Relationship Between the DFT and the DTFT

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Table of Content
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** How are the DTFT and the DFT related? 00:00
** An example to highlight the relation between DTFT and DFT 12:58
** Using the DFT as a proxy for the DTFT 27:38
In this video lecture, undergraduate students learn about the relationship between the discrete Fourier transform and the discrete-time Fourier transform. The focus here is on the basic understanding and not on the formal derivation of the relationship.

[@chandanjaiswal7011]

At the interval 16:30 in the above video , you plotted the DTFT from -pi to +pi , but the value of omega(w) can vary from - infinity to +infinity , why ? Aslo please explain why we are sampling omega in 2*pi intervals only as it is a angular frequency and can vary upto infinity in a given ramdom signal , though I heard that DTFT is periodic and continuous but don't know the reason .

[@SigProcessing]

The reason we focus only on any 2\pi interval of discrete frequencies is because discrete frequencies are only unique up to addition of integer multiples of 2\pi. That is, a frequency \omega_0 and \omega_0 + 2\pi k for any integer k means the same frequency. The easiest way to see this is by noting that e^{j (\omega_0 + 2\pi k)n} = e^{j \omega_0 n} e^{j 2\pi k n} = e^{j \omega_0 n}, since e^{j 2\pi k n} = 1 for all integers k and n.
These topics are discussed in an earlier lecture on discrete-time Fourier transform and you may want to watch them for further explanation.

[@developingstrong8386]

How do you convert DTFT signal into DFT signal

[@SigProcessing]

The DTFT of a signal can be converted to the DFT of that signal by replacing \omega with 2\pi k / N (i.e., sampling the DTFT every 2\pi/N radians)

[@TheExecutioner21]

Please rerecord these videos if you use them for the next classes

[@SigProcessing]

That's the plan.

[@benbenny1438]

how the DFT can be obtained from the DTFT?

[@SigProcessing]

+mahinder singh As we discuss in the video, DFT is simply samples of the DTFT every 2\pi/N radians/sample, starting from 0 and going up to 2\pi (N-1)/N radians/sample. So if you are given DTFT, you compute the DTFT at these discrete set of frequencies and the answer would be the DFT sequence.

[@uanlsoporte5332]

​@@SigProcessing Hi sir! as i understand DTFT is just the correlation of x[n] with a CONTINUOUS basis complex exponentials? am i right?

[@SigProcessing]

@@uanlsoporte5332 There are different ways of interpreting DTFT, depending upon one's background. Signal processing folks might interpret it differently than harmonic analysis people, just because of the technical language one is used to. The interpretation you are thinking about is correct (this correlation can also be thought of as a projection step) from the signal processing engineer's perspective. You are correlating your signal to a continuum of frequencies and recording the answer returned by this correlation for each \omega.

[@jameshopkins3541]

NO ME GUSTAN LOS GARABATOS