1
$\begingroup$

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;

figure
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.

$\endgroup$
4
  • $\begingroup$ Your first equation $X(z)=\ldots$ is messed up, please correct it. Are you sure you get a double complex pole? $\endgroup$
    – Matt L.
    Commented 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
    Commented 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.
    Commented 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
    Commented Nov 30, 2015 at 19:14

1 Answer 1

2
$\begingroup$

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:

$$X(z)=\frac{A}{1-pz^{-1}}+\frac{A^*}{1-p^*z^{-1}}$$

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

$\endgroup$
3
  • $\begingroup$ 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
    Commented Nov 30, 2015 at 18:09
  • $\begingroup$ @JohnMarvin: I've added the final solution and its derivation to my answer. $\endgroup$
    – Matt L.
    Commented 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
    Commented Dec 1, 2015 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.