0
\$\begingroup\$

What should the transfer function be for the RCL circuit?

enter image description here

When I derive the TF (using complex impedance method) for Output voltage/Input voltage I get the following:

$$\dfrac{Eo(s)}{Ei(s)}=\dfrac{1}{s^2CL+sRC+1} $$ (1)

When I derive the TF (using Laplace transform) for Charge/Input voltage I get the following:

$$\dfrac{Q(s)}{E(s)}=\dfrac{1}{Ls^2+Rs+(1/C)} $$ (2)

Update: I have realised my error. My question revolved around the fact that I didn't recognise that equations (1) and (2) were the same, except for the denominator being divided through by C.

This question is no longer relevant and I would like to delete it! But, I'm not able to!!

\$\endgroup\$
4
  • 5
    \$\begingroup\$ That's kind of like asking "when working on a car, which wrench should I use?" The one that fits and does what you want it to. \$\endgroup\$
    – John D
    Mar 24, 2021 at 16:49
  • \$\begingroup\$ Is there an assumed level of knowledge for asking questions on here, so that one can avoid the ridicule and/or deducted points for asking one? \$\endgroup\$
    – aLoHa
    Mar 24, 2021 at 19:23
  • \$\begingroup\$ Sorry if my comment came off as ridicule, it was not intended that way. It was just an analogy to point out that it's difficult to answer your question without knowing what exactly your requirements and goals are for your analysis. \$\endgroup\$
    – John D
    Mar 24, 2021 at 19:27
  • 2
    \$\begingroup\$ For a bandpass filter transfer function you would typically look at the transfer function of the output voltage to the input voltage. But there could by scenarios where you would want to look at something different. \$\endgroup\$
    – John D
    Mar 24, 2021 at 19:29

1 Answer 1

2
\$\begingroup\$

A filter necessarily processes some sort of signal, so the transfer function that makes the most sense is the one that describes the filter's processing of the signal of interest. If the input and output signals are both voltages (e.g. the filter input is from, say, a voltage amplifier, and the filter output serves as the input to a voltage buffer or amplifier) then the transfer function of interest is

$$H(s) = \frac{E_{\text{out}}(s)}{E_{\text{in}}(s)}$$

The reason is that you will presumably want to determine the transfer function of the entire signal chain, in which case you can most easily combine the filter's transfer function with the transfer functions of the input and output stage(s) if you've determined the filter's transfer function in the form that matches the input and output stages' transfer functions.

Many filters operate on voltage signals so in the absence of any other information about the signals (e.g. you are analyzing the filter in isolation) I would use this transfer function. Note that loads at the input and output will often affect the transfer function regardless of the signal of interest -- e.g. a finite load impedance at the filter's output will reduce \$E_{\text{out}}(s)\$ compared to an infinite load impedance -- so it may not be particularly helpful to analyze the transfer function of a circuit in isolation anyway.

If the input and output signals are both currents then the transfer function

$$H(s) = \frac{I_{\text{out}}(s)}{I_{\text{in}}(s)}$$

would be more useful.

While the signal of interest is often in the same form at both the input and output (e.g. both input and output are voltages) it is possible for the input to be of a different form than the output. For example, if the filter input is connected to the output of a voltage amplifier and the filter output is connected to the input of a transimpedance amplifier (which uses current as its input signal) then the transfer function

$$H(s) = \frac{I_{\text{out}}(s)}{E_{\text{in}}(s)}$$

would make the most sense.

The transfer function

$$H(s) = \frac{Q_{\text{out}}(s)}{E_{\text{in}}(s)}$$

would be useful if the filter input was a voltage signal and the filter output was a charge signal, but a charge \$Q(s)\$ is not a common signal (at least not directly -- perhaps the filter is charging a capacitor, but in that case I'd still expect \$I(s)\$ to be considered the output "signal").

All this applies to any filter (not just a bandpass) or, for that matter, any circuit block: the most useful transfer function is that which matches the input and output signals of interest.

\$\endgroup\$
1
  • \$\begingroup\$ Any chance of a vote for my question? ;) \$\endgroup\$
    – aLoHa
    Feb 13, 2023 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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