I've been tasked with finding the transfer function of the RLC filter shown below. I converted to the S-domain and used the voltage divider rule to find the output, and came up with:

$$H(s) = \frac{R}{Ls + \frac{1}{sC} + R}$$

I multiplied the numerator/denominator by \$sC\$ which resulted in:

$$H(s) = \frac{RsC}{CLs^2 + RsC + 1}$$

I plugged in the values of R = 220, C = 0.1 µF, and L = 100 mH, then used Octave to generate the transfer function with:

F = tf([220*.1E-6 0],[.1*.1E-6 220*.1E-6 1])

When I do the impulse or step response of the function I get a result of 0. I simulated the circuit in LTspice and approximated the impulse response with a sharp pulse and got something similar to a sinusoid. I know the simulation won't be exactly the same as the theoretical result, but it seems like my transfer function isn't correct.

Bandpass Filter


2 Answers 2


Your transfer function is correct. I have not used Octave before, so I can't guess what has gone wrong, but when I use MATLAB with the following code

R = 220;
L = 100*10^(-3);
C = 0.1*10^(-6);
H = tf([R*C,0],[L*C,R*C,1])

H =
          2.2e-05 s
  1e-08 s^2 + 2.2e-05 s + 1
Continuous-time transfer function.


I get the following responses: -

enter image description here

But here is a question to you: How can you see from the transfer function, that your step response will be a decaying oscillation? (Hint: Damping ratio)


Well, the output voltage when applying an impulse response is:

\begin{equation} \begin{split} \text{V}_\text{o}\left(t\right)&=\int\limits_0^t\delta\left(t-\tau\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle\text{R}}{\displaystyle\text{R}+\text{sL}+\frac{\displaystyle1}{\displaystyle\text{sC}}}\right]_{\left(\tau\right)}\space\text{d}\tau\\ \\ &=\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle\text{R}}{\displaystyle\text{R}+\text{sL}+\frac{\displaystyle1}{\displaystyle\text{sC}}}\right]_{\left(t\right)} \end{split}\tag1 \end{equation}

  • \$\begingroup\$ Thanks for the response. In this class we've typically solved the inverse Laplace transform using partial fraction expansion and a table with Laplace transforms so I'm a bit confused by that representation. I was mainly trying to verify that I had the correct transfer function, the impulse response should be the inverse Laplace transform of the transfer function correct? Using the impulse function in Octave didn't seem to return the right result when used on the transfer function, which made me suspect I made an error. \$\endgroup\$ Commented Nov 11, 2021 at 19:53

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