Larger phase margin means less overshoot?

Question

Why is it that the larger phase margin means the system has less overshoot? I have heard that the larger phase margin means greater damping, but I don't how they are connected to each other. Could someone give a physical example that would show their relationship, like the low pass filter I mentioned above( if my example is right )? And in many system, they require that the phase margin must be higher than 45°. How do they know the number 45° can get their requirement?

Answer 1

Let me start by saying that the step response of a system is determined only by the position of the poles and zeroes of the transfer function describing the system.

The terms: bandwidth, phase margin, crossover frequency, etc. are things we can determine when looking at the open loop Bode Plot of a system (closed loop for bandwidth). An important thing to remember is that a Bode Plot is only a tool we can use to predict the response of the system but in the end, the system response is determined by the pole/zero placements.

Wikipedia derives a connection between bandwidth and rise-time of a system \$t_r=\frac{0.34}{BW} \$ and a connection between overshoot and damping ratio here \$PO = 100e^{\frac{-\zeta\pi}{\sqrt{1-\zeta^2}}} \$.

Example

Let's take a simple second order transfer function \$G_\text{plant} = \frac{100}{s^2+2s+8} \tag1\$ Let's check the phase margin and the step response of the closed loop systems \$G_\text{cl} = \frac{G_\text{plant}}{1+G_\text{plant}}\tag2\$ and \$G_\text{cl2} =\frac{G_\text{plant} \cdot 0.1}{1+G_\text{plant} \cdot 0.1} \tag3\$ The transfer function of the closed loop system is also shown: -

Clearly, we see a faster rise time for the system with the high bandwidth (the left one). However, we also see many oscillations and large overshoot.

For the system with high phase margin, we see less oscillations and smaller overshoot. But we also see a slower response, due to the lower bandwidth. Looking at the transfer functions $$G_{cl} = \frac{100}{s^2+2s+108} \: \: \: \text{and} \: \: \: G_{cl2} = \frac{10}{s^2+2s+18} \tag4$$ it's clear that the damping ratio is largest for the slower system. The poles for these systems are: -

Observations

When we lower the bandwidth the response of the system becomes slower, because the system poles move closer to the origin.

When we increase the phase margin the damping ratio becomes larger, because the poles move closer to the real axis in the complex plane.

Answer 2

I too when I was in the university, was told that 45° of phase margin was the way to go, without justification and I confess it was frustrating. The thing to realize is that you set a margin for a system studied with an open-loop ac-response. However, in the end, you will close the loop and that system will operate in closed-loop, obviously exhibiting different characteristics now that a feedback path exists. The missing link for me was the relationship between the open-loop phase margin and closed-loop quality factor of the regulating system.

To determine this link, look at the below figure in which you see a compensated voltage-mode buck converter - a second-order system - offering a certain phase margin at the considered crossover frequency \$f_c\$. This plot is the result of cascading the converter ac-response with that of the selected compensator:

Now, let's apply a magnifier circa \$f_c\$ and reduce the entire transfer function (TF) to a simple 2nd-order system shown in the equation. The subscript OL indicates it is open-loop. If you now include this expression in a unity-return system, you obtain a closed-loop expression featuring the parameters of the open-loop TF. This is what I described in my APEC 2012 seminar on loop control:

You now see that we have open- and closed-loop parameters among which, \$Q_c\$ is the closed-loop quality factor we want. Going further with the maths, we can obtain a relationship between the open-loop phase margin (PM) and the resulting quality factor of the closed-loop system:

This graph shows the relationship between the PM and the response of the closed-loop system. When the open-loop phase is 76°, \$Q_c=0.5\$ meaning that for a 2nd-order response, the roots are real and coincident: there is no ringing in the time-domain response. A 45° PM would induce a closed-loop \$Q_c\$ of \$\approx 1.2\$. Now, it is important to realize that this approach is based on a simplified view, focused on the response of a loop gain around crossover. Studying the exact response of a higher-order system would be over-complicated - but doable of course - and would perhaps not bring the insight we have here. I have explored these concepts deeper in my book on control loops.

This is the explanation you need when selecting the phase margin: you choose it based on the transient response you (or your customer) want and not solely on a kind of magic number 45° would be. If you want a fast and nervous system, while accepting a bit of overshoot, then you can shoot for a lower phase margin, between 45 and 60° knowing that it will vary or drift with production and lifetime. If your customer tells you that he can't tolerate any overshoot and recovery speed is not his major concern but rock-solid long-term stability is, then you could shoot for a PM up to 90° for instance. In that case, the recovery is sluggish but there is no overshoot.

The below graph shows the transient response of a closed-loop system to a load step with a constant 5-kHz crossover while the phase margin has been changed to different values. The undershoot does not change much but the recovery time does, as expected.

Answer 3

1. There is a direct relationship between -3dB and risetime. Although there are variations for high order filters using 10% to 90% rather than T=0 to 64% of Vss you can estimate as follows;

\$t_{R_{(10-90)}}=0.35/f_{-3db}\$ {approx}

Note here for a 1st order filter T(0to64%)=10us=RC

Imagine the cap current as your control system with step load currents, but the voltage is lagging and is now triangular and attenuated from the source resulting in lower error correction and with a 2nd order system, more overshoot.

2. If we could make all control systems 1st order they would be very stable but due to stored energy in all electro-mechanical-magnetic-dielectric components, this is impossible when there is more than 1 component. Remember this: We always strive for 1st order feedback e.g. use current feedback to control current and torque, RPM feedback to control Velocity, accelerometers to control g and position sensors to control position. The signal processing of integration adds a pole and thus 90 deg phase shift above the breakpoint. So we try to improve phase margin by adding partial derivative gain a.k.a. lead-lag compensation to make the phase slope "look more like" a 1st order filter at the critical unity gain breakpoint. Gain is necessary to correct the error between input and output and the "margin" for phase or gain is directly related to the amount of 2nd order overshoot. In higher than 2nd order systems, this method requires additional analysis of Root Locus.

If there is insufficient gain, i.e. low margin, the error correction is ineffective and the system rings or oscillations from the stored energy and lag in response time. (but you can have too much gain too)

Thus when filtering out noise, in a controlled system, it is important to understand the relationship between damping factor and Q ratio is the ratio of reactive and real impedance or power.

\$\zeta=\dfrac{1}{2Q}\$ described in numerous places.

To answer 2) consider that -45 deg and the -3dB break point are one and the same for a 1st order system. This is the intuitive goal for a closed loop behaviour at unity gain to make it behave a towards a critically damped 1st order filter but not quite.

Simplified relevant explanation

For Op Amps the open-loop gain must be internally compensated because after so many stages of amplification e.g. >6 and each having junction capacitance and thus some energy storage, they have converted a very high order LPF high gain amplifier into a unity gain stable amplifier with a 1st order filter near 10 Hz and ideally end up with 60 deg Phase margin.

• axiom

The more stages of conversion with energy storage, latency or mass, the lower the BW and longer the response time must be, to keep the system unity-gain stable with a minimal overshoot to a step load change.

e.g. In very complex systems (Saturn V 5 liftoff stability), You must be able to predict the outcome and response to every stimulus and wind factor in the use of vector thrust engines and pulsed radial gas jets. The pulsed radial jets add radial thrust to vector direction which when integrated over time, results in a time again to achieve a new vector elevation angle at a certain time, so this thrust must be reversed before it gets there. The same process is used to seek in an embedded servo HDD with < 5% overshoot in the fastest time using 1st order feedback for (a,v,x). You don't want the drive to say "Ontrack or seek complete" too soon with overshoot, just as you are ready to write a sector and write it off track. (although smart drives may have built-in error recovery built-in using off-track control)