Suppose we have the complex signal $x(t)= \exp(j\omega_0 t)$. Using the properties of Fourier transform we can prove its CTFT is Dirac $\delta$ function.

If any one ask me about the spectrum of $x(t)$ "Does $x(t)$ has continuous spectrum or discrete spectrum", my answer will be "The spectrum of $x(t)$ is discrete".

Now, if I apply a rectangular window to the complex exponential $x(t)$ in the time domain and then take CTFT I will end up with $\mathrm{sinc}$ function. Now the spectrum of the windowed complex exponential is continuous and not discrete.Is this interpretation true?

  • $\begingroup$ I'd suggest asking a specific question, and also editing the title to summarize your question. $\endgroup$
    – MBaz
    Commented Feb 12, 2017 at 1:51
  • $\begingroup$ This wasn't on hold when I started answering, and so I gave it an attempt. After trying to answer it, I agree now that you should refine your title and question. I hope my attempt to answer will help you in that regard. I think I have a pretty good idea of what you are asking now, so if you want me to help clean up the question, I probably can. I don't want to do too much without permission from you (the OP) though, because it would require quite a bit of alteration. $\endgroup$
    – hops
    Commented Feb 12, 2017 at 17:16

1 Answer 1


If you don't understand the difference between the Continuous Time Fourier Transform (CTFT), the Discrete Time Fourier Transform (DTFT) and the Discrete Fourier Transform (DFT), now would be a good time to read about them. The very short version is that the DTFT yields the (continuous-valued) spectrum of a sequence (i.e., a sampled signal). The DFT computation results in a sampled version of the DTFT. To apply the DFT requires a finite number of samples (i.e., a time-domain window) whereas such restrictions are not placed on the DTFT in general. On the other hand, the CTFT deals with continuous time signals. There is a lot more to all of this, and I recommend you read more.

It is true that a complex-valued signal $x(t) = \exp\left(j \omega_0 t\right)$ and the Dirac delta function form a CTFT pair. I can agree with you that the spectrum of this signal is discrete (nonzero for a finite number of frequencies, in this case a single frequency).

After applying a rectangular window in time to the complex-valued signal $x(t)$, the CTFT is a frequency-translated sinc function centered at $\omega_0$ with a lobe width inversely proportional to the size of the time domain window. This shows us that the result of truncating a time domain signal with infinite support in time and a discrete spectrum in frequency can lead to a new time domain signal with finite support and a continuous spectrum. So, the answer is yes, the interpretation given in the question is true.

  • $\begingroup$ Thanks for your explanation. I understood carefully, but I still have a question. The Fourier transform of continuous signal x(t)=ejω0t is dirac function and DTFT of its sampled version is sinc function. I think they differs because we are dealing with finite number of samples. Is this interpretation true??? $\endgroup$
    – Mahdi
    Commented Feb 12, 2017 at 10:33
  • $\begingroup$ If I'm understanding your question, then yes. The sinc-like response of the DTFT is a direct result of applying a rectangular window to the complex exponential in the time domain and sampling it. If you apply this window and then take the CTFT, you do end up with a sinc response (not sinc-like). When the spectrum of this signal with infinite bandwidth becomes $2\pi$-periodic after downsampling, you obtain the sinc-like response alluded to in the answer. $\endgroup$
    – hops
    Commented Feb 12, 2017 at 16:53
  • $\begingroup$ Now that I understand your confusion, I will add that to my response. $\endgroup$
    – hops
    Commented Feb 12, 2017 at 16:58
  • $\begingroup$ The question was restated without reference to the DFT or DTFT, so I have removed the parts of the answer that referred to sampling (except the links which I still think are relevant). This results in an answer based on continuous time signals only. $\endgroup$
    – hops
    Commented Feb 13, 2017 at 21:01

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