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In my systems and signals course I had been asked a question about finding the convolution of two sets. I was given:

\begin{align} x[n] &= \{3,2,1\}\\ h[n] &= \{1,-2,3\}\\ \text{Find}\quad y[n] & = x[n]\star h[n] \end{align}

I have no idea on how to find the convolution of two sets. Any help is appreciated.

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  • $\begingroup$ I must say that I really don't understand this question. Have you tried to just use the definition of convolution? Which problems have you encountered? $\endgroup$ – Matt L. Jan 16 at 12:02
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    $\begingroup$ Sets cannot be convolved in the usual sense. Sequences can. And this is also what you have. It's very important in mathematical subjects to get the names of objects right because they have precisely defined meanings. $\endgroup$ – Jazzmaniac Jan 16 at 12:40
  • $\begingroup$ If you really have no idea, you need to change your approach towards taking this course. It's guaranteed that this has been covered in class. If you really have no idea, you have missed the essential introduction. $\endgroup$ – Hilmar Jan 16 at 15:29
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If it is difficult for you to remember or calculate the convolution of two sequences then you may try doing it as polynomial multiplication.
Think of x[n] and h[n] as polynomial coefficients. So we have

Px = 3x^2 + 2*x + 1
Ph = 1x^2 - 2*x + 3

Remember that linear convolution of two sequences is polynomial multiplication. Therefore

Py = Px * Ph
Py = (3x^2 + 2*x + 1) * (1x^2 - 2*x + 3)
Py = 3x^4 + 2x^3 - 6x^3 + x^2 + 9x^2 - 4x^2 + 6x - 2x + 3
Py = 3x^4 - 4x^3 +6x^2 + 4x + 3

Now writing the polynomial coefficients back in sequence format

y[n] = {3, -4, 6, 4, 3}
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A matrix method is far simpler I suppose:

h  x   x1      x2      x3
h1  ⌈x1*h1   x2*h1    x3*h1 ⌉
h2  |x1*h2   x2*h2    x3*h2|
h3  ⌊x1*h3   x2*h3    x3*h3 ⌋

h x 3     2    1
1   ⌈3    2    1 ⌉
-2  |-6  -4   -2|
 3  ⌊9    6    3 ⌋

Now just add up the diagonals:

y[0] = 3
y[1] = -4
y[2] = 6
y[3] = 4
y[4] = 3

p.s I don't know how to type in a matrix here.

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