# 2E2 Tutorial sheet 7 Solution [Wednesday December 6th, 2000] 1. Find the ```2E2 Tutorial sheet 7 Solution
[Wednesday December 6th, 2000]
1. Find the Z transfer function and impulse response for the difference equation
xk+2 + 3xk+1 − 4xk = vk
(with zero initial conditions).
Solution: Taking Z transforms of both sides gives
∞
(z 2 + 3z − 4)Z [(xk )∞
k=0 ] (z) = Z [(vk )k=0 ] (z)
or
Z [(xk )∞
k=0 ] (z) =
z2
1
Z [(vk )∞
k=0 ] (z)
+ 3z − 4
and the Z transfer function is
1
z 2 + 3z − 4
the impulse response if the sequence with Z transform equal to Y (z) and we can find this
via partial fractions.
Y (z) =
1
1
=
z 2 + 3z − 4
(z + 4)(z − 1)
B
A
=
+
z+4 z−1
1 = A(z − 1) + B(z + 4)
z = −4 :
1 = −5A
1
A=−
5
z=1:
1 = 5B
1
B=
5
1
1 1
1 1
= −
+
2
z + 3z − 4
5z +4 5z −1
z times this would be the Z transform of the sequence with k th term
1
1
1
1
− (−4)k + 1k = − (−4)k +
5
5
5
5
Without the z factor, we must delay this by one. So the impulse repsonse has k th term
0
for k = 0
xk =
− 15 (−4)k−1 + 51 for k &gt; 0
2. Find the convolution of the sequence (1)∞
k=0 = (1, 1, 1, 1, . . .) with itself.
Solution: We could work directly with the definition of convolutions of two sequences.
∞
th term
The convolution of sequences (xk )∞
k=0 and (yk )k=0 has k
k
X
xj yk−j
j=0
and when both xj = 1 and yk−j = 1 this turns out to be
k
X
1=k+1
j=0
So the answer is the sequence (k + 1)∞
k=0 = (1, 2, 3, . . .).
Another possible solution would make use of the fact that the Z transform of the convolution is the product of the Z transforms, that is
z
z
z
&times;
=z
z−1 z−1
(z − 1)2
z
∞
Now (z−1)
2 is the Z transform of the sequence (k)k=0 and the z factor advances it by one,
resulting in the same answer again.
Richard M. Timoney
2
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