
#1
Jun511, 08:58 PM

P: 1

Hello. This technically isn't homework, however I am studying for an upcoming final. I feel like I must be missing something. I'm trying to calculate the phase margin of a system so my first step is to find the gain crossover frequency (GCF).
1. The problem statement, all variables and given/known data Transfer function: [tex]H = \frac{H}{s(s^2+7s+140)}[/tex] 2. Relevant equations GCF occurs where the magnitude of the gain = 1 , or [tex] H(j \omega_{GC})=1 [/tex] where [tex]s= j \omega_{GC}[/tex] 3. The attempt at a solution For the sake of brevity I'll skip most of the details of the algebra. I start with, [tex] \left \frac{20}{j \omega ((j \omega)^2 + 7 j \omega + 140)} \right = 1 [/tex] By definition of GCF I'm interested only in magnitude. After factoring the 2nd order term , crossmultiplying to get 20/1, and taking the norm of each root the above equation leads to, [tex] 20 = (\sqrt{\omega^2}) \left( \sqrt{ \left( j \omega + j \frac{\sqrt{511}}{2} \right)^2 + \left( \frac{7}{2} \right)^2} \right) \left( \sqrt{ \left(j \omega  j \frac{\sqrt{511}}{2} \right)^2 + \left( \frac{7}{2} \right)^2 } \right) [/tex] At this point I just crank through the algebra (which I'll skip) and end up with, [tex]400 = \omega^6 + 280 \omega^4  \frac{53361 \omega^2}{4} [/tex] Finally solving for omega I get, [tex] \omega = \pm j 0.173106[/tex] [tex] \omega = \pm 7.80531 [/tex] [tex] \omega = \pm 14.8023 [/tex] Now we finally arrive at my problem. Plotting the original function shows the magnitude decreasing for all time. This makes sense as there are no zeros and 3 poles. So I should only have a single crossover frequency (also shown on a bode plot). So the question is how do I know which of the above values for omega is the actual gain crossover frequency? I'm trying to solve this without the use of any plots. Any help would be greatly appreciated!! Thanks. :) Edit: I should say that I do recognize that a negative frequency doesn't make sense so I only have 3 options and not 6. I think I can eliminate the option with the j term for the same reason. Yes, no? 



#2
Oct1613, 10:38 AM

P: 22

I realize this is 2 years late... but better late than never right?
Your math is wrong. You should have ended up with [itex]\omega^{6}  231\omega^{4}+19600\omega^{2} = 400[/itex] There is only one real, positive solution to that equation, which is the gain crossover frequency. 


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