# Problem with performances of a control scheme

I am trying to prove with Matlab that if I have an improper system and I place poles at higher and higher frequencies the performances of the system improves. In particular I am considering the following two degree of freedom scheme:

where $$C_{f}= \frac{(1+s)(1+0.05s)^{2}]}{(1+\tau s)^{3}}$$

My code is the following:

s = tf('s');
P = 1/[(1+s)*(1+0.05*s)^2];
C = (s+1)/s;

tau_1 = 0.1;
CF_1 = [(1+s)*(1+0.05*s)^2]/((1+tau_1*s)^3);

tau_2 = 0.01;
CF_2 = [(1+s)*(1+0.05*s)^2]/((1+tau_2*s)^3);

tau_3 = 0.001;
CF_3 = [(1+s)*(1+0.05*s)^2]/((1+tau_3*s)^3);

T1 = (C+CF_1)*P/(1+P*C);
T2 = (C+CF_2)*P/(1+P*C);
T3 = (C+CF_3)*P/(1+P*C);

figure;
bodemag(T1,'r',T2,'b',T3,'g'),grid
legend('tau = 0.1','tau = 0.01','tau = 0.001')

so what I expected is that the performances with respect to the reference tracking increse as tau gets smaller, but if I do the Bode plot, what I get is:

from which I don't really see much of an improvement. Moreover if I change some values:

which is somenthing that to me does not makes sense because I should have that with $$\tau =1$$ I should have better performances than with $$\tau =10$$, this because with $$\tau =1$$ the pole is at higher frequencies that with $$\tau =10$$.

[EDIT] If I plot the step responses I see the same problem:

[EDIT 2]For completeness, I post the image of the step response for the first choise of tau's:

for these values of $$\tau$$ there is a clear improvement for overshoot and a faster response.

I also tried other values smaller than 1 for $$\tau$$ and all of these show what I expected in the step response. While for values bigger than 1, I obtain something similar at the other situation.

Does somenone know why this happens? Thanks.

[EDIT 3] With the last one, so values of $$\tau$$ smaller than 1, I have also noticed an increase in phase margin, so this should mean that the system is performing better.

While, if I consider the values $$\tau=1$$ $$\tau=10$$ i get that for the first one the phase margin is 125 deg and for the second is 170 deg. So this should be in according to the fact that $$\tau=10$$ performs better.

• Your Bode plots are of the final system transfer function, and generally a closed-loop Bode plot that stays at 0dB farther out in frequency is better. So your performance is increasing with decreasing $\tau$. How are you reading your plots? Nov 19 '19 at 19:17
• Thanks for your answers. I was looking for a lower peak of resonance in the complementary sensitivity function and less overshoot in the step response to see the increase of performances. I think the point that I am missing is why if the bandwidth is at higher frequencies the performances are better. Thanks.
– J.D.
Nov 20 '19 at 7:37
• @fibonatic this is what comes out if I do manually the computations of the input-output transfer function. Do you think it is wrong? Thank you in advance.
– J.D.
Nov 20 '19 at 7:48
• Your equations are correct, I jumped to conclusion when I saw the sum of the two controllers in the numerator (but they should also be summed in the denominator in order to have what I initially thought). Nov 20 '19 at 12:10
• Phas margin is a property of the closedloop, but $\tau$ only affects the feedforward transfer function and thus the feedback loop should remain unaltered so the phase margin should remain unaltered also. Nov 21 '19 at 9:05

Your closedloop crossover frequency (when the magnitude of $$P(s)\,C(s)$$ is equal to one) lies at roughly 1 rad/s. This means that the feedback controller already causes the system to track reference signals that have a frequency content sufficiently below that. So adding feedforward that only approximates the inverse of the plant ($$C_f(s)\,P(s)\approx1$$) below the crossover frequency should not aid in the tracking performance, so that is why for both $$\tau=1$$ and $$\tau=10$$ the magnitude of the overall transfer functions both start to drop off around 1 rad/s, similar to if you wouldn't use any feedforward ($$C_f(s)=0$$).
However, for $$\tau=1$$ the magnitude of $$C_f(s)\,P(s)$$ near 1 rad/s is still close to 0 dB, but the phase at 1 rad/s has already dropped to -135°, so the actual feedforward is actually doing partially the opposite of what the ideal feedforward would do and thus steering away from better reference tracking. For $$\tau=10$$ the magnitude of $$C_f(s)\,P(s)$$ near 1 rad/s is already close to -60 dB. So, even though the phase is already way lower compared when using $$\tau=1$$, the magnitude is so low that the feedback controller can counteract the little disturbance the feedforward term is causing near that frequency. So that is why $$\tau=10$$ performs better (less overshoot) than $$\tau=1$$. Though, it can be noted that for $$\tau=10$$ there is a 0.3 dB peak near 0.06 rad/s and an overshoot of 0.03 after 20 seconds, while the system with only feedback and no feedforward has no overshoot.