I have an analog circuit containing unknown resistance, capacitance, and inductance. I have measured its complex impedance across the frequency range of interest. Is there an equation that will allow me to compute the frequency response (complex voltage gain as a function of frequency) of the circuit given the impedance? Clearly lower impedance magnitude would result in greater voltage gain magnitude, but what is the exact mathematical relationship?

As an aside, I am aware that I could directly measure the frequency response e.g. by taking the Fourier transform of its impulse response. This question is not about how to measure the frequency response, it is about the mathematical relationship between the impedance and frequency response.

  • 1
    $\begingroup$ Plot your impedance, gain and phase at various frequencies, you'll have your frequency response $\endgroup$
    – Ben
    Apr 15 '18 at 6:50
  • $\begingroup$ Impedance measured where in the circuit? What is the output voltage measured across? You need to specify these things in the question for us to be able to give a meaningful answer. Even if the resistor, capacitor, inductor component values are left unspecified, does the circuit have a specific topology with a circuit diagram you can include with just R,L,C elements in it? $\endgroup$
    – Robert L.
    Oct 12 '18 at 22:39

this is more about electronics, but i think it can be asked here.

when you are looking for a frequency response of an impedance, i believe this means you have a two-terminal (a.k.a. one-port) device. so, instead of an input voltage and output voltage to relate by a transfer function (and from the transfer function you get a frequency response), what you have is a voltage applied to the two terminals and a current measured, or a current shoved into (and out of) the terminals and the resulting voltage measured.

the complex impedance of the two-terminal device, as a function of frequency, is the frequency response the device assuming current is the input and voltage is the output. you can plot it on a dB scale, but without a reference resistance, you don't have a reference for 0 dB on that frequency response plot. but you have relative dB for the different impedances at different frequencies.

maybe ask the EE guys this question. i dunno.


Depends on how you define transfer function.

A transfer function is a property of a "system" which means you need to have an input signal and an output signal. If you define "input = voltage", "output = current", you can indeed calculate the transfer function: it's simply the admittance (inverse of the impedance).

However "voltage gain" is more difficult. A circuit that consists only of a singel impedance only has one voltage.

In order to get an input and an output voltage you need to embed the impedance into a larger circuit and the details of that circuit matter. Typically would attach a voltage source and a "load". Both source and load impedance will impact the voltage gain of the overall circuit


To have a frequency response (as you have defined it, voltage gain as a function of frequency), you need an input voltage $x(t)$ and an output voltage $y(t)$. Right now all you have is an input voltage.

If $X(f)$ denotes the Fourier transform of the input and $Y(f)$ denotes the Fourier transform of the output, then the frequency response $H(f)$ is given by: $$ H(f) = \frac{Y(f)}{X(f)} $$ This formulation also assumes that the source impedance of the input signal generator is zero and the load impedance is infinite. This is equivalent to having a voltage buffer at both the input and output of the system $H$.

An example of a circuit with both inputs and outputs is a low-pass filter: lowpass filter Note that the output is defined to be the voltage across the capacitor. This is not an arbitrary choice, but rather, an engineering design decision based on the fact that it produces the desired result (filtering out high frequency noise).

To draw a connection between impedance and voltage gain, you need to specify what the output voltage is defined as being.


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