Solution of ECE 315 Test 2 F06
1.
x n
A system is described by the equation y n = e in which x is the
excitation and y is the response. Identify its properties by circling the
appropriate answer, Yes or No.
Linear
Stable
Causal
2.
Yes
Yes
Yes
No
No
No
Time Invariant
Dynamic
Invertible
Yes
Yes
Yes
No
No
No
In the system below a = 7 and b = 3 .
(a)
Write the differential equation describing it.
(1 / 3) y (t ) = x (t ) (7 / 3) y (t )
(b)
()
()
The impulse response can be written in the form h t = Ket u t .
Find the numerical values of K and .
K = 3 , = -7
x(t)
+ -
∫
a
b
()
()
()
()
()
()
y(t)
y t + 7 y t = 3x t
h t + 7 h t = 3 t
Eigenvalue = 7 . Integrating once through zero.
h 0 h 0 + 7 h t dt = 3 u 0+ u 0 K = 3
0
=1
=K
=0
=0
( ) ( )
+
0+
()
=0
( ) ( )
() () ()
()
(
)
()
()
If y t = x t h t and x t = 4 rect t 1 and h t = 3rect t , what is
3.
(
)
the numerical value of y 1 / 2 ?
(
)
y 1/ 2 = 6
( ) x ( ) h (t ) d = 4 rect ( 1) 3rect (t ) d
y t =
(
)
(
) (
)
(
) (
)
y 1 / 2 = 12 rect 1 rect 1 / 2 d = 12 rect 1 rect 1 / 2 d
( )
rect ( 1 / 2 ) is 1 between 0 and 1 and zero elsewhere
rect ( 1) rect ( 1 / 2 ) is 1 between 1/2 and 1 and zero elsewhere
rect 1 is 1 between 1/2 and 3/2 and zero elsewhere
(
1
)
y 1 / 2 = 12 d = 6
4.
1/ 2
If y n = x n h n and x n = ramp n and
h n = 2 n 3 n 4 , what is the numerical value of y 10 ?
y 10 = 2
(
y n = ramp n 2 n 3 n 4 )
y n = 2 ramp n n 3ramp n n 4 y n = 2 ramp n 3ramp n 4 y 10 = 2 ramp 10 3ramp 6 = 20 18 = 2
5.
A system with an impulse response h1 n = a n u n is cascade connected
with a second system with impulse response h 2 n = bn u n . If
a = 1 / 2 and b = 2 / 3 , and h n is the impulse response of the overall
cascade-connected system what is the numerical value of h 3 ?
h 3 = 0.8102
N 1
N
, r = 1
rn = N
1
r
, r 1
n=0
1 r
a
h n = h1 n h 2 n =
m= m
u m bn m u n m n
n
m=0
m=0
(
h n = bn a m b m = bn a / b
h n = b
n
)
n+1
1/ 2
n+1
1 a / b
( 2 / 3)
h n =
(
1 a / b
(
1/ 6
(
)
m
bn a n+1 / b bn+1 a n+1
=
=
b a
1 a / b
)
n+1
(
)
= 6 2 / 3
) (
)
n+1
(
1/ 2
4
4
h 3 = 6 2 / 3 1 / 2 = 0.8102
)
n+1
Solution of ECE 315 Test 2 F06
1.
(
Linear
Stable
Causal
2.
)
A system is described by the equation y n = ln x n 1 in which x is
the excitation and y is the response. Identify its properties by circling the
appropriate answer, Yes or No.
Yes
Yes
Yes
Time Invariant
Dynamic
Invertible
No
No
No
Yes
Yes
Yes
No
No
No
In the system below a = 2 and b = 4 .
(a)
Write the differential equation describing it.
(1 / 4) y (t ) = x (t ) (1 / 2) y (t )
(b)
()
()
The impulse response can be written in the form h t = Ket u t .
Find the numerical values of K and .
K = 4 , = -2
x(t)
+ -
∫
a
b
()
()
()
()
()
()
y(t)
y t + 2 y t = 4 x t
h t + 2 h t = 4 t
Eigenvalue = 2 . Integrating once through zero.
h 0 h 0 + 2 h t dt = 4 u 0+ u 0 K = 4
0
=1
=K
=0
=0
( ) ( )
+
0+
()
=0
( ) ( )
() () ()
()
(
)
()
()
If y t = x t h t and x t = 2 rect t + 1 and h t = 5rect t , what is
3.
(
)
the numerical value of y 1 / 2 ?
(
)
y 1 / 2 = 5
( ) x ( ) h (t ) d = 2 rect ( + 1)5rect (t ) d
y t =
(
)
(
) (
)
(
) (
)
y 1 / 2 = 10 rect + 1 rect 1 / 2 d = 10 rect + 1 rect + 1 / 2 d
( )
rect ( + 1 / 2 ) is 1 between 0 and -1 and zero elsewhere
rect ( + 1) rect ( + 1 / 2 ) is 1 between -1/2 and -1 and zero elsewhere
rect + 1 is 1 between -1/2 and -3/2 and zero elsewhere
(
1/ 2
)
y 1 / 2 = 10
4.
d = 5
1
( )
n
If y n = x n h n and x n = 0.9 u n and
h n = 2 n 3 n 4 , what is the numerical value of y 3 ?
y 3 = 1.458
( )
(
n
y n = 0.9 u n 2 n 3 n 4 ( )
( )
n
)
n
y n = 2 0.9 u n n 3 0.9 u n n 4 ( )
( )
n
y n = 2 0.9 u n 3 0.9
( )
3
( )
y 3 = 2 0.9 u 3 3 0.9
1
n 4
u n 4 ( )
3
u 1 = 2 0.9 = 1.458
5.
A system with an impulse response h1 n = a n u n is cascade connected
with a second system with impulse response h 2 n = bn u n . If
a = 1 / 4 and b = 1 / 3 , and h n is the impulse response of the overall
cascade-connected system what is the numerical value of h 3 ?
h 3 = 0.1013
N 1
N
, r = 1
rn = N
1
r
, r 1
n=0
1 r
h n = h1 n h 2 n =
a
m= m
u m bn m u n m n
n
m=0
m=0
(
h n = bn a m b m = bn a / b
h n = b
n
(1 / 4)
h n =
(
1 a / b
)
n+1
1 a / b
n+1
( )
1/ 3
1 / 12
)
m
bn a n+1 / b bn+1 a n+1
=
=
b a
1 a / b
n+1
( )
= 12 1 / 3
( ) (
)
n+1
(
1/ 4
4
4
h 3 = 12 1 / 3 1 / 4 = 0.1013
)
n+1
Solution of ECE 315 Test 2 F06
1.
(
Linear
Stable
Causal
2.
)
A system is described by the equation y n = ln x n + 1 in which x is
the excitation and y is the response. Identify its properties by circling the
appropriate answer, Yes or No.
Yes
Yes
Yes
Time Invariant
Dynamic
Invertible
No
No
No
Yes
Yes
Yes
No
No
No
In the system below a = 8 and b = 5 .
(a)
Write the differential equation describing it.
(1 / 5) y (t ) = x (t ) (8 / 5) y (t )
(b)
()
()
The impulse response can be written in the form h t = Ket u t .
Find the numerical values of K and .
K = 5 , = -8
x(t)
+ -
∫
a
b
()
()
()
()
()
()
y(t)
y t + 8 y t = 5x t
h t + 8 h t = 5 t
Eigenvalue = 8 . Integrating once through zero.
h 0 h 0 + 8 h t dt = 5 u 0+ u 0 K = 5
0
=1
=K
=0
=0
( ) ( )
+
0+
()
=0
( ) ( )
3.
() () ()
()
()
the numerical value of y (1) ? y (1) = -3
()
( )
If y t = x t h t and x t = 2 rect t and h t = 3rect t / 2 , what is
( ) x ( ) h (t ) d = 2 rect ( )( 3) rect ((t ) / 2) d
y t =
()
( ) ((
) )
( ) ((
) )
y 1 = 6 rect rect 1 / 2 d = 6 rect rect 1 / 2 d
()
rect (( 1) / 2 ) is 1 between 0 and 2 and zero elsewhere
rect ( ) rect ( / 2 ) is 1 between 0 and 1/2 and zero elsewhere
rect is 1 between -1/2 and 1/2 and zero elsewhere
()
1/ 2
y 1 = 6 d = 3
4.
0
( )
n
If y n = x n h n and x n = 0.9 u n and
h n = 2 n 3 n 2 , what is the numerical value of y 3 ?
y 3 = -1.242
( )
(
n
y n = 0.9 u n 2 n 3 n 2 ( )
( )
n
)
n
y n = 2 0.9 u n n 3 0.9 u n n 2 ( )
n
( )
y n = 2 0.9 u n 3 0.9
( )
3
( )
n 2
u n 2 ( )
3
( )
y 3 = 2 0.9 u 3 3 0.9 u 1 = 2 0.9 3 0.9 = 1.242
5.
A system with an impulse response h1 n = a n u n is cascade connected
with a second system with impulse response h 2 n = bn u n . If
a = 1 / 5 and b = 1 / 3 , and h n is the impulse response of the overall
cascade-connected system what is the numerical value of h 2 ?
h 2 = 0.2178
N 1
N
, r = 1
rn = N
1
r
, r 1
n=0
1 r
h n = h1 n h 2 n =
a
m
m= u m bn m u n m For n < 0 , h n = 0 .
For n 0 ,
n
n
m=0
m=0
(
h n = bn a m b m = bn a / b
h n = b
(1 / 5)
h n =
(
1 a / b
n
)
n+1
1 a / b
n+1
( )
1/ 3
m
bn a n+1 / b bn+1 a n+1
=
b a
1 a / b
=
n+1
2 / 15
)
( )
= 7.5 1 / 3
n+1
( )
1/ 5
For all time
( )
h n \ = 7.5 1 / 3
n+1
( )
1/ 5
( ) ( )
n+1
u n
3
3
h 2 = 7.5 1 / 3 1 / 5 u 2 = 0.2178
n+1