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

Question

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}\$

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'