1
$\begingroup$

Given, $$X(z) = \frac{z}{3z^2 - 4z + 1}$$ Question 1

I need to calculate inverse z-transform for ROC $|z|>1$

When I try to calculate inverse $z$-transform using partial fraction of $X(z)$ and using long division method, I seem to get two different answers.

Let us use partial fraction first. Since $X(z)$ is already in its proper form, the partial fraction evaluates to: $$ X(z) = \frac{\frac 3 2}{z-1} - \frac{\frac 1 2 }{z-\frac 1 3}$$ On rearranging, $$X(z) = \frac 3 2 \frac{z^{-1}}{1-z^{-1}}-\frac 1 2 \frac{z^{-1}}{1-\frac 1 3 z^{-1}}$$ Taking inverse z-transform, $$x(n) = \frac 3 2 u(n-1) - \frac 1 2 (\frac 1 3)^{n-1}u(n-1) \tag{1}$$

Now on using long division method, the first few terms of the division are: $$\frac 1 3 z^{-1} + \frac 4 9 z^{-2} + ...$$ This suggests that the signal in time domain is: $$x(n) = \{0, \frac 1 3, \frac 4 9 , ...\}$$

But when I substitute $ n = 0,1,2,...$ in equation (1), I get the sequence: $$x(n) = \{0,1,\frac 4 3,...\}$$

Why am I getting two different sequences for the same function $X(z)$?

Question 2

The notes my teacher provided me had this exercise done by finding partial fraction for $\frac{X(z)}{z}$ rather than $X(z)$ itself. As the given function is already in the proper form (order of denominator is greater than the order of numerator), is it incorrect to calculate partial fraction of $X(z)$ itself?

This has eaten almost three hours of my time. Any help would be appreciated. Thanks.

EDIT

For when the ROC of $X(z)$ is $|z| < \frac 1 3$, $x(n)$ should be anti-causal. While shifting $x(n)$ due to presence of $z^{-1}$ in the numerator, which one among the following two is correct?

$$x(n) = - \frac 3 2 u(-n) + \frac 1 2 (\frac 1 3)^{n-1} u(-n) \tag{2}$$ or $$ x(n) = - \frac 3 2 u(-n-2) + \frac 1 2 (\frac 1 3)^{n-1} u(-n-2) \tag{3}$$

$\endgroup$
1
$\begingroup$

Q1: You should ask yourself if

$$\frac{z}{3z^2 - 4z + 1}\stackrel{?}{=}\frac{\frac 3 2}{z-1} - \frac{\frac 1 2 }{z-\frac 1 3}$$

really holds. You'll find out that you forgot to scale correctly.

Q2: There is no reason to consider $X(z)/z$ instead of $X(z)$ in this case.

And concerning your question about the anti-causal sequence, a multiplication by $z^{-1}$ always corresponds to replacing $n$ by $n-1$, no matter if a sequence is right-sided or left-sided.

$\endgroup$
  • $\begingroup$ You mean equation (2) is the correct one, right? $\endgroup$ – bikalpa Sep 11 at 7:00
  • $\begingroup$ What I meant was the equation $x(n) = - \frac 3 2 u(-n) + \frac 1 2 (\frac 1 3)^{n-1} u(-n)$ labelled as equation (2) in my question. But I think I got my answer. I'm convinced it is correct. $\endgroup$ – bikalpa Sep 11 at 9:32
  • 1
    $\begingroup$ @bikalpa: Yes, but your scaling is still off. $\endgroup$ – Matt L. Sep 11 at 9:37
  • $\begingroup$ I got the part on scaling. I didn't edit the question to reflect the correct scaling so that other readers would not get confused to see your answer and not identify what actually the error was by looking at the question. Thanks :) $\endgroup$ – bikalpa Sep 11 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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