The response of a system is for t greaterthanorequalto 0 yt

The response of a system is for t greaterthanorequalto 0 y(t) = y(0)e^-t + integral_0^t e^-(t - tau)x (tau) d tau and zero otherwise. Consider y(0) = 0, is the system linear? If y(0) neq 0, is the system linear? Explain. If x(t) = 0, what is the response of the system called? If y(0) = 0. what is the response of the system to any input x(t) called? Let y(0) = 0, find the response due to delta(t). What is this response called? When y(0) = 0 and x(t) = u(t). the corresponding output of the system Is called unit step response s(t). Find s(t) and calculate ds(t)/dt. what does this correspond to from the above results.

Solution

(a) If y(0) = 0, then the first term of the right hand side becomes 0 and the second term can be represented linearly using laplace transform and convolution theorem. But if y(0) is nonzero, then first term will result into a nonlinear term and therefore, system will be nonlinear.

(b) if x(t) = 0, then system becomes y(t) = y(0)e^(-t). This is called as Natural Response. If y(0) = 0, then response is called Linear Time Invariant Response.

(c) If y(0) = 0. the system becomes y(t) = e^(-t)*x(t)

On taking laplace transform:

Y(s) = L{e^(-t)}L{x(t)}

Y(s) = 1/(s+1)L{delta(t)} = 1/(s+1)(1) = 1/(s+1)

On taking inverse laplace transform, we get;

y(t) = e^(-t)

This is known as impluse response.

(d)

When x(t) = u(t)

X(s) = 1/s

Therefore, Y(s) = (1/(s+1))(1/s) = 1/s - 1/(s+1)

On taking inverse laplace, we get:

s(t) = 1 - e^(-t)

ds/dt = d/dt(1) - d/dt(e^(-t)) = 0 - (-e^(-t)) = e^(-t)

ds/dt = e^(-t)