Need to find the transfer function of this band rejection filter via its differential equation but cannot figure it out since it was some time ago I studied electrical circuits.

Band rejection filter

I know that I should use Kirchoff´s laws as well as the differential equations for the different components:

Inductor: $$ V=L \frac{di}{dt}$$ Capacitor: $$ V = \frac{1}{C}\int i dt$$ But I´m not sure in how to move on. Any help on this matter is would be much appreciated.

Another way to get hold of the transfer function would be by going via the circuits impedance, which I thought I knew how to do but after simulating my TF´s Bodeplot in MATLAB with the same in LTSpice I must be doing something wrong. First of all, the parallel combination should result in: $$ Z_{tot}=\frac{Z_C Z_L}{Z_C+Z_L}=\frac{\frac{L}{C}}{\frac{1}{jwC}+jwL}=\frac{jwL}{1+j^2w^2 CL}=\frac{sL}{1+s^2CL} $$

As usual: $$Z_C=\frac{1}{jwC}, Z_L=jwL, Z_R=R$$

And the voltage divider should give: $$ H=\frac{V_O}{V_I}=\frac{Z_2}{Z_1+Z_2}=\frac{Z_2}{Z_{tot}+Z_2} = \frac{Z_R}{\frac{sL}{1+s^2CL}+Z_R} \frac{R}{\frac{sL}{1+s^2CL}+R} = \frac{R(1+s^2CL)}{sL+R(1+s^2CL)} $$

And breaking out RCL I get: $$ \frac{RCL(s^2+\frac{1}{CL})}{RCL(s^2+\frac{s}{RC}+\frac{1}{CL})} = \frac{s^2+\frac{1}{CL}}{s^2+\frac{s}{RC}+\frac{1}{CL}}$$

But after making a Bode of this in MATLAB with the values C=20nF, L=50mH and R=5k I do not get the same cut-off frequency in both softwares. What is it I´m doing wrong?

  • \$\begingroup\$ Right click the image and press "Open in new tab" to see the image. Don´t know why it didn´t work. \$\endgroup\$ Commented Sep 23, 2014 at 16:02
  • \$\begingroup\$ Thanks @Andyaka, how did you do that? \$\endgroup\$ Commented Sep 23, 2014 at 21:50
  • \$\begingroup\$ Your transfer function looks correct to me. Clearly, when \$j\omega = \frac{j}{\sqrt{LC}}\$, the numerator is zero so this is a notch filter - the output goes to 1 at very low and very high frequencies - so I'm not sure if you mean the zero is different for the two simulations or not. \$\endgroup\$ Commented Sep 24, 2014 at 2:46
  • \$\begingroup\$ @user2069136 - just do an edit of your question and see the script I used - then abandon the edit. \$\endgroup\$
    – Andy aka
    Commented Sep 24, 2014 at 7:13
  • \$\begingroup\$ @Andyaka Yes, saw that when I was editing. Thanks. \$\endgroup\$ Commented Sep 24, 2014 at 14:30

3 Answers 3


You must raise the equilibrium equations of the circuit (applying LKV, LKC and Ohm's Law), and then apply the Laplace Transform, considering that the initial conditions must be zero.
Next, you must operate to find an expression for \$V_o(s)\$ depending on \$V_i(s)\$.

Then, the trasfer function is:

$$ G(s) = \dfrac{V_o(s)}{V_i(s)} $$

Remember that

$$ L\quad\overset{\text{Laplace domain}}{\to}\quad s\cdot L\\ C\quad\overset{\text{Laplace domain}}{\to}\quad \dfrac{1}{s\cdot C} $$


You need to apply the transform properties for differentiation and integration.

Assuming, for the moment, zero initial conditions,

Given that x(t) transforms to X(s), then d(x(t))/dt transforms to sX(s), and the integral of x(t)dt transforms to X(s)/s. Given this, and the Ohm's law formulation of V=iR, when you force your expressions into the V=iZ form, you'll find that Z for inductor L is Ls, and Z for capacitor C is 1/sC


If you really want to derive the transfer function H(s) starting in the time domain with the differential equation you must do the following:

1.) Based on the general voltage-current relation of all components (attention: NOT for sinus signals using sL and 1/sC) you can find the step response g(t) of your circuit - as a solution of the corresponding diff. equation.

2.) Apply the Laplace transformation to the function g(t). Then, multiply the result G(s) with "s" to get H(s)=s*G(s).


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