# MATH 267 Problem Set 11 Solutions The notation ```MATH 267 Problem Set 11 Solutions
The notation
u[n] =
n
1
0
o
if n ≥ 0
,
if n &lt; 0
δ[n] =
1 if n = 0
0 if n 6= 0
will be used throughout this problem set.
1. The output y[n] generated by a system S for the input x[n] is
∞
X
y[n] =
x[k]g[n − 2k]
k=−∞
where g[n] = u[n] − u[n − 4].
(a) Determine y[n] when x[n] = δ[n − 1].
(b) Determine y[n] when x[n] = δ[n − 2].
(c) Is the system LTI?
(d) Determine y[n] when x[n] = u[n].
P∞
P∞
Solution. (a) y[n] = k=−∞ x[k]g[n − 2k] = k=−∞ δ[k − 1]g[n − 2k] = g[n − 2] = u[n − 2] − u[n − 6]
P∞
P∞
(b) y[n] = k=−∞ x[k]g[n − 2k] = k=−∞ δ[k − 2]g[n − 2k] = g[n − 4] = u[n − 4] − u[n − 8]
(c) No. If S were LTI, the answer to (b) would have to be the answer to (a) with a time shift of 1, rather
than 2.
(d)
∞
∞
∞
X
X
X
y[n] =
x[k]g[n − 2k] =
u[k]g[n − 2k] =
g[n − 2k]
=
=
k=−∞
∞
X
∞
X
k=−∞
k=0
k=0
u[n − 2k] −
X
1−
0≤k≤ n
2
X
k=0
u[n − 2k − 4]
1=
X
n −2&lt;k≤ n
2
2
k≥0
0≤k≤ n
2 −2

0 if n &lt; 0


1 if n = 0
1=

 1 if n = 1
2 if n ≥ 2
2. Determine the z–transforms for each of the following signals. Sketch the pole–zero plot and indicate the
region of convergence. If the unit circle, |z| = 1, is in the ROC, find the signal’s Fourier Transform as
well. n
(a) 12
u[n + 4] − u[n − 5]
(b) (−2)n u[n − 3]
|n|
π
(d) 4n cos 2π
(c) 12
6 n + 4 u[−n − 1]
Solution. (a) Note that u[n + 4] is one for n ≥ −4 and zero otherwise and u[n − 5] is one for n ≥ 5 and
zero otherwise. So u[n + 4] − u[n − 5] is one for −4 ≤ n &lt; 5 and zero otherwise. Hence
H(z) =
4
X
n=−4
1 n −n
z
2
=
4
X
(2z)−n = (2z)4 +(2z)3 +(2z)3 +(2z)+1+(2z)−1 +(2z)−2 +(2z)−3 +(2z)−4
n=−4
(∗)
The region of convergence consists of the entire complex plane, with the exception of z = 0, where there
p+1
is a pole (because of the negative powers of z). To determine the zeroes, use a + ar + &middot; &middot; &middot; + arp = a 1−r
1−r
with r = (2z)−1 , a = (2z)4 and p = 8. Except when r = (2z)−1 = 1, i.e. except when z = 12 , we have
−9
9
1−(2z)
−4 (2z) −1
H(z) = (2z)4 1−(2z)
−1 = (2z)
(2z)−1
1
So the zeroes of H(z) are 21 times the various 9th roots of unity, except z = 12 itself. (When z = 12 , H(z)
is a sum of nine strictly postive terms and so cannot be zero. See (∗).) So the zeroes are z = 21 e2kπi/9 for
iω 9
) −1
k = 1, 2, 3, 4, 5, 6, 7, 8. Because |z| = 1 is in the region of convergence ĥ(ω) = H(eiω ) = (2eiω )−4 (2e
(2eiω )−1 .
(b) By definition
∞
X
X(z) =
x[n]z −n =
n=−∞
∞
X
−
2 n
z
=
(−2/z)3
1+2/z
8
= − z2 (z+2)
n=3
provided − z2 | &lt; 1. So the region of convergence is |z| &gt; 2 . There are poles at z = 0, −2 and no
zeroes. (The pole at z = 0 is not even on the boundary of the region of convergence and so can be
legitimately discarded.) In this part |z| = 1 is not in the region of convergence.
(c)
∞
X
H(z) =
1 |n| −n
z
2
=
n=−∞
∞
X
1 n −n
z
2
−1
X
+
1 −n −n
z
2
=
n=−∞
n=0
∞
X
(2z)
−n
−
n=0
∞
X
z n
2
n=1
We made the substitution in n → −n in the second sum. Continuing,
1
1−(2z)−1
H(z) =
−
z/2
1−z/2
=
2z
2z−1
+
z
z−2
4z−5
= z 2(z−2)+(2z−1)
(2z−1)(z−2) = z (2z−1)(z−2)
1
&lt; 1, z &lt; 1 or 1 &lt; |z| &lt; 2. There are poles at z = 1 , 2 and zeroes at
with region of convergence 2z
2
2
2
iω
−5
z = 0, 54 . Because |z| = 1 is in the region of convergence ĥ(ω) = H(eiω ) = eiω (2eiω4e
−1)(eiω −2) .
(d) The z–transform is
H(z) =
∞
X
h[n]z −n =
n=−∞
=
1 −i π
4
2e
−1
X
4n cos
2π
6 n
+
π
4
z −n =
n=−∞
∞
X
m=1
m
π
ei 3 z4
+
∞
X
1
2
π
π
π
π
ei 3 m−i 4 + e−i 3 m+i 4
z m
4
where m = −n
m=1
1 iπ
4
2e
∞
X
m
π
e−i 3 z4
m=1
=
π z
1 −i π
4 ei 3
2e
4
π
1 − ei 3 z4
+
π z
1 iπ
4 −i 3
2e e
4
π
1 − e−i 3 z4
π
π
π
π
π
π π
π
z ei 12 1 − e−i 3 z4 + e−i 12 1 − ei 3 z4
z ei 12 + e−i 12 − z4 ei 4 + e−i 4
=
=
π
π
π
π
8
8
1 − ei 3 z4 1 − e−i 3 z4
1 − ei 3 z4 1 − e−i 3 z4
π
− z4 cos π4
cos 12
z
=
π z
π
4 1 − ei 3 4 1 − e−i 3 z4
π π
provided e&plusmn;i 3 z4 &lt; 1. That is, provided |z| &lt; 4. There are poles at e&plusmn;i 3 z4 = 1 or equivalently, at
π
cos 12
π
z = 4e&plusmn;i 3 . There are zeroes at z = 0 and z = 4
. Because |z| = 1 is in the region of convergence
cos π4
ĥ(ω) = H(eiω ) =
π
cos 12
− 14 eiω cos π4
1 iω
e
π
π
4
1 − 14 ei 3 eiω 1 − 14 e−i 3 eiω
3. Let
x[n] = (−1)n u[n] + αn u[−n − n0 ]
Determine the constraints imposed on the complex number α and the integer n0 by the requirement
that the z–transform of x[n] has region of convergence 1 &lt; |z| &lt; 2.
2
Solution. The z–transform of x[n] is
X(z) =
=
∞
X
n=−∞
∞
X
x[n]z −n =
(−z)−n +
n=0
∞
X
(−z)−n +
−n
X0
α n
z
n=−∞
n=0
∞
X
=
∞
X
(−z)−n +
∞
X
α −m
z
where m = −n
m=n0
n=0
z m
α
m=n0
The first sum converges for (−z)−1 &lt; 1 or |z| &gt; 1. The second sum converges for αz &lt; 1 or |z| &lt; |α|.
So we need |α| = 2 . There is no constraint placed on n0 .
4. Use partial fractions and the fact that the z–transform of an u[n] is
|z| &gt; |a|, to find the inverse z–transform of
X(z) =
1 − 13 z −1
,
(1 − z −1 )(1 + 2z −1 )
1
1−az −1 ,
with region of convergence
|z| &gt; 2
Solution. Write ζ = z −1 . Then
X(z) =
1 − 13 ζ
A
B
A(1 + 2ζ) + B(1 − ζ)
=
+
=
(1 − ζ)(1 + 2ζ)
1−ζ
1 + 2ζ
(1 − ζ)(1 + 2ζ)
with A and B determined by
A(1 + 2ζ) + B(1 − ζ) = 1 − 31 ζ ⇐⇒ A + B = 1, 2A − B = − 13 ⇐⇒ A = 29 , B =
Hence
X(z) =
2 1
9 1−ζ
+
7 1
9 1+2ζ
⇒
3
x[n] = 29 u[n] + 79 (−2)n u[n]
7
9
```