The DTFT is given by:

$$X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$$

The IDTFT is given by:

$$x[n]=\frac{1}{2\pi}\int_{0}^{2\pi}X(e^{j\omega})e^{j\omega n}d\omega$$

I have been able to show by substitution of the DTFT into the IDTFT that the transform and a subsequent inverse transform return $x[n]$:

$$\begin{align} x[n]&=\frac{1}{2\pi}\int_{0}^{2\pi}X(e^{j\omega})e^{j\omega n}d\omega\\ &=\frac{1}{2\pi}\int_{0}^{2\pi} \left( \sum_{k=-\infty}^{\infty}x[k]e^{-j\omega k} \right)e^{j\omega n}d\omega\\ \end{align}$$

Swap the order of integration and summation:

$$x[n]=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}\int_{0}^{2\pi}x[k]e^{j\omega (n-k)}d\omega$$

Argue that given $e^{j\omega (n-k)}$ is an odd function, it will only evaluate to anything other than 0 when $k=n$:

$$\begin{align} x[n]&=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}\int_{0}^{2\pi}x[k]e^{j\omega (n-k)} d\omega \ \delta[n-k]\\ &=\frac{1}{2\pi}\int_{0}^{2\pi}x[n]d\omega \\ &=\frac{2\pi}{2\pi}x[n]\\ \end{align}$$

However, I have been unable to show the dual case: that the inverse transform (IDTFT) substituted into the forward transform (DTFT) gives $X(e^{j\omega})$. How can we show this?

  • 2
    $\begingroup$ i hope you don't mind that i "cleaned" some of the notational convention a little. i added a Kroenecker delta $\delta[n-k]$ to it. $\endgroup$ Feb 12, 2019 at 0:49
  • $\begingroup$ Thanks, if it makes it clearer then that’s a good thing! $\endgroup$
    – Resquiens
    Feb 12, 2019 at 1:13
  • 1
    $\begingroup$ Note that all integrals should be over one period, not over two. I've changed the formulas accordingly. $\endgroup$
    – Matt L.
    Feb 12, 2019 at 8:07
  • $\begingroup$ oh yeah, that's right. $\endgroup$ Feb 12, 2019 at 10:46

1 Answer 1



where I've used the identity


  • $\begingroup$ This is perfect! I think my own confusion arose when using the same $\omega$ inside and outside the integral so became confused... using the different notations cleared up the misunderstanding. $\endgroup$
    – Resquiens
    Feb 12, 2019 at 12:31
  • $\begingroup$ @Resquiens: Yes, you have to be careful with variables inside and outside the integral, and you need to recognize the identity I used for the Dirac impulse. $\endgroup$
    – Matt L.
    Feb 12, 2019 at 12:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.