Equivalent "capacitative reactance" for calculating rms current under mixed-frequency AC voltage waveform

Question

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:

$$f'(0)<\infty$$

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

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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.

Answer 1

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.

Answer 2

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.