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:

enter image description here

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);

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:

enter image description here

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

enter image description here

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 $\tau =10$.

Can somebody please help me solving this problem?

Thanks in advance.

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

enter image description here

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

enter image description here

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.

  • 1
    $\begingroup$ 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? $\endgroup$
    – TimWescott
    Commented Nov 19, 2019 at 19:17
  • $\begingroup$ 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. $\endgroup$
    – J.D.
    Commented Nov 20, 2019 at 7:37
  • $\begingroup$ @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. $\endgroup$
    – J.D.
    Commented Nov 20, 2019 at 7:48
  • $\begingroup$ 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). $\endgroup$
    – fibonatic
    Commented Nov 20, 2019 at 12:10
  • $\begingroup$ 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. $\endgroup$
    – fibonatic
    Commented Nov 21, 2019 at 9:05

1 Answer 1


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.

  • $\begingroup$ Thanks for the answer. But the fact that gives me doubts is the fact that tau=10 should be worse with reapect to performances than tau = 1. And moreover, if I choose values of tau smaller than 1 I get what I expected from the step response. I have edited the question with this. Thanks. $\endgroup$
    – J.D.
    Commented Nov 21, 2019 at 7:37
  • $\begingroup$ @J.D. my second paragraph of my answer already tries to explain that difference. Though, it will probably also depend on the frequency content of reference. For example the impulse and ramp response contain more and less higher frequencies than the step response respectively. $\endgroup$
    – fibonatic
    Commented Nov 21, 2019 at 9:20

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