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i was wondering, for convolution, when we do the graph shifting for h(t-tou) we flip the graph on the y axis and then if t = 0.5, then shouldn't we shift the graph left by 1/2? In the examples I am seeing, they shift it to the right by 1/2.... why do they do this? don't you shift to the right only if it was -t?

thanks

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If your convolution integral is $$y(0.5) = \int_{-\infty}^{\infty} x(\tau)h(0.5-\tau)\mathrm dt,$$ then, as a function of $\tau$, $x(0)$ is being multiplied by $h(0.5)$ and $x(0.5)$ by $h(0)$, right? So, after "flipping the graph over" (which keeps $h(0)$ where it is at $0$), which way do you need to shift the flipped graph to make sure that $h(0)$ moves over to be in the same place as $x(0.5)$? – Dilip Sarwate Oct 1 '12 at 18:46

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