Inverse Z-Transform of F(z) How to set ROC?

The same F(z) expression have different discrete sequence when they have different ROC, How do i set the ROC in following expression when calling iztrans? like 0.8<|z|<1.25

syms n z;
h=1/((1-0.8*z^(-1))*(1-0.8*z));
r=iztrans(h,z,n)

[yair edery]

did you mange to find the answer?

Answer 1

[Paul]

No, we can't specify the ROC for iztrans, which always assumes the time domain sequence is causal. Same issue with ilaplace.

However, we can use z-transform properties to obtain the desired result

syms n integer
syms z;

Define the discrete time unit step function (assumes default sympref)

u(n) = kroneckerDelta(n)/2 + heaviside(n);

Define H(z)

H(z) = 1/((1-0.8*z^(-1))*(1-0.8*z)) % ROC: 0.8 < |z| < 1.25

H(z) = 

Partial fraction expansion of H(z)

H(z) = partfrac(H(z))

H(z) = 

c = children(H(z));
H1(z) = c{1}

H1(z) = 

H2(z) = c{2}

H2(z) = 

Based on the ROC, we know that H2(z) with a pole at z = 0.8 that is inside the ROC corresponds to a right-sided, causal sequence. We can use iztrans directly

h2(n) = iztrans(H2(z))*u(n);

Based on the ROC, we know that H1(z) with a pole at z = 1.25 that is outside the ROC corresponds to a left-sided, noncausal sequence.

Define w[n] = h1[-n]. w[n] is right-sided and causal, and its z-transform is

W(z) = H1(1/z); % ROC |z| > 1.25

Get w[n]

w(n) = iztrans(W(z),z,n)*u(n);

Then h1[n] must be

h1(n) = w(-n);

Therefore h[n] is

h(n) = h1(n) + h2(n)

h(n) = 

Plot h[n] for some values of n

nval = -10:10;
figure
stem(nval,h(nval))

Had we done this by hand using z-transform tables, we'd get

hbyhand(n) = 25/9*( (4/5)^n*u(n) + (5/4)^(n)*u(-n-1) )

hbyhand(n) = 

It doesn't look like the same answer, but we can make it look closer:

rewrite(h(n),'piecewise')

ans = 

rewrite(hbyhand(n),'piecewise')

ans = 

Verify these are actually the same signals

simplify(rewrite(h(n),'piecewise') - rewrite(hbyhand(n),'piecewise'))

ans =