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I have the input signal $x(t)$ enter image description here

And impulse response $h(t)=20 e^{-1000t} u(t)$ in which u(t) is the unit step function.

When I try a convolution, I thought the solutions would be something like:

$ \begin{array}{ll} \int\limits_{0}^{t} 2\cdot 20e^{-1000(t-\tau)}d\tau = \frac{1}{25}(1-e^{-1000t}),& 0 \le t < 2 \\ \int\limits_{0}^{2} 2\cdot 20e^{-1000(t-\tau)}d\tau - \int\limits_{2}^{t} 20e^{-1000(t-\tau)}d\tau = \frac{3}{50}e^{2000-1000t}-\frac{e^{-1000t}}{25}-\frac{1}{50},& 2\le t < 3 \\ \int\limits_{0}^{2} 2\cdot 20e^{-1000(t-\tau)}d\tau - \int\limits_{2}^{3} 20e^{-1000(t-\tau)}d\tau = \frac{3}{50}e^{2000-1000t}-\frac{1}{50}e^{3000-1000t}-\frac{e^{-1000t}}{25},& t \geq 3 \\ \end{array} $

But this did not look like the answer in the solution manual, so did I do something wrong here?

Answer in solution manual: enter image description here

Alternative solution in solution manual: enter image description here

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  • $\begingroup$ Hints: $\frac{2}{50}=\frac{1}{25}$ and so $\displaystyle\frac{2}{50}(1-e^{-1000t})=\frac{1}{25}(1-e^{-1000t})$ as you obtained. $$\frac{2}{50}(1-e^{-1000t})-\frac{3}{50}(1-e^{-1000(t-2)})=-\frac{1}{50}-\frac{1}{25}e^{-1000t}+\frac{3}{50}e^{2000-1000t}$$ which is what you obtained by simply expanding out the expression. $\endgroup$ Jan 11 '15 at 14:27
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You just multiplied the two functions and integrated them but you didn't convolve them. You must compute

$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau=2\int_0^2h(t-\tau)d\tau-\int_2^3h(t-\tau)d\tau$$

Alternatively, you can compute the step response

$$a(t)=\int_{-\infty}^th(\tau)d\tau$$

Because $x(t)=2u(t)-3u(t-2)+u(t-3)$ the output can be written in terms of the step response:

$$y(t)=2a(t)-3a(t-2)+a(t-3)$$

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  • $\begingroup$ How about the edited version? $\endgroup$
    – Hadrudsp
    Jan 11 '15 at 7:06
  • $\begingroup$ @Hadrudsp: Now it's the same now as in the solution manual, but I guess you can see that yourself. If my answer helped, you can accept it by clicking the check mark to its left. $\endgroup$
    – Matt L.
    Jan 11 '15 at 8:28

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