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


Thinking in terms of convolution with shifted impulses help. Multiplication in the time domain corresponds to convolution in the frequency domain. Your example is a classic showcase of frequency/band shifting using a "carrier" such as a cosine (or sine) or a complex exponential. Now, note, that the Fourier transform of these carriers are actually shifted impulses. When you convolve a function x(t) with a shifted impulse, say, D(t-T), the result that pops out is x(t-T), i.e., basically a shifted form of x(t). Note that the amount of shift is exactly equal to the shift in the impulse (the actual math is quite easy). Likewise, in the frequency domain, convolution with a shifted impulse results in your Fourier transform of the original signal getting shifted accordingly - For an impulse shifted to the right of origin, it feels as if the impulse "pushes all the bands in the spectrum to the right" exactly by the amount it itself is shifted from the origin. Vice-versa analogy holds for left shifted impulses. This is the mathematical background of this if you needed one. And yes, those results look correct to me. Cheers!

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