Let S1 and S2 be 2 LTI systems with responses to the unit impulse given by \$g_1(t)\$ and \$g_2(t)\$.Consider that we associate S1 and S2 .Knowing that \$g_1(t)=e^{-t}u(t)\$ (u is the heavside function) and \$g_2(t)=e^{-\alpha t}u(t), \alpha >0 \$ and real, determine the response to the unit impulse of the association of these systems. Use the Laplace transform and its properties.

My doubt: I don't understand why to use Laplace transform properties. Isn't the answer the convolution between \$g_1\$ and \$g_2\$? This should be 0 for \$t \leq 0 \$ and \$ \int_0^t e^{t-\tau} e^{\alpha \tau} d\tau \$ for \$t > 0\$

What am I doing wrong?



1 Answer 1


Assuming that with "associate" you mean "put in series" that is you hook S1 output to S2 input then you are not doing anything wrong. Problem is that sometimes (plenty of times) solving a convolution integral is not a piece of cake thus you prefere the Laplace transform.

Let's start by computing \$G_1(s)\$ and \$G_2(s)\$:

\$g_2(t)=e^{-\alpha t}u(t)\$, but \$U(s)=\mathcal{L}\{u(t)\}=\frac{1}{s}\$ and you can use the frequency shifting property that goes: $$ e^{at}f(t) \leftrightarrow F(s-a) $$ finally: $$ G_2(s)= \frac{1}{s+\alpha} $$

That's called exponential decay and is usually provided in Laplace transforms tables. Of course \$G_1(s)=G_2(s)|_{\alpha=1}\$ so that: $$ G_1(s) = \frac{1}{s+1} $$ Now we're searching for $$G(s)\triangleq G_1(s)G_2(s)=\frac{1}{s+1}\frac{1}{s+\alpha}=\frac{A}{s+1}+\frac{B}{s+\alpha}=\frac{s(A+B)+A\alpha+B}{(s+1)(s+\alpha)}$$ From the last step you can find out \$A=-B=\frac{1}{\alpha-1}\$, now we're at: $$ G(s) = \frac{1}{\alpha-1}\left(\frac{1}{s+1}-\frac{1}{s+\alpha}\right) $$

The anti transform is quite simple now since both terms in parenthesis are known transforms:

$$ g(t) = u(t)\frac{1}{\alpha-1}\left(e^{-t}-e^{-\alpha t}\right) $$

Could you get this result from your integral? Probably. Did you? Nope, and that's because integrals are so boring. Plus your book/teacher probably wanted you to use the whole A and B thing, also known as partial fraction decomposition.

The Laplace transform is a great weapon, learn how to use it and you will be rewarded.

  • 1
    \$\begingroup\$ The A and B thing is partial fraction. \$\endgroup\$
    – rioraxe
    Commented Mar 15, 2015 at 20:31

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