# Relation between the DTFT and the spectrum of a sampled signal

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?

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.

• Now Matt, maybe we should clear up the relationship between $X(j\Omega)$ and $X_c(j\Omega)$ Aug 17, 2022 at 15:12

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 $\left[-\frac{F_s}{2}, \frac{F_s}{2}\right]$.
• Thank you for the answer.Above answer explains me about sampling theorm.My question is what is the relation between sampled signal and a discrete aperiodic signal?. Because both have periodic spectrum, I am curious whether they are related somehow? Mar 2, 2016 at 9:13
• equation 1 says both are same, where i dont think that is true because sampled signal has spectrum with period of 'Fs' where a discrete signal has spectrum which periodic with period '2*pi' Mar 2, 2016 at 9:15
• @spectre Ok, let me edit my answer. And could you edit the title to fit more your statement (question) in the first comment ? Mar 2, 2016 at 10:33
• When we sample continuos signal, its values at 'nTs' is multiplied by impulse fn.But in descrete aperiodic signal it is of finite height( where impulse has infinite height). How can they be similar in time domain? Mar 2, 2016 at 13:49
• @spectre Please see my edit. Btw, I still think you need to edit your title. Mar 2, 2016 at 15:08