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.

 

The function simplifies to \(\frac{1}{s (0.25s + \frac{1}{2})}\)

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

 

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.

 

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.

thanks t_n_k,

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

 

thanks t_n_k,

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

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.

 

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.

thanks t_n_K, i got it