+9519 Any convolution is defined as
\((f*g)(t)=\displaystyle\int^{\infty}_{-\infty}f(x)g(t-x)dx=\int^{\infty}_{-\infty}g(x)f(t-x)dx\)
Trying to do it... Never done this type of question before.
\((f*h)(3)=\displaystyle\int^{\infty}_{-\infty}f(x)h(3-x)dx=\int^{\infty}_{\infty}h(x)f(3-x)dx\)
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?
+169 \(\text{The definition of the discrete convolution is:}\)
\(f*h[n]=\sum\limits^\infty_{m=-\infty}f[m]g[n-m].\)
\(\text{We require the solution for the case }n=3:\)
\(f*h[3]=\sum\limits^1_{m=-1}f[m]g[3-m],\)
\(\text{since } f[m]=0, \text{for: } m\neq-1,0,1.\)
\(\text{Our solution is:} \)
\(\begin{align*} f*h[3]&=f[-1]g[4]+f[0]g[3]+f[1]g[2]\\ &=aI_4+bI_3+cI_2. \end{align*}\)
.+169 \(\text{An intuitive approach of solving discrete convolution is sliding an inverted version of }\\f\text{ over }h. \text{ For }n=0\text{ we have:} \)
\(\begin{array}{c| c c} m&-1&0&1&2&3&4&5\\\hline h&0&0&I_1&I_2&I_3&I_4&I_5\\ f&c&b&a&0&0&0&0\\\hline f*h[0]&0&0&aI_1&0&0&0&0 \end{array}\)
\(\text{the convolution product in this case is the product of each column.}\\ \text{Now we slide }f\text{ three places:}\)
\(\begin{array}{c| c c} m&-1&0&1&2&3&4&5\\\hline h&0&0&I_1&I_2&I_3&I_4&I_5\\ f&0&0&0&c&b&a&0\\\hline f*h[3]&0&0&0&cI_2&bI_3&aI_4&0 \end{array}\)
\(\text{and we have the same solution.}\)
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