Picture shows the electrical system. Analytically determine and draw system response $u_2(t)$ for the step input shown (use Laplace transform) (tp>>0).

So, I know to solve RC system transfer function and get G(s)=1/(RCs+1), but I don't understand how to do this problem, we have never done it in class and professor often asks this question. If anyone could help me out, thank you!

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Because this is a linear and time-invariant system, you can use the principle of superposition.

If the input to $u_i(t)$ is $y_i(t)$, then

  • The response to $\alpha u_i(t) $ is $\alpha y_i(t) $
  • The response of $u_i(t)+u_j(t)$ is $y_i(t)+y_j(t)$
  • The response of $u(t-t_p)$ is $y(t-t_p)$

The response of the system to a unit step input $\theta (t)$ is $1-e^{-\frac{t}{R C}}$. (I assume you have no problem with this.)

The input to the system can be split into two.

  • $5 \theta (t)$ when $0\leq t\leq t_p$. The response of the system to this input is $5(1-e^{-\frac{t}{R C}})$.
  • $5 \theta (t)-\theta(t-t_p)$ when $t_p < t$. The response is $5(1-e^{-\frac{t}{R C}})-(1-e^{-\frac{t-\text{tp}}{R C}})$.

The final response is a sum of these two piecewise functions.

I have plotted the result for $R=1$, $C=1$ and various values of $t_p$. The dashed line shows the response if the second step was not applied.

enter image description here

  • $\begingroup$ Thanks a lot! How would I determine gain only from system repsonse ( Transient function), I know that T is when function reaches 63% ... $\endgroup$ – DomVl Dec 13 '16 at 22:05
  • $\begingroup$ The gain comes from the steady-state response to a unit-step function. In this case it is 1. $\endgroup$ – Suba Thomas Dec 13 '16 at 22:13
  • $\begingroup$ Is that input/(where function settles +1) ? $\endgroup$ – DomVl Dec 13 '16 at 22:23
  • $\begingroup$ The input is a unit step. It is where the output settles. If you are using a unit step of magnitude m, it is ((steady-state value of output)/m). $\endgroup$ – Suba Thomas Dec 13 '16 at 22:31

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