Derivatice of Complex Impedance

In summary, the student attempted to solve a problem involving taking the derivative of the impedance of a parallel RLC circuit with respect to the frequency ω of the applied voltage, but ended up getting an answer that was imaginary.
  • #1
Opus_723
178
3

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:
Physics news on Phys.org
  • #2
Opus_723 said:

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
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
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
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!
 

FAQ: Derivatice of Complex Impedance

1. What is complex impedance?

Complex impedance is a measure of the resistance in an electrical circuit that contains both resistance and reactance components. It is represented by a complex number that includes both a real (resistive) and imaginary (reactive) part.

2. How do you calculate the derivative of complex impedance?

The derivative of complex impedance is calculated by taking the derivative of the real and imaginary parts separately. The derivative of the real part is equal to the derivative of resistance, while the derivative of the imaginary part is equal to the derivative of reactance multiplied by the imaginary unit, i.

3. Why is the derivative of complex impedance important?

The derivative of complex impedance is important because it helps us understand how the impedance changes with respect to frequency. By analyzing the derivative, we can determine the resonant frequency and other important characteristics of the circuit.

4. What is the physical significance of the derivative of complex impedance?

The physical significance of the derivative of complex impedance is that it represents the rate of change of impedance with respect to frequency. This can tell us about the behavior of the circuit at different frequencies and how it responds to different types of signals.

5. Are there any applications of the derivative of complex impedance?

Yes, there are many applications of the derivative of complex impedance in the field of electrical engineering. It is used in circuit analysis and design, as well as in the development of electronic devices and systems. It is also important in understanding and optimizing the performance of electrical networks and communication systems.

Back
Top