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Tutorial II
1. Evaluate the following discrete-time convolution sums:
(i) y[n] = u[n+3] * u[n-3]
(ii) 3nu[-n+3] * u[n-2]
Tutorial II
EKT 230 Signals & Systems
Tutorial II
EKT 230 Signals & Systems
(iii) y[n] = (1/4)n u[n] * u[n+2]
for n+2 < 0, n < -2
y[n] = 0
for n+2 >= 0, n >= -2
k

1
y[n]     u[k ].u[n  k  2]
k    4 
n2
1
  
k 0  4 
k
n  2 1
1
1  
4

1
1
4
Tutorial II
Tutorial II
EKT 230 Signals & Systems
n
4   1   1 
 1     
3   4   64 
Thus,
 4   1  n  1 
 1
y[n]   3   4   64 , n  2



0, n  2

2. Evaluate the convolution sum for a system with the input signal, x[n] and impulse
response, h[n] as in Figure 1. Draw the response y[n]. Use graphical and cross
multiplication methods.
Cross multiplication method
x[n]
h[n]
1
2
3
-1
-1
-2
-3
-1
-1
-2
-3
2
2
4
6
1
1
2
3
1
1
2
3
Ans : y[n] = [-1 -3 -3 2 9 5 3]
Tutorial II
Tutorial II
EKT 230 Signals & Systems
h[-k]
x[k]
1
0
n=0
2
1
1
3
2
3
4
5
h[0-k]
1
-2
2
-1
0
-1
2
1
-3
1
1
-1
y[0] = -1
-3
-2
-1 0
-1
n=1
1
-1
y[1] = -2-1
= -3
h[1-k]
1
-3
2
1
-2
-1 0
1
-1
n=2
y[2] = -3-2+2
= -3
h[2-k]
1
-3
n=3
-1
2
1
-2
-1 0
1
-1
h[3-k]
1
-3
1
-2
2
-1 0
1
-1
n=4
-1
h[4-k]
1
-3
-2
1
y[4] = 6+2+1
=9
2
-1 0
-1
Tutorial II
y[3] = -3+4+1
=2
-1
1
-1
Tutorial II
EKT 230 Signals & Systems
h[k]
x[k]
1
0
n=5
2
1
3
2
3
4
6
-3
2
1
-2
-1 0
-1
y[5] = 3+2
=5
1
-1
h[6-k]
y[6] = 3
1
-3
2
1
-2
-1 0
1
-1
n=7
-1
y[7] = 0
h[7-k]
1
-3
-2
1
2
-1 0
-1
y[n]=[-1 -3 -3 2 9 5 3]
Tutorial II
7
h[5-k]
1
n=6
5
1
-1
Tutorial II
EKT 230 Signals & Systems
3. Compute the convolution y[n] = x[n] * h[n] of the following signals. Draw the output
signal, y[n].
¼ for n = 0
½ for n = 1
¾ for n = 2
x[n] =
1
0
for n = 3
elsewhere
h[n] =
1
1
0
1
1
0
for n = 2
for n = 3
for n = 4
for n = 5
for n = 6
elsewhere
y[n] = x[n] * h[n]

=
 x[k ]h[n  k ]
k  
= x[0]h[n] + x[1]h[n-1] + x[2]h[n-2] + x[3]h[n-3]
= (1/4)h[n] + (1/2)h[n-1] + (3/4)h[n-2] + h[n-3]
Thus,
y[2] = ¼
y[3] = ¼ + ½ = ¾
y[4] = ½ + ¾ = 5/4
y[5] = ¼ + ¾ + 1 = 2
y[6] = ¼ + ½ + 1 = 7/4
y[7] = ½ + ¾ = 5/4
y[8] = ¾ + 1 = 7/4
y[9] = 1
y[10] = 0
Tutorial II
Tutorial II
EKT 230 Signals & Systems
2
y[n]
7/4
5/4
7/4
5/4
1
3/4
1/4
Or
0
1
2
3
4
5
6
7
8
9
10
x[n]
h[n]
¼
½
¾
1
1
¼
½
¾
1
1
¼
½
¾
1
0
0
0
0
0
1
¼
½
¾
1
1
¼
½
¾
1
y[n] = [ ¼ ¾ 5/4 2 7/4 5/4 7/4 1]
n=2
4. . A linear time invariant system has an impulse response, h(t) and input signal, x(t). Use
convolution to find the response, y(t) for following signal:

Flip.
Tutorial II
Tutorial II

For t < 1,

For 1=< t < 2,
EKT 230 Signals & Systems
y (t )  0
t 1
y (t )   (2)e ( ) d
0
 2e ( )

t 1
0
 2e (t 1)  2
For 2=< t < 3,
t 2
y (t ) 
 ( 2) e
 ( )
d 
 (2)e
 ( )
d
t 2
0
 2e
t 1
 ( ) t  2
1
 2e ( )
t
 2e ( t  2 )  2e (1)  2e ( t )  2e ( 2 )
2
 4e ( t  2 )  2e ( t 1)  2

For 3=< t < 4,
t 2
y (t ) 
 ( )
 ( 2) e d 
t 3
  2e ( )
t 2
t 3
2
 ( 2 ) e
 ( )
 2e ( )
2
t 2
 4 e  ( t  2 )  2 e  ( t  3 )  2
Tutorial II
d
t 2
 2 e  ( t  2 )  2 e  ( t  3 )  2 e  ( 2 )  2 e  ( t  2 )
Tutorial II

EKT 230 Signals & Systems
For 4=< t < 5,
2
y (t ) 
 ( 2) e
 ( )
d
t 3
  2e ( )

For 5 < t,
2
t 3
 2e ( 2 )  2e ( t 3)
y (t )  0
0


 2  2e (t 1)

(t  2 )
 2e (t 1)  2
  4e
y (t )  
(t  2 )
 2e (t 3)  2e  2
 4e

2e (t 3)  2e  2


0
t 1
1 t  2
2t 3
3t 4
4t 5
t 5
5. A linear time invariant system has an impulse response, h(t) and input signal, x(t).
Use convolution to find the response, y(t) for following signals:
x(t )  cos(t )u (t  1)  u (t  3) 
h(t )  u (t )
Convolution, y(t) = x(t)*h(t)
Tutorial II
Tutorial II

For t < -1, y(t) = 0.

For -1 =< t<0
EKT 230 Signals & Systems
t
y (t )   (1) cos( )d
1


For -1 =< t<0
Tutorial II
sin(  )

sin( t )


t

1
sin( t )


sin(  )

Tutorial II
EKT 230 Signals & Systems
3
y (t )   (1) cos( )d
1

sin(  )

3

1
sin( 3 )

0
 sin( t )

, 1  t  1
y (t )   
 0,
otherwise
6. Evaluate the following convolution integral:
m(t) = y(t) * z(t)
Tutorial II

sin(  )

Tutorial II
Tutorial II
EKT 230 Signals & Systems