I am doing the following question (it is not homework, I am preparing for an exam).

The time discrete signals $x_1(n)$ and $x_2(n)$ is created by sampling the continuous signal $x_a(t) = \cos (2 \pi 300t) + \cos (2 \pi 600 t)$ with sampling frequency $F_s = 1000\textrm{ Hz}$, $x_1$ with anti aliasing filter and $x_2$ without aliasing filter.

  • a) Calculate $X_a(f)$ for the time continuous signal $x_a(t)$.
  • b) Calculate the magnitude spectrum for the time discrete signals $x_1(n)$ and $x_2(n)$ i.e $\lvert X_1(f)\rvert$ and $\lvert X_2(f)\rvert$, with $-1<f<1$.

I have done a), I used the table for Fourier Transform to conclude that $$X_a(f) = \frac{1}{2}\Big(\delta(f+300) + \delta(f-300) + \delta(f+600) + \delta(f-600)\Big).$$

But I can not solve b), I believe that I can use the information/answer from a) that is why I provided it.

  • 1
    $\begingroup$ Hi Olba, a) is good work; I unified your usage of $F$ vs $f$; you were mixing that, and as frequency variable, we typically use $f$. It's totally OK to use Fourier tables, but it's also very important to understand from the depths of your heart why $\mathcal{F}\left\{ \cos (2\pi f_0 t) \right\}(f)=\frac 12 \left(\delta (f-f_0)+\delta (f+f_0) \right)$: this cosine is a pure tone of frequency $f_0$, and hence, it has spectral components exactly at $\pm f_0$ and nowhere else; the factor of $\frac12$ is pretty much energy conservation. $\endgroup$ Commented Aug 17, 2016 at 9:13

1 Answer 1


This looks like a HW problem, so I can only give you a hint. Recall how the sampling process works - it's a convolution with a spike train i.e. a sequence of Dirac delta functions spaced at $F_s$. So the analog spectrum (that you derived in part (a) of the problem) gets replicated many many times centered around integer multiples of $F_s$. Also be careful about aliasing - if the largest frequency in your analog signal exceeds $F_s/2$ it means that this replication process will leak signal energy into the lower frequencies (draw a picture and convince yourself of this). Finally, analog frequencies in the interval $[-F_s/2, F_s/2]$ get mapped to digital frequencies in the interval $[-1/2, 1/2]$.

  • $\begingroup$ It is now HW, I am prepairing for an exam. "if the largest frequency in your analog signal exceeds $F_s/2$". Well I have a signal with frequency 600 right? So there is going to happend something with that signal. But the signal with frequency 300 will be mapped directly to the intervall $-1/2, 1/2$? $\endgroup$
    – Olba12
    Commented Aug 17, 2016 at 13:48
  • $\begingroup$ That's right. Since 600 > 1000/2=500, you'll see aliasing. $\endgroup$
    – Atul Ingle
    Commented Aug 17, 2016 at 14:04
  • $\begingroup$ That is where the problem occours for me, I do not know how to handle that. $\endgroup$
    – Olba12
    Commented Aug 17, 2016 at 14:05
  • $\begingroup$ I recommend drawing a picture of the spectrum. The original analog signal has spikes at $\pm 300$ and $\pm 600Hz$. After sampling, this pattern will replicate around integer multiples of 1000. Then note how the replicated 600Hz spike ends up at a spot lower than 500 Hz. $\endgroup$
    – Atul Ingle
    Commented Aug 17, 2016 at 14:08
  • 1
    $\begingroup$ Yes! Now note that the 600 component shows up as its alias at 400 in the absence of the antialiasing filter. Also note that after sampling, you can only identify frequencies up to 500 Hz. So really, $x_1$ will only have a 300 Hz component (and what you call 700 is just a replica of this 300 Hz component centered at 1000). $\endgroup$
    – Atul Ingle
    Commented Aug 17, 2016 at 14:24

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