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Nov 21, 2019 at 10:41 vote accept J.D.
Nov 21, 2019 at 9:05 comment added fibonatic 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, 2019 at 7:53 history edited J.D. CC BY-SA 4.0
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Nov 21, 2019 at 7:27 history edited J.D. CC BY-SA 4.0
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Nov 20, 2019 at 12:50 answer added fibonatic timeline score: 4
Nov 20, 2019 at 12:10 comment added fibonatic 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, 2019 at 7:48 comment added J.D. @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.
Nov 20, 2019 at 7:37 comment added J.D. 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.
Nov 19, 2019 at 19:17 comment added TimWescott 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, 2019 at 18:23 history edited J.D. CC BY-SA 4.0
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Nov 19, 2019 at 17:49 history edited J.D. CC BY-SA 4.0
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Nov 19, 2019 at 17:15 history undeleted J.D.
Nov 19, 2019 at 17:15 history deleted J.D. via Vote
Nov 19, 2019 at 17:13 history edited J.D.
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Nov 19, 2019 at 16:49 history asked J.D. CC BY-SA 4.0