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$$G(z)=\displaystyle \frac{\frac{1}{z}}{1+\frac{5}{6z}+\frac{1}{6}z^{-2}}$$

I found: $$g(k)=\displaystyle \left(\frac{-1}{3}\right)^k - \left(\frac{-1}{2}\right)^k$$

I don't understand how I can find the fourier transform of $g(k)$ from $G(z)$.

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  • $\begingroup$ You have $G(z)$, your other question that was answered here can help you answer this one. $\endgroup$ – Gilles Sep 1 '15 at 22:14
  • $\begingroup$ is it gonna be G(f)=(1/(e^jw))/1+5/(6e^jw)+1/(6e^2jw)? $\endgroup$ – angie Sep 2 '15 at 6:50
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First of all it's very important to realize that without any further information it is not possible to uniquely identify the corresponding time-domain sequence from the given $G(z)$. The reason is that you need to know the associated region of convergence (ROC). If the ROC is not known then for the given $G(z)$ there are three possible time-domain sequences. One of them is causal, another one is anti-causal, and the third one is two-sided.

If the ROC is given as $|z|>\frac12$, or, equivalently, if you know that you're looking for a causal sequence, then the inverse transform is uniquely defined. It is similar to the one you found, but there is an important difference:

$$g[k]=\left[\left(-\frac13\right)^k+\left(-\frac12\right)^k\right]u[k]$$

You need the unit step $u[k]$, which makes the sequence causal. The sequence you found extends to infinity for positive as well as for negative indices $k$, and does not even have a $\mathcal{Z}$-transform.

Now that you know that the ROC includes the unit circle, you know from the answers to this question that in order to find the Fourier transform you can simply evaluate the $\mathcal{Z}$-transform on the unit circle, i.e. set $z=e^{j\omega}$. What they didn't tell you is that this is only possible because the unit circle is included in the ROC. Otherwise you can't do that because the Fourier transform wouldn't even exist.

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  • $\begingroup$ If |z|>1/2 how does that include the unit circle? The unit circle is not part of the ROC in this case. $\endgroup$ – angie Sep 7 '15 at 12:58
  • $\begingroup$ @angie: The unit circle is $|z|=1$, so it is included in $|z|>1/2$. $\endgroup$ – Matt L. Sep 7 '15 at 13:30

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