DFT vs. DTFT: Understanding the Difference

[CoolDude420]

Homework Statement

I'm kind of confused between DFT and DTFT. Here is my understanding:

Okay, so let's say we have time domain, continuous, analogue signal from a sensor - $x(t)$

1. We sample this signal, giving us something like the following with an impulse train

Now this is a discrete-time sequence $x_n$. So we can apply the Discrete-Time Fourier Transform to obtain the spectrum of this discrete sequence $x_n$ - let's call that $X_s(w)$. The DTFT is stated as follows,

Thus, the DTFT gives us the spectrum of the SAMPLED original signal. Correct?

Now, to use the DTFT in a computer, we cannot have an inifnite summation, thus we change to using N because it is periodic with a period of 2pi. So my lecture notes is beginning to derive the DFT from the DTFT, so it says this now,

Note that this is not the full DFT, it's in the process of deriving.

Here is my confusion. How come this equation has $X(w)$, i.e the spectrum of the very original signal. But the DTFT formula has $X_s(w)$, the spectrum of the sampled signal?? Is there a mistake in the notes? How did we magically just get back the spectrum of the original signal when we were infact dealing with the spectrum of the sampled signal?

Homework Equations

The Attempt at a Solution

 

[jasonRF]

Thus, the DTFT gives us the spectrum of the SAMPLED original signal. Correct?

Correct.

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Now, to use the DTFT in a computer, we cannot have an inifnite summation, thus we change to using N because it is periodic with a period of 2pi.

Actually, the fact is that you have a finite number of numbers in your sequence. There is more than one thing you can conceivably do with that (for example, you could assume that the function is zero everywhere else), but if you assume that your finite sequence is a single period of a periodic function then the DFT is the natural frequency domain representation.

​

Here is my confusion. How come this equation has $X(w)$, i.e the spectrum of the very original signal. But the DTFT formula has $X_s(w)$, the spectrum of the sampled signal?? Is there a mistake in the notes? How did we magically just get back the spectrum of the original signal when we were infact dealing with the spectrum of the sampled signal?

The $X(\omega)$ above is not the spectrum of the original continuous time function. You do not get the spectrum of the original signal by sampling a finite length of it and then assuming it is periodic.