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Obtain the response of the following models with zero initial conditions. Note: Treat the unit step input us(t) as a constant value of 1. Zero initial conditions means x(0) = X(0) = 0 a. č + 4x + 8x = 2us(t) b. * + 8x + 12x = 2us(t) C. ï + 4x + 4x = 2us(t)

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a. The characteristic equation of the system is s^3 + 4s^2 + 8s = 0, which has roots at s = 0 and s = -2 ± 2i. Since the system is stable (all poles have negative real parts), we can use partial fraction expansion to obtain the response:

X(s) = (1/2s) - (1/(s+2+2i)) + (1/(s+2-2i))

Taking the inverse Laplace transform, we get:

x(t) = (1/2) - e^(-2t)cos(2t) - (1/2)e^(-2t)sin(2t)

b. The characteristic equation of the system is s^3 + 8s^2 + 12s = 0, which has roots at s = 0 and s = -4 ± 2√3i. Since the system is stable, we can use partial fraction expansion to obtain the response:

X(s) = (1/2s) - (1/(s+4+2√3i)) + (1/(s+4-2√3i))

Taking the inverse Laplace transform, we get:

x(t) = (1/2) - e^(-4t)cos(2√3t) - (1/2)e^(-4t)sin(2√3t)

c. The characteristic equation of the system is s^3 + 4s^2 + 4s = 0, which has a double root at s = -2 and a simple root at s = 0. Since the system is marginally stable (one pole has zero real part), we need to use a different approach to obtain the response. The transfer function of the system is:

G(s) = (2/(s+2)^2)

Using the final value theorem, we can find the steady-state response to a unit step input:

x_ss = lim(s->0) sX(s)G(s) = lim(s->0) (2s/(s+2)^2) = 0

Therefore, the response of the system is:

x(t) = x_ss + x_p(t) = 0 + 2tu(t) = 2t, for t >= 0.
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User Mraliks
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