Latest revision as of 07:22, 6 May 2011

Problem

Compute the convolution

$ z(t)=x(t)*y(t) \ $

between

$ x(t) = sin(t)u(t + \pi) \ $

and

$ y(t) = cos(t)u(t-\pi) \ $.

My Solution

$ \begin{align} z(t) &= sin(t)u(t + \pi) * cos(t)u(t-\pi) \\ &= \int_{-\infty}^{\infty} sin(\tau)u(\tau + \pi)cos(t - \tau)u(t - \tau -\pi) \mathrm{d}\tau \\ &= \int_{-\pi}^{t-\pi} sin(\tau) cos(t - \tau) \mathrm{d}\tau \\ &= \int_{-\pi}^{t-\pi} \frac{e^{j\tau} - e^{-j\tau}}{2j} \frac{e^{j(t-\tau)} + e^{-j(t-\tau)}}{2} \mathrm{d}\tau \\ &= \int_{-\pi}^{t-\pi} \frac{e^{j\tau} - e^{-j\tau}}{2j} \frac{e^{j t} e^{-j\tau} + e^{-j t}e^{+j\tau}}{2} \mathrm{d}\tau \\ &= \int_{-\pi}^{t-\pi} \frac{e^{j\tau}e^{j t}e^{-j\tau}}{4j} + \frac{e^{j\tau}e^{-j t}e^{+j\tau}}{4j} - \frac{e^{-j\tau}e^{j t}e^{-j\tau}}{4j} - \frac{e^{-j\tau}e^{-j t}e^{+j\tau}}{4j}\mathrm{d}\tau \\ &= \int_{-\pi}^{t-\pi} \frac{e^{j t}}{4j} + \frac{e^{j2\tau}e^{-j t}}{4j} - \frac{e^{-j2\tau}e^{j t}}{4j} - \frac{e^{-j t}}{4j}\mathrm{d}\tau \\ &= \begin{cases} \frac{t e^{j t}}{4j} + \frac{e^{-j t}}{4j}\int_{-\pi}^{t-\pi}e^{2j\tau}\mathrm{d}\tau - \frac{e^{j t}}{4j}\int_{-\pi}^{t-\pi}e^{-2j\tau}\mathrm{d}\tau - \frac{t e^{-j t}}{4j} &, \text{when } t> 0 \\ 0 &, \text{else} \end{cases}\\ &= \begin{cases} \frac{t e^{j t}}{4j} + \frac{e^{-j t}}{4j} \left( \frac{e^{2j\tau}}{2j}\Big|_{-\pi}^{t-\pi} \right)- \frac{e^{j t}}{4j} \left( \frac{e^{-2j\tau}}{-2j}\Big|_{-\pi}^{t-\pi} \right)- \frac{t e^{-j t}}{4j} &, \text{when } t> 0 \\ 0 &, \text{else} \end{cases}\\ &= \begin{cases} \frac{t e^{j t}}{4j} -\frac{e^{jt}}{8} + \frac{e^{-jt}}{8} - \frac{e^{-jt}}{8} + \frac{e^{jt}}{8} - \frac{t e^{-j t}}{4j} &, \text{when } t> 0 \\ 0 &, \text{else} \end{cases}\\ &= \begin{cases} \frac{t e^{j t}}{4j} -\frac{t e^{-j t}}{4j} &, \text{when } t> 0 \\ 0 &, \text{else} \end{cases}\\ \end{align} $

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