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Important Properties of the z-Transform 111
with no change in the ROC. Applying the multiplication by a ramp property,
{
X(z) =z −2 −z dZ[(0.5)n cos( π n)u(n)] }
3
dz
with no change in the ROC. Now the z-transform of (0.5) n cos( π n)u(n) from
3
Table 4.1 is
Hence
[
Z (0.5) n cos
X(z) =−z −1 d
dz
{
= −z −1
=
( ) ] πn
u(n) =
3
{
=
1 − (0.5 cos π 3 )z−1
1 − 2(0.5 cos π ; |z| > 0.5
3 )z−1 +0.25z−2 1 − 0.25z −1
; |z| > 0.5
1 − 0.5z −1 +0.25z−2 }
1 − 0.25z −1
, |z| > 0.5
1 − 0.5z −1 +0.25z −2
}
−0.25z −2 +0.5z −3 − 0.0625z −4
, |z| > 0.5
1 − z −1 +0.75z −2 − 0.25z −3 +0.0625z −4
0.25z −3 − 0.5z −4 +0.0625z −5
, |z| > 0.5
1 − z −1 +0.75z −2 − 0.25z −3 +0.0625z−4 MATLAB verification: To check that this X(z) isindeed the correct expression,
let us compute the first 8 samples of the sequence x(n) corresponding to X(z),
as discussed before.
>> b = [0,0,0,0.25,-0.5,0.0625]; a = [1,-1,0.75,-0.25,0.0625];
>> [delta,n]=impseq(0,0,7)
delta =
1 0 0 0 0 0 0 0
n =
0 1 2 3 4 5 6 7
>> x = filter(b,a,delta) % check sequence
x =
Columns 1 through 4
0 0 0 0.25000000000000
Columns 5 through 8
-0.25000000000000 -0.37500000000000 -0.12500000000000 0.07812500000000
>> x = [(n-2).*(1/2).^(n-2).*cos(pi*(n-2)/3)].*stepseq(2,0,7) % original sequence
x =
Columns 1 through 4
0 0 0 0.25000000000000
Columns 5 through 8
-0.25000000000000 -0.37500000000000 -0.12500000000000 0.07812500000000
This approach can be used to verify the z-transform computations.
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