0

Problem

I'm given the following convolution problem. I distribute the values that correspond to each delta, then add the time shifts. I get the correct solution which is D

But, is it possible to work these problems in a calculator? This problem in particular creates a 4x3 matrix. Would it be possible to turn the input into a series of complex numbers then multiply the real part while adding the time shifts.

For example (I'm using the asterisk (*) to denote convolution)

(2δ[n]-δ[n-1])*(δ[n-1]-δ[n-2])

Would go to 

(2+0i, -1-i) * (1-i, -1-2i)

So my convolution would be 

(2-i, -2-2i, -1-2i, 1-3i) or 2δ[n-1]-2δ[n-2]-δ[n-2]+δ[n-3]

So, ultimately what I'm trying to ask is, can I multiply one part of a number while adding the other. Or, is there a better way to solve convolutions all together?

Bill Tudor
  • 3
  • With the help of $z$-transfom, all you need to do is to multiply two functions. But then you will need to take inverse $z$-transform – polfosol Sep 28 '16 at 06:54

1 Answers1

0

Take $z$ transform: $$(2\cdot 1-z^{-1})(z^{-1}-z^{-2})=2z^{-1}-3z^{-2}+z^{-3}$$ Now take inverse $z$-transform to get: $$2\delta[n-1]-3\delta[n-2]+\delta[n-3]$$

polfosol
  • 9,245