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

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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*}$

  • $\begingroup$ shouldn't that be -dx $\endgroup$ – Drew U Oct 10 '16 at 7:57
  • $\begingroup$ I was really over complicating things $\endgroup$ – Drew U Oct 10 '16 at 7:57
  • $\begingroup$ I switched the lower and upper bound at the first step, so cancelled the minus sign. $\endgroup$ – m-agag2016 Oct 10 '16 at 7:59

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