fourier transform - Relation between the DTFT and the spectrum of a sampled signal

In the $\rm DTFT$ (Discrete Time Fourier Transform) the spectrum is periodic with period of $2\pi$ . A continuous signal when sampled has a spectrum which is a repeated version of its original spectrum before sampling with a period of sampling frequency. Both have periodic spectrum and both are discrete in nature.When we sample continuous signal, its values at 'nTs' is multiplied by impulse fn.But in discrete aperiodic signal it is of finite height( where impulse has infinite height)

Question is:

1. Can both signals be considered similar because their spectra are having a kind of similarity?


2. Is there any application for the above observation?

Answer

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.