Validity of taking an inverse $\mathcal{Z}-$ transform instead of taking an inverse DTFT

Question

I have the following problem: I am using the Convolution Theorem and have got an expression of $H(z)X(z)$ and now I need to take $\text{DTFT}^{-1}(H(z)X(z))$, namely I have to take the inverse DTFT of $\displaystyle\frac{1}{1-\frac{1}{12e^{i\omega}}}\cdot\frac{1}{1-\frac{1} {10e^{i\omega}}}$ which is a horrendous integral to compute. Can't I, however, take the inverse $\mathcal{Z}-$transform, as the DTFT is a special case of this? If I take $\mathcal{Z}^{-1}$ which is much easier to compute with residues, I end up with $y(n)=60^{-n}(6^{n+1}-5^{n+1})$. Is this correct? Can I do this? If so, is it always applicable? Thanks, in advance.

Answer 1

Let's first clear up a misunderstanding: the inverse DTFT and the inverse $\mathcal{Z}$-transform are equally simple or difficult to compute; the integrals are the same. If you use the contour $|z|=1$ in the formula for the inverse $\mathcal{Z}$-transform (assuming that $|z|=1$ is inside the ROC) you just compute the inverse DTFT.

So of course you can use the inverse $\mathcal{Z}$-transform to compute the inverse DTFT, even though - as mentioned above - it doesn't make any difference. But note that when computing the inverse $\mathcal{Z}$-transform you need to define the ROC in order to be able to choose the appropriate time domain sequence. If you're interested in the inverse DTFT, you need to choose the ROC that includes the unit circle.

In your example, you could use partial fractions to rewrite the given transfer function as

$$H(z)=\frac{A}{1-az^{-1}}+\frac{B}{1-bz^{-1}}\tag{1}$$

From $(1)$ you can directly write down the corresponding time domain sequence without solving any integrals.