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I got my Y(t) to be : $$12 \, e^{-4} \, e^{-2s} \, [\frac{1}{12(s+2)} + \frac{1}{4(s-2)} - \frac{1}{3(s-1)}] + \frac{1}{(s-2)} - \frac{1}{(s-1)}.$$

so i assume I need to use t shifting for the laplace inverse/transformation (not sure of the term to use), but i'm not sure what to do with the first terms. Could someone help me here.

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I don't agree with your $Y(s)$. Here comes a suggestion on how to get $Y(s)$ and on how to get your function back in the end:

Taking Laplace transform of the left-hand side, invoking the initial value conditions, gives $$ (s^2-3s+2)Y(s)-1. $$ The right-hand side can be written $$ 12\theta(t-3)e^{-2t}. $$ The Laplace transform of $12\theta(t)$ is $12/s$, and so the Laplace transform of $12\theta(t-3)$ is $12e^{-3s}/s$. Using the shift rule, we find that the Laplace transform of the right-hand side is $$ \frac{12e^{-3(s+2)}}{s+2}. $$ Thus, your equation for $Y(s)$ should be (since $s^2-3s+2=(s-1)(s-2)$) $$ (s-1)(s-2)Y(s)-1=\frac{12e^{-3(s+2)}}{s+2}. $$ Thus, $$ Y(s)=\frac{1}{(s-1)(s-2)}+\frac{12e^{-3(s+2)}}{(s-1)(s-2)(s+2)} $$ If I understood you correct, the first part of this expression is not a problem, so let us concentrate on how to calculate the inverse Laplace transform of the second one. We first do a partial fraction decomposition, to write $$ \frac{12}{(s-1)(s-2)(s+2)}=-\frac{4}{s-1}+\frac{3}{s-2}+\frac{1}{s+2}. $$ I show you how to calculate the inverse Laplace transform of $$ -\frac{4e^{-3(s+2)}}{s-1}, $$ and leave the other terms to you. This is like working with the right-hand side, but backwards. We write $$ -\frac{4e^{-3(s+2)}}{s-1}=-4e^{-9}\frac{e^{-3(s-1)}}{s-1}. $$ This is the $1$-shift of $$ -4e^{-9}\frac{e^{-3s}}{s} $$ which should give a $e^t$ factor in the end. Now, the term $e^{-3s}$ implies that we should have the inverse Laplace transform of $$ -4e^{-9}\frac{1}{s} $$ shifted $3$ steps. But this we recognize as the Laplace transform of $\theta(t)$. Hence, we get $$ -4e^{-9}\theta(t-3)e^t, $$ or, maybe a bit cleaner $$ -4e^{t-9}\theta(t-3). $$ You do the same with the other terms.

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  • $\begingroup$ I never got a laplace inverse that had two variables, or is θ(t-3) considered a function. $\endgroup$ Commented Sep 2, 2015 at 5:39
  • $\begingroup$ $\theta(t)$ is the Heaviside step function. It is zero for $t<0$ and one for $t>0$. I should have mentioned that in the solution. $\endgroup$
    – mickep
    Commented Sep 2, 2015 at 5:42
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I would suggest doing it one time interval at a time to keep your work neat. It gets quite messy with all the partial fractions sometimes.

Start with the interval $0<t<3$, with $y(0)=0$ and $y'(0)=1$ as initial conditions.

$\frac{d^2y}{dt^2}-3\frac{dy}{dt}+2y=0$

$(s^2Y-sy(0)-y'(0))-3(sY-y(0))+2Y=0$

$Y={1 \over s^2-3s+2}=\frac{1}{s-2}-\frac{1}{s-1}$

$y(t)=e^{2t}-e^t \ \ \ | \ \ 0<t<3$

Now do the part for $t>3$. Initial conditions are $y(3)=e^6-e^3=c_1$ and $y'(3)=2e^6-e^3=c_2$. I used $c_1$ and $c_2$ to make the work neater.

$(s^2Y-sc_1-c_2)-3(sY-c_1)+2Y=\frac{12}{s+2}$

$(s-1)(s-2)Y=\frac{12}{s+2}+(c_1s+c_2+3c_1)$

$Y=\frac{12}{(s+2)(s-1)(s-2)}+\frac{c_1s+c_2+3c_1}{(s-1)(s-2)}$

$Y=\frac{1}{s+2}+\frac{3}{s-2}-\frac{4}{s-1}+\frac{-4c_1-c_2}{s-1}+\frac{5c_1+c_2}{s-2}$

$Y=\frac{1}{s+2}+\frac{3+5c_1+c_2}{s-2}-\frac{4-4c_1-c_2}{s-1}$

$y(t)=e^{-2(t-3)}+(3-6e^3+7e^6)e^{(t-3)}+(4+5e^3-6e^6)e^{2(t-3)} \ \ \ | \ \ t>3$

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  • $\begingroup$ just a quick question, I understand the rest, but where did the IC y(3) and y'(3) come from. $\endgroup$ Commented Sep 2, 2015 at 5:37
  • $\begingroup$ $y(3)$ is simply from substituting $t=3$ into the equation $y(t)=e^{2t}-e^t$, which is the value of the signal right at the beginning of the $12e^{-2t}$ input. The $y'(3)$ is obtained by differentiating the above function and plugging in 3. $\endgroup$
    – Ud779
    Commented Sep 2, 2015 at 13:40

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