0
\$\begingroup\$

I'm trying to write and solve a differential equation for this simple RLC notch filter and I'm not sure what I'm doing wrong.

schematic

simulate this circuit – Schematic created using CircuitLab

So first, I apply KVL to (1) the loop source-resistor-capacitor-source and (2) the loop source-resistor-capacitor-source to get what's essentially an RC and RL circuit:

\$ RC\frac{dI_R}{dt} + I_c = C\frac{dV}{dt}\$

\$ L\frac{d^2I_L}{dt^2} + I_R = V\$

Adding these, I get

\$ L\frac{d^2I_L}{dt^2} + 2R\frac{dI_r}{dt} + \frac{I_C}{C} = 2\frac{dV}{dt} \$

Using KCL I know that

\$ I_R = I_C + I_L \$

Using impedances, I get

\$ Z_C = \frac{1}{j\omega C} \$ and \$ Z_L = j\omega L \$

Based on the current divider equation, I have

\$ I_L = \frac{Z_C}{Z_L}I_R \$ and \$I_C \frac{Z_L}{Z_C}I_R \$

\$ I_C = -I_R \omega^2 LC \$

\$ I_L = -\frac{I_R}{\omega^2 LC} \$

If I put this into the differential equation, I get

\$ \frac{d^2I_R}{dt^2} -2RC \omega^2\frac{dI_R}{dt} + \omega^4LCI_R = -2\frac{dV}{dt} \$

If I try to solve this, I get positive exponentials, which I know are wrong since current won't increase without bound. I'm not sure how else to go about this. Any help is appreciated. Thanks!

\$\endgroup\$
  • 1
    \$\begingroup\$ Phasor analysis requires that time domain transients have decayed to zero; differential equation analysis concentrates on the transients. The two cannot be mixed in a single expression as you are trying to do. \$\endgroup\$ – Chu Aug 27 at 6:52
  • \$\begingroup\$ Thanks to both of you! I'll have to look at the phasor diagrams more carefully then. \$\endgroup\$ – khaaaaan Aug 28 at 1:13
0
\$\begingroup\$

You can’t use phasor impedances with differential equations. Phasors are used to convert a sinusoidal input into the s domain via Laplace transform or frequency domain via Fourier transform which takes a differential equation and converts it into an algebra problem, so you can solve the equation using differential equation technique and completely drop the impedances or convert it using the Laplace Transform.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.