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Derivatice of Complex Impedance

  • Thread starter Opus_723
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  • #1
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Homework Statement



[Edit: Sorry for the typo in the thread title]

I'm doing a problem that involves taking the derivative of the impedance of a parallel RLC circuit with respect to the frequency ω of the applied voltage. I have to then set that derivative equal to zero to find the frequency at which the maximum impedance occurs. I'm pretty sure I did everything mostly right, but my answer ended up being imaginary. I ended up with ω = [itex]\frac{i}{\sqrt{LC}}[/itex]. I'm sure the coefficient is right, since that's the resonant frequency of the undamped circuit. But should it really be imaginary? That seems odd to me. I've never actually taken the derivative of anything involving i before, so there might be something different you have to do that I don't know about. Anyway, I'm going to post my work. If someone could point out where I messed up (or explain why it's okay for the answer to be imaginary) it would be greatly appreciated. Thanks!

Homework Equations




Z = [itex]\frac{1}{\frac{1}{R}+iωC -\frac{i}{ωL}}[/itex]

The Attempt at a Solution



[itex]\frac{dZ}{dω} = -1(\frac{1}{R}+iωC -\frac{i}{ωL})^{-2}(iC+\frac{i}{ω^{2}L}[/itex])

-iC-[itex]\frac{i}{ω^{2}L}[/itex] = 0

[itex]ω^{2} = \frac{-1}{LC}[/itex]

ω = [itex]\frac{i}{\sqrt{LC}}[/itex]
 
Last edited:

Answers and Replies

  • #2
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
36

Homework Statement



[Edit: Sorry for the typo in the thread title]

I'm doing a problem that involves taking the derivative of the impedance of a parallel RLC circuit with respect to the frequency ω of the applied voltage. I have to then set that derivative equal to zero to find the frequency at which the maximum impedance occurs. I'm pretty sure I did everything mostly right, but my answer ended up being imaginary. I ended up with ω = [itex]\frac{i}{\sqrt{LC}}[/itex]. I'm sure the coefficient is right, since that's the resonant frequency of the undamped circuit. But should it really be imaginary? That seems odd to me. I've never actually taken the derivative of anything involving i before, so there might be something different you have to do that I don't know about. Anyway, I'm going to post my work. If someone could point out where I messed up (or explain why it's okay for the answer to be imaginary) it would be greatly appreciated. Thanks!

Homework Equations




Z = [itex]\frac{1}{\frac{1}{R}+iωC -\frac{i}{ωL}}[/itex]

The Attempt at a Solution



[itex]\frac{dZ}{dω} = -1(\frac{1}{R}+iωC -\frac{i}{ωL})^{-2}(iC+\frac{i}{ω^{2}L}[/itex])

-iC-[itex]\frac{i}{ω^{2}L}[/itex] = 0

[itex]ω^{2} = \frac{-1}{LC}[/itex]

ω = [itex]\frac{i}{\sqrt{LC}}[/itex]

I guess you do have to be careful when differentiating something with 'i' in it, because apparently $$\frac{d}{d\omega}\left(\frac{-i}{\omega L}\right) = \frac{i}{\omega^2 L}$$No negative sign.
 
  • #3
176
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I'm not sure I understand. That looks like what I did. I know there's no negative sign.

[itex]\frac{d}{dω}[/itex]([itex] \frac{-i}{L}ω^{-1}[/itex]) = -1([itex]\frac{-i}{L}[/itex])ω[itex]^{-2}[/itex] = [itex]\frac{i}{Lω^{2}}[/itex]

This is what I did above.
 
  • #4
115
2
There is no less/greater relation for complex number so complex function cannot have extremum. Actually you are seeking extremum of impedance modulus:
[itex]|Z|=\frac{1}{\sqrt{\frac{1}{R^2}+\left(\omega C-\frac{1}{\omega L}\right)^2}}[/itex]
 
  • #5
176
3
That makes sense. I get the right answer taking that derivative instead. I suppose I should have realized that a maximum of a complex function doesn't make a lot of sense.

Funny thing is, i tried the modulus first, but accidentally skipped a step in the derivative and so I lost the real root of the equation, and only got the complex one, which is the same thing i got above. I wonder if finding the zeros of the complex function always gives you the complex roots of the modulus?

Anyway, thanks a lot!
 

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