I've been reading some papers in signal proccesing and I'm very confused about the issue in the title of my question. Consider a continuous function of time $t$, $f(t)$, that I sample at uneven times $t_k$, where $k=1,2,...,N$. To me, it makes sense that the sampled function is: $$f_s(t)=\sum_{k=1}^N\delta_{t,t_k}f(t),\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ where $\delta_{t,t_k}$ is Kronecker's delta (equals $1$ when $t=t_k$, zero elsewhere). However, in this paper, the author defines the sampled signal as: $$f_s(t)=\frac{1}{N}\sum_{k=1}^{N}f(t)\delta(t-t_k),\ \ \ (2)$$ where $\delta(t-t_k)$ is Dirac's delta function and I really don't get why the $1/N$ appears here (the author claims that the sampling function is actually a weighted sum of delta functions $$s(t)=C\frac{\sum_{k=1}^{N}w_k\delta(t-t_k)}{\sum_{k=1}^{N}w_k},$$ and here he choose $C=w_k=1$. I really didn't understand why). This last statement doesn't make much sense to me: the sampled signal would have infinite amplitude at $t=t_k$!

Despite of all this, it is much easier to define the Fourier Transform of $f_s(t)$ in the second case (equation $(2)$), because it is just the convolution of the window function (the FT of the Dirac comb) and the FT of the continuous signal $f(t)$, while on equation $(1)$ the FT is a bit more complicated because we have an integer function (Kronecker's delta) multiplied by a continuous function ($f(t)$). Any highlights on this?


1 Answer 1


Modeling of the sampling process via multiplication of a continuous-time signal by a train of Dirac impulses is the most common interpretation in my experience. If you dig into it deeply enough, you'll find some disagreement about the mathematical precision of this approach*, but I wouldn't worry about it; it's just a convenient model for the process. There are no impulse generators inside your cell phone's ADC generating periodic bolts of lightning that multiply their analog inputs.

As you noted, you can't calculate the continuous-time Fourier transform of the Kronecker delta function, as its domain isn't continuous (it is limited to the integers). The Dirac delta function, in contrast, has a simple Fourier transform, and the effect of multiplying a signal by a train of Dirac impulses is easy to show due to its sifting property.

*: As an example, if you're going to be mathematically precise, you would say that the Dirac delta isn't a function at all, but a distribution instead. But at an engineering level, these issues are really just semantics.

Edit: I'll address the comment below. You gave your mental model of the sampling process as:

$$ f_s(t) = \sum_{k=1}^N \int_{t_k-\epsilon_k}^{t_k+\epsilon_k} f(t) \delta(t-t_k) dt. $$

The problem with this interpretation is that the typical ideal sampling model does not have that integration built in. Instead, it's a pure multiplication of the input signal by a Dirac impulse train. If you look more closely at the equation you showed for $f_s(t)$, you'll see that the right hand side actually has no independent variable; $t$ is the dummy variable of integration. For any $\epsilon_k > 0$ in the above, according to the Dirac impulse's sifting property, you would get:

$$ f_s(t)=\sum_{k=1}^N f(t_k), $$

which isn't correct. Instead, the model for the sampled signal is:

$$ f_s(t)=\sum_{k=-\infty}^{\infty} f(t) \delta(t-kT) $$

Which is very similar to the above, except generalizing for an infinitely long impulse train along the time axis and assuming that the data is uniformly sampled at time instants $t_k = kT$. The Fourier transform of the resulting signal is:

$$ \begin{align} F_s(\omega) &= \int_{-\infty}^{\infty}f_s(t) e^{-j \omega t} dt \\ &= \int_{-\infty}^{\infty} \sum_{k=-\infty}^{\infty} f(t) \delta(t-kT) e^{-j \omega t} dt \\ &= \sum_{k=-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) \delta(t-kT) e^{-j \omega t} dt \\ &= \sum_{k=-\infty}^{\infty} f(kT) e^{-j \omega kT} \end{align} $$

If we define the discrete, sampled version of the signal $f(t)$ to be $x[n] = f(nT)$, then you are left with:

$$ F_s(\omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-j \omega n} $$

which is exactly the definition of the discrete-time Fourier transform.

  • $\begingroup$ How would you address the fact that the amplitude is "infinite"? What I have usually thought is that you actually don't "sample" the signal at a discrete time $t_k$, but rather you integrate the signal for a given time $\Delta t_k$. However, this interpretation would violate any form of calculating the Fourier Transform by the same reason as the Kronecker delta. Also...why does the author of the paper in the link I give divides the Dirac comb by $N$? That doesn't make any sense to me. $\endgroup$
    – Néstor
    Commented Jul 23, 2012 at 0:35
  • 1
    $\begingroup$ In practice, you're right. There is always some effective "integration time" of the analog signal owing to the finite bandwidth of any ADC's analog front end. However, the theoretical construct isn't limited by such concerns. Roughly speaking, the "infinite height" of the impulse is balanced by its "zero width", such that it integrates to unity. If you apply your short-time integration interpretation in this case (multiplying by an impulse, integrating for an infinitestimally-short time period), then by the sifting property, you would get $x[n] = x(nT)$, as is usually presented. $\endgroup$
    – Jason R
    Commented Jul 23, 2012 at 1:25
  • $\begingroup$ Yeah, but my concern is not with the interpretation, but rather with taking the fourier transform of it. Suppose I write the sampling process that we are talking about as: $$f_{s}(t)=\sum_{k=1}^N \int_{t_k-\epsilon_k}^{t_k+\epsilon_k} f(t)\delta(t-t_k)dt,$$ how would you take the fourier transform of that? I know the trick when $\epsilon_k \to 0$, but that doesn't make much sense to me (and it's even harder to do the FT!). Even supposing that is the way, I don't get the same window function as in the paper of Roberts et al. that I cited. And I insist...that $1/N$ doesn't make any sense to me. $\endgroup$
    – Néstor
    Commented Jul 23, 2012 at 2:17
  • $\begingroup$ Ok with your edit and with my error in my comment above. However, I still can't make up my mind about the fact that the impulse has infinite amplitude at $t=t_k$ thanks to Dirac's Delta, i.e., $f(t=t_k)\to \infty$, while what we actually observe (and want to model) is $f(t=t_k)=f(t_k)$ $\endgroup$
    – Néstor
    Commented Jul 25, 2012 at 4:48
  • 1
    $\begingroup$ I agree with Nestor. Zero width for the impulse is not the issue, as it just means the infinite amplitude is just going to be considered at the 'exact' time - hence the zero width. The issue is with the infinite amplitude being multiplied by f(kT), which results in an infinite amplitude. $\endgroup$
    – Kenny
    Commented Jul 25, 2023 at 7:51

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