How can I solve the convolution for this example in steps? 

Guest Jun 17, 2017

Any convolution is defined as

Trying to do it... Never done this type of question before.


But f(x) is only defined in {-1<=x<=1|x\(\in\mathbb{Z}\)}

h(x) is only defined in {1<=x<=5|x\(\in\mathbb{Z}\)}

So Idk what does it mean by integrating f(x)h(3-x) from infinity to negative infinity... How on Mars could I do that?

MaxWong  Jun 18, 2017

\(\text{The definition of the discrete convolution is:}\)




\(\text{We require the solution for the case }n=3:\)




\(\text{since } f[m]=0, \text{for: } m\neq-1,0,1.\)


\(\text{Our solution is:} \)


\(\begin{align*} f*h[3]&=f[-1]g[4]+f[0]g[3]+f[1]g[2]\\ &=aI_4+bI_3+cI_2. \end{align*}\)

Honga  Jun 18, 2017

\(\text{An intuitive approach of solving discrete convolution is sliding an inverted version of }\\f\text{ over }h. \text{ For }n=0\text{ we have:} \)


\(\begin{array}{c| c c} m&-1&0&1&2&3&4&5\\\hline h&0&0&I_1&I_2&I_3&I_4&I_5\\ f&c&b&a&0&0&0&0\\\hline f*h[0]&0&0&aI_1&0&0&0&0 \end{array}\)


\(\text{the convolution product in this case is the product of each column.}\\ \text{Now we slide }f\text{ three places:}\)


\(\begin{array}{c| c c} m&-1&0&1&2&3&4&5\\\hline h&0&0&I_1&I_2&I_3&I_4&I_5\\ f&0&0&0&c&b&a&0\\\hline f*h[3]&0&0&0&cI_2&bI_3&aI_4&0 \end{array}\)


\(\text{and we have the same solution.}\)

Honga  Jun 18, 2017
edited by Honga  Jun 18, 2017
edited by Honga  Jun 18, 2017

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