Writing convolution in other terms

Question

Could I get some help with this question. I just started convolution and I'm still not clear on a lot of things

Try a change of variable of some form. I can't say more than this, it's a one liner. — Sarvesh Ravichandran Iyer, Oct 10 '16 at 07:46

score 2 · Accepted Answer · answered Oct 10 '16 at 07:52

2

Use the change of variables, let $x=-\tau-b$, then we have $\tau=-x-b$ and $dx=-d\tau$. Hence

$ \begin{eqnarray*} \nu(t)&=&\int_{-\infty}^\infty\chi(x) h(-x-b+at)\ dx\\ &=&y(at-b). \end{eqnarray*}$

answered Oct 10 '16 at 07:52

m-agag2016

shouldn't that be -dx – Drew U Oct 10 '16 at 07:57
I was really over complicating things – Drew U Oct 10 '16 at 07:57
I switched the lower and upper bound at the first step, so cancelled the minus sign. – m-agag2016 Oct 10 '16 at 07:59

1 Answers1