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:
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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.