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Sagot :
Solution :
Given :
[tex]$H(S) =\frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
Transfer function, [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
[tex]$Y(S) \left(S^2+\frac{10}{3}S+1\right) = (2S-2) \times (S)$[/tex]
[tex]$S^2Y(S) + \frac{10}{3}(SY(S)) + Y(S) = 2(S \times (S)) - 2 \times (S)$[/tex]
Apply Inverse Laplace Transforms,
[tex]$\frac{d^2y(t)}{dt^2} + \frac{10}{3} \frac{dy(t)}{dt} + y(t)=2 \frac{dx(t)}{dt} - 2x(t)$[/tex]
The above equation represents the differential equation of transfer function.
Given : [tex]$x(t)=e^{-t} u(t) \Rightarrow X(S) = \frac{1}{S+1}$[/tex]
We have : [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
[tex]$Y(S) = X(S) \times \frac{6S-6}{3S^2+10 S + 3} = \frac{6S-6}{(S+1)(3S+1)(S+3)}$[/tex]
[tex]$Y(S) = \frac{A}{S+1}+\frac{B}{3S+1} + \frac{C}{S+3}[/tex]
[tex]$A = Lt_{S \to -1} (S+1)Y(S)=\frac{6S-6}{(3S+1)(S+3)} = \frac{-6-6}{(-3+1)(-1+3)} = 3$[/tex]
[tex]$B = Lt_{S \to -1/3} (3S+1)Y(S)=\frac{6S-6}{(S+1)(S+3)} = \frac{-6/3-6}{(1/3+1)(-1/3+3)} = \frac{-9}{2}$[/tex]
[tex]$C = Lt_{S \to -3} (S+3)Y(S)=\frac{6S-6}{(S+1)(3S+1)} = \frac{-18-6}{(-3+1)(-9+1)} = \frac{-3}{2}$[/tex]
So,
[tex]$Y(S) = \frac{3}{S+1} - \frac{9/2}{3S+1} - \frac{3/2}{S+3}$[/tex]
[tex]$=\frac{3}{S+1} - \frac{3/2}{S+1/3} - \frac{3/2}{S+3}$[/tex]
Applying Inverse Laplace Transform,
[tex]$y(t) = 3e^{-t}u(t)-\frac{3}{2}e^{-t/3}u(t) - \frac{3}{2}e^{-3t} u(t)$[/tex]
[tex]$=\frac{-3}{2}e^{-\frac{1}{3}t}u(t) + \frac{3}{1}e^{-t}u(t)-\frac{3}{2}e^{-3t} u(t)$[/tex]
where, a = 2
b = 1
c= 2
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