Suppose that I have two signals $x[n] = \left\{2,4,1\right\}$ and $p[n] = \left\{5,1,8\right\}$ and I want to multiply them.

  • How do you do that?
  • How different is it from convolving two signals?

I understand that multiplication in one domain is equal to convolution in other domain. How do you choose as to what to use : multiplication or convolution?

  • 2
    $\begingroup$ How different are apples and oranges? This part of the question is too broad, please specify what kind of difference interests you. $\endgroup$
    – chirlu
    Aug 23, 2013 at 7:02
  • 2
    $\begingroup$ This is like asking "What's better, an umbrella or a hammer?" Depends what you're trying to do. $\endgroup$
    – endolith
    Aug 23, 2013 at 14:19
  • $\begingroup$ Will Normal Time domain convolution result in frequency domain multiplication ? or it should be CIRCULAR CONVOLUTION ! Somewhere I read Property of Discrete Fourier Transform [DFT] if x(n) and h(n) are two time-domain signals, circular convolution between them x(n)©h(n) is equivalent to multiplication in the frequency domain. Because this property used in cyclic prefix for OFDMA systems. $\endgroup$
    – rajez79
    Aug 26, 2013 at 16:45

1 Answer 1


Yes, you are correct. Multiplication in time domain means convolution in frequency domain and vice versa. Multiplying your signals $x[n]$ and $y[n]$ will give an output:

\begin{align} z[n]&=\{2\cdot 5, 4\cdot 1, 1\cdot 8\}\\ &= \{10, 4, 8\}\end{align}

Remember that this output is in time domain. When you convolve $x[n]$ and $y[n]$, you will get $z[n]$ in time domain as:

\begin{align} z[n]&=\{5\cdot 2, 5\cdot 4+1\cdot 2, 5\cdot 1+1\cdot 4+8\cdot 2, 1\cdot 1+8\cdot 4, 8\cdot 1\}\\ &= \{10,22,25,33,8\} \end{align}

Just flip one of the signals around zero and start moving right one place at a time. Multiply the corresponding points as you go along. The output has a larger sequence because convolution output has ${\rm length}(x)+{\rm length}(y)-1$ points.

To answer your query about where to use multiplication & convolution, assume you want to pass signal $x(n)$ through filter $y(n)$. The output of the filter $z(n)$ will be convolution of $x(n)$ and $y(n)$.
Now assume that you first converted from time to frequency domain, i.e. $X(e^{i\omega})$ and $Y(e^{i\omega})$ are frequency domain representation of $x(n)$ and $y(n)$. Now, to find $Z(e^{i\omega})$(output in frequency domain), you have to multiply $X(e^{i\omega})$ and $Y(e^{i\omega})$. To get the output in time domain i.e. $z(n)$ you have to apply inverse transform.

So that's how you use convolution and multiplication.

  • 1
    $\begingroup$ So when to use multiplication and when to use convolution? From what you explained convolution is equal to polynomial multiplication. $\endgroup$ Aug 23, 2013 at 7:24
  • $\begingroup$ @user2381442 edited the answer to include your query. And yes, convolution is a polynomial multiplication operation $\endgroup$ Aug 23, 2013 at 8:17
  • 1
    $\begingroup$ Convolution is filtering in the spatial domain, multiplication is filtering in the Fourier domain. What to use depends on what you are trying to do. $\endgroup$ Aug 23, 2013 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.