[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?
 

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.
Click to expand...
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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The correct answer is 1.732 x 10^4. But I didn't get that answer.
Click to expand...

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?
 

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

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
 
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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