Given a continuous time signal $f(t)$, we can sample it signal by multiplying with a Dirac comb (impulse train) $$\bar{f}(t) = \sum_{n=-\infty}^{\infty} f(nT) \delta(t-nT) \tag{1}$$ where each impulse is spaced by $T$ (Lathi. pp 319).

I have a concern about this claim. When sampling a signal, we are interested in storing values of the signal at and only at discrete instances $t=nT, n \in \mathbb{Z}$.

In other words, a sampled signal is a collection of points $f[n] = \{1,2,-1,...\}$ with uniform time spacing, where $f[n]$ represents the signal strength of $f(nT)$ for any particular $n$.

According to the sampling property of a unit impulse we have

$$\int_{-\infty}^{\infty} \phi(t)\delta(t-T) \ \text{d}t = \phi(T).$$

So shouldn't the sampled signal be represented by something like this equation instead

$$\bar{f}(t) = \sum_{n=-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) \delta(t-nT) \ \text{d}t \tag{2}.$$

I know equation (2) is wrong, but I couldn't derive an appropriate expression. In any case, wouldn't multiplying $f(t)$ by the impulse train (as stated in $(1)$) only yield a sum of scaled impulses, and not collection of points?

Don't we need some kind of integration to recover the signal strength of $f(t)$ at the time instance $t=nT$?


2 Answers 2


Sampling with an impulse train is a mathematical model that is sometimes convenient. Note that the sampled signal


cannot be evaluated at the sample instants $t=nT$ because the distribution $\delta(t)$ is not a function. The expression $(1)$ only makes sense under an integral.

The reason why the representation $(1)$ is useful is because it can be treated as a continuous-time signal. E.g., it can be the input to a continuous-time LTI system with impulse response $h(t)$. The corresponding output signal is


which is a practically relevant signal, and which can be evaluated for any value of $t$. Retrieving the sample values $x(nT)$ from $(2)$ is possible if $h(t)$ satisfies $h(0)=1$ and $h(nT)=0$ for $n\neq 0$. Such a pulse $h(t)$ is called a Nyquist pulse.

Furthermore, we can compute the continuous-time Fourier transform of $(1)$, and the result is identical with the discrete-time Fourier transform of the sequence $x_d[n]=x(nT)$.

In sum, the expression $(1)$ is a mathematical model and not a physical model. If you're only interested in the sample values $x(nT)$ then you don't need the model $(1)$.

  • $\begingroup$ "... we can compute the continuous-time Fourier transform ..." Which means that this view of sampling unifies continuous time and uniformly spaced discrete sampling. Then, instead of juggling for disparate flavors of the Fourier transform, you have, at bottom, just one. $\endgroup$
    – TimWescott
    Commented Mar 14 at 23:01

When we multiply a continuous time signal $f(t)$ by a Dirac comb (an impulse train), what we are actually doing is creating a new signal that consists of a series of impulses, each located at a sample point $t=nT$ and scaled by the value of the signal at that point, $f(nT)$. This is precisely what is described by equation $(1)$:

$$\bar{f}(t) = \sum_{n=-\infty}^{\infty} f(nT) \delta(t-nT).$$

Each term in the sum is an impulse at time $t=nT$ scaled by $f(nT)$. This isn't a "sum of scaled impulses" in the sense that we add their values together at each point in time. Rather, it's a sum of impulses in the sense that we are creating a signal that consists of a series of impulses each representing the value of the original continuous signal at the discrete times $t=nT$.

The result of multiplying the continuous signal by the Dirac comb is a signal that is only "non-zero" at the sample points—in the sense that each impulse represents the signal strength at that point. It is important to note that the Dirac delta function is not a function in the traditional sense but rather a distribution, and hence, we use it to model the concept of sampling in a mathematical way.

  • 1
    $\begingroup$ Thank you for your answer, and I completely agree with it, but I'm still missing something. Let me give an example. Say $f(t) = t, T=1$ and I sample at $n=1,2,3$. Then, according to your formula I get $\bar{f}(t) = \delta(t-1) + 2\delta(t-2) + 3\delta(t-3)$. If I then evaluate $f(1)$ I get $\delta(0)$ which is undefined. Don't I need some kind of integration to get my signal strength back? $\endgroup$
    – Carl
    Commented Mar 13 at 9:32

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