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[SOLVED] Help me solve this problem- Frequency for the given phase shift calculation.

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iVenky

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Given the function is

A(s)= [ 10 / (1+s/10000) ] ^3

What is the frequency at which the phase shift is 180 degree.

I tried using tan inverse formula but I didn't get the correct answer.

What's your answer?
 

jimito13

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You should write the transfer function in the form Re{A(s)}+jIm{A(s)} with s=jω.

Next step is to get the arg of the latter function to get the phase response,thus arg[Re{A(s)}+jIm{A(s)}]=arctan[Im{A(s)}/Re{A(s)}].

Finally equal the latter expression to 180 and solve for ω.

I tried using tan inverse formula but I didn't get the correct answer.
How do you know the correct answer since you didn't solve it?Is it a homework exercise and you just have the answer and try to confirm it?

After a little math and with the aid of mathematica i saw that there is no ω that satisfies your equation!
 
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jimito13

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The correct answer is 1.732 x 10^4. But I didn't get that answer.
So,what answer did you get and in which way you tried to solve your problem?Give some clues to disguss and help you find the proper solution,with one-word answers what do you expect?
 

FvM

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It seems like the "correct answer" is related to tan(60°).
 

rsashwinkumar

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Taking the transfer function in frequency domain,

H(jw) = 10/(1+jw/10000)^3)

Now, coming to basics of complex-algebra,

a complex number a + jb has a phase tan-1(b/a); extending this (a+jb)^3 has a phase 3*tan-1(b/a); you can prove this easily using euler's notation of a complex variable (a+jb=exp(jw)*r where w=tan-1(b/a); and so raising to the power 3, on both sides, you can find net phase = 3*tan-1(b/a));

similarly 1/(a+jb)=1/r*exp(-jw); and so the phase here, w'=-w=-tan-1(b/a);

Now just apply all these to your H(jw).

so the phase is -3*tan-1(w/10000) and equating it to -180

tan-1(w/10000)=60 degrees.,
so you get w=10000*sqrt(3)=1.732*10000 rad/s
 

iVenky

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Hello RS.

This is Venky.

For some strange reasons I had equated that to 1 instead of Pi. Should have slept while doing this.

Thanks anyway.
 
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