Consider the system with unitary negative feedback such that the open loop transfer function is $$G(s) = \frac{as + 1}{s^2}$$

a) Determine the value of $a$ such that the phase margin is 45º

b) Determine the stationary state error for the unitary ramp input

c) For $a > 0$, what is the gain margin

My attempt: My doubt is about the item (c). That what I've made:

$GM = \frac{1}{|G(j\omega)|}$ for $\omega$ such that $\text{phase}(G(j\omega)) = -180º$. In this case $$\text{phase}(G(j\omega)) = -180º \iff \text{arctan}(a \omega) - 180º = -180º \iff \text{arctan}(a \omega) = 0 \iff \omega = 0$$

But for $\omega \to 0$, $|G(j\omega)| \to \infty \Rightarrow GM \rightarrow 0$

Is that correct?


  • 1
    $\begingroup$ Are you trying to answer part c? Or are you asking a more general question? $\endgroup$ Commented Nov 24, 2015 at 18:37
  • $\begingroup$ @willpower2727 trying to answer C $\endgroup$
    – Giiovanna
    Commented Nov 25, 2015 at 8:49
  • $\begingroup$ Do you calculate gain margin for the open or closed loop transfer function? $\endgroup$ Commented Nov 25, 2015 at 15:20
  • $\begingroup$ Open, for G(s). $\endgroup$
    – Giiovanna
    Commented Nov 26, 2015 at 9:25

1 Answer 1



The two poles at the origin contribute to a phase of -180${}^{\circ}$.

The zero at $-\frac{1}{a}$ contributes a phase ranging from 0 to 90${}^{\circ}$, with a phase 45${}^{\circ}$ at the corner frequency $\frac{1}{a}$.

Thus at the corner frequency the phase is -135${}^{\circ}$. The phase margin can be 45${}^{\circ}$ if at this frequency the magnitude is 1.

$$\left|\frac{\frac{i a}{a}+1}{\left(\frac{i}{a}\right)^2}\right| = 1$$

$$\sqrt{2} a^2=1$$



The steady-state error for a ramp input is zero because it is a type 2 system.


The gain margin is infinite. This is because the Bode phase plot is always above -180${}^{\circ}$ for $a>0$.


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