# Fourier transform of a time discrete signal

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) $$x[n]= \sum_{k=} {X_k \cdot e^{j2 \pi \nu_0 kn}}$$ where $$X_k = \frac{1}{N_0} \cdot \sum_{n=} {x[n] \cdot e^{-j2 \pi \nu_0 kn}}$$ 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?

• Yes, you're right, I have corrected. Apr 19 at 14:07
• "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. Apr 19 at 15:12
• @Hilmar hello, isn't the DFT the Fourier transform of a periodic signal, while the DTFT is the Fourier transform of an aperiodic signal? Apr 19 at 15:57
• 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. Apr 19 at 17:49

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.
• The DTFT is always periodic with period $2\pi$ Apr 20 at 21:28