The open loop transfer function of a unity feedback control system is given as \$G(s) = \frac{as+1}{s^2}\$. What value of 'a' will give a phase margin of 45° ?

\$G(s) = \frac{as+1}{s^2}\$

\$Transfer\$ \$function\$, \$T(s)=\frac{G(s)}{1+G(s)*1}\$




Let, \$\omega=\sqrt{1-\frac{a^2}{4}}\$



  • \$\begingroup\$ It could be helpful to add some text to explain the passages. I can't understand at first sight what they lead to. \$\endgroup\$
    – clabacchio
    Apr 10, 2013 at 15:20
  • \$\begingroup\$ 1. Why are you analyzing the closed-loop transfer function and 2. Why have you made that strange substitution for ω? \$\endgroup\$
    – MikeJ-UK
    Apr 10, 2013 at 15:51

1 Answer 1


The phase margin of a closed loop system is defined at it's gain crossover frequency which is calculated for it's open loop gain (as pointed out by MikeJ-UK). The gain crossover frequency is when the magnitude gain of the open system is unity. Positive values of gain and phase margins would indicate that the given open-loop system is stable when a feedback loop is added to it. $$|G_{open}(j\omega)| =1 $$

Now you need to obtain the angle at which this happens. For that just take the inverse tangent of the numerator and denominator angles and subtract them, $$ \theta = \arctan (N(j\omega)) - \arctan (D(j\omega)) $$

Now to calculate the phase margin use the following equation: $$ \phi = 180^\circ + \theta $$

When you calculate this way, you'll get the value of 'a'

  • \$\begingroup\$ You have correctly said that the magnitude of the loop gain is 1 at the frequency of interest, but on the next line you have analyzed the closed-loop gain. \$\endgroup\$
    – MikeJ-UK
    Apr 10, 2013 at 16:12
  • \$\begingroup\$ The closed loop gain is used for calculating crossover frequencies and hence the stability of the closed-loop system, isn't it? \$\endgroup\$ Apr 10, 2013 at 16:20
  • \$\begingroup\$ Look at the last line of the first paragraph, then the second paragraph of that wikipedia article, where it references the loop gain \$\endgroup\$
    – MikeJ-UK
    Apr 10, 2013 at 16:24
  • \$\begingroup\$ Yes, I think you're correct. Figure I've embarrassingly forgotten some basics. I'll edit my answer. \$\endgroup\$ Apr 10, 2013 at 16:33

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