# The open loop transfer function of a unity feedback control system is given as $G(s) = \frac{as+1}{s^2}$

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° ?

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

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

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

$T(s)=\frac{as+1}{(s+\frac{a}{2})^2+1-\frac{a^2}{4}}$

$T(s)=\frac{as+1}{(s+\frac{a}{2})^2+\left(\sqrt{1-\frac{a^2}{4}}\right)^2}$

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

$T(s)=\frac{as+1}{(s+\frac{a}{2})^2+{\omega}^2}$

$T(s)=\frac{a(s+\frac{a}{2})+(1-\frac{a^2}{2})}{(s+\frac{a}{2})^2+{\omega}^2}$

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

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: $$\phi = 180^\circ + \theta$$