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I am trying to understand the math. The inverse $\mathcal Z$-transform is given by:

$$x[n] = \displaystyle\frac{1}{j2\pi} \int_cX(z)z^{n-1}dz$$

where $\displaystyle \int_c$ is a contour integral. The inverse Fourier transform is given by:

$$x[n] = \displaystyle\frac{1}{2\pi}\int_{-\pi}^\pi X(e^{j\omega})e^{j\omega n}d\omega$$

My textbook claims, and I agree, that evaluating the inverse z-transform at $z=e^{j\omega}$ will result in the inverse Fourier transform. However, I can't get the math to show this. Substituting $z=e^{j\omega}$, I get as far as:

$$x[n] = \displaystyle \frac{1}{j2\pi}\int_cX(e^{j\omega})e^{j\omega n}e^{-j\omega}d(e^{j\omega})$$

Can somebody tell me intuitively how to simplify this equation to look like the inverse Fourier transform? Is my confusion due to a lack of understanding of the contour integral?

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Hint:

With $z = e^{j\omega}$, you have:

\begin{align} \displaystyle \frac{dz}{d\omega} &= je^{j\omega}\\ \Rightarrow dz &= je^{j\omega} d\omega \end{align}

With $-\pi \leq \omega \leq \pi$, your contour integral goes from $-\pi$ to $\pi$.

Can somebody tell me intuitively how to simplify this equation to look like the inverse Fourier transform?

Plugging all the above should do the job.

Is my confusion due to a lack of understanding of the contour integral?

It is more the change of variables.

EDIT:

Note that $\omega$ is taken in the fundamental interval $\left[-\pi, \pi\right]$ because $X(\omega)$ is periodic with period $2\pi$. So, $\displaystyle \int_c$ becomes $\displaystyle \int_{2\pi}$

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  • $\begingroup$ You're welcome. See the edit on the choice of $2\pi$ for the contour. $\endgroup$ – Gilles Jul 24 '15 at 23:04

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