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Calculate Inverse Discrete Time Fourier Transform of the following where $|a| < 1$:

$$ X(e^{j\omega}) = \frac{1-a^2}{(1-ae^{-j\omega})(1-ae^{j\omega})} $$

Plugging this directly into the IDTFT equation, I get:

\begin{align*} x[n] &= \frac{1}{2\pi} \int_{-\pi}^\pi X(e^{j\omega}) e^{j \omega n} d\omega \\ x[n] &= \frac{1}{2\pi} \int_{-\pi}^\pi \frac{(1-a^2)e^{j \omega n}}{(1-ae^{-j\omega})(1-ae^{j\omega})} d\omega \\ \end{align*}

I am having trouble getting started. I'm not sure what to try. None of the standard Fourier Transform property laws seem to directly apply to this.

(This is problem 2.57 from Oppenheim textbook on Discrete Time Signal Processing)

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  • $\begingroup$ The lower integral bounds should all be $-\pi$, not $\pi$. $\endgroup$ Commented Sep 7, 2015 at 20:43
  • $\begingroup$ @AlexR, you are right. I fixed it. That was obviously a typo... I'm still stuck on what to do :( $\endgroup$ Commented Sep 7, 2015 at 21:59

1 Answer 1

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Writing $z=e^{j\omega}$ and using partial fraction expansion, you can rewrite $X(z)$ as

$$X(z)=\frac{a}{z-a}+\frac{1}{1-az}\tag{1}$$

The two terms in $(1)$ are DTFTs (or $\mathcal{Z}$-transforms) of basic sequences:

$$\frac{a}{z-a}\Longleftrightarrow a^nu[n-1]\\ \frac{1}{1-az}\Longleftrightarrow a^{-n}u[-n]\tag{2}$$

where $u[n]$ is the unit step, and where $|a|<1$ has been taken into account. Combining the expressions in $(2)$ results in

$$x[n]=a^nu[n-1]+a^{-n}u[-n]=a^{|n|}$$

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