Discrete Time Fourier Transform (DTFT) cross correlation property

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.

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

$$\sum_{n=-\infty}^{\infty}x[n]y^*[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})Y^*(e^{j\omega})d\omega\tag{1}$$

which is just Parseval's theorem.

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

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

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.

• Thank you Matt L. Yeah, so it only holds when $y[n]$ is real and even. Oct 17 '19 at 0:40