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I will quote an answer of Matt L. from this post (I didn't comment there because I can't)

If you have a continuous-time signal $x(t)$, then the two signals you're talking about are

$$\begin{align} x_c(t) &=x(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT) \\ &=\sum_{n=-\infty}^{\infty}x(t)\delta(t-nT) \\ &=\sum_{n=-\infty}^{\infty}x(nT)\delta(t-nT) \\ \tag{1} \end{align}$$

and you define

$$x_d[n] \triangleq x(nT)\tag{2}$$

The first signal given by $(1)$ is technically a continuous-time signal, even though it is only non-zero at discrete times $t=nT$. The reason why it is considered a continuous-time signal is because it can and must be transformed using the continuous-time Fourier transform (CTFT). So $(1)$ is the continuous-time representation of a sampled signal. Eq. $(2)$ is the discrete-time representation of the same signal. Here the sampled signal is represented as a sequence of numbers. You can't apply the CTFT to $(2)$, but you must use the discrete-time Fourier transform (DTFT).

The nice thing is now that the CTFT of $x_c(t)$ given by $(1)$ and the DTFT of $x_d[n]$ given by $(2)$ are identical. So if the CTFT

$$\begin{align} X_c(j\Omega) &= \int_{-\infty}^{\infty}x_c(t)e^{-j\Omega t}dt \\ &= \int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}x(nT)\delta(t-nT) e^{-j\Omega t}dt \\ &= \sum_{n=-\infty}^{\infty}x(nT) \int_{-\infty}^{\infty}\delta(t-nT) e^{-j\Omega t}dt \\ &= \sum_{n=-\infty}^{\infty}x(nT) e^{-j\Omega nT} \\ &= \sum_{n=-\infty}^{\infty}x_d[n] e^{-j n(\Omega T)} \\ \end{align}$$

and the DTFT:

$$X_d(e^{j\omega})=\sum_{n=-\infty}^{\infty}x_d[n]e^{-jn\omega}$$

we have:

$$X_d(e^{j\omega})=X_c\left(\tfrac{j\omega}{T}\right)\tag{3}$$

In sum, the signals $(1)$ and $(2)$ are just two different representations of the same signal, and their spectra (one defined by the CTFT, the other defined by the DTFT) are identical.

My question arises due to a question I got when the period of the impulse train is not the same as the period of the sample:

$$x_c(t)=\sum_{n=-\infty}^{\infty}x(nT)\delta\big(t-n(\tfrac{2}{3}T)\big)\tag{1}$$

and

$$x_d[n]=x(nT)\tag{2}$$

does we still get the connection:

$$X_d(e^{j\omega})=X_c\left(\tfrac{j\omega}{T}\right)\tag{3}$$

if yes, does the $T$ in the connection imply to the impulse train period ($\frac{2}{3}T$) or to the sample time ($T$) and why?

When I set in $X_d(e^{j\omega})$ the value $\omega=\frac{2}{3}T\omega$ I got an answer which should be true. but I don't know if it's by luck or really true and I don't know why.

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We are given two signals:

$$x_c(t)=\sum_{n=-\infty}^{\infty}x(nT_1)\delta(t-nT_2)\tag{1}$$

and

$$x_d[n]=x(nT_1)\tag{2}$$

where $x(t)$ is some known signal whose Fourier transform $X(j\Omega)$ exists, and $T_1>0$ and $T_2>0$, but generally $T_1\neq T_2$.

We want to express the DTFT $X_d(e^{j\omega})$ of $x_d[n]$ in terms of the CTFT $X_c(j\Omega)$ of $x_c(t)$. The DTFT $X_d(e^{j\omega})$ is an aliased version of $X(j\Omega)$:

$$X_d(e^{j\omega})=\frac{1}{T_1}\sum_{k=-\infty}^{\infty}X\left[j\left(\frac{\omega-2\pi k}{T_1}\right)\right]\tag{3}$$

In order to compute the CTFT $X_c(j\Omega)$ of $x_c(t)$ we define a compressed ($T_1>T_2$) or expanded ($T_1<T_2$) version of $x(t)$:

$$\tilde{x}(t)=x\left(\frac{T_1}{T_2}t\right)\tag{4}$$

The CTFT of $\tilde{x}(t)$ is

$$\tilde{X}(j\Omega)=\frac{T_2}{T_1}X\left(j\frac{T_2}{T_1}\Omega\right)\tag{5}$$

With $(4)$ we have $x(nT_1)=\tilde{x}(nT_2)$, and $x_c(t)$ can be written as

$$x_c(t)=\sum_{n=-\infty}^{\infty}\tilde{x}(nT_2)\delta(t-nT_2)\tag{6}$$

The CTFT of $(6)$ is

$$X_c(j\Omega)=\frac{1}{T_2}\sum_{k=-\infty}^{\infty}\tilde{X}\left[j\left(\Omega-\frac{2\pi k}{T_2}\right)\right]\tag{7}$$

and with $(5)$ this becomes

$$X_c(j\Omega)=\frac{1}{T_1}\sum_{k=-\infty}^{\infty}X\left[j\left(\frac{T_2}{T_1}\Omega-\frac{2\pi k}{T_1}\right)\right]\tag{8}$$

Comparing $(3)$ and $(8)$ we finally obtain

$$X_d(e^{j\omega})=X_c\left(j\frac{\omega}{T_2}\right)\tag{9}$$

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