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Let $$y[k]=x[k]\cdot \cos(2\pi\nu_0k) .\tag{1}$$

Then, given a signal $x[k]$ with the DTFT $X(\nu)$ according to the following figure

frequency domain of signal x

what will the frequency domain for $Y(\nu)$ look like for a given $\nu_o$ in the range $0<\nu_0<a/2$?


According to my calculation, the resulting function $Y(\nu)$ is

$$Y(\nu)= \frac{1}{2}\left ( X(\nu-\nu_0) +X(\nu+\nu_0) \right ). \tag{2}$$

This since the DTFT of cosine is two unit impulses in $\pm\nu_0$.

Is this correct? And how can this resulting $Y(\nu)$ be drawn in the frequency domain? Thank you! Some explanation with mathematical elements would be appreciated.

Would the following figure for $Y(\nu)$ be correct?

resulting function in frequency domain

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