# System response with initial condition

I have a small problem figuring out how to properly solve that. I could use help.

Say a system with a transfer function $H(s) = \frac{100}{0.4s + 1}$ is excited with an excitation $u(t) = 8V$ since very long (steady state). At time $t_1$, $u(t)$ is abruptly set to 0V. What is the value of y(t) at time $t_1$ + 0.1?

I calculate steady state value (before time $t_1$) using laplace final value theorem: $$U_0(s) = 8V \\ Y_0(s) = H(s) \cdot U_0(s) = \frac{800}{s(0.4s + 1)} \\ y_0 = \lim_{s \to 0} sY_0(s) = 800$$

I have a first order linear system, so the form is (including initial conditions): $$Y(s) = H(s)U(s) + C(s) = \frac{K}{\tau s + 1}U(s) + \frac{\tau y_0}{\tau s + 1}$$

Since at time $t_1$, u(t) goes from 8V to 0V abruptly, I conclude the excitation at this instant is a negative step of 8V, i.e. $U_1(s) = -\frac{8}{s}$. So I guess I have to calculate the response at time 0.1 of a negative step of -8V with initial condition $y_0 = 800V$ to obtain the correct answer.

So the general response $Y_1(s)$ at $t = t_1$ will be: $$Y_1(s) = H(s)U_1(s) + C(s) = \frac{-800}{s(0.4s + 1)} + \frac{320}{0.4s + 1}$$

Doing the inverse Laplace of this expression I get: $$y_1(t) = 1600 e^{- 2.5t} - 800$$

Knowing this, finding y(t) at $t = t_1 + 0.1$ would correspond to $y_1(0.1)$, which gives 446.08.

Well the answer should be 623.04. Could someone explain me what I'm doing wrong? I feel I did everything correctly.

The transfer function corresponds to a first order system, you can either assume that it's an RL or RC network with input voltage.

For unit step input of 8v,

Y(s) = 800/s.(0.4s+1) = 250 /s.(s+2.5).

By partial fractions and taking inverse laplace,

y(t) = 800 (1 - exp(-2.5.t) ).

At steady steady state y(t) =800 V.

Now, this can thought as an RC circuit with 800 volts input, so that at steady state the capacitor is charged to 800 volts.

Once at t = to, say you short the source ( Vin = 0 ),the capacitor discharges through the resistor R .

We already know from equation of y(t) that the time constant (RC) = 1/2.5 = 0.4.

So,the capacitor discharges as follows:

y(t) = 800 exp(-2.5t).

At t = 0.1 (after source is shorted);

y(0.1) = 800 x 0.7788 = 623.04 volts.