I would like some help to better understand the Fourier transform of a discrete time signal. My doubts are:

  1. The sampling of a signal can be seen as $x_s(t)=x(t) \cdot \sum_{k=-\infty}^{+\infty} \delta(t-kT_s)$ so its Fourier transform is $X_s(f)= f_s \cdot \sum_{k=-\infty}^{+\infty} X(f-k f_s)$ where $f_s$ is the sampling rate. If $x(t)$ is aperiodic, then the result is pretty straightforward; what if $x(t)$ is a periodized or a periodic signal?
  2. If $x[n]$ is a periodic discrete signal, it can be expressed as a Fourier Series (DFS) $x[n]= \sum_{k=<N_0>} {X_k \cdot e^{j2 \pi \nu_0 kn}}$ where $X_k = \frac{1}{N_0} \cdot \sum_{n=<N_0>} {x[n] \cdot e^{-j2 \pi \nu_0 kn}} $ and $\nu_0= \frac{1}{N_0}$. Since the spectrum of a periodic signal is discrete, can I apply to the DFS the Fourier transform (like I would do with a continuous time signal) and write it as a sum of Dirac pulses ?

Let's take as example following exercise : The signal $x(t)= 12\text{sinc}^2(4t) + \cos(4 \pi t)$ is sampled with a rate $f_c = 3$. It then becomes $$y[n]=y(nT_c)= 12\text{sinc}^2(\frac{4}{3} n) + \cos(\frac{4 \pi}{3} n) = y_{1s}(t) + y_{2s}(t) $$. What is the Fourier transform of $y[n]$? Sampling an aperiodic signal in the time domain brings us a periodicity in the frequency domain, so $$Y_{1s}(f) = f_c \cdot \sum_{k=-\infty}^{+\infty} {Y_1(f) * \delta(f-kf_c)}= 3 \cdot \sum_{k=-\infty}^{+\infty} {tr(\frac{f-3k}{4})}$$

Now, $y_{2}[n]=4 \cos(\frac{4}{3} \pi n) $ is a periodic signal: to obtain its Fourier transform do I use the same method as before, getting $$Y_{2s}(f)= \frac{3}{2} \cdot \sum_{k=-\infty}^{+\infty} {\delta(f-2-3k) + \delta(f+2-3k)}$$ ? Is it correct?

If I then decide to send $Y(f)$ in input to a low pass filter with frequency response $H_{LP}(f)= \frac{1}{f_c} \cdot rect(\frac{f}{f_c})$, I should have $Z(f)=Y(f) \cdot H_{LP}(f)= 3 tr(\frac{f}{4}) \cdot rect(\frac{f}{3}) + 2[\delta(f-1) + \delta(f +1)] $. Now I have to get $z[n]$ from this. We know that $\mathscr{F}[x[n]]= \sum_{n=-\infty}^{+\infty}{x[n] \cdot e^{-j2 \pi fT_c n}}$ and $x[n]= T_c \cdot \int_{\frac{-f_c}{2}}^{\frac{f_c}{2}} {X(f) \cdot e^{j2 \pi fT_c n}}df$ , so $z[n]=12sinc^2(4n)*sinc(3n)+ \frac{4}{3}\cos(2 \pi n)$. Is there any mistake?

Note: is the way this post is written ok, or should the various questions each have a dedicated post?

  • 1
    $\begingroup$ Yes, you're right, I have corrected. $\endgroup$ Commented Apr 19, 2023 at 14:07
  • $\begingroup$ "I want the Fourier transform of y[n]": which one? The DFT or the DTFT ? There are two Fourier Transforms for discrete time signals and they have significant differences. $\endgroup$
    – Hilmar
    Commented Apr 19, 2023 at 15:12
  • $\begingroup$ @Hilmar hello, isn't the DFT the Fourier transform of a periodic signal, while the DTFT is the Fourier transform of an aperiodic signal? $\endgroup$ Commented Apr 19, 2023 at 15:57
  • $\begingroup$ Sort of. The main difference is related. For the DFT the spectrum is also discrete, for the DTFT the spectrum is continuous. The main implication her (roughly speaking) is that you can only do DTFT on paper (or equivalent). Everything that involves vectors of numbers in a computer needs to use the DFT. So it all depends a bit on what exactly you are trying to do with the result. Your signal is not bandlimited, so you'll end up with frequency domain aliasing. It's not time limited either, so if you use the DFT you'll get time domain aliasing as well. $\endgroup$
    – Hilmar
    Commented Apr 19, 2023 at 17:49

1 Answer 1


The OP wants to learn about the various Fourier Transforms. Nothing will bypass reading a few chapters in a good DSP book (along with requisite chapters if those are confusing), so I don't recommend skipping that step. I can't reduce it to a short post here but will offer the following companion graphics I have that help illuminate the differences and features of each, hoping that this will be helpful to the OP's educational pursuits (and reading those chapters!):

Fourier Transform

With the FSE below, I also use this to demonstrate the property that what is periodic in one domain is discrete in the other domain. With the FSE we can restrict the time domain to be just $0$ to $T$, or we can extend the time axis to $\pm \infty$ and in either case we get the same relative values in frequency (so an implied periodicity due to mathematical equivalence). But to be clear, it is important to note that the waveform extending to $\pm infty$ in time would actually be "continuous" but zero everywhere except for the discrete frequencies shown. "Discrete" as shown in these plots means discrete non-zero values.

Fourier Series Expansion

Discrete Time Fourier Transform

Discrete Fourier Transform

  • $\begingroup$ Hello and thank you for your answer. I think I understand the continuous FT and FS, my problem is with the discrete one. If I have a continuous time periodic signal, its Fourier transform is a sum of infinite Dirac pulses; what about a discrete periodic one? Why do I get a finite sum with no explicit Dirac delta? $\endgroup$ Commented Apr 20, 2023 at 6:47
  • 1
    $\begingroup$ The DTFT is always periodic with period $2\pi$ $\endgroup$ Commented Apr 20, 2023 at 21:28
  • $\begingroup$ @Maghreb_1911 Look at the DTFT graphic I gave above along with the formula-- the key point was to show that it is that it is not periodic in time. Because of that, it is continuous in frequency. Periodicity in one domain is Discrete (discrete for the non-zero values) in the other domain. $\endgroup$ Commented Apr 20, 2023 at 23:42
  • $\begingroup$ The result for the DFT in contrast, similar to the Fourier Series Expansion has a result in frequency that is consistent with the time domain waveform being periodic. We can process the signal in time over the fixed time interval 0 to N-1, or we can periodically repeat that to infinity and we will get the same non-zero result (with proper scaling). This is why many say loosely "The Fourier Transform assumes the time domain waveform is periodic"-- it is simply mathematically equivalent. $\endgroup$ Commented Apr 20, 2023 at 23:44
  • $\begingroup$ @DanBoschen I think things look clearer than before. I realize that the way my question was written was quite generic. I edited it, see if it's better now, please. $\endgroup$ Commented Apr 21, 2023 at 14:19

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