Consider the system described by the following second-order differential equation. dy(t) dt d²y(t) dt² — 6y(t) = x(t) Suppose that the following input signal x(t) was fed into the system. Suppose further that at t = 0, y(t) = 4 and y' (t) = 0.5. + t>0 x(t) = {4€ ²,20} t<0 The solution of the forced response is Yp(t) = Ce-2t, where C = The solution of the total response is y(t) = K₁e³¹t + K₂e³2t + Ce-2¹, where K1 = and K2 = Note: Assume K₁ ≤ K₂

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Consider the system described by the following second-order differential equation.
dy(t)
dt
d²y(t)
dt²
– 6y(t) = x(t)
Suppose that the following input signal ä(t) was fed into the system.
Suppose further that at t = 0, y(t) = 4 and y' (t) = 0.5.
+
x(t) = {4€ ²,20}
4e
t<0
The solution of the forced response is
Yp(t) = Ce-2t, where C =
The solution of the total response is
y(t) = K₁e³¹t + K₂e2t + Ce-2t,
where K1 =
and K2 =
Note: Assume K₁ ≤ K₂
1
Transcribed Image Text:Consider the system described by the following second-order differential equation. dy(t) dt d²y(t) dt² – 6y(t) = x(t) Suppose that the following input signal ä(t) was fed into the system. Suppose further that at t = 0, y(t) = 4 and y' (t) = 0.5. + x(t) = {4€ ²,20} 4e t<0 The solution of the forced response is Yp(t) = Ce-2t, where C = The solution of the total response is y(t) = K₁e³¹t + K₂e2t + Ce-2t, where K1 = and K2 = Note: Assume K₁ ≤ K₂ 1
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