- #1

evolosophy

- 1

- 0

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).

Transfer function: [tex]H = \frac{H}{s(s^2+7s+140)}[/tex]

GCF occurs where the magnitude of the gain = 1 , or [tex] |H(j \omega_{GC})|=1 [/tex] where [tex]s= j \omega_{GC}[/tex]

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 , cross-multiplying 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]

I'm trying to solve this without the use of any plots.

Any help would be

## Homework Statement

Transfer function: [tex]H = \frac{H}{s(s^2+7s+140)}[/tex]

## Homework 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]

## 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 , cross-multiplying 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?
Last edited: