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Impulse Response Problems March 9, 2014 1
Impulse Response Problems
Given x n [ ] = sin 4 π n and h n [ ] = δ[ ] n − 2 [δ n − 1], Find y n [ ] = x n [ ] ∗ h n [ ]. Express your answer as a single cosine of
the form y n [ ] = A cos( n + θ).
► Solution: a)
4 (^ )
4 4 4 4 4 4 4 4 4 4 4 4 4
[ ] sin [ ] 2 [ 1] sin [ ] 2 sin [ 1] sin 2 sin ( 1) sin 2 sin cos 2 cos sin sin sin cos cos
y n n n n n n n n n n n n n n n n n
π π π π π π π π π π π π π π
For each of the following pairs of x n [ ] and h n [ ] , plot y n [ ] = x n [ ] ∗ h n [ ].
► Solution:
Impulse Response Problems March 9, 2014 2
2.
Given a filter with impulse response (^) h n [ ] and output (^) y n [ ] , we observe that when the input
is x n [ ] = δ [ ] n − δ[ n − 1] + δ[ n − 2], the output is y n [ ] = δ [ ] n + δ[ n − 1] − 2 [δ n − 2] + 3 [δ n − 3] − δ[ n − 4]. Find the output
when x n [ ] = δ [ ] n + δ[ n − 1] + δ[ n − 2].
► Solution: By deconvolution:
1 2 1 1 1 1 1 1 2 3 1 1 1 1 2 3 3 2 2 2 1 1 1 1
So, h n [ ] = δ [ ] n + 2 [δ n −1] − δ[ n − 2].
Then, when x n [ ] = δ [ ] n + δ[ n − 1] + δ[ n − 2], y n [ ] = δ [ ] n + 3 [δ n −1] + 2 [δ n − 2] + δ[ n − 3] − δ[ n − 4].
2.
Given the system shown in Figure P9-1, with
1 2 3
[ ] [ ] [ 1] [ 2]
[ ] [ 1] [ 1]
[ ] [ 3]
h n n n n h n n n h n n
a) Find the equivalent impulse response of the system, (^) h n [ ]. b) Find the response of the system when (^) x n [ ] = δ[ n + 1] + 2 [ ]δ n + δ[ n − 1].
Impulse Response Problems March 9, 2014 4
h n 1 [ ] ∗ h 2 [ ] n a 2 a + b − a + 2 b + c − b + 2 c − c h 3 (^) [ ] n d^ e^ f h n [ ] a 2 a + b − a + 2 b + c − b + 2 c + d − c + e f
Matching with (^) h n [ ] , we get (^) a = 1 , (^) b = 1 c = 1 , (^) d = 2 e = 3 f = 2 and (^) m = 2.
2.
Given the system shown in Figure P9-9, with
1
[ ] [ ] [ 1]
[ ] [ 1] [ ]
[ ] 2 [ ] 4 [ 1] [ 2] 3 [ 3]
x n n n h n n n y n n n n n
Find (^) h 2 (^) [ ] n.
h 1 [n]
x[n] (^) + y[n]
h 2 [n]
Figure 9-
► Solution:
Let h n [ ] = h n 1 [ ] + h 2 [ ] n. Then y n [ ] = x n [ ] ∗ h n [ ]. Deconvolution gives h n [ ] = 2 [ ] δ n + 2 [δ n − 1] − 3 [δ n − 2].
Hence, h 2 (^) [ ] n = h n [ ] − h n 1 [ ] = 3 [ ] δ n + δ[ n − 1] − 3 [δ n − 2].
2.
We observe that when x n [ ] = [1 __ 2 1]and h n [ ] = [ 1 __ __ ], (assuming all sequences start at n = 0 ), then y n [ ] = x n [ ] ∗ h n [ ] = [1 2 0 __ __ 1 ]. Find all possible outputs.
► Solution: Let the unknown values of x n [ ] and h n [ ] be denoted by a , b and c. Then y n [ ] is found by convolution, which can be written as
( ) ( ) ( ) ( )
[1 2 1]
[1 ]
a b c c ac c c b ab b b a a + b ab + c + ac + b + c + b c
Comparing with the known output, we immediately conclude that c = 1 and a + b = 2. Hence, the output becomes
Impulse Response Problems March 9, 2014 5
[1 2 (^) ( − a^2 + 2 a + (^3) ) ( 5 − a (^) ) ( 4 − a )1].
The quadratic − a^2 + 2 a + 3 = 0 solves to give a = − 1 or 3. Hence, there are two possible solutions: [1 2 0 6 5 1] or [1 2 0 2 1 1].
2.
The input, x n [ ] , and impulse response, h n [ ] , are shown in Figure P9-11. Find y [1] , y [6] and y [10].
Figure P9-
► Solution: y [1] = y [6] = y [10] = 3
