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Signals & Systems
(Classroom Practice Questions)
03. Signum function: Sgn(t)
1. Introduction
01. Unit step function, u(t): is usually employed to
switch other signals ON or OFF
u(t)
1;
u (t) = (
0;
t>0
t<0
Z] 1;
t>0
]]
Sgn (t) = [
]] 0; t = 0
] − 1; t < 0
\
Sgn (t)
1
1
t
0
0
u(t) at t = 0 is discontinuous,
u(0) = (1/2).
t; t > 0
r (t) = )
0; t = 0
with “k” any positive integer
Property 2:
t0
u (at − t 0) = u (t − a ), a ! 0, a > 0
An example of unit step function is the output
of 1V dc voltage source in series with a switch
that is turned on at t = 0
02. Rectangular (or Gate) Pulse: A rect(t/2a)
A
0
–1
04. Unit ramp function: r(t)
Property 1:
k
u (t − t 0) = [u (t − t 0)]
−a
t
a
t
The rectangular function is the result of an
ON - OFF switching operation of a constant
voltage source in an electrical circuit.
r(t)
t
0
r (t) = tu (t)
An example of a ramp function is the linear sweep waveform of a cathode - ray tube.
05. Triangular function tri(t)
Width = 2
Height = 1
Area = 1
1
–1 0
1
t
2
06. Sampling function:
sin x
sin πx
Sa (x) = x
Sinc (x) = πx
Signals & Systems
(iii)
δ (t) dt = 1:
-3
Area under unit impulse function is one.
Sa(x)
1
3
#
(iv) δ(–t) = δ(t)
(v) Product property
-2π
0
-π
Sinc(x)
π
2π
x
x(t) δ(t – t0) = x(t0) δ(t – t0)
if x(t) is continuous at t = t0
Eg: (2t + 1) δ(t – 3) = 7δ(t – 3)
(vi) Sifting property
t2
1
x (t) δ (t - t 0) dt = )
#
t1
x (t 0); t1 # t 0 # t2
0 ; elsewhere
4
# (t + t2) δ (t − 3) dt = 3 + 32 = 12
Eg:
–2
0
–1
1
2
-2
x
Energy and Power signals:
07. Unit impulse function (or) Dirac delta function:
•
signal if total energy has a non-zero finite
δ(t)
The impulse function is not a function in the
ordinary sense, since it is zero at every point
except t0, where it is unbounded. However the
area under a unit impulse function is equal to
unity.
Many physical phenomenon such as point
sources, point charges, voltage sources acting
for very short time can be modeled as delta
functions.
A signal x(t) (or) x(n) is called an energy
value 0 < Ex < ∞
•
A signal is called a power signal if it has
non-zero finite power i.e, 0 < Px < ∞.
•
A signal can’t be both an energy and
power signal simultaneously.
•
The term instantaneous power is reserved
for the true rate of change of energy in
a system. In most cases, when the term
power is used it refers to average power
i.e, the average rate of energy utilization,
(i) δ(0) → ∞
a constant quantity independent of time.
(ii) δ(t) = 0, t ≠ 0
T
E x (t) = Lt
δ (t)
#
x (t) 2 dt
T"3
-T
1
T " 3 2T
Pavg x (t) = Lt
0
t
E x (n) = Lt
N
/
N " 3 n =-N
T
#
x (t) 2 dt
-T
x (n)
2
1
/ x (n)
Pavg x (n) = Lt
2N + 1 n = - N
N"3
N
2
3
Periodic And Non Periodic
(Or Aperiodic) Signals:
A periodic function is one which has been
repeating an exact pattern for an infinite and
will continue to repeat that exact pattern for
an infinite time.
A signal is periodic if g(t) = g(t + nT) for any
integer “n” T → period of a function.
x(t)
1
----2
2
0
T=2
4
--t
y(t)
t
T
An analog sinusoid or harmonic signals is always
periodic & unique for any choice of period or
frequency.
Hint:
The sum of harmonic signals
y(t) = x1(t) + x2(t) + x3(t) + - - - - - is periodic with
overall period
T = LCM (T1, T2, T3, ….)
• A discrete
signal x(n) is periodic
if x ⇒ x[n] = x[n + N] ;
where N → periodic of x[n]
Hint:
For finding fundamental period of discrete
sinusoid (or) Complex sinusoids always use the
equation ω0/2π ratio is a rational number.
Signals & Systems
4
Signals & Systems
Sampled version of certain continuous Signals
1
0
0
1
0.5
1.5
2
2.5
3.5
3
t
4
–1
(a) Sampling x(t) = sin(3πt) with sample period T = 0.25
1
0
0
1
2
t
4
3
–1
(b) Sampling x(t) = sin(3πt) with sample period T = 1/π
1
1
1
0
0
0
–1
–1
–1
0
1
0.5
1.5
(a) cos(πnT), T = 0.25
2
0
1.5
1
0.5
(b) cos(2πnT), T = 0.25
2
0
1
1
1
0
0
0
–1
–1
–1
0
1
0.5
1.5
(d) cos(4πnT), T = 0.25
2
0
2
1
1.5
0.5
(e) cos(5πnT), T = 0.25
0
1
1
1
0
0
0
–1
0
1
1.5
0.5
(g) cos(7πnT), T = 0.25
2
–1
0
2
1
0.5
1.5
(h) cos(8πnT), T = 0.25
–1
0
2
1
0.5
1.5
(c) cos(3πnT), T = 0.25
1.5
2
1
0.5
(f) cos(6πnT), T = 0.25
1.5
1
0.5
(i) cos(9πnT), T = 0.25
2
5
Classification of Systems:
Signals & Systems
4.
1. Linear and Non Linear Systems:
For the system to be linear, it should satisfy
2 properties
(A) x1(t)→ y1(t) and x2(t) → y2(t)
Then x1(t) + x2(t) → y1(t) + y2(t) → Additivity
Ex: y(n) = (n+1) x(n) is static
y(n) = x(n+3) is dynamic
(B) cx(t) → cy(t) → Scaling (or) Homogenity
(or)
ax1(t) + bx2(t) → ay1(t) + by2 (t)→ Superposition
Static (or) Memoryless & Dynamic
(With Memory) System:
A system is static if its output at t = t0 depends
only on the value of the input at t = t0 and no
other value of the input signal. The response of
a dynamic system depends on past (and/or
future) inputs.
5.
Stable And Unstable System:
•
A system is said to be BIBO stable if and
Example: y(n) = x(n –2)
y(n) = nx(n) are linear systems
2.
only if every bounded input results in a
bounded output If x (t) # M x 1 3 then
y (t) # M y 1 3
Time-Invariant (Shift-Invariant) & Time-Variant
•
(Shift-Dependent) Systems:
A system is T.I. if the input output characteristic
don’t change with time. T.I. implies that the
shape of the response y(t) depends only on
the shape of the input x(t) and not on the time
when it is applied. If the input is delayed by “t0”
the output is also delayed by the same amount
and is simply shifted replica of the original
output.
2
Ex: y (n) = x (n)
are time invariant
y (n) = x (n − 5)
For T.I system if x(t) → y(t);
Then x(t – t0)  y(t – t0 )
3.
Causal & Non Causal System:
A causal (or) non – anticipating system is one
whose present response does not depend on
future values of the input (or) A system is causal
if its output at t = t0 depends on the values of
the input in the past t ≤ t0 and doesn’t require
future value of input (t > t0) systems whose
present response requires knowledge of future
values of the input are termed non causal.
Ex: y(n) = 2n+1x(n) is causal
y(n) = 2nx(n+1) is Non-causal
6.
A bounded signal has an amplitude
remains finite.
Invertible & Inverse System:
A system is said to be invertible if the input of
the system can be recovered from the output
x(t)
T {-}
y(t)
T-1 {-}
x(t)
T.T-1 = I
In any event, a system is not invertible unless
distinct inputs applied to the system produces
distinct outputs
6
Signals & Systems
Practice Questions
Useful Mathematical Relations
n
∫ x dx =
x n +1
n +1
∫ Sin (ax) dx =
01. An aperiodic discrete time signal of length 5 is
shown in Fig. The maximum value of
- Cos ax
a
x(n)
1
Sin ax
∫ Cos (ax) dx = a
∫a
2
−2 −1
1
1
 bx 
dx =
Tan -1  
2 2
ab
+b x
 a 
∫ x Sin (ax) dx =
Sin (ax) - ax Cos (ax)
a2
∫ x Cos (ax) dx =
Cos (ax) + ax Sin (ax)
a2
N -1
∑αn =
K =0
+∞
∑αk =
K =0
1- α N
1- α
K =1
−1
(A) x(n) + 2x(–n)
(B) 5x(n) x(n–1)
(C) x(n) x(–n–1)
(D) 4x(2n)
(a) A > D > B > C
(b) C > D > A > B
(c) B > D > A > C
(d) D > A > C > D
(a) 6
(b) 3
(c) 1.5
(d) 2
03. Find the value of the following integrals
N ( N + 1)
2
2
# (t + t2) δ (t − 4) dt
(a)
-1
N ( N + 1) (2 N + 1)
K2 =
∑
6
K =1
N
3
# (t + cos πt) δ (t − 1) dt
(b)
-2
∞
-a x
∫e d x =
0
2
n
−
elsewhere. Then the signal y (t) = 2x ; 2 t E is
2
nonzero for duration of
1
,| α |< 1
1- α
α
K αk =
,| α | < 1
∑
(1 - α) 2
K =1
∑K =
2
02. A signal x(t) is nonzero for –1 ≤ t ≤ 2 and zero
+∞
N
1
0
1
2
π
,a>0
a
(c)
3
#
cos t u (t − 3) δ (t) dt
0
3
∞
∞
-∞
-∞
∫ sin cx dx = ∫ sin c
#
(d)
2
e(t - 2) δ (2t − 4) dt
-2
x dx = 1
(e)
2r
#
0
t sin t δ (π/2 − t) dt
7
04. Consider the D.T. signal
x(n) = 1 −
3
/ δ [n − 1 − k]
k=3
Find the values of M and n0 so that
x(n) = u[Mn – n0 ]
05. Sketch the wave forms of the following signals?
(a) x(t) = u(t+1) – 2 u(t) + u(t–1)
(b) x(t) = r(t+2) – r (t+1) – r(t–1) + r(t–2)
where r(t) is unit ramp function
Classification of signals
06. Determine whether the following signals are
energy (or) power signals?
(a) x(t) = e-t u(t)
(b) x(t) = A
(c)
y(t)
A
Ae
These signals are sampled with a sampling
period of T = 0.25 seconds to obtain discrete
time signals x1[n] and x2[n], respectively. Which
one of the following statements is true?
(a) The energy of x1[n] is greater than the
energy of x2[n].
(b) The energy of x2[n] is greater than energy of
x2[n].
(c) x1[n] and x2[n] have equal energies.
(d) Neither x1[n] nor x2[n] is a finite-energy signal.
08. Find the conjugate anti-symmetric part of
x (n) = &1 + j2, 2- , j5 0
09. Give in figure are the parts of a signal x(t) and
its even part xe(t) for t ≥ 0 only; that is x(t) & xe(t)
for t < 0 are not given. Complete the plots of x(t)
& x0(t).
xe(t)
x(t)
2
2
–t
0
t
(d) x(t) = A cos(ωt + θ)
(e) x(t) = tu(t)
(f) x(t) = e3tu(t)
07. (i) Two sequences x1[n] and x2[n] have the same
energy. Suppose x1[n] = α(0.5)nu[n], where α is
a positive real number and u[n] is the unit step
sequence. Assume
x2[n] =
Signals & Systems
 1.5 for n = 0,1

other wise.
 0
Then the value of α is ________.
(ii) Consider the two continuous-time signals
defined below:
t , −1 # t # 1
x1 (t) = )
0,
other wise
1− t , −1 # t # 1
x2 (t) = )
0,
other wise
0
t
1
0
2
t
10. Determine which of the following signals are
periodic, if periodic find the fundamental
period ?
(a) x(t) = cos(18πt) + sin (12πt)
(b) x(t) = sin (2πt/3)cos(4πt/5)
(c) x(t) = cos 3t + sin 5πt
(d) x(t) = jej10t
(e) x(t) = cos(5t)u(t) (f) x(t) = Ev[cos(2πt)u(t)]
(g) x(n) = sin : 5πn D
3
πn
πn
πn π
(h) x (n) = 2 cos : 4 D + sin : 8 D − 2 cos : 2 + 6 D
(i) x(n) = ej7πn
πn
(j) x(n) = cos(n/4) sin : 8 D
(k) x(n) = u(n) + u(– n)
(l) x(n) = (- 1) n
(m) x (n) =
/ ^δ (n − 4k) − δ (n − 1 − 4k)h
3
k =-3
8
11.
A) A signal x(t) = 2 cos (150πt + 300) is sampled at
200Hz. Find the fundamental period of discrete
signal?
B) A periodic discrete time signal x(n) is given by
x(n) = cos (3πn) + sin (7πn) + cos (2.5πn). The
term sin (7πn) in x(n) corresponds to
(a) 14th harmonic
(b) 7th harmonic
th
(c) 6 harmonic
(d) 5th harmonic
C) For each periodic signal shown in figure,
evaluate the average value xav, the energy E
in one period, the signal power P, and the rms
value xrms.
4
1
–2
t
2
-5
-2
-2 2 5
7
t
Signal 3
x(t)
2
1
5
Classification of systems
12. Determine which of the following systems is
linear?
(a) y(t) = x(t) x(t – 2)
(b) y(t) = sin{x(t)}
d
(c) y (t) =
(x (t)) dt
(d) y(t) = 2x(t) + 3
t
(e) y (t) = # x (τ) dτ
(n) y(n) = ex(n)
13. Test the following systems for time - invariance?
(a) y(t) = tx(t) + 3
(b) y(t) = ex(t)
(c) y(t) = x(t) cos3t
(d) y(t) = sin{x(t)}
d
(e) y(t) =
x (t) dt
(f) y(t) = x2(t)
d2 y (t)
+ y (t) y (3t) = x (t)
dt2
d2 y (t)
+ 3y (t) = 2x (t)
(k)
dt2
(j)
14. Consider an LTI system whose response to the
input signal x1(t) is y1(t) as shown in figure. Find
the response of the system due to the Input
x2(t) and x3(t)?
x1(t)
y1(t)
1
2
(g) y(t) = x(t) cosω0t
0
(j) y(n) = x*(n)
(k) y(n) = Median {x[n]}
(l) y (n) =
x (n)
x (n − 1)
0
t
-1-
t
2
x3(t)
x2(t)
1
(i) y(n) = |x[n]|
2
0
(f) y(t) = x2(t)
(h) y(n) = log {x[n]}
d2 y (t) 5dy (t)
+
+ 2 = x (t)
dt
dt2
t
2
-3
(m)
(g) y(t) = x(2t)
(h) y(n) = 2x(n) x(n)
(i) y(n) = x(n + 2) – x(7 – n)
x(t) Signal 2
x(t) Signal 1
Signals & Systems
1
2
4 t
2
-1 0
1
2
t
9
15. Check whether the following systems are
causal (or) non causal?
(a) y(t) = (2t + 3) x(t)
(b) y(t) = x2(t)
(c) y(t) = x(t)sin5t
(d) y(t) = x{sin(t)}
(e) y(t) = |x(t)|
(f) y(n) = ex(n)
(g) y (n) =
n
/ x [k],
"n0 " is finite
k = n0
(h) y (n) =
n + n0
/ x [k],
"n 0 " is finite
k = n - n0
n
(i) y (n) = / x [k]
k =- 3
n
(j) y (n) = / x (k)
k=0
(k) y (n) =
1
/ x (n − k)
2m + 1 k =- m
m
(l) y(t) = x(αt)
(m) yl (t + 4) + 2y (t) = x (t − 2) (n) yl (t) + 2y (t) = x (t + 3)
16. Check whether the following systems are static
or dynamic?
(a) y(t) = 3x(t) + 5
(b) y(t) = x2(t)
(c) y(n) = ex(n)
(d) y(n) = cos{x(n)} (e) y(n) = x[3n]
(f) y (t) = d x (t)
dt
t
(g) y (t) =
# x (τ) dτ
-3
17. Find whether the following systems are linear,
T.I. and dynamic
(a)
2tdy (t)
+ 4y (t) = 2tx (t)
dt
(b)
d2 y (t)
+ 4y (t) = 2x (t)
dt2
Signals & Systems
(c) 4
(d) ;
d2 y (t) 2dy (t)
+
+ y (t) = 3d x (t)
dt
dt
dt2
4dx (t)
dy (t) 2
E + 2t y (t) =
dt
dt
18. Match the following
List I (System) (a) y(n+2)+ y(n+1) + y(n) = 2x(n+1) + x(n)
(b) n2y2(n) + y(n) = x2(n)
(c) y(n + 1) + ny(n) = 4nx(n)
(d) y(n + 1)y(n) = 4 x(n)
List II (system category)
(1) Linear, T.V., dynamic
(2) Linear, T.I., dynamic
(3) N.L., T.V., dynamic
(4) N.L., T.I., dynamic
(5) N.L., T.V., memory less.
19. Check whether the following systems are stable
(or) not?
(a) y(t) = x2(t)
(b) y(t) = x(t) cos 3t
(c) y(t) = x(t–3)
(d) y (t) = d x (t)
dt
t
(e) y (t) =
# x (τ) dτ
-3
(f) y(t) = sin{x(t)}
(g) y(t) = tx(t)
(h) y(n) = ex(n)
(i) y(n) = x[3n]
n
(j) y (n) = / x [k], "n0 " is finite
k = n0
20. Determine which of the following systems are
invertible?
(a) y(t) = x2(t)
(b) y(t) = x(t) 
(c) y(t) = x(t – 3)
d
(d) y (t) = dt x (t)
10
Signals & Systems
24. Statement (I): A memory less system is causal
(e) y(n) = x[n]x[n – 1]
(f) y(n) = nx(n)
Statement (II): A system is causal if the output
(g) y(n) = x[n] – x[n – 1]
at any time depends only on values of input at
(h) y (n) =
n
/ x [k]
that time and in the past.
k =- 3
21. Consider the feedback system shown in figure,
Key for Practice Questions
assume that y[n] = 0 for n < 0
x(n)
+_
y(n)
+
01. Ans: (c)
Unit delay
02. Ans: (a)
03. (a). Ans: 0
Find y(n) when the input is
(i) x(n) = δ[n]
22. Statement (I):
(ii) x(n) = u[n]
(c). Ans: 0
A discrete time system with
input x(n) and output y(n) and given by the
relation y(n) = nx(n–2) + 2 is non linear.
Statement (II):
If x(n) is delayed by 2 units,
y(n) is not delayed by 2 units.
(a) Both Statement (I) and Statement (II) are
individually true and Statement (II) is the
correct explanation of Statement (I)
(b) Both Statement (I) and Statement (II) are
individually true but Statement (II) is not
the correct explanation of Statement (I)
(c) Statement (I) is true but Statement (II) is
(b). Ans: 0
(d). Ans: 0.5
(e). Ans: π/2
04. Ans: M = -1, n0 = -3
1
08. Ans: 2 %1 + j7, 0- , − 1 + j7 /
10. Ans: a, b, d, f, g, h, i, l, m are periodic
signals.
12. Ans: c, e, g are linear
13. Ans: b, d, e, f, h, k are time invariant
15. Ans: a, b, c, e, f, i, m are causal
false
(d) Statement (I) is false but Statement (II) is
true
23. Statement
(I):
The
discrete
time
system
described by y[n] = 2 x[n] + 4 x[n - 1] is unstable,
(here y[n] is the output and x[n] the input)
Statement (II): It has an impulse response with
a finite number of non-zero samples.
16. Ans: a, b, c, d are static
18. (a). Ans: -(2),
(c). Ans: -(1),
(b). Ans: -(5),
(d). Ans: -(4)
19. Ans: a, b, c, f, h, i are stable
21. (i) y(n) = {0, 1, -1, 1,-------}
(ii)
y(n)
=
{0,
1,
0,
1,-------
11
Signals & Systems
2. LTI (LSI Systems)
Convolution is a formal mathematical operation, just as multiplication, addition, and integration. Addition
takes two numbers and produces a third number, while convolution takes two signals and produces a third
signal. Convolution is used in the mathematics of many fields, such as probability and statistics
A linear system's characteristics are completely specified by the system's impulse response, as governed
by the mathematics of convolution. For example: Digital filters are created by designing an appropriate
impulse response. Enemy aircraft are detected with radar by analyzing a measured impulse response.
Echo suppression in long distance telephone calls is accomplished by creating an impulse response that
counteracts the impulse response of the reverberation.
2.1 Continuous & Discrete Convolution
x(t)
x(n)
y(t) = x(t)*h(t)
L.T.I
System h(t)
3
y (t) = # x (τ) h (t − τ) dτ
y (n) =
3
#
=
x (t − τ) h (τ) dτ 1. x(t) → x(τ), h(t) → h(τ)
2.
x(–τ)
x(t – τ)
h(t – τ)
4. Multiplication
5.
Integration
1. x[n] → x[k], h[n] → h[k]
2. Folding
h(–τ)
3. Shifting
3
/ x [n − k] h [k]
Steps:
Steps:
Folding
=
k =-3
-3
3
/ x [k] h [n − k]
k =-3
-3
y(n) = x(n)*h(n)
L.T.I
System h(n)
x(t – τ)h(τ)
h(t – τ)x(τ)
3. Shifting
x[–k]
h[–k]
x[n – k]
h[n – k]
4. Multiplication
5. Summation
x[k] h[n – k]
x[n – k] h[k]
12
•
Signals & Systems
An L.T.I System is always considered w.r.t
Practice Questions
impulse response denoted as h(t) or h(n).
•
If the Input is impulse, then the output is impulse
response.
•
SIFTING property states that any signal can be
produced as a combination of impulses.
•
Convolution may be regarded as a method of
finding the zero-state response of a relaxed LTI
system.
•
Any DT signal is the sum of scaled and shifted
01.(a) Find the convolution of the signals
x(t) = e–3t u(t) & h(t) = u(t – 2)
(b)Find the convolution of the signals shown in
figure?
x(t)
h(t)
1
1
unit impulses.
•
Convolution may be treated as flip - shift -
0
multiply - time - area method.
•
1
t
0
t
In the convolution integral time “t” determines
relative location
of
h(t – τ) w. r. t. x(τ). The
convolution will yield a non zero result only
for those value of “t” over which h(t - τ) & x(τ)
overlap.
02. The impulse response and the excitation
function of a linear time invariant causal
system are shown in Fig. a and b respectively.
The output of the system at t = 2 sec. is equal
to
2.2 Properties of L.T.I System
x(t)
h(t)
1
Causality:
Stability:
1/2
+3
h(t) = 0; t < 0
#
h (x) dx < 3
+3
/ h (n)
<3
n =- 3
Memory less :
h(t) = 0 for t ≠ 0
h(n) = 0 for n ≠ 0
Invertibility & Inverse:
h(t) ∗ hinv(t) = δ(t)
h(n) ∗ hinv(n) = δ(n)
t (sec) 0
6
2
Fig. a
Fig. b
(a) 0
(c) 3/2 (b) 1/2
(d) 1
-3
h(n) = 0; n < 0
6
0
t (sec)
03. An L.T.I system is having impulse response
h(t) = u(–t–1) for which the input signal
applied is shown in figure. Find the output at
t = 4 & t = 0.5?
x(t)
3
1
0
1
2
6
t
13
04. The impulse response of a continuous time
system is given by h(t) = δ(t - 1) + δ(t - 3). The
value of the step response at t = 2 is
(a) 0
(c) 2
(d) 3
Suppose z (t) =
#
10. An Input signal x(t) shown in figure is applied to
the system with impulse response
h (t) =
3
/ δ (t − kT) .
k =-3
(b) 1
3
05.
Signals & Systems
x (− τ + a) h (t + τ) dτ
Find the output, for the values of
i) T = 4
x(t)
ii) T = 2
-3
Express z(t) in terms of y(t) = x(t) ∗ h(t)
06. Find the following terms
(a) x(t + 5) * δ(t – 7) = _________
x(t)
1
2
*
–1
0
1 t
–3
=
3
t
07. Explain the difference between each of the
following operations?
(a) [e-t u(t)]δ(t – 1)
3
#
(b)
0
1
t
11. (a) Find the convolution of
7u (t + 3) - u (t - 1)A with 7u (t + 1) - u (t - 1)A
(b) x(t) * δ(at + b) = __________
(c)
–1
e -t u (t) δ (t − 1) dt
(b) The impulse response of a system is
h(t) = t u(t). For an input u(t – 1), the output
is
(a)
t2
u (t)
2
(b)
t (t - 1)
u (t - 1)
2
(c)
(t - 1) 2
u (t - 1)
2
(d)
t2 - 1
u (t - 1)
2
-3
(c) e-tu(t) ∗ δ(t – 1)
08. Let x(t) = u(t–3) – u(t – 5) & h(t) = e–3tu(t).
Find
d
x (t) ) h (t)
dt
09. Signals that replicate under self-convolution
include the impulse, Sinc, Gaussian, and
Lorentzian. For each of the following known
results, determine the constant A using the
area property of convolution.
(a) δ(αt) ∗ δ(αt) = Aδ(αt)
(b) Sinc(αt) ∗ Sinc(αt) = ASinc(αt)
(c) e - rt ) e - rt = Ae - rt /2
2
(d)
2
2
1
1 =
A
)
1 + t2 1 + t2 1 + 0.25t2
(c) Consider a signal averaging filter whose
−
1
impulse response is h (t) = T rect b t 0.5T l .
T
Find the response of the filter due to the
input x(t) = u(t)?
1, 0 < t < 1
12. Suppose that x (t) = )
0, elsewhere
and h (t) = x b αt l,
0<α#1
dy (t)
Let y(t) = x(t) ∗ h(t). If
has to contain
dt
only three discontinuities, the value of α is
(a) 1
(b) 2
(c) 0.5
(d) – 1
14
t
13. Given x (t) = 5rect b 2 l and h (t) = / 2δ (t − 2k)
k =-3
then x (t) ) h (t) is
(a) 10 ∀t
3
(b)
(c)
3
/
k =-3
5rect c t 2 m
2
/ 10rectc t 2-k2 m 3
Signals & Systems
18. Two discrete -time signals x[n] and h[n] are
both non-zero only for n = 0,1,2, and are zero
otherwise. It is given that x[0] = 1, x[1] = 2,
x[2]=1, h[0] = 1, Let y[n] be the linear
convolution of x[n] and h[n]. Given that
y[1] = 3 and y[2] = 4, the value of the expression
(10y[3] + y[4]) is _______
k =-3
(d) None of these
3
14. The
integral
#
x (τ) h (τ − t) dτ
-3
can
be
interpreted as the Convolution of
(a) x(τ) and h(τ)
(b) x(t) and h(t)
(c) x(-τ) and h(-τ)
(d) x(t) and h(-t)
Properties of L.T.I system
19. The range of ‘a’ and ‘b’ for the impulse
a n; n $ 0
response h (n) = ) n
b ; n<0
to be stable is _______
(a) |a|<1, |b| < 1
(b) |a|>1, |b| < 1
(c) |a|>1, |b| > 1
Discrete Convolution
15. A linear system with Input x(n) & output y(n)
related 3as
y (n) =
/ x (k) g (n - 2k)
k =-3
where g(n) = u(n) – u(n – 4). Find y(n) when
x(n) = δ(n – 2)
16. The I.R of a D.T LTI system is given by
h(n)=(0.5)n u(n) of the input is x (n) = 2δ (n) + δ (n − 3).
Find the output at n = 1 & n = 4?
(d) |a|<1, |b| > 1
20. Given h(t) = eαt u(t) + eβt u(–t). For what values of
α and β system is stable?
(a) α < 0, β < 0
(b) α < 0, β > 0
(c) α > 0, β > 0
(d) α > 0, β < 0
21. Consider the system in figure.
x(n)
17. Given x = [a, b, c, d] as the Input to an LTI
system produces an
output
y = [x, x, x, x, … repeated N times].
The impulse response of the system is
N-1
(a) /
δ [n − 4i]
i=0
+
h2(n) = αnu(n)
h1(n) = βδ(n – 1)
(a) Find I.R. of overall system.
(b) u(n) – u(n – N)
(b) Is this system causal ? Under
(c) u(n) – u(n–N–1)
What condition the system is stable.
(d)
N-1
/ δ [n − i]
i=0
y(n)
15
Signals & Systems
28. If step responses of 2 L.T.I systems are s1(t) &
22. Consider a D.T system ‘s1’, with I.R
h(n) = (1/5) u(n)
n
(a) Find ‘A’ such that h(n) – Ah(n–1) = δ(n)
(b) Using result from part (a), determine the I.R
g(n) of an LTI system s2 which is inverse of s1
23. For the interconnected system shown in Fig.
s2(t) respectively, how the cascaded step
response sc(t) is related interms of s1(t) &
s2(t)?
Key for Practice Questions
find the overall impulse response.
x(n)
y1[n] = x(n) – ½ x(n-1)
n
h2 [n] = (½ ) u[n]
y(n)
- -3 (t - 2)
01. (a) Ans: 1 e3
u (t - 2)
04. Ans: (b)
05. Ans: z(t) = y(t + a)
24. Determine whether each of the following
statements are TRUE (or) FALSE.
Justify your answer.
(1)
The cascade of a non causal LTI system
with casual one is necessarily noncausal
(2)
If an LTI system is causal, it is stable
(3)
If h(t) is the I.R of an LTI system which is
periodic & nonzero, the system is unstable
(4)
The inverse of a causal system is always
causal
25. If the unit step response of a system is
(1 - e - at) u (t) , then its unit impulse response is
_______
(a) αe
- at
(b) α e
-1
u (t)
- at
u (t)
(c) (1 − α ) e - at u (t)
-1
(d) (1 − α) e - at u (t)
26. Find the step response of the system if the
impulse response is h(n) = (0.5)n u(n)
27. An LTI system with Input u(n) produces the
output as δ(n), then find the output due to
the input nu(n)?
06. a. Ans: x(t – 2),
1 b +bl
b. Ans: a
x t a
07. a. Ans: e -1 δ (t − 1) ,
c. Ans: e–(t–1). u(t–1)
b. Ans: e -1 ,
08. Ans: e -3 (t - 3) .u (t - 3) - e -3 (t - 5) .u (t - 5)
11. b). Ans: (c)
13. Ans: (a)
14. Ans: (d)
15. Ans: y(n) = g (n-4)
16. Ans: y(1) = 1, y(4) = 5/8
17. Ans: (a)
19. Ans: (d)
20. Ans: (b)
21. Ans: h (n) = α n u (n) + βα n - 1 u (n − 1) is causal
for any value of α, β and stable if |α| < 1,
and any value of β.
22. Ans: A = 15
23. Ans: h (n) = δ (n)
24. Ans: 1-false, 2-false, 3-true, 4-false.
25. Ans: (a)
26. Ans: s (n) = 2 ;1 − b 1 l E u (n)
2
n+1
27. Ans: u(n-1)
16
Signals & Systems
However, musical instruments, such as a piano,
3. Fourier Series
are made of many strings all vibrating at once.
The question that intrigued Fourier was: How do
Historical perspective:
you evaluate the waveforms from a number of
Jean Baptiste Joseph Fourier (1768 – 1830)
strings all vibrating at once? As a product of his
Joseph Fourier was born in Auxerre, France
research, Fourier realized that the sound heard
on March 21, 1768 and died in Paris on May 4,
by the ear is actually the arithmetic sum of
1830.
each of the individual waveforms. This is called
Motivation:
•
signals in the frequency domain
•
the principle of superposition.
Representation of continuous time, periodic
•
Representing CT signals as superposition of
Periodic signals occur frequently - motion
complex exponentials leads to frequency –
of planets and their satellites, vibration of
domain characterizations. eg :- A human ear is
oscillators, electric power distribution, beating
sensitive to audio signals within the frequency
of the heart, vibration of vocal chords, etc.
range 20Hz to 20 kHz. Typically, musical note
Introduction:
occupies a much wider frequency range.
•
The Fourier series is named after the French
Therefore the human ear processor frequency
mathematician Joseph Fourier.
components within the audible range &
In this chapter we will consider approximating
rejects other frequency components. In such
a function by a linear combination of basis
applications,
functions, which are simple functions that can
provides a convenient means of solving for the
be generated in a laboratory. Joseph Fourier
response of L.T.I. systems to arbitrary input.
(1768–1830) developed the mathematical
theory of heat conduction using a set of
•
•
Sinusoidal signals arise in describing motion of
be represented by its Fourier series.
planets & periodic behavior of earth’s climate.
Fourier series and the Fourier transform are
A.C. sources generate sinusoidal voltages &
basics to mathematics and science, especially
currents.
to the theory of communications. For example,
a phoneme in a speech signal is smooth and
wavy. A linear combination of a few sinusoidal
functions would approximate a segment of
speech within some error tolerance.
•
By using F.S, a non-sinusoidal periodic function
functions.
form we now call Fourier series. He established
•
analysis
can be expressed as an infinite sum of sinusoidal
trigonometric (sine and cosine) series of the
that an arbitrary mathematical function can
frequency-domain
Fourier and a number of his contemporaries
were interested in the study of vibrating
strings. In the simple case of just one naturally
vibrating string the analysis is straightforward:
the vibration is described by a sine wave.
•
There are 2 reasons for evaluating the F.S.
1. To obtain an expression for f(t) that applies
everywhere, rather than only over a single
period.
2. To obtain phasors, which indirectly tell how
much power is available at each harmonic
of the waveform.
17
3.1 ANALOGY BETWEEN VECTORS & SIGNALS
•
•
Signals are not just like vectors.
A vector can be represented as a sum of
its components, depending on the choice
of coordinate system. A signal can also be
represented as a sum of its components.
We know that an arbitrary M-dimensional
vector can be represented in terms of
M orthogonal co-ordinates.
A vector is specified by its magnitude &
direction.
f
e
θ
FIG (a)
CX
f
Signals & Systems
v
Length of the component vf along X
v = vf Cosθ
= CX
v
v 2 = vf . X
v Cosθ = vf .X
CX
vf .X
v
C= v 2
X
•
2 vectors vf & X
v are orthogonal if inner (or)
v =0
scalar product vf : X
•
If we consider 2 basis vectors vi & vj
orthogonality vi=
: vj
v
unit magnitude i =
: vi
vj=
: vi 0
4 orthonormal Property
vj=
: vj 1
Component of a Signal
g(t)
X
x(t) = sin t
+1
0
e1
π
2π
t
−1
FIG (b)
X
C1 X
g(t) ≅ C x(t) ; t1 < t < t2
t
# g (t) x (t) dt
t
=
C
t
2
e2
f
FIG (c)
C2 X
1
X
v as shown in figure.
Consider 2 vectors vf & X
Let the component of f along X be CX.
(Geometrically the component f along X is the
projection of f on X)
v +e
v
From Fig (a) vf = CX
From Fig (b) & (c)
vf = C1 X
v +e
v +e
v 1 = C2 X
v2
#
v
If we approximate f by CX, vf , CX
Error in the approximation e = f – CX
v
Minimize the error vector, such that vf and X
are approximated
2
x2 (t) dt
t1
Note:
2 signals g(t) & x(t) are said to be orthogonal
over the interval (t1, t2)
t2
# g (t) x* (t) dt
if =
t1
t2
=
# x (t) g* (t) dt 0
t1
They are also said to be orthonormal if they
satisfies
t2
=
# x (t) x* (t) dt
t1
t2
=
# g (t) g* (t) dt 1 (unit magnitude)
t1
18
Signals & Systems
Examples
(b) For orthonormality 2T = 1 ⇒ T = 1/2
(c) x(t) = C1∅1(t) + C2∅2(t) + C3 ∅3(t)
Example 01:
1
Ci = E
Q
For the three continuous functions shown in
figure
1
T t
0
φ3(t)
-T
0
-T
T t
JK T/2
NO
T
KK
−
=
# A dtOOOO = 0
C2 1 K # A dt
K0
T/2
L
P
Similarly C3 = – A/2
∴ x(t) = A/2 [∅1(t) – ∅3(t)]
1
3.2 Trigonometric F.S
T
0
t
•
Any practical periodic function of frequency
ω0 can be expressed as an infinite sum of sine
(or) cosine functions that are integral multiples
-1
(a) Show that the functions form an orthogonal
set
of ω0
g(t) = ao+ a1cosω0t + a2cos2ω0t + - - - - - + b1sinω0t + b2sin2ω0t + - - - - - - - -
(b) Find the value of T that makes 3 functions
orthonormal
g (t) = a 0 +
(c) Express the signal
x (t) = )
.
dc
A for 0 # t # T
0 elsewhere
in (a)?
T
#
Q1 (t) dt =
2
T
Q2 (t) dt =
2
#
Q3 (t) dt = 2T
2
-T
-T
T
1-444444444444444444444444444
42444444444444444444444444444
43
Unit - Magnitude Property
Orthogonality
T
-T
T
Q1 (t) Q2 (t) dt =
#
-T
T
Q2 (t) Q3 (t) dt =
n=1
------ (I)
ac
a0, an, bn → T.F.S. coefficients
1
a0 = T
Sol: (a) For all the 3 signals
T
3
/ a14444444444444
n cos nω 0 t + b n sin nω 0 t
42444444444444443
ω0 → Fundamental frequency
in terms of orthogonal set determined
#
t1
0
φ2(t)
#
x (t) Q*i (t) dt but EQ = 2T
T
=
# A (1) A/2
C1 1=
φ1(t)
-T
t2
#
#
-T
Q1 (t) Q3 (t) dt = 0
2
an = T
2
bn = T
T
#
g (t) dt " d.c (or) Average value
0
T
#
g (t) cos nω 0 t dt
0
T
#
0
g (t) sin nω 0 t dt
19
Signals & Systems
Effect of symmetry on Fourier coefficients:
Symmetry
Condition
Even
g(t) = g(–t)
a0
?
g(t) = – g(–t)
0
0
g(t) = – g(t ± T/2)
0
= 0 ; n even
= ? ; n odd
Odd
Half-wave (or)
Rotational
an
?
bn
0
?
= 0 ; n even
= ? ; n odd
Notes:
•
•
•
•
•
•
a0, the DC offset term, can be non-zero even though all the other an’s are zero
An odd-harmonic function is one where the second half of its period is the negative of the first half.
An even-harmonic function is one where the second half of its period is exactly the same as the first
half. Therefore, any function that is even-harmonic is actually a regular periodic function whose period
has been labeled twice what it should be. In other words, there is nothing special about even-harmonic
functions.
Shifting a signal left/right in time does not affect whether or not it is odd-harmonic.
Shifting a signal up/down (adding a DC offset) does not affect whether it is odd-harmonic, other than
adding a term in the Fourier series at Zero frequency.
An odd-harmonic function does not have to be odd.
t
even
odd
t
t
even and odd - harmonic
odd-harmonic
odd and odd
- harmonic
t
t
even and odd - harmonic w/ DC offset
t
20
3.3 Exponential (or) Complex F.S
5.
Another form of the F.S that involves complex
exponential functions can be obtained by
substituting the complex exponential forms of
sin ω0t and cos ω0t into (I)
| Cn |
Signals & Systems
2-sided
spectrum
Differentiation in time:x(t) ↔ Cn
dx (t)
* (jnω 0) Cn
dt
6.
Parseval’s Power theorem:The total average power in a periodic signal
equal to sum of squared amplitude of each
harmonic
x(t) ↔ Cn
1
then T
- 2ω0
- ω0
0
ω0
3
t
1
g (t) = / Cn e jn~Where
Cn = T
0
n = -3
nω0
2ω0
T
#
g (t) e -jn~ t dt
0
0
Convergence Of F.S: (Dirichlet Conditions)
(1) x(t) is absolutely integrable i.e.,
T
#
x (t) dt < 3
0
(2) x(t) has only a finite number of maxima &
minima
(3) The number of discontinuities in x(t) must be
finite.
3.4 Properties of F.S
1. Linearity: x1(t) ↔ Cn
x2(t) ↔ dn
then αx1(t) + βx2(t) ↔ αCn + βdn
2.
Time Shift: x (t - t 0) * Cn e -jn~0 t0
When we shift in the time-domain, it changes
the phase of each harmonic in proportion to its
frequency nω0.
3. Frequency Shift:
x(t) ↔ Cn
then x (t) e j~0 Mt * Cn - M
4.
Time-Scaling:
x (t) * Cn then x (at) * Cn
Time-Compressing by α changes frequency
from ω0 to αω0
#0
T
x (t) 2 dt =
3
/
n =-3
Cn
2
21
Signals & Systems
(R)
Practice Questions
g (t)
+V
01. A periodic signal is given by
x(t) = 3 sin (4t + 30o) – 4 cos (12t – 60o).
Find the amplitude of second harmonic?
-2π - π
02. Which of the following signal is not the
representation of F.S.?
(a) Cos3t + sin12t
(b) 1 + sinπt
(c) ej6t
(d) cos2πt + sin6t
03. For the Periodic waveforms shown in figure
what frequency components are present
in trigonometric series expansion.
(P)
0
1
t
2
(a) DC and cosine terms
(b) DC and sine terms
(c) Cosine & sine terms
(d) None of these
(d) DC, Cosine and Sine terms
(S)
f(t)
m(t)
2
–1
–1
1
2
4
t
–2
V
-V
t
(a) DC, Cosine terms
(b) DC, Cosine terms with even harmonics
(c) DC, Cosine terms with odd harmonics
(d) None of these
–2
f(t)
0
1
1
(Q)
-π
t
(a) Sine terms (b) Sine terms with odd harmonics
(c) Sine terms with even harmonics
(T)
2π
–π -π/2 0 π/2 π
2
-1
π
-V
x(t)
-2
0
π
2π
3π
(a) Sine and Cosine terms
(b) Only odd harmonics
(c) Sine terms with odd harmonics
(d) DC and Sine terms
t
(a) DC, Cosine and Sine terms
(b) DC, Cosine terms with even harmonics
(c) DC, Cosine terms with odd harmonics
(d) None of these
04. Consider the signal
x(t) = 10cos(10πt + π/7) + 4sin(30πt + π/8).
It’s power lying within the frequency band
10 Hz to 20 Hz is ____
(a) 4W
(b) 8W
(c) 50W (d) 58W
22
05. Consider the trigonometric series, which holds
true∀t, given by
x(t) = sinω0 t + 1/3 sin 3ω0t + 1/5 sin 5ω0t
+ 1/7 sin 7ω0 t + - - - - - - - - - At ω0t = π/2 the series converges to
(a) 0.5
(b) π/4
(c) π/2
(d) 2
06. x(t) is a real valued function of a real variable
with period T. Its trigonometric. Fourier Series
expansion contains no terms of frequency
ω = 2π(2k)/T; k = 1, 2….……. Also, no sine terms are
present. Then x(t) satisfies the equation
(a) x(t) = –x(t – T)
(b) x(t) = x(T – t) = –x(–t)
(c) x(t) = x(T – t) = –x(t –T/2)
(d) x(t) = x(t – T) = x(t – T/2)
07. The fundamental frequency of the composite
signal
x(t) = 2 + 3cos (0.2t) + cos (0.25t + π/2)
+ 4cos (0.3t – π) is
(a) 0.05 rad/sec
(b) 0.1 rad/sec
(c) 0.2 rad/sec
(d) 0.25 rad/sec
08. The average value of the periodic signal x(t)
shown in figure is
Signals & Systems
09. f(t), shown in figure is represented by
f (t) = a 0 +
3
/ an Cosnt + bn Sinnt
n=1
The value of a0 is
f(t)
1.5
+1
–π
π
0
3π
t
-1.5
(a) 0
(c) π
(b) π/2
(d) 2π
10. A periodic rectangular signal x(t) has the
waveform shown in fig frequency of the fifth
harmonic of its spectrum is
x(t)
–4 –2
0
2
4
t (msec)
(a) 40 Hz
(b) 200 Hz
(c) 250 Hz
(d) 1250 Hz
11. A periodic signal in a duration of one period is
shown. Its Fourier series expansion Includes
x(t)
3
2π
3
1
– 6 – 4 – 3– 2
(a) 5/6
(b) 1
(c) 5
(d) 6
0 2
3
4
6
8 9 10
t
0
3
4
−1
5
t
(a) cosine terms of odd harmonics
(b) sine terms of odd harmonics
(c) sine terms of even harmonics
(d) cosine terms of even harmonics
23
12. The rms value of the periodic waveform shown
in figure is
6A
0
T
T/2
Exponential F.S
16. For the periodic signal
2rt
5rt
x (t) = 2 + cos : 3 D + 4 sin : 3 D
Find the E.F.S. coefficients?
t
–6A
(a) 2 6 A 4 (c)
A
3
Signals & Systems
17. Obtain the E.F.S. representation of periodic
signal shown, hence find the T.F.S?
(b) 6 2 A
(d)1.5 A
x(t)
1
13. A half - wave rectified sinusoidal waveform has
a peak voltage of 10 V. Its average value and
the peak value of the fundamental component
are respectively given by :
20
10
r V, 2 V 10
(c) r V, 5V (a)
10
10
r V, 2 V
20
(d) r V, 5V
(b)
14. A periodic signal with period T = 2 is defined as
x (t) = )
−1
t
x(t)
t, 0 # t # 1
2 - t, 1 # t # 2
-2
r2
2
18. The F.S. coefficient of the signal x(t) shown in
fig (a) are C0 = 1/π, C1 = –j0.25, Cn = 1/π(1 – n2)
(n even). Find F.S. coefficients of y(t), f(t) and
g(t)?
1
If A cos (πt ) is one of the terms it contains A is
equal to
4
2
(a) 2
(b) 2
r
r
(c)
1
0
(d)
-1
2
1
0
Fig(a)
3
t
y(t)
-4
r2
1
15. A periodic input signal x(t) shown below is
applied to an L.T.I. system with frequency
response
1; ω < 4π
H (ω) = )
0; ω > 4π
Which frequency components are present in
the output?
x(t)
0
-1
1
2
3
2
3
t
f(t)
1
0
- 2 1
1
t
-1
g(t)
10
1
-2
-1
0
1
2
3
t
-1
0
1
2
t
24
19. Let x(t) be a periodic signal with period T and
Signals & Systems
F.S coefficient Cn
1
w1
w2
1
Let y(t) = x(t–t0) + x(t + t0). The F.S. coefficient of
y(t) is dn.
(a) T/8
(c)T/2
(b) T/4
t
T
0 T/2
If dn= 0 ∀ odd n then t0 can be
–1
(d) 2T
20. By using derivative method, find F.S. coefficient
of the signal shown in figure?
–1
(a) |n-3| & |n-2|
(c) |n-1| & |n-2|
1
Magnitude
2
t
T
-3 -2 -1 0 1 2 3
Phase
21. Consider a periodic signal x(t) as shown below
x(t)
−3−2−1 0 1 2 3 4 5 6
300
600
f(Hz)
900
-3 -2 -1 0 1 2 3
1
t
−1
Fig.
It has a Fourier series representation
x (t) =
(b) |n-2| & |n-3|
(d) |n-4| & |n-2|
23. The magnitude & phase spectra of a periodic
signal x(t) are shown in figure
x(t)
-T/2 -d/2 0 d/2 T/2
t
T
T/2
0
3
–900
–30
–600
f(Hz)
0
(a) Write the polar form of TFS.
(b) Sketch the magnitude & phase spectrum
of f(t) = x(2t), g(t) = x(t - 1/6) & h(t) = xl (t)
/ ak e j(2r/T)kt
k =- 3
Which one of the following statement is TRUE?
(a) ak = 0, for k odd integer and T = 3
(b) ak = 0, for k even integer and T = 3
(c) ak = 0, for k even integer and T = 6
(d) ak = 0, for k odd integer and T = 6
22. One period (0, T) of 2 periodic waveforms
w1 & w2 are shown in Fig. then Fourier series
Coefficient are respectively proportional to
____
24.
(a) Find the power up to second harmonic for the
periodic signal shown in figure?
1
–1
0
1
2
3
t
25
Signals & Systems
(b) For the periodic signal x(t) shown below with
∠Cn
period T = 8 s, the power in the 10th harmonic is
π/4
π/2
x(t)
–4
8
4
0
t
(c)
1 b 4 l2
2 10π
0
−1
2
1
3
4
−π/4
(c) Assume that the fundamental frequency
of 10Hz, find the trigonometric form of the
signal x(t)?
(d) If this signal is passed through an ideal LPF
with cut-off frequency 25Hz, what would
be the steady-state response
4 2
(d) b 10π l
25. The F.S Coefficients, of
a periodic signal x(t) is
3
expressed as x (t) =
/ Cn e jn~ t
0
n =-3
are given by
C–2 = 2– j1; C–1= 0.5 + j0.2; C0 = j2;
27.
Match the following
List – I (Periodic function)
A. Impulse train
C1 = 0.5 – j0.2; C2 = 2 + j1;
Cn = 0 for |n| >2.
-T
-2T
Which of the following is TRUE?
(a) x(t) has finite energy because only finitely
many coefficients are non zero
0
periodic
B. Full-wave rectified sine wave
D.
1
(c) the imaginary part of x(t) is constant
(d) the real part of x(t) is even.
0
26. For the following spectrum shown in figure.
|Cn|
−3
−2
−1 0
4
2
1
2
1
3
t
T
C. 2sin(2πt/6) cos(4πt/6)
(b) x(t) has zero average value because it is
−4
n
(a) Find the power in the periodic signal
(b) Is this signal odd symmetric, even
symmetric & neither
(a) 0 1 b 2 l2
2 10π
−2
−π/2
–1
(b)
−3
−4
+1
0.5
4 n
T/2
T
t
-1
List-II (Properties of spectrum)
1. Only even harmonics are present
2. Impulse train with strength 1/T
3. C3 = 1/2j
C-3 = –1/2j
C1 = –1/2j
C-1 = 1/2j
4. Only odd harmonics are present
5. Both even & odd harmonics are present
26
28. The systems S1 & S2 shown in figure satisfy the
following properties
sin ωt
(1/4) sin10t
(a)
(b)
(c)
(d)
Signals & Systems
Key for Practice Questions
5 cos (ωt + 30°)
S1
S2
cos (5t)
Both S1 and S2 are LTI systems
S1 is LTI, but S2 is not LTI system
S1 is not LTI system, S2 is LTI system
Both S1 & S2 are not LTI systems
29. Consider the following statements related to
Fourier series of a periodic waveform:
1. It expresses the given periodic waveform
as a combination of d.c. component,
sine and cosine waveforms of different
harmonic frequencies.
2. The amplitude spectrum is discrete.
3. The evaluation of Fourier coefficients gets
simplified if waveform symmetries are used
4. The amplitude spectrum is continuous
Which of the above statements are
correct? (a) 1, 2 and 4
(b) 2, 3 and 4
(c) 1, 3 and 4
(d) 1, 2 and 3
30. Statement (I): In the exponential Fourier
representation of a real-valued periodic
function f(t) of frequency f0, the coefficients of
the terms e j 2 π n f0 t and e - j 2 π n f0 t are negatives
of each other.
Statement (II): The discrete magnitude
spectrum of f(t) is even and the phase spectrum
is odd.
01. Ans: 0
02. Ans: (d)
03. P. b, Q. b, R. b, S. c, T. d
04. Ans: (b)
05. Ans: (b)
07. Ans: (a)
08. Ans: (a)
09. Ans: (a)
10. Ans: (d)
11. Ans: (b)
13. Ans: (c)
14. Ans: (d)
20
20
15. Ans: y (t) = 5 + r sin rt +
sin 3rt
3r
16. C 0 = 2, C 2 =
1
1
,C
, C - 2j, C -5 = 2j
2 -2 = 2 5 =
19. Ans: (b)
22. Ans: (c)
24. (a) Ans: 0.45 Watt.
27. A – 2, B. –1, C. –3, D
27
Signals & Systems
4. Fourier Transform
4.1 Introduction
•
•
•
•
•
Fourier Transform (F.T.) provides a frequency domain description of time domain signals and is extension
of F.S to non-periodic signals.
CTFT expresses signals as linear combination of complex Sinusoids
Transformation makes the analysis of signal much easier because certain features which may be
obscure in one form may be obvious in other form
Spectrum of F.T. is continuous whereas spectrum of F.S is discrete.
F.T (or) spectrum of a signal x(t) is
3
= X (ω) =
#
x (t) e -j~t dt
........(1)
-3
•
Inverse F.T. (I.F.T) is
3
1 #
x (t) =
X (ω) e j~t dω
2π
......(2)
-3
•
X (f) =
x (t) =
# x (t) e-j2rft dt
# X (f) e j2rft df Convergence of F.T.
1.
F.T. is defined for all stable signals i.e.,
3
# x (t) dt < 3
-3
2.
3.
Periodic signals, which are neither absolutely integrable nor square integrable over an infinite interval,
can be considered to have F.T. if impulse functions are permitted in the transform.
x(t) have a finite number of discontinuities and finite number of maxima and minima within any finite
interval.
F.T. of Standard Signals:
1.
Decaying exponential: e -at u (t) *
1
;a>0
a + j~
1/a
1
|X(ω)|
π/2
1
a 2
π/4
0
t
-a
- π/4
-π/2
0
a
ω
∠X(ω)
28
2.
Increasing exponential
1
0
3.
Signals & Systems
e at u (− t ) ↔
t
1
a − jω
C.T. impulse function: δ(t) ↔ 1
1
t
0
4.
0
ω
Rectangular (or) Gate function:~T
~T
x (t) = A rect (t/T) (or) Ar (t/T) * ATSa b 2 l (or) ATSinc b 2r l
| X(ω) |
AT
x(t)
−T/2
A
0
A
T/2
t
A
−
4π
T
−
0
2π
T
2π
T
4π
T
ω
∠X(ω)
π
−
4π
T
−
2π
T
0
A
-π
2π
T
4π
T
ω
29
Signals & Systems
Spectral Width
In the frequency domain, all signals may be classified as follows:
•
A broadband signal is the one, which spectrum is distributed over a wide range of frequencies as it is
shown in Fig. 4(a)
•
A bandlimited signal is limited in the frequency domain with some maximum frequency as it is shown in
Fig. 4(b).
•
A narrowband signal has a spectrum that is localized about a frequency f0 that is illustrated in Fig. 4(c).
•
A baseband signal has a spectral contents in a narrow range close to zero (Fig. 4(d)). Accordingly, a
spectrum beginning at 0 Hz and extending contiguously over an increasing frequency range is called a
baseband spectrum.
(c)
(d)
(a)
(b)
0
f0
Fig (4): Types of signals: (a) broadband,
(b) bandlimited, (c) narrowband, and
(d) baseband.
f
30
Signals & Systems
Operational Properties of the Fourier Transform
Property
x(t)
X(f)
X(ω)
(1). Similarity
X(t)
x(-f)
2πx(-ω)
(2). Time Scaling
x(αt)
(3). Folding
x(-t)
X(-f)
X(-ω)
(4). Time Shift
x(t - α)
e-j2πfα X(f)
e-jωα X(ω)
(5). Frequency shift
ej2παt x(t)
X(f - α)
X(ω - 2πα)
(6). Convolution
x(t) ∗ h(t)
X(f) H(f)
X(ω) H(ω)
(7). Multiplication
x(t) h(t)
X(f) ∗ H(f)
(8). Modulation
x(t) cos(2παt)
0.5[X(f + α) + X(f- α)]
0.5[X(ω + 2πα) + X(ω - 2πα)]
(9). Derivative
x′(t)
j2πf X(f)
jωX(ω)
(10). Times-t
-j2πt x(t)
X′(f)
2πX′(ω)
(11). Integration
1 bωl
X α
α
1 bfl
X a
a
t
1
X (ω) ) H (ω)
2π
1
X (f) + 0.5X (0) δ (f)
j2πf
# x (t) dt
-3
1
X (ω) + πX (0) δ (ω)
jω
(12). Conjugation
x∗(t)
X∗(-f)
X∗(-ω)
(13). Correlation
x(t) ∗∗ y(t)
X(f) Y∗(f)
X(ω) Y∗(ω)
(14). Autocorrelation
x(t) ∗∗ x(t)
X(f) X∗(f) = |X(f)|2
X(ω) X∗(ω) = |X(ω)|2
(15). Central ordinates
x (0) =
3
#
1
2π
X (f) df =
-3
3
(16). Parseval’s theorem
E=
#
3
-3
-3
x (t) y) (t) dt =
#
3
(17). Plancheral’s theorem
#
-3
#
x (t) 2 dt =
3
#
3
X (ω) dω
-3
X (f) Y) (f) df =
#
-3
X (f) 2 df = 1
2π
3
-3
X (0) =
3
#
X (ω) 2 dω
-3
1
2π
3
#
-3
X (ω) Y) (ω) dω
x (t) dt
31
Signals & Systems
Some Useful Fourier Transform Pairs
Entry
x(t)
X(f)
X(ω)
1
δ(t)
1
1
2
rect(t)
sinc(f)
ω
sin c b 2π l
3
tri(t)
sinc2(f)
ω
sin c2 b 2π l
4
Sinc(t)
rect(f)
ω
rect b 2π l
5
cosω0t
[δ(f – f0) + δ(f + f0)]/2
π[δ(ω–ω0) + δ(ω+ω0)]
6
sinω0t
[δ(f – f0) – δ(f + f0)]/2j
π
[δ(ω–ω0) – δ(ω+ω0)]
j
7
e-αtu(t)
1
a + j2rf
1
a + j~
8
te-αtu(t)
1
(a + j2rf) 2
1
(a + j~) 2
9
e- a t
2a
a2 + 4r2 f2
2a
a2 + ~2
10
e- π t
11
sgn(t)
12
u(t)
13
e-αt cos(2πβt) u(t)
14
e-αt sin(2πβt) u(t)
15
/ δ (t − nT)
2
e- π f
0.5δ (f) +
3
xp (t) =
3
0
2
/ 4π
2
jω
1
j2πf
πδ (ω) +
1
jω
a + j2rf
2
(a + j2rf) 2 + ^2rbh
α + jω
2
(α + jω) 2 + ^2πβ h
2rb
2
(a + j2rf) 2 + ^2rbh
2πβ
2
(α + jω) 2 + ^2πβ h
1 / bf − k l
T
T k =-3 δ
3
2π / b ω − 2πk l
T
T k =-3 δ
3
/ Cn e j2rf t
n =- 3
e-ω
1
jr f
n =- 3
16
2
3
/ Cn δ (f − nf0)
n =-3
3
/ 2πCn δ (ω − nω0)
n =-3
32
4.2 DISTORTIONLESS TRANSMISSION
In
several
application
such
as
Signals & Systems
•
If the gain is not constant over the required
frequency range, we have amplitude distortion.
If the phase shift is not linear with frequency, we
have phase distortion as the signal undergoes
different delays for different frequencies.
•
Ideal filters are non causal, unstable and
physically unrealizable in the sense that their
characteristics can not be achieved with a
finite number of elements
•
For a physically realizable system, h(t) must be
causal i.e., h(t) = 0 for t < 0. In the frequency
domain, this condition is known as Paley –
Wiener Criterion which states that the necessary
and sufficient condition for the amplitude
response |H(ω)| to be realizable is
signal
amplification or message signal transmission
over communication channel, we require that
the output waveform be a replica of the Input
waveform.
•
Transmission is said to be distortionless if the
input and output have identical waveshapes
within a multiplicative constant (or) a delayed
output that retains the input waveform is
considered to be distortionless.
•
For distortionless transmission, the input x(t) and
output y(t) satisfies the condition
y(t) = k x(t – t0)
Y (ω)
H (ω) =
= Ke -j~t
X (ω)
+3
#
0
-3
k
•
ω
Phase delay tp(ω) is the delay occurring at a
Single frequency. As the signal propagates
from source to destination the amount of delay
caused is known as phase delay.
θ (ω )
t p (ω ) = − ω
|H(ω)|= k
∠H(ω) = θ(ω) = –ωt0
•
ω
Slope = –t0
•
ln H (ω)
dω < 3
1 + ω2
For distortion less transmission, magnitude
response must be a constant, phase response
must be a linear function of ω with slope –t0,
where t0 is delay in output with respective to
input.
Phase delay is not necessarily the true signal
delay. A steady Sinusoidal signal doesn’t carry
information. Information can be transmitted
only by applying some appropriate change to
Sinusoidal wave. Suppose that a slowly varying
signal is multiplied by a Sinusoidal wave so that
resulting modulated wave consists of a narrow
group of frequencies. When this modulated
wave is transmitted through the channel, we
find that there is a delay between envelope
of Input and received signal. This is known as
envelope (or) group delay (True signal delay)
t g (ω ) = −
dθ (ω)
dω
33
Example :
Signals & Systems
The following figure illustrates the effect of ideal filters using the input voltage
v i (t) = 0.8 sin ω 0 t + 0.5 sin 4ω 0 t + 0.2 sin 16ω 0 t V
1.5v
1.5v
0v
1.5v
−1.5v
1.5v
0v
1.5v
0v
1.5v
0v
1.5v
Low pass
High pass
Band pass
Band reject
0v
−1.5v
1.5v
−1.5v
1.5v
−1.5v
1.5v
−1.5v
Frequency
Time
As an example, Shown in figure at the left are the spectra that we would observe with a spectrum
analyzer; shown at the right are the waveforms that we would observe with an oscilloscope. The
spectrum and waveform at the top pertain to the input signal, and those below pertain, respectively,
to the low-pass, high-pass. Band-pass, and band-reject outputs. For instance, if we send Vi(t) through a
low-pass filter with ωc somewhere between 4ω0 and 16ω0, the first two components are multiplied by 1
and thus passed, but the third component is multiplied by 0 and is thus blocked: the result is
v i ( t ) = 0.8 Sinω0 t + 0.5Sin 4ω0 t V
Example :
A channel has the frequency response
]Z]
rf
f
]]4rect b 40 l e -j 30 for f # 15Hz
H (f) = ][
]] 4rect b f l e -j r2 for f > 15Hz
]
40
\
Draw the phase delay tp(f) & group delay tg(f). For what values of f does tp(f) = tg(f)?
Ans:
Z] rf
]] ] 30 ; f # 15
i (f) = [] r
]] - ; f > 15
] 2
\
Z] 1
]]
] 60 ; f # 15
Phase delay t p (f) = [] 1
]] ; f > 15
] 4f
\
34
Properties of H.T:
1. H.T. doesn’t change the domain of a signal
2. H.T. doesn’t alter the amplitude spectrum of a
signal
3. If xt (t) is H.T. of x(t), then H.T. of xt (t) is – x(t)
4. x(t) and xt (t) are orthogonal to each other.
1
; f # 15
Group delay t g (f) = * 60
0; f > 15
∴ tp(f) = tg(f) for f # 15
tg(f)
tp(f)
−15
1
60
0
15
−15
f
Signals & Systems
0
15
Example : Find H.T. of
(1) x(t) = cosω0t
(2) x(t) = sinω0t
(3) x (t) =
n=1
4.3 HILBERT TRANSFORM: (H.T)
(4) x(t)cos(2πfct), where x(t) represents a
signal band limited to B, fc > B
The Hilbert transform is an operation that shifts
the phase of x(t) by – π/2, while the amplitude
spectrum of the signal remains unaltered.
An ideal H.T. is an all pass 900 phase shifter.
Example :
For the system shown in figure if x(t) = cost and
h(t) = (1/πt), then the output y(t) is
x(t)
H.T. is used in number of application such as
representation of band pass signals, phase shift
(a) cos t
(c) sin t
modulators, generation of SSB.
x(t)
1
xˆ ( t ) = x( t ) ∗  
 πt 
1
πt
•
|H(ω)|
•
ω
∠H(ω)
π /2
•
ω
– π /2
h(t)
h(t)
h(t)
y(t)
(b) – cos t
(d) – sin t
4.4 CORRELATION
Frequency response of the H.T.
= –jsgn(ω)
0
3
/ an cos nω0 t + bn sin nω0 t
It provides a measure of the similarity between
2 waveforms as the function of search
parameter.
An application of correlation to signal detection
in a radar, where a signal pulse is transmitted in
order to detect a suspect target. If a target is
present, the pulse will be reflected by it. If the
target is not present, there will be no reflection
pulse, just noise. By detecting the presence or
absence of the reflected pulse we confirm the
presence or absence of the target.
In digital communication, the important
thing is that 1s and 0s in the data stream be
distinguishable from each other so the receiver
can reproduce the bit pattern that was
transmitted.
35
•
Auto correlation function of an energy signal
x(t) is
3
R x (x) =
#
x (t + x) x (t) dt
-3
T"3
1
2T
1
= Lt 2T
T"3
01. If x(t) is a voltage waveform, then what are the
units of X(f) ?
ACF of power signals is
R x (x) = Lt
Practice Questions
3
#
x (t) x (t - x) dt =
-3
•
Signals & Systems
T
#
x (t) x (t - x) dt
02. For the signal x(t) shown in figure, find
-T
(a) X(0)
T
#
x (t + x) x (t) dt
(b)
-T
+3
#
X (ω) dω
-3
Properties of ACF:
x(t)
x(t)
1. ACF is an even function of τ i.e.,
Rx(τ) = Rx(-τ)
1
2. ACF at origin indicates either energy (or) power
in the signal
3. Maximum value of ACF occurs at origin i.e.,
2
0
-1
1
2
t
3
|Rx(τ)| ≤ |Rx(0)| ∀ τ
4. Rx(τ) = x(τ) ∗ x(-τ)
5. F.T. of ACF is known as ESD (or) PSD
Properties of F.T
Rx(τ) ↔ Sx(ω) ESD / PSD
6. For an LTI system
Linearity, duality & scaling
Y (ω) = X (ω) H (ω)
|Y(ω)|2 = |X(ω)|2 |H(ω)|2
SY(ω) = SX(ω) |H(ω)|2
x(t)
L .T.I
System h(t)
output S.D = [input S.D] [|H(ω)|2]
03. Find the F.T of the signals
i) x (t) = e -a t [Two sided exponential]
y(t)
ii) x(t) = Sgn(t) [Signum function]
04. The F.T. of a function g(t) is given as
ω2 + 21
. Find g(t)?
G (ω) = 2
ω +9
05. Find the F.T of the signal shown in figure Find
Y(f) at f = 1 ?
y(t)
2
1
-2
-1
0
A
1
2
t
36
06. The magnitude of F.T. X(ω) of a function
in fig (a) the magnitude of F.T Y(ω) of other
function y(t) is shown below in fig (b). The phases
X(ω) and Y(ω) are zero for all ω. The magnitude
and frequency units are identical in both the
figures. The function y(t) can be expressed in
terms of x(t) as ______
X(ω)
1
–100
0
Y(ω)
3
100 ω
-50
0
(a)
2 btl
x
3 2
(b)
3 ]2tg
x
2
(c)
2 ]2tg
x
3
(d)
3 btl
x
2 2
Signals & Systems
11. The F.T of a triangular pulse f(t) shown in figure
e j~ − jωe j~ − 1
is F(ω) =
using this find the F.T of
ω2
the signals shown in figure? f(t)
1
-1
f1(t)
-1/2
1
(a + jt)
2a
(iii) x (t) = 2 2
(a + t )
(iv) x (t) = 1
rt
0
08. Let x(t) = rect( t – ½ )
{where rect(x) = 1 for –1/2 ≤ x ≤ 1/2 } then if
sin rx
sinc(x) = = rx
F.T. of x(t) + x(–t) will be given by ____
i) x(t) = e -3t u(t-1)
ii) z(t) = π [(t – 1) / 2]
iii) y (t) = e -2 t - 2
1/2 t
0
2
x(t)
1
(ii) x (t) =
10. Find the F.T. of the signals
1.5
t
12. If x(t) as shown in Fig. (a) has Fourier transform
X(f), then the Fourier transform of g(t) as shown
in Fig.(b) is
(i) x(t) = 1
Time & frequency shifting
f2(t)
1
50 ω
07. Find F.T. of the following signals:
09. What is the I.F.T. of u(ω) ?
t
0
0
1 Fig. (a)
t
g(t)
1
0
1
2
3
4
5
Fig. (b)
(a) e-j6πfX(f ) – e-j4πfX(–f ) (b) [e-j6πf – e-j4πf]X(f)
(c) [e-j6πf – e-j4πf]X(–f )
(d) e-j6πf X(–f ) – e-j4π f X(f )
13. Find the F.T. of the following signals?
i) cosω0t
ii) sinω0t
iii) e–at sinωctu(t)
iv) A rect(t/T) cosω0t
t
37
14. Find the F.T. of y(t) = Sinc(t) cos10πt
15. i) The inverse F.T of X(4ω+3) in terms of x(t)
is_______
ii) I.F.T of X(ω) = 2πδ(ω) + πδ(ω - 4π)
+ πδ(ω + 4π) is
(a) 1+ cos 4πt
(b) π(1– cos 4 πt)
(c) 2π(1– cos 4 πt)
(d) 2π(1+cos 4 πt)
16. The Fourier spectrum X(f) of a signal x(t) is
shown. The phase angle of x(t) at
t = 1/(8fo) is equal to
X(f)
1
f
fo
(a) –90° (c) –45° (b) 180°
(d) 45°
17. F.T. of x(t) is 2rect(0.25f). Then F.T. of
x(t)cos(2πt) is
(a)
1
- 4
x(t) cos(ωct)
H(ω)
Filter
y(t)
cos(ωCt +θ)
1
(a) x(t) (b) x (t)
2
(c) – x(t) (d) x(t) cos(θ)
Time & Frequency Differentiation
19. A differentiable non constant even function
x(t) has a derivative y(t), and their respective
Fourier Transforms are X(ω) and Y(ω). Which of
the following statements is TRUE ?
(a) X(ω) and Y(ω) are both real.
(b) X(ω) is real Y(ω) is imaginary
(c) X(ω) and Y(ω) are both imaginary
(d) X(ω) is imaginary Y(ω) is real
X(ω)
j π
(b)
2
-1
1
-3
The output, y(t) is equal to
20. For the spectrum X(ω) shown in figure,
d
find
x (t) at t = 0 ?
dt
f
4
0
Signals & Systems
-1
(c)
0
1
4
-3
0
(d) None of these
3
1
ω
-j π
f
3
0
T
f
18. A signal x(t) with bandwidth B is put on a carrier
cos(ωct) with ωc >> B. The modulated signal x(t)
cos(ωct) is then applied to a system shown in
fig with filter frequency response given by
2, − B < ω < B
H (ω) = )
0, all other ω
21. Find the F.T. of te - t ,hence find the transform
4t
of
( 1 + t 2) 2
22. Given x(t)↔X(ω), express the F.T. of the following
signals in terms of X(ω) ?
(i) x1(t) = x(2 – t) + x(– t – 2 )
(ii) x2(t) = x( 3t – 6 )
(iii) x3 (t) =
d2
x (t - 3) dt2
(iv) x4 (t) =
tdx (t)
dt
38
23. Given
Signals & Systems
sin ω t sin ω t
25. The input signal x (t) = πt 1 + πt 2 ω1< ω2
is applied to the system shown in
figure.
2 sin ω
x (t) = *1; t < 1 ) ω
0;
elsewhere
Find the F. T of the following signals?
(a) y1(t)
1
0
−2
2 t
y3(t)
1
(c)
(d)
0
−1
1
2
-ωf
t
1
0
26. Let
ω
a signal whose
X (ω) = δ (ω) + δ (ω − π) + δ (ω − 5) & Let
y5(t)
(f)
y6(t)
1
2
−1
0
t
1
(h)
(g) y7(t)
1
−1
−1
t
y8(t)
1
1
0
4
t
1
−2 −1 0
2 t
−1
y9(t)
(i)
y10(t)
1
0
t −2 −1
Convolution
2
0
x(t)
be
F.T.
is
h(t) = u(t) – u(t–2)
(a) is x(t) periodic? (b) is x(t)∗h(t) periodic?
(c) can the convolution of two aperiodic
signals be periodic?
27. Using convolution property of F.T. find the
convolution of following signals.
(a) y1 (t) = rect (t) ) cos (rt)
(b) y2 (t) = rect (t) ) cos (2rt)
t
(c) y3 (t) = sin c (t) ) sin c b 2 l
(d) y4 (t) = sin c (t) ) e j3rt sin c (t)
28. (a) Find the output of a system having impulse
response h (t) = 8 sin c 68 ]t - 1g@ when the
input applied is x(t) = cosπt
(j)
1
−2
ωf
0
(c) ω2 < ωf
t
−1
(e)
1
find y(t) if
(a) 0 < ωf < ω1
(b) ω1 < ωf < ω2
Sin πt
y4(t)
−1
t
H(ω)
y2(t)
2
(b)
(b) Let g (t) = e - rt , and h(t) is a filter matched
to g(t). If g(t) applied as input to h(t), then
the Fourier transform of the output is
(a) e - rf (b) e - rf /2
(c) e - r f (d) e -2rf
2
1
2
t
24. An L.T.I system is having Impulse response
sin 4t
for which the input applied is
h (t) = rt
x(t) = cos2t + sin 6t, find the output?
2
2
2
39
29. The time function x(t) corresponding to the
Fourier spectrum:
X (f) = e - r (f - 1) is
(a) real and even symmetric
(b) conjugate odd symmetric
(c) conjugate even symmetric (d) real and odd symmetric
2
30. Find the F.T. of
(a) y(t) = x(t)cos ω0t ?
t
Sint Sin b 2 l
(b) x (t) =
rt 2
t
#
31. Find the F.T. of
-3
Signals & Systems
2a
36. For the signal g (t) = a2 + t2 , determine the
essential B.W. B Hz of g(t) such that the energy
contained in the spectral components of g(t) of
frequencies below B Hz is 99% of signal energy
in g(t)?
37. The magnitude of the Fourier transform of a
signal x(t) is shown in figure.
The energy of the signal is
|X(f)|
sin (2rt)
dt
rt
-104
Parseval’s Power theorem
32. Find the energy in the signal
sin at
x (t) = rt
10-6
104
0
f(Hz)
(a) 32 # 10-8
(b) 32 # 10-8
(c) 13 # 10-8
(d) 10-10
Distortionless Transmission
33. Find the energy in the spectrum shown in fig.?
X(ω)
π
38. Consider a transmission system H(ω) with
magnitude and phase response as shown
in figure. If an input signal
x(t) = 2Cos 10πt + Sin 26πt is given to the system
the output will be ________
|H(ω)|
π
2
2
1
-1
-0.5
0
0.5
1
ω
- 40π −20π
34. An input signal x(t) = e -2t u(t) is applied to an ideal
L.P.F with frequency response characteristics
H (ω) = 1; ω < ω c
= 0; ω > ω c
Find ωc, such that energy in the output is half
that of Input energy?
35. Find the value of the integral
+3
# ^ω2 8+ 4h2 dω
-3
20π
ω (rad/sec)
40π
∠H(ω)
π/2
30π
- 30π
ω
- π/2
(a) 4 cos 10πt + sin 26πt
(b) 8 cos 10πt + sin 26πt
(c) 4 cos (10πt – π/6) + sin (26πt – [13π/30])
(d) 8 cos (10πt – π/2) + sin (26πt – π/2)
40
39. For a linear phase channel, what is tp & tg?
40. The system under consideration is an RC LPF
with R = 1 k Ω & C = 1 μF
i) Let H(f) denote the frequency response of RC
LPF. Let f1 be the highest frequency component
H (f1)
such that 0≤|f|≤f1,
$ 0.95
H (0)
then f1 (in Hz) is _____
(a) 327.8
(b) 163.9
(c) 52.2 (d) 104.4
(ii) Let tg(f) denote the group delay of RC LPF and
f2 = 100 Hz, then tg(f2 ) in msec, is ____
(a) 0.717
(b) 7.17
(c) 71.7 (d) 4.505
Signals & Systems
43. Consider an LTI system with magnitude
response
]Z]
] − f
=
H (f) ][1 20 , f # 20
]] 0,
f > 20
\ response Arg[H(f)] = –2f.
and phase
If the input to the system is
π
π
π
x (t) = 8 cos b 20πt + l + 16 sin b 40πt + l + 24 cos b 80πt + l
4
8
16
Then the average power of the output signal
y(t) is ______
44. An analog filter has the magnitude and phase
characteristics shown in figure. If the following
Inputs are applied to this filter, we got the
following steady - state outputs. Tell what kind
of distortion has occurred?
|H(ω)|
2
41. The input to a channel is a bandpass signal. It
1
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
1
y (t) =
cos (100t - 10 -6) cos (106 t - 1.56)
100
The group delay (tg) & the phase delay (tp) in
ω
0
−200
200
∠H(ω)
π
2
seconds, of the channel are
(a) tg = 10−6, tp = 1.56
(b) tg = 1.56, tp = 10−6
−100
(c) tg = 10−8, tp = 1.56×10−6 (d) tg = 10−8, tp = 1.56
42. Suppose a transmission system has the
frequency response as shown in figure. For
what range of frequency there is no distortion?
|H(f)|
1
0
Arg H(f)
20
30
50
f, kHz
f
-900
−
Input
100
ω
π
2
S.S. Output
x1(t)=cos20t+cos 60t
3r
r
y1 (t) = cos b 20t - 10 l + cos b60t - 10 l
x2(t)=cos20t+cos140t
r
r
y2 (t) = cos b 20t - 10 l + cos b140t - 2 l
x3(t)=cos20t+cos220t
r
r
y3 (t) = cos b 20t - 10 l + 2 cos b 220t - 2 l
41
Sampling Theorem
Correlation
45. Find the auto correlation and power in the
signal x(t) = 6 cos(6πt + π/3)
46. Find the ACF of x(t) = e-3t u(t)
1
47. Consider a filter with H (ω) =
and input
1 + jω
-2t
x(t) = e u(t)
(a) Find the ESD of the output?
(b) Show that total energy in the output is
one-third of the input energy?
48. Let g(t) = exp(–8t)u(t)
Define x(t) = convolution of g(t) with it self and
y(τ) = correlation of g(t) with it self
i) The value of x(t) at t = 1/16 is
1
1
(a)
(b)
^8 e h
^16 e h
1
(c)
(d) 1/(16e)
^1 e h
ii) The energy density spectrum at f = 0 is
(a) 8
(b) 1/8
(c) 1/64
(d) 64
iii) y(τ) value at τ = 0, is
1
(a)
8
1
(c)
4
(b)
Signals & Systems
1
16
(d) 16
49. Find the C.C.F of x(t) = e-tu(t) and
y(t) = e-3tu(t) ?
50. The signal x(t) = sinc10t is the input to a system
with frequency response
ω
H (ω) = 3rect b 8π l e -j2~
Find the output energy?
51. Find the Nyquist rate & Nyquist interval for each
of the following signals?
sin 200rt l
(a) x1 (t) = b
rt
2
sin
200
r
t
l
(b) x2 (t) = b
rt
(c) x3(t) = 5cos1000πt cos4000πt
(d) x4 (t) = e -6t u (t) ) sin at
rt
(e) x5 (t) = sin c (100t) + 3 sin c2 (60t)
πt
sin b 2 l + 3
(f) x (t) =
) / δ (t − 10n)
n =-3
b πt l
2
52. Let x(t) be a signal with Nyquist rate ωo.
Determine the Nyquist rate for each of the
following signals.
d
(a) x(t) + x(t–1)
(b)
x (t)
dt
(c) x(3t) (d) x(t) cosωot
53. Two signals x1(t) & x2(t) are band limited to 2 kHz
& 3 kHz respectively, find the Nyquist rate of the
following signals?
(a) x1(2t)
(b) x2(t – 3)
(c) x1(t) + x2(t)
(d) x1(t)x2(t)
(e) x1(t)∗ x2(t)
(f) x1(t) cos(1000πt)
54. A signal x(t) = 100 cos(24π×103t) is ideally
sampled with a sampling period of 50 µsec
and then passed through an ideal low pass
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
42
55. A signal represented by x(t) = 5cos(400πt)
is sampled at a rate of 300Hz. The resulting
samples are passed through an ideal LPF
with cut-off frequency of 150 Hz. Which of the
following will be contained in the output of LPF?
(a) 100 Hz
(b) 100 Hz, 150 Hz
(c) 50 Hz, 100Hz
(d) 20,100,150Hz
56. The frequency spectrum of a signal is shown in
figure, if this signal is ideally sampled at intervals
of 1 msec, then the frequency spectrum of the
sampled signal will be _________
57. A signal x(t) = 6cos10πt is sampled at a rate
of 14 Hz to recover the original signal, cut-off
frequency of the LPF should be _________
(a) 5 < fc < 9
(b) 9
(c) 10
(d) 14
58. The spectrum of a bandlimited signal after
sampling is shown in figure. The value of
sampling interval is
−100
0 100 150
350 400
(a) 1msec
(c) 4msec
U (f)
–1
Signals & Systems
0
1
f (kHz)
(a)
600
f(Hz)
(b) 2msec
(d) 8msec
59. Let x(t) = 2 cos(800πt) + cos(1400πt) and x(t) is
sampled with the rectangular pulse train shown
in fig. The only spectral components (in kHz)
present in the sampled signal in the frequency
range 2.5kHz to 3.5kHz.
p(t)
f
3
(b)
f
–T0
(c)
–T0/6
0
T0/6
T0 = 10–3sec
(a) 2.7, 3.4
(b) 3.3, 3.6
(c) 2.6, 2.7, 3.3, 3.4, 3.6
(d) 2.7, 3.3
f
(d)
f
T0
t
43
60.
(i) The output y(t) of the following system is to be
sampled, so as to reconstruct it from its samples
uniquely. The required minimum sampling
rate is
X(ω)
−1000π
1000π
ω
h(t) =
x(t)↔ X(ω)
sin(1500πt)
πt
y(t)
cos(1000πt)
(a) 1000 samples/s
(b) 1500 samples/s
(c) 2000 samples/s
(d) 3000 samples/s
r
(ii) The signal cos b10rt + 4 l is ideally sampled
at a sampling frequency of 15 Hz. The
sampled signal is passed through a filter
with impulse response b sin ]rtg l cos b 40rt - r l .
rt
2
The filter output is
r
15
(a)
cos b 40rt - 4 l
2
r
15 c sin (rt) m
(b)
cos b10rt + 4 l
2
rt
r
15
(c)
cos b10rt - 4 l
2
r
15 c sin (rt) m
(d)
cos b 40rt - 2 l
2
rt
Signals & Systems
Hilbert Transform & Complex Envelope
61. The complex envelope of the bandpass signal
x (t) = - 2 c
sin (rt/5)
m sin b rt - r l ,
4
r t/ 5
1
centered about f =
Hz, is
2
(a) c
r
sin (rt/5)
m e j 4 rt/5
(b) c
sin (rt/5) -j r4
me
rt/5
(c) 2 c
r
sin (rt/5)
m e j 4 rt/5
(d) 2 c
sin (rt/5) -j r4
me
rt/5
62. A modulated signal is given by
s(t) = e-atcos[(ωc+∆ω)t]u(t), where a, ωc are
positive constants, and ωc >> ∆ω. The complex
envelope of s(t) is given by
(a) exp(-at) exp[j(ωc+ ∆ω)t]u(t)
(b) exp(-at) exp(j∆ωt)u(t)
(c) exp(j∆ωt)u(t)
(d) exp[j(ωc + ∆ω)t]
63. The input 4sinc(2t) is fed to a Hilbert transformer
to obtain y(t), as shown in the figure below:
4sinc(2t)
Hilbert
Transform
y(t)
sin (πx)
Here sin c (x) = πx . The
value (accurate to
3
two decimal places) of # y (t) 2 dt is __________
-3
64. A modulated signal is y(t) = m(t)cos(40000πt),
where the baseband signal m(t) has frequency
components less than 5 kHz only. The minimum
required rate (in kHz) at which y(t) should be
sampled to recover m(t) is ______.
44
65. Statement (I): Aliasing occurs when the
sampling frequency is less than twice the
maximum frequency in the signal.
Statement (II): Aliasing is a reversible process.
66. Statement (I): Sampling in one domain makes
the signal to be periodic in the other domain.
Statement (II): Multiplication in one domain is
the convolution in the other domain.
Signals & Systems
Key for Practice Questions
01. Ans: Volt/Hz
02. (a). Ans: 7,
(b). Ans: 4π
04. Ans: g (t) = δ (t) + 2e -3 t
05. Ans: 0
15.
i. Ans:
06. Ans: (d)
12.Ans: (a)
1 - 34 jt b t l
e x 4
4
ii. Ans: (a)
17. Ans: (b)
18. Ans: (d)
24. Ans: cos(2t)
25. a). Ans:
b). Ans:
2 sin ]ω f tg
,
πt
sin (ω1 t) sin (ω f t)
+
πt
πt
c). Output = Input
26. Ans: a is non - periodic, b is periodic, c may
be periodic.
2
27. a). Ans: y1 (t) = r cos rt ,
b). Ans: y2 (t) = 0
t
c). Ans: Sinc b 2 l
d). Ans:0
28. (a). Ans: cos π (t - 1), (b). Ans: d
29. Ans: (c)
a
32. Ans: r
35.
Ans: π/2
36. Ans: B =
37.
Ans: (a)
2.302
a rad/ sec
38. Ans: (c)
45
40.
Ans: (i). c, (ii). a,
45.
Ans: rxx (x) = 18 cos (6rx) , power = 18
46.
Ans:
5. Laplace Transform
•
e -3 x
6
L.T expresses signals as linear combination
of complex exponentials, which are eigen
48.
Ans: (i). b,
(ii). c,
51.
Ans:
a) 200 Hz,
d) 2a
b) 400 Hz,
e) 120 Hz.
52.
Signals & Systems
functions of D.E which describe continuous –
(iii). b
time L.T.I systems.
c) 5 KHz,
Ans:
a). ω0, b). ω0, c). 3 ω0, d). 3 ω0,
•
The primary role of the L.T in engineering is
the transient & stability analysis of Causal L.T.I
systems.
•
L.T provides a broader characterization of
systems & their interaction with signals than is
55.
possible with F.T.
Ans: (a)
56. Ans: (b)
57.
Ans: (a)
58.
Ans: (c)
•
In addition to its simplicity, many design
techniques in circuits, filters & Control systems
have been developed in L.T. domain.
•
Consider applying an INPUT of the form x(t) = est
(where s = σ +jω then the output is y(t) = estH(s)
where H(s) is transfer function of the system.
60. (ii) Ans: (a)
•
L.T. of a general signal x(t) is
3
X (s) =
#
x (t) e -st dt --------- (1)
-3
•
= F.T {x(t)e−σt}
e-σt may be decaying (or) growing depending
on whether ‘σ’ is +ve (or) –ve.
x (t) ) X (s)
5.1 Region of Convergence (R.O.C) of L.T:
The range of values
of ‘S’ for which eq(1) is
3
satisfied i.e.
R.O.C of L.T.
•
#
|x(t)e-σt|dt < ∞,
is known as
-3
L.T. calculated on the jω-axis (σ = 0) is F.T.
46
L.T. of Standard Signals
1.
Signals & Systems
6.
1
x (t) = e -at u (t), a > 0 ) s + a ; Re ! s + > - a
Note: If the L.T. X(s) of x(t) is rational, then if x(t) is
right sided the ROC is the region in the s-plane
jω
to the right of the rightmost pole and if x(t) is left
1
sided, ROC is left of the left most pole.
σ
-a
0
2.
1
unit step function u (t) ) s ; Re ! s + > 0
5.2 Properties of L.T
t
1.
1
x (t) = - e -at u (- t) ) s + a ; Re ! s + < - a
Linearity:
If x1(t) ↔ X1(s) with ROC = R1
jω
x2(t) ↔ X2(s) with ROC = R2
Then a x1(t) + b x2(t) ↔ aX1(s) + bX2(s)
0
with ROC = R1 ∩ R2
t
-1
-a
σ
2.
Time-shifting:
x(t) ↔ X(s), ROC = R
3.
then x (t - t 0) ) e -st X (s)
1
x (t) = e at u (t) ) - ; Re ! s + > a
s a
0
with ROC = R
jω
3.
1
4.
x(t) ↔ X(s) with ROC = R
a
0
Shift in S-domain
σ
then x (t) e s t ) X (s - s 0)
0
t
with ROC = R + Re(s0)
1
x (t) = - e at u (- t) ) - ; Re ! s + < a
s a
4.
Time-reversal:
x(t) ↔ X(s)
jω
then x(–t) ↔ X(–s), ROC = – R
0
t
5.
a
σ
unit impulse δ(t)↔1; ROC: entire s-plane
5.
Differentiation in time:
x(t) ↔ X(s) with ROC = R
dx (t)
then
) sX (s) with ROC = R
dt
47
UNILATERAL L.T
Signals & Systems
Ans: (a) 5cos (2t+30°)
3
X (s) =
#
j2 + 2
ω 0 = 2, H (jω 0) = −
4 + 10j + 4
x (t) e -st dt
0
Differentiation in time:
d
x (t) * sX (s) - x (0)
dt
d2
x (t) * s2 X (s) - sx (0) - xl (0)
dt2
Example:
Suppose that an LTI system has the following
8
T.F H (s) =
. Compute the system response
s+4
due to following inputs. Identify the steady state & transient solution?
(a) x1(t) = u(t)
y1(t) = x1(t)*h(t)
Y1(s) = X1(s)H(s) =
Take I.L.T.
y1 (t) = 2u (t) - 2e -4t u (t)
Z 1444442444443
Transient
Y2(s) = X2(s)H(s) =
y2 (t) =
1 2 1 2 2
8
=
s + s2
s2 (s + 4) s + 4
1 -4t
1
e u (t) - u (t) + 2tu (t)
2
1444442444443 12444444442444444443
Transient
S.S
(c) x3 (t) = 2 sin 2tu (t) 8 5 8 5 s 16 2
4
8
+ ; 2 E
Y3 (s) = : s + 4 D ; 2 E = s 4 - 2
+
s +4
s +4 s s +4
8 - 4t - 8
16
e
cos 2t + 5 sin 2tH
y3 (t) = >1244424443 124444444444424444444444
43 u (t)
Transient
1 j
=5-5
2
1
1
H (ω 0) =
25 + 25 = 5
−1 5
−
o= π
H (jω 0) = tan -1 e
4
1 5
2
y (t) = 5 # 5 cos (2t + 30 o - 45 o)
=
2 cos (2t - 15 o)
(b) Ans: 2 2 sin (2t)
(b) x2(t) = tu(t)
j2 + 2 j + 1
=
10j
5j
1 1
H (jω 0) = 5 +
5j
=
8
2
2
= s -s 4
+
s (s + 4)
s.s
S.S
Note that the form of the transient remains the
same no matter what the input is !
Example:
For an L.T.I system described by the transfer
s+2
function H (s) = 2
, find the response
s + 5s + 4
due to
(a) 5cos(2t+300)
(b) 10sin(2t+450)
2
Sol: ω0 = 2, |H(jω0)| = 5
π
H (jω 0) = −
4
2
y (t) = 10 # 5 sin (2t + 45 o - 45 o)
y (t) = 2 2 sin (2t)
48
06. Find the L.T of the waveform shown in figure?
Practice Questions
x(t)
01. Find the L.T. of the following signals with R.O.C?
1. x1(t) = e-t u(t) + e-3t u(t)
2. x2(t) = e-2t u(t) + e4t u(-t)
3. x3(t) = e-t u(-t) + e5t u(t)
4. x4(t) = 1 ∀ t
5. x5(t) = sgn(t)
6. x6(t) = e-|t|
02. Consider the signal x(t) = e u(t) + e u(t) & its L.T.
is X(s). What are the constraints placed on the
real & imaginary parts of β if the R.O.C of X(s) is
Re{s} > - 3 ?
–5t
Signals & Systems
–βt
10
-5
0
2
4
t(secs)
07. L.T. of the waveform shown in fig
1^
1 + Ae -s + Be -4s + Ce -6s + De -8sh then D
s2
______
is
1
03. How many possible ROCs are there for the
pole-zero plot shown in fig ?
6
0
jω
is
1 2 3 4
7
5
8
t
-1
-3
-1
1
Fig.
σ
2
(a) –0.5
(c) 0.5
Fig
Time shifting and shift in S - domain
04. Find the I.L.T. of
e -3s
Y (s) =
, σ > -1
(s + 1) (s + 2)
(b) –1.5
(d) 2
08. Let x(t) be a signal that has a rational L.T. with
exactly 2 poles located at s = −1 and s = −3.
If g(t) = e2tx(t) and G(ω) converges, determine
whether g(t) is
(a) Left-sided 05. Consider the signal x(t) = e u(t−1) with L.T. X(s)
−5t
(a) Find X(s) with R.O.C.?
(b) Find the values of ‘A’ &‘t0’ such that the
L.T. G(s) of g(t) = Ae-5tu(-t-t0) has same
algebraic form as X(s). What is the R.O.C
corresponding to G(s)?
(b) right-sided
(c) two-sided (d) finite-duration.
09. Let g(t) = x(t) + αx(–t) where
x(t) = βe–tu(t) and
s
G (s) = 2
; - 1 < Re ! s + < 1
s -1
Find α and β?
49
Differentiation
10. Consider 2 right-sided signals x(t) & y(t) related
through the equation
dy (t)
dx (t)
− 2y (t) + δ (t)&
2x (t)
dt =
dt =
Find X(s) & Y(s) with ROCs?
11. Find the ILT of
(a) X (s) =
4
(s + 2) (s + 1) 3
dd 1 n
(b) X (s) = e-2s ds
(s + 1) 2
Convolution
12. Solve the following equation.
3
y (t) +
#
y (τ) x (t − τ) dτ = x (t) + δ (t) ?
0
13. Consider a signal y(t) = x1(t–2) ∗ x2(–t + 3) where
x1(t) = e-2t u(t) & x2(t) = e-3tu(t).
Find Y(s) with ROC?
14. Find the impulse response of a linear causal
system describe by the equation.
t
dy (t)
4y (t) + 3 # y (x) dx = x (t)
+
dt
-3
Also determine response to an excitation
x(t) = u(t) + δ(t) ?
15. An Input x(t) = exp(–2t) u(t) + δ(t – 6) is applied to
an L.T.I 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)
16. Consider an LTI system with impulse response
h(t) = e−5t u(t). If the output of the system is
y(t) = e−3t u(t) − e−5t u(t) then the input, x(t), is given
by
(a) e−3t u(t)
(b) 2e−3t u(t)
(c) e−5t u(t)
(d) 2e−5t u(t)
Signals & Systems
17. For the interconnected system shown in figure,
the combined Casual impulse response is
x(t)
v ( t ) + v ( t ) = x ( t )
v(t)
y ( t ) + y ( t ) = v ( t )
y(t)
Fig.
(a) e-tu(t)(b) te-tu(t)
(c) δ(t )(d) u(t)
18. The running integrator, given by
t
y (t) =
# x (t') dt'
-3
(a) has no finite singularities in its double sided
Laplace Transform Y(s)
(b) produces a bounded output for every
causal bounded input
(c) produces a bounded output for every
anticausal bounded input
(d) has no finite zeros in its double sided
Laplace Transform Y(s)
19. A causal LTI system has zero initial conditions
and impulse response h(t). Its input x(t) and
output y(t) are related through the linear
constant - coefficient differential equation
d2 y (t)
dy (t)
+ a dt + a2 y (t) = x (t)
2
dt
Let another signal g(t) be defined as
t
g (t) = a2
# h (x) dx + dhd(t t) + ah (t) .
0
If G(s) is the Laplace transform of g(t), then the
number of poles of G(s) is _______.
50
Unilateral L.T
20. A system described by a linear, constant
coefficient, ordinary, first order differential
equation has an exact solution given by y(t) for
t > 0, when the forcing function is x(t) and the
initial condition is y(0). If one wishes to modify
the system so that solution becomes –2y(t) for
t > 0, we need to
(a) change the initial condition to –y(0) and
the forcing function to 2x(t)
(b) change the initial condition to 2y(0) and
the forcing function to –x(t)
(c) change the initial condition to j 2 y (0)
and the forcing function to j 2 x (t)
(d) change the initial condition to –2y(0) and
the forcing function to –2x(t)
21. Find the initial & final values for the following
L.T.?
2s + 5
s2 + 5s + 6
4s + 5
(b) X (s) =
2s + 1
12 (s + 2)
(c) X (s) =
s (s2 + 4)
(a) X (s) =
(d) X (s) = e -s <
-2
F
s (s + 2)
22. A LTI, Causal continuous time system has a
rational transfer function with simple poles at
s = -2 and s = -4 and one of the simple zero at
s = -1. A unit step u(t) is applied as the input of
the system. At steady state, the output has a
constant value of 1. Find the impulse response?
23. Consider a system with transfer function
s-2
.
Find
the
steady-state
H (s) = 2
s + 4s + 4
response when the input applied is 8cos2t?
Signals & Systems
24. An LTI system is having transfer function
s2 + 1
and input x(t) = sin(t+1) is in steady
2
s + 2s + 1
state. The output is sampled at ωs rad/sec to
obtain final output {y(k)} which of the following
is true?
(a) y(•) = 0 for all ωs (b) y(•) ≠ 0 for all ωs
(c) y(•) ≠ 0 for ωs > 2 but zero for ωs < 2
(d) y(•) = 0 for ωs >2 but nonzero for ωs < 2
25.
(i) What is the output as t → ∞ for a system that
2
has T.F., G (s) = 2
when subjected to a
s -s-2
step input?
(a) –1
(b) 1
(c) 2
(d) unbounded
(ii) The transfer function of a causal LTI system
is H(s) = 1/s. If the input to the system is
x(t) = [sin(t)/πt]u(t); where u(t) is a unit step
function. The system output y(t) as t→∞ is ____
26. Let a signal a1sin(ω1t + φ1) be applied to a stable
LTI system. Let the corresponding steady state
output be represented as a2F(ω2t + φ2). Then
which of the following statement is TRUE?
(a) F is not necessarily a “sine” or “cosine”
function but must be peroidic & ω1 = ω2
(b) F must be “sine” or “cosine” with a1= a2
(c) F must be “sine”, ω1 = ω2, a1 ≠ a2
(d) F must be “sine” (or) “cosine” functions
with ω1 = ω2.
51
Causality & Stability
27. Consider an LTI system for which we are given
the following information
s+2
X (s) = − and x(t) = 0, t > 0 and output is
s 2
y (t) = −
2 2t − + 1 -t
e u ( t)
e u (t)
3
3
(a) Find T.F & R.O.C.?
(b) Find the output if the input is
x(t) = e ∀ t using part (a)?
3t
28. Consider an LTI system with input x(t) and
Signals & Systems
31. The Laplace transform of a causal signal
s+2
Y (s) = + . The value of the signal y(t) at
s 6
t = 0.1 sec is _______ Key for Practice Questions
1
1
01. (1) = + + + , σ > − 1
s 1 s 3
(2) =
1 − 1 −
2<σ<4
s+2 s−4
(3) No ROC, No L.T
(4) No ROC, No L.T
(5) No ROC, No L.T
output y(t) related as
dy (t)
d2 x (t) dx (t)
3y (t) =
+
+ dt - 2x (t)
dt
dt2
Find the T.F. of inverse system. Does a stable &
02. Ans: Re[β] = 3, Img[β] any value
Causal inverse system exist?
04. Ans: y (t) = e - (t - 3) u (t - 3) - e -2 (t - 3) u (t - 3)
29. 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 jω 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 jω
axis.
30. The output of a causal all-pass system is
y(t) = e-2tu(t), for which the system function is
s-1
H (s) = s + 1 . Then the corresponding stable
input signal is
2
1
(a) − e t u (− t) + e -2t u (t)
3
3
2 t
1 - 2t
(b) e u (t) + e u (t)
3
3
1
2
-2t
(c) − e u (t) + e t u (− t)
3
3
1
2
(d) e t u (- t) - e -2t u (t) 3
3
e - ( s + 5)
05. Ans: s + 5 , σ > −5
A = −1, to = −1
06. Ans: X (s) =
5 - 5e -2s e -2s 5e -4s
15. s + s
2
2
s
s
07. Ans: (a) 08.
Ans: (c)
1
09. Ans: a = - 1, b =
2
10. Ans: X (s) =
s
,v > 0
s2 + 4
Y (s) =
2
,v > 0
s2 + 4
12. Ans: y(t) = δ(t)
13. Ans: Y (s) =
e -5s
−2 < σ < 3
(s + 2) (3 - s)
14. Ans: h (t) =
- 1 -t
3
e u (t) - e -3t u (t)
2
2
15. Ans: (d)
17. Ans: (b)
22. Ans: h(t) = −4e-2tu(t) + 12e-4tu(t)
52
Signals & Systems
23. Ans: y(t) = 2.8288 cos(2t + 45°)
24.
Ans: (a)
27.
Ans:
(a) H (s) =
6. DTFT
•
The DTFT describes the spectrum of discrete
signals & formalizes that discrete signals have
s
, σ > −1
(s + 1) (s + 2)
(b) y (t) = 3 e3t
20
periodic spectra. The frequency range for a
discrete signal is unique over ( –π, +π) (or)(0, 2π)
3
X (e j~) =
/ x (n) e-j~n
n =-3
28. Ans:
s+3
, it can’t be both stable
H inv (s) =
(s + 2) (s - 1)
and causal.
•
x (n) =
1
2π
#
X (e j~) e j~n dω
< 2r >
X(ejω) is decomposition of x(n) into its frequency
components.
6.1 Convergence of DTFT
A sufficient condition for existence of DTFT is
+3
/
x (n) < 3
n =-3
•
Some sequences are not absolutely summable,
but they are square summable.
•
There are signals that are neither absolutely
summable nor have finite energy, but still have
DTFT.
1
anu(n), |a| < 1 ↔ 1 ae -j~
δ(n) ↔ 1
Periodicity property:
X(e j(ω+2π)) = X(ejω)
Time – Shift:
x(n) ↔ X(ejω)
then x(n-n0) ↔ e -j~n X(ejω)
0
Frequency – shift:
x(n) ↔ X(ejω)
then x (n) e j~ n * X (e j (~ - ~ ))
0
0
53
Signals & Systems
6.2 Oversampling and Sampling Rate Conversion
the same signal sampled at NS Hz extends only to
In practice, different parts of a DSP system are often
B/NS Hz, and the spectrum of the signal sampled
designed to operate at different sampling rates
at S/M Hz extends farther out to BM/S Hz. After an
because of the advantages it offers. For example,
analog signal is first sampled, all subsequent sampling
oversampling the analog signal prior to digital
rate changes are typically made by manipulating
processing can reduce the errors (Sinc distortion)
the signal samples (and not re-sampling the analog
caused by zero-order-hold sampling devices. Since
signal). The key to the process lies in interpolation
real-time digital filters must complete all algorithmic
and decimation.
operations in one sampling interval, oversampling
can impose an added computational burden.
It is for this reason that the sampling rate is often
reduced
(by
decimation)
before
performing
DSP operations and increased (by interpolation)
before reconstruction. Oversampling prior to signal
reconstruction allows us to relax the stringent
requirements for the design of reconstruction filters.
Zero Interpolation and Spectrum Compression:
A property of the DTFT that forms the basis for signal
interpolation and sampling rate conversion is that
M fold zero interpolation of a discrete-time signal
x[n] leads to an M-fold spectrum compression and
replication, as illustrated in Figure (ii).
Figure (i) shows the spectrum of sampled
signals obtained from a band-limited analog signal
whose spectrum is X(f) at a sampling rate S, a higher
rate NS, and a lower rate S/M.
X(f)
B
SX(f)
f
B
S
f
nonzero only if n = kN, k = 0, ±1,± 2, ......(i.e., if n is an
integer multiple of N). The DTFT Yp(F) of y[n] = y[kN]
may be expressed as
Sampling rate =S
B
Sampling rate =NS
NSX(f)
The zero-interpolated signal y[n] = x[n/N] is
f
S
Sampling rate =S/M
SX(f)M
B
S
=
/
/ y]kNge
y 5n? e
=
Yp ]F g
=
3
3
-j2rnF
n =- 3
-j2rkNF
k =- 3
/ x]k ge
3
-j2rkNF
k =- 3
= X p ]NFg
f
Figure (i) Spectra of a signal
sampled at three sampling rates
x[n] Discrete-time signal
X[F] Spectrum of
discrete-time signal
n
1234
The spectrum of the oversampled signal shows
0.5
y[n] Fourfold zero interpolation
a gain of N but covers a smaller fraction of the
principal period. The spectrum of a signal sampled
at the lower rate S/M is a stretched version with
a gain of 1/M In terms of the digital frequency F,
the period of all three sampled versions is unity.
One period of the spectrum of the signal sampled
at S Hz extends to B/S, whereas the spectrum of
n
1234
1
F
Y[F] Fourfold spectrum compression
0.5/4
Figure (ii) Zero interpolation of a signal
leads to spectrum compression
0.5
1
F
54
Signals & Systems
This describes Yp(F) as a scaled (compressed)
Practice Questions
version of the periodic spectrum Xp(F) and leads to
N-folds spectrum replication. The spectrum of the
interpolated signal shows N compressed images in
the principal period |F| ≤ 0.5 with the central image
occupying the frequency range |F| ≤ 0.5/N. This is
exactly analogous to the Fourier series result where
spectrum zero interpolation produced replication
(compression). The spectrum of the interpolated
signal y(n) shows N compressed images per period
1
centered at F = N
F
1
x (Mn) * Yp ]F g = M X p b M l
The factor 1/M ensures that we satisfy Parseval’s
01. Determine whether or not the DT systems with
these frequency response are causal?
7ω
sin b 2 l
(a) H (ω) =
ω
sin b 2 l
3ω
sin b 2 l
(b) H (ω) =
e -j~
ω
sin b 2 l
(c) H(ω) = e –j3ω + e +j2ω
02. (a) Let x(n) = (1/2)n u(n), y(n) = x2(n) and
Y(e jω) be the F.T of y(n). Then Y(ej0) is
(b) Given X(ejω) = cos3(3ω), then find the sum
S=
relation. If the spectrum of the original signal x[n]
extends to |F|≤ 0.5/M in the central period, the
spectrum of the decimated signal extends over
|F| ≤ 0.5,
3
/ (- 1) n x (n) n =-3
(c) What is the d.c & high - frequency gain of
the filter described by h(n) = {1, 2, 3, 4}
and there is no aliasing or overlap
between the spectral images.
03. (i)
Find the inverse DTFT of
X(ejω) = 1 + 2 cosω + 3 cos2ω
(ii) The DTFT of x(n) = 2δ[n+3] − 3δ[n−3],
can be expressed in the form
X(ejω) = asin(bω) + cejdω Find a, b, c & d.
04. Find the signal corresponding to the spectrum
shown in figure?
Y (ejω)
1
-3π/4 -π/4
π/4
3π/4
ω
05. (a) Let h(n) is the impulse response of ideal L.P.F
with cutoff frequency ωc, what type of filter
has unit impulse response as
(
g n) = ]- 1gn h (n)
55
Signals & Systems
Convolution
(b) A discrete system with input x(n) & output
y(n) are related as
y(n) = x(n) + (−1)n x(n)
If the input spectrum X(ejω) is shown below
the o/p spectrum values at ω = 0 & ω = π
are
X(ejω)
−π −π/2
11. An L.T.I system is having impulse response
]Z] 4 2 ; n = 2, - 2
]]
h (n) = [] - 2 2 ; n = 1, - 1
]]
] 0 ; elsewhere
\
Find the output when input applied is
1
0
π/2 π
ω
x(n) = ejnπ/4 ?
Scaling & Frequency Differentiation
1
06. Find the I.F.T of Y (e j~) =
?
1
1 - e -j10~
2
07. Given the signal x (n) = %1, 2, 3- , 2, 1, 0 / .
rn
rn
10. Consider x (n) = sin b 8 l - 2 cos b 4 l
Find the output if the impulse response is
rn
sin b 6 l
h (n) =
rn
12. For each of the following pair of signals
determine whether or not the system is LTI if
such system exists find the frequency response
n
1 ln
=
b 1 l u (n)
(a) x1 (n) b=
u
(
n
),
y
1 (n)
2
4
jnr/4
=
(b) x2 (n) e=
,
y2 (n) 0.5e jnr/4
Then Fourier transform of x[2n] is
sin (nπ/4)
,
nπ
(c) x3 (n) =
y3 (n) =
sin (nπ/2)
nπ
(a) 3 + 2cos 2ω + 4 cosω
13. Design a 3 point FIR filter with impulse response
h (n) = {a, b, a} and the amplitude response
(b) 3 + 2cosω
(c) 3 + 2cos2ω -
(d) 3 + 2cosω + 2cos(ω/2)
08. The spectrum of signal x[n] is X(F) = 2tri(5F).
Sketch X(F) and the spectra of the following
signals and explain how they are related to
X(F).
D.C gain of the filter?
14. Consider
the
system
described
by
the
equation y(n) = ay(n–1) + bx(n) + x(n–1), where
‘a’ & ‘b’ such that
ii. d[n] = x[2n]
x [n];
0;
frequency f = 1/8 with unity gain. What is the
‘a’ & ‘b’ are real, find the relation between
i. y[n] = x[n/2]
iii. g (n) = )
blocks the frequency f = 1/3 & passes the
|H(ejω)| = 1∀ω ?
even "n"
odd "n"
09. Find the F.T of x (n) = ne
jnr
8
15. A filter is described by
a n - 3 u (n - 3)
y(n) = x(n) + αx(n–1)+ βx(n–2). Then the values of
α & β, such that the input x(n) = 1 + 4 cos (nπ)
results in the output y(n) = 4, are
(a) α = 2, β = 1
(b) α = 1, β = 2
(c) α = β = 1
(d) α = 1, β = ± 2
56
16. An input x(n) with length 3 is applied to a LTI
system having an impulse response h(n) of
length 5, and Y(ω) is the DTFT of the output y(n)
of the system. If |h(n)| ≤ L & |x(n)|≤ B∀n, the
maximum value of Y(0) can be ….
(a) 15 LB
(b) 12 LB
(c) 8 LB
(d) 7 LB
17. An L.T.I filter is described by the difference
equation y(n) = x(n) + 2x(n-1) + x(n-2)
(a) Obtain the magnitude & phase response?
(b) Find the o/p when the input is
rn r
x (n) = 10 + 4 cos : 2 + 4 D ?
18. The impulse response of a causal linear phase
discrete time system of five samples is given
by h(0) = 2, h(1)= -3, h(2) = 0, h(3) = 3 and
h(4)= k. The value of k and slope of the phase
curve are respectively
(a) 2, –2
(b) –2, –2
(c) 0, 2
(d) –3, 2
Signals & Systems
21. Let h[n] be the impulse response of a discretetime linear time invariant (LTI) filter. The impulse
1
1
response is given by h (0) = ; h 51? = ;
3
3
1
h 52? = ; and h[n] = 0 for n < 0 and n > 2.
3
Let H (ω) be the Discrete-Time Fourier transform
(DTFT) of h[n], where ω is the normalized angular
frequency in radians. Given that H(ω0)= 0 and
0 < ω0 < π, the value of ω0 (in radians) is equal to
______.
22. Find the energy in the signal
sin ω n
x (n) = πnc
23. Find the value of
x(n)
2
1
-3 -2
-1 0
1 2 3 4 5
Fig.
(a) X(ej0)
(b) X(ejπ)
r
# X (e j~) dω
(c)
-r
#
(e)
2
X (e j~) dω
-r
r
(f)
#
-r
(g)
r
(d)
# X (e j~) e j2~ dω
-r
r
20. A signal x(n) is defined by
x(n) = 1, for n = ±1, 3 & 5
= 2, for n = 0 & 4
= 0, elsewhere
Its phase spectrum at ω = 0.25π is
(a) 0.25π
(b) – 0.5π
(c) 0.5π
(d) π
n
6 7
-1
rn r
(b) cos : 2 - 8 D
(d)
nr
nr
sin b 4 l sin b 3 l
5rn
2rn
24. For the signal shown in fig., find the following
quantities without calculating DTFT?
(a) cos : r2n + r4 D
rn r
2 cos : 2 + 3 D
/
n =-3
19. The frequency response of a discrete LTI system
is H(ejω) = e -jω/4, -π<ω≤ π.
Then the output of the system due to the input,
5rn
x 5n? = cos : 2 D is
3rn r
(c) cos : 2 + 6 D
+3
2
d
X (e j~) dω
dω
X (e j~)
57
25. Given h(n) = [1, 2, 2], f(n) is obtained by
convolving h(n) with itself and g(n) by
correlating h(n) with itself. Which one of the
following statements is TRUE?
(a) f(n) is causal and its maximum value is 9
(b) f(n) is non-causal
(c) g(n) is causal and its maximum value is 9
(d) g(n) is non causal and maximum value is 9
26. A continuous time signal x(t) is to be filtered to
remove frequency component in the range
5kHz ≤ f ≤ 10 kHz. The maximum frequency
present in x(t) is 20 kHz. Find the minimum
sampling frequency & find frequency response
of ideal digital filter that will remove the desired
frequencies from x(t)?
rn
27. The sequence x 5n? = cos : 4 D was obtained
by sampling a continuous signal, x(t) = cos(Ω0t)
at a sampling rate of 1000 Hz. The two possible
values of Ω0 that have resulted in the sequence
x[n] are respectively
(a) 250π and 2250π
(b) 125π and 2250π
(c) 250π and 1125π
(d) 125π and 1125π
Signals & Systems
Key for Practice Questions
01. (a). Ans: Non causal
(b). Ans: causal
(c). Ans: Non causal
4
02. (a). Ans: 3 (b).Ans: -1
(c). Ans: DC gain = 10, HF gain = -2
nr
04. Ans: y (n) = 2 cos b 2 l .
nr
sin b 4 l
nr
05. (a). Ans: Ideal High pass filter
(b). Ans: 1
07. Ans: (b)
nr
10. Ans: sin b 8 l
11. Ans: y (n) = - 4e
jnr
4
13. Ans: DC gain =
3
1+ 2
14. Ans: b = -a
15. Ans: (a)
16. Ans: (a)
28.
A continuous time signal x(t) is sampled at
fs = 9kHz is passed through a digital filter with
17. Ans: (b)
impulse response h(n) = 0.5[δ[n] + δ[n-1]].
Find the 3 dB cut-off frequency of the analog
filter?
18. Ans: (b)
19. Ans: (b)
23. Ans:
1
40
25. Ans: (d)
26. Ans: fs = 40 kHz,
27. Ans: (a)
π
π
#ω#
4
2
58
7. Z-transform
Signals & Systems
Z.T. of standard Signals:
a n u (n), 0 < a < 1 *
7.1 INTRODUCTION TO Z-TRANSFORM
•
Discrete-time counterpart of L.T. is Z.T.
•
For a D.T.L.T.I system with impulse response h(n),
1
z
or - ; z > a
z a
1 - az -1
the response y(n) of the system to a complex
exponential Input of the form zn is y(n) = znH(z)
where H(z) is known as transfer function of the
0
n
1 2 3 4
Im{z}
system.
Z.T. of a general D.T. signal x(n) is
X (z) =
3
/ x (n) z-n ------------ (1)
n =- 3
where z = rejω → Complex Variable
z
x (n)
a
X (z)
1
Re{z}
X(z) = F{x(n)r-n}
•
Z.T. calculated on the unit circle is D.T.F.T.
- a n u (- n - 1) *
Im{z}
z-plane
1
z
or - ; z < a
z a
1 - az -1
−4 −3 −2 −1
n
1 Re{z}
Unit circle
•
The range of values of ‘z’ for which eq (1) is
Im
defined
=
3
/
n =- 3
•
x (n) r -n < 3G is R.O.C. of Z.T.
0
a
Re
DTFT is defined only for stable signals where as
Z.T. is defined for unstable signals also.
δ(n)↔1; ROC : entire z-plane
•
The primary role of Z.T. in engineering are the
study of system characteristics & the derivation
of computational structures for implementing
discrete systems on computers.
u(n) ↔ 1/1-z-1 ; |z| > 1
59
Note:
1.
2.
3.
4.
For a finite length signal, ROC is entire
z-plane, except perhaps for z = 0 and/or
z=∞
For a right-sided signal ROC is outside
a circle whose radius is largest pole in
magnitude.
For a left sided signal, ROC is inside a
circle whose radius is smallest pole in
magnitude.
For a two-sided signal, ROC is annulus
bounded by largest & smallest pole
radius.
Signals & Systems
Practice Questions
01. Let x(n) = (-1)n u(n) + αn u(-n-n0). Determine the
constraints on “α” & “n0”,
given that the ROC of X(z) is 1<|z|<2
02. Find the ROC of the following signals without
finding Z.T.
(a) x1 (n) = %1, 2, 3- , - 1 /
(b) x2(n) = (1/2)n[u(n) - u(n-10)]
(c) x3(n) = {(1/2)n + (3/4)n}u(n-10)
(d) x4 5n? = (1 3) n - (1 2) n u 5n?
03. Given x(n) = anu(n) – b2nu(–n– 1). What condition
must hold on ‘a’ and ‘b’ for the
z – transform to exist?
(a) |a| < |b2|
(b)|a| > |b2|
(c) |a2| < |b|
(d) |a2| > |b|
04. i) The R.O.C for Z – transform of the signal
x(n) = an u(n) + bn u(n) + cn u(-n - 1)
fora<b<c is
(a)a<z<c
(b)b<z<c
(c)z<c
(d)z>a
ii) Consider the signal
x[n] = αn u[n] + βn u[–n–1]. Find its Z-transform X(z). Will X(z) represent a valid
transform for the following cases?
(a) α > β
(b) α < β
(c) α = β
05. The Z transform of a sequence x(n) is given by
X (z) =
1
1 - z -1
, z >b2l
1 l -2
b
1 4 z
The value of x(n) at n = 2 is given by
1
1
(a)
(b) 2
2
(c)
1
4
1
(d) 4
60
Signals & Systems
06. The Z.T. X(z) of a real & right-sided sequence
Im(z)
x(n) has exactly 2 poles and one of them is at
z = ejπ/2 and there are 2 zeros at the origin.
If X(1)=1, which one of the following is TRUE ?
(a) X (z) =
1
2z2
; < z <1
(z - 1) 2 + 2 2
(b) X (z) =
2z2 z
1
; >
2
z +1
(c) X (z) =
2z2
; z >1
(z - 1) 2 + 2
(d) X (z) =
2z2
; z >1
z +1
×
0.5 ×
–0.5
×
0.5
0.5 ×
Re(z)
2
(a) h(n) is real for all n
(b) h(n) is purely imaginary for all n
2
(c) h(n) is real for only even n
(d) h(n) is purely imaginary for odd n
07. The discrete-time signal
x 5n? * X (z) =
3
/
n=0
3 n 2n
z , where ↔ denotes a
2+n
transform-pair relationship, is orthogonal to the
signal
n
(a) y1 5n? * Y1 (z) = / 3n = 0 b 32 l z-n
(b) y2 5n? * Y2 (z) = /
3
^ n
n=0
5 - nh z
- (2n + 1)
(c) y3 5n? * Y3 (z) = / 3n =- 3 2- n z-n
(d) y4[n] ↔ Y4(z) = 2z-4 + 3 z-2 + 1
Time shifting, scaling & differentiation
1 n
10. Find the Z.T. of y (n) = b - 3 l u 6- n - 2@
11. Consider the signal
1; 0 # n # 5
x (n) = )
0; elsewhere
Let g(n) = x(n) - x(n-1). Find G(z) with ROC?
12. Given X ] z g =
z3 − 2z
and x(n) is left-sided, find
z−2
x(n)?
08. Let H1(z) = (1–pz–1)–1, H2(z) = (1–qz–1)–1,
H(z) = H1(z) + rH2(z). The quantities p, q, r are
1
1
real numbers. Consider, p = , q = - , r < 1.
2
4
If the zero H(z) lies on the unit circle, then
r = ______
13. Consider a causal and stable LTI system
with rational transfer function H(z), whose
corresponding impulse response begins at
5
n = 0. Further more, H(1) =
.
4
−
The poles of H(z) are Pk = 1 exp c j (2k 1) π m
2
4
for k = 1, 2, 3, 4. The zeros of H(z) are all at
z = 0. Let g(n) = jnh(n). The value of g(8) equals
09. The pole-zero diagram of a causal and stable
discrete-time system is shown in the figure. The
zero at the origin has multiplicity 4. The impulse
response of the system is h[n]. If h[0] = 1, we
can conclude.
________ . (Give the answer up to three decimal
places).
14. Find the Z-transform & ROC of
5 n
10 n
x (n) = b 4 l u (n) + b 7 l u (- n)
61
Convolution
15. A sequence x(n)↔X(z) = z4 + z2 - 2z + 2 - 3z-4
is applied as an input to a LTI system
with
impulse response h(n) = 2δ(n-3). The output at
n = 4 is__________
16. Consider a signal
Where
Find
y(n) = x1(n + 3) ∗ x2( - n + 1)
x1(n) = (1/2)nu(n) and
x2(n) = (1/3)nu(n)
Y(z) with ROC?
17. Find the response of the system
y(n) = 5/6 y(n-1) - 1/6 y(n-2) + x(n) to the Input
signal, x(n) = δ(n) - 1/3δ(n-1)
18. G(z) = αz-1 + βz-3 represents a digital LPF with
linear phase if and only if
(a) α = β
(b) α = β1/3
(c) α = -β
(d) α = -β1/3
19. A system with T.F. H(Z) has I.R. h(n) defined as
h(2) = 1, h(3) = –1 and h(k) = 0
otherwise, consider the following statements
S1: H(Z) is a LPF
S2: H(Z) is a FIR filter
(a) only S2 is true
(b) both are false
(c) both are true and S2 is the reason for S1
(d) both are true but S2 is not a reason for S1
20. The following is known about a D.T.L.T.I system
with Input x(n) & output y(n)
1. If x(n) = (-2)n ∀n, then y(n) = 0∀n
2. If x(n) = (1/2)nu(n)∀n,
then y(n) = δ(n) + a(1/4)nu(n) ∀n
Where ‘a’ is a constant.
(a) Find the value of ‘a’
(b) Find the response y(n) if the Input is
x(n) = 1 ∀n.
Signals & Systems
21. Let y(n) denote the convolution of h(n) &
1 n
g(n) where h (n) = b 2 l u (n) & g(n) is a causal
sequence. If y(0) = 1 & y (1) = 1 , then g(1)
2
equals
1
(a) 0
(b)
2
3
(c) 1
(d)
2
22. An analog signal is sampled at 9 kHz. The
sequence so obtained is filtered by an FIR
filter with T.F. H(z) = 1 – z-6. One of the analog
frequencies for which the magnitude response
of the filter is zero is
(a) 0.75 kHz
(b) 1 kHz
(c) 1.5 kHz
(d) 2 kHz
Initial & final value
0.5
. It is given that the ROC
1 - 2z -1
of X(z) includes the unit circle. Then x(0) is
_________
23. Given X (z) =
1; n even
24. Apply F.V.T. for x (n) = )
.
0; n odd
Assume the signal is Causal?
25. A digital filter has an impulse response
δ (n) + δ (n − 1) + δ (n − 2)
h (n) =
10
(a) How many finite poles & zeros are there
in its transfer function?
(b) If the excitation is unit step sequence,
what is final value of output?
26. An input signal x(t) = 2 + 5sin(100πt)u(t) is
sampled
with
400
and
Hz
a
sampling
applied
to
frequency
the
of
system
whose transfer function is represented by
1 1 − z -N
H (z) = N ; − -1 E
1 z
where N represents the
number of samples per cycle. The output y(n)
of the system under steady state is ________.
(a) 0
(b) 1
(c) 2
(d) 5
62
Signals & Systems
1 n
27. If h 5n? = Aδ 5n? + b 3 l u (n) is the unit sample
31. The discrete-time transfer function
response of a LSI system and S[n] is the step
1 - 2z -1
is
1 - 0.5z -1
(a) non-minimum phase and unstable
response, then the value of A that will make
(b) minimum phase and unstable
steady-state step response as zero is
(c) minimum phase and stable
(a) – 1
(b) 0
32. Consider the system with transfer function,
(c) – 3/2 H (z) =
(d) – 2/3
Causality & Stability
28. A casual LTI system is described by the
D.E 2y(n) = αy[n–2] – 2x[n] + βx[n–1]
The system is stable only if
(a) |α|= 2, |β| < 2
(b) |α|> 2, |β| > 2
(c)|α|< 2, any β
(d) |β|< 2, any α
29. Consider an LTI system whose pole-zero pattern
is shown in figure
Im {z}
X
−3
X
0
−0.5 1
X
2
Re {z}
(a) Find the ROC of system function, if it is
known to be stable?
(b) Is it possible for the given pole-zero plot to
be a causal & stable system?
(c) How many possible ROC’s are there?
30. The impulse response h(n) of a LTI system is real.
The transfer function H(z) of the system has only
one pole and it is at z = 4/3. The zeros of H(z)
are non-real and located at |z| = 3/4. The
system is
(a) stable & causal
(b) unstable & anticausal
(c) unstable & causal
(d) stable & anticausal
(d) non-minimum phase and stable
z -1
Then the corresponding stable
1 - 2z -1
impulse response is
(a) -0.5 δ(n) - 0.5 (2)n u[-n - 1]
(b) 2n-1 u[n - 1]
(c) 0.5 δ(n) + 0.5 (2)n-1 u[n - 1]
(d) 0.5 δ(n) + 2n u[-n - 1]
33. Suppose x[n] is an absolutely summable
discrete-time signal. Its z-transform is a rational
function with two poles and two zeros. The
poles are at z = ! 2j . Which one of the following
statements is TRUE for the signal x[n]?
(a) It is a finite duration signal.
(b) It is a causal signal.
(c) It is a non-causal signal.
(d) It is a periodic signal.
34. Assertion (A): A linear time-invariant discretetime system having the system function
z
H]zg =
1 is a stable system.
z+
2
Reason (R): The pole of H(z) is in the left-half
plane for a stable system.
63
35. Consider the following statements regarding a
linear discrete - time system
H(z) =
z2 + 1
(z + 0.5) (z - 0.5)
Signals & Systems
Realization Structures
P0 + P1 z -1 + P3 z -3
using
1 + d 3 z -3
DFI and DFII realizations of IIR the number
39. A DF having T.F. H (z) =
1.
The system is stable.
of delay units required in DFI and DFII are,
2.
The initial value h(0) of the impulse
respectively …….
(a) 6 & 6
(c) 3 & 3
response is - 4.
3.
The steady-state output is zero for a
sinusoidal discrete time input of frequency
equal
to
one-fourth
the
sampling
frequency.
Which of these statements are correct?
(a) 1, 2 and 3
(b) 1 and 2
(c) 1 and 3
(d) 2 and 3
36. Assertion (A): The stability of the system is
assured if the Region of Convergence (ROC)
includes the unit circle in the Z-plane.
Reason (R): For a causal stable system all the
poles should be outside the unit circle in the
Z-plane.
37. Statement (I): The system function
z3 − 2z2 + z
H]zg =
1
1 is not causal.
z2 + z +
4
8
Statement (II): If the numerator of H(z) is of
lower order than the denominator, the system
may be causal.
38. Statement (I):
Z-transform approach is used to analyze the
discrete time systems and is also called as pulse
transfer function approach.
Statement (II):
The sampled signal is assumed to be a train of
impulses whose strengths, or areas, are equal
to the continuous time signal at the sampling
instants.
(b) 6 & 3
(d) 3 & 2
40. 2 systems H1(z) & H2(z) are connected in
cascaded 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
y(n)
x(n)
1 − 0.4 z −1
H1 ( z ) =
1 − 0.6 z −1
H2(z)
41. A direct form implementation of an LTI system
1
with H (z) =
is shown in fig. the
1 - 0.7z -1 + 0.13z -2
values of a0, a1 & a2 are respectively _________
Output
Input
a0
a1
a2
(a) 1.0, 0.7 and - 0.13
(b) -0.13, 0.7 and 1.0
(c) 1.0, -0.7 and 0.13
(d) 0.13, -0.7 and 1.0
z−1
z−1
64
42. In the IIR filter shown below, a is a variable gain.
For which of the following cases, the system will
transit from stable to unstable condition?
x(n)
Σ
Signals & Systems
45. The nature of the above filter is
(a) Band pass
(b) All pass
(c) Low pass
(d) High pass
46. For the causal filter structure shown in figure,
the range of k for system stability is ____
y(n)
x(n) +
a
+
z -1
Fig.
43. Consider the causal digital filter structure shown
in fig. For what values of k the system is stable?
+
∑
+
y[n]
− k/3
1
(a) :- 1, 2 D 1
(b) : 2 , 1D
(c) 6- 1, 1@ 1
(d) : 2 , 2D
47. Consider an All-pass system shown in figure.
z -1 − 0.54
H (z) =
1 − 0.54z -1
x(n)
d
y(n)
Common Data for Questions 44 & 45
A digital filter realization is shown:
-1
z -1
Y
-1
Y (z)
of this filter is
X (z)
3 + z -1
(b)
1 + 3z -1
2z - 1
(d)
z+2
44. The transfer function H (z) =
(a)
0.2 + z -1
1 + 2z -1
(c)
2z + 1
z+2
z −1
c
Find b, c & d such that SFG is a realisation
of H(z)
− k/4
3
y(n)
k
b
z–1
X
Z-1+ Z-2
(a) 0.1 < a < 0.5
(b) 0.5 < a < 1.5
(c) 1.5 < a < 2.5
(d) 2 < a < ∞
x[n]
+
65
Signals & Systems
8. Digital Filter Design
Key for Practice Questions
01. Ans: α = ±2, n0 is any value
In the design of frequency-selective filters, the
desired filter characteristic are specified in the
frequency domain in terms of the desired magnitude
and phase response of the filter. In the filter design
process, we determine the coefficients of a causal
FIR or IIR filter that closely approximates the desired
frequency response specifications.
02. (a). Ans: 0<|z|<∞
(b). Ans: |z|>0
(c). Ans: |z| > 3/4
(d). Ans:
1
< z <3
2
03. Ans: (a)
05. Ans: (c)
10. Ans: Y (z) =
In practice FIR filters are employed where there is a
requirement for linear phase characteristic with the
pass band of the filter.
06. Ans: (d)
3z
z <1
1 -1 ,
3
1+ z
3
An IIR filter has lower side lobes in the stop band than
an FIR filter having same number of parameters.
11. Ans: G(z) = 1-z-6, |z|>0
Design of IIR Filters from Analog Filters:
12. Ans: x(n) = δ(n+2) + 2δ(n+1) - 2(2)n u(-n-1)
15.
Ans: 0
16.
Ans: Y (z) =
z -2
b1 - 1 z -1 l b1 - 1 z l
2
3
20. (a). Ans: a = -
9
8
(b). Ans: y (n) =
-1
4
21. Ans: (a)
23.
Ans: 0
24. Ans:
29. (a). Ans: 0.5 <|z|< 2
(b). Ans: No
(c). |z|< 0.5,|z|>3,
1
2
To convert analog filter to digital filter the following
properties are desirable.
1. The jΩ axis in the s-plane should map into the
unit circle in the z-plane. Thus there will be a
direct relationship between the two frequency
variables in the two domains
2. The left-half plane (LHP) of the s-plane should
map into the inside of the unit circle in the
z-plane. Thus a stable analog filter will be
converted to a stable digital filter.
Converting an analog filter to a digital filter
1
< z < 2, 2 < z < 3
2
40.
Ans:
z -1 (1 - 0.6z -1)
(1 - 0.4z -1)
41.
Ans: (a)
43.
Ans: |k|<3
44.
Ans: (c)
45.
Ans: (b)
Analog
Low-pass
transformation
analog
prototype HP(s)
ΩC = 1 rad/s
Analog
Filter H(s)
s →z
mapping
Digital
filter H(z)
Indirect conversion of an analog filter to a digital
filter.
s→z
Low-pass
mapping
analog
prototype HP(s)
ΩC = 1 rad/s
D2D
Low-pass
transformation Digital
digital
filter H(z)
prototype HP(z)
ωC = 1 rad/s
66
Steps in the design of IIR filters:
1. Convert digital filter specifications to analog
filter specifications.
2. Find the order (n) & cut-off frequency (ΩC) of
the prototype L.P.F.
3. Find the normalized analog transfer function
using frequency transformations.
4. Convert analog transfer function to digital
transfer function.
5. Draw the filter structure.
8.1 IIR filter design by approximation of Derivatives
One of the simplest methods for converting an
analog filter into a digital filter is to approximate the
differential equation by an equivalent difference
equation.
For the derivative dy(t)/dt at time t = nT, we substitute
the backward difference
y (nT) - y (nT - T)
T
y (nT) - y (nT - T)
dy (t)
=
T
dt t = nT
=
y (n) - y (n - 1)
T
Where T represents the sampling interval and
y(n) ≡ y(nT). The analog differentiator with
output
dy(t)/dt
has
the
system
function
H(s) = s, while the digital system that produces
y (nT) - y (nT - T)
has the system function
T
-1
(1 - z )
.
H (z) =
T
(1 - z -1)
Therefore s =
……. (1)
T
- -1 k
For kth derivative s k = b 1 z l …… (2)
T
1
From equation (1) z = - . If we substitute s = jΩ,
1 sT
the output
we can observe that as Ω varies from –∞ to ∞, the
corresponding locus of points in the z-plane is a
circle of radius 1/2 and center at (1/2, 0)
Signals & Systems
jΩ
Unit circle
z plane
s plane
σ
1
2
Therefore this technique is limited to the design of
low pass and band pass filter only.
8.2 IIR filter Design by Impulse Invariance
In this technique impulse response of digital system
is sampled version of impulse response of analog
system so the frequency response of digital system
is aliased version of frequency response of analog
system.
I.L.T
H (s)
h (t)
h 5nT? $ H (z)
t = nT
ana log T.F
digital T.F
Example:
1
H (s) = s + a
. I.L.T
h (t) = e -at u (t)
. t = nT
h 5nT? = e -anT u (nT)
Z.T
H (z) =
1
1 - e -aT z -1
1
1
.T →
.I
I
1 - e - aT z -1
s+a
(- 1)m-1 . d m-1 . 1
1
(m - 1)! d a m-1 1 - e -aT z -1
(s + a )m
→
s+a
(s + a )2 + b 2
b
(s + a )
2
+b
2
cos bT )z
→ 1 - 2e1 -(ecos (bT
)z + e
- aT
-aT
→ 1 - 2e
-1
- 2 aT - 2
z
(sin bT )z
(cos bT )z -1 + e -2aT z -2
e
-aT
-1
- aT
-1
67
Let us consider the mapping of points from the
s-plane to the z-plane implied by the relation
z = esT
If we substitute s = σ + jΩ and express the complex
variable z in polar form as z = rejω
rejω = eσT ejΩT
clearly, we must have
r = eσT
ω = ΩT
Signals & Systems
8.3 IIR Filter Design by Bilinear Transformation
The bilinear transformation is a conformal mapping
that transforms the jΩ-axis into the unit circle in
the z-plane only once, thus avoiding aliasing of
frequency components. Furthermore, all points
in the LHP of s-plane are mapped inside the unit
circle in the z-plane and all points in the RHP of s are
mapped into corresponding points outside the unit
circle in the z-plane.
2 1 - z -1 m
…………… (1)
s= Tc
1 + z -1
Consequently, σ < 0 implies that 0 < r < 1 and
σ > 0 implies that r > 1. When σ = 0, we have
r = 1. Therefore, the LHP in s is mapped inside the
unit circle in z and the RHP in s is mapped outside
the unit circle in z.
However, the mapping of the jΩ-axis into the unit
circle is not one-to-one. Since ω is unique over the
range (–π, π) the mapping ω = ΩT implies that the
interval –π /T ≤ Ω ≤ π/T maps into the corresponding
values of –π ≤ ω ≤ π. Even the frequency interval
π/T ≤ Ω ≤ 3π/T also maps into the interval –π ≤ ω ≤ π.
Thus the mapping from analog frequency Ω to the
frequency variable ω in the digital domain is many
to one, which reflect the affect of aliasing due to
sampling.
jΩ
s-plane
z - plane
z=e
sT
From equation (1) we note that if
r < 1, then σ < 0,
r > 1, then σ > 0,
When r = 1, then σ = 0,
2
ω
………………..(2)
X = T tan
2
Which show that
Ω=0
Ω→∞
Ω → –∞
⇒ω=0
⇒ω→π
⇒ ω → –π
…….. (3)
The non-linear relationship between ω and Ω
in eq (2) is known as frequency warping. The
bilinear transformation converts H(jΩ) to H(ejω) by
compressing the continuous time frequency axis
according to eq (3).
π
T
ω
π
 ΩT 
2 tan −1 

 2 
σ
Unit
circle
0
π
–
T
–
3π
T
Due to the presence of aliasing, the impulse
invariance method is appropriate for the design of
low-pass and band-pass filters only.
Ω
-π
1.
We note that for ω less than about 0.3π, the
relation between Ω and ω is approximately
linear (recall that tanφ ≈ φ for small φ). Thus, any
shape of magnitude response in this range is
preserved.
68
2.
Frequency
warping
does
not
distort
“flat” magnitude responses because any
“rearrangement” of equal values will preserve
the flatness of the shape. Thus, equiripple
Frequency
warping
distorts
“non
flat”
magnitude response. Thus, a continuoustime
differentiator
cannot
be
Example:
Design a single-pole lowpass digital filter with
a 3-dB bandwidth of 0.2π, using the bilinear
transformation applied to the analog filter
bands are mapped into equiripple bands.
3.
Signals & Systems
converted
to a discrete-time one using the bilinear
transformation.
H (s) =
Where Ωc is the 3-dB bandwidth of the analog
filter.
Sol: The digital filter is specified to have its –3dB gain
at ωc = 0.2π. In the frequency domain of the
analog filter ωc = 0.2π corresponds to
0.65
2
X c = T tan 0.1r = T
To avoid warping we are using pre-warping that is
converting digital filter specification to analog filter
specification. The bilinear transformation maps the
point s = ∞ into the point z = –1.
We conclude form the previous discussion that the
bilinear transformation is most appropriate for filters
X
s + Xc
Thus the analog filter has the system function
0.65/T
H (s) =
s + 0.65/T
This represents our filter design in the analog
domain. By applying bilinear transformation
we will get H (z) =
with piecewise-constant magnitude responses,
such as lowpass, highpass and bandpass filter.
Example:
Convert the analog filter with system function
s + 0.1
H a (s) =
(s + 0.1) 2 + 16
Into a digital IIR filter by means of the bilinear
transformation. The digital filter is to have a
Sol:
resonant frequency of ωr = π/2
Analog filter frequency Ω = 4. This frequency
is to be mapped into ω = π/2 by selecting the
value of T.
r
2
2
=
4 T=
Tan a 22 k & T 4
1−z
s = 4 ; + -1 E
1 z
-1
The resulting digital filter has the system function
H (z) =
0.128 + 0.006z -1 - 0.122z -1
1 + 0.0006z -1 + 0.975z -1
0.245 (1 + z -1)
.
1 - 0.509z -1
Types of Analog Filters:
1. Butterworth Filter (Maximally flat)
2. chebyshev
3. Elliptical (cauer)
8.4 Butterworth Filter
Lowpass
Butterworth
characterized
by
filters
are
all-pole
filters
the
magnitude-squared
frequency response
1
H (X) 2 =
………… (1)
1 + ]X/X cg2n
Where n is the order of filter Ωc is its –3 dB frequency.
Eq (1) can be adjusted as
H (s) H (- s) =
1
…….. (2)
1 + (- s2 /X2c) n
The poles of H(s) H(–s) occur on a circle of radius Ωc
at equally spaced points.
From eq (2) poles can be obtained by using
S k = X c e jr/2 e j (2k + 1) r/2n , k = 0, 1, ….., n –1
69
Signals & Systems
Pole positions for Butterworth filters
From eq (1) we can verify
1.
2.
2
1
At X = X c , H (X) = for all n
2
π π
+
2 8
From eq (1) it is clear that |H(Ω)|2 decreases
monotonically as Ω increases.
|H(Ω)|2
−
Poles of
H(s)
Ideal characteristic
n =18
n =14
n =10
0.5
n =6
0
π π
+
2 10
RS
V
SS 12 − 1 WWW
Sδ
WW
1
log SS 1s
0.1d dB
-1
2
SS − 1 WWW 1 log <10
F
2
0
S δp
W
2
10 .1d dB - 1
…… (3)
n=
T
X=
Ω
Ω
log c Ω s m
log c Ω s m
p
p
s
p
1
1
d 2 - 1n
dp
2n
……… (4)
Example:
Determine the order of a lowpass Butterworth
filter that has a –3 dB bandwidth of 500 Hz and
an attenuation of 40 dB at 1000 Hz.
Sol: The
critical
frequencies
are
the
–3
dB
frequency Ωc and the stop-band frequency
Ωs which are
Ωc = 1000π
Ωs = 2000π
n=
log10 (10 4 - 1)
2 log10 2
= 6.64 , 7
8.5 Chebyshev Filter
Butterworth polynomials
Order
N=5
Ω
Ωc
Xp
Poles of
H(–s)
Poles of
H(s)
n =2
From the above figure as the order of the filter
increases the Butterworth filter characteristic is
more close to ideal characteristics.
The order & cut off frequency of the Butterworth
prototype filter is obtained using
Xc =
Poles of
H(–s)
N=4
1.0
3.
π π
−
2 8
Factors
1
s+1
2
s2 + 2 s + 1
3
(s2 + s + 1) (s + 1)
4
(s2 + 0.76536s + 1) (s2 + 1.848s + 1)
5
(s + 1) (s2 + 0.6180s + 1) (s2 + 1.6180s + 1)
6
(s2 + 0.5176s + 1) (s2 + 2 s + 1)
(s2 + 1.9318s + 1)
There are two types of chebyshev filters.
Type-I chebyshev filter are all-pole filters that
exhibit equiripple behavior in the pass-band and
a monotonic characteristic in the stop-band. On
the other hand, the family of type-II chebyshev
filters contains both poles and zeros and exhibits
a monotonic behavior in the pass-band and an
equiripple behavior in the stop-band. The zeros of
this class of filters lie on the imaginary axis in the
s-plane.
70
|H(Ω)|2
1
Few observations can be made from the above
function:
i. For |Ω| ≤ 1, |H (jΩ)|2 varies between 1 and
1
. This is the pass-band description of
1 + f2
1
|H(jΩ)|. i.e., 1 # H (jΩ) #
for Ω # 1
1 + ε2
|H(Ω)|2
1
1
1+ ∈2
Signals & Systems
1
1+ ∈2
1
1 + f2
ii. At Ω = 1, C2n (1) = 1 always. This can be easily
verified for the polynomials for Ω = 1 eq (2) can
be as follows:
and, ripple in pass-band = 1 -
Ωp
Ωp
Ω
n odd type I
Ω
n even type I
The chebyshev approximation is implemented
with the help of chebyshev polynomials. They are
defined as follows.
C n (X) = cos (n cos -1 (X)) X # 1
= cosh (n cosh -1 (X)) X > 1
For n = 0 ⇒ C0(Ω) = cos(0) = 1
For n = 1 ⇒ C1(Ω) = cos(1cos–1 (Ω)) = Ω
The higher order chebyshev polynomials are
obtained by following recursive formula.
Cn(Ω) = 2Ω Cn–1 (Ω) – Cn-2 (Ω)
n
Chebyshev polynomials
0
1
1
Ω
2
2Ω2 – 1
3
4Ω3 – 3Ω
4
8Ω – 8Ω + 1
5
16Ω5 – 20Ω3 + 5Ω
6
32Ω6 – 48Ω4 + 18Ω2 – 1
4
Some of the properties of these polynomials are
1.
|Cn(Ω)| ≤ 1 for all |Ω| ≤ 1
2.
Cn(1) = 1 for all n
3.
All the roots of the polynomials Cn(Ω) occurs in
the interval –1≤ Ω ≤ 1
The squared magnitude function of the chebyshev
filter is given as,
H (jX) 2 =
Thus at cutoff frequency Ω = 1, magnitude is
1
=
1 + f2
iii. In the stopband |Ω| > 1, and f2 C2n (X >> 1) .
Hence eq (2) can be written as,
H (jX) 2 = 2 12
f C n (X)
or
1
……….. (2)
1 + f2 C2n (X)
H (jX) ,
1
fC n (X)
The above equation can be written in decibels
also i.e.,
1
E
H (jX) in dB = 20 log10 ;
fC n (X)
2
1
1
or H (jX) =
1 + f2
1 + f2
2
H (jX) =
= –20 log10 [εCn (Ω)]
As Ω is large Cn(Ω) is mainly represented by its
first term. Hence Cn ≅ 2n-1 Ωn for large Ω
Hence the above equation can be written as
|H(j(Ω)| in dB = –20 log10 [ε.2n-1 Ωn]
= –20 log ε – 20 log 2n-1 – 20log Ωn
= –20 log ε – 20(n – 1) log 2 – 20n log Ω
= –20 log ε – 6(n –1) – 20n log Ω (norm)
f = 610 0.1d dB - 1@
p
1 2
The following points can be noted about transfer
function of the chebyshev filter:
i.
The transfer function of the chebyshev filter is
an all pole function like Butterworth filter.
ii.
The numerator is constant and there are no
finite zeros.
71
iii.
iv.
v.
Poles of the transfer function lie on an ellipse
[for Butterworth filter they lie on the circle]. The
major axis of the ellipse lies along imaginary
axis of the s – plane and minor axis lies along
real axis of the s-plane.
The narrower the ellipse, the closer will be the
poles to the imaginary axis and hence each
individual pole will have stronger impact. This
means that ripples will be more pronounced.
The ripple magnitude will have a strong effect
on the locations of the poles of the transfer
function.
The transfer function of the chebyshev filter is
given as
K
H (s) = n
s + b n - 1 s n - 1 + f + b1 s + b 0
The constant K in the numerator in the above
equation is adjusted to provide a loss of 0 dB at
the pass band minima.
Z] b for odd "n"
0
]]
K = [] b 0
for even "n"
]]
1 +! 2
\
Poles location for a chebyshev filter
r2
r1
Type II chebyshev filter
|H(Ω)|2
1
δ 22
Ωp Ωs
n odd
Ω
Signals & Systems
|H(Ω)|2
1
δ 22
Ω
Ωp Ωs
n even
Frequency Transformations for Analog Filters:
(Prototype low-pass filter)
Prototype
Required
LPF
s
HPF
s
s/Ωc
s2 + X20
Xc s
s
Xc s
s2 + X20
BPF
BSF
X20 = X P X P
Xc = XP - XP
1
2
2
1
Ωc/s
s
72
Signals & Systems
Frequency Transformations for Digital Filters
(Prototype LPF has band-edge frequency ωP)
Type of transformation
Transformation
Low-pass
z -1 - a
z "
1 - az -1
-1
High-pass
z -1 "
z -1 + a
1 + az -1
Parameters
ω'p = band edge frequency of new
sin 7^ω p − ωl p h /2A
filter a =
sin 6^ω p + ωl ph /2@
ω'p = band edge frequency of new
l
filter a = − cos 6^ω p + ω p h /2@
cos 7^ω p − ωl p h /2A
ω , = lower band edge frequency
ωu = upper band edge frequency
a1 = 2αK / (K + 1)
Band-pass
z -1 " -
a2 = (K – 1) / (K + 1)
z - a1 z + a2
a2 z -2 - a1 z -1 + 1
-2
-1
cos 6]ω u + ω ,g /2@
cos 6^ω u − ω ,h /2@
^ω u − ω ,h
ωp
K = cot
tan
2
2
α=
ω , = lower band edge frequency
ωu = upper band edge frequency
Band-stop
z -2 - a1 z -1 + a2
z "
a2 z -2 - a1 z -1 + 1
-1
a1 = 2α / (K + 1)
a2 = (1 – K) / (1 + K)
cos 6]ω u + ω ,g /2@
cos 6^ω u − ω ,h /2@
^ω u − ω ,h
ωp
K = tan
tan
2
2
α=
73
Signals & Systems
8.6 FIR Filter Design
Type 1
An FIR filter of length N with input x(n) and output
y(n) is described by the difference equation
n
–1
–0.5
F
0.5
1
Even symmetry about
F = 0 and F = 0.5
Center of symmetry
y(n) = b0x(n) + b1x(n–1) +…+ bN–1x(n–N+1)
=
N-1
/ b k x (n - k )
Type 2
k=0
n
Where {bk} is the set of filter coefficients. Alternatively,
we can express the output sequence as the
Center of symmetry
y (n) =
/ h (k) x (n - k)
k=0
………. (1)
–0.5
1F
0.5
Odd symmetry about F = 0.5
convolution of the unit sample response h(n) of the
system with the input signal. Thus we have
N-1
–1
Type 3
n
–1
–0.5
F
0.5
1
Odd symmetry about F = 0.5
Where the lower and upper limits on the convolution
sum reflect the causality and finite-duration
characteristics of the filter.
H (z) =
N-1
/ h (k) z-k
Center of symmetry
Type 4
………….. (2)
n
k=0
–1
–0.5
F
0.5
1
The roots of this polynomial constitute the zero of
the filter.
An FIR filter has linear phase if its unit sample
response satisfies the condition
h(n) = ± h(N–1–n), n = 0, 1, …., N –1
Type
Symmetry
Length
1
Even
Odd
2
Even
Even
3
Odd
Odd
4
Odd
Even
Center of symmetry
Odd symmetry about F = 0.5
74
Signals & Systems
Window Functions
Name
of window
Bartlett
(triangular)
Blackman
Hamming
Hanning
Kaiser
Time-domain sequence
h(n), 0 ≤ n ≤ N –1
2 n- N 1
2
1N-1
0.42 - 0.5 cos
2rn
4rn
0.08 cos N-1 +
N 1
0.54 - 0.46 cos
2rn
N-1
1 b1 - cos 2rn l
N-1
2
- 2
- 2
I 0 <a c N 1 m - cn - N 1 m F
2
2
I 0 ;a c N 1 mE
2
Comparison of window characteristics
Type of window
Approximate
transition
width of
main lobe
4π/N
Peak
sidelobe
level (dB)
Transition
Width
Minimum S.B
Attenuation (dB)
– 13
–21
Bartlett
8π/N
– 25
Hanning
8π/N
– 31
Hamming
8π/N
– 41
Blackman
12π/N
– 57
0.9π
N
3π
N
3.1π
N
3.3π
N
5.5π
N
Rectangular
–25
–44
–53
–74
75
The following example illustrates low-pass filter
design using rectangular (a), Hamming (b),
Blackman (c) and Kaiser window with α = 4 for a
length of 61 samples.
Signals & Systems
Location of zeros of a linear phase FIR filter:
1
z1*
1
z*3
z1
z3
z*3
1
z3
(a)
1
z2
z2
z1*
1
z1
Unit
circle
If “z0” is a zero of H(z) in eq (2)
Then “ z 0-1 ” is a zero of H(z)
If z1 = –1 then z1-1 = 1
If z2 is a real zero then z2-1 = 1/z2
If z3 is a complex zero 6 z3 = 1@ then z3-1 = z*3
If z4 is a complex zero [|z4| ≠ 1] then other zeros are
z 4-1, z*4 ^z*4h
-1
(b)
8.7 FIR Filter Design using windowing method
1.
2.
Choose the desired frequency response
Hd(ejω)
r
1 #
Obtain hd 5n? =
Hd (e j~n) e j~n dω → non
2π
-r
causal impulse response.
3.
h 5n? = hd 5n? w 5n? is the causal Impulse
response of FIR filter.
(c)
4.
Find H (z) =
/ h5n? z-n
N-1
n=0
5.
Draw filter structure.
Design of FIR Differentiators:
Differentiators are used in many analog and digital
systems to take the derivative of a signal. An ideal
differentiator has a frequency response that is
linearly proportional to frequency. Similarly, an ideal
digital differentiator is defined as one that has the
(d)
frequency response
Hd(ω) = jω –π ≤ ω ≤ π
76
The unit sample response corresponding to Hd(ω) is
hd (n) =
1
2π
r
-r
r
1
= 2π
cos rn =
n , 3 < n < 3, n ! 0
#
jω e j~n dω
=
-r
Design of Hilbert Transformers
An ideal Hilbert transformer is an all-pass filter that
imparts a 90o phase shift on the signal at its input.
Hence the frequency response of the ideal Hilbert
transformer is specified as
−j 0 < ω # π
Hd (ω) = )
j −π < ω < 0
Hilbert transformers are frequently used in
communication systems and signal processing, as,
for example, in the generation of single-side band
modulated signals, radar signal processing, and
speech signal processing.
The unit sample response of an ideal Hilbert
transformer is
r
1 #
hd (n) =
H d (ω) e j~n dω
2π
-r
0
#
r
j e j~n dω −
-r
#
j e j~n dω
0
2
2 sin (rn/2)
, n!0
= *r
n
, n=0
0
Frequency Sampling Method
The transfer function of FIR filter is
(
H z) =
N-1
/ h (n) z-n
n=0
In the method we specify the desired frequency
response at a set of equally spaced frequencies
and solve for that unit impulse response from these
equally spaced frequency specifications.
N-1
N-1
1
j rkn/N
3 z -n ….. (1)
H (z) = / ) N / H (k) e 2
k=0
k=0
1
H (z) = N / H (k) ) / e j2rkn/N z n 3
n=0
k=0
N-1
N-1
/ ^e j2rk/N z-1hn = 1 --^e j2rk/Nz -1h
N-1
n=0
Putting the above result in eq (1)
N-1
1
1 - z -N
H (z) = N / H (k) .
1 - e j2rk/N z -1
k=0
# Hd (ω) e j~n dω
1
= 2π
Signals & Systems
j2rk/N -1 N
1 e
z
N-1
H (k)
1 - z -N /
N k = 0 1 - e j2rk/N z -1
77
Signals & Systems
Normalized Chebyshev Polynomials
(a) Ripple = 0.5 dB i.e ε = 0.349
N
b0
1
2.8627752
b1
b2
b3
b4
b5
b6
b7
b8
b9
2
1.5162026
1.4265245
3
0.7156938
1.5348954
1.2529130
4
0.3790506
1.0254553
1.7168662
1.1973856
5
0.1789234
0.7525181
1.3095747
1.9373675
1.1724909
6
0.0947626
0.4323669
1.1718613
1.5897635
2.1718446
1.1591761
7
0.0447309
0.2820722
0.7556511
1.6479029
1.8694079
2.4126510
1.1512176
8
0.0236907
0.1525444
0.5735604
1.1485894
2.1840154
2.1492173
2.6567498
1.1460801
9
0.0111827
0.0941198
0.3408193
0.9836199
1.6113880
2.7814990
2.4293297
2.9027337
1.425705
10
0.0059227
0.2372688
0.2372688
0.6269689
1.5274307
2.1442372
3.4409268
2.7097415
3.1498757
b7
b8
1.1400664
(b) Ripple = 1 dB i.e ε = 0.508
N
b0
b1
b2
b3
b4
b5
b6
1
1.9652267
2
1.1025103 1.0977343
3
0.4913067 1.2384092 0.9883412
4
0.2756276 0.7426194 1.4539248 0.9528114
5
0.1228267 0.5805342 0.9743961 1.6888160
0.9368201
6
0.0689069 0.3070808 0.9393461 1.2021409
1.9308256 0.9282510
7
0.0307066 0.2136715 0.5486192 1.3575440
1.4287930 2.1760778
0.9231228
8
0.0172267 0.1073447 0.4478257 0.8468243
1.8369024 1.6551557
2.4230264 0.9198113
9
0.0067767 0.0706048 0.2441864 0.7863109
1.2016071 2.3781188
1.8814798 2.6709468 0.9175474
10 0.0043067 0.0344971 0.1824512 0.4553892
1.2444914 1.6129856
2.9815094 2.1078524 2.9194657
b9
0.9159320
(c) Ripple = 2 dB i.e ε = 0.764
N
b0
b1
b2
b3
b4
b5
b6
b7
b8
1
1.3075603
2
0.6367681 0.8038164
3
0.3268901 1.0221903 0.7378216
4
0.2057651 0.5167981 1.2564819 0.7162150
5
0.0817225 0.4593491 0.6934770 1.4995433
0.7064606
6
0.0514413 0.2102706 0.7714618 0.8670149
1.7458587 0.7012257
7
0.0204228 0.1660920 0.3825056 1.1444390
1.0392203 1.9935272
0.6978929
8
0.0128603 0.0729373 0.3587043 0.5982214
1.5795807 1.2117121
2.2422529 0.6960646
9
0.0051076 0.0543756 0.1684473 0.6444677
0.8568648 2.0767479
1.3837474 2.4912897 0.6946793
10 0.0032151 0.0233347 0.1440057 0.3177560
1.0389104 1.1585287
2.6362507 1.5557424 2.7406032
b9
0.6936904
78
Signals & Systems
(d) Ripple = 3dB i.e ε = 0.997
N
b0
b1
b2
b3
1
1.0023773
2
0.7079478 0.6448996
3
0.2505943 0.9283480 0.5972404
4
0.1769869 0.4047679 1.1691176 0.5815799
5
0.626391
6
0.0442497 0.1634299 0.6990977 0.6906098
1.6628481 0.5706979
7
0.0156621 0.1461530 0.3000167 1.0518448
0.8314411 1.6628481
0.5684201
8
0.0110617 0.0564813 0.3207646 0.4718990
1.4666990 0.9719473
2.1607148 0.5669476
9
0.0039154 0.0475900 0.1313851 0.5834984
0.6789075 1.9438443
1.1122863 2.4101346 0.5659234
10 0.0027654 0.0180313 0.1277560 0.2492043
0.9499208 0.9210659
2.4834205 1.2526467 2.6597378
0.4079421 0.5488626 1.4149847
b4
b5
b6
b7
b8
b9
0.5744296
0.5652218
79
Practice Questions
1
s+2
(a) Convert H(s) to a digital filter H(z) using
01. Consider the analog filter H (s) =
impulse invariance. Assume that the
sampling frequency is 2 Hz.
(b) Will the impulse response h[n] match the
impulse response h(t) of the analog filter at
Signals & Systems
04. We are given the analog low-pass filter
1
H (s) = s + 1 whose cutoff frequency is known
to be 1 rad/s. It is required to use this filter as the
basis for designing a digital filter by the impulse
invariant transformation. The digital filter is to
have a cutoff frequency of 50 Hz and operate
at a sampling frequency of 200 Hz.
What is the transfer function H(z) of the digital
filter if no gain matching is used?
05. Consider
the
low-pass
3
.
H (s) = 2
s + 3s + 3
the sampling instants? Should it? Explain.
(c) Will the step response s[n] match the step
sampling instants? Should it? Explain.
1
.
s+2
(a) Convert H(s) to a digital filter H(z) using step
02. Consider the analog filter H (s) =
invariance at a sampling frequency of 2Hz.
impulse response h(t) of the analog filter at
the sampling instants? Should it? Explain.
(c) Will the step response s[n] match the step
response s(t) of the analog filter at the
sampling instants? Should it? Explain.
1
.
s+2
(a) Convert H(s) to a digital filter H(z) using
03. Consider the analog filter H (s) =
ramp invariance at a sampling frequency
of 2Hz.
(b) Will the impulse response h[n] match the
impulse response h(t) of the analog filter at
the sampling instants? Should it? Explain.
(c) Will the step response s[n] match the step
response s(t) of the analog filter at the
sampling instants? Should it? Explain.
filter
(a) Use the bilinear transformation to convert
this analog filter H(s) to a digital filter H(z) at
a sampling rate of 2 Hz.
(b) Use H(s) and the bilinear transformation to
design a digital lowpass filter H(z) whose
gain at f = 20 kHz matches the gain of H(s)
at Ω = 3 rad/s. The sampling frequency is
80 kHz.
response s(t) of the analog filter at the
(b) Will the impulse response h[n] match the
analog
06.
s
.
s2 + s + 1
(a) What type of filter does H(s) describe?
(b) Use H(s) and the bilinear transformation
to design a digital filter H(z) operating at
1 kHz such that its gain at f0 = 250 Hz
matches the gain of H(s) at ωa = 1 rad/s.
What type of filter does H(z) describe?
Consider the analog filter H (s) =
07. A digital low-pass filter is required to meet the
following specifications:
Passband ripple: ≤ 1 dB
Passband edge: 4 kHz
Stopband edge: ≥ 40 dB
Stopband edge: 6 kHz
Sample rate: 24 kHz
The filter is to be designed by performing a
bilinear transformation on an analog system
function. Determine what order Butterworth,
Chebyshev analog design must be used
to meet the specifications in the digital
implementation.
80
08. An IIR digital lowpass filter is required to meet
the following specifications:
Passband ripple
(or peak-to-peak ripple): ≤ 0.5 dB
Passband edge: 1.2 kHz
Stopband attenuation: ≥ 40 dB
Stopband edge: 2.0 kHz
Sample rate: 8.0 kHz
Determine the required filter order for
(a) A digital Butterworth filter
(b) A digital chebyshev filter
09. Determine the system function H(z) of the
lowest-order chebyshev digital filter than meets
the following specifications:
(a) 1-dB ripple in the pass-band 0 ≤|ω|≤ 0.3π
(b) At least 60 dB attenuation in the
stop-band 0.35π ≤ |ω| ≤ π. Use the bilinear
transformation.
10. Determine the system function H(z) of the
lowest-order chebyshev digital filter that meets
the following specification.
1 (a)
dB ripple in the pass-band
2
0 ≤ |ω| ≤ 0.24π
(b) At least 50 dB attenuation in the stop-band
0.35π ≤ |ω| ≤ π.
Use the bilinear transformation.
11. A linear phase FIR filter has one of the zero at
1
z = e jr/3 . Find the remaining zeros?
2
12. Transfer function of a IVth order linear phase
F.I.R filter is given by H(z) = (1 + 2z–1 + 3z–2)G(z)
then G(z) is _____
(a) 3 + 2z–1 + z–2 1
(b) 1 + z -1 + z -2
2
1
(c)
2z -1 + z -2
3+
(d) 1 + 2z–1 + 3z–2
Signals & Systems
13. Find H(z) and H(F) for each sequence and
establish the type of FIR filter it describes by
checking values of H(F) at F = 0 and F = 0.5
(a) h 5n? = %1, 0, 1 /
0
(b) h 5n? = %1, 2, 2, 1 /
0
(c) h 5n? = %1, 0, - 1 /
0
(d) h 5n? = %1, - 2, 2, - 1 /
0
14. The first few values of the impulse response
sequence
of
a
linear-phase
filter
are
h[n] = {2, –3, 4, 1, ….. }. Determine the complete
sequence (assuming the smallest length for
h[n] if the sequence is to be:
(a) type 1
(b) type 2
(c) type 3
(d) type 4
15. Partial details of various filters are listed. Zero
locations are in keeping with linear-phase and
real coefficients. Assuming the smallest length,
identify the sequence type and find the transfer
function of each filter.
(a) zero location: z = 0.5ej0.25π
(b) zero location: z = ej0.25π
(c) zero location: z = 1, z = ej0.25π
(d) zero locations: z = 0.5, z = –1; odd symmetry
(e) zero locations: z = 0.5, z = 1, z = –1; even
symmetry
16. Design the symmetric FIR LPF whose desired
frequency response is given as
π
e -j3~ for ω #
Hd (e j~) = *
4
0 else where
use Hamming window.
81
Signals & Systems
Properties of DFT
9. DFT & FFT
01. Circular shift:
x 6n - n 0@N * e
9.1 Introduction to DFT
•
The Fourier series describes periodic signals by
discrete spectra, whereas the DTFT describes
discrete signals by periodic spectra. As a result,
signals that are both discrete and periodic in
one domain are also periodic & discrete in the
other. This is the basis for the formulation of the
DFT.
•
Sampled version of D.T.F.T. spectrum is D.F.T.
•
The N point DFT of a signal x(n) is
X (k) =
N-1
/ x (n) e
-
j2rkn
N
k = 0, 1 ……..(N – 1)
n=0
j2rkn
1
x (n) = N / X (K) e N n = 0, 1 ………. (N – 1)
k=0
N-1
The DFT & its IDFT are also periodic with period
N, and it is sufficient to compute the results for
only one period (0 to N-1).
X (k) =
(or) W Nkn X (K)
N
x :n - 2 D * (- 1) k X 5k?
0
02. Modulation:
* X 6k - k 0@
D
]- 1gn x (n) * X :k - N
2
x (n) e
j2rk 0 n
N
03. Circular Convolution:
x(n) Ⓝ h(n) ↔ X(k) H(k)
04. Central ordinates:
(a) X 50? =
N-1
/ x (n)
n=0
1
(b) x 50? = N / X (k)
k=0
N-1
N
(c) X : 2 D = / ]- 1gn x (n), N even
n=0
N-1
N
1
(d) x : 2 D = N / ]- 1gk X (k)
k=0
05. Parseval’s relation:
N-1
/
n=0
N-1
/ x (n) WNkn
n=0
1
x (n) = N / X (k) W N-kn
k=0
N-1
Where WN= e -j2r/N is the phase factor.
Periodicity: W NK + N = W NK
N
K+ 2
Symmetry: W N
Note:
X 5k?
N-1
IDFT of X(k) is
•
-j2r
N kn0
= - W NK
For direct calculation of N point DFT, we
require N2complex multiplications and
N(N-1) additions.
x (n) 2 = 1 / X (k)
N k=0
N-1
2
82
Signals & Systems
9.2 Linear convolution using DFT
The circular convolution of 2 sequences, of lengths N1 & N2, respectively, can be made equal to the
linear convolution of the 2 sequences by zero padding both sequences, so that they both consist of (N1
+ N2–1) samples.
x1(n)
N1 samples
x2(n)
N2 samples
Zero padding
x e1 (n)
(N2 – 1) samples
Zero padding
(N1 – 1) samples
x e 2 (n)
DFT
X1(k)
X2(k)
Y(k)
DFT
I DFT
y(n)
x1(n) ∗ x2(n)
Let us consider the computational efficiency of calculating a convolution using the DFT rather than
the direct method. In calculating the convolution of two N - element sequences using DFT method, we
required 3N log22N + 2N Complex multiplications where as direct convolution of 2 sequences requires N2
complex multiplication. ∴ DFT method is more efficient for N ≥ 32.
(a) Picket - fence Effect:
The picket - fence effect is caused by the
approximation of the continuous frequency
spectrum of the DTFT using a finite number of
frequency points. The spectrum is observed
very much like looking through a picket fence
with the exact value of the spectrum known
only as integer multiples of the frequency
at frequencies other than ±f0 because the
periodic extension of the sampled portion of
the periodic signal doesn’t match the original
signal but describes a different signal altogether
as shown in fig. suppose we sample the signal
x(t) = sin(2πt) at fs=16 Hz. Then the sampled
signal frequency (i.e, digital frequency)
F0 =
ana log frequency
1
= 16
sampling rate
resolution. The peak of a particular frequency
component in a signal could be hidden from
view because it is located between 2 adjacent
frequency points in the spectrum. To reduce
this effect, the number of frequency points
must be increased, since this enables more
frequency components of a signal to coincide
with the more closely spaced frequency points.
(b) Spectral leakage:
Leakage is present in the DFT results if a periodic
signal x(t) is not sampled for an integer number
of periods. The DFT shows nonzero components
If we choose N = 8 the DFT spectral spacing
= fs/N = 2 Hz. In other words there is no DFT
component at 1 Hz, the frequency of the sine
wave ! where should we expect to see the DFT
components?
If we express
K
1
F0 =
as F0 = Nf
16
We obtain kf = NF0 = 0.5. Thus F0 corresponds
to the fractional index kf = 0.5 & the largest
DFT components should appear at the integer
indices closest to kf at k = 0 (d .c), k = 1(2Hz)
83
Sinusoid sampled for one
full period &its periodic
extension
t
Sinusoid sampled for half
period & its periodic
extension
Signals & Systems
Example:
Let x(n) = {1, 2, 3, 4, 5} and h(n) = {1,1,1} using
Overlap-add method find y(n)?
Sol: L = 6 & N = 3.
x0(n) = {1, 2, 3}
h(n) = {1, 1, 1}
x1(n) = {3, 4, 5}
y0(n) = x0(n) * h(n) = {1, 3, 6, 5, 3}
y1(n) = x1(n) * h(n) = {3, 7, 12, 9, 5}
Shifting & superposition results in the required
convolution
y(n) = y0(n) + y1[n – 3]
y0(n)
t
To minimize leakage sample for the longest
duration possible or for integer periods.
Convolution of Long Sequences
1. Overlap-Add method
2. Overlap-Save method
Some times we have to process a long stream
of incoming data by a filter whose impulse
response is much shorter than that of incoming
data. The convolution of a short sequence h(n)
of length N with a very long sequence x(n) of
length L > > N can involve large amount of
computation & memory.
1.
Overlap - Add method:
Suppose h(n) is of length N, and the length of
x(n) is L = mN (if not, we can always zero pad it
to this length). We partition x[n] into m segments
x0(n), x1(n) …xm-1(n), each of length N. We find
the regular convolution of each section with
h(n) to give partial results y0(n), y1(n), … ym-1(n).
y(n) = y0[n] + y1[n–N] + ym-1[n–(m–1)N]
Since each regular convolution contains
(2N – 1) samples, we zero – pad h(n) and each
section xk[n] with [N – 1] zeros before finding
yk[n] using the FFT. Splitting x(n) into equal –
length segments is not a strict requirement.
→
1
3
6
5 3
3 7
y1[n–3] →
y(n) = {1, 3, 6, 8, 10, 12, 9, 5}
12 9 5
2. Overlap-Save method:
If L > N and we zero - pad the second sequence
to length L, their periodic convolution has
(2L – 1) samples. Its first (N -1) samples are
contaminated by wraparound, and the rest
correspond to the regular convolution. eg. Let
L = 16 & N = 7.
If we pad N by 9 zeros, their regular convolution
has 31(or 2L-1) samples with 9 trailing zeros
(L-N = 9). For periodic convolution 15 samples
(L-1 = 15) are wrapped around. Since the last
nine [or L-N] are zeros, only the first 6 samples
of the periodic convolution are contaminated
by wraparound, which is the basis idea of this
method.
First, we add (N – 1) leading zeros to the longer
sequence x(n) & section it into k overlapping
{by N – 1} segments of length M. Typically, we
choose M ≈ 2N.
Next, we zero-pad h(n) { with trailing zeros} to
length M, and find the periodic convolution
of h(n) with each section of x(n). Finally, we
discard the first (N–1) [contaminated] samples
from each convolution & glue(concatenate)
the results to give the required convolution.
84
Signals & Systems
Example:
Find convolution of x(n) = {1, 2, 3, 4, 5} and h(n) = {1, 1, 1} using Overlap Save method?
Sol: First add (N – 1) = 2 zeros to
x(n) = {0, 0, 1, 2, 3, 3, 4, 5}
Take M = 2N – 1 = 5, section into K overlapping segments of length M(5)
x0(n) = {0, 0, 1, 2, 3}
Zero pad h(n) to length M = 5 samples
x1(n) = {2, 3, 3, 4, 5}
h(n) = {1, 1, 1, 0, 0}
x2(n) = {4, 5, 0, 0, 0}
x0(n)⊛h(n) = {5, 3, 1, 3, 6}
x1(n)⊛h(n) = {11, 10, 8, 10, 12}
x2(n)⊛ h(n) = {4, 9, 9, 5, 0}
We discard the first 2 samples from each convolution & give the results to obtain
y(n) = {1, 3, 6, 8, 10, 12, 9, 5, 0}
9.3 FFT
Fast algorithm reduce the problem of calculating an N-point DFT to that of calculating many smallersize DFTs. The computation is carried out separately on even - indexed and odd-indexed samples to
reduce the computational effort. All algorithms allocate for computed results. The less the storage
required, the more efficient is the algorithm.
Many FFT algorithms reduce storage requirements by performing computations in place by storing
results in the same memory locations that previously held the data.
3 stages in an 8-point DIT-FFT:
x[0] 2 point
x[4]
DFT
X[0]
Combine
X[1]
2 point
DFTs
x[2]
2 point
Combine
x[6]
DFT
4 point
DFTs
x[1] 2 point
Combine
x[5]
DFT
2 point
DFTs
x[3] 2 point
X[7]
DFT
x[7]
Stage 3
Stage 2
Stage 1
Typical butterfly for DIT – FFT algorithm:
A + BWr
A
B
Wr
⊗
–1
A - BWr
85
Signals & Systems
DIT algorithms for N = 2, 4, 8:
DIF algorithm for N = 2, 4, 8:
Feature
N-point DFT
N-point FFT
Algorithm
Solution of N
equations in N
unknowns
N/2 butterflies/stage
m stages
Total butterflies = m(N/2)
Multiplication
N per equation
1 per butterfly
Addition
N–1 per equation
2 per butterfly
Total multiplication
N
N
log2N
2
Total additions
N(N–1)
N log2N
2
86
Bit - reversal using Bruneman’s algorithm:
(1)
(2)
Start with {0, 1} multiply by 2 to get {0, 2}
Add 1 to the list of numbers obtained above.
Now it is {1, 3}.
(3) Append the list in step2 to that in step1 to get
{0, 2, 1, 3}
(4) The list obtained in step3 now becomes the
starting list in step1. The steps are repeated until
the desired length of list is obtained.
0 1



↓×2
+1
0
2 →
13


↓×2
+1
→1 5 3 7
0 4 2 6 


↓×2
0 8 4 12 2 10 6 14
+1
→
1 9 5 13 3 11 7 15
Signals & Systems
Practice Questions
01. Analog data to be spectrum analyzed are
sampled at 10kHz & DFT of 1024 samples
are computed. Find the frequency spacing
between spectral samples?
02. Find the 4 point DFT of x(n) = {0, 1, 2, 3}?
03. Let x (n)
X 5k? , then prove the following
statements.
(i) If x(n) = – x[N – 1 – n], then X(0) = 0
(ii) If x(n) = x[N – 1 – n], with N even,
N
then X : 2 D = 0
N pt
04. Fig shows a finite length sequence x(n).
6
5
4
2
2
3
2
2
0 1 2 3
2 2
Fig
x(n)
n
Sketch the signals
(a) x([n – 2])4
(b) x([n + 1])4
(c) x([– n])4
05. The first five points of an 8 point DFT of a
real-valued sequence are
{0.25, 0.125 - j0.3018, 0, 0.125 - j0.0518, 0}.
Find other 3 points?
06. The DFT of a vector [a b c d] is the vector
[α β γ δ]. Consider the product
RS
V
SSa b c dWWW
Sd a b cWW
W
6p q r s@ = 6a b c d@SSS
SSc d a bWWW
SSb c d aWW
T
X
The DFT of the vector [p q r s] is a scaled
version of
87
2 2 2 2
(a) 7α β γ δ A 10. Find the 10-point inverse DFT of
3, k = 0
X (k) = )
1, 1 # k # 9
(b) 7 α β γ δ A
(c) 6α + β β + δ δ + γ γ + α@ (d) [α β γ δ]
07. Given x(n) = {1, –2, 3, –4, 5, –6}, without
calculating DFT, find the following quantities
(a) X(0)
(b)
5
/ X (k)
(c) X(3)
(d)
k=0
5
/
X (k)
2
(e)
5
/ (- 1) k X (k)
k=0
k=0
08. X(k) is the discrete Fourier Transform of a 6-point
real sequence x(n).
If X(0) = 9 + j0, X(2) = 2 + j2,
X(3) = 3 – j0, X(5) = 1 – j1, x(0) is
(a) 3
(b) 9
(c) 15
(d) 18
09.
(i) The two 8-point sequences x1(n) & x2(n) shown
in fig. have DFTs X1(k) and X2(k) respectively.
x2[n]
c
c
x1[n]
b
b
d
d
a
e
e
a
0 1 2 3 4 5 6 7 n
0
1 2 3 4 5 6 7 n
Find the relation between X1(k) & X2(k)?
(ii). Consider the sequence x(n) shown in fig. Find
y(n) whose six-point DFT is
Y (k) = W64k X (k) ,
Where X(k) is the six-point DFT of x(n).
4
2 3 2
2 2 1
2
0 1 2 34 5
2 22 2
Fig.
x(n)
n
Signals & Systems
11. A signal x(t) is bandlimited to 10 kHz is sampled
with a sampling frequency of 20 kHz. The DFT of
N = 1000 samples x(n) is then calculated.
(a) What is the spacing between the spectral
samples?
(b) To what analog frequency does the index
k = 150 correspond? What about k = 800?
12. Consider the real finite-length sequence
x[n] shown in Fig., whose six-point DFT is X[k].
If Q[k] = X[2k], k = 0, 1, 2 represents the 3-point
DFT, then q [n] is
4
x[n]
3
2
1
0
1
2
Fig.
3
4
5
n
(a) {5, 3, 2}
(b) {4, 3, 2}
(c) {2, 3, 4}
(d) {1, 2, 3}
13. Given x[n] = {A, 2, 3, 4, 5, 6, 7, B} is having 8
point DFT X[k]. If X(0) = 20 & X(4) = 0, find A & B?
14. Given X[K] = K+1; 0 ≤ K ≤ 7 is 8-point DFT of x[n].
Then the value of
3
/ x [2n]
is ____
n=0
15. Let x[n] be a real 8 point sequence and let X(k)
be its 8 point DFT
(A) Evaluate
7
1 /
X (k) e j (2r/8) kn n = 9 in terms of x (n)
8 k=0
(B) Let w(n) be a 4 point sequence for
0 ≤ n ≤ 3 and W(k) be its 4 point DFT.
If W(k) = X(k) + X(k+4), express w(n) in
terms of x(n)
88
(C) Let y(n) be an 8 point sequence for
0 ≤ n ≤ 7 and Y(k) be 8 point DFT.
If Y (k) = )
2X (k) k = 0, 2, 4, 6
0 for k = 1, 3, 5, 7
express y(n) in terms of x(n).
16. The four point DFTS of x(n) and y(n) is
X(K) = {22, – 4 + j2, – 6, – 4 – j2} and
Y(K) = {8, – 2 – j2, 0, – 2 + j2}. The circular
convolution of x(n) and y(n) is w(n), then w(2)
is
(a) 38
(b) 30
(c) 12
(d) 60
17. Consider x[n] = {1, 2, 3, 4} with DFT
X(K) = {10, -2+2j, -2, -2-2j}. If the sampling rate
is 10Hz
(i) Determine the sampling period, time index
and the sampling instant for a digital sample
x(3) in time domain
(ii) Determine the frequency resolution,
frequency bin number and frequency for
each of the DFT coefficients X(1) and X(3).
18. We wish to sample a signal of 1-s duration, and
band-limited to 100Hz, in order to compute
its spectrum. The spectral spacing should not
exceed 0.5 Hz. Find the minimum number N
of samples needed and the actual spectral
spacing ∆f if we use
(a) The DFT
(b) The radix-2 FFT
19. We wish to sample the signal,
x(t) = cos(50πt) + sin(200πt) at 800 Hz and
compute the N-Point DFT of the sampled
signal x[n]
(a) Let N = 100. At what indices would you
expect to see the spectral peaks? Will the
peaks occur at the frequencies of x(t)?
(b) Let N = 128. At what indices would you
expect to see the spectral peaks? Will the
peaks occur at the frequencies of x(t)?
Signals & Systems
20. Let X 5k? = %1, - 2, 1 - j, j2, 0f / be the 8-point DFT
of a real signal x[n]
(a) Determine X[k] in its entirety.
(b) What is the DFT Y[k] of the signal
y[n] = (–1)n x[n]?
(c) What is the DFT G[k] of the zero-interpolated
signal g[n] = x[n/2]
0
21. The Discrete Fourier Transform (DFT) of the
4-point sequence
x[n] = {x[0], x[2], x[3]} = { 3, 2, 3, 4} is
X[k] = {X[0], X[1], X[2], X[3]}
= {12,2j, 0, -2j}.
If X1[k] is the DFT of the 12-point sequence
x1[n]= {3, 0, 0, 2, 0, 0, 3, 0, 0, 4, 0, 0} the value
X1 58?
of
is
X1 511?
22. Assume that a complex multiply takes 1µs and
that the amount of time to compute a DFT is
determined by the amount of time it takes to
perform all of the multiplications.
(a) How much time does it take to compute a
1024-point DFT directly?
(b) How much time is required if an FFT is used
23.
Speech data is sampled at a rate of 10 kHz is
to be processed in real time using FFT. Part of
the computations required involve collecting
blocks of 1024 speech values and computing
a 1024-point DFT and a 1024 point inverse DFT.
If it takes 1 µs for each real multiply, how much
time remains for processing the data after the
DFT and inverse DFT are computed?
89
Key for Practice Questions
01. Ans:
10 # 103
1024
02. Ans: { 6, -2+2j, -2, -2 -2j}
04.
(a). Ans: [4, 3, 6, 5]
(b). Ans: [5, 4,3,6]
(c). Ans: [6, 3,4,5]
05.
Ans:
06.
Ans: (a)
X (5) = 0.125 + j0.0518
X(6) = 0, X(7) = 0.125 + j0.3018
09. (i) Ans: X2(k) = (-1)kX1(k)
(ii) Ans: y(n) = x((n-4))6
= {2, 1, 0, 0, 4, 3}
12. Ans: (a)
16. Ans: (a)
Signals & Systems
90
Signals & Systems
10. Tables
Operational Properties of the Laplace Transform
Note: x(t) is to be regarded as the causal signal x(t)u(t).
Entry
Property
x(t)
X(s)
1
Superposition
αx1(t) + βx2(t)
αX1(s) + βX2(s)
2
Times-exp
e-αt x(t)
X(s + α)
3
Times-cos
cos(αt)x(t)
0.5[X(s + jα) + X(s - jα)]
4
Times-sin
sin(αt)x(t)
j0.5[X(s + jα) - X(s - jα)]
5
Time Scaling
x(αt), α > 0
6
Time Shift
x(t - α) u(t - α), α > 0
e-αs X(s)
7
Times-t
t x(t)
-
tn x(t)
]- 1gn ds n
x′(t)
sX(s) - x(0-)
10
x′′(t)
s2X(s) – sx(0–) – x′(0–)
11
x(n)(t)
snX(s) – sn–1 x(0–) –….– xn-1(0–)
8
9
Derivative
Integral
# x (t) dt
X (s)
s
0-
13
14
15
Convolution
Switched periodic,
x1(t) = first period
Initial value
dX (s)
ds
d n X (s)
t
12
1 bsl
aX a
x(t)∗h(t)
X(s)H(s)
xp(t)u(t)
X1 (s)
, T = time period of x (t)
1 - e -sT
Laplace Transform Theorems
x (0 +) = lim 6sX (s)@ (if X (s) is strictly proper)
s"3
16
Final value
x (t)
t"3
= lim 6sX (s)@ (if poles of X (s) lie in LHP)
s"0
91
Signals & Systems
A Short Table of Laplace Transform
Entry
x(t)
X(s)
Entry
x(t)
X(s)
s2 - b2
^s2 + b2h2
1
δ(t)
1
10
tcos(βt)u(t)
2
u(t)
1
s
11
tsin(βt)u(t)
2sb
^s2 + b2h2
3
r(t) = tu(t)
1
s2
12
cos2(βt)u(t)
s2 + 2b2
s ^s2 + 4b2 h
4
t2u(t)
2
s3
13
sin2(βt)u(t)
5
tnu(t)
n!
sn + 1
14
e-αtcos(βt)u(t)
6
e-αtu(t)
1
s+a
15
e-αtsin(βt)u(t)
7
te-αtu(t)
1
(s + a) 2
16
te-αtcos(βt)u(t)
8
tn e-αtu(t)
n!
(s + a) n + 1
17
te-αtsin(βt)u(t)
9
cos(βt)u(t)
s
s2 + b2
18
sin(βt)u(t)
2b2
s ^s + 4b2 h
2
s+a
]s + ag2 + b2
b
]s + ag2 + b2
(s + a) 2 - b2
6]s + ag2 + b2@2
2b (s + a)
6]s + ag2 + b2@2
b
s2 + b2
92
Signals & Systems
SFG for 8-Point DIT-FFT algorithm:
X(0)
x(0)
x(4)
W 80
–1
x(2)
x(6)
W8
0
X(1)
W 80
W8
X(2)
–1
2
–1
–1
W8
X(3)
0
x(1)
x(5)
W8
0
W 81
–1
x(3)
x(7)
W 80
W8
0
W8
–1
W 82
–1
X(4)
–1
X(5)
–1
2
X(6)
–1
W 83
X(7)
–1
–1
SFG for 8-Point DIF-FFT algorithm:
X(0)
x(0)
W8
x(1)
x(2)
–1
x(3)
W8
x(4)
x(5)
x(6)
x(7)
–1
–1
–1
–1
0
W8
0
W8
2
0
–1
X(4)
X(2)
–1
W8
–1
0
X(6)
X(1)
W8
W 81
W8
2
W 83
W8
–1
–1
W8
0
0
X(5)
–1
X(3)
2
W8
–1
0
X(7)
93
Signals & Systems
SFG for computation of IDFT using inverse Radix-2 DIT-FFT:
1/8
X(0)
-
W 80
X(4)
-
W 80
X(2)
–1
X(6)
X(1)
–1
X(5)
–1
X(7)
-
-2
8
W
W 80
–1
–1
1/8
1/8
W
-3
8
-0
8
–1
-
W
1/8
W
-
W 81
–1
X(3)
–1
W 82
–1
-
W 80
1/8
W
-0
8
1/8
–1
-2
8
1/8
W
-0
8
1/8
–1
–1
SFG for computation of IDFT using inverse Radix-2 DIF-FFT:
1/8
X(0)
-
X(1)
W 80
1/8
–1
X(2)
X(3)
W
-0
8
1/8
-
W 80
W
-2
8
–1
–1
–1
1/8
W
-0
8
X(4)
X(5)
W
-0
8
-
W 81
–1
X(6)
X(7)
W
-0
8
W
W
–1
-0
8
-2
8
W
–1
–1
-2
8
-
W 83
1/8
–1
1/8
–1
1/8
–1
1/8
–1
x(0)
x(4)
x(2)
x(6)
x(1)
x(5)
x(3)
x(7)
x(0)
x(1)
x(2)
x(3)
x(4)
x(5)
x(6)
x(7)