3.6 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 e-2s Y1(s)= (s +2)² +2 (s +2)³ 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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3.6 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
-25
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
-s? – s+1
s(s² + 3s + 2)
Y2(s)=
239
232
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
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).
Answer: (a) yı(t) = sin(t – 2)u(t – 2) + e-2ª u(1) +1?e-2'u(t);
dy2(0)/dt = 2; (c) transient y, (t) = [-(1/5)e¬ cos(2t) – (1/10)e¬' sin(2t)]u(t).
(b) y2(0) = –1
and
Transcribed Image Text:3.6 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 -25 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 -s? – s+1 s(s² + 3s + 2) Y2(s)= 239 232 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 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). Answer: (a) yı(t) = sin(t – 2)u(t – 2) + e-2ª u(1) +1?e-2'u(t); dy2(0)/dt = 2; (c) transient y, (t) = [-(1/5)e¬ cos(2t) – (1/10)e¬' sin(2t)]u(t). (b) y2(0) = –1 and
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