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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.

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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.

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

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