RLC bandpass filter transfer function

Question

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.

Answer 1

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

$$\text{V}_\text{o}\left(t\right)=\int_0^t\delta\left(t-\tau\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{R}}{\text{R}+\text{sL}+\frac{1}{\text{sC}}}\right]_{\left(\tau\right)}\space\text{d}\tau\tag1$$

Using your values:

$$\text{V}_\text{o}\left(t\right)=\int_0^t\delta\left(t-\tau\right)\cdot$$ $$\frac{2200\exp\left(-1100\tau\right)\left(\sqrt{9879}\cos\left(100\sqrt{9879}\tau\right)-11\sin\left(100\sqrt{9879}\tau\right)\right)}{\sqrt{9879}}\space\text{d}\tau\tag1$$

Answer 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.

subplot(1,2,1)
step(H)
subplot(1,2,2)
impulse(H)

I get the following responses: -

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)