I have the following system (see picture). When calculating gain and phase margins, should I consider $L=GK$ to be the open loop transfer function? If so, what difference does the transfer function $H$ applied on the reference $r$ make to the gain and phase margins?

Closed loop system

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    $\begingroup$ Please give us a little more background on your problem so that we may more accurately help you. Also, this reads like a homework problem. If so, please identify it as such in the body of your question. $\endgroup$
    – grfrazee
    Aug 6, 2015 at 20:37
  • $\begingroup$ Firstly, this is not a homework problem - I am just trying to deepen my understanding of the subject. For the background - I just happen to have a system which looks like a block diagram in the picture, and I am interested in its stability, and just wandering what effect the H transfer function (if any) has on the result. $\endgroup$
    – montyynis
    Aug 6, 2015 at 21:48
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    $\begingroup$ @montyynis - Homework problems are allowed, but we ask for them to be noted as it can affect the answers that are provided. Real world solutions tend to be a bit messier and must account for additional factors that idealized problems do not have to deal with. $\endgroup$
    – user16
    Aug 7, 2015 at 1:26

2 Answers 2


No, you do not need to consider $H$ when thinking about the loop stability. Doing the loop algebra shows why: $$ y= KG(\eta H-y)\qquad\rightarrow\qquad y=\left(\frac{KG}{1+KG}\right)\ \eta H $$ The part in parentheses is just the standard transfer function of the loop. The concept of gain and phase margin comes from considering that the overall gain becomes infinite (or very large) if $KG$ equals (or gets close to) -1. Writing $$ KG=Ae^{i\phi} $$ shows that this can happen if the gain $A$ is too close to 1 when the phase $\phi$ is too close to 180°. $H$ and $\eta$ play no role in this consideration.

  • $\begingroup$ @Suba I (perhaps obviously) disagree with you. Could you elaborate on what you mean by 'the more basic definition of gain and phase margins'. Simply saying that it has to do with phase and gain crossover frequencies doesn't say much. $\endgroup$ Aug 7, 2015 at 20:55
  • $\begingroup$ The definitions (from Ogata): "The gain margin is the reciprocal of the magnitude at the frequency at which the phase angle is -180 degrees." "The phase margin is that amount of additional phase lag at the gain crossover frequency required to bring the system to the verge of stability." Your analysis starting with $1+G K=0 $ is valid when $H = 1$, not otherwise. $\endgroup$ Aug 7, 2015 at 21:04
  • $\begingroup$ Those definitions are just re-wordings (with fancier language) of the ones that I give. Your analysis is, unfortunately, incorrect. $\endgroup$ Aug 7, 2015 at 21:13
  • $\begingroup$ I'm sorry, but you haven't given any definitions. You haven't shown how your analysis follows from Ogata's definition. And, you haven't show how my analysis is wrong either. :) $\endgroup$ Aug 7, 2015 at 21:23
  • $\begingroup$ I have not updated my answer and comments. I agree that your conclusion is correct (only because of pole-zero cancellations), and so I do not agree that you can completely ignore $H$ when trying to interpret the significance of the the margins that are computed. $\endgroup$ Aug 8, 2015 at 18:19

To compute the gain and phase margins, we first need to determine the loop gain from $r\to y$. The block diagram is equivalent to the following one.

enter image description here

It is now straightforward to compute the loop gain from $r\to y$ and it turns out as $L=\frac{H K G }{H}$, and with pole-zero cancellations simplifies to $L=K G$.

So $H$ does not affect the stability margins, because it cancels itself. However, $H$ does affect stability. If you want to determine whether it is minimum phase or not, use $L=\frac{H K G}{H}$ before cancellation. For example, if $H$ is non-minimum phase and $G$ and $K$ are minimum phase and the gain and phase margins turn out to be positive, just don't overlook that you are dealing with a non-minimum phase system.

  • $\begingroup$ You are arguing that the characteristics of the reference signal $r$ affect the stability of the loop which is not true for a linear time-invariant system. What if $r$ is a signal which came from another filter before being passed to $H$, i.e. $r=Qr_0$? Would $Q$ also need to be included in the stability analysis? $\endgroup$ Aug 7, 2015 at 20:59
  • $\begingroup$ I have made no reference to $r$. $\endgroup$ Aug 7, 2015 at 21:05
  • $\begingroup$ But what about $Q$ in my example; is it also relevant to the loop stability? $\endgroup$ Aug 7, 2015 at 21:07
  • $\begingroup$ Depends on what we are analyzing. The OP has asked for stability margins for the system $r \to y$. If you what to analyze from some other upstream input which includes Q, then yes, you have to consider that. Your analysis essentially starts from the output signal of $H$. $\endgroup$ Aug 7, 2015 at 21:15
  • $\begingroup$ "Your analysis essentially starts from the output signal of $H$." You've hit on exactly my point. How does the loop (which does not include $H$) have information about what the signal looks like before $H$? What is the difference between applying a signal $r$, a signal $rH$ or a signal $rQH$? They are all out of the loop. $\endgroup$ Aug 7, 2015 at 21:21

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