CE 40763
DIGITAL SIGNAL PROCESSING
Homework 3 Solution
1. Suppose the system is causal:
2.
3.
1 n
a) x n = − 3
u −n − 2
∞
x n z −n =
X z =
n=−∞
∞
n=−∞
1
−
3
−2
n
u −n − 2 z
−n
=
−
n=−∞
1
3
n
z −n =
∞
1
−
3
n=2
∞
−n
n
z =
n=0
1
−
3
−n−2
z
n+2
9z 2
3z
1
=
=
, z <
1 + 3z 1 + 1 z −1
3
3
The fourier transform doesn′ t exist because the ROC doesn′ t include the unit circle.
1 n
b) x n = 2n u −n +
4
u n−1
0
∞
n
n −n
x1 n = 2 u −n → X1 z =
2 z
=
n=−∞
1
4
x2 n =
2
−2z −1
z =
, z <2
1 − 2z −1
−n n
n=0
n
u n − 1 → X2 z
∞
=
n=1
−2z −1
X z = 1−2z −1 +
z −1
4
1
4
1
[
∞
n
1
1− z −1
4
z
−n
=
n=0
1
4
n+1
z −n−1 =
z −1
1
1
[
], z >
4 1 − 1 z −1
4
4
1
],4 < z < 2
The fourier transform exist because the ROC includes the unit circle.
c) x n = n
1
2
n
=n
1 n
2
u n − n2n u −n − 1 = nx1 n − nx2 n
1
1
, z >
1
2
1 − z −1
2
−1
X2 z =
, z <2
1 − 2z −1
X1 z =
d
d
X z = −z dz X1 z + z dz X2 z =
1
2
1
(1− z −1 )2
2
− z −1
2z −1
1
− (1−2z −1 )2 , 2 < z < 2
The fourier transform exist
4.
(a) Since limz→∞ H(z)=1, H(z) has no poles at infinity. Furthermore, since h[n] is
given to be right-sided, h[n] has to be causal.
(b) Since h[n] is causal, the numerator and denominator polynomials of H(z) have
the same order. Since H(z) is given to have two zeros, we may conclude that it
also has two poles. Since h[n] is real, the poles must occur in conjugate pair. It is
3
given that one of the poles lie on the circle defined by z = 4, Therefore, the
other pole also lies on the same circle. Clearly the ROC for H(z) will be of the
3
form z > 4 and will include the unit circle. Therefore the system is stable.
5.
a) The z-transform of x1 n = x[−n]:
∞
∞
x −n z −n =
X1 z =
n=−∞
n=−∞
1
x n z n = X( )
z
1
Therefore if x n = x −n , then X z = X(z ).
1
1
b) If z0 is a pole then X(z ) =0, from part (a) we know that X(z0 ) = X(z ) therefore
0
1
X(z 0 )
=
1
1
z0
X( )
0
=0. This implies that there is a pole at
1
z0
.
6. We know that x[n] is real. Therefore the poles and zeros of X(z) have to occur in
π
conjugate pairs. X(z) has a pole at z = 0.5ej 3 we can conclude that X(z) must have
π
another pole is at z = 0.5e−j 3 . Now since X(z) has no more poles we have to assume
that X(z) has 2 or less zeros. If X(z) has more than 2 zeros then X(z) must have poles
at infinity.
Since X(z) has 2 zeros at the origin, we know that X(z) must be of the form:
X z =
X 1 =
Az 2
π
π
z − 0.5e−j 3 (z − 0.5ej 3 )
8
3
We may conclude that A = 2 And since X(z) is right-sided, the ROC must
1
be z > 2.
7.
a) Taking the z-transform of both sides of the given difference equation and
simplifying, we get
𝑌(𝑧)
𝑧 −1
𝐻 𝑧 =
=
𝑋(𝑧) 1 − 𝑧 −1 − 𝑧 −2
1
The poles of H(z) are at 𝑧 = 2 ±
1
ROC for H(z) has to be 𝑧 > 2 +
5
2
. H(z) has a zero at z=0. Since h[n] is causal the
5
2
.
b) The partial fraction of H(z) is
1
1
5
5
𝐻 𝑧 =−
+
1+ 5
1− 5
1 − ( 2 )𝑧 −1 1 − ( 2 )𝑧 −1
ℎ𝑛 =
−1 1 + 5 𝑛
1 1− 5 𝑛
(
) 𝑢𝑛 +
(
) 𝑢𝑛
2
2
5
5
c) The ROC of a stable (non causal) unit sample response that satisfies the difference equation is
5 1
1
5
− < 𝑧 < +
2
2
2
2
So
ℎ𝑛 =
1 1+ 5 𝑛
1 1− 5 𝑛
(
) 𝑢 −𝑛 − 1 +
(
) 𝑢𝑛
2
2
5
5