I want to analyze a very simple circuit subject to a not-so-simple driving AC voltage waveform. In particular, my circuit consists simply of a single capacitor with capacitance \$C\$ and an AC voltage source \$V\$. Now, if \$V\$ were operating at a fixed angular frequency \$\omega\$, then I could calculate the the capacitative reactance \$X_c\$ very simply as:

$$X_c(\omega) = \frac{1}{\omega C}$$

However, what if my voltage source waveform is composed of a mixture of frequencies given by a spectral density function i.e., Fourier transform):

$$f(\omega):\int_{0}^{\infty} f(\omega)d\omega = 1$$

Question:I was wondering if there exists a way to get an "equivalent capacitative reactance" \$X_{c,eqiv}\$ such that:

$$I_{rms} = \frac{V_{rms}}{X_{c,eqiv}} $$


My initial reaction is that \$X_c(\omega)\$ is additive across the frequencies, and so we get the functional :

$$X_{c,equiv}= \int_{0}^{\infty} \frac{f(\omega)}{\omega C}d\omega$$

With the requirement that $$\lim_{t\to 0}\frac{f(t)}{t} < \infty$$

to ensure that the improper integral converges.

If \$f(0)=0\$ then we can use L'Hospital Rule to strengthen this to:


Question: Is this the correct approach to getting \$X_c\$ for mixed-frequency circuits?

Response to Andy aka comment

Andy requested a specific scenario. Below is a example of a setup that I am analyzing:


simulate this circuit – Schematic created using CircuitLab

The voltage source waveform \$V(t)\$ has the following Fourier transform in the frequency domain (\$f\$ in kHz): $$S(f) = \frac{e^{-\frac{1}{2} \log^2(f)}}{\sqrt{2 \pi} f} $$

I will be monitoring the current at the point indicated and calculating the rms value of the resulting current waveform.

That's a pretty typical setup, although the specific values will change, or I may use a different distribution over frequencies.

  • \$\begingroup\$ To clarify your question, if you take the simple case where \$f(\omega)\$ is composed of two discrete sinusoids at, say, \$\omega_1\$ and \$\omega_2\$, that will give two values of \$X_c\$. In what sense would these two reactance values be additive? In other words, what would \$X_c(f)\$ look like? \$\endgroup\$
    – Chu
    Commented Jun 30, 2016 at 7:29
  • \$\begingroup\$ @Chu in that case, \$f(\omega)\$ would have the form \$f(\omega)=\alpha 1(\omega)_{\omega_1} + (1-\alpha) 1(\omega)_{\omega_2}\$ where \$0\leq \alpha \leq 1\$ and the resulting capacitance would be \$X_c(f) = \alpha X_c(\omega_1)+(1-\alpha) X_c(\omega_2)\$ \$\endgroup\$
    – user115412
    Commented Jun 30, 2016 at 11:42
  • \$\begingroup\$ @Chu ah, thanks so much. Yes, I was wondering where additivity came into play. I was hoping to be able to reduce a complex situation like I describe into an "rms-equivalent" circuit driven by a single-frequency AC voltage and some capacitance. From both your and Andy's comments, it looks like this integration has to happen at the actual RMS value itself, and not at the underlying reactance (i.e., maybe no analog of Thevenin's theorem for dc circuits for this situation). \$\endgroup\$
    – user115412
    Commented Jun 30, 2016 at 12:24
  • \$\begingroup\$ @Chu where did you comment go? It was helpful. \$\endgroup\$
    – user115412
    Commented Jun 30, 2016 at 12:25
  • \$\begingroup\$ @Bey Your ammeter will not observe anything if it is not in series with the current you would like to measure. \$\endgroup\$ Commented Jun 30, 2016 at 12:29

2 Answers 2


Since you've now specified a circuit to analyze lets look at the normal method of analysis using transfer functions. The normal method of analysis for this type of circuit is called transfer function analysis, i.e., finding the dependence on the output current, which I will take to be across the capacitor, on the input voltage. That is to say the transfer function is defined by: $$ H(j\omega) = \frac{I_{out}(j\omega)}{V_{in}(j\omega)}~(S) $$ For the circuit provided this gives the voltage -> current transfer function (with units of Siemens) of: $$ I_{out}(j\omega) = \frac{V_{in}(j\omega)}{\frac{1}{j\omega C}}=j\omega C V_{in}(j\omega)\\ H(j\omega) = j\omega C $$ Since you've decided that the input frequency distribution is going to be log-squared distributed: $$ V_{in}(j\omega) = \frac{e^{-\frac{1}{2}\log^2(j\omega)}}{j\omega\sqrt{2\pi}} $$ $$ I_{out}(j\omega) = H(j\omega)V_{in}(j\omega) \\ = \frac{e^{-\frac{1}{2}\log^2(j\omega)}}{j\omega\sqrt{2\pi}}\cdot j\omega C \\ = \frac{Ce^{-\frac{1}{2}\log^2(j\omega)}}{\sqrt{2\pi}} $$ If you would like to see what this response looks like in the time domain you can find the real time domain current by taking the inverse Fourier transform: $$ I_{out}(t) = \frac{1}{2\pi}\int_{-\infty}^\infty I_{out}(j\omega)e^{j\omega t}d\omega $$ The constant affixed to the Fourier transform will depend on which "form" of the transform you are using. Typically in EE we will attach the constant on the inverse transform.

  • \$\begingroup\$ Thank you! I was trying to find out the distribution of the current if the voltage source were white noise (gaussian). So it looks there is no simple "thevenin-equivalent" circuit operating at fixed frequency (which is what I was hypothesizing in my post). \$\endgroup\$
    – user115412
    Commented Jun 30, 2016 at 12:37
  • \$\begingroup\$ Specifically, I was hoping that I could calculate my "equivalent reactance" \$X_c(f)\$ and get the rms current as \$I_{rms}=\frac{V_{rms}}{X_c(f)}\$ \$\endgroup\$
    – user115412
    Commented Jun 30, 2016 at 12:42
  • 1
    \$\begingroup\$ @Bey You can calculate Thevenin equivalent circuits in the frequency domain using the same method, i.e., finding the short circuit current and open circuit voltage between two nodes. Your Thevenin equivalent circuit will just have a complex impedance instead of a real resistance. \$\endgroup\$ Commented Jun 30, 2016 at 12:52
  • \$\begingroup\$ Thanks again! So if I know my rms Voltage, can I calculate an equivalent capacitance as alluded to in my previous comment (note, I've created a new question just about this issue:electronics.stackexchange.com/questions/243674/…) \$\endgroup\$
    – user115412
    Commented Jun 30, 2016 at 12:57

My initial reaction is that Xc(ω) is additive across the frequencies

Take the case of a simple capacitor of 1uF at 1kHz - it has a reactance of 159 ohms. It's as simple as that.

Yes it has a reactance of 15.9 ohms at 10 kHz but nobody goes on to say that the reactance is 159 ohms in parallel (or series) with 15.9 ohms. To do so is to miss the whole point.

  • \$\begingroup\$ Thanks, but I'm interested in the response of my simple circuit to a mixed-frequency signal. So yes, I can talk about its reactance pointwise, but my circuit is responding to the particular mix of frequencies. For example, what is the rms current of the above circuit if it's AC spectral density has, say, a Lognormal(0,1) distribution? \$\endgroup\$
    – user115412
    Commented Jun 30, 2016 at 11:39
  • \$\begingroup\$ You are missing the point when you think you can add reactances at different frequencies and I'm not going to do some hypothetical example to prove the uselessness of your belief that somehow adding reactances is significant in any way. \$\endgroup\$
    – Andy aka
    Commented Jun 30, 2016 at 11:45
  • \$\begingroup\$ Look, I'm trying to understand the response of this simple circuit to a mixed-frequency input. My initial guess is what is stated above. I'm not claiming that this is correct. That is my question, and I would be appreciative if you could point out the correct way to think about this. A concrete example is calculating the resulting rms current of the circuit given the input waveform. \$\endgroup\$
    – user115412
    Commented Jun 30, 2016 at 11:49
  • \$\begingroup\$ I guess a broader (too broad, I think) question is how to analyze circuit behavior if it is driven by something other than a nice sine wave voltage. \$\endgroup\$
    – user115412
    Commented Jun 30, 2016 at 11:51
  • \$\begingroup\$ Draw/describe a circuit and state the input waveform. Then indicate the signal you want to "understand" that has been produced by a reactive component reshaping the input waveform. It's impossible to generalize on a concept that I believe you are missing the point of. \$\endgroup\$
    – Andy aka
    Commented Jun 30, 2016 at 11:54

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