Partial Fraction Expansion for Inverse Fourier Transform

Question

In many textbooks, I've seen the application of Partial Fraction Expansion (PFE) to find an inverse Fourier Transform. Let's stick to the discrete time case, and let me give you an example.

Let's say that $$X(e^{j\omega}) = \frac{-\frac{5}{6}e^{-j\omega} + 5}{1 + \frac{1}{6}e^{-j\omega} - \frac{1}{6}e^{-j2\omega}}$$ and we want to find the signal $x[n]$ that has the above DTFT. We can apply PFE and get $$X(e^{j\omega}) = \frac{A}{1 + \frac{1}{2}e^{-j\omega}} + \frac{B}{1 - \frac{1}{3}e^{-j\omega}} \tag{1}$$ where $B$ (and similarly, $A$) is given by $$B = X(e^{j\omega})\Big(1 - \frac{1}{3}e^{-j\omega}\Big)\Big|_{e^{-j\omega}=3}$$

I know now that we can set $u = e^{-j\omega}$ and turn this into

$$B = X(u)\Big(1 - \frac{1}{3}u\Big)\Big|_{u=3}$$

However, setting $u = e^{-j\omega} = 3$ does not look good to me, because this equation has no real solution for $\omega$.

What do I miss in this application of PFE?

Answer 1

You should indeed replace $e^{j\omega}$ by the more general complex variable $z$ because when talking about poles and zeros of a rational function, you must consider the whole complex plane. The fact that $e^{-j\omega}=3$ doesn't have a real solution just means that the corresponding pole does not lie on the unit circle.

After having performed the partial fraction expansion, there is nothing that keeps you from setting $z=e^{j\omega}$.