Question 5 of 8: PI Controller — Laplace Transform and Output for a Pulse Train

Problem 5: PI Controller — Laplace Transform and Output for a Pulse Train (20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A staircase error signal into an ideal PI controller:

Find. (a) $E(s)$; (b) the controller output $p(t)$ and its plot.

Approach. Write $e(t)$ as a sum of shifted unit steps to get $E(s)$ by the time-shift theorem, then apply the ideal PI law $p=K_c e+(K_c/\tau_I)\int e\,dt$ interval by interval (the proportional part copies $e$; the integral part accumulates its running area).

1. (a) Decompose the error into steps. Each level change is a step of height $\pm0.5$: $$e(t)=0.5\,u(t)-0.5\,u(t-1)-0.5\,u(t-2)+0.5\,u(t-3).$$ The pattern $+,-,-,+$ builds the up-block on $[0,1)$ and the down-block on $[2,3)$.
2. (a) Transform. Using $\mathcal L\{u(t-a)\}=e^{-as}/s$, $$\boxed{E(s)=\frac{0.5}{s}\left(1-e^{-s}-e^{-2s}+e^{-3s}\right).}$$
3. (b) Controller law. The ideal PI controller is $p(t)=K_c\,e(t)+\dfrac{K_c}{\tau_I}\displaystyle\int_0^t e\,dt'$, with $K_c=2$ and $K_c/\tau_I=2/0.5=4$. The proportional term jumps with $e$; the integral term is the running area of $e$ scaled by 4.
4. (b) Interval $0\le t<1$. Here $e=0.5$ and $\int_0^t e=0.5t$, so $p=2(0.5)+4(0.5t)=\boxed{1+2t}$ — a ramp from $p(0)=1$ up to $p(1^-)=3$.
5. (b) Interval $1\le t<2$. Now $e=0$, so the proportional part drops out and the integral freezes at its accumulated value $0.5$: $p=0+4(0.5)=\boxed{2}$ (flat). The output steps down from $3$ to $2$ at $t=1$ as the proportional contribution vanishes.
6. (b) Interval $2\le t<3$. With $e=-0.5$ the running area falls, $\int_0^t e=0.5-0.5(t-2)$, giving $p=2(-0.5)+4[0.5-0.5(t-2)]=\boxed{1-2(t-2)}$ — a ramp from $p(2)=1$ down to $p(3^-)=-1$. At $t=2$ the output steps down from $2$ to $1$.
7. (b) Interval $t\ge3$. The net area of $e$ is zero (equal $+$ and $-$ blocks) and $e=0$, so $p=0$. The integral’s memory has been perfectly cancelled, and the controller returns to rest (stepping up from $-1$ to $0$ at $t=3$).