I have the following expression:

$$X(z) = \frac{16}{15}\frac{1}{1-\frac14z^{-1}} - \frac{16}{15}\frac{1}{1-4z^{-1}}$$

According to my understanding this should become:

$$x(n) = \frac{16}{15}\left(\frac14\right)^n u(n) - \frac{16}{15} 4^n u(n)$$

But according to my source it is:

$$x(n) = \frac{16}{15}\left(\frac14\right)^n u(n) + \frac{16}{15} 4^n u(-n-1)$$

Are these expressions equal?

If they are, how can $-u(n) = u(-n-1)$ ?

If not, why?

I used $a^nu(n)\Longleftrightarrow \frac{1}{1-az^{-1}}$.


1 Answer 1


This is probably homework, so I'll just give you a few hints to get you on the right track. First of all, given an expression $X(z)$, the corresponding sequence is usually not unique. Your solution is correct if the region of convergence (ROC) of $X(z)$ is $|z|>4$, which gives you an exponentially increasing sequence. If this were the impulse response of a discrete-time system, then the system would be causal but unstable.

The other solution assumes that the ROC is $\frac14<|z|<4$, which includes the unit circle. This corresponds to a two-sided decaying sequence. As an impulse response it would correspond to a non-causal but stable system.

So, to answer your question, both solutions are correct but they are not identical, because they are inverse transforms of $X(z)$ assuming different regions of convergence.

  • $\begingroup$ Are you sure about this? No information is given about the region of convergence. I simply want to calculate the auto correlation of a signal resulting in the provided X(Z). Then I need to inverse-transform that. $\endgroup$
    – user264230
    Commented May 28, 2015 at 15:31
  • $\begingroup$ I mean, how can you see this? $\endgroup$
    – user264230
    Commented May 28, 2015 at 15:41
  • $\begingroup$ @user264230: OK, if this is supposed to be an autocorrelation sequence, then it's clear that it must be symmetric, so the only option is the two-sided solution, which implies the ROC I mentioned in my answer. So that was the additional information that is needed to choose the correct sequence. $\endgroup$
    – Matt L.
    Commented May 28, 2015 at 16:02
  • $\begingroup$ Okey, provided that this is clear, how can −u(n)=u(−n−1)? Is this a rule that pepole should know or do you visualize this?:) Or you mean that they are different soulutions? How can i get the correct one, using inverse transform gives me the first solution with invalid ROC? $\endgroup$
    – user264230
    Commented May 28, 2015 at 16:20
  • $\begingroup$ @user264230: That equality doesn't hold. As I said in my answer, the two solutions are not identical. $\endgroup$
    – Matt L.
    Commented May 28, 2015 at 16:22

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.