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This is a problem from Michael Lindeburg's FE prep book - find the convolution sum v[n] = x[n] * y[n]. I am familiar with the graphical method of convolution. However, I am not familiar with convolution when the signals are given as data sets (see picture).

I tried solving this using the tabular method that I learned from online tutorials - the solution involves multiplying the inputs one for one, and adding the products in diagonal. The answer I get is incorrect. The answer, according to the book, is 3. Where did I mess up?

Also, the problem doesn't specify where the given values lie on the axis. So given the signal x[n] = 3,2,1,0; do I just assume that 3 is at n = 0; 2 is at n = 1, and so on.

Please advise.

enter image description here

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    \$\begingroup\$ Is "convolution sum" synonymous with calculating the convolution? The convolution is correct at least. I don't do the square diagonal method you're talking about but it's the same as me reversing y from left to right and sliding the rows of numbers past each other. Like this: en.wikipedia.org/wiki/Convolution#/media/… But it doesn't make sense to me what you mean when you say the answer is three. To me, that's like saying 1+1=pizza. Are you sure the answer isn't just cutoff? What is the original material? Is it a poor quality photocopy or scan? \$\endgroup\$
    – DKNguyen
    Dec 27, 2021 at 2:05
  • \$\begingroup\$ I agree. It didn't make sense to me either. The question is from a book - FE Practice Problems electrical and computer by Michael Lindeburg. \$\endgroup\$
    – vasiqshair
    Dec 27, 2021 at 18:08

1 Answer 1

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GIVEN:
x[n] = [3,2,1,0]
y[n] = [1,2,3,0]

v[n] = x[n] * y[n] is...

v[0]= x[0]y[0] = 3 * 1 = 3
v[1]= x[0]y[1] + x[1]y[0] = 3 * 2+2 * 1 = 8
v[2]= x[0]y[2] + x[1]y[1] + x[2]y[0] = 3 * 3+2 * 2+1 * 1 = 14
v[3]= x[0]y[3] + x[1]y[2] + x[2]y[1] + x[3]y[0] = 3 * 0+2 * 3+1 * 2+0 * 1 = 8
v[4]= x[0]y[4] + x[1]y[3] + x[2]y[2] + x[3]y[1] + x[4]y[0] = 3 * 0+2 * 0+1 * 3+0 * 2+0 * 1 = 3
v[5]= x[0]y[5] + x[1]y[4] + x[2]y[3] + x[3]y[2] + x[4]y[1] + x[5]y[0] = 3 * 0+2 * 0+1 * 0+0 * 3+0 * 2+0 * 1 = 0
v[6]= x[0]y[6] + x[1]y[5] + x[2]y[4] + x[3]y[3] + x[4]y[2] + x[5]y[1] + x[6]y[0] = 3 * 0+2 * 0+1 * 0+0 * 0+0 * 3+0 * 2+0 * 1 = 0

So the answer appears to be v[n] = [3, 8, 14, 8, 3, 0, 0...]

Which is exactly what you wrote down.

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