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The following signal $$ f(t) = 2\cos(20 \pi t) + 3\cos(80 \pi t) + 4\cos(200 \pi t) $$ is sampled at a frequency $F_s = 100\text{ Hz}$, then it is reconstructed using an ideal lowpass filter (no frequency given). I am asked to determine the signal coming out of the reconstruction filter.

By my understanding on ideal filters, a cutoff frequency of $ 100\text{ Hz}$ should be enough (since the signal has $ F_{max} = 100 \text{ Hz}) $.

So the result that I am expecting is the same that I started with, however the result that I am given is:

$$ f(t) = 4 + 2\cos(20 \pi t) + 3\cos(80 \pi t) $$

Why is it that there's no $\cos(200 \pi t)$ and there's only a $4$?

Lorenzo B.
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Jafar Eldeni
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1 Answers1

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If the sampling frequency is $ F_s = 100$ Hz, the maximum frequency you can "capture" is 50 Hz. The last term $ 4 cos (200 \pi t) = 4 cos (2\pi t \cdot 100) $ is therefore rejected.

Duns
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  • oh i see! But why is it that there's only the 4? Isn't it the amplitude of the $ cos $ function? – Jafar Eldeni Jun 09 '19 at 13:27
  • If you sample cos (2 \pi \cdot 100) at 100 Hz, you will always get samples of the same amplitude, as if you had sampled a constant signal. – Duns Jun 09 '19 at 13:37
  • and $\cos(200 \pi t)$ is sampled always at $\cos(0)$, that's why you get an average of $4 \cdot 1$ – G Cab Jun 09 '19 at 13:41