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In the following system: Multirate system

The signal x(n) is upsampled by 2 and then filtered. The signal and the filter have the following spectrum:

enter image description here enter image description here

After upsampling the spectrum of X would look the same except the x axis would be divided by 2. So the limit of the spectrum would be at pi/4. Since the filter has amplitude 2 would the amplitude of X be increased from 1 to 2 or will the amplitude of X remain unchanged?

Finally, for the multiplication of the cosine it would be equivalent to multiply with 2 deltas located at -pi/2 and pi/2. Would the multiplication with this deltas cause a kind of mirroring effect on the spectrum?

Thank you

edit:

I do not see how to obtain Y1 and Y2 complete system enter image description here

and provided answer: enter image description here

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You are on the right track.

After upsampling the spectrum of X would look the same except the x axis would be divided by 2. So the limit of the spectrum would be at pi/4.

Yes, that's correct.

Since the filter has amplitude 2 would the amplitude of X be increased from 1 to 2 or will the amplitude of X remain unchanged?

Recall that in the frequency domain, the application of a linear system transfer function $H_B(\Omega)$ is done via straight multiplication of the input signal's spectrum. So yes, since the filter has a magnitude of 2, the amplitude of its output is doubled.

Finally, for the multiplication of the cosine it would be equivalent to multiply with 2 deltas located at -pi/2 and pi/2. Would the multiplication with this deltas cause a kind of mirroring effect on the spectrum?

Here's where I will make a small correction: multiplication by a cosine in the time domain yields convolution with a pair of Dirac impulse functionss in the frequency domain. Convolution with a Dirac impulse in the frequency domain is equivalent to a shift along the frequency axis. So, you end up with 2 copies of the filter's output spectrum, centered at $\omega = \pm \frac{\pi}{2}$.

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  • $\begingroup$ Thank you! This confirms my reasoning. I updated the main post with the complete system and the provided answer, but I do not understand why there are some mirrored parts in the answer. It seems that the provided answer is not correct $\endgroup$ – karagedon Jul 20 '19 at 16:15

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