I was computing an inverse z-transform here, and I am facing some problems.

So, the z-transform is:

$$ X(z) = \frac{2+3z^{-1}}{1 - z^{-1} + 0.81z^{-2}} , |z| > 0.9 $$

I found the following poles:

$$ p_1 = 0,5 + j0,7483 $$ $$ p_2 = 0,5 - j0,7483 $$

and then, the following residues:

$$ R_1 = 1 - j2,6727 $$ $$ R_2 = 1 + j2,6727 $$

Rewriting $ X(z) $ with the poles in polar coordinates:

$$ X(z) = \frac{1 - j2,6727}{1 - 0,9 \angle 56,25^{\circ}} + \frac{1 + j2,6727}{1 - 0,9 \angle -56,25^{\circ}} $$

Through the magnitude of the poles, I can see that they are both inside the ROC, then, I can write $ x(n) $ as:

$$ x(n) = (1 - j2,6727)(0,9 \angle 56,25^{\circ})^{n}u(n) + (1 + j2,6727)(0,9 \angle -56,25^{\circ})^{n}u(n) $$

I know that it is possible to write $ x(n) $ in a form without complex numbers. I know this, because it was asked and I was able to plot the figure in Matlab using the following code:

delta = zeros(30, 1);
delta(1) = 1;

a = [1 -1 0.81];
b = [2 3];

x = filter(b, a, delta);
n = 0:29;

stem(n, x);

My question is, what I did up to now is right? If it is, How can I compute $ x(n) $ without complex numbers? Is there any tips that you can give me to follow?

Thanks in advance.

  • $\begingroup$ Your first equation $X(z)=\ldots$ is messed up, please correct it. Are you sure you get a double complex pole? $\endgroup$
    – Matt L.
    Nov 30, 2015 at 17:44
  • $\begingroup$ Hi, thank you. I made the correction. Sorry, it was not a double complex pole. They are conjudate. I wrote wrong. Sorry. $\endgroup$
    – JohnMarvin
    Nov 30, 2015 at 18:06
  • $\begingroup$ That all seems correct. The only thing you might want to do next is to change from $0.9\angle 56.25$ to polar coordinates $0.9 e^{j\frac{5\pi}{32}}$ and then use the fact that you have complex conjugate entries to find $\cos$ or $\sin$ expressions which will be real-valued only. $\endgroup$
    – Peter K.
    Nov 30, 2015 at 18:46
  • 2
    $\begingroup$ Hi @Peter K. thanks for your help. I am gonna give it a try and give you a feedback. $\endgroup$
    – JohnMarvin
    Nov 30, 2015 at 19:14

1 Answer 1


Since the given $\mathcal{Z}$-transform

$$X(z)=\frac{2+3z^{-1}}{1-z^{-1}+0.81z^{-2}},\quad |z|>0.9\tag{1}$$

has real-valued coefficients, its poles and zeros must be either real-valued or they must occur in complex conjugate pairs. The poles of $(1)$ are a complex conjugate pair:

$$p_1=0.5 + 0.74833j=re^{j\phi}=p\\ p_2=0.5 - 0.74833j=re^{-j\phi}=p^*$$

with $r=0.9$ and $\phi=0.98177$.

You can proceed from the partial fraction expansion of $X(z)$, which you correctly obtained:


With the given region of convergence (ROC) $|z|>0.9$, we know that we're looking for a causal sequence $x[n]$:

$$x[n]=\left(Ap^n+A^*(p^*)^n\right) u[n]\tag{3}$$

With $p=re^{j\phi}$ and $A=|A|e^{j\phi_A}$, $(3)$ can be rewritten as

$$x[n]=|A|r^n\left(e^{j(n\phi+\phi_A)}+e^{-j(n\phi+\phi_A)}\right)u[n]= 2|A|r^n\cos(n\phi+\phi_A)u[n]\tag{4}$$

which is of course real-valued.

With the given values of $p$ and $A$ the impulse response looks like this:

enter image description here

  • $\begingroup$ Hi, @Matt, thanks for your answer. I am really sorry, but I wrote the poles in the question wrong. It is not a double pole, they are conjugate poles. So, I am still stuck in it. $\endgroup$
    – JohnMarvin
    Nov 30, 2015 at 18:09
  • $\begingroup$ @JohnMarvin: I've added the final solution and its derivation to my answer. $\endgroup$
    – Matt L.
    Dec 1, 2015 at 12:59
  • $\begingroup$ You have given me an enormous help. I am gonna try the algebraic procedures by myself using your answer as reference. Thanks a lot! $\endgroup$
    – JohnMarvin
    Dec 1, 2015 at 18:35

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