Electronic – Closed Loop Transfer Function – PD Control

The integrator in the PI controller guarantees zero steady state error - you can check this by letting s=0 in the closed loop transfer function, G(s), which gives G(0)=1. If you haven't been given any constraint on the response speed then the simplistic solution is to let Kp=0 and the transfer function reduces to the standard 2nd order form and then you can design Ki to give a damping coefficient, zeta=1

With your added constraint of minimising rise time, it's perhaps instructive to look at the root locus (I assume you've covered this in your course). The open loop TF can be reduced to

$$G_o(s) = \frac{K(s+z)}{s(s+a)}$$

where K=bKp. and z=Ki/Kp. We can now design for zero overshoot and examine the speed of response.

The open loop poles are at s=0 and s=-a, and the open loop zero is at s=-z. So let's place the zero between the two OL poles, this will guarantee that the root locus lies entirely on the negative real axis and therefore there will be no oscillatory closed loop poles. The dominant CL pole (ie the one that determines overall speed of response) is the one whose locus tracks from the pole at s=0 to the zero at s=-z. As this CL pole moves to the left, the overall response gets faster. The other pole is more distant, moving further to the left as K increases, hence it's a lot faster but its impact on the overall response reduces as it moves further away.

So as K increases, the response speed increases; but remember that for a physically realisable system, K is limited by practical considerations, e.g. current limit for motors, which serves to constrain acceleration. That is, the price you pay for improved performance is increased control effort, and that equates to more bucks!

Try doing the analysis with z=1 and a=2. Matlab might be useful.

Ignoring the semantics of "root locus," you can certainly plot the roots of a polynomial as its coefficients change. The following script plots the set of poles of your system as \$RL\$ varies from 0 to 10000 with \$IL\$ fixed at 1:

IL = 1;
RLvec = linspace(0,10e3,1e5);
rts = zeros(1e5,4);
for k = 1:length(RLvec);
  RL = RLvec(k);
  D = [8*IL, 28*IL+8*RL+3, 4*(7*IL+7*RL+3), 4*(2*IL+7*RL+3), 8*RL+4];
  r = roots(D);
  rts(k,:) = r';
end;
figure
hold on;
for b = 1:4;
  plot(real(rts(:,b)),imag(rts(:,b)));
end;
xlim([-5,0]); grid on;

It executes for me in under 5 seconds.

Admittedly, it does not show how the roots move as both parameters change, but that task is complicated by having a two-dimensional domain and a two-(real)-dimensional range.