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

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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.

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  • $\begingroup$ 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

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