V = 0
and
X2 = 4U2 = 1
also,
Y2 = 9V2 = 1
Therefore, X - Y is also a normal random variable with
X-Y = 0
Hence,
P ^X - Y $ 0h = P ^X - Y # 0h = 1
2
or, we can say
P ^2U - 3V # 0h = 1
2
Thus,
P ^3V $ 2U h = 1
2
8.3
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8.2
Hence, for ISI free transmission, minimum bandwidth is
W = Rb = R kHz
2
10
Option (B) is correct.
Given, random variables U and V with mean zero and variances 1
4
and 1
9
i.e.,
U =V=0
su2 = 1
4
1
2
and
sv =
9
so,
P ^U $ 0h = 1
2
1
and
P ^V $ 0h =
2
The distribution is shown in the figure below
8.4
Option (C) is correct.
The mean of random variables U and V are both zero
i.e.,
U =V=0
Also, the random variables are identical
i.e.,
fU ^u h = fV ^v h
or,
FU ^u h = FV ^v h
i.e., their cdf are also same. So,
FU ^u h = F2V ^2v h
i.e., the cdf of random variable 2V will be also same but for any
instant
2V $ U
Therefore,
G ^x h = F ^x h
but,
x G ^x h $ xF ^x h
or,
6F ^x h - G ^x h@x # 0
Option (C) is correct.
P ^U =+ 1h = P ^U =- 1h = 1
2
where U is a random variable which is identical to V i.e.,
P ^V =+ 1h = P ^V =- 1h = 1
2
So, random variable U and V can have following values
U =+ 1, - 1; V =+ 1, - 1
Therefore the random variable U + V can have the following values,
U+V
- 2 When U = V =- 1
= *0 When U = 1,V = 1 or u =- 1, v = 1
2 When U = V = 1
Hence, we obtain the probabilities for U + V as follows
Given,
P ^U + V h
U+V
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1
1=1
2#2 4
1
1
1
1
1
b2 # 2l+b2 # 2l = 2
-2
0
fu ^u h =
fv ^v h =
1 e -u
2s
2p su2
2
u
1 e -v
2s
2p sv2
We can express the distribution in standard form by assuming
X = u - 0 = u = 2U
su
Y2
v
0
and
= v = 3V
Y =
sv
Y3
2
v
1
1=1
2#2 4
Therefore, the entropy of the ^U + V h is obtained as
2
H ^U + V h =
/ P^U + V h log
1
' P ^U + V h 1
= 1 log 2 4 + 1 log 2 2 + 1 log 2 4
2
4
4
2
1
2
= + +
4 2 4
2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 199
= 1 ;2 b 1 # 1 # 103 # 6 l + 400E
p
2
= 1 66000 + 400@ = 6400
p
p
E [X (t)] is the absolute value of mean of signal X (t) which is also
equal to value of X (w) at (w = 0).
From given PSD
=3
2
8.5
Option (D) is correct.
For the shown received signal, we conclude that if 0 is the transmitted
signal then the received signal will be also zero as the threshold is
1 and the pdf of bit 0 is not crossing 1. Again, we can observe that
there is an error when bit 1 is received as it crosses the threshold.
The probability of error is given by the area enclosed by the 1 bit
pdf (shown by shaded region)
SX (w) w = 0 = 0
SX (w) = X (w) 2 = 0
X (w) 2w = 0 = 0
X (w) w = 0 = 0
8.8
P (error when bit 1 received) = 1 # 1 # 0.25 = 1
8
2
or
P b received 1 l = 1
8
transmitted 0
Since, the 1 and 0 transmission is equiprobable:
i.e.,
P ^ 0 h = P ^1 h = 1
2
Hence bit error rate (BER) is
BER
0
received
received 1 P 1
= Pb
l P 0 + P b transmitted
0l ^ h
transmitted 1 ^ h
= 0+1 #1
8
2
1
=
16
8.6
8.7
Option (B) is correct.
The optimum threshold is the threshold value for transmission as
obtained at the intersection of two pdf. From the shown pdf. We
obtain at the intersection
(transmitted, received) = b 4 , 1 l
5 5
we can obtain the intersection by solving the two linear eqs
pdf of received bit 0
x+y = 1
y = 0.5 x pdf of received bit 1
2
4
Hence for threshold = , we have
5
BER
= P b received 1 l P ^0 h + P b received 0 l P ^1 h
transmitted 0
transmitted 1
= b1 # 1 # 1l# 1 +b1 # 4 # 1l# 1
2
5
2
2
2
5
5
2
= 1 <(BER for threshold = 1)
20
Hence, optimum threshold is 4
5
Option (A) is correct.
The mean square value of a stationary process equals the total area
under the graph of power spectral density, that is
E [X 2 (t)] =
or,
or,
#S
3
-3
E [X 2 (t)] = 1
2p
X
Option (C) is correct.
For raised cosine spectrum transmission bandwidth is given as
a " Roll of factor
BT = W (1 + a)
BT = Rb (1 + a) Rb " Maximum signaling rate
2
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3500 = Rb (1 + 0.75)
2
Rb = 3500 # 2 = 4000
1.75
8.9
Option (D) is correct.
Entropy function of a discrete memory less system is given as
H =
3
-3
X
k
k
where Pk is probability of symbol Sk .
For first two symbols probability is same, so
N-1
H = P1 log b 1 l + P2 log b 1 l +
Pk log b 1 l
P1
P2
Pk
k=3
/
=-e P1 log P1 + P2 log P2 +
N-1
/ P log P o
k
k
k=3
=-e 2P log P +
N-1
/ P log P o
k
k
(P1 = P2 = P)
k=3
P1 = P + e, P2 = P - e
Now,
So,
Hl =-=(P + e) log (P + e) + (P - e) log (P - e) +
N-1
/ P log P G
k
k
k=3
By comparing,
8.10
Hl < H , Entropy of source decreases.
Option (B) is correct.
Probability density function of uniformly distributed variables X
and Y is shown as
(w) dw
3
E [X 2 (t)] = 2 # 1 SX (w) dw (Since the PSD is even)
2p 0
= 1 [area under the triangle + integration of delta function]
p
/ P log b P1 l
k=0
(f ) df
#S
N-1
#
P &[max (x, y)] < 1 0
2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Since X and Y are independent.
P &[max (x, y)] < 1 0
2
PbX < 1 l
2
Similarly for Y :
P bY < 1 l
2
So
P &[max (x, y)] < 1 0
2
= P b X < 1 l P bY < 1 l
2
2
= shaded area = 3
4
=3
4
=3#3= 9
4
4 16
Alternate Method:
From the given data since random variables X and Y lies in the
interval [- 1, 1] as from the figure X , Y lies in the region of the
square ABCD .
Page 200
where 0 # t # T , E is the transmitted energy per bit.
General function of local oscillator
2 sin (w t), 0 # t < T
f1 (t) =
c
T
But here local oscillator is ahead with 45c. so,
2 sin (w t + 45c)
f1 (t) =
c
T
The coordinates of message points are
s11 =
=
T
# s (t) f (t) dt
#
1
2E sin w t
c
T
T
0
=
2E
T
=
2E
T
= 1
T
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1
0
E
Similarly,
Signal space diagram
2 sin (w t + 45c) dt
c
T
T
# sin (w t) sin (w t + 45c) dt
c
0
2
T
#
0
T
#
0
T
c
1 [sin 45c + sin (2w t + 45c)] dt
c
2
1 dt + 1 E Tsin (2w t + 45c) dt
c
T
0
2
1444444
42
4
44444 3
0
= E
2
#
s21 =-
E
2
Now here the two message points are s11 and s21 .
The error at the receiver will be considered.
When : (i) s11 is transmitted and s21 received
(ii) s21 is transmitted and s11 received
So, probability for the 1st case will be as :
P b s21 received l = P (X < 0) (as shown in diagram)
s11 transmitted
= P _ E/2 + N < 0i
= P _N < - E/2 i
Taking the Gaussian distribution as shown below :
8.11
Probability for max 6X, Y @ < 1/2 : The points for max 6X, Y @ < 1/2
will be inside the region of square AEFG .
So,
P &max 6X, Y @ < 1 0 = Area of 4AEFG
2
Area of square ABCD
3
3
#2
2
=
= 9
2#2
16
Option (B) is correct.
In a coherent binary PSK system, the pair of signals s1 (t) and s2 (t)
used to represent binary system 1 and 0 respectively.
s1 (t) = 2E sin wc t
T
s2 (t) =- 2E sin wc t
T
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 201
Note: For phase lead compensator zero is nearer to the origin as
compared to pole, so option (C) can not be true.
Mean of the Gaussian distribution =
E/2
Variance = N 0
2
Putting it in the probability function :
P bN < -
Taking,
E =
2l
#
0
=
#
0
1
`x + E/2 j
2N 0 /2
f = tan-1 a w k - tan-1 a w k
a
b
1/a
1/b
df
=
=0
2
2
dw
1 +awk 1 +awk
a
b
1 + w2 = 1 + 1 w2
a ab2
b b a2
1 - 1 = w2 1 - 1
a b
ab b a b l
dx
0
-3
x + E/2
=t
N 0 /2
N 0 dt
2
dx =
So, P _N < - E/2 i =
#
E/N 0
w = ab = 1 # 2 =
E
N0 m
1 e- t2 dt Q
c
2p
2
3
8.16
where Q is error function.
Since symbols are equiprobable in the 2 nd case
So,
P b s11 received l = Q c E m
N0
s21 transmitted
So the average probability of error
= 1 ;P b s21 received l + P b s11 received lE
2
s11 transmitted
s21 transmitted
= 1 =Q c
2
8.12
8.13
Option (A) is correct.
2
e-
2p N 0
2
2
1 e- `x + NE/2 j dx
pN 0
-3
8.15
E +Q
c
N0 m
E =Q
c
N 0 mG
E
N0 m
Option (D) is correct.
Quantized 4 level require 2 bit representation i.e. for one sample
2 bit are required. Since 2 sample per second are transmitted we
require 4 bit to be transmitted per second.
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Option ( ) is correct.
Option (B) is correct.
General equation of FM and PM waves are given by
fFM (t) = Ac cos ;wc t + 2pk f
8.17
t
# m (t) dtE
0
fPM (t) = Ac cos [wc t + k p m (t)]
For same maximum phase deviation.
t
k p [m (t)] max = 2pk f ; m (t) dtE
#
0
max
k p # 2 = 2pk f [x (t)] max
where,
x (t) =
2 rad/ sec
t
# m (t) dt
8.18
0
Option (B) is correct.
In FM the amplitude is constant and power is efficient transmitted.
No variation in power.
There is most bandwidth efficient transmission in SSB- SC. because
we transmit only one side band.
Simple Diode in Non linear region ( Square law ) is used in
conventional AM that is simplest receiver structure.
In VSB dc. component exists.
Option (A) is correct.
We have
Sx (f) = F {Rx (t)} = F {exp (- pt2)}
2
= e- pf
The given circuit can be simplified as
So,
Power spectral density of output is
[x (t)] max = 4
k p # 2 = 2pk f # 4
kp
= 4p
kf
Sy (f) = G (f) 2 Sx (f)
= j2pf - 1 2 e- pf
= ( (2pf) 2 + 1) 2 e- pf
or
8.14
Option (A) is correct.
jw + a
GC (s) = s + a =
s+b
jw + b
Phase lead angle
f = tan-1 a w k - tan-1 a w k
a
b
Jw - wN
-1 K a
b O = tan-1 w (b - a)
f = tan
c ab + w 2 m
2
O
KK
1+ w O
ab P
L
For phase-lead compensation f > 0
b-a > 0
b >a
2
8.19
Sy (f) = (4p2 f 2 + 1) e- pf
2
Option (B) is correct.
Highest frequency component in m (t) is fm = 4000p/2p = 2000 Hz
Carrier frequency
fC = 1 MHz
For Envelope detector condition
1/fC << RC << 1/fm
1 μs << RC << 0.5 ms
8.20
2
Option (D) is correct.
Four phase signal constellation is shown below
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 202
Where m < 1, for no over modulation.
In option (C)
x (t) = AC :1 + 1 m (t)D cos (2pfC t)
4
Thus m = 1 < 1 and this is a conventional AM-signal without over4
modulation
8.23
2
2
1
d = r +r
d2 = 2r 12
Now
2
1
Option (B) is correct.
(6) 2
= 18 W
P =
2
Power
r1 = d/ 2 = 0.707d
8.24
Option (C) is correct.
Impulse response of the matched filter is given by
h (t) = S (T - t)
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8.25
q = 2p = 2p = p
8
4
M
Applying Cooine law we have
1,
H (f ) = *
0,
d2 = r 22 + r 22 - 2r 22 cos p
4
= 2r 22 - 2r 22 1/ 2 = (2 d
r2 =
= 1.3065d
2- 2
or
8.21
Option (B) is correct.
Let response of LPF filters
Noise variance (power) is given as
2 ) r 22
P = s2 =
#0
Option (D) is correct.
Here Pe for 4 PSK and 8 PSK is same because Pe depends on d .
Since Pe is same, d is same for 4 PSK and 8 PSK.
or
8.26
#0
fo
2
H (f ) No df = 22 (given)
a
2 # 10-20 df = 22
a
-20
6
2 # 10 # 10 = 22
a
2
a = 1014
a = 107
1 # 106
Option (D) is correct.
Probability of error is given by
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Additional Power SNR
= (SNR) 2 - (SNR) 1
= 10 log b ES2 l - 10 log b ES1 l
No
No
= 10 log b ES2 l
ES1
Pe = 1 [P (0/1) + P (1/0)]
2
P (0/1) =
Additional SNR = 5.33 dB
Option (C) is correct.
Conventional AM signal is given by
x (t) = AC [1 + mm (t)] cos (2pfC t)
a/2
#- 3 0.5e- a n - a dn = 0.5e-10
where a = 2 # 10-6 V and a = 107 V - 1
P (1/0) =
2
= 10 log a r2 k & 20 log a r2 k = 20 log 1.3065d
r1
r1
0.707d
8.22
f < 1 MHz
elsewhere
#a/32 0.5e- a n dn
= 0.5e-10
Pe = 0.5e-10
8.27
Option (C) is correct.
S (t) = sin c (500t) sin c (700t)
S (f ) is convolution of two signals whose spectrum covers f 1 = 250 Hz
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 203
and f 2 = 350 Hz . So convolution extends
f = 25 + 350 = 600 Hz
Nyquist sampling rate
X2 = Sxi2 pi (x)
= 1 # 0.1 + 4 # 0.2 + 9 # 0.4 + 16 # 0.2 + 25 # 0.1
= 0.1 + 0.8 + 3.6 + 3.2 + 2.5 = 10.2
N = 2f = 2 # 600 = 1200 Hz
8.28
Option (D) is correct.
For the given system, output is written as
y (t) = d [x (t) + x (t - 0.5)]
dt
dx (t) dx (t - 0.5)
+
y (t) =
dt
dt
Taking Laplace on both sides of above equation
sx2 = X2 - ^X h2
= 10.2 - (3) 2 = 1.2
Variance
8.32
Option (C) is correct.
m (t) = 1 cos w1 t - 1 sin w2 t
2
2
Modulation index
Y (s) = sX (s) + se-0.5s X (s)
Y (s)
= s (1 + e-0.5s)
H (s) =
X (s)
H (f ) = jf (1 + e-0.5 # 2pf ) = jf (1 + e- pf )
Power spectral density of output
SY (f ) = H (f ) 2 SX (f ) = f 2 (1 + e- pf ) 2 SX (f )
For SY (f ) = 0 ,
1 + e- pf = 0
f = (2n + 1) f0
or
f0 = 1 KHz
8.29
Option (C) is correct.
cos (2pfm t) cos (2pfc t)
cos (2pfc t)
cos [2p (fc + fm) t]
[1 + cos (2pfm t) cos (2pfc t)]
8.30
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Option (B) is correct.
C1 = B log2 `1 + S j
N
. B log2 ` S j
N
As S >> 1
N
If we double the S ratio then
N
C2 . B log2 ` 2S j
N
. B log2 2 + B log2 S . B + C1
N
8.34
P (A + B)
P (B)
Event P (A + B) happen when X + Y = 2 and X - Y = 0 . It is only
the case when X = 1 and Y = 1.
Thus
P (A + B) = 1 # 1 = 1
4 4 16
Now event P (B) happen when
X - Y = 0 It occurs when X = Y , i.e.
X = 0 and Y = 0 or
X = 1 and Y = 1 or
X = 2 and Y = 2
Thus
P (B) = 1 # 1 + 1 # 1 + 1 # 1 = 6
2 2 4 4 4 4 16
1/16
P (A + B)
Now
=
=1
6
P (B)
6/16
8.31
# 100% = 20%
We have
X+Y = 2 $ A
X-Y = 0 $ B
P (X + Y = 2 X - Y = 0) =
2
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p (X = 0) = p (Y = 0) = 1
2
p (X = 1) = p (Y = 1) = 1
4
p (X = 2) = p (Y = 2) = 1
4
Let
and
Now
1 2
2j
1 2
+
2j
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8.33
Option (C) is correct.
We have
`
`
sAM (t) = [1 + m (t)] cos wc t
m (t) max
=
Vc
2
2
m = `1j +`1j = 1
2
2
2
2
h = m
# 100%
m2 + 2
Option (B) is correct.
The mean is
X = Sxi pi (x)
= 1 # 0.1 + 2 # 0.2 + 3 # 0.4 + 4 # 0.2 + 5 # 0.1
= 0.1 + 0.4 + 1.2 + 0.8 + 0.5 = 3.0
Option (C) is correct.
We have
SNR = 1.76 + 6n
or
43.5 = 1.76 + 6n
6n = 43.5 + 1.76
6n = 41.74 $ n . 7
No. of quantization level is
27 = 128
Step size required is
= VH - VL =
128
5 - (- 5)
= 10
128
128
= .078125 . .0667
8.35
Option (B) is correct.
For positive values step size
s+ = 0.05 V
For negative value step size
s- = 0.1 V
No. of quantization in + ive is
= 5 = 5 = 100
s+
0.05
Thus
2n + = 100 $ n+ = 7
No. of quantization in - ve
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 204
Q1 = 5 = 5 = 50
s0.1
= 1 [2W + 2W + 4W + 6W] = 7W
2
2n = 50 $ n - = 6
-
Thus
8.42
+
S
` N j+ = 1.76 + 6n = 1.76 + 42 = 43.76 dB
S
` N j- = 1.76 + 6n = 1.76 + 36 = 37.76 dB
Best
8.36
Maximum frequency deviation is
S
` N j0 = 43.76 dB
3wmax = 2p (5 # 1500 + 7.5 # 1000)
3 fmax = 15000
3f
Modulation index is = max = 15000 = 10
fm
1500
Option (A) is correct.
We have
xAM (t) = Ac cos wc + 2 cos wm t cos wc t
= AC c1 + 2 cos wm t m cos wc t
Ac
For demodulation by envelope demodulator modulation index
must be less than or equal to 1.
2 #1
Thus
Ac
8.43
Option (C) is correct.
8.44
Option (B) is correct.
fm = 4 KHz
fs = 2fm = 8 kHz
Bit Rate
Rb = nfs = 8 # 8 = 64 kbps
The minimum transmission bandwidth is
BW = Rb = 32 kHz
2
Ac $ 2
Hence minimum value of Ac = 2
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8.37
8.38
8.45
8.46
m
bits
8.47
p1 = p2 = ... = pn = 1
n
8.48
H =
n
/ n1 log n = log n
i=1
8.39
Option (C) is correct.
PSD of noise is
8.49
N0 = K
2
...(1)
The 3-dB cut off frequency is
fc = 1
2pRC
8.50
...(2)
= N0 = c N0 m 1 = Kpfc
4RC
2 2RC
Option (D) is correct.
At receiving end if we get two zero or three zero then its error.
Let p be the probability of 1 bit error, the probability that
transmitted bit error is
= Three zero + two zero and single one
= 3 C3 p3 + 3C2 p2 (1 - p)
= p3 + p2 (1 - p)
8.41
Option (D) is correct.
Bandwidth of TDM is
= 1 (sum of Nyquist Rate)
2
Option (C) is correct.
Autocorrelation is even function.
Option (B) is correct.
Power spectral density is non negative. Thus it is always zero or
greater than zero.
Option (A) is correct.
The variance of a random variable x is given by
E [X2] - E2 [X]
Option (A) is correct.
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Output noise power is
8.40
Option (B) is correct.
Noise \ 12
L
To reduce quantization noise by factor 4, quantization level must
be two times i.e. 2L .
Now
L = 2n = 28 = 256
Thus
2L = 512
i=1
Since
We have n = 8
As
Option (A) is correct.
The entropy is
/ pi log2 p1i
Option (C) is correct.
S0
c N m = 1.76 + 6n dB
0
= 1.76 + 6 # 8 = 49.76 dB
Option (A) is correct.
CDF is the integration of PDF. Plot in option (A) is the integration
of plot given in question.
H =
Option (B) is correct.
We have
qi = 2p105 t + 5 sin (2p1500t) + 7.5 sin (2p1000t)
wi = dqi = 2p105 + 10p1500 cos (2p1500t) + 15p1000 cos (2p1000t)
dt
A Hilbert transformer is a non-linear system.
8.51
8.52
Option (D) is correct.
Slope overload distortion can be reduced by increasing the step size
3 $ slope of x (t)
Ts
Option (C) is correct.
We have
p (t) =
sin (4pWt)
4pWt (1 - 16W2 t2)
at t = 1 it is 0 form. Thus applying L' Hospital rule
0
4W
4pW cos (4pWt)
p( ) =
4pW [1 - 48W2 t2]
1
4W
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
=
8.53
Page 205
cos (4pWt)
= cos p = 0.5
1-3
1 - 48W2 t2
Option (B) is correct.
The block diagram is as shown below
6
Rb max = RC = 1.2288 # 10 = 12.288 # 103 bps
Gmin
100
8.57
Option (B) is correct.
Energy of constellation 1 is
Eg1
2
2 a) 2 + ( 2 a) 2 + (- 2 2 a) 2
= 2a2 + 2a2 + 2a2 + 8a2 = 16a2
Energy of constellation 2 is
= (0) + (-
2
2 a) + (-
Eg2 = a2 + a2 + a2 + a2 = 4a2
2
E
Ratio = g1 = 16a2 = 4
Eg2
4a
Here
M1 (f) = Mt (f)
8.58
Y1 (f) = M (f) c e
j2 p B
Y2 (f) = M1 (f) c e
j 2p B
Noise Power is same for both which is N0 .
2
Thus probability of error will be lower for the constellation 1 as it
has higher signal energy.
+ e -j 2 p B
m
2
- e -j2pB
m
2
Y (f) = Y1 (f) + Y2 (f)
All waveform is shown below
Option (A) is correct.
8.59
Option (A) is correct.
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Area under the pdf curve must be unity
Thus
2a + 4a + 4b = 1
...(1)
2a + 8b = 1
For maximum entropy three region must by equivaprobable thus
...(2)
2a = 4b = 4b
From (1) and (2) we get
b = 1 and a = 1
12
6
8.60
8.61
8.54
Option (C) is correct.
By Binomial distribution the probability of error is
n
r
n-r
pe = Cr p (1 - p)
Probability of at most one error
= Probability of no error + Probability of one error
= n C0 p0 (1 - p) n - 0 + n C1 p1 (1 - p) n - 1
Option (*) is correct.
Option (B) is correct.
A LPF will not produce phase distortion if phase varies linearly
with frequency.
i.e.
8.62
f (w) \ w
f (w) = kw
Option (B) is correct.
Let m (t) is a low pass signal, whose frequency spectra is shown
below
= (1 - p) n + np (1 - p) n - 1
8.55
8.56
Option (B) is correct.
Bandwidth allocated for 1 Channel = 5 M Hz
Average bandwidth for 1 Channel 5 = 1 MHz
5
Total Number of Simultaneously Channel = 1M # 8 = 40 Channel
200k
Option (A) is correct.
Chip Rate
RC = 1.2288 # 106 chips/sec
Data Rate
Rb = RC
G
Since the processing gain G must be at least 100, thus for Gmin we
get
Fourier transform of g (t)
3
1
d (f - 20 # 103 k)
0.5 # 10-4 k =- 3
Spectrum of G (f ) is shown below
G (t) =
/
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 206
= 0.25 log2 4 + 0.25 log2 4 + 0.5 log2 2
= 0.5 + 0.5 + 1 = 3 bits/symbol
2
2
Rb = 3000 symbol/sec
= Rb H
= 3 # 3000 = 4500 bits/sec
2
Average bit rate
Now when m (t) is sampled with above signal the spectrum of
sampled signal will look like.
When sampled signal is passed through a LP filter of BW 1 kHz,
only m (t) will remain.
8.63
Option (C) is correct.
8.67
We can say that RC depends on W , thus
RC < 1
W
8.68
8.69
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8.70
Option (B) is correct.
We have
8.71
x (t) = 125t [u (t) - u (t - 1)] + (250 - 125t) [u (t - 1) - u (t - 2)]
The slope of expression x (t) is 125 and sampling frequency fs is
32 # 1000 samples/sec.
Let 3 be the step size, then to avoid slope overload
3 $ slope x (t)
Ts
3 fc $ slope x (t)
3# 32000 $ 125
3 $ 125
32000
Option (A) is correct.
The sampling frequency is
fs =
1 = 33 kHz
0.03m
Option (B) is correct.
We have p1 = 0.25 , p2 = 0.25 and p3 = 0.5
H =
3
/ p1 log2 p11
Option (C) is correct.
After the SSB modulation the frequency of signal will be fc - fm i.e.
1000 - 10 kHz . 1000 kHz
The bandwidth of FM is
Option (A) is correct.
We have
p (t) = u (t) - u (t - 1)
g (t) = p (t)* p (t)
s (t) = g (t) - d (t - 2)* g (t) = g (t) - g (t - 2)
All signal are shown in figure below :
The impulse response of matched filter is
h (t) = s (T - t) = s (1 - t)
Here T is the time where output SNR is maximum.
Since fs $ 2fm , the signal can be recovered and are correlated.
8.66
Option (B) is correct.
When 3 /2 is added to y (t) then signal will move to next quantization
level. Otherwise if they have step size less than 3 then they will be
2
on the same quantization level.
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3 = 2- 8
8.65
Option (B) is correct.
BW = 2 (b + 1) 3 f
For NBFMb << 1, thus
BWNBFM . 2 3 f = 2 (109 - 106) . 2 # 109
The highest frequency signal in x (t) is 1000 # 3 = 3 kHz if expression
is expanded. Thus minimum frequency requirement is
f = 2 # 3 # 103 = 6 # 103 Hz
8.64
Option (A) is correct.
The diagonal clipping in AM using envelop detector can be avoided
if
1 << RC < 1
wc
W
1 $ Wm sin Wt
But from
RC
1 + m cos Wt
bits/symbol
i=1
= p1 log2 1 + p2 log2 1 + p3 log2 1
p1
p2
p3
= 0.25 log2 1 + 0.25 log2 1 + 0.5 log2 1
0.25
0.25
0.5
8.72
Option (A) is correct.
We have
xAM (t) = 10 [P (t) + 0.5g (t)] cos wc t
where
p (t) = u (t) - u (t - 1)
and
g (t) = r (t) - 2r (t - 1) + r (t - 2)
For desired interval 0 # t # 1, p (t) = 1 and g (t) = t , Thus we
have,
xAM (t) = 100 (1 - 0.5t) cos wc t
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
8.73
Page 207
Hence modulation index is 0.5
The transfer function of matched filter is
Option (A) is correct.
We know that
SYY (w) = H (w) 2 .SXX (w)
Now SYY (w) = 16 2 and SXX (w) = 1 white noise
16 + w
h (t) = x (t - t) = x (2 - t)
The output of matched filter is the convolution of x (t) and h (t) as
shown below
16
= H (w) 2
16 + w2
4
H (w) =
16 + w2
H (s) = 4
4+s
Thus
or
or
which is a first order low pass RL filter.
8.74
Option (A) is correct.
R
= 4
4+s
R + sL
We have
8.81
Option (B) is correct.
H (f) = 2e - jwt
We have
R
L
= 4
4+s
+s
Comparing we get L = 1 H and R = 4W
or
8.75
8.76
R
L
Option (C) is correct.
We have
xAM (t) = 10 (1 + 0.5 sin 2pfm t) cos 2pfc t
The modulation index is 0.5
Carrier power
Pc =
(10) 2
= 50
2
Side band power
Ps =
(10) 2
= 50
2
Side band power
2
(0.5) 2 (50)
= 6.25
Ps = m Pc =
2
2
Option (B) is correct.
Mean noise power = Area under the PSD curve
= 4 ; 1 # B # No E = BNo
2
2
H (f) = 2
G0 (f) = H (f) 2 Gi (f)
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8.82
$ Amplitude modulation
$ DSB-SC modulation
$ Phase Modulation
$ Frequency Modulation
Option (A) is correct.
Let x (t) be the input signal where
x (t) = cos (cos t + b1 cos wm t)
cos (2wc t + 2b1 cos wm t)
y (t) = x2 (t) = 1 +
2
2
3f
Here
= 90 = 18
b = 2b1 and b1 =
fm
5
BW = 2 (b + 1) fm = 2 (2 # 18 + 1) # 5 = 370 kHz
8.80
Now mean square value is
sv2 =
Option (C) is correct.
+3
#- 3
v2 p (v) dv
v2 ` v j dv
8
3
4
v
=
c 8 m dv = 8
0
=
#0
4
as p (v) = 1 v
8
#
8.83
Option (C) is correct.
VSB $ fm + fc
DSB - SC $ 2fm
SSB $ fm
AM $ 2fm
Thus SSB has minimum bandwidth and it require minimum power.
8.79
Option (C) is correct.
As the area under pdf curve must be unity
1 (4 # k) = 1 $ k = 1
2
2
Option (D) is correct.
{1 + km (t)} A sin (wc t)
dm (t) Asin (wc t)
A sin {cos t + km (t)}
A sin [wct + k] t- 3 m (t) dt
8.78
= 4No W/Hz
= 4No # B
The noise power is
The ratio of average sideband power to mean noise power is
Side Band Power = 6.25 = 25
Noise Power
N0 B
4No B
8.77
d
Option (D) is correct.
The phase deviation is
b =
3f
= 10 = 10
fm
1
If phase deviation remain same and modulating frequency is
changed
BW = 2 (b + 1) fm' = 2 (10 + 1) 2 = 44 kHz
8.84
8.85
Option (B) is correct.
As the area under pdf curve must be unity and all three region are
equivaprobable. Thus are under each region must be 13 .
2a # 1 = 1 $ a = 2
3
3
4
Option (A) is correct.
Nq =
+a
#- a
x2 p (x) dx = 2
Substituting a = 2 we have
3
Nq = 4
81
#0
3 a
3
x $ 1 dx = 1 ; x E = a
2 3 0
6
4
a 2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
8.86
Option (C) is correct.
When word length is 6
2#6
12
S
` N jN = 6 = 2 = 2
8.90
Option (B) is correct.
Carrier frequency
Modulating frequency
or
8.91
Option (A) is correct.
The SNR at transmitter is
10 - 3
= 109
10 - 20 # 100 # 106
In dB
SNRtr = 10 log 109 = 90 dB
Cable Loss
= 40 db
At receiver after cable loss we have
SNRRc = 90 - 40 = 50 dB
fc = 1 # 106 Hz
For an envelope detector
8.92
2pfc > 1 > 2pfm
Rc
1 < RC < 1
2pfc
2pfm
1 < RC < 1
2pfc
2pfm
1
1 < RC <
6
2p10
2 # 103
1.59 # 10 - 7 < RC < 7.96 # 10 - 5
so, 20 msec sec best lies in this interval.
Option (B) is correct.
The impulse response of matched filter is
h (t) = x (T - t)
Since here T = 4 , thus
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h (t) = x (4 - t)
The graph of h (t) is as shown below.
From graph it may be easily seen that slope between 3 < t < 4 is
- 1.
8.93
Option (B) is correct.
SAM (t) = Ac [1 + 0.1 cos wm t] cos wm t
sNBFM (t) = Ac cos [wc t + 0.1 sin wm t]
s (t) = SAM (t) + SNB fm (t)
For QPSK,
Thus
+ Ac cos wc t cos (0.1 sin wm t) - Ac sin wc t. sin (0.1 sin wm t)
As
0.1 sin wm t ,+ 0.1 to - 0.1
so,
cos (0.1 sin wm t) . 1
As when q is small cos q . 1 and sin q , q, thus
- Ac 0.1 sin wm t sin wc t
= 2Ac cos wc t + 0.1Ac cos (wc + wm) t
1 44 2 44 3 1 4444
4 2 4444
43
cosec
USB
Thus it is SSB with carrier.
Option (A) is correct.
Consecutive pulses are of same polarity when modulator is in slope
overload.
Consecutive pulses are of opposite polarity when the input is
constant.
M = 4 = 2n $ n = 2
B2 = 2Rb = 10 kHz
2
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8.94
sin (0.1 sin wm t)
= 0.1 sincos wc t cos wm t + Ac cos wc t
Option (C) is correct.
The required bandwidth of M array PSK is
BW = 2Rb
n
where 2n = M and Rb is bit rate
For BPSK,
M = 2 = 2n $ n = 1
Thus
B1 = 2Rb = 2 # 10 = 20 kHz
1
= Ac [1 + 0.1 cos wm t] cos wc t + Ac cos (wc t + 0.1 sin wm t)
= Ac cos wc t + Ac 0.1 cos wm t cos wc t
8.89
P (X = 1) = P (X = 1+) - P (X = 1 -)
= 0.55 - 0.25 = 0.30
SNRtr = Ptr
NB
fm = 2 # 103 Hz
8.88
Option (D) is correct.
F (x1 # X < x2) = p (X = x2) - P (X = x1)
When word length is 8
2#8
16
S
` N jN = 8 = 2 = 2
S
16
^ N hN = 8
Now
= 212 = 2 4 = 16
S
^ N hN = 6
2
Thus it improves by a factor of 16.
8.87
Page 208
Option (C) is correct.
We have
MHz
fc = 100 MHz = 100 # 106 and fm = 1
= 1 # 106
The output of balanced modulator is
VBM (t) = [cos wc t][ cos wc t]
= 1 [cos (wc + wm) t + cos (wc - wm) t]
2
If VBM (t) is passed through HPF of cut off frequency fH = 100 # 106
, then only (wc + wm) passes and output of HPF is
VHP (t) = 1 cos (wc + wm) t
2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 209
V0 (t) = VHP (t) + sin (2p # 100 # 106) t
Now
Probability of error of 1
P (0 # X # 0.2) = 0.2
Probability of error of 0 :
= 1 cos [2p100 # 106 + 2p # 1 # 106 t] + sin (2p # 100 # 106) t
2
= 1 cos [2p108 + 2p106 t] + sin (2p108) t
2
P (0.2 # X # 0.25) = 0.05 # 2 = 0.1
P (0 # X # 0.2) + P (0.2 # X # 0.25)
2
= 0.2 + 0.1 = 0.15
0
Average error =
= 1 [cos (2p108 t) t cos (2p106 t)] - sin [2p108 t sin (2p106 t) + sin 2p108 t]
2
8.97
= 1 cos (2p106 t) cos 2p108 t + `1 - 1 sin 2p106 t j sin 2p108 t
2
2
This signal is in form
Option (B) is correct.
The square mean value is
#- 3 (x - xq) 2 f (x) dx
=
#0 (x - xq) 2 f (x) dx
= A cos 2p108 t + B sin 2p108 t
Now
GATE
ELECTRONICS & COMMUNICATION
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RK Kanodia & Ashish Murolia
in 3 Volumes
Fully Revised
=
=
=
8.95
1 cos2 (2p106 t) + 1 + 1 sin2 (2p106 t) - sin (2p106 t)
4
4
1 + 1 - sin (2p106 t)
=
4
5 - sin (2p106 t)
=
4
Option (A) is correct.
The pdf of transmission of 0 and 1 will be as shown below :
3 0.3
3
2 1
= ; x E + ; x + 0.49x - 14 x E
3 0
3
2 0.3
s2 = 0.039
RMS =
8.98
8.99
s2 =
0.039 = 0.198
Option (C) is correct.
FM
DM
PSK
PCM
$ Capture effect
$ Slope over load
$ Matched filter
$ m - law
Option (C) is correct.
Since fs = 2fm , the signal frequency and sampling frequency are as
follows
fm1 = 1200 Hz $ 2400 samples per sec
fm2 = 600 Hz $ 1200 samples per sec
fm3 = 600 Hz $ 1200 samples per sec
Thus by time division multiplexing total 4800 samples per second
will be sent. Since each sample require 12 bit, total 4800 # 12 bits
per second will be sent
Thus bit rate
Rb = 4800 # 12 = 57.6 kbps
3
8.96
0.1
#0.3 (x - 0.7) 2 f (x) dx
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or
Option (A) is correct.
s (t)
= A cos [2p10 # 10 t] + A cos [2p10.1 # 103 t]
1
Here
T1 =
= 100m sec
10 # 103
1
and
= 99m sec
T2 =
10.1 # 103
Period of added signal will be LCM [T1, T2]
Thus
T = LCM [100, 99] = 9900m sec
Thus frequency
f = 1 = 0.1 kHz
9900m
(x - 0) 2 f (x) dx +
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A2 + B2
2
6
6 2
1
1
` 2 cos (2p10 t)j + `1 - 2 sin (2p10 t j
0. 3
1
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The envelope of this signal is
=
#0
3
s2 =
8.100
Option (B) is correct.
The input signal X (f) has the peak at 1 kHz and - 1 kHz. After
balanced modulator the output will have peak at fc ! 1 kHz i.e. :
10 ! 1 $ 11 and 9 kHz
10 ! (- 1) $ 9 and 11 kHz
9 kHz will be filtered out by HPF of 10 kHz. Thus 11 kHz will remain. After passing through 13 kHz balanced modulator signal will
have 13 ! 11 kHz signal i.e. 2 and 24 kHz.
Thus peak of Y (f) are at 2 kHz and 24 kHz.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
8.101
8.102
8.103
Option (A) is correct.
The input is a coherent detector is DSB - SC signal plus noise.
The noise at the detector output is the in-phase component as the
quadrature component nq (t) of the noise n (t) is completely rejected
by the detector.
8.106
Ed
2h m
Since Pe of Binary FSK is 3 dB inferior to binary PSK
8.107
Option (D) is correct.
The pdf of Z will be convolution of pdf of X and pdf of Y as shown
below.
Now
p [Z # z] =
p [Z #- 2] =
W = Y-Z
E [W2] = E [Y - Z] 2 = E [Y2] + E [Z2] - 2E [YZ]
= sw2
E [X2 (t)] = Rx (10)
= 4 [e - 0.2 0 + 1] = 4 [1 + 1] = 8
E [Y2] = E [X2 (2)] = 8
E [Z2] = E [X2 (4)] = 8
E [YZ] = RXX (2) = 4 [e-0.2 (4 - 2) + 1] = 6.68
E [W2] = sw2 = 8 + 8 - 2 # 6.68 = 2.64
Option (B) is correct.
Pe = 1 erfc c
2
Option (C) is correct.
We have
Option (C) is correct.
The noise at the input to an ideal frequency detector is white. The
PSD of noise at the output is parabolic
We have
8.104
Page 210
Option (C) is correct.
Step size d =
2mp
= 1.536 = 0.012 V
L
128
2
(0.012) 2
Quantization Noise power = d =
12
12
z
#- 3 fZ (z) dz
= 12 # 10-6 V2
-2
#- 3fZ (z) dz
8.108
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Option (D) is correct.
The frequency of pulse train is
f 1- 3 = 1 k Hz
10
The Fourier Series coefficient of given pulse train is
-T /2
Cn = 1
Ae-jnw t dt
To -T /2
#
o
o
o
= 1
To
#
-To /6
-To /6
Ae-jhw t dt
o
A
[e-jw t] --TT //66
To (- jhwo)
A
=
(e-jw t - e jhw T /6)
(- j2pn)
= A (e jhp/3 - e-jhp/3)
j2pn
or
Cn = A sin ` np j
pn
3
From Cn it may be easily seen that 1, 2, 4, 5, 7, harmonics are present
and 0, 3, 6, 9,.. are absent. Thus p (t) has 1 kHz, 2 kHz, 4 kHz, 5 kHz,
7 kHz,... frequency component and 3 kHz, 6 kHz.. are absent.
The signal x (t) has the frequency components 0.4 kHz and 0.7
kHz. The sampled signal of x (t) i.e. x (t)* p (t) will have
1 ! 0.4 and 1 ! 0.7 kHz
2 ! 0.4 and 2 ! 0.7 kHz
4 ! 0.4 and 4 ! 0.7 kHz
=
= Area [z #- 2]
= 1 # 1 #1 = 1
2 6
12
o
o
o
o
o
o
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8.105
Thus in range of 2.5 kHz to 3.5 kHz the frequency present is
2 + 0.7 = 2.7 kHz
4 - 0.7 = 3.3 kHz
Option (D) is correct.
We have
RXX (t) = 4 (e - 0.2 t + 1)
RXX (0) = 4 (e - 0.2 0 + 1) = 8 = s2
or
mean
Now
s =2 2
m =0
P (x # 1) = Fx (1)
Given
X-m
at x = 1
s m
= 1 - Qc 1 - 0 m = 1 - Qc 1 m
2 2
2 2
= 1 - Qc
8.109
Option (C) is correct.
vi = Ac1 cos (2pfc t) + m (t)
v0 = ao vi + avi3
v0
= a0 [Ac' cos (2pfc' t) + m (t)] + a1 [Ac' cos (2pfc' t) + m (t)] 3
= a0 Ac' cos (2pfc' t) + a0 m (t) + a1 [(Ac' cos 2pfc' t) 3
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 211
then for coherent binary PSK r = 0.5 and s is required.
+ (Ac' cos (2pfc') t) 2 m (t)
+
3Ac' cos (2pfc' t) m2 (t)
3
+ m (t)]
= a0 Ac' cos (2pfc' t) + a0 m (t) + a1 (Ac' cos 2fc' t) 3
+ 3a1 Ac'2 ;
1 + cos (4pfc' t)
E m (t)
2
= 3a1 Ac' cos (2pfc' t) m2 (t) + m3 (t)
The term 3a1 Ac' ( cos 42pf t ) m (t) is a DSB-SC signal having carrier
frequency 1. MHz. Thus 2fc' = 1 MHz or fc' = 0.5 MHz
8.117
= 1.76 + 6.02 # 8 = 49.8 dB
8.118
'
c
8.110
Option (D) is correct.
2
PT = Pc c1 + a m
2
2
P (0.5) 2
Psb = Pc a = c
2
2
Psb = 1
Pc
8
or
8.111
Option (D) is correct.
AM Band width = 2fm
Peak frequency deviation = 3 (2fm) = 6fm
6f
Modulation index b = m = 6
fm
The FM signal is represented in terms of Bessel function as
xFM (t) = Ac
3
Option (B) is correct.
Bit Rate = 8k # 8 = 64 kbps
(SNR)q = 1.76 + 6.02n dB
Option (C) is correct.
The frequency of message signal is
fc = 1000 kHz
1 The frequency of message signal is
1
= 10 kHz
fm =
100 # 10 - 6
Here message signal is symmetrical square wave whose FS has only
odd harmonics i.e. 10 kHz, 30 kHz 50 kHz. Modulated signal contain fc ! fm frequency component. Thus modulated signal has
fc ! fm = (1000 ! 10) kH = 1010 kHz, 990 kHz
fc ! 3fm = (1000 ! 10) kH = 1030 kHz, 970 kHz
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/Jn (b) cos (wc - nwn) t
n =- 3
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3
wc + nwm = 2p (1008 # 10 )
2p10 + n4p # 103 = 2p (1008 # 103), n = 4
Thus coefficient = 5J4 (6)
6
8.112
8.113
Option (B) is correct.
Ring modulation $ Generation of DSB - SC
VCO $ Generation of FM
Foster seely discriminator $ Demodulation of fm
mixer $ frequency conversion
Thus, there is no 1020 kHz component in modulated signal.
8.119
We have
8.114
10 cos (8p # 103) t # 5 # 10 - 6
100 # 10 - 6
= 5 # 10 - 1 cos (8p # 103) t
=
8.120
8.121
Option (D) is correct.
Eb = 10 - 6 watt-sec
No = 10 - 5 W/Hz
o
(SNR) matched filler = E
=
N
106 = .05
2 # 10 - 5
2
(SNR)dB = 10 log 10 (0.05) = 13 dB
o
8.115
Option (B) is correct.
3 fs
2pfm
This is satisfied with Em = 1.5 V and fm = 4 kHz
For slopeoverload to take place Em $
8.116
Option (A) is correct.
If
s " carrier synchronization at receiver
r " represents bandwidth efficiency
+3
/ d (t - nTs)
n =- 3
fmax = 1650 + 450 = 2100 kHz
fmin = 550 + 450 = 1000 kHz
1
or
f =
2p LC
frequency is minimum, capacitance will be maximum
or
y (t) = 5 # 10 - 6 x (t)
x (t) = 10 cos (8p # 103) t
Ts = 100m sec
The cut off fc of LPF is 5 kHz
We know that for the output of filter
x (t) y (t)
=
Ts
Option (A) is correct.
f2
= (2.1) 2
R = Cmax = max
2
Cmin
fmin
R = 4.41
fi = fc + 2fIF = 700 + 2 (455) = 1600 kHz
Option (C) is correct.
8.122
Option (C) is correct.
Transmitted frequencies in coherent BFSK should be integral of bit
rate 8 kHz.
Option (B) is correct.
For best reception, if transmitting waves are vertically polarized,
then receiver should also be vertically polarized i.e. transmitter and
receiver must be in same polarization.
Option (D) is correct.
s (t) = cos 2p (2 # 106 t + 30 sin 150t + 40 cos 150t)
= cos {4p106 t + 100p sin (150t + q)}
Angle modulated signal is
s (t) = A cos {wc t + b sin (wm t + q)}
Comparing with angle modulated signal we get
Phase deviations
b = 100p
Frequency deviations
3 f = bfm = 100p # 150 = 7.5 kHz
2p
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 212
8.126
8.123
Option (*) is correct.
We have
m (t) s (t) = y1 (t)
2 sin (2pt) cos (200pt)
=
t
sin (202pt) - sin (198pt)
=
t
8.127
Option (A) is correct.
For any PDF the probability at mean is 1 . Here given PDF is
2
Gaussian random variable and X = 4 is mean.
8.128
y1 (t) + n (t)
= y2 (t) = sin 202pt - sin 198pt + sin 199pt
t
t
y2 (t) s (t) = u (t)
8.129
=
Option (A) is correct.
An ideal low - pass filter with appropriate bandwidth fm is used to
recover the signal which is sampled at nyquist rate 2fm .
[sin 202pt - sin 198pt + sin 199pt] cos 200pt
t
Option (C) is correct.
We require 6 bit for 64 intensity levels because 64 = 26
Data Rate = Frames per second # pixels per frame # bits per pixel
= 625 # 400 # 400 # 6 = 600 Mbps sec
Option (C) is correct.
We have
sin c (700t) + sin c (500t) =
Here the maximum frequency component is 2pfm = 700p i.e.
fm = 350 Hz
Thus Nyquist rate
fs = 2fm
= 2 (350) = 700 Hz
Thus sampling interval
= 1 sec
700
= 1 [sin (402pt) + sin (2pt) - {sin (398pt) - sin (2pt)}
2
+ sin (399pt) - sin (pt)]
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8.130
Option (D) is correct.
Probability of error = p
Probability of no error = q = (1 - p)
Probability for at most one bit error
= Probability of no bit error
After filtering
sin (2pt) + sin (2pt) - sin (pt)
y (t) =
2t
sin (2pt) + 2 sin (0.5t) cos (1.5pt)
=
2t
sin
sin
.
t
2
p
0
5pt cos 1.5pt
=
+
t
2t
8.124
+ probability of 1 bit error
= (1 - p) n + np (1 - p) n - 1
8.131
and power is
Pg = 1
2p
3
fm = 24p10 = 12 kHz
2p
Ts = 50m sec " fs = 1 = 1 # 106 = 20
50
Ts
d (n) = x (n) - x (n - 1)
E [d (n)] 2 = E [x (n) - x (n - 1)] 2
or
E [d (n)] 2
= E [x (n)] 2 + E [x (n - 1)] 2 - 2E [x (n) x (n - 1)]
or
sd2 = sx2 + sx2 - 2Rxx (1)
2
As we have been given sd2 = sx , therefore
10
or
G (w)
Sg (w) = G (w) 2
Now
PSD of ag (t) is
or
sx2 = s2 + s2 - 2R (1)
x
x
xx
10
2Rxx (1) = 19 sx2
10
Rxx = 19 = 0.95
20
sx2
ag (t)
3
#- 3Sg (w) dw
FT
aG (w)
Sag (w) = a (G (w)) 2
= a2 G (w) 2
Sag (w) = a2 Sg (w)
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Option (A) is correct.
or
FT
g (t)
then PSD of g (t) is
After sampling signal will have fs ! fm frequency component i.e. 32
and 12 kHz
At filter output only 8 kHz will be present as cutoff frequency is
15 kHz.
8.125
Option (A) is correct.
If
Option (B) is correct.
The signal frequency is
kHz
sin (700pt) sin (500pt)
+
700pt
500pt
Similarly
8.132
as k = 1
8.133
8.134
Pag = a2 Pg
Option (C) is correct.
The envelope of the input signal is [1 + ka m (t)] that will be output
of envelope detector.
Option (D) is correct.
Frequency Range for satellite communication is 1 GHz to 30 GHz,
Option (B) is correct.
Waveform will be orthogonal when each bit contains integer number
of cycles of carrier.
Bit rate
Rb = HCF (f1, f2)
= HCF (10k, 25k)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
= 5 kHz
Tb = 1 = 1 = 0.2 msec = 200 msec
Rb
5k
Thus bit interval is
8.135
Page 213
8.141
g (t)
= 6 # 10 sin c (400t) ) 10 sin c3 (100t)
Let
g1 (t) = 6 # 10 4 sin c2 (400t)
g2 (t) = (106) sin c3 (100t)
We know that g1 (t) ) g2 (t) ? G1 (w) G2 (w) occupies minimum of
Bandwidth of G1 (w) or G2 (w)
Band width of G1 (w)
= 2 # 400 = 800 rad/ sec or = 400 Hz
Band width of G2 (w)
= 3 # 100 = 300 rad/ sec or 150 Hz
Sampling frequency
= 2 # 150 = 300 Hz
Option (D) is correct.
4
Pm = m2 (t)
We have
The input to LPF is
x (t) = m (t) cos wo t cos (wo t + q)
m (t)
[cos (2wo t + q) + cos q]
2
m (t) cos (2wo t + q) m (t) cos q
=
+
2
2
=
The output of filter will be
y (t) =
m (t) cos q
2
Power of output signal is
8.142
2
Py = y2 (t) = 1 m2 (t) cos2 q = Pm cos q
4
4
8.136
Option (A) is correct.
Hilbert transformer always adds - 90c to the positive frequency
component and 90c to the negative frequency component.
Hilbert Trans form
Thus
8.137
cos wt " sin wt
sin wt " cos wt
cos w1 t + sin w2 t " sin w1 t - cos w2 t
Option (A) is correct.
We have
8.139
8.140
2
6
Option (B) is correct.
For a sinusoidal input SNR (dB) is PCM is obtained by following
formulae.
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x (t) = Ac cos {wc t + b sin wm t}
y (t) = {x (t)} 3
= Ac2 cos (3wc t + 3b sin wm t) + 3 cos (wc t + b sin wm t)
Thus the fundamental frequency doesn’t change but BW is three
times.
BW = 2 (3 f') = 2 (3 f # 3) = 3 MHz
8.138
Option (B) is correct.
Given function
SNR (dB) = 1.8 + 6n
n is no. of bits
n =8
SNR (dB) = 1.8 + 6 # 8 = 49.8
Here
So,
8.143
Option (D) is correct.
We know that matched filter output is given by
g 0 (t) =
t = T0
Option (C) is correct.
Option (C) is correct.
This is Quadrature modulated signal. In QAM, two signals having
bandwidth. B 1 & B 2 can be transmitted simultaneous over a
bandwidth of (B 1 + B 2) Hz
so
B.W. = (15 + 10) = 25 kHz
Option (B) is correct.
A modulated signal can be expressed in terms of its in-phase and
quadrature component as
6g 0 (t)@max =
=
# g (l) g (T - t + l) dl at
3
-3
0
# g (l) g (l) dl = #
3
3
-3
-3
#
1 # 10-4
0
g 2 (t) dt
[10 sin (2p # 106) 2] dt
[g 0 (t)] max = 1 # 100 # 10-4 = 5 mV
2
8.144
Option (B) is correct.
Sampling rate must be equal to twice of maximum frequency.
f s = 2 # 400 = 800 Hz
8.145
S (t)
= S1 (t) cos (2pfc t) - SQ (t) sin (2pfc t)
Here
S (t)
-at
at
= [e cpsDwt cos wc t - e sin Dwt sin wc t] m (t)
Option (C) is correct.
The amplitude spectrum of a gaussian pulse is also gaussian as
shown in the fig.
-y 2
fY (y) = 1 exp c
2 m
2p
= [e-at cos Dwt] cos 2pfc t - [e-at sin Dwt] sin 2pfc t
= S1 (t) cos 2pfc t - SQ (t) sin 2pfc t
Complex envelope of s (t) is
S (t) = S1 (t) + jSQ (t)
= e-at cos Dwt + je-at sin Dwt
= e-at [cos Dwt + j sin Dwt]
= exp (- at) exp (jDwt) m (t)
8.146
Option (C) is correct.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 214
2#8
(SNR) 1
= 22 # 9 = 22 = 1
4
(SNR) 2
2
so SNR will increased by a factor of 4
Let the rectangular pulse is given as
8.149
Auto correlation function is given by
T/2
Rxx (t) = 1 x (t) x (t - t) dt
T -T/2
When x (t) is shifted to right (t > 0), x (t - t) will be shown as dotted
line.
8.150
#
8.151
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8.152
8.153
8.154
8.155
8.156
8.157
Rxx (t) = 1
T
#
T
+t
2
A2 dt
T
- +t
2
2
2
= A :T + T - tD = A :T - tD
2
T 2
T 2
(t) can be negative or positive, so generalizing above equations
2
Rxx (t) = A :T - t D
T 2
Rxx (t) is a regular pulse of duration T .
8.147
8.148
Option (A) is correct.
In flat top sampling an amplitude distortion is produced while
reconstructing original signal x (t) from sampled signal s (t). High
frequency of x (t) are mostly attenuated. This effect is known as
aperture effect.
Option (A) is correct.
Carrier C (t) = cos (we t + q)
Modulating signal = x (t)
DSB - SC modulated signal = x (t) c (t) = x (t) cos (we t + q)
envelope = x (t)
Option (D) is correct.
In Quadrature multiplexing two baseband signals can transmitted
or modulated using I 4 phase & Quadrature carriers and its quite
different form FDM & TDM.
Option (A) is correct.
Fourier transform perform a conversion from time domain to
frequency domain for analysis purposes. Units remain same.
Option (A) is correct.
In PCM, SNR is depends an step size (i.e. signal amplitude) SNR
can be improved by using smaller steps for smaller amplitude. This
is obtained by compressing the signal.
Option (C) is correct.
Band width is same for BPSK and APSK(OOK) which is equal to
twice of signal Bandwidth.
Option (A) is correct.
The spectral density of a real value random process symmetric about
vertical axis so it has an even symmetry.
Option (A) is correct.
Option (C) is correct.
It is one of the advantage of bipolar signalling (AMI) that its
spectrum has a dc null for binary data transmission PSD of bipolar
signalling is
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Option (B) is correct.
Selectivity refers to select a desired frequency while rejecting all
others. In super heterodyne receiver selective is obtained partially
by RF amplifier and mainly by IF amplifier.
Option (C) is correct.
In PCM, SNR a
22n
so if bit increased from 8 to 9
8.158
Option (A) is correct.
Probability Density function (PDF) of a random variable x defined
as
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Px (x) =
K =
so here
8.159
8.160
8.161
1 e-x /2
2p
1
2p
Page 215
factor of 22 (n + 1)/n
2
8.165
Option (C) is correct.
Here the highest frequency component in the spectrum is 1.5 kHz
[at 2 kHz is not included in the spectrum]
Minimum sampling freq. = 1.5 # 2 = 3 kHz
R (t) =
put
Let
Option (B) is correct.
We need a high pass filter for receiving the pulses.
Option (D) is correct.
Power spectral density function of a signal g (t) is fourier transform
of its auto correlation function
8.162
Option (C) is correct.
For a signal g (t), its matched filter response given as
h (t) = g (T - t)
so here g (t) is a rectangular pulse of duration T .
3
# x (t) x (t + t) dt
-3
3
R (- t) = # x (t) x (t - t) dt
-3
t-t = a
dt = da
R (- t) =
change variable a " t
F
Sg (w)
Rg (t)
here Sg (w) = sin c2 (f)
so Rg (t) is a triangular pulse.
f [triang.] = sin c2 (f)
Option (D) is correct.
The auto correlation of energy signal is an even function.
auto correlation function is gives as
R (- t) =
3
# x (a + t) x (a) da
-3
3
# x (t) x (t + t) dt = R (t)
-3
R (- t) = R (t) even function
8.166
Option (D) is correct.
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output of matched filter
y (t) = g (t) ) h (t)
if we shift g (- t) for convolution y (t) increases first linearly then
decreases to zero.
8.163
8.164
Option (C) is correct.
The difference between incoming signal frequency (fc) and its image
frequency (fc) is 2I f (which is large enough). The RF filter may
provide poor selectivity against adjacent channels separated by a
small frequency differences but it can provide reasonable selectivity
against a station separated by 2I f . So it provides adequate suppression
of image channel.
Option (C) is correct.
In PCM SNR is given by
SNR = 3 22n
2
if no. of bits is increased from n to (n + 1) SNR will increase by a
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 216
UNIT 9
p # 10 ^x + z h
(C) 45c and E 0 ^atx - atz h e-j 3 2
V/m
2
4
ELECTROMAGNETICS
p # 10 z
(D) 60c and E 0 ^atx - atz h e-j 3 V/m
2
The expression for Evr is
p # 10 ^x - z h
(A) 0.23 E 0 ^atx + atz h e-j 3 2
V/m
2
4
9.5
4
2013
9.1
v ^rvh. The closed loop line integral A
v : dlv
Consider a vector field A
can be expressed as
vh : dsv over the closed surface bounded by the loop
(A)
^d # A
vh dv over the closed volume bounded by the loop
(B)
^d : A
p # 10 z
(B) - E 0 ^atx + atz h e j 3 V/m
2
(C)
p # 10 ^x + z h
3
(D) E 0 ^atx + atz h e-j
V/m
2
4
#
(D)
9.2
ONE MARK
##
###
### ^d : Avhdv over the open volume bounded by the loop
## ^d # Avh : dsv over the open surface bounded by the loop
p # 10 ^x - z h
(C) 0.44 E 0 ^atx + atz h e-j 3 2
V/m
2
4
4
v = xatx + yaty + zatz is
The divergence of the vector field A
(A) 0
(B) 1/3
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(C) 1
2012
9.6
-
(D) 3
The return loss of a device is found to be 20 dB. The voltage
standing wave ratio (VSWR) and magnitude of reflection coefficient
are respectively
(A) 1.22 and 0.1
(B) 0.81 and 0.1
(C) – 1.22 and 0.1
(D) 2.44 and 0.2
9.8
2013
TWO MARKS
Statement for Linked Answer Questions 52 and 53:
A monochromatic plane wave of wavelength l = 600 mm is propagating in the direction as shown in the figure below. Evi , Evr and
Evt denote incident, reflected, and transmitted electric field vectors
associated with the wave.
A
plane
wave
propagating
in
air
with
E = (8ax + 6ay + 5az ) e j (wt + 3x - 4y) V/m is incident on a perfectly
conducting slab positioned at x # 0 . The E field of the reflected
wave is
(A) (- 8ax - 6ay - 5az ) e j (wt + 3x + 4y) V/m
(B) (- 8ax + 6ay - 5az ) e j (wt + 3x + 4y) V/m
(C) (- 8ax - 6ay - 5az ) e j (wt - 3x - 4y) V/m
(D) (- 8ax + 6ay - 5az ) e j (wt - 3x - 4y) V/m
9.7
9.3
ONE MARK
The electric field of a uniform plane electromagnetic wave in free
space, along the positive x direction is given by E = 10 (ay + jaz ) e-j 25x
. The frequency and polarization of the wave, respectively, are
(A) 1.2 GHz and left circular
(B) 4 Hz and left circular
(C) 1.2 GHz and right circular
(D) 4 Hz and right circular
A coaxial-cable with an inner diameter of 1 mm and outer diameter
of 2.4 mm is filled with a dielectric of relative permittivity 10.89.
-9
Given m0 = 4p # 10-7 H/m, e0 = 10 F/m , the characteristic
36p
impedance of the cable is
(A) 330 W
(C) 143.3 W
9.9
(B) 100 W
(D) 43.4 W
The radiation pattern of an antenna in spherical co-ordinates is given
by F (q) = cos 4 q ;
0 # q # p/2 . The directivity of the antenna is
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(A) 10 dB
(C) 11.5 dB
2012
9.10
9.4
The angle of incidence qi and the expression for Evi are
p # 10 ^x + 2h
(A) 60c and E 0 ^atx - atz h e-j 3 2
V/m
2
4
p # 10 z
(B) 45c and E 0 ^atx + atz h e-j 3 V/m
2
4
(B) 12.6 dB
(D) 18 dB
TWO MARKS
A transmission line with a characteristic impedance of 100 W is used
to match a 50 W section to a 200 W section. If the matching is to be
done both at 429 MHz and 1 GHz, the length of the transmission
line can be approximately
(A) 82.5 cm
(b) 1.05 m
(C) 1.58 cm
(D) 1.75 m
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
9.11
Page 217
The magnetic field among the propagation direction inside a
rectangular waveguide with the cross-section shown in the figure is
(A) 0.8 # 108 m/s
(C) 1.6 # 108 m/s
Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t - bz)
9.16
The phase velocity v p of the wave inside the waveguide satisfies
(A) v p > c
(B) v p = c
(C) 0 < v p < c
(D) v p = 0
Statement for Linked Answer Question 7 and 8 :
An infinitely long uniform solid wire of radius a carries a uniform
dc current of density J
9.12
9.13
The magnetic field at a distance r from the center of the wire is
proportional to
(A) r for r < a and 1/r 2 for r > a (B) 0 for r < a and 1/r for r > a
(C) r for r < a and 1/r for r > a (D) 0 for r < a and 1/r 2 for r > a
A hole of radius b (b < a) is now drilled along the length of the wire
at a distance d from the center of the wire as shown below.
(B) 1.2 # 108 m/s
(D) 3 # 108 m/s
The modes in a rectangular waveguide are denoted by TE mn where
TM mn
m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following
statements is TRUE?
(A) The TM 10 mode of the waveguide does not exist
(B) The TE 10 mode of the waveguide does not exist
(C) The TM 10 and the TE 10 modes both exist and have the same
cut-off frequencies
(D) The TM 10 and the TM 01 modes both exist and have the same
cut-off frequencies
2011
9.17
TWO MARKS
A current sheet Jv = 10uty A/m lies on the dielectric interface x = 0
between two dielectric media with er 1 = 5, mr 1 = 1 in Region-1
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(x < 0) and er2 = 2, mr2 = 2 in Region-2 (x 2 0). If the magnetic field
in Region-1 at x = 0- is Hv1 = 3utx + 30uty A/m the magnetic field in
Region-2 at x = 0+ is
The magnetic field inside
(A) uniform and depends
(B) uniform and depends
(C) uniform and depends
(D) non uniform
2011
9.14
the hole is
only on d
only on b
on both b and d
(A) Hv2 = 1.5utx + 30uty - 10utz A/m
(B) Hv2 = 3utx + 30uty - 10utz A/m
(C) Hv2 = 1.5utx + 40uty A/m
(D) Hv2 = 3utx + 30uty + 10utz A/m
ONE MARK
Consider the following statements regarding the complex Poynting
vector Pv for the power radiated by a point source in an infinite
homogeneous and lossless medium. Re(Pv ) denotes the real part of
Pv, S denotes a spherical surface whose centre is at the point source,
and nt denotes the unit surface normal on S . Which of the following
statements is TRUE?
(A) Re(Pv ) remains constant at any radial distance from the source
(B) Re(Pv ) increases with increasing radial distance from the
9.18
9.19
source
(C)
##s Re (Pv) : nt dS
remains constant at any radial distance from
##s Re (Pv) : nt dS
decreases with increasing radial distance from
the source
(D)
the source
9.15
A transmission line of characteristic impedance 50 W is terminated
by a 50 W load. When excited by a sinusoidal voltage source at 10
GHz, the phase difference between two points spaced 2 mm apart on
the line is found to be p/4 radians. The phase velocity of the wave
along the line is
A transmission line of characteristic impedance 50 W is terminated
in a load impedance ZL . The VSWR of the line is measured as 5 and
the first of the voltage maxima in the line is observed at a distance
of l/4 from the load. The value of ZL is
(A) 10 W
(B) 250 W
(C) (19.23 + j 46.15) W
(D) (19.23 - j 46.15) W
The electric and magnetic fields for a TEM wave of frequency 14
GHz in a homogeneous medium of relative permittivity er and
relative permeability mr = 1 are given by Ev = E p e j (wt - 280py) utz V/m
and Hv = 3e j (wt - 280py) utx A/m . Assuming the speed of light in free space
to be 3 # 108 m/s , the intrinsic impedance of free space to be 120p
, the relative permittivity er of the medium and the electric field
amplitude E p are
(B) er = 3, E p = 360p
(A) er = 3, E p = 120p
(C) er = 9, E p = 360p
(D) er = 9, E p = 120p
2010
9.20
If the scattering matrix [S ] of a two port network is
ONE MARK
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 218
0.2+0c 0.9+90c
, then the network is
[S ] = >
0.9+90c 0.1+90cH
(A) lossless and reciprocal
(B) lossless but not reciprocal
(C) not lossless but reciprocal
(D) neither lossless nor reciprocal
9.21
9.22
A transmission line has a characteristic impedance of 50 W and a
resistance of 0.1 W/m . If the line is distortion less, the attenuation
constant(in Np/m) is
(A) 500
(B) 5
(C) 0.014
(D) 0.002
The electric field component of a time harmonic plane EM wave
traveling in a nonmagnetic lossless dielectric medium has an
amplitude of 1 V/m. If the relative permittivity of the medium is 4,
the magnitude of the time-average power density vector (in W/m2
) is
(A) 1
(B) 1
30p
60p
(C) 1
(D) 1
120p
240p
(A) 1.00
(C) 2.50
(B) 1.64
(D) 3.00
2009
9.26
ONE MARK
Two infinitely long wires carrying current are as shown in the figure
below. One wire is in the y - z plane and parallel to the y - axis.
The other wire is in the x - y plane and parallel to the x - axis.
Which components of the resulting magnetic field are non-zero at
the origin ?
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2010
9.23
v = xyatx + x 2 aty , then
If A
is
TWO MARKS
(A) x, y, z components
(C) y, z components
# Av $ dlv over the path shown in the figure
o
C
9.27
(A) 0
(C) 1
9.24
9.25
In the circuit shown, all the transmission line sections are lossless.
The Voltage Standing Wave Ration(VSWR) on the 60 W line is
Which of the following statements is true regarding the fundamental
mode of the metallic waveguides shown ?
(A) Only P has no cutoff-frequency
(B) Only Q has no cutoff-frequency
(C) Only R has no cutoff-frequency
(D) All three have cutoff-frequencies
(B) 2
3
(D) 2 3
A plane wave having the electric field components
Evi = 24 cos ^3 # 108 - by h atx V/m and traveling in free space is
incident normally on a lossless medium with m = m0 and e = 9e0
which occupies the region y $ 0 . The reflected magnetic field
component is given by
(A) 1 cos (3 # 108 t + y) atx A/m
10p
(B) 1 cos (3 # 108 t + y) atx A/m
20p
(C) - 1 cos (3 # 108 t + y) atx A/m
20p
(D) - 1 cos (3 # 108 t + y) atx A/m
10p
(B) x, y components
(D) x, z components
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2009
9.28
If a vector field V is related to another vector field A through
V = 4# A , which of the following is true? (Note : C and SC refer to
any closed contour and any surface whose boundary is C . )
(B) A $ dl =
(A) V $ dl =
V $ dS
A $ dS
(C)
9.29
TWO MARKS
#S #C
#S #C
#C
#C
#C D # V $ dl = #S #C D # A $ d S (D) #C D # V $ dl = #S #CV $ d S
A transmission line terminates in two branches, each of length l ,
4
as shown. The branches are terminated by 50W loads. The lines are
lossless and have the characteristic impedances shown. Determine
the impedance Zi as seen by the source.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 219
(C) 35 dB
(D) 45 dB
2007
9.38
(A) 200W
(C) 50W
9.30
A magnetic field in air is measured to be
y
xt m
B = B0 c 2 x 2 yt - 2
x + y2
x +y
What current distribution leads to this field ?
[Hint : The algebra is trivial in cylindrical coordinates.]
t
t
(A) J = B0 z c 2 1 2 m, r ! 0
(B) J =- B0 z c 2 2 2 m, r ! 0
m0 x + y
m0 x + y
t
(C) J = 0, r ! 0
(D) J = B0 z c 2 1 2 m, r ! 0
m0 x + y
2008
9.31
9.32
(B) 100W
(D) 25W
ONE MARK
For a Hertz dipole antenna, the half power beam width (HPBW) in
the E -plane is
(A) 360c
(B) 180c
(C) 90c
(D) 45c
ONE MARK
A plane wave of wavelength l is traveling in a direction making an
angle 30c with positive x - axis and 90c with positive y - axis. The
E field of the plane wave can be represented as (E0 is constant)
p
p
3p
3p
t 0 e j c wt - l x - l z m
t 0 e jc wt - l x - l z m
(A) E = yE
(B) E = yE
t 0 e jc wt +
(C) E = yE
9.39
3 p x+ p z
l
l m
p
t 0 e jc wt - l x +
(D) E = yE
If C is code curve enclosing a surface S , then magnetic field intensity
H , the current density j and the electric flux density D are related
by
2D
(A)
H $ ds =
c j + 2t m $ d t
S
c
##
(B)
(C)
##
#S H $ d l = ##S c j + 22Dt m $ dS
##S H $ dS = #C c j + 22Dt m $ d t
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For static electric and magnetic fields in an inhomogeneous sourcefree medium, which of the following represents the correct form of
Maxwell’s equations ?
(A) 4$ E = 0 , 4# B = 0
(B) 4$ E = 0 , 4$ B = 0
(C) 4# E = 0 , 4# B = 0
(D) 4# E = 0 , 4$ B = 0
#C H $ d l # = ##S c j + 22Dt m $ ds
(D)
c
2007
2008
9.33
9.34
9.35
9.36
9.37
TWO MARKS
9.40
A rectangular waveguide of internal dimensions (a = 4 cm and
b = 3 cm) is to be operated in TE11 mode. The minimum operating
frequency is
(A) 6.25 GHz
(B) 6.0 GHz
(C) 5.0 GHz
(D) 3.75 GHz
One end of a loss-less transmission line having the characteristic
impedance of 75W and length of 1 cm is short-circuited. At 3 GHz,
the input impedance at the other end of transmission line is
(A) 0
(B) Resistive
(C) Capacitive
(D) Inductive
At 20 GHz, the gain of a parabolic dish antenna of 1 meter and 70%
efficiency is
(A) 15 dB
(B) 25 dB
TWO MARKS
The E field in a rectangular waveguide of inner dimension a # b is
given by
2
wm
E = 2 ` l j H0 sin ` 2px j sin (wt - bz) yt
a
h 2
Where H0 is a constant, and a and b are the dimensions along the
x - axis and the y - axis respectively. The mode of propagation in
the waveguide is
(A) TE20
(B) TM11
(C) TM20
(D) TE10
9.41
A uniform plane wave in the free space is normally incident on
an infinitely thick dielectric slab (dielectric constant e = 9 ). The
magnitude of the reflection coefficient is
(A) 0
(B) 0.3
(C) 0.5
(D) 0.8
In the design of a single mode step index optical fibre close to upper
cut-off, the single-mode operation is not preserved if
(A) radius as well as operating wavelength are halved
(B) radius as well as operating wavelength are doubled
(C) radius is halved and operating wavelength is doubled
(D) radius is doubled and operating wavelength is halved
3pz
l m
A load of 50 W is connected in shunt in a 2-wire transmission line of
Z0 = 50W as shown in the figure. The 2-port scattering parameter
matrix (s-matrix) of the shunt element is
(A) >
1
2
1
1
2
2
(C) >
2
3
1
2
3
3
- 12
- 13
9.42
H
0 1
(B) =
1 0G
H
(D) >
-
1
4
3
4
- 43
1
4
H
The parallel branches of a 2-wirw transmission line re terminated in
100W and 200W resistors as shown in the figure. The characteristic
impedance of the line is Z0 = 50W and each section has a length of
l . The voltage reflection coefficient G at the input is
4
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 220
(C) 2
2006
9.47
(A) - j 7
5
(C) j 5
7
9.43
(B) - 5
7
(D) 5
7
The H field (in A/m) of a plane wave propagating in free space is
given by H = xt 5 3 cos (wt - bz) + yt`wt - bz + p j .
h0
2
The time average power flow density in Watts is
h
(A) 0
(B) 100
100
h0
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(C)
9.44
9.45
9.48
50h20
(D) 50
h0
9.50
An air-filled rectangular waveguide has inner dimensions of 3 cm
# 2 cm. The wave impedance of the TE20 mode of propagation in
the waveguide at a frequency of 30 GHz is (free space impedance
h0 = 377 W )
(A) 308 W
(B) 355 W
(C) 400 W
(D) 461 W
A l dipole is kept horizontally at a height of l0 above a perfectly
2
2
conducting infinite ground plane. The radiation pattern in the lane
of the dipole (E plane) looks approximately as
9.51
ONE MARK
The electric field of an electromagnetic wave propagation in the positive
direction is given by E = atx sin (wt - bz) + aty sin (wt - bz + p/2). The
wave is
(A) Linearly polarized in the z -direction
(B) Elliptically polarized
(C) Left-hand circularly polarized
(D) Right-hand circularly polarized
A transmission line is feeding 1 watt of power to a horn antenna
having a gain of 10 dB. The antenna is matched to the transmission
line. The total power radiated by the horn antenna into the free
space is
(A) 10 Watts
(B) 1 Watts
(C) 0.1 Watts
(D) 0.01 Watt
2006
9.49
(D) 3
TWO MARKS
When a planes wave traveling in free-space is incident normally on a
medium having the fraction of power transmitted into the medium
is given by
(B) 1
(A) 8
2
9
(C) 1
(D) 5
3
6
A medium of relative permittivity er2 = 2 forms an interface with
free - space. A point source of electromagnetic energy is located in
the medium at a depth of 1 meter from the interface. Due to the
total internal reflection, the transmitted beam has a circular crosssection over the interface. The area of the beam cross-section at the
interface is given by
(A) 2p m 2
(B) p2 m 2
(C) p m 2
(D) p m 2
2
A rectangular wave guide having TE10 mode as dominant mode is
having a cut off frequency 18 GHz for the mode TE30 . The inner
broad - wall dimension of the rectangular wave guide is
(B) 5 cm
(A) 5 cm
3
(D) 10 cm
(C) 5 cm
2
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9.52
9.46
A medium is divide into regions I and II about x = 0 plane, as
shown in the figure below.
A right circularly polarized (RCP) plane wave is incident at an angle
60c to the normal, on an air-dielectric interface. If the reflected wave
is linearly polarized, the relative dielectric constant xr2 is.
(A)
2
(B)
3
An electromagnetic wave with electric field E1 = 4atx + 3aty + 5atz is
incident normally on the interface from region I . The electric file
E2 in region II at the interface is
(B) 4atx + 0.75aty - 1.25atz
(A) E2 = E1
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) 3atx + 3aty + 5atz
9.53
(D) - 3atx + 3aty + 5atz
2005
ONE MARK
The magnetic field intensity vector of a plane wave is given by
H (x, y, z, t) = 10 sin (50000t + 0.004x + 30) aty
Refractive index of glass is 1.5. Find the wavelength of a beam of
light with frequency of 1014 Hz in glass. Assume velocity of light is
3 # 108 m/s in vacuum
(A) 3 mm
(B) 3 mm
(C) 2 mm
(D) 1 mm
2005
9.56
9.57
9.58
Voltage standing wave pattern in a lossless transmission line with
characteristic impedance 50 and a resistive load is shown in the
figure.
(D) 20p2
where aty , denotes the unit vector in y direction. The wave is
propagating with a phase velocity.
(B) - 3 # 108 m/s
(A) 5 # 10 4 m/s
(C) - 1.25 # 107 m/s
(D) 3 # 108 m/s
9.55
Statement of Linked Answer Questions 9.46 & 9.47 :
A mast antenna consisting of a 50 meter long vertical conductor
operates over a perfectly conducting ground plane. It is base-fed at
a frequency of 600 kHz. The radiation resistance of the antenna is
Ohms is
2
2
(A) 2p
(B) p
5
5
2
(C) 4p
5
9.54
Page 221
9.59
9.60
The value of the load resistance is
(A) 50 W
(B) 200 W
(C) 12.5 W
(D) 0
The reflection coefficient is given by
(A) - 0.6
(B) - 1
(C) 0.6
(D) 0
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TWO MARKS
Which one of the following does represent the electric field lines
for the mode in the cross-section of a hollow rectangular metallic
waveguide ?
Characteristic impedance of a transmission line is 50 W. Input
impedance of the open-circuited line when the transmission line a
short circuited, then value of the input impedance will be.
(A) 50 W
(B) 100 + j150W
(C) 7.69 + j11.54W
(D) 7.69 - j11.54W
9.61
Many circles are drawn in a Smith Chart used for transmission line
calculations. The circles shown in the figure represent
(A) Unit circles
(B) Constant resistance circles
(C) Constant reactance circles
(D) Constant reflection coefficient circles.
2004
9.62
Two identical and parallel dipole antennas are kept apart by a
distance of l in the H - plane. They are fed with equal currents
4
but the right most antenna has a phase shift of + 90c. The radiation
pattern is given as.
9.63
The phase velocity of an electromagnetic wave propagating in a
hollow metallic rectangular waveguide in the TE10 mode is
(A) equal to its group velocity
(B) less than the velocity of light in free space
(C) equal to the velocity of light in free space
(D) greater than the velocity of light in free space
Consider a lossless antenna with a directive gain of + 6 dB. If 1 mW
of power is fed to it the total power radiated by the antenna will be
(A) 4 mW
(B) 1 mW
(C) 7 mW
(D) 1/4 mW
2004
9.64
ONE MARK
TWO MARKS
A parallel plate air-filled capacitor has plate area of 10 - 4 m 2 and
plate separation of 10 - 3 m. It is connect - ed to a 0.5 V, 3.6 GHz
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
source. The magnitude of the displacement current is ( e =
F/m)
(A) 10 mA
(B) 100 mA
(C) 10 A
(D) 1.59 mA
9.65
1
36p
Page 222
10 - 9
the medium should be
(A) 120p W
(C) 600p W
9.70
Consider a 300 W, quarter - wave long (at 1 GHz) transmission line
as shown in Fig. It is connected to a 10 V, 50 W source at one end
and is left open circuited at the other end. The magnitude of the
voltage at the open circuit end of the line is
(B) 60p W
(D) 24p W
A lossless transmission line is terminated in a load which reflects a
part of the incident power. The measured VSWR is 2. The percentage
of the power that is reflected back is
(A) 57.73
(B) 33.33
(C) 0.11
(D) 11.11
2003
9.71
(A) 10 V
(C) 60 V
9.66
(B) 5 V
(D) 60/7 V
9.72
In a microwave test bench, why is the microwave signal amplitude
modulated at 1 kHz
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9.73
9.68
If E = (atx + jaty) e jkz - kwt and H = (k/wm) (aty + katx ) e jkz - jwt , the timeaveraged Poynting vector is
(A) null vector
(B) (k/wm) atz
(C) (2k/wm) atz
(D) (k/2wm) atz
Consider an impedance Z = R + jX marked with point P in an
impedance Smith chart as shown in Fig. The movement from point
P along a constant resistance circle in the clockwise direction by an
angle 45c is equivalent to
9.74
9.75
(B) Ampere/meter
(D) Ampere-meter
The depth of penetration of electromagnetic wave in a medium
having conductivity s at a frequency of 1 MHz is 25 cm. The depth
of penetration at a frequency of 4 MHz will be
(A) 6.25 dm
(B) 12.50 cm
(C) 50.00 cm
(D) 100.00 cm
2003
(A) To increase the sensitivity of measurement
(B) To transmit the signal to a far-off place
(C) To study amplitude modulations
(D) Because crystal detector fails at microwave frequencies
9.67
The unit of 4# H is
(A) Ampere
(C) Ampere/meter 2
ONE MARK
TWO MARKS
Medium 1 has the electrical permittivity e1 = 1.5e0 farad/m and
occupies the region to the left of x = 0 plane. Medium 2 has the
electrical permittivity e2 = 2.5e0 farad/m and occupies the region to
the right of x = 0 plane. If E1 in medium 1 is E1 = (2ux - 3uy + 1uz )
volt/m, then E2 in medium 2 is
(A) (2.0ux - 7.5uy + 2.5uz ) volt/m
(B) (2.0ux - 2.0uy + 0.6uz ) volt/m
(C) (2.0ux - 3.0uy + 1.0uz ) volt/m
(D) (2.0ux - 2.0uy + 0.6uz ) volt/m
If the electric field intensity is given by E = (xux + yuy + zuz ) volt/m,
the potential difference between X (2, 0, 0) and Y (1, 2, 3) is
(A) + 1 volt
(B) - 1 volt
(C) + 5 volt
(D) + 6 volt
A uniform plane wave traveling in air is incident on the plane
boundary between air and another dielectric medium with er = 4 .
The reflection coefficient for the normal incidence, is
(A) zero
(B) 0.5+180c
(B) 0.333+0c
(D) 0.333+180c
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9.76
(A) adding an inductance in series with Z
(B) adding a capacitance in series with Z
(C) adding an inductance in shunt across Z
(D) adding a capacitance in shunt across Z
9.69
A plane electromagnetic wave propagating in free space is incident
normally on a large slab of loss-less, non-magnetic, dielectric material
with e > e0 . Maxima and minima are observed when the electric
field is measured in front of the slab. The maximum electric field is
found to be 5 times the minimum field. The intrinsic impedance of
9.77
If the electric field intensity associated with a uniform plane
electromagnetic wave traveling in a perfect dielectric medium is
given by E (z, t) = 10 cos (2p107 t - 0.1pz) V/m, then the velocity of
the traveling wave is
(B) 2.00 # 108 m/sec
(A) 3.00 # 108 m/sec
(C) 6.28 # 107 m/sec
(D) 2.00 # 107 m/sec
A short - circuited stub is shunt connected to a transmission line as
shown in fig. If Z0 = 50 ohm, the admittance Y seen at the junction
of the stub and the transmission line is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 223
(C) elliptically polarized
9.84
(D) unpolarized
Distilled water at 25c C is characterized by s = 1.7 # 10 - 4 mho/m
and e = 78eo at a frequency of 3 GHz. Its loss tangent tan d is
( e = 10
36p F/m)
-9
(A) 1.3 # 10-5
(C) 1.3 # 10-4 /78
9.85
(B) 1.3 # 10-3
(D) 1.3 # 10-5 /78e0
The electric field on the surface of a perfect conductor is 2 V/m.
The conductor is immersed in water with e = 80eo . The surface
charge density on the conductor is ( e = 10
36p F/m)
2
(A) 0 C/m
(B) 2 C/m 2
-9
(A) (0.01 - j0.02) mho
(C) (0.04 - j0.02) mho
9.78
9.79
(C) 1.8 # 10 - 11 C/m 2
(B) (0.02 - j0.01) mho
(D) (0.02 + j0) mho
9.86
A rectangular metal wave guide filled with a dielectric material of
relative permittivity er = 4 has the inside dimensions 3.0 cm # 1.2
cm. The cut-off frequency for the dominant mode is
(A) 2.5 GHz
(B) 5.0 GHz
(C) 10.0 GHz
(D) 12.5 GHz
Two identical antennas are placed in the q = p/2 plane as shown
in Fig. The elements have equal amplitude excitation with 180c
polarity difference, operating at wavelength l. The correct value
of the magnitude of the far-zone resultant electric field strength
normalized with that of a single element, both computed for f = 0
, is
(D) 1.41 # 10 - 9 C/m 2
A person with receiver is 5 Km away from the transmitter. What
is the distance that this person must move further to detect a 3-dB
decrease in signal strength
(A) 942 m
(B) 2070 m
(C) 4978 m
(D) 5320 m
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2001
9.87
A transmission line is distortonless if
(B) RL = GC
(A) RL = 1
GC
(C) LG = RC
(A) 2 cos b 2ps l
l
(C) 2 cos a ps k
l
(B) 2 sin b 2ps l
l
(D) 2 sin a ps k
l
9.88
2002
9.80
9.81
9.82
The VSWR can have any value between
(A) 0 and 1
(B) - 1 and + 1
(C) 0 and 3
(D) 1 and 3
9.89
In in impedance Smith movement along a constant resistance circle
gives rise to
(A) a decrease in the value of reactance
(B) an increase in the value of reactance
(C) no change in the reactance value
(D) no change in the impedance
The phase velocity for the TE10
waveguide is (c is the velocity of
(A) less than c
(C) greater than c
2002
9.83
ONE MARK
9.90
-mode in an air-filled rectangular
plane waves in free space)
(B) equal to c
(D) none of these
TWO MARKS
jp/2
jwt - jkz
t )e
A plane wave is characterized by E = (0.5xt + ye
wave is
(A) linearly polarized
(B) circularly polarized
. This
(D) RG = LC
2
2
If a plane electromagnetic wave satisfies the equal d E2x = c2 d E2x ,
dZ
dt
the wave propagates in the
(A) x - direction
(B) z - direction
(C) y - direction
(D) xy plane at an angle of 45c between the x and z direction
The plane velocity of wave propagating in a hollow metal waveguide
is
(A) grater than the velocity of light in free space
(B) less than the velocity of light in free space
(C) equal to the velocity of light free space
(D) equal to the velocity of light in free
The dominant mode in a rectangular waveguide is TE10 , because this
mode has
(A) the highest cut-off wavelength
(B) no cut-off
(C) no magnetic field component
(D) no attenuation
2001
9.91
ONE MARK
TWO MARKS
A material has conductivity of 10 - 2 mho/m and a relative permittivity
of 4. The frequency at which the conduction current in the medium
is equal to the displacement current is
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) 45 MHz
(C) 450 MHz
9.92
9.93
9.94
(l1 /l2) is
(A) er1 /er2
(B) 90 MHz
(D) 900 MHz
A uniform plane electromagnetic wave incident on a plane surface
of a dielectric material is reflected with a VSWR of 3. What is the
percentage of incident power that is reflected ?
(A) 10%
(B) 25%
(C) 50%
(D) 75%
A medium wave radio transmitter operating at a wavelength of
492 m has a tower antenna of height 124. What is the radiation
resistance of the antenna?
(A) 25 W
(B) 36.5 W
(C) 50 W
(D) 73 W
In uniform linear array, four isotropic radiating elements are spaced
l apart. The progressive phase shift between required for forming
4
the main beam at 60c off the end - fire is :
(A) - p
(B) - p2 radians
(C) - p4 radians
(D) - p8 radians
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2000
9.95
9.96
9.97
A TEM wave is incident normally upon a perfect conductor. The E
and H field at the boundary will be respectively,
(A) minimum and minimum
(B) maximum and maximum
(C) minimum and maximum
(D) maximum and minimum
If the diameter of a l dipole antenna is increased from l to l
2
100
50
, then its
9.99
9.100
(D)
9.102
Two coaxial cable 1 and 2 are filled with different dielectric constants
er1 and er2 respectively. The ratio of the wavelength in the cables
For an 8 feet (2.4m) parabolic dish antenna operating at 4 GHz, the
minimum distance required for far field measurement is closest to
(A) 7.5 cm
(B) 15 cm
(C) 15 m
(D) 150 m
ONE MARK
An electric field on a place is described by its potential
V = 20 (r-1 + r-2)
where r is the distance from the source. The field is due to
(A) a monopole
(B) a dipole
(C) both a monopole and a dipole (D) a quadruple
9.103
9.104
9.105
Assuming perfect conductors of a transmission line, pure TEM
propagation is NOT possible in
(A) coaxial cable
(B) air-filled cylindrical waveguide
(C) parallel twin-wire line in air
(D) semi-infinite parallel plate wave guide
Indicate which one of the following will NOT exist in a rectangular
resonant cavity.
(A) TE110
(B) TE 011
(C) TM110
(D) TM111
Identify which one of the following will NOT satisfy the wave
equation.
(A) 50e j (wt - 3z)
(B) sin [w (10z + 5t)]
2
(C) cos (y + 5t)
(D) sin (x) cos (t)
1999
9.106
9.107
TWO MARKS
In a twin-wire transmission line in air, the adjacent voltage maxima
are at 12.5 cm and 27.5 cm . The operating frequency is
(A) 300 MHz
(B) 1 GHz
(C) 2 GHz
(D) 6.28 GHz
A transmitting antenna radiates 251 W isotropically. A receiving
antenna, located 100 m away from the transmitting antenna, has an
effective aperture of 500 cm2 . The total received by the antenna is
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(A) 10 mW
(C) 20 mW
9.108
2
A rectangular waveguide has dimensions 1 cm # 0.5 cm. Its cut-off
frequency is
(A) 5 GHz
(B) 10 GHz
(C) 15 GHz
(D) 12 GHz
(B) er2 /er1
(D) er2 /er1
1999
TWO MARKS
A uniform plane wave in air impings at 45c angle on a lossless
dielectric material with dielectric constant dr . The transmitted
wave propagates is a 30c direction with respect to the normal. The
value of dr is
(B) 1.5
(A) 1.5
(C) 2
9.101
(B) bandwidth decrease
(D) gain decreases
2000
9.98
(C) er1 /er2
ONE MARK
The magnitudes of the open-circuit and short-circuit input
impedances of a transmission line are 100 W and 25 W respectively.
The characteristic impedance of the line is,
(A) 25 W
(B) 50 W
(C) 75 W
(D) 100 W
(A) bandwidth increases
(C) gain increases
Page 224
9.109
(B) 1 mW
(D) 100 mW
In air, a lossless transmission line of length 50 cm with L = 10 mH/m
, C = 40 pF/m is operated at 25 MHz . Its electrical path length is
(A) 0.5 meters
(B) l meters
(C) p/2 radians
(D) 180 deg rees
A plane wave propagating through a medium [er = 8, vr = 2, and s = 0]
t - (z/3) sin (108 t - bz) V/m . The
has its electric field given by Ev = 0.5Xe
wave impedance, in ohms is
(A) 377
(B) 198.5+180c
(D) 133.3
(C) 182.9+14c
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
1998
9.110
9.111
9.112
ONE MARK
Page 225
9.119
The intrinsic impedance of copper at high frequencies is
(A) purely resistive
(B) purely inductive
(C) complex with a capacitive component
(D) complex with an inductive component
The Maxwell equation V # H = J + 2D is based on
2t
(A) Ampere’s law
(B) Gauss’ law
(C) Faraday’s law
(D) Coulomb’s law
9.120
All transmission line sections shown in the figure is have a
characteristic impedance R 0 + j 0 . The input impedance Zin equals
9.121
An antenna in free space receives 2 mW of power when the incident
electric field is 20 mV/m rms. The effective aperture of the antenna
is
(A) 0.005 m2
(B) 0.05 m2
(C) 1.885 m2
(D) 3.77 m2
The maximum usable frequency of an ionospheric layer at 60c
incidence and with 8 MHz critical frequency is
(A) 16 MHz
(B) 16 MHz
3
(C) 8 MHz
(D) 6.93 MHz
A loop is rotating about they y -axis in a magnetic field
Bv = B 0 cos (wt + f) avx T. The voltage in the loop is
(A) zero
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(A) 2 R 0
3
(C) 3 R 0
2
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(B) R 0
(D) 2R 0
1998
9.113
9.114
9.115
9.116
9.117
9.118
(B) due to rotation only
(C) due to transformer action only
(D) due to both rotation and transformer action
TWO MARKS
The time averages Poynting vector,
Ev = 24e j (wt + bz) avy V/m in free space is
(A) - 2.4 avz
(B)
p
(C) 4.8 avz
(D)
p
in W/m2 , for a wave with
9.122
2.4 av
p z
- 4.8 avz
p
The wavelength of a wave with propagation constant (0.1p + j0.2p) m-1
is
2 m
(A)
(B) 10 m
0.05
(C) 20 m
(D) 30 m
1997
9.123
The depth of penetration of wave in a lossy dielectric increases with
increasing
(A) conductivity
(B) permeability
(C) wavelength
(D) permittivity
The polarization of
Ev = E 0 e j^wt + bz h ^avx + avy h is
(A) linear
(C) left hand circular
wave
with
electric
field
vector
9.124
(B) elliptical
(D) right hand circular
The vector H in the far field of an antenna satisfies
(A) d $ Hv = 0 and d # Hv = 0
(B) d $ Hv ! 0 and d # Hv ! 0
(C) d $ Hv = 0 and d # Hv ! 0
(D) d $ Hv ! 0 and d # Hv = 0
The radiation resistance of a circular loop of one turn is 0.01 W. The
radiation resistance of five turns of such a loop will be
(B) 0.01 W
(A) 0.002 W
(C) 0.05 W
(D) 0.25 W
The far field of an antenna varies with distance r as
(A) 1
(B) 12
r
r
(C) 13
(D) 1
r
r
9.125
ONE MARK
A transmission line of 50 W characteristic impedance is terminated
with a 100 W resistance. The minimum impedance measured on the
line is equal to
(A) 0 W
(B) 25 W
(C) 50 W
(D) 100 W
A rectangular air filled waveguide has cross section of 4 cm #10 cm
. The minimum frequency which can propagate in the waveguide is
(A) 0.75 GHz
(B) 2.0 GHz
(C) 2.5 GHz
(D) 3.0 GHz
A parabolic dish antenna has a conical beam 2c wide, the directivity
of the antenna is approximately
(A) 20 dB
(B) 30 dB
(C) 40 dB
(D) 50 dB
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
1997
9.126
9.127
Page 226
TWO MARKS
A very lossy, l/4 long, 50 W transmission line is open circuited at
the load end. The input impedance measured at the other end of the
line is approximately
(A) 0
(B) 50 W
(C) 3
(D) None of the above
(A) 0%
(B) 4%
(C) 20%
(D) 100%
9.133
The skin depth at 10 MHz for a conductor is 1 cm. The phase velocity
of an electromagnetic wave in the conductor at 1, 000 MHz is about
(A) 6 # 106 m/ sec
(B) 6 # 107 m/ sec
(C) 3 # 108 m/ sec
(D) 6 # 108 m/ sec
9.134
1996
9.128
ONE MARK
A lossless transmission line having 50 W characteristic impedance
and length l/4 is short circuited at one end and connected to an
ideal voltage source of 1 V at the other end. The current drawn from
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the voltage source is
(A) 0
(C) 3
9.129
9.130
9.131
A 1 km long microwave link uses two antennas each having 30 dB
gain. If the power transmitted by one antenna is 1 W at 3 GHz, the
power received by the other antenna is approximately
(A) 98.6 mW
(B) 76.8 mW
(C) 63.4 mW
(D) 55.2 mW
Some unknown material has a conductivity of 106 mho/m and a
permeability of 4p # 10-7 H/m . The skin depth for the material at
1 GHz is
(A) 15.9 mm
(B) 20.9 mm
(C) 25.9 mm
(D) 30.9 mm
(B) 0.02 A
(D) none of these
The capacitance per unit length and the characteristic impedance of
a lossless transmission line are C and Z 0 respectively. The velocity
of a travelling wave on the transmission line is
(A) Z 0 C
(B) 1
Z0 C
(C) Z 0
(D) C
Z0
C
A transverse electromagnetic wave with circular polarization is
received by a dipole antenna. Due to polarization mismatch, the
power transfer efficiency from the wave to the antenna is reduced
to about
(A) 50%
(B) 35.3%
(C) 25%
(D) 0%
A metal sphere with 1 m radius and a surface charge density of
10 Coulombs/m2 is enclosed in a cube of 10 m side. The total
outward electric displacement normal to the surface of the cube is
(A) 40p Coulombs
(B) 10p Coulombs
(C) 5p Coulombs
(D) None of these
1996
9.132
9.135
The critical frequency of an ionospheric layer is 10 MHz. What is
the maximum launching angle from the horizon for which 20 MHz
wave will be reflected by the layer ?
(A) 0c
(B) 30c
(C) 45c
(D) 90c
TWO MARKS
A uniform plane wave in air is normally incident on infinitely thick
slab. If the refractive index of the glass slab is 1.5, then the percentage
of incident power that is reflected from the air-glass interface is
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 227
SOLUTION
Also, direction of propagation is
v
v
avk = ax + az
2
x
+
z
So,
k =
2
Substituting it in equation (1), we get
p # 10 ^x + z h
Evi = Eo _avx - avz i e-j 3 2
2
4
9.1
Option (D) is correct.
v around
Stoke’s theorem states that the circulation a vector field A
v over
a closed path l is equal to the surface integral of the curl of A
the open surface S bounded by l .
vh : dsv
v : dlv =
^d # A
i.e.,
A
9.5
##
#
Here, line integral is taken across a closed path which is denoted
by a small circle on the integral notation where as, the surface invh is taken over open surface bounded by the loop.
tegral of ^d # A
9.2
Option (A) is correct.
We obtain the reflection coefficient for parallel polarized wave (since,
electric field is in the plane of wave propagation) as
h cos qt - h1 cos qi
Gz = 2
h2 cos qt + h1 cos qi
...(1)
As we have already obtained
Option (D) is correct.
Given, the vector field
qi = 45c, qt = 19.2c
m
1 =
h2 =
= h0
e
4.5
Also,
v = xavx + yavy + zavz
A
so,
v (Divergence of A
v ) = 2Ax + 2Ay + 2Az
d$A
2x
2y
2z
= 1+1+1 = 3
9.3
Option (A) is correct.
Given, the return loss of device as 20 dB
i.e.,
G in dB =- 20 dB (loss)
^
or,
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h
20 log G =- 20
&
G = 10-1 = 0.1
Therefore, the standing wave ration is given by
1+ G
VSWR =
1- G
m
= h0 1 = h0
1
e
Substituting these in eq. (1) we get
G z = cos 19.2c - 4.5 cos 45c
cos 19.2c + 4.5 cos 45c
=- 0.227
.- 0.23
Therefore, the reflected field has the magnitude given by
Ero
= T 11'
Eio
or
Ero = G z Eio =- 0.23 Eio
Hence, the expression of reflected electric field is
p # 10
Evr =- 0.23 Eo _- avx - avz i e-j 3 k
2
(2)
Option (C) is correct.
For the given incidence of plane wave, we have the transmitting
angle
qt = 19.2c
From Snell’s law, we know
n1 sin qi = n 2 sin qt
4
c m1 e1 sin qi = c m2 e2 sin qt
...(1)
For the given interfaces, we have
m1 = m2 = 1
e1 = 1, e2 = 4.5
So, from Eq. (1)
sin qi = 4.5 sin 19.2
or,
qi . 45c
v
Now, the component of Ei can be obtained as
Evi = _Eox avx - Eoz avz i e-jbk
(observed from the shown figure)
Since, the angle qi = 45c so,
Eox = Eoz = Eo
2
E
v
o
v
Therefore,
Ei =
_ax - avz i e-jbk
2
...(1)
Now, the wavelength of EM wave is
l = 600 mm
So,
b = 2p = p # 10 4
3
l
h1 =
and
= 1 + 0.1 = 1.1 = 1.22
1 - 0.1 0.9
9.4
h0
4.5
Again, we have the propagation vector of reflected wave as
v
v
avk = ax - az
2
x
z
or,
k =
2
Substituting it in Eq. (2), we get
p # 10 x - z
Evr =- 0.23 Eo _- avx - avz i e-j 3 b 2 l
2
jp # 10 ^x - z h V
E
m
Evr = 0.23 o _avx + avz i e- 3 2
2
4
4
9.6
Option (C) is correct.
Electric field of the propagating wave in free space is given as
Ei = (8ax + 6ay + 5az ) e j (wt + 3x - 4y) V/m
So, it is clear that wave is propagating in the direction (- 3ax + 4ay)
.
Since, the wave is incident on a perfectly conducting slab at x = 0 .
So, the reflection coefficient will be equal to - 1.
i.e.
Er = (- 1) Ei =- 8ax - 6ay - 5az
Again, the reflected wave will be as shown in figure.
0
0
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 228
l = (2m + 1) l
4
3 # 108
c =
= 0.174 m
f1 # 4 429 # 106 # 4
8
f2 = 1 GHz ,
l2 = c = 3 # 10
= 0.075 m
9
f2 # 4 1 # 10 # 4
Only option (C) is odd multiple of both l1 and l2 .
(2m + 1) = 1.58 = 9
l1
(2m + 1) = 1.58 - 21
l2
f1 = 429 MHz,
i.e. the reflected wave will be in direction 3ax + 4ay . Thus, the
electric field of the reflected wave will be.
Ex = (- 8ax - 6ay - 5az ) e j (wt - 3x - 4y) V/m
9.7
Option (A) is correct.
The field in circular polarization is found to be
Es = E 0 (ay ! jaz ) e-jbx propagating in + ve x -direction.
where, plus sign is used for left circular polarization and minus
sign for right circular polarization. So, the given problem has left
circular polarization.
b = 25 = w
c
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25 =
9.8
9.9
2pf
c
9.11
Substituting above values,
b =
- j261
9.12
# H : dl
For r < a ,
#
cos 4 q sin qdqdfE
p/2
5
= 1 ;2p b- cos q lE = 1 # 2p :- 0 + 1 D
5
5
4p
4p
0
= 1 # 2p = 1
5
10
4p
D = 1 = 10
10
9.10
D (in dB) = 10 log 10 = 10 dB
Option (C) is correct.
Since
Z0 =
Z1 Z 2
100 = 50 # 200
This is quarter wave matching. The length would be odd multiple
of l/4 .
Ienclosed =
Io = (pa2) J
J (pr 2) Jr 2
= 2
a
pa2
# H : dl
= Ienclosed
2
H # 2pr = Jr2
a
H = Jr 2
2pa
H \ r , for r < a
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# #
0
= Ienclosed
H # 2pr = (pa2) J
H = Io
2pr
H \ 1 , for r > a
r
Fmax = 1
Favg = 1 F (q, f) dW
4p
2p
2p
= 1 ;
F (q, f) sin qdq dfE
4p 0 0
or,
Option ( ) is correct.
For r > a ,
Ienclosed = (pa2) J
So,
p/2
6.283 # 1010 2 - (2.0942 + 2.6182) 10 4
#
c 3 108 m
#
b is imaginary so mode of operation is non-propagating.
vp = 0
Option (A) is correct.
The directivity is defined as
D = Fmax
Favg
# #
w 2 - (b 2 + b 2)
x
y
c2
b =
Option (B) is correct.
Let
b " outer diameter
a " inner diameter
Characteristic impedance,
m0
4p # 10-7 # 36p ln 2.4 = 100 W
Z0 =
ln b b l =
b 1 l
e0 er
a
10-9 # 10.89
2p
= 1 ;
4p 0
Option (D) is correct.
Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t - bz)
bx = 2.094 # 102
by = 2.618 # 102
w = 6.283 # 1010 rad/s
For the wave propagation,
8
f = 25 # c = 25 # 3 # 10 = 1.2 GHz
2p
2 # 3.14
&
l1 =
9.13
Option (B) is correct.
Assuming the cross section of the wire on x -y plane as shown in
figure.
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 229
Further if
H z = 1.5utx + Auty + Buz
Then from Boundary condition
10ut
(3utx + 30uty) utx = (1.5utx + Auty + Butz ) xt + v y
J
=- 30utz =- Autz + Buty + 10uty
Comparing we get A = 30 and B =- 10
So
H z = 1.5utx + 30uty - 10utz A/m
9.18
Since, the hole is drilled along the length of wire. So, it can be
assumed that the drilled portion carriers current density of - J .
Now, for the wire without hole, magnetic field intensity at point P
will be given as
Hf1 (2pR) = J (pR2)
Hf1 (2pR) = JR
2
Since, point o is at origin. So, in vector form
H1 = J (xax + yay)
2
Again only due to the hole magnetic field intensity will be given as.
9.19
9.14
9.15
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9.17
From the expressions of Ev & Hv , we can write,
b = 280 p
2p = 280 p & l = 1
or
140
l
v
E
Wave impedance,
Zw = E = p = 120 p
3
v
er
H
again,
y = yl
H2 = - J [(x - d) ax + yay]
2
Now
So, total magnetic field intensity = H1 + H2 = J dax
2
So, magnetic field inside the hole will depend only on ‘d ’.
or
Option (C) is correct.
Power radiated from any source is constant.
Now
or
9.20
Option (C) is correct.
We have d = 2 mm and f = 10 GHz
Phase difference = 2p d = p ;
4
l
Option (A) is correct.
TM11 is the lowest order mode of all the TMmn modes.
or
or
Option (C) is correct.
For a lossless network
(not lossless)
S12 = S21 = 0.9 90c (Reciprocal)
9.21
Option (A) is correct.
From boundary condition
Bn1 = Bn2
m1 Hx1 = m2 Hx2
Hx2 = Hx1 = 1.5
2
Hx2 = 1.5utx
f = 14 GHz
8
3
l = C = 3 # 10 9 =
er f
er 14 # 10
140 er
3
1
=
140
140 er
er = 9
Ep
= 120p = E p = 120p
3
9
S11 2 + S21 2 = 1
For the given scattering matrix
S11 = 0.2 0c , S12 = 0.9 90c
S21 = 0.9 90c , S22 = 0.1 90c
Here,
(0.2) 2 + (0.9) 2 ! 1
Reciprocity :
= l = 8d = 8 # 2 mm = 16 mm
v = fl = 10 # 109 # 16 # 10-3
= 1.6 # 108 m/ sec
or
9.16
Option (D) is correct.
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(Hf2) (2pr) =- J (pr 2)
Hf2 = - Jr
2
Again, if we take Ol at origin then in vector form
H2 = - J (xlax + ylay)
2
where xl and yl denotes point ‘P ’ in the new co-ordinate system.
Now the relation between two co-ordinate system will be.
x = xl + d
So,
Option (A) is correct.
Since voltage maxima is observed at a distance of l/4 from the
load and we know that the separation between one maxima and
minima equals to l/4 so voltage minima will be observed at the
load, Therefore load can not be complex it must be pure resistive.
Now
G = s-1
s+1
also RL = R 0 (since voltage maxima is formed at the load)
s
RL = 50 = 10 W
5
Option (D) is correct.
For distortion less transmission line characteristics impedance
Z0 = R
G
Attenuation constant
a =
So,
9.22
Option (C) is correct.
RG
a = R = 0.1 = 0.002
50
Z0
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 230
tan bl = tan b 2p : l l = 3
4
l
R ZL
V
S tan bl + jZo W
2
W = Z 0 = 60 W
Z in = Zo S
S Zo + jZL W ZL
S tan bl
W
T
X
For length of l/8 transmission line
Z + jZo tan bl
Z in = Zo ; L
Zo + jZL tan bl E
Intrinsic impedance of EM wave
m
m0
=
= 120p = 60p
h =
e
2
4e0
Time average power density
2
1
= 1
Pav = 1 EH = 1 E =
2
2 h
2 # 60p 120p
9.23
So,
Option (C) is correct.
v = xyatx + x 2 aty
A
v = dxatx + dyaty
dl
v = # (xyatx + x 2 aty) : (dxatx + dyaty)
# Av : dl
= # (xydx + x 2 dy)
Zo = 30 W, ZL = 0 (short)
tan bl = tan b 2p : l l = 1
8
l
C
C
Z in = jZo tan bl = 30j
Circuit is shown below.
C
=
2/ 3
#1/
3
xdx +
1/ 3
#2/
3
3xdx +
#1
3
4 dy +
3
#3
1
1 dy
3
= 1 : 4 - 1 D + 3 :1 - 4 D + 4 [3 - 1] + 1 [1 - 3] = 1
2 3 3
2 3 3
3
3
9.24
Option (A) is correct.
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Reflection coefficient
60 + 3j - 60
t = ZL - Zo =
=
60 + 3j + 60
ZL + Zo
1+ t
VSWR =
= 1 + 17 = 1.64
1- t
1 - 17
9.26
In the given problem
9.27
Reflection coefficient
h - h1
t = 2
= 400p - 120p =- 1
h2 + h 1
2
40p + 120p
t is negative So magnetic field component does not change its
direction Direction of incident magnetic field
atE # atH = atK
atZ # atH = aty
atH = atx ( + x direction)
So, reflection magnetic field component
Hr = t # 24 cos (3 # 108 + by) atx , y $ 0
h
= 1 # 24 cos (3 # 108 + by) atx , y $ 0
2 # 120p
So,
9.25
8
b = w = 3 # 108 = 1
vC
3 # 10
Hr = 1 cos (3 # 108 + y) atx , y $ 0
10p
Option (B) is correct.
For length of l/4 transmission line
Z + jZo tan bl
Z in = Zo ; L
Zo + jZL tan bl E
ZL = 30 W ,
Zo = 30 W, b = 2p , l = l
4
l
9.28
1
17
Option (D) is correct.
Due to 1 A current wire in x - y plane, magnetic field be at origin
will be in x direction.
Due to 1 A current wire in y - z plane, magnetic field be at origin
will be in z direction.
Thus x and z component is non-zero at origin.
Option (A) is correct.
Rectangular and cylindrical waveguide doesn’t support TEM modes
and have cut off frequency.
Coaxial cable support TEM wave and doesn’t have cut off frequency.
Option (B) is correct.
We have
V = 4# A
By Stokes theorem
# A $ dl
=
## (4 # A) $ ds
...(1)
...(2)
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From (1) and (2) we get
# A $ dl
9.29
=
Option (D) is correct.
##V $ ds
The transmission line are as shown below. Length of all line is l
4
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 231
2
2
Zi1 = Z01 = 100 = 200W
ZL1
50
Reflection coefficient
2
2
Zi2 = Z02 = 100 = 200W
ZL2
50
G=
Substituting values for h1 and h2 we have
ZL3 = Zi1 Zi2 = 200W 200W = 100W
Zi =
9.30
Z02
ZL3
= 50 = 25W
100
er = 9
x a - y a
y
xm
c 2
x + y2
x2 + y2
To convert in cylindrical substituting
x = r cos f and y = r sin f
Now
9.31
9.32
+
mo
eo
mo
eo
= 11+
er = 1 er
1+
9
9
since
=- 0.5
...(1)
9.36
ax = cos far - sin faf
ay = sin far + cos faf
Bv = Bv0 af
v
Bv a
Hv = B = 0 f
m0
m0
Jv = 4# Hv = 0
mo
eo er
mo
eo er
t =
2
Option (C) is correct.
We have
Bv = B0
and
In (1) we have
h2 - h1
h2 + h1
Option (C) is correct.
In single mode optical fibre, the frequency of limiting mode increases
as radius decreases
Hence
r \ 1
f
So. if radius is doubled, the frequency of propagating mode gets
halved, while in option (D) it is increased by two times.
9.37
constant
since H is constant
Option (C) is correct.
The beam-width of Hertizian dipole is 180c and its half power beamwidth is 90c.
Option (D) is correct.
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Option (D) is correct.
Maxwell equations
4- B = 0
4$ E = r/E
4# E =- B
4# Ht = D + J
For static electric magnetic fields
8
l = c = 3 # 10 9 = 3
f
200
20 # 10
2
2
Gp = hp2 ` D j = 0.7 # p2 c 13 m = 30705.4
l
100
Gain
= 44.87 dB
4$ B = 0
9.38
4$ E = r/E
Option (A) is correct.
g = b cos 30cx ! b sin 30cy
4# E = 0
S
4# H = J
9.33
= 2p 3 x ! 2p 1 y
l 2
l 2
= p 3 x! py
l
l
Option (A) is correct.
Cut-off Frequency is
fc = c
2
For TE11 mode,
m 2
n 2
` a j +`b j
3 # 1010
fc =
2
9.34
9.39
1 2
1 2
` 4 j + ` 3 j = 6.25 GHz
For ZL = 0 ,
Zin = iZo tan (bl)
The wavelength is
8
l = c = 3 # 109 = 0.1 m or 10 cm
f
3 # 10
bl = 2p l = 2p # 1 = p
10
l
5
Thus
Zin = iZo tan p
5
Thus Zin is inductive because Zo tan p is positive
5
Option (C) is correct.
m
We have
h =
e
Maxwell Equations
## 4# H $ ds = ## `J + 22Dt j .ds
Integral form
## `J + 22Dt j .ds
Stokes Theorem
s
s
ZL + iZo tan (bl)
Zo + iZL tan (bl)
p 3 x! p y
l
l mE
Option (D) is correct.
4# H = J + 2D
2t
Option (D) is correct.
Zin = Zo
9.35
E = ay E0 e j (wt - g) = ay E0 e j;wt - c
# H $ dl
=
s
9.40
Option (A) is correct.
2
wm p
H sin ` 2px j sin (wt - bz) yt
2 `2j 0
a
h
This is TE mode and we know that
mpy
Ey \ sin ` mpx j cos `
a
b j
E =
Thus m = 2 and n = 0 and mode is TE20
9.41
Option (C) is correct.
The 2-port scattering parameter matrix is
S11 S12
S ==
S21 S22 G
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 232
(ZL Z0) - Zo
(50 50) - 50
=
=- 1
(ZL Z0) + Zo
(50 50) + 50
3
2 (ZL Zo)
2 (50 50)
S12 = S21 =
=
=2
(ZL Zo) + Zo
(50 50) + 50
3
(Z Z ) - Zo
(50 50) - 50
S22 = L o
=
=- 1
(ZL Zo) + Zo
(50 50) + 50
3
The Brewster angle is
S11 =
9.42
Option (D) is correct.
The input impedance is
Zin1
tan 60c =
if l = l
4
2
2
= Zo1 = 50 = 25
ZL1 100
9.48
ZL = Zin1 Zin2
25 12.5 = 25
3
Zs =
(50) 2
= 300
25/3
9.49
For free space
watts
9.44
Option (A) is correct.
mo
eo er
mo
eo er
+
mo
eo
mo
eo
Pt = (1 - G2) Pi = 1 - 1 = 8
9
9
Pt = 8
Pi
9
Option (D) is correct.
sin q = 1 = 1
er
2
p
q = 45c =
4
Hy2
or
The configuration is shown below. Here A is point source.
m 2
n 2
` a j +`b j
Since the mode is TE20, m = 2 and n = 0
8
fc = c m = 3 # 10 # 2 = 10 GHz
2 2
2 # 0.03
ho
377
=
h' =
= 400W
10
2
fc 2
1-c m
1 - c 10 10 m
f
3 # 10
Option (B) is correct.
Using the method of images, the configuration is as shown below
Now
From geometry
AO = 1 m
BO = 1 m
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= pr2 = p # OB = p m 2
Thus area
9.51
Option (C) is correct.
The cut-off frequency is
fc = c
2
Here d = l, a = p, thus bd = 2p
Array factor is
9.46
h2 - h1
=
h2 + h1
Option (C) is correct.
The cut-off frequency is
fc = c
2
9.45
Prad = 10 Watts
or
2
2
2
= c 5 3 m + c 5 m = c 10 m
H =
ho
ho
ho
2
2
ho H
E
2
h
=
= o c 10 m = 50
P =
2 ho
ho
2ho
2
We have
Option (A) is correct.
We have
10 log G = 10 dB
or
G = 10
Now gain
G = Prad
Pin
or
10 = Prad
1W
= 1 + er = 1 - 4 =- 1
3
1 + er
1+ 4
The transmitted power is
9.50
Option (D) is correct.
Hx2 +
, thus wave is left hand circularly polarized.
G=
G = ZS - Zo = 300 - 50 = 5
ZS + Zo
300 + 50
7
2
p
2
or
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9.43
Option (C) is correct.
We have
E = atxx sin (wt - bz) + aty sin (wt - bz + p/2)
Here Ex = Ey and fx = 0, fy = p2
Phase difference is
2
2
Zin2 = Zo2 = 50 = 12.5
ZL2
200
Now
er2 = 3
or
9.47
2
Zin = Zo ;
ZL
er2
er1
er2
1
tan qn =
Option (D) is correct.
bd cos y + a
E
2
2p cos y + p
= cos ;
E = sin (p cos y)
2
= cos ;
m 2
m 2
` a j +` b j
Since the mode is TE30 , m = 3 and n = 0
fc = c m
2 a
or
or
8
18 # 109 = 3 # 10 3
2 a
a = 1 m = 5 cm
40
2
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
9.52
Option (C) is correct.
We have
E1 = 4ux + 3uy + 5uz
Since for dielectric material at the boundary, tangential component
of electric field are equal
E21 = E1t = 3aty + 5atz
at the boundary, normal component of displacement vector are
equal
i.e.
Dn2 = Dn1
or
e2 E2n = e1 E1n
or
4eo E2n = 3eo 4atz
or
E2n = 3atx
Thus
E2 = E2t + E2a = 3atx + 3aty + 5atz
9.53
Option (C) is correct.
w = 50, 000 and b =- 0.004
4
vP = w = 5 # 10 - 3 = 1.25 # 107 m/s
b
- 4 # 10
Phase Velocity is
9.55
9.61
9.62
9.63
Option (C) is correct.
Refractive index of glass m = 1.5
Frequency
f = 1014 Hz
c = 3 # 108 m/sec
9.57
Option (D) is correct.
ZZC
9.58
bd sin q + a
l
2
p
2
The option (A) satisfy this equation.
9.59
4pU (q, f) = Prad GD (q, f) = 1 m # 3.98 = 3.98 mW
9.64
Option (D) is correct.
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The capacitance is
- 12
-4
C = eo A = 8.85 # 10 - 3 # 10 = 8.85 # 10 - 13
d
10
The charge on capacitor is
Q = CV = 8.85 # 10 - 13 = 4.427 # 10 - 13
Displacement current in one cycle
Q
I = = fQ = 4.427 # 10 - 13 # 3.6 # 109 = 1.59 mA
T
9.65
Option (C) is correct.
VL = ZO
Vin
Zin
or
VL = ZO Vin = 10 # 300 = 60 V
Zin
50
9.66
Option (D) is correct.
9.67
Option (A) is correct.
Ravg = 1 Re [E # H*]
2
p
p
m = cos ` sin q + j
4
2
Option (C) is correct.
From the diagram, VSWR is
s = Vmax = 4 = 4
Vmin
1
When minima is at load ZO = s.ZL
or
ZL = Zo = 50 = 12.5W
s
4
9.60
4pU (q, f)
Prad
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Option (A) is correct.
The array factor is
Here b = 2p , d = l and a = 90c
l
4
2p l
sin q +
Thus
A = cos c l 4
2
GD (q, f) =
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Zo2 = ZOC .ZSC
2
= Zo = 50 # 50 = 50
ZOC
100 + j150
2 + 3j
50 (2 - 3j)
=
= 7.69 - 11.54j
13
A = cos b
Option (A) is correct.
Prad = Pin
Here we have
Prad = Pin = 1 mW
and
10 log GD (q, f) = 6 dB
or
GD (q, f) = 3.98
Thus the total power radiated by antenna is
-6
lg = a = 3 # 10 = 2 # 10 - 6 m
1.5
m
Option (D) is correct.
Option (D) is correct.
We know that vp > c > vg .
For lossless antenna
8
l = c = 3 # 10
= 3 # 10 - 6
14
f
10
wavelength in glass is
9.56
Option (C) is correct.
The given figure represent constant reactance circle.
We have
Option (C) is correct.
Since antenna is installed at conducting ground,
2
2
2
50
Rrad = 80p2 ` dl j = 80p2 c
= 4p W
m
3
5
l
0.5 # 10
9.54
Page 233
Option (A) is correct.
The reflection coefficient is
G = ZL - ZO = 12.5 - 50 =- 0.6
ZL + ZO
125. + 50
Thus
9.68
E # H* = (atx + jaty) e jkz - jwt # k (- jatx + aty) e-jkz + jwt
wm
= atz ; k - (- j) (j) k E = 0
wm
wm
Ravg = 1 Re [E # H*] = 0
2
Option (A) is correct.
Suppose at point P impedance is
Z = r + j (- 1)
If we move in constant resistance circle from point P in clockwise
direction by an angle 45c, the reactance magnitude increase. Let us
consider a point Q at 45c from point P in clockwise direction. It’s
impedance is
Z1 = r - 0.5j
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 234
or
Z1 = Z + 0.5j
Thus movement on constant r - circle by an +45c in CW direction
is the addition of inductance in series with Z .
9.69
or
Option (D) is correct.
We have
VSWR
or
G
Thus
G
Now
G
-2
3
or
1- G
= Emax = 5 =
Emin
1+ G
=2
3
=- 2
3
h - h1
= 2
h2 + h1
h - 120p
= 2
h2 + 120p
Thus
9.74
or
Thus
or
=
2=
1- G
1+ G
1
+
0
0
#2 ydyutz + #3 zdzuzt
0
2 0
y2
+z G
2 2 2 3
Option (D) is correct.
m
e
h =
Reflection coefficient
t =
h2 - h1
h2 + h1
Substituting values for h1 and h2 we have
t =
er = 4
G =1
3
mo
eo er
mo
eo er
+
m0
eo
mo
eo
= 11+
er = 1 er
1+
4
4
since
= - 1 = 0.333+180c
3
Pref
= G2= 1
Pinc
9
Pref = Pinc
9
9.76
9.77
Thus 4# H has unit of current density J that is A/m2
Option (B) is correct.
or
xdxutx +
=- 1 [22 - 12 + 02 - 22 + 02 - 32] = 5
2
9.75
4# H = 2D + J
2t
Thus
2
2 2
Option (C) is correct.
By Maxwells equations
We know that
#1
Y
#XE.dl
=-= x
2
i.e. 11.11% of incident power is reflected.
9.73
Option (C) is correct.
We have
E = xux + yuy + zuz
dl = utx dx + uty dy + utz dz
VXY =-
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9.72
E2 = E2t + E2n =- 3uy + uz + 1.2ux
Option (D) is correct.
The VSWR
9.71
e1 E1n = e2 E2n
1.5eo 2ux = 2.5eo E2n
E2n = 3 ux = 1.2ux
2.5
h2 = 24p
or
9.70
or
or
d \ 1
f
d2 =
d1
d2 =
25
d2 =
f1
f2
1
4
1 # 25 = 12.5 cm
4
Option (C) is correct.
We have
E1 = 2ux - 3uy + 1uz
E1t = - 3uy + uy and E1n = 2ux
Since for dielectric material at the boundary, tangential component
of electric field are equal
(x = 0 plane)
E1t =- 3uy + uy = E2t
E1n = 2ux
At the boundary the for normal component of electric field are
D1n = D2n
Option (B) is correct.
We have
E (z, t) = 10 cos (2p # 107 t - 0.1pz)
where
w = 2p # 107 t
b = 0.1p
7
Phase Velocity
u = w = 2p # 10 = 2 # 108 m/s
b
0.1p
Option (A) is correct.
The fig of transmission line is as shown below.
[Z + jZo tan bl]
We know that
Zin = Zo L
[Zo + jZL tan bl]
For line 1, l = l and b = 2p , ZL1 = 100W
l
2
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[ZL + jZo tan p]
= ZL = 100W
[Zo + jZL tan p]
For line 2, l = l and b = 2p , ZL2 = 0 (short circuit)
l
8
[0 + jZo tan p4 ]
Thus
= jZo = j50W
Zin2 = Zo
[Zo + 0]
Y = 1 + 1 = 1 + 1 = 0.01 - j0.02
Zin1 Zin2
100 j50
Thus
Zin1 = Zo
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 235
9.84
Option (A) is correct.
1.7 # 10 - 4
tan a = s =
we
2p # 3 # 109 # 78eo
Loss tangent
-4
9
= 1.7 # 10 9# 9 # 10 = 1.3 # 10 - 5
3 # 10 # 39
9.85
Option (D) is correct.
The flux density is
s = eE = e0 er E = 80 # 8.854 # 10 - 12 # 2
s = 1.41 # 10 - 9 C/m 2
or
9.86
9.78
9.79
P \ 12
r
P1 = r22
Thus
P2
r12
3 dB decrease $ Strength is halved
P1 = 2
Thus
P2
Option (A) is correct.
8
u = c = 3 # 10 = 1.5 # 108
2
e0
In rectangular waveguide the dominant mode is TE10 and
fC = v ` m j2 + ` n j2
2
a
b
8
1 2 + 0 2 = 1.5 # 108 = 2.5 GHz
= 1.5 # 10
` 0.03 j ` b j
2
0.06
Option (D) is correct.
Normalized array factor = 2 cos
y
2
Option (B) is correct.
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y = bd sin q cos f + d
q = 90c,
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d =
Now
2 s,
f = 45c,
d = 180c
bd sin q cos f + d
y
= 2 cos ;
2 cos
E
2
2
= 2 cos 8 2p 2 s cos 45c + 180 B
l. 2
2
= 2 cos 8 ps + 90cB = 2 sin ` ps j
l
l
9.80
Option (D) is correct.
VSWR
Substituting values we have
or
Distance to move
9.87
9.88
s = 1+G
1-G
where G varies from 0 to 1
9.82
Option (B) is correct.
Reactance increases if we move along clockwise direction in the
constant resistance circle.
9.89
Option (A) is correct.
In wave guide vp > c > vg and in vacuum vp = c = vg
where
vp $ Phase velocity
c $ Velocity of light
vg $ Group velocity
f 2
1-c c m
f
9.90
Option (C) is correct.
We have
Option (B) is correct.
VC
When wave propagate in waveguide fc < f $ VP > VC
9.83
Option (C) is correct.
A transmission line is distortion less if LG = RC
d2 Ex = c2 d2 Ex
dz2
dt2
This equation shows that x component of electric fields Ex is
traveling in z direction because there is change in z direction.
Option (C) is correct.
Phase velocity
VP =
p
2
t j ) e j (wt - kz)
E = (0.5xt + ye
9.91
Ex = 0.5e j (wt - kz)
jp
2
Ey = e e
Option (A) is correct.
In a wave guide dominant gives lowest cut-off frequency and hence
the highest cut-off wavelength.
Option (A) is correct.
or
or
j (wt - kz)
Ex
p
= 0.5e- 2
Ey
Ex
Since
! 1, it is elliptically polarized.
Ey
= 7071 - 5000 = 2071 m
We have
Thus s varies from 1 to 3.
9.81
2
2 = r22
5
r2 = 5 2 kM = 7071 m
9.92
Ic = Id
sE = jw d E
s = 2pfeo er
-2
9
s
= 2s = 9 # 10 # 2 # 10
2p # eo er
4peo er
4
or
f =
or
f = 45 # 106 = 45 MHz
Option (B) is correct.
w = 2pf and e = er e0
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
or
or
Now
VSWR = 1 + G
1-G
3 = 1+G
1-G
Page 236
9.99
G = 0.5
Pr = G2 = 0.25
Pi
Option (A) is correct.
We have
Distance between elements
or
Because of end fire
y =0
q = 60c
0 = 2p # l cos 60c + d = p # 1 + d
2 2
l
4
d =- p
4
9.96
9.97
ZOC .ZSC = 100 # 25 = 10 # 5 = 50W
BW \
1
(Diameter)
Option (C) is correct.
The fig is as shown below :
As per snell law
sin qt =
1
sin qi
er
sin 30c = 1
or
sin 45c
er
1
2
= 1
1
er
2
l1 =
l2
e2
e1
Option (D) is correct.
2
l
` 2 jd = l
8
l = c = 3 # 10 9 = 3 m
f
40
4 # 10
2
3
` 40 # 2 j d = (2.4)
or
9.102
d =
80 # (2.4) 2
. 150 m
3
Option (C) is correct.
We know that for a monopole its electric field varies inversely with
r 2 while its potential varies inversely with r . Similarly for a dipole
its electric field varies inversely as r 3 and potential varies inversely
as r 2 .
In the given expression both the terms a _ r1 + r1 i are present, so
this potential is due to both monopole & dipole.
-1
9.103
9.104
As diameter increases Bandwidth decreases.
9.98
9.101
Option (C) is correct.
As the impedance of perfect conductor is zero, electric field is
minimum and magnetic field is maximum at the boundary.
Option (B) is correct.
l \ 1
e
we get
Option (B) is correct.
Zo =
For air
b = 2p = w me
l
l = 2p
w me
Phase Velocity
Thus
Thus
m 2
n 2
` a j +`b j
Option (B) is correct.
Option (C) is correct.
The array factor is
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9.95
9.100
or
y = bd cos q + d
d =l
4
vp
2
= 15 GHz
l = 492 m
25 W.
where
fc =
For rectangular waveguide dominant mode is TE01
8
v
Thus
fc = p = 3 # 10- 2 = 15 # 109
2a
2 # 10
vp = 3 # 108
and height of antenna
= 124 m . l
4
It is a quarter wave monopole antenna and radiation resistance is
9.94
Option (C) is correct.
Cutoff frequency
Thus 25% of incident power is reflected.
9.93
er = 2
or
9.105
-2
Option (D) is correct.
In TE mode Ez = 0 , at all points within the wave guide. It implies
that electric field vector is always perpendicular to the waveguide
axis. This is not possible in semi-infine parallel plate wave guide.
Option (A) is correct.
Option (C) is correct.
A scalar wave equation must satisfy following relation
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2 2 E - m 22 2 E = 0
...(1)
2t 2
2z 2
Where
m = w (Velocity)
b
Basically w is the multiply factor of t and b is multiply factor of
z or x or y .
In option (A)
E = 50e j (wt - 3z)
m =w=w
3
b
We can see that equations in option (C) does not satisfy equation
(1)
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
9.106
Option (B) is correct.
We know that distance between two adjacent voltage maxima is
equal to l/2 , where l is wavelength.
l = 27.5 - 12.5
2
Frequency
9.107
9.108
9.110
Option (D) is correct.
Power received by antenna
-4
PR = PT 2 # (apeture) = 251 # 500 # 102 = 100 mW
4pr
4 # p # (100)
Option (C) is correct.
Electrical path length = bl
Where
b = 2p , l = 50 cm
l
We know that
l =u =1# 1
au=
f
f
LC
1
1
=
6 #
-6
25 # 10
10 # 10 # 40 # 10-12
9.115
1
LC
Option (D) is correct.
In a lossless dielectric (s = 0) median, impedance is given by
mr
m0 mr
m
= 120p #
h =
0c =
er
e0 er
e
= 120p # 2 = 188.4 W
8
d\ 1 \ l
f
so depth increases with increasing in wavelength.
9.116
or
#l H $ dl
=
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along a line that makes an angle f with x -axis as shown below.
9.117
(ampere's law)
9.118
9.119
Option (A) is correct.
9.113
Option (B) is correct.
9.114
Power Re ceived
Polynting vector of incident wave
A =W
P
=
r = a + ib = 0.1p + j0.2p
b = 2p = 0.2p
l
2
P = E h0 = 120p is intrinsic impedance
h0
of space
Based on continuity equation
So
A
-6
= 2 # 10
E2
c h0 m
Option (B) is correct.
Propagation constant
here
Option (C) is correct.
We have
#s Jds
4# H = J
then it is modified to
4# H = J + 2D
2t
9.112
Option (D) is correct.
Radiation resistance of a circular loop is given as
Rr = 8 hp3 :ND2 S D
3
l
2
Rx \ N N " no. of turns
So,
Rr 2 = N 2 # Rr 1
= (5) 2 # 0.01 = 0.25 W
Aperture Area
Applying curl theorem
=
Option (C) is correct.
v
v
Hv = 1 4
A
m #
v is auxiliary potential function.
where A
So
4: H = 4: (4 # A) = 0
4# H = 4# (4 # A) =
Y 0
#s Jds
#s (4 # H) $ ds
...(1)
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Option (A) is correct.
This equation is based on ampere’s law as we can see
= I enclosed
Option (A) is correct.
Given
j (wt + bz) v
a x + e0 e j (wt + bz) avy
E (z, t) = Eo e
...(2)
E (z) = avx E1 (z) + avy E2 (z)
Comparing (1) and (2) we can see that E1 (z) and E2 (z) are in space
quadrature but in time phase, their sum E will be linearly polarized
Option (D) is correct.
Impedance is written as
#l H $ dl
Option (C) is correct.
The depth of penetration or skin depth is defined as –
1
d=
pfms
Generalizing
jwm
h =
s + jwe
Copper is good conductor i.e. s >> we
jwm
wm
So
h =
=
45c
s
s
Impedance will be complex with an inductive component.
9.111
l = 2 = 10 m
0.2
l = 2 # 15 = 30 cm
10
u = C = 3 # 10 = 1 GHz
30
l
7
= 5 # 10 6 = 2 m
25 # 10
Electric path length
= 2p # 50 # 10-2 = p radian
5
2
9.109
Page 237
=
-6
2 # 10
# 120 # 3.14
(20 # 10-3) 2
-6
3.14 = 1.884 m2
= 2 # 10 # 12 -#
400 # 10 6
9.120
Option (B) is correct.
Maximum usable frequency
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
fo
sin Ae
fm = 8MHz =
sin 60c
Page 238
fm =
9.121
9.122
8
= 16 MHz
3
3
c 2 m
or
Option (D) is correct.
When a moving circuit is put in a time varying magnetic field educed
emf have two components. One for time variation of B and other
turn motion of circuit in B .
f 2 = 1000 MHz is
4pf 2
V =
ms
Option (A) is correct.
V
Option (B) is correct.
Z in min = Z 0
S
where S = standing wave ratio
1 + GL
S =
1 - GL
=
9.128
ZL = 0
(Short circuited)
Z 0 = 50 W
0 + j50 tan p/2
=3
Zin = 50 =
50 + j0 tan p/2G
and
9.129
Option (A) is correct.
The cutoff frequency is given by
ml
m 2+ n 2
fc =
2 a a k a2k
Here a < b , so minimum cut off frequency will be for mode TE 01
a ml = c
2
*
c = 3 # 108
Option (B) is correct.
For lossless transmission line, we have
Velocity
V =w= 1
b
LC
Characteristics impedance for a lossless transmission line
Z0 = L
C
From eqn. (1) and (2)
1
= 1
V =
C (Z 0 C ) Z 0 C
9.130
Option (C) is correct.
9.131
Option (A) is correct.
9.132
Option (C) is correct.
Option (B) is correct.
Option (A) is correct.
For any transmission line we can write input impedance
Z + jZ 0 tanh lg
Zin = Z 0 ; L
Z 0 + jZL tanh lg E
R
jZ tanh lg V
S1 + 0
W
Z0
ZL
W=
so
Zin = Z 0 lim S
Z " 3S Z 0
j
tanh
lg
W
S ZL + j tanh lg W
T
X
Option (A) is correct.
We know that skin depth is given by
1
s =
= 1 # 10-2 m
pf1 ms
...(1)
...(2)
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Reflected power
Here given ZL = 3 (open circuited at load end)
So
L
9.127
Z 0 = Charateristic impedance of line
ZL = Load impedance
b = 2p
l = length
l
bl = 2p l = p
2
l 4
Thus infinite impedance, and current will be zero.
m = 0, n = 1
9.126
Option (A) is correct.
Input impedance of a lossless transmission line is given by
Z + jZ 0 tan bl
Zin = Z 0 ; L
Z 0 + jZL tan bl E
so
1+1
3 =2
S =
1
13
50
Z in min =
= 25 W
2
9.125
4 # p # 1000 # 10 # p - 6 106 m/ sec
#
10-3
so here
GL = ZL - Z 0 = 100 - 50 = 50 = 1
ZL + Z 0 100 + 50 150 3
1
(10 # 10-12)
3 # 108
= 0.75 GHz
=
2 # 2 # 10 # 10-2
6
where
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8
fc = 3 # 10
2#2
-3
ms = 10 in above equation
p
Put
GL = reflection coefficient
9.124
f 1 = 10 MHz
Now phase velocity at another frequency
Far field \ 1
r
9.123
1
= 10-2
6
p # 10 # 10 # m s
-3
ms = 10
p
or
9.133
Er = GEi
G = Reflection coefficient
h - h1
G = 2
= 1.5 - 1 = 1
h 2 + h1
1.5 + 1
5
Er = 1 # Ei
5
Er = 20%
Ei
Ei " Incident power
Option (B) is correct.
We have maximum usable frequency formulae as
f0
fm =
sin Ae
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
20 # 106 = 10 # 10
sin Ae
sin Ae = 1
2
Page 239
6
Ae = 30c
9.134
Option (C) is correct.
9.135
Option (A) is correct.
Skin depth
d=
1
pfms
Putting the given value
d=
= 15.9 mm
1
3.14 # 1 # 109 # 4p # 10-7 # 106
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
UNIT 10
(C)
10.8
GENERAL APTITUDE
YEAR 2013
10.1
10.2
Page 240
ONE MARK
Choose the grammatically CORRECT sentence:
(A) Two and two add four
(B)
Two and two become four
(C) Two and two are four
(D) Two and two make four
Statement: You can always give me a ring whenever you need.
Which one of the following is the best inference from the above
statement?
(A) Because I have a nice caller tune.
(B) Because I have a better telephone facility
(C) Because a friend in need is a friend indeed
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10.9
10.10
10.4
10.5
10.1
In the summer of 2012, in New Delhi, the mean temperature of
Monday to Wednesday was 41°C and of Tuesday to Thursday was
43cC . If the temperature on Thursday was 15% higher than that of
Monday, then the temperature in cC on Thursday was
(A) 40
(B) 43
(C) 46
(D) 49
10.11
They were requested not to quarrel with others.
Which one of the following options is the closest in meaning to the
word quarrel?
(A) make out
(B) call out
(C) dig out
(D) fall out
10.7
10.14
Find the sum to n terms of the series 10 + 84 + 734 + ...
9 ^9n - 1h
9 ^9n + 1h
(B)
(A)
+1
+1
8
10
Statement: There were different streams of freedom movements
in colonial India carried out by the moderates, liberals, radicals,
socialists, and so on.
Which one of the following is the best inference from the above
statement?
(A) The emergence of nationalism in colonial India led to our Independence
(B) Nationalism in India emerged in the context of colonialism
(C) Nationalism in India is homogeneous
(D) Nationalism in India is heterogeneous
The set of values of p for which the roots of the equation
3x2 + 2x + p ^p - 1h = 0 are of opposite sign is
(A) ^- 3, 0h
(B) ^0, 1h
(C) ^1, 3h
(D) ^0, 3h
What is the chance that a leap year, selected at random, will contain
53 Sundays?
(A) 2/7
(B) 3/7
(C) 1/7
(D) 5/7
ONE MARK
If (1.001) 1259 = 3.52 and (1.001) 2062 = 7.85, then (1.001) 3321
(B) 4.33
(A) 2.23
(C) 11.37
(D) 27.64
Choose the most appropriate alternate from the options given below
to complete the following sentence :
If the tired soldier wanted to lie down, he..................the mattress
out on the balcony.
(A) should take
(B) shall take
(C) should have taken
(D) will have taken
Choose the most appropriate word from the options given below to
complete the following sentence :
Give the seriousness of the situation that he had to face, his........
was impressive.
(A) beggary
(B) nomenclature
(C) jealousy
(D) nonchalance
Which one of the following options is the closest in meaning to the
word given below ?
Latitude
(A) Eligibility
(C) Coercion
TWO MARKS
A car travels 8 km in the first quarter of an hour, 6 km in the
second quarter and 16 km in the third quarter. The average speed
of the car in km per hour over the entire journey is
(A) 30
(B) 36
(C) 40
(D) 24
9 ^9n - 1h
+ n2
8
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Option (D) is correct.
They were requested not to quarrel with others.
Quarrel has a similar meaning to ‘fall out’
YEAR 2013
10.6
10.12
10.13
Complete the sentence: Dare .................. mistakes.
(A) commit
(B) to commit
(C) committed
(D) committing
(D)
2012
(D) Because you need not pay towards the telephone bills when
you give me a ring
10.3
9 ^9n - 1h
+n
8
10.15
(B) Freedom
(D) Meticulousness
One of the parts (A, B, C, D) in the sentence given below contains
an ERROR. Which one of the following is INCORRECT ?
I requested that he should be given the driving test today instead
of tomorrow.
(A) requested that
(B) should be given
(C) the driving test
(D) instead of tomorrow
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
2012
10.16
10.17
10.18
10.19
TWO MARKS
One of the legacies of the Roman legions was discipline. In the
legious, military law prevailed and discipline was brutal. Discipline
on the battlefield kept units obedient, intact and fighting, even when
the odds and conditions were against them.
Which one of the following statements best sums up the meaning
of the above passage ?
(A) Through regimentation was the main reason for the efficiency
of the Roman legions even in adverse circumstances.
(B) The legions were treated inhumanly as if the men were animals
(C) Disciplines was the armies inheritance from their seniors
(D) The harsh discipline to which the legions were subjected to led
to the odds and conditions being against them.
Raju has 14 currency notes in his pocket consisting of only Rs. 20
notes and Rs. 10 notes. The total money values of the notes is Rs.
230. The number of Rs. 10 notes that Raju has is
(A) 5
(B) 6
(C) 9
(D) 10
There are eight bags of rice looking alike, seven of which have
equal weight and one is slightly heavier. The weighing balance is
of unlimited capacity. Using this balance, the minimum number of
weighings required to identify the heavier bag is
(A) 2
(B) 3
(C) 4
(D) 8
Category
Amount (Rs.)
Food
4000
Clothing
1200
Rent
2000
Savings
1500
Other Expenses
1800
10.23
10.24
The question below consists of a pair of related words followed by
four pairs of words. Select the pair that best expresses the relations
in the original pair :
Gladiator : Arena
(A) dancer : stage
(B) commuter : train
(C) teacher : classroom
(D) lawyer : courtroom
Choose the most appropriate word from the options given below to
complete the following sentence :
Under ethical guidelines recently adopted by the Indian Medical
Association, human genes are to be manipulated only to correct
diseases for which...................treatments are unsatisfactory.
(A) similar
(B) most
(C) uncommon
(D) available
Choose the word from the from the options given below that is most
opposite in meaning to the given word :
Frequency
(A) periodicity
(B) rarity
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(C) gradualness
10.25
(D) persistency
Choose the most appropriate word from the options given below to
complete the following sentence :
It was her view that the country’s had been ............. by foreign
techno-crafts, so that to invite them to come back would be counter-productive.
(A) identified
(B) ascertained
(C) exacerbated
(D) analysed
2011
10.26
TWO MARKS
The fuel consumed by a motor cycle during a journey while travelling
at various speed is indicated in the graph below.
A and B are friends. They decide to meet between 1 PM and 2 PM
on a given day. There is a conditions that whoever arrives first will
not wait for the other for more than 15 minutes. The probability
that they will meet on that days is
(A) 1/4
(B) 1/16
(C) 7/16
(D) 9/16
2011
10.21
10.22
The data given in the following table summarizes the monthly
budget of an average household.
The approximate percentages of the monthly budget NOT spent
on savings is
(A) 10%
(B) 14%
(C) 81%
(D) 86%
10.20
Page 241
ONE MARK
There are two candidates P and Q in an election. During the
campaign, 40% of voter promised to vote for P , and rest for Q
. However, on the day of election 15% of the voters went back on
their promise to vote for P and instead voted for Q . 25% of the
voter went back on their promise to vote for Q and instead voted
for P . Suppose, P lost by 2 votes, then what was the total number
of voters ?
(A) 100
(B) 110
(C) 90
(D) 95
The distance covered during four laps of the journey are listed in
the table below
Lap
Distance (km)
Average speed (km/hour)
P
15
15
Q
75
45
R
40
75
S
10
10
From the given data, we can conclude that the fuel consumed per
kilometre was least during the lap
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) P
(C) R
10.27
10.28
10.29
(B) Q
(D) S
Page 242
10.34
The horse has played a little known but very important role in the
field of medicine. Horses were injected with toxins of disease until
their blood build up immunities. Then a serum was made from their
blood. Serums to fight with diphteria and tetanus were developed
this way.
It can be inferred from the passage, that horses were
(A) given immunity to diseases
(B) generally quite immune to diseases
(C) given medicines to fight toxins
(D) given diphtheria and tetanus serums
The sum of n terms of the series 4 + 44 + 444 + ........
(A) (4/81) [10n + 1 - 9n - 1]
(B) (4/81) [10n - 1 - 9n - 1]
(C) (4/81) [10n + 1 - 9n - 10]
(D) (4/81) [10n - 9n - 10]
10.35
10.36
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10.30
10.37
(B) - 1
(D) 2
Three friends R, S and T shared toffee from a bowl. R took 1/3 rd
of the toffees, but returned four to the bowl. S took 1/4 th of what
was left but returned three toffees to the bowl. T took half of the
remainder but returned two back into the bowl. If the bowl had 17
toffees left, how many toffees were originally there in the bowl ?
(A) 38
(B) 31
(C) 48
(D) 41
10.38
10.39
2010
10.31
10.32
10.33
ONE MARK
Which of the following options is the closest in meaning to the word
below ?
Circuitous
(A) Cyclic
(B) Indirect
(C) Confusing
(D) Crooked
The question below consist of a pair of related words followed by
four pairs of words. Select the pair that best expresses the relation
in the original pair.
Unemployed : Worker
(A) Fallow : Land
(B) Unaware : Sleeper
(C) Wit : Jester
(D) Renovated : House
Choose the most appropriate word from the options given below to
complete the following sentence :
If we manage to ........ our natural resources, we would leave a better planet for our children.
(A) unhold
(B) restrain
(C) cherish
(D) conserve
25 persons are in a room 15 of them play hockey, 17 of them play
football and 10 of them play hockey and football. Then the number
of persons playing neither hockey nor football is
(A) 2
(B) 17
(C) 13
(D) 3
2010
Given that f (y) = y /y, and q is any non-zero real number, the
value of f (q) - f (- q) is
(A) 0
(C) 1
Choose the most appropriate word from the options given below to
complete the following sentence :
His rather casual remarks on politics..................his lack of seriousness about the subject.
(A) masked
(B) belied
(C) betrayed
(D) suppressed
10.40
TWO MARKS
Modern warfare has changed from large scale clashes of armies to
suppression of civilian populations. Chemical agents that do their
work silently appear to be suited to such warfare ; and regretfully,
their exist people in military establishments who think that chemical
agents are useful fools for their cause.
Which of the following statements best sums up the meaning of
the above passage ?
(A) Modern warfare has resulted in civil strife.
(B) Chemical agents are useful in modern warfare.
(C) Use of chemical agents in ware fare would be undesirable.
(D) People in military establishments like to use chemical agents in
war.
If 137 + 276 = 435 how much is 731 + 672 ?
(A) 534
(B) 1403
(C) 1623
(D) 1531
5 skilled workers can build a wall in 20 days; 8 semi-skilled workers
can build a wall in 25 days; 10 unskilled workers can build a wall
in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled
workers, how long will it take to build the wall ?
(A) 20 days
(B) 18 days
(C) 16 days
(D) 15 days
Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how much distinct 4 digit numbers
greater than 3000 can be formed ?
(A) 50
(B) 51
(C) 52
(D) 54
Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers
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and sisters.) All were born on 1st January. The age difference between
any two successive siblings (that is born one after another) is less
than 3 years. Given the following facts :
1. Hari’s age + Gita’s age > Irfan’s age + Saira’s age.
2. The age difference between Gita and Saira is 1 year. However,
Gita is not the oldest and Saira is not the youngest.
3. There are no twins.
In what order were they born (oldest first) ?
(A) HSIG
(B) SGHI
(C) IGSH
(D) IHSG
***********
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 243
SOLUTIONS
10.1
10.2
10.3
S2 =
10.8
Option (D) is correct
Two and two make four
Option (C) is correct.
You can always given me a ring whenever you need. Because a friend
is need is a friend indeed
Option (C) is correct.
Let the temperature on Monday, Tuesday, Wednesday and Thursday
be respectively as TM , TTU , TW , TTH
So, from the given data we have
TH + TTU + TW = 41
....(1)
3
TTU + TW + TTH = 43
....(2)
and
3
also, as the temperature on Thursday was 15% higher than that of
Monday
....(3)
i.e.
TTH = 1.15 TM
solving eq (1), (2) and (3), we obtain
10.9
S3
9 ^93 - 1h
=
+ 32 = 828
8
Option (D) is correct.
Nationalism in India is heterogeneous
Option (B) is correct.
Given, the quadratic equation
3x2 + 2x + P ^P - 1h = 0
It will have the roots with opposite sign if
P ^P - 1h < 0
So it can be possible only when
P < 0 and P - 1 > 0
or
P > 0 and P - 1 < 0
The 1 st condition tends to no solution for P .
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TTH = 46cC
10.4
10.5
10.6
Option (B) is correct.
Dare to commit mistakes
Hence, from the second condition, we obtain
Option (D) is correct.
They were requested not to quarrel with others.
Quarrel has a similar meaning to ‘fall out’
Option (C) is correct.
Given, the distance travelled by the car in each quarter intervals
as
Distance
0 <P<1
i.e., P is in the range ^0, 1h
10.10
Time Duration
8 km
1
4
hr
6 km
1
4
hr
16 km
1
4
hr
Therefore, the total time taken = 1 + 1 + 1 + 3 hr
4 4 4 4
Hence,
10.7
Total distance travelled = 8 + 6 + 16 = 30 km
average speed = Total distance travelled
Total time taken
= 30 = 40 km/hr
3/4
10.11
Option (D) is correct.
It will be easy to check the options for given series. From the given
series.
Option (A) is correct.
In a leap year, there are 366 days So, 52 weeks will have 52 saturdays
and for remaining two days ^366 - 52 # 7 = 2h. We can have the
following combinations
Saturday, Sunday
Sunday, Monday
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Out of these seven possibilities, only two consist a saturday.
Therefore, the probability of saturday is given as
P =2
7
Option (D) is correct.
Let
1.001 = x
So in given data :
10 + 84 + 734 + ......
We get
Sum of 1 term = S1 = 10
Sum of 2 terms = S2 = 10 + 84 = 94
and sum of 3 terms = S 3 = 10 + 84 + 734 = 828
Checking all the options one by one, we observe that only (D) option satisfies as
9 ^9n - 1h
+ n2
Sn =
8
9 ^92 - 1h
so,
S1
+ 22 = 10
8
9 ^9 - 1h
+ 22 = 94
8
Again
10.12
Option (C) is correct.
10.13
Option (D) is correct.
10.14
Option (B) is correct.
x1259 = 3.52
x2062 = 7.85
x3321 = x1259 + 2062
= x1259 x2062
= 3.52 # 7.85
= 27.64
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
10.15
Option (B) is correct.
10.16
Option (A) is correct.
10.17
Option (A) is correct.
Let no. of notes of Rs.20 be x and no. of notes of Rs. 10 be y .
Then from the given data.
Page 244
= 9000 # 100% = 86%
10500
10.20
Option (S) is correct.
The graphical representation of their arriving time so that they
met is given as below in the figure by shaded region.
x + y = 14
20x + 10y = 230
Solving the above two equations we get
x = 9, y = 5
So, the no. of notes of Rs. 10 is 5.
10.18
Option (A) is correct.
We will categorize the 8 bags in three groups as :
(i) A1 A2 A 3 , (ii) B1 B2 B 3 , (iii) C1 C2
Weighting will be done as bellow :
1st weighting " A1 A2 A 3 will be on one side of balance and B1 B2 B 3
on the other. It may have three results as described in the following
cases.
Case 1 :
A1 A 2 A 3 = B1 B 2 B 3
So, the area of shaded region is given by
Area of 4PQRS - (Area of TEFQ +
Area of TGSH )
= 60 # 60 - 2 b 1 # 45 # 45 l
2
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This results out that either C1 or C2 will heavier for which we will
have to perform weighting again.
2 nd weighting " C1 is kept on the one side and C2 on the other.
if
then C1 is heavier.
C1 > C 2
then C2 is heavier.
C1 < C 2
Case 2 :
A1 A 2 A 3 > B1 B 2 B 3
it means one of the A1 A2 A 3 will be heavier So we will perform next
weighting as:
2 nd weighting " A1 is kept on one side of the balance and A2 on the
other.
it means A 3 will be heavier
if
A1 = A 2
then A1 will be heavier
A1 > A 2
then A2 will be heavier
A1 < A 2
Case 3 :
A1 A 2 A 3 < B 1 B 2 B 3
This time one of the B1 B2 B 3 will be heavier, So again as the above
case weighting will be done.
2 nd weighting " B1 is kept one side and B2 on the other
if
B 3 will be heavier
B1 = B 2
B1 > B 2
B1 will be heavier
B1 < B 2
B2 will be heavier
So, as described above, in all the three cases weighting is done
only two times to give out the result so minimum no. of weighting
required = 2.
10.19
Option (D) is correct.
Total budget = 4000 + 1200 + 2000 + 1500 + 1800
= 1575
10.21
Now Voter for P
40 - 6 = 34
Also, 25% changed form Q to P (out of 60%)
25
Changed voter from Q to P
60 = 15
100 #
Now Voter for P
34 + 15 = 49
Thus P P got 49 votes and Q got 51 votes, and P lost by 2 votes,
which is given. Therefore 100 voter is true value.
10.22
Option (A) is correct.
A gladiator performs in an arena. Commutators use trains. Lawyers
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performs, but do not entertain like a gladiator. Similarly, teachers
educate. Only dancers performs on a stage.
10.23
10.24
= 10, 500
The amount spent on saving = 1500
So, the amount not spent on saving
= 10, 500 - 1500 = 9000
So, percentage of the amount
So, the required probability = 1575 = 7
3600 16
Option (A) is correct.
Let us assume total voters are 100. Thus 40 voter (i.e. 40 %) promised
to vote for P and 60 (rest 60 % ) promised to vote fore Q.
Now, 15% changed from P to Q (15 % out of 40)
15
Changed voter from P to Q
40 = 6
100 #
10.25
Option (D) is correct.
Available is appropriate because manipulation of genes will be done
when other treatments are not useful.
Option (B) is correct.
Periodicity is almost similar to frequency. Gradualness means
something happening with time. Persistency is endurance. Rarity is
opposite to frequency.
Option (C) is correct.
The sentence implies that technocrats are counterproductive
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(negative). Only (C) can bring the same meaning.
10.26
10.27
10.28
Option (B) is correct.
Since fuel consumption/litre is asked and not total fuel consumed,
only average speed is relevant. Maximum efficiency comes at 45 km/
hr, So least fuel consumer per litre in lap Q
Page 245
10.33
10.34
10.35
Option (B) is correct.
Option B fits the sentence, as they built up immunities which helped
humans create serums from their blood.
Option (C) is correct.
4 + 44 + 444 + .............. 4 (1 + 11 + 111 + .......)
= 4 (9 + 99 + 999 + ............)
9
10.37
= 4 [10 (1 + 10 + 102 + 103) - n]
9
n
= 4 :10 # 10 - 1 - nD
9
10 - 1
= 4 610n + 1 - 10 - 9n@
81
10.29
Option (D) is correct.
y
y
-y
f (- y) =
=- f (y)
y
f (q) - f (- q) = 2f (q) = 2
Option (C) is correct.
Betrayed means reveal unintentionally that is most appropriate.
Option (D) is correct.
Number of people who play hockey n (A) = 15
Number of people who play football n (B) = 17
Persons who play both hockey and football n (A + B) = 10
Persons who play either hockey or football or both :
n (A , B) = n (A) + n (B) - n (A + B)
= 15 + 17 - 10 = 22
Thus people who play neither hockey nor football = 25 - 22 = 3
10.36
= 4 [(10 - 1) + (100 - 1) + ........]
9
Option (D) is correct.
Here conserve is most appropriate word.
Option (D) is correct.
Option (C) is correct.
Since 7 + 6 = 13 but unit digit is 5 so base may be 8 as 5 is the
remainder when 13 is divided by 8. Let us check.
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f (y) =
Now
or
10.30
Option (C) is correct.
Let total no of toffees be x . The following table shows the all calculations.
Friend
Bowl Status
R
= x -4
3
= 2x + 4
3
S
= 1 :2x + 4D - 3
4 3
= x +1-3 = x -2
6
6
= 2x + 4 - x + 2
3
6
= x +6
2
T
= 1 a x + 6k - 2
2 2
= x +1
4
= x +6-x -1
2
4
= x +5
4
Now,
or
x + 5 = 17
4
x = 17 - 5 = 12
4
x = 12 # 4 = 48
10.31
10.32
Option (B) is correct.
Circuitous means round about or not direct. Indirect is closest in
meaning to this circuitous
(A) Cyclic
: Recurring in nature
(B) Indirect
: Not direct
(C) Confusing
: lacking clarity of meaning
(D) Crooked
: set at an angle; not straight
Option (B) is correct.
A worker may by unemployed. Like in same relation a sleeper may
be unaware.
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137 8
276 8
435
10.38
Thus here base is 8. Now
731 8
672 8
1623
Option (D) is correct.
Let W be the total work.
=W
20
Per day work of one skill worker
= W =W
5 # 20 100
Similarly per day work of 1 semi-skilled workers
= W = W
8 # 25 200
Similarly per day work of one semi-skill worker = W
= W
10 # 30 300
Thus total per day work of 2 skilled, 6 semi-skilled and 5 unskilled
workers is = 2W + 6W + 5W = 12W + 18W + 10W = W
100 200 300
600
15
Therefore time to complete the work is 15 days.
Per day work of 5 skilled workers
10.39
Option (B) is correct.
As the number must be greater than 3000, it must be start with 3
or 4. Thus we have two case:
Case (1) If left most digit is 3 an other three digits are any of 2, 2,
3, 3, 4, 4, 4, 4.
(1) Using 2, 2, 3 we have 3223, 3232, 3322 i.e. 3! = 3 no.
2!
(2) Using 2, 2, 4 we have 3224, 3242, 3422 i.e. 3! = 3 no.
2!
(3) Using 2, 3, 3 we have 3233, 3323, 3332 i.e. 3! = 3 no.
2!
(4) Using 2, 3, 4 we have 3! = 6 no.
(5) Using 2, 4, 4 we have 3244, 3424, 3442 i.e. 3! = 3 no.
2!
(6) Using 3, 3, 4 we have 3334, 3343, 3433 i.e. 3! = 3 no.
2!
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 246
(7) Using 3, 4, 4 we have 3344, 3434, 3443 i.e. 3! = 3 no.
2!
(8) Using 4, 4, 4 we have 3444 i.e. 3! = 1 no.
3!
Total 4 digit numbers in this case is
1 + 3 + 3 + 3 + 6 + 3 + 3 + 3 + 1 = 25
Case 2 : If left most is 4 and other three digits are any of 2, 2, 3, 3,
3, 4, 4, 4.
(1) Using 2, 2, 3 we have 4223, 4232, 4322 i.e. . 3! = 3 no
2!
(2) Using 2, 2, 4 we have 4224, 4242, 4422 i.e. . 3! = 3 no
2!
(3) Using 2, 3, 3 we have 4233, 4323, 4332 i.e. . 3! = 3 no
2!
(4) Using 2, 3, 4 we have i.e. . 3! = 6 no
(5) Using 2, 4, 4 we have 4244, 4424, 4442 i.e. . 3! = 3 no
2!
(6) Using 3, 3, 3 we have 4333 i.e 3! = 1. no.
3!
(7) Using 3, 3, 4 we have 4334, 4343, 4433 i.e. . 3! = 3 no
2!
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(8) Using 3, 4, 4 we have 4344, 4434, 4443 i.e. . 3! = 3 no
2!
(9) Using 4, 4, 4 we have 4444 i.e. 3! = 1. no
3!
Total 4 digit numbers in 2nd case
= 3 + 3 + 3 + 6 + 3 + 3 + 1 + 3 + 1 = 26
Thus total 4 digit numbers using case (1) and case (2) is = 25 + 26 = 51
10.40
Option (B) is correct.
Let H , G , S and I be ages of Hari, Gita, Saira and Irfan respectively.
Now from statement (1) we have H + G > I + S
Form statement (2) we get that G - S = 1 or S - G = 1
As G can’t be oldest and S can’t be youngest thus either GS or
SG possible.
From statement (3) we get that there are no twins
(A) HSIG : There is I between S and G which is not possible
(B) SGHI : SG order is also here and
S > G > H > I and G + H > S + I which is possible.
(C) IGSH : This gives I > G and S > H and adding these both
inequalities we have I + S > H + G which is not possible.
(D) IHSG : This gives I > H and S > G and adding these both
inequalities we have I + S > H + G which is not possible.
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