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I'm trying to understand how I can start from the CTFT of a signal and end up with a DTFT.

For example if I take a basic example: $$ x(t) = cos(\omega_x \cdot t) = \frac{1}{2} \cdot \left( e^{j\omega_x t} + e^{-j\omega_x t} \right) $$ $$ \implies X(\omega) = \pi \cdot (\delta(\omega - \omega_x) + \delta(\omega+\omega_x)) $$ $$ x_c(t) = x(t) \cdot \sum_{n=-\infty}^{\infty}\delta(t - nT_s) = \sum_{n=-\infty}^{\infty}x(nT_s)\delta(t-nT_s) $$

$$ \implies X_c(\omega) = X(\omega) * \left( \omega_s \sum_{n=-\infty}^{\infty} \delta(\omega - n\omega_s)\right) $$ $$ X_c(\omega) = \omega_s \pi \sum_{n=-\infty}^{\infty} \left( \delta(\omega - \omega_x - n\omega_s) + \delta(\omega + \omega_x - n\omega_s) \right) \tag{1} \label{1} $$

From there I'm lost and everything crumbles. I'm only trying to get the DTFT of a cosine which is: $$ cos(\Omega_0 n) \Leftrightarrow \pi \sum_{n=-\infty}^{\infty} \left( \delta(\Omega - \Omega_0 - n2\pi) + \delta(\Omega + \Omega_0 - n2\pi) \right) \tag{2} \label{2} $$

How can I obtain $\eqref{2}$ starting from $\eqref{1}$? I hope what I'm trying even makes sense. After all the DTFT with infinite period is the CTFT so I suppose there's a link we can make between these equations?

Thanks

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You're almost there. First, there's a small scaling error in the transform $X_c(\omega)$. Since the Fourier transform of an impulse train is given by

$$\mathcal{F}\left\{\sum_{n=-\infty}^{\infty}\delta(t-nT)\right\}=\omega_s\sum_{k=-\infty}^{\infty}\delta(\omega-k\omega_s)\tag{1}$$

with $\omega_s=2\pi/T$, and since multiplication in the time domain corresponds to convolution in the frequency domain with a scaling of $1/2\pi$ (if you use $\omega$ as the independent frequency variable), the transform of the sampled signal is

$$X_c(\omega) = \frac{\omega_s}{2} \sum_{n=-\infty}^{\infty} \big[ \delta(\omega - \omega_x - n\omega_s) + \delta(\omega + \omega_x - n\omega_s) \big]\tag{2}$$

Now you just need to use the relation between $\omega$ and $\Omega$

$$\omega=\frac{\omega_s}{2\pi}\Omega\tag{3}$$

and the scaling property of the Dirac Delta impulse

$$\delta(ax)=\frac{1}{|a|}\delta(x)\tag{4}$$

which for your example results in

$$\delta(\omega)=\frac{2\pi}{\omega_s}\delta(\Omega)\tag{5}$$

and you'll arrive at the correct expression for the DTFT of a sampled cosine.

Also take a look at this answer to a related question.

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  • $\begingroup$ Beautifully explained, thanks a lot, you made my day. I wonder though... starting from the CTFT cosine equation, do you have an idea how to also derive the windowed equation DTFT $W(\Omega) = e^{-j\Omega(L-1)/2}\cdot \frac{sin(\Omega L /2)}{sin(\Omega / 2}$ for a length L rectangular window given a complex exponential $e^{j\omega_x t}$? I'm not too sure it's even doable given the CTFT rect() function is not the aliased sinc() but the normalized sinc(). Deriving it from the finite geometric series is easy, but starting from the CTFT seems harder. $\endgroup$ – Yannick Aug 26 '18 at 4:14
  • $\begingroup$ @Yannick: Note that this function (also called the Dirichlet kernel) is also referred to as aliased sinc function. So that's what it is, even though it might not be trivial to show this equivalence directly. $\endgroup$ – Matt L. Aug 26 '18 at 18:37

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