As we are all taught in signal processing classes, the Dirac delta is not a normal function but a generalized function. It is sometimes defined as either the limit of some normal function whose support goes to zero as some parameter goes to zero, while keeping unit area. This apparently leads to the usual representation of the Dirac delta function as being defined and only having meaning under the so-called sifting integral, $$ \intop_{-\infty}^{\infty}x\left(\tau\right)\delta\left(\tau-t\right)d\tau=x\left(t\right) $$

Yet, pick up any book or paper on signal sampling and you will see the sampling process defined as a modulation process with the delta standing alone as a normal function, multiplying the analog signal to be sampled, like this: $$ x\left(nT\right)=\sum_{n=-\infty}^{\infty}x\left(t\right)\delta\left(t-nT\right) $$ Why is this legal? Why isn't the sampling process modeled as $$ x\left(nT\right)=\sum_{n=-\infty}^{\infty}\intop_{-\infty}^{\infty}x\left(\tau\right)\delta\left(\tau-nT\right)d\tau $$ or $$ x\left(nT\right)=\intop_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}x\left(\tau\right)\delta\left(\tau-nT\right)d\tau\text{?} $$

Similarly, in the frequency domain, we know that a linear system's response to a sinusoidal input is a common measurement. Using the common definitions for $X$, $Y$, and $H$, we know that $$ Y\left(f\right)=X\left(f\right)H\left(f\right). $$ If $$x\left(t\right)=\exp\left(j2\pi f_{0}t\right)$$ is a sinusoid, then $$ X\left(f\right)=\delta\left(f-f_{0}\right) $$ and $$ Y\left(f\right)=H\left(f_{0}\right) $$ where we have once again used the stand-alone delta as a sampling operator.

Can someone please clearly explain my confusion on this matter? Why do we seem to be so cavalier about how we use the Dirac delta when modeling sampling?

(While we're in the neighborhood, maybe someone could explain why we're even allowed to think about the Fourier transform of a sinusoid since it it is neither absolutely integrable nor square integrable.)

By the way, I find the discussion at Why is dirac delta used in continuous signal sampling? to be unsatisfying. Can someone do better?

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    $\begingroup$ Speaking solely for myself, I try to avoid using the Dirac delta as much as possible. You can find an excellent (if not the best) book on digital communications, which includes the FT and sampling, and never uses a single delta: "A foundation in digital communication", by Amos Lapidoth (free online from the author). $\endgroup$ – MBaz Jan 31 '18 at 0:33
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    $\begingroup$ no one is saying this: $$ x(nT)=\sum_{n=-\infty}^{\infty}x(t)\delta(t-nT) $$ what's on the left is something that is not a function of $t$ (but sorta looks like a function of $n$) and what is on the right is a function of $t$. what we do say is $$\begin{align} x(t) \sum_{n=-\infty}^{\infty}\delta(t-nT) &= \sum_{n=-\infty}^{\infty} x(t) \delta(t-nT) \\ &= \sum_{n=-\infty}^{\infty} x(nT) \delta(t-nT) \\ \end{align}$$ but there are still "naked" dirac impulses in that. i usually just suggest that we think of the dirac delta as having width of on Planck time. $\endgroup$ – robert bristow-johnson Jan 31 '18 at 8:36
  • $\begingroup$ if you want a "proof" of the sampling theorem that uses no dirac impulses, i can take the Poisson summation formula and rearrange variables and that, plus an assumption of being bandlimited, will prove the sampling theorem. $\endgroup$ – robert bristow-johnson Jan 31 '18 at 8:40
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    $\begingroup$ @robertbristow-johnson In sampling by multiplying a function with a very narrow pulse, one still must integrate to get a number (the sample). Otherwise, all you get is a bunch of narrow pulses with different heights and slopy hats. The latter is an OK approximation for computing the DTFT but doesn't yield the sample, such as would fall out of a ADC, quantization notwithstanding. Right? Which sort of makes us think what we mean by sampling: getting a sequence of numbers or a sequence of deltas weighted by those numbers. $\endgroup$ – Oscar Feb 2 '18 at 5:04
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    $\begingroup$ but the non-zero width pulse can be treated as a "normal function" instead of a distribution. the non-zero width pulse can exist nake outside of the integral and the mathematicians will not complain. $\endgroup$ – robert bristow-johnson Feb 2 '18 at 7:14

The OP's "equation" $$ x\left(nT\right)=\sum_{n=-\infty}^{\infty}x\left(t\right)\delta\left(t-nT\right)\tag{1} $$ is incorrect; the sum on the right side of $(1)$ is the sampled signal $x_s(t)$, not $x(nT)$ at all. That is, $$x_s(t) = \sum_{n=-\infty}^{\infty}x\left(t\right)\delta\left(t-nT\right).\tag{2}$$ This is not an ordinary function since it is a weighted sum of Dirac deltas, but if we pretend that it is indeed an ordinary function, then the "value" of $x_s(t)$ at $t=mT$ is $$x_s(mT) = \sum_{n=-\infty}^{\infty}x\left(nT\right)\delta\left(mT-nT\right) = x(mT)\delta(0),$$ while if $t$ is not an integer multiple of $T$, then the "value" of $x_s(t)$ is $0$. Note the quotation marks around the word value and avoid the usual nonsense of saying that $\delta(0) = \infty$ that is found in so many low-level DSP books. $x_s(t)$ is not a function and it does not have any value at any real number $t$. What the sampled signal $x_s(t)$ as defined in $(2)$ is is an impulse train; a collection of impulses spaced $T$ seconds apart with coefficient or amplitude of the impulse at time $t=mT$ being $x(mT)$, that is, the sampled value of $x(t)$ at time $mT$. As stated earlier, $x_s(t)$ is not an ordinary function of time; it is, as all the other answers have pointed out, a functional, that is, a function that operates on functions as a whole, not on individual specific values that the function being operated on takes on as ordinary functions do. Now, it is a standard notion that $$x(t)\delta(t-t_0) = x(t_0)\delta(t-t_0)$$ provided that $x(t)$ is continuous at $t=t_0$, and so if we re-write $(2)$ as $$x_s(t) = \sum_{n=-\infty}^\infty x\left(nT\right)\delta\left(t-nT\right) \tag{3}$$ we see that $(3)$ formally resembles the commonly used sum $$x_m(t) = \sum_{n=-\infty}^{\infty}x\left(nT\right)g\left(t-nT\right)\tag{4}$$ which describes the modulation of a stream of pulses (of the form $g(t)$ spaced $T$ seconds apart) by the sample values of $x(t)$ spaced $T$ seconds apart. Small wonder then that textbook writers notice the analogy between $x_s(t)$ as given in $(3)$ and $x_m(t)$ in $(4)$ and impulsively describe the process that leads to $(2)$ or $(3)$ as a modulation process instead of a multiplication process. (At this point, it is worth reflecting that several modulation processes (e.g. AM-DSBSC which can be described as $x(t) \mapsto x(t)\cos(2\pi f_ct)$) are indeed just multiplication of a carrier signal by a baseband signal).

So, why is this rigamarole useful to anyone? Well, let's look at the Fourier transform of $x_s(t)$:

\begin{align} \int_{-\infty}^\infty x_s(t) \exp(-j2\pi ft) \,\mathrm dt &= \int_{-\infty}^\infty \sum_{n=-\infty}^{\infty}x\left(t\right)\delta\left(t-nT\right) \exp(-j2\pi ft) \,\mathrm dt\\ &= \sum_{n=-\infty}^{\infty}\int_{-\infty}^\infty x\left(t\right)\delta\left(t-nT\right) \exp(-j2\pi ft) \,\mathrm dt &{\scriptstyle{\text{sifting property of Dirac delta}}}\\ &= \sum_{n=-\infty}^{\infty} x(nT)\exp(j2\pi f(nT)) \tag{5} \end{align} which is the DTFT of the discrete-time sequence $\big\{x(nT)\big\}_{n=-\infty}^\infty$ of sample values of $x(t)$ spaced $T$ seconds apart. More generally, if we have a (linear time-invariant BIBO) filter whose impulse response (dirty word!) is $g(t)$, then applying the input $x_s(t)$ to this filter, we get that the output is \begin{align} \int_{-\infty}^\infty \left( \sum_{n=-\infty}^{\infty}x\left(t\right)\delta\left(t-nT\right)\right) g(\tau-t) \,\mathrm dt &= \sum_{n=-\infty}^{\infty}\int_{-\infty}^\infty x\left(t\right)\delta\left(t-nT\right) g(\tau-t) \,\mathrm dt\\ &= \sum_{n=-\infty}^{\infty} x(nT) g(\tau-nT)\\ &= x_m(\tau)& \text{as defined in }(4) \end{align} So, Equation $(2)$ does have some utility whereas the jury is still out on the OP's $(1)$.

The OP asks:

Why isn't the sampling process modeled as $$ x\left(nT\right)=\sum_{n=-\infty}^{\infty}\intop_{-\infty}^{\infty}x\left(\tau\right)\delta\left(\tau-nT\right)d\tau ?? $$

Well, quite apart from the royal confusion that the OP is creating by using $n$, a parameter on the left side of the OP's alleged definition, as an index of summation, note that the sifting integral property says that the value of that integral is $x(nT)$. Consequently, the OP's purported definition reduces to $$x(nT) = \sum_{n=-\infty}^\infty x(nT),$$ which is nonsensical, and matters don't improve very much if a different symbol is used for the parameter on the left side because all one gets is $$x(mT) = \sum_{n=-\infty}^\infty x(nT),$$ which suggests that $x(nT) = 0$ for all integers $n$ that are not equal to the chosen $m$. That's no way to run a railroad!

Turning to the frequency domain, the OP asserts that

If $$x\left(t\right)=\exp\left(j2\pi f_{0}t\right)$$ is a sinusoid, then $$ X\left(f\right)=\delta\left(f-f_{0}\right) \tag{4} $$ and $$ Y\left(f\right)=H\left(f_{0}\right).\tag{5} $$

While $(4)$ is correct, $(5)$ is not. Actually, $$Y(f) = H(f)X(f) = H(f)\delta(f-f_0) = H(f_0)\delta(f-f_0)\tag{6}$$ which does not equal $H(f_0)$ as the OP claims. Indeed, in $(5)$ the left side is a function of $f$ while the right side is a constant so that the standard interpretation of $(5)$ is that $Y(f)$ is a constant, which in turn implies that $y(t)$ is an impulse or Dirac delta! Since $H(f)$ is arbitrarily chosen, what $(5)$ seems to be implying is that the response of any filter to a complex sinusoid is an impulse, which is manifestly nonsense. What is true is that if we assume the system with transfer function $H(f)$ (a.k.a frequency response) has input $\exp(j2\pi f_0 t)$, then the output is $$y(t) = \mathcal F^{-1}(Y(f)) = \int_{\infty}^\infty H(f)\delta(f-f_0)\exp(j2\pi tf)\, \mathrm df = H(f_0)\exp(j2\pi f_0t)$$ which is the epitome of the definition of the frequency response function:

If for each real number $f_0$, the complex sinusoidal input $\exp(j2\pi f_0t)$ at frequency $f_0$ produces complex sinusoidal output $H(f_0)\exp(j2\pi f_0t) = |H(f_0)|\exp\big(j(2\pi f_0t+\angle H(f_0))\big)$ (which is also at frequency $f_0$ but has amplitude $|H(f)|$ and phase $\angle H(f_0)$), then $H(f)$ is called the frequency response function or transfer function of the linear time-invariant system.

In short, the OP's whole question is based on massaging "alternative facts" (that do not withstand scrutiny) to arrive at incorrect conclusions.

  • $\begingroup$ Thank you for your response. "My 'equation'" is not mine. As I stated in my post, I am merely copying what appears in countless texts on signal analysis and signal processing and as such is the basis of my question. $\endgroup$ – Oscar Feb 1 '18 at 23:57
  • $\begingroup$ You offer the expnression $\endgroup$ – Oscar Feb 2 '18 at 0:01
  • $\begingroup$ Please cite some textbooks where you found the expression $$ x\left(nT\right)=\sum_{n=-\infty}^{\infty}x\left(t\right)\delta\left(t-nT\right), $$ (which I copied and pasted into my answer from your question) preferably with scans of the relevant pages even if the textbooks are "so famous that everyone on dsp.SE should have it on their bookshelf." $\endgroup$ – Dilip Sarwate Feb 2 '18 at 0:08
  • $\begingroup$ I should have typed $$s\left(t\right)=\sum_{n=-\infty}^{\infty}x\left(t\right)\delta\left(t-nT\right)$$ as the common textbook equation. The problem is on the RHS, not the LHS, and remains with this correction. $\endgroup$ – Oscar Feb 2 '18 at 4:26
  • $\begingroup$ I claim neither $Y\left(f\right)=H\left(f-f_{0}\right)$ nor $Y\left(f\right)\neq H\left(f-f_{0}\right)$. $\endgroup$ – Oscar Feb 2 '18 at 4:31

What you call the “so called”sifting integral is the identity operation of continuous time convolution. It defines an input to output relationship of an ideal linear time invariant system that passes a signal with no distortion and without delay. While not realizable, for a baseband audio signal, it isn’t a bad model for a short segment of wire.

In general, the convolution integral (LTI) is written as $$ y(t)=\int_{\infty}^{\infty} x(\tau) h(t-\tau) d \tau $$ where $y(t)$ is the output signal, $x(t)$ is the input signal, and $h(t)$ is called the impulse response. If $x(t)=\delta(t)$ then $y(t)=h(t)$ which is a good reason to call $h(t)$ an impulse response. So we have this notion of a naked Dirac delta as a probe to a LTI system that can live outside of an integral. Heaviside approached this probe idea with the unit step function which seems more reasonable but if you differentiate the unit step, one interpretation is the derivative is the Dirac Delta.

The physicists let one assume that all the mass of a rigid object can be concentrated at a single point in several well accepted equations in classic gravity, which is essentially another idealized naked delta. Monopole radiators fall into the same class, Green’s Functions, Orthogonality of continuous functions, mixed continuous and discrete random variables. ...

There are a lot of physical systems where one needs to represent a discrete infinitely compact object in a continuous valued coordinate system and the Dirac Delta is often used, so when encountered in sampling theory, it is no more esoteric than a point mass.

Sampling lends itself to some profound observations if one is so inclined. A continuous waveform defined on an uncountable infinite number of points, constrained by bandwidth can be reduced to a countable infinite set of discrete samples with no loss of information, and the converse is also true.

  • $\begingroup$ +1 This is a great answer. I would only quibble that in your impulse example the $\delta$ still "lives inside the integral". The discussion in the link I gave is about the DTFT of the unit step function which has the $\delta$ firmly outside any integral. $\endgroup$ – Cedron Dawg Jan 31 '18 at 4:38

BTW, if anyone is interested, I've been pestering mathematicians about this for decades.

My spin on it is simple. If you want to look at the dirac delta as a limit of nascent delta functions, such as:

$$ \delta(t) \ \triangleq \ \lim_{\sigma \to 0^+} \tfrac{1}{ \sigma} \Pi \left( \tfrac{t}{\sigma} \right) $$

where $ \Pi(x) \triangleq \begin{cases} 1 \quad |x|<\frac{1}{2} \\ \frac{1}{2} \quad |x|=\frac{1}{2} \\ 0 \quad |x|>\frac{1}{2} \\ \end{cases} $

(sometimes $\Pi(\cdot)$ is called "$\operatorname{rect}(\cdot)$".)


$$ \delta(t) \ \triangleq \ \lim_{\sigma \to 0^+} \tfrac{1}{\sigma} e^{-\pi \left(\frac{t}{\sigma}\right)^2} $$

then just stop the limit of the width $\sigma$ at 1 Planck time, $t_\text{P}$. If $\sigma>0$, then $\delta(t)$ remains an ordinary real function, even for the mathematicians, and for the engineers, there will be no essential numerical or physical difference between $\sigma = t_\text{P}$ and any smaller positive $\sigma$.


Engineers and physicists think about $\delta$ as a function on $\mathbb{R}$, but this is mathematically wrong. The Dirac delta is not a function of numbers but of functions.

We think of $\delta$ as a sequence of functions $f_n$, with total area under the graph 1. Formally speaking, in the language of distributions: $$\lim_n f_n = \delta$$

But... The fact that $\lim_n f_n$ equals $\delta$ has nothing to do with the limit at some or many $t\in\mathbb{R}$. When we write that limit, the exact meaning is that for all functions $f$ which are infinitely many times differentiable and have compact support we have $$\lim_n \int_{-\infty}^\infty f(t) f_n(t) dt = f(0)$$

To put it even clearer, $\delta$ is an operator that, when applied to some function $f(t)$, does the following:


Thus, the use of the Dirac delta without an integral and as a function of $t\in\mathbb{R}$ is not correct in mathematically rigorous terms... But it's useful for us, so we use it anyway (we are not mathematicians after all).

Regarding your question about the Fourier transform of a sinusoid, note that the "function" you get is not a normal one - it is a generalized function (a distribution, the Dirac delta). Therefore there are diverging integrals there, so nothing converges in the classical way. Maybe the comments in this question can help. Also check Dilip's answer here, which is as clear as it gets.

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    $\begingroup$ I would think that following notational convention $\delta(f(t))$ could be inferred as the zeros of $f(t)$ $\endgroup$ – Stanley Pawlukiewicz Jan 31 '18 at 17:16
  • $\begingroup$ @StanleyPawlukiewicz Maybe with brackets instead of parentheses would be better? $\endgroup$ – Tendero Jan 31 '18 at 17:22
  • $\begingroup$ maybe brackets, or a tilde over it, or something like that. $\endgroup$ – Stanley Pawlukiewicz Jan 31 '18 at 17:25
  • $\begingroup$ @Tendero I like the first part of your answer which seems consistent with my recent reading. But I feel that simply treating it as an operator, while convenient, sweeps most of the mathematical detail that I was looking for under the rug. $\endgroup$ – Oscar Feb 2 '18 at 4:39
  • $\begingroup$ actually, Ten, the support of this nascent delta: $$ \delta(t) \ \triangleq \ \lim_{\sigma \to 0^+} \tfrac{1}{\sigma} \tfrac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{t}{\sigma}\right)^2} $$ never shrinks down to zero. $\endgroup$ – robert bristow-johnson Feb 3 '18 at 7:01

Because Engineers and Mathematicians look at math differently. In many contexts, using $\delta$ informally works, so it gets used.

This discussion will probably drive you nuts.

I've had similar difficulty, particularly on comp.dsp, convincing Engineers that knowing that a solution is "exact" has significance in the meanings of the equations.


  • $\begingroup$ Looks like someone downvoted this answer. That's the reason for the -2. Rep changes +10 for an upvote on an answer, but only -2 on a downvote. So even if every answer was upvoted and downvoted, you'd still wind up with 8 rep per answer. $\endgroup$ – Peter K. Feb 2 '18 at 21:40
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    $\begingroup$ @ Peter K, Thanks, I figured that out which is why I deleted my question on Meta. I can see somebody getting pissy about my last sentence, but oh well. Based on the other answers in this thread, I think my first sentence hit the nail on the head quite succinctly. $\endgroup$ – Cedron Dawg Feb 2 '18 at 22:05
  • $\begingroup$ Ced, i think your first sentence (about how engineers and mathematicians look at mathematics) is most certainly accurate. i dunno how well it answers the OP's question about "Why is this [use of naked dirac deltas] legal?" it's not a bad answer, but i can see why someone might ding you for lacking a satisfying answer to that question. $\endgroup$ – robert bristow-johnson Feb 3 '18 at 7:20
  • $\begingroup$ @robert bristow-johnson, Nobody is going to jail over this, so I think a better word than "legal" is "proper", perhaps "mathematically proper". I read the crux of the OP's question to be the matter of being "cavalier" about it, not whether it is proper. I share the OP's puzzlement over the use in Dilip Sarwate's second equation which boils down to $x_s(mT) = x(mT)\delta(0)$. This seems to be the (continuous) Kronecker Delta, not the Dirac Delta. Then, there is also the issue of "$\delta[]$" vs "$\delta()$" $\endgroup$ – Cedron Dawg Feb 3 '18 at 14:32

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