The dynamic behavior of a system is described by the following ODE. x(t) = -lamb

Question

The dynamic behavior of a system is described by the following ODE. x(t) = -lambda x(t) + u(t) x(0)=0 Find T(s) = x(s)/u(s) Find x(t) for u(t) = lambda U_s(t). Apply the final value theorem (FVT) to find the steady-state value of x(t) for u(t) = lambda U_s(t). Does this agree with what you found in part b? Find x(t) for u(t) = sin(omega t). For a = 0.1/sec. and omega = 0.0796 Hz., plot x(t) versus t. 0 lessthanorequalto t lessthanorequalto 60 sec. and delta t for plotting = 0.1 sec. [Note that 1 Hz. = 2 pi rad./sec. and phi = tan^-1(omega/lambda)]. You are encouraged to use Excel. Plot x(t) and u(t) for part d on the same plot. The time limits and delta t are the same as those in d. For X = 0.1/sec. and omega = 0.0796 Hz., what is the maximum steady-state value of x(t) and what is phi? Also find |T(omega) and phi(omega) using your result from part a. How do the quantities compare?

Explanation / Answer

solution:

lambda=n

1) here applyinfg laplace transform to ODE we get

SX(s)+(n)*X(s)-u(s)=0

here T(s)=X(s)/u(s)=(1/(S+n))

2) for u(t)=n*Us(t)

we get

Sx(s)+nx(s)-nu(s)=0

hence we get

X(s)/u(s)=(n/(s+n))

3) here on takinf final value thereom which state that

X(infinity)=Sx(s) at s=0

hence on putting value in above function we get

at s=0

T(s)=1/n

4) where for solution we have to take inverse laplace so we get

x(t)=laplace^-1((1/n)*(1/s+n))u(s) for first case

on solving we get

x(t)=(1/n)exp^(-nt)u(t)   for first case

and x(t)=(n/n)exp^(-nt)u(t)=exp^(-nt)u(t) for second case

4) for u(t)=sinwt

we have w=.0796 hz hence in rad/s as

w=.50014 rad/s

hence for first case solution we get

x(t)=(1/n)exp^(-nt)u(t)   for first case

x(t)=(1/n)exp^(-nt)sinwt

on putting value we get

x(t)=(10)exp^(-0.1*t)sin.5t

6) in this way for third case x(t) and u(t) both plot are sinusoidal

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