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\text{Let } \gamma &= \tau - 5 \\ | \text{Let } \gamma &= \tau - 5 \\ | ||
\mathrm{d}\gamma &= \mathrm{d}\tau \\ | \mathrm{d}\gamma &= \mathrm{d}\tau \\ | ||
| − | \text{Since } \int u(\gamma)\mathrm{d}\gamma &= \ | + | \text{Since } \int u(\gamma)\mathrm{d}\gamma &= \gamma u(\gamma) \\ |
| − | \int_a^b u(\gamma)\mathrm{d}\gamma &= [\ | + | \int_a^b u(\gamma)\mathrm{d}\gamma &= [\gamma u(\gamma)] \Big|_a^b\\ |
\text{Replace } \tau -5 &= \gamma \\ | \text{Replace } \tau -5 &= \gamma \\ | ||
\int_a^b u(\tau -5)\mathrm{d}\tau &= [(\tau - 5)u(\tau - 5)] \Big|_a^b\\ | \int_a^b u(\tau -5)\mathrm{d}\tau &= [(\tau - 5)u(\tau - 5)] \Big|_a^b\\ | ||
Latest revision as of 09:42, 30 April 2011
Problem
Compute the convolution
$ z(t)=x(t)*y(t) \ $
between
$ x(t) = u(t) - u(t-1) \ $
and
$ y(t) = u(t+2) - u(t-2) \ $.
My Solution
$ \begin{align} z(t) &= y(t) * x(t) \\ &= \int_{-\infty}^{\infty}x(\tau) y(t-\tau)\mathrm{d}\tau \\ &= \int_{-\infty}^{\infty}\left( u(\tau) - u(\tau-1) \right) \left( u(t-\tau+2) - u(t-\tau-2) \right)\mathrm{d}\tau \\ &= \int_{0}^{1}\left( u(t-\tau+2) - u(t-\tau-2) \right)\mathrm{d}\tau \\ &= \left[ (t-\tau+2)u(t-\tau+2)\right]\big|_0^1 - \left[ (t-\tau-2)u(t-\tau-2)\right]\big|_0^1 \\ &= \left[ (t+1)u(t+1) - (t+2)u(t+2)\right] + \left[ -(t-3)u(t-3) + (t-2)u(t-2)\right] \end{align} $
Proof of $ \int_a^b u(\tau -5)\mathrm{d}\tau = [(\tau - 5)u(\tau - 5)] \Big|_a^b $
$ \begin{align} \int_a^b u(\tau -5)\mathrm{d}\tau &= [(\tau - 5)u(\tau - 5)] \Big|_a^b\\ \text{Let } \gamma &= \tau - 5 \\ \mathrm{d}\gamma &= \mathrm{d}\tau \\ \text{Since } \int u(\gamma)\mathrm{d}\gamma &= \gamma u(\gamma) \\ \int_a^b u(\gamma)\mathrm{d}\gamma &= [\gamma u(\gamma)] \Big|_a^b\\ \text{Replace } \tau -5 &= \gamma \\ \int_a^b u(\tau -5)\mathrm{d}\tau &= [(\tau - 5)u(\tau - 5)] \Big|_a^b\\ \end{align} $
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