How can I solve for u and ultimately find ω in this equation?

[baby_1]

Hello
Could you help me to solve this problem? i want to find w

asnwer is 151660
but i want to know the approach to solve it.

 

[HallsofIvy]

Well, what have you done on this? I can see a few obvious things to do to start- like multiplying through by -1! Have you done that? And taking the tangent of both sides seems obvious. What is tan(180)? Do you know the formula for tan(a+b)? Can you extend that to tan(a+ b+ c)?

 

[baby_1]

Yes, i do 2 ways but none of them work fine...could you write the equation because its a start point of G.M(Gain margin) and Phase margin ampilfier.

 

[Curious3141]

Hello
Could you help me to solve this problem? i want to find w

asnwer is 151660
but i want to know the approach to solve it.

Multiply throughout by -1.

Bring one of the arctan terms over to the RHS (to the same side as the 180 degree term). Suggest the one with 10⁴ in the denominator.

Take tangent of both sides. LHS can be resolved with angle sum formula. RHS is easy.

That should be enough to start you out. Show your work (in detail), then you'll get more help.

 

[Saitama]

Is 180 in degrees?
It cannot be in degrees because the range of arctan lies between -90 and +90 (degrees) and even if you simplify the LHS, you get tan^-1x=180, (x is anything which remains after applying identities) and 180 degrees is out of range. Is my reasoning correct?

 

[pcm]

range of phi is -1.5pi to 1.5 pi.so obviously 180 is degree

 

[Saitama]

range of phi is -1.5pi to 1.5 pi.so obviously 180 is degree

Range of arctan is -pi/2 to pi/2.
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

 

[pcm]

what i meant was phi can't be greater than 1.5pi and can't be less than -1.5pi.see there are 3 arctan terms.

 

[Saitama]

what i meant was phi can't be greater than 1.5pi and can't be less than -1.5pi.see there are 3 arctan terms.

Ah, thanks for rectifying. I was just confused by the solution i was doing for this problem.

 

[SammyS]

Is 180 in degrees?
It cannot be in degrees because the range of arctan lies between -90 and +90 (degrees) and even if you simplify the LHS, you get tan^-1x=180, (x is anything which remains after applying identities) and 180 degrees is out of range. Is my reasoning correct?

Suppose you have three arctangents, all with equal argument, A. Then you have:

$-tan^{-1}(A)-tan^{-1}(A)-tan^{-1}(A)=-180^\circ\ .$

Then you have tan^-1(A) = 60° .

No problem with that!

 

[SammyS]

Hello
Could you help me to solve this problem? i want to find w

asnwer is 151660
but i want to know the approach to solve it.

If you let u = ω/10⁵, then ω/10⁴ = 10u and ω/(2∙10⁵) = u/2 .

That leaves you with

$\displaystyle -\tan^{-1}(10u)-\tan^{-1}(u)-\tan^{-1}(u/2)=-180^\circ\ .$

That looks a bit less intimidating.

WolframAlpha gives an exact solution for u, but doesn't show a method.