I'm teaching myself some DSP for future projects and my method is mostly learn by doing. I have this time-domain function with its correspondent frequency-domain transform:

$$s_1(t) = (t-2)e^{-t}u(t-2) $$

$$S_1(f)=\frac{e^{-2(1+j2\pi f)}}{(1+j2\pi f)^2} $$

Plotting the frequency-domain function in Matlab, for a frequency vector ranging from [-50,50) in steps of 0.1 (1000 values of frequency).

magnitude plot of the frequency-domain function

The question I don't know how to solve, is about the period sample. I know each sample is one of those dots, and that despite it can be arbitrary (choosed by us), there is criteria to follow in order to pick the best sampling rate possible to avoid aliasing.

If I set the frequency vector like I did, that means I already determined the sample rate, am I right? That also means, I can compute the sampling period as:

Ts=1/fs; fs=0.1 being the sample frequency.

Or should I go with another one? Should I try Nyquist (since its a continuous signal)? If so, how could I apply it?

Thanks for any suggestion you may have on this.


1 Answer 1


I think you are mixing up two things here. By "best sampling rate", I suppose your concern is how to select the $\Delta f$ in the spectrum such that the time-domain signal could be fully reconstructed from the sampled spectrum. This can be answered by the dual of the sampling theorem.

Consider the Fourier transform pair $s(t)\xrightarrow{\mathcal{F}}S(f).$ Let me first review the sampling theorem:

Sampling in time with sampling rate $F_s=\frac{1}{\Delta t}$ causes replication in the frequency domain at every $F_s$ (i.e. frequency aliasing). To prevent frequency aliasing, $s(t)$ should be bandlimited to $B$, and the sampling interval should satisfy $$\Delta t=\frac{1}{F_s}<\frac{1}{2B}$$ the signal's spectrum is recovered by filtering the replicated spectrum.

The dual theorem states that If the sampling interval in frequency domain is not short enough, it will result in time aliasing:

Sampling in frequency with rate $\frac{1}{\Delta f}$ causes replication in the time domain at every $\frac{1}{\Delta f}$ (i.e. time aliasing). To prevent time aliasing, $s(t)$ should be time-limited to $T$ and the sampling interval should satisfy $$\Delta f<\frac{1}{2T}$$ The signal at the time domain can be recovered by time-windowing the replicated signal.

Getting back to your signal $s_1(t)$, since it is not time-limited (is not theoretically zero outside a time horizon), you should truncate it somewhere (e.g. at $T$). Then, it will satisfy the criteria and you can choose $\Delta f=\frac{1}{2T}$.

  • $\begingroup$ Thanks @msm. Let's say I truncate my signal from 2.5s to 3s (T=0.5s), then I say Δf=1? For this problem, I said: T=t1(2)-t1(1). And also tried other intervals without seeing much difference in the signals plot. $\endgroup$ Sep 16, 2016 at 5:23
  • $\begingroup$ In a nutshell, if your signal in the time domain is zero for $T>0$ then $\Delta f=\frac{1}{2T}$ is a good choice. Make sure to understand the word "truncate" in here). $\endgroup$
    – msm
    Sep 16, 2016 at 12:06

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