I came across this property of the Discrete Time Fourier Transform (DTFT) and I am having a tough time proving it.

In general, consider two real signals $x[n] \: \& \: y[n]$. If $$ x[n] \leftrightarrow X(e^{jw}) \\ y[n] \leftrightarrow Y(e^{jw}) $$ Then, $$\boxed{ \sum_{n=-\infty}^{\infty} x[n]y[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{jw})Y(e^{jw})dw} $$

However, when I tried proving this, I am getting an extra conjugate:

$$ x[n]*y[-n] = \sum_{l=-\infty}^{\infty} x[l]y[n+l] \; $$ (Cross Correlation between $x[n]$ and $y[n]$)

$$Also, \; y[-n] \leftrightarrow Y(e^{-jw}).\: As \: x^{*}[n] = x[n], we \: have \; \; Y^{*}(e^{jw}) = Y(e^{-jw})$$

Thus from the convolution property of DTFT, we get:

$$x[n]*y[-n] = \sum_{l=-\infty}^{\infty} x[l]y[n+l] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{jw})Y(e^{-jw})e^{jwn}dw $$

Setting n = 0 and using the conjugation property as mentioned above (for real signals), we get: $$\sum_{l=-\infty}^{\infty} x[l]y[l] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{jw})Y^{*}(e^{jw})dw $$

Which is nothing but a result of the cross correlation property of DTFT.

Does the property (mentioned above in the box) hold only for special cases of $x[n]$ and/or $y[n]$ (like the signals being real and even etc.,), or am I making a mistake somewhere?

Thank you.


1 Answer 1


Your first formula in your question is generally wrong, that's why you can't prove it. The correct formula is


which is just Parseval's theorem.

If $x[n]$ and $y[n]$ are real-valued, $(1)$ can be written as


which is the form you've correctly obtained.

In the special case that $y[n]$ is not only real-valued but also even, i.e., $y[n]=y[-n]$, then $Y(e^{j\omega})=Y(e^{-j\omega})$ holds, and $(2)$ turns into the first formula in your question.

  • $\begingroup$ Thank you Matt L. Yeah, so it only holds when $y[n]$ is real and even. $\endgroup$ Oct 17, 2019 at 0:40

Your Answer

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

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