In the following problems we use the inverse Laplace transform and the relation between input and output of LTI systems. (a) The Laplace transform of the output of a system is (s +2)² + 2 (s +2)³ e-2s Y1 (s) = s2 +1 find yı (t), assume it is causal. (b) The Laplace transform of the output y2(t) of a second-order system is -s2 – s+1 s (s2 + 3s + 2) Y2(s)=

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In the following problems we use the inverse Laplace transform and the relation between input
and output of LTI systems.
(a) The Laplace transform of the output of a system is
(s +2)² + 2.
+
-2s
Y1 (s)=
s2 +1
(s +2)3
find y1 (t), assume it is causal.
(b) The Laplace transform of the output y2(t) of a second-order system is
-s2 – s+1
s (s² + 3s + 2)'
Y2(s) =
23
32
CHAPTER 3 THE LAPLACE TRANSFORM
If the input of this system is x2(t) =u(t), find the ordinary differential equation that rep
resents the system and the corresponding initial conditions y2(0) and dy2(0)/dt.
(c) The Laplace transform of the output y(t) of a system is
%3D
1
Y (s) =
s(s +1)2 +4)*
Assume y(t) to be causal. Find the steady-state response yss (t), and the transient y; (t).
(b) y2(0) = -1
dy2(0)/dt =2; (c) transient y; (t) =[-(1/5)e¬ cos(2t) – (1/10)e¬ sin(2t)]u(t).
Answer:
(a) yı(t) = sin(t – 2)u(t – 2) +e¯2ªu(t) +t²e-2ª u (t
anc
Transcribed Image Text:In the following problems we use the inverse Laplace transform and the relation between input and output of LTI systems. (a) The Laplace transform of the output of a system is (s +2)² + 2. + -2s Y1 (s)= s2 +1 (s +2)3 find y1 (t), assume it is causal. (b) The Laplace transform of the output y2(t) of a second-order system is -s2 – s+1 s (s² + 3s + 2)' Y2(s) = 23 32 CHAPTER 3 THE LAPLACE TRANSFORM If the input of this system is x2(t) =u(t), find the ordinary differential equation that rep resents the system and the corresponding initial conditions y2(0) and dy2(0)/dt. (c) The Laplace transform of the output y(t) of a system is %3D 1 Y (s) = s(s +1)2 +4)* Assume y(t) to be causal. Find the steady-state response yss (t), and the transient y; (t). (b) y2(0) = -1 dy2(0)/dt =2; (c) transient y; (t) =[-(1/5)e¬ cos(2t) – (1/10)e¬ sin(2t)]u(t). Answer: (a) yı(t) = sin(t – 2)u(t – 2) +e¯2ªu(t) +t²e-2ª u (t anc
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