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I have two equations $$x(t) = u(t) - 2u(t-2) + u(t-5)$$ $$h(t) = e^{2t}u(1-t)$$ where $u(t)$ is the unit step function.

I'm attempting to find the convolution of the two: $$y(t) = h(t)*x(t)$$

I decided that I would do the convolution of just u(t) and sum the transformations of that to get the total convolution. For this I get $$y(t) = \int^{\infty}_{-\infty}{u(\tau)e^{2(t-\tau)}u(1-t+\tau)d\tau}$$

I can break this into a piecewise function and remove u(t): $$y(t) = \cases{0 &\text{ if } t\lt 0\cr \int^{\infty}_{0}{e^{2(t-\tau)}u(1-t+\tau)d\tau} &\text{ if } t\ge 0}$$

It's at this point that I feel more on shaky ground. Looking at the remaining step function, I would say that $u(\tau)$ dominates when $t \ge 1$ and $u(t-1+\tau)$ dominates for $t \lt 1$

Thus I get a new expression... $$y(t) = \cases{\int^{\infty}_{0}{e^{2(t-\tau)d\tau}} &\text{ if } t\ge 1 \cr \int^{\infty}_{t-1}{e^{2(t-\tau)d\tau}} &\text{ if }0 \lt t\lt 1 \cr 0 &\text{ if } t\le 0}$$

From there I can do indefinite integration on each integral for a solution.

But I'm not sure that I developed the second piecewise function correctly...

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1 Answer 1

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$u(1-t+\tau) > 0$ for all $1-t + \tau \gt 0 \implies \tau \gt t-1$, but we also need $\tau \gt 0$.

Thus for $t \ge 1$ we have $\int_{t-1}^{\infty} \mathbb{e}^{2(t-\tau)} \,\mathbb d \tau$ and for $0 \lt t \lt 1$ we have $\int_{0}^{\infty} \mathbb{e}^{2(t-\tau)}\,\mathbb d \tau$

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  • $\begingroup$ that makes sense. The 0 for t<= 0 piece is correct though, right? $\endgroup$
    – Daniel B.
    Commented Sep 18, 2014 at 1:26
  • $\begingroup$ Actually, no. For $t \le 0$ we have $\tau \gt - 1$ as the region where $u(1-t+\tau) > 0$. But, again, we want $\tau \gt 0$. $\endgroup$
    – user164587
    Commented Sep 18, 2014 at 2:02
  • $\begingroup$ I see. Thank you, you've been a big help. Hopefully this'll stick, it's been a hard problem ^_^ $\endgroup$
    – Daniel B.
    Commented Sep 18, 2014 at 2:04

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