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?

##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.

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?

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.

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 circuit driven by a single frequency at some capacitance that will give the same rms current as would an analysis of the actual circuit under the complex waveform.

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

$$X_c(f)= \int_{0}^{\infty} \frac{f(\omega)}{\omega C}d\omega$$

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?

##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.

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?

##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.

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 circuit driven by a single frequency at some capacitance that will give the same rms current as would an analysis of the actual circuit under the complex waveform.

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

$$X_c(f)= \int_{0}^{\infty} \frac{f(\omega)}{\omega C}d\omega$$

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?

##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.

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 circuit driven by a single frequency at some capacitance that will give the same rms current as would an analysis of the actual circuit under the complex waveform.

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

$$X_c(f)= \int_{0}^{\infty} \frac{f(\omega)}{\omega C}d\omega$$

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?

##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.