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Magnitude and Phase of Transfer Function

by zonedestruct
Tags: function, magnitude, phase, transfer
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zonedestruct
#1
Feb24-12, 04:33 AM
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1. The problem statement, all variables and given/known data

I am given the transfer function H(s) = 10/(s(s^2 + 80s +400)) where s = jω [j is the imaginary unit i] and I am trying to get it into its magnitude and phase components.






3. The attempt at a solution

I rearranged it to 1/(40jω(1+ 4jω/20 + (jω/20)^2)) which is the standard form for transfer functions. I am wanting to plot the bode plots so I took the 20 * log base 10 of the entire transfer function and got for the magnitude: -20log(40) - 20log(1/(jω))- 20log(1+ 4jω/20 + (jω/20)^2)
and for the phase I got: -90 - tan^-1((ω/5)/(1-ω^2/400))

but in my textbook when they show the bode plots for the phase it has - tan^-1(ω/(1-ω^2/400)) for one of the phase factors and I am not sure why my numerator is ω/5 and there one is just ω.

If you don't study electrical engineering and are not sure of some of the stuff I said then basically H(s) = 10/(s(s^2 + 80s +400)) where s = jω is a complex number and j is the same as the imaginary unit i, so I need help to put this complex number into its magnitude and phase, I am pretty sure i got the magnitude part right.

Please can anyone help me by at least showing me how to get the phase part of this complex number
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gneill
#2
Feb24-12, 04:03 PM
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Since the numerator of your transfer function is purely real we can concentrate on the denominator; it's phase will be the negative of (180 shifted) the transfer function's phase.

## s ( s^2 + 80 s + 400 ) → \omega j (-\omega^2 + 80 j \omega + 400) ##

## -80 \omega^2 + j (400 \omega - \omega^3) ##

Divide through by 400ω:

## -\frac{\omega}{5} + j \left( 1 - \frac{\omega^2}{400} \right) ##

Looks like the book's solution is not correct.
rude man
#3
Feb26-12, 04:47 PM
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The phase part is just tan-1{(imaginary part)/(real part)}.

As gneill pointed out, your transfer function phase angle will be the negative of the numerator phase angle, i.e.

ψ = - tan-1{(imaginary part)/(real part)} of the denominator.

{However, -ψ is not the same as ψ shifted by 180. E.g if ψ = 30 deg, -ψ is the same as 360 - 30 = 330 deg, not 30 - 180 or 30 + 180.}

Also, be careful to preserve numerator and denominator signs in the arc tan expression. So arc tan(-a/b) is not the same angle as arc tan(a/-b).

gneill
#4
Feb26-12, 06:27 PM
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Magnitude and Phase of Transfer Function

Quote Quote by rude man View Post
{However, -ψ is not the same as ψ shifted by 180. E.g if ψ = 30 deg, -ψ is the same as 360 - 30 = 330 deg, not 30 - 180 or 30 + 180.}
Good catch, rude man; That was a brain fart on my part I hate when that happens
rude man
#5
Feb27-12, 02:25 AM
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Quote Quote by gneill View Post
Good catch, rude man; That was a brain fart on my part I hate when that happens
Don't feel bad! After writing that I wasn't so sure myself that what I wrote was really right! It's really something to think about - after all -V does = V at 180! Subtle business, electricity!

No to mention that what I wrote wasn't quite right anyway - the angle of the transfer function, not that of the numerator, is the negative of the angle of the denominator, as you of course had already pointed out. Oy vey!


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