Bridging CTFT and DTFT for a cosine 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:

$$\begin{aligned}
x(t) &amp;= \cos(\omega_x \cdot t) = \frac{1}{2} \cdot \left( e^{j\omega_x t} + e^{-j\omega_x t} \right) \\ \implies X(\omega) &amp;= \pi \cdot (\delta(\omega - \omega_x) + \delta(\omega+\omega_x)) \\ x_c(t) &amp;= x(t) \cdot \sum_{n=-\infty}^{\infty}\delta(t - nT_s) = \sum_{n=-\infty}^{\infty}x(nT_s)\delta(t-nT_s) \\ \implies X_c(\omega) &amp;= X(\omega) * \left( \omega_s \sum_{n=-\infty}^{\infty} \delta(\omega - n\omega_s)\right) \\
\end{aligned}$$



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

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

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
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.