Continue to Site

Welcome to

Welcome to our site! is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

multiplication of two impulses!

Not open for further replies.


Advanced Member level 4
Oct 21, 2005
Reaction score
Trophy points
Activity points
hi can anybody clarify me whether the multiplication of two impulses defined?

ie., i got this context because first i wanted to know how to convolve two DC signals. For example

\[ $x(t) = 2$\] and \[ $h(t) = 1$\].

Corresponding FT are
\[$4\pi \delta(omega)$\]
\[$ 2\pi \delat{omega}\]

So i have to multiply these two impulses to get convolution of 2 & 1. If the multiplication of impulses is not defined then can we infer that convolution of two DC signals also is not defined?

pls clarify me


Last edited by a moderator:

I don't know if what you are asking for is possible or not. According to
The main problem of the theory of distributions (and hyperfunctions) is that it is a purely linear theory, in the sense that the product of two distributions cannot consistently be defined (in general), as has been proved by Laurent Schwartz in the 1950's.

Yes it is possible to convolute two impulse(or delta) functions. and the output of these changes depending on wether its a discrete or continous time impulse.

i am asking not the convolution of delta functions but about their multiplication



You are trying to multiply two impulses in the time domain -- correct?

Well, the multiplication two signals in the time-domain is equivalent to the convolution of their transforms in the frequency domain. So let us say that you have two impluse functions occuring at different times \[f_1(t) = \delta (t-t_1)\] and \[f_2(t)=\delta (t-t_2)\]. We want to compute the product \[f(t) = f_1(t) f_2(t)\]. It seems that the product would be zero unless \[t_1=t_2\].

If these are two impulse functions occuring at time \[t_1\] and \[t_2\], respectively. The fourier transforms for these function are \[F_1(\omega) = e^{-j \omega t_1}\] and \[F_2(\omega) = e^{-j \omega t_2}\]. The convolution of these signals in the frequency domain would be
\[F(\omega) = F_1(\omega) \ast F_2(\omega)=\int_{-\infty}^{\infty} F_1(x) F_2(\omega-x) dx\]
where \[x\] is a dummy variable. You may want to continue along these lines to see if you get any insights. Any comments?
Last edited by a moderator:

Note that the two functions \[$F_1(\omega)$ \] and \[$F_1(\omega)$\] are orthogonal signals. Thus the integration turns out to be 0.

Actually I want to find out what happens if i multiply two impulses at the same instant whether it is time domain or frequency domain. i.e., i want to know is it possible to get the product of two impulses if they are at same instants?i.e.,
\[$\delta(t) \times \delta(t)=$ \]????



Added after 1 minutes:

Note that the two functions \[F_1(\omega) \] and \[F_1(\omega)\] are orthogonal signals. Thus the integration turns out to be 0.

Actually I want to find out what happens if i multiply two impulses at the same instant whether it is time domain or frequency domain. i.e., i want to know is it possible to get the product of two impulses if they are at same instants?i.e.,
\[\delta(t) \times \delta(t)= \]????


Last edited by a moderator:

HI Purnapragna,

You seem to have got it wrong, you want multiplication of 2 impulses(obviously time domain) is called CONVOLUTION........ because in signal theory, you can only convolute not multiply. but in frequency domain, you can think of multipying 2 signals!!!

Yes you can multiply two signals in a time domain -- I don't think there is anything wrong with that. I think the problem is with the fact that the thing that are being multiplied are not nice, continuous function but rather functions that are described as distributions.

By the way, the convolution operation is not limited to time-domain, it can be done in the frqequency domain as well. I already mentioned in a previous post.

i dont know what arjun meant. Did he eman both convolution and multiplication are same? Then he is obviously wrong.

but i did not get the answer for what i want!


Purnapragma, V_C has explained about multiplication and convolution.
I think you do not understand your own question, and moreover you cannot undersand these things without knowledge of Fourier or Laplace transform.

Your "impulses" are impulses( delta functions) or step functions or something else?

If it is delta function, then their multiplication is ZERO unless t1=t2 !!!

We are doing convolution in time domain because it represents a physical nature
of signals, although we can do multiplication for a specific reason.

Therefore, if we want to do a convolution of two signals in a time domain then we find their transformation(Fourier or Laplace) , multiply them and find inverse transform to get result in a time domain.

Convolution in a time domain <=> Multiplication in Fourier or Laplace domain
Multiplication in a time domain <=> Convolution in Fourier or Laplace domain

About convolution
**broken link removed**

Mr zox i m very clear about my question u r getting confused by giving unnecessary answer.

I have asked the question is it possible to have \[\delta(t) \times \delta(t)\]. Thats all


Last edited by a moderator:

I think \[\delta(t) \delta(t)=\delta(t)\].

Why? Here's are my thoughts ...

Think about what the delta function represents. Let's say we are going to represent a current that is impulsive. We may elect to write it as a delta function if its duration is much much less than any other time constant in the circuit. For example we may write \[i(t) = Q \delta (t)\]. Now why did I write that \[Q\] in front of the delta function. That is because the thing that multiplies the delta function represents the weight of the delta function and it is physically the area under the \[i(t)\] vs. time curve. And in the example that I am showing you, the area under the current versus time curve is simply the charge. So when you write something like \[\delta (t)\], you are implicitly talking about a unit area (area=1). So when you take two of the same functions with unit area and multiply them you get another function with a unit area.

Is there anything physical you are trying to tie this to or is this just a thought exercise. I am trying to tie it to something physical so we can talk about something concrete like a circuit, instead of something more abstract like pure mathematics.

Best regards,
Last edited by a moderator:

OK !!!!

δ(t1)×δ(t2)=0 for t1≠t2,
δ(t1)×δ(t2)=δ(t1) for t1=t2

Continuous case
δ(t)=∞ for t=0
δ(t)=0 for t≠0

Discrete case
δ(n)=1 for n=0
δ(n)=0 for n≠0

Mr V_c why i got this question was that i want to find the convolution of two periodic signals not with the help of Fourier Series but with the help of Fourier Transform. So, there i found that i have to multiply the weighted impulses. But according to you if \[\delta(t)\times \delta(t)=\delta(t)\] it fails if i try to convolve two signals like \[x(t)=\cos(t)\] and \[h(t)=\sin(t)\]. This is because FT of \[x(t)\] is \[\pi[\delta(\omega +1)+\delta(\omega-1)]\] and that of \[h(t)\] is \[j\pi[\delta(\omega +1)-\delta(\omega-1)]\]. So according to you the answer is \[y(t)=-\frac{1}{2}\sin(t)\] which is wrong. To be more easy, let \[x(t)=\cos(t)\] and \[h(t)=\sin(2t)\]. According to you the output should be \[0\] but it is not the case as you know. So i feel that \[\delta(t)\times \delta(t)=\delta(t)\] is not a valid answer.


Last edited by a moderator:

The dirac delta function δ(t) was initiated for a physical shock but finally formated by a strict mathematical concept, i.e., a linear continuous functional. It looks pretty funny to me when you want to make sense of the product of the two linear functionals, especially when it is almost trivial to convolve cos(t) and sin(t) ... yuck.. have a nice sleep.

Of course,the output is zero.
What does mean,for example, x(t)=cost(t)?
It means that you consider a signal from t=-∞ to t=+∞.
Your signal are not cos(t) and sin(2t). Your signals are something else!
Like cost(t)[u(t1)-u(t2)]. Where , u(t) is unit step function.

Mr steve can u be more clear abt the answer? It seems a reasonable answer and basically as i m not a math student i m unable to follow u.



I thought I had provided enough and clear information for you to proceed. I guess you either have to do it by yourself or someone else has to be willing to help you. Good night.

u can multiply two signals in frequency domain and transform to time domain

Not open for further replies.

Part and Inventory Search

Welcome to