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How can we prove that the Gain Margin $[GM]$ of $2nd$ order system is $\infty\quad ?$

My Approach:
Let us consider a $2nd$ order open loop system : $$G(s)= \frac{k}{(s+1)(s+2)}$$
Now, we know: $GM(in\quad dB)=-20log_{10} |G(jw_p)|$
$$ For \quad w_p \quad : \quad \angle GM(jw)=-180$$ $$\implies -tan^{-1}(w)\quad-tan^{-1}(w/2)=-180$$ $$\implies tan^{-1}(w)\quad+tan^{-1}(w/2)=180$$ $$\implies w=0 \quad or \quad w_p=0$$ $$\therefore |G(jw_p)|=k/(1 \times 2)=k/2$$ $$\implies GM(in\quad dB)=-20log_{10} (k/2) \neq \infty$$

so where is my mistake and how can i proof this results....??please help

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  • $\begingroup$ This is actually not true, namely non minimum phase second order systems can have a finite gain margin, for example $$\frac{1-s}{s^2+s+1}$$ $\endgroup$ – fibonatic Apr 2 '18 at 10:19
  • $\begingroup$ That might actually not the best example, since it has a gain margin of minus infinity dB (passes though the minus one point). But the following second order system does has a finite gain margin: $$\frac{1-s}{s^2+2\,s+1}$$ $\endgroup$ – fibonatic Apr 2 '18 at 10:56
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For a second order transfer function

$$G(s)=\frac{k}{s^2+2\zeta \omega s +\omega^2}$$

What are the real part of the poles?

$$\Re\{s_{p1},s_{p2}\}=-\zeta\omega$$

The closed loop transfer function from a constant feedback $h$ is

$$G(s)=\frac{k}{s^2+2\zeta \omega s +\omega^2+kh}$$

$$s_{p1},s_{p2}=\frac{-2\zeta\omega\pm\sqrt{-4\omega^2(1-\zeta^2)-4kh}}{2}$$

$$s_{p1},s_{p2}=-\zeta\omega\pm\sqrt{-\omega^2(1-\zeta^2)-kh}=-\zeta\omega\pm\sqrt{\Delta}$$

What are the real part of the poles?

For a system with poles which have imaginary parts $$\Re\{s_{p1},s_{p2}\}=-\zeta\omega$$

It means that it doesn't matter how big $k$ is.

But if the poles are pure real numbers, as $h>0$, by increase of the k, $\Delta$ decreases and only the upper bound of the poles reduce. So, the poles do not go to the right hand side of s-plane.

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