Relation between the DTFT and the spectrum of a sampled signal

Question

In the $\rm DTFT$ (Discrete Time Fourier Transform) the spectrum is periodic with a 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 equal to the sampling frequency. Both have periodic spectra and both are discrete in nature.

When we sample a continuous signal, its values at $nT$ are 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 1

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.

Answer 2

The relationship between an analog (continuous-time) signal $x_a(t)$ sampled uniformly at $t = nT$ to its discrete-time signal $x(n)$ and their spectrum relationship is: \begin{align}x(n)&=x_a(nT), \quad-\infty \leq n\leq \tag{1}\infty\\X(F)&=F_s\sum_{k=-\infty}^\infty X_a(F - kF_s)\tag{2}\end{align}

With: $T = \frac 1{F_s} \Rightarrow t=nT=\frac{n}{F_s}$.

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

Clearly one is continuous time, whilst the other discrete time. They are equal only at those discrete and uniformly sampled times $t=nT$; or $x(n) \ne x_a(nT)$ when $t\ne nT$. Put another way:

$$x(n) = \begin{cases} x_a(nT), &\text{ if } t=nT, (\text{ with } n \in \mathbb Z) \\[2ex] \text{undefined}, & \text{elsewhere} \tag{3}\\ \end{cases}$$

In the frequency domain, and with no aliasing, you have:

$$X_a(F) = \begin{cases} \frac{1}{F_s}X(F), &\left|F\right|\leq \frac{F_s}{2}\\[2ex] 0, &\left|F\right| > \frac{F_s}{2}\tag{4}\\ \end{cases}$$

Is there any application for the above observation?

It is the basis of the Sampling theorem and reconstruction. That is, if $x_a(t)$ is band-limited with $B\text{ Hz}$ the highest frequency, it can be uniquely recovered from its samples if $F_s \geq B$. So, from $(2)$, a lowpass filter of gain $T$ with cutoff frequency $F_c > B$ is used to remove the spectral periodicity to get the spectrum of the continuous-time signal. You may have a look at interpolation in $\rm DACs$, this happens there.

EDIT:

• The relationship between the sampled signal $x(n)$ and the discrete aperiodic signal $x_a(nT)$ is none other that the relationship in equation $(3)$. And this is by definition: Sampling uniformly the function $x_a(t)$ at $t=nT$, or: $$x(n) \triangleq x_a(nT)\tag{5}$$ You don't have a different/separate signal $x(n)$ you're trying to equate to $x_a(nT)$, no. You have $x_a(t)$ and "extract" $x(n)$ by ideal continuous-time-to-discrete-time conversion as defined in equation $(3)$. So, by definition in the equality in equation $(5)$ we're talking about an ideal conversion of a continuous-time signal to a discrete-time signal. In practice this involves $\rm ADCs$, that's another story.
• The spectrum of $x_a(t)$ (in its continuous form) given by the $\rm CTFT$ is a continuous aperiodic function of frequency: $$\displaystyle X_a(F) = \int_{-\infty}^{\infty}x_a(t)e^{-j2\pi Ft}dt, \quad -\infty\leq F\leq \infty\tag{6}$$
• The spectrum of $x(n)$, obtained by sampling uniformly $x_a(t)$, given by the $\rm DTFT$ is a continuous and periodic function of $\omega$ with period $2\pi$: $$\displaystyle X(\omega) = \sum_{n=-\infty}^{\infty}x(n)e^{-j\omega n}, \quad -\infty\leq n\leq \infty\tag{7}$$
• With the relation $t = \frac{n}{F_s}$, equation $(5)$, and inverse transform $(6)$: $$\displaystyle x(n) \equiv x_a(nT)= \int_{-\infty}^{\infty}X_a(F)e^{j2\pi nF/F_s}dF\tag{8}$$
• With the uniform sampling, you have the following relationship with frequency variables: $$\omega = 2\pi f = 2\pi \frac{F}{F_s}\tag{9}$$ The periodicity of $2\pi$ in $\omega$, or chunks of $\left[-\pi, \pi\right]$, is equivalent to a periodicity of $F_s$ in $F$, or chunks of $\left[-\frac{F_s}{2}, \frac{F_s}{2}\right]$.