Discussion in 'Homework Help' started by YukiWong, Dec 13, 2011.

1. ### YukiWong Thread Starter New Member

Dec 5, 2011
27
1
Hi everybody ,

I got a transfer function which is 2/[s*(0.5s+1)] and figure below show the bode plot of the system. My quesiton is, why is the gain margin of this system is infinite?

Thank you very much.

2. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
The function simplifies to $\frac{1}{s (0.25s + \frac{1}{2})}$

No clue why it would be infinite, it should converge to 1

Last edited: Dec 14, 2011
3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
It is infinite - the total phase shift tends to exactly 180° as the frequency tends to infinity at which point the transfer gain is infinitesimal.

4. ### YukiWong Thread Starter New Member

Dec 5, 2011
27
1
thanks t_n_k,

erm, is that every second order system will cause infinity gain margin?

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Consider a [2nd order?] transfer function like

$G(s)=\frac{(10-s)}{s(s+5)}$

Which has a non-infinite gain margin but has an unbounded step response.

Considered in isolation, gain margin of itself doesn't tell us much about system stability - so a system could have an infinite gain margin but be unstable or unbounded in transient response.

For instance the transfer function

$G(s)=\frac{0.3}{{(s-0.001)}^2+0.09}$

has infinite gain margin but has an unstable step response.

YukiWong likes this.
6. ### YukiWong Thread Starter New Member

Dec 5, 2011
27
1
thanks t_n_K, i got it