Digital Signal Processing Reference
In-Depth Information
and
Ae
ju
2
A
2
l
u Q A = 2 ƒ k
1
ƒ ;
k
1
=
=
u = arg k
1
.
(11.50)
Thus, we calculate and from the poles, and
A
and from the partial-fraction
expansion. We can then express the inverse transform as the sinusoid of (11.47). An
illustrative example is given next.
©
Æ
u
Inverse
z
-transform with complex poles
EXAMPLE 11.11
We find the inverse
z
-transform of the function
-2.753z
-2.753z
(z - 0.550 - j0.550)(z - 0.550 + j0.550)
Y(z) =
=
z
2
- 1.101z + 0.6065
k
*
z
z - 0.550 + j0.550
.
k
1
z
z - 0.550 - j0.550
+
=
Dividing both sides by
z
, we calculate
k
1
:
-2.753
(z - 0.550 - j0.550)(z - 0.550 + j0.550)
B
R
k
1
= (z - 0.550 - j0.550)
z= 0.550 + j0.550
-2.753
2(j0.550)
= 2.50
l
90°
.
=
From (11.49) and (11.50),
p
1
= 0.550 + j0.550 = 0.7778
l
45°
,
p
4
,
©=ln ƒp
1
ƒ = ln (0.7778) =-0.251;
Æ=arg p
1
=
and
p
2
.
A = 2 ƒ k
1
ƒ = 2(2.50) = 5;
u = arg k
1
=
Hence, from (11.47),
p
4
n +
p
2
y[n] = Ae
©n
cos(Æn + u) = 5e
-0.251n
cos
¢
≤
.
We can verify this result by finding the
z
-transform of this function, using Table 11.2. The
partial-fraction expansion can be verified with the following MATLAB program:
n = [0 0 -2.753];
d = [1 -1.101 0.6065];
[r,p,k]=residue(n,d)
result: r=0+2.5001i 0-2.5001i
p=0.5505+0.5509i 0.5505-0.5509i
■
