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Capacitors block DC signals and pass AC signals. Their resistance, ideally, is purely reactive (no real part to its impedance). For a sine wave of frequency f passing through a capacitor, the reactance is given by the equation:

\$X_C\$ =1/( j \$\omega\$C) = -j/(\$\omega\$C)

|\$X_C\$| =1/(2*\$\pi\$*f*c)

\$\omega\$=angular frequency (rad/s)

C=capacitance(Farads)

f=frequency (Hz)

Similarly, how to calculate the reactance for a capacitor when a square wave is passing through it. What is the formula?

Shabab
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nrnw
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  • Don't forget, \$X_C\$ is real,i.e., the impedance of a capacitor is \$Z_C = -jX_C \$ or \$X_C = \frac{1}{\omega C} \$ – Alfred Centauri Oct 05 '13 at 12:20
  • aa... so i have to calculate reactance for first harmonic then for second harmonic and add these two readctances right? then calculate for third harmonic and add that to previously added reactances. but i have nearly infinite harmonics inside the square wave each of them having different amplitudes so.. i guess calculating for 1st harmonic, 3rd harmonic and 5th harmonic may give me more or less accurate approximation of the reactance in square wave conditions right? is that the correct approach? –  Jul 24 '14 at 07:12

3 Answers3

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Similarly, how to calculate the reactance for a capacitor when a square wave is passing through it. What is the formula?

There isn't a capacitive reactance associated with a square wave. The very concept of reactance depends on the context of sinusoidal excitation.

When we solve AC circuits in the phasor domain, it is taken for granted that the circuit is in sinusoidal steady state, i.e., all sources are sinusoidal of the same frequency and all transients have decayed away.

That fact is this: one cannot meaningfully sum phasors or reactances for sinusoids of different frequencies.

Now, that doesn't mean that you can't apply the concept of reactance to find the capacitor voltage across for a square wave current through.

Since (ideal) capacitors are linear, we can decompose the square wave into sinusoidal components, find the associated sinusoidal voltage for each component, and then sum to voltage components to find the total voltage.

Recall the fundamental phasor domain relationship for capacitor voltage and current:

$$\vec V_c = \dfrac{1}{j \omega C}\vec I_c $$

where \$\omega\$ is the angular frequency of the associated sinusoid.

Now, let

$$i_C(t) = a_1 \cos(\omega t + \phi_1) + a_2 \cos(2\omega t + \phi_2) + a_3 \cos(3\omega t + \phi_3) + ...$$

For each sinusoidal component, there is an associated phasor. For example, for the first component, the associated phasor is

$$\vec I_{c_1} = a_1 e^{j\phi_1}$$

Thus

$$\vec V_{c_1} = \dfrac{a_1 e^{j\phi_1}}{j \omega C}$$

so that

$$v_{C_1}(t) = \dfrac{a_1}{\omega C}\cos(\omega t +\phi_1 - \frac{\pi}{2}) $$

Repeat for each term in the series and then sum to find the total capacitor voltage.

Note that we have not defined a reactance to the entire current waveform nor can we define such a thing. Instead, we

(1) found the reactance to each sinusoidal component

(2) converted each resulting phasor voltage back into the time domain

(3) summed the individual time domain voltage components

Alfred Centauri
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Maybe you've already come across the answer and you're looking for a simpler formula, but I'll toss in my two cents. The reactance for the capacitor is expressed in terms of a continuous sinusoid and its capacitance or \$ |X_{C}| = \frac {1}{2 \pi f C}\$, as previously stated. However, a square wave being passed through a capacitor is not continuous.

Let's consider the square wave as a Fourier series of continuous sinusoids. Since I'm lazy and don't want to go into all the details at 2:00AM local time, I'll just copy MathWorld by _Wolfram's answer. This states that a square wave of amplitude 1 and offset of zero can be expressed as some function \$ f(t) = \frac{4}{\pi} \sum\limits_{n = odd}^\infty \frac{1}{n} sin(\frac{n \pi t}{T}) \$ where T is the period. So your reactance should be for all frequencies \$ \omega = \frac{n \pi}{T}\$. So you should have n different reactances for n different signals (for some very large n). Once again, \$ X_{C} = \infty \$ for low frequencies and \$X_{C} = 0\$ for higher frequencies, or the higher harmonics. This is why you'll see capacitors as coupling elements in DC power sub-circuits. They actually serve to filter out the noise/transience so you can have a pure DC signal output.

In summation, the reactance is high for low frequencies and negligible at high frequencies.

The real question is where is the capacitor in your circuit and what purpose does it serve? Then you might get a clearer answer.

On a final note, I would like to apologize for bouncing around with my logic. Please let me know in the comments if I can clean this answer up a bit.

Shabab
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  • "So reactance would be for all signals and therefore be a sum of all reactances." This is false, one cannot sum the reactance to sinusoids of differing frequencies to find an "overall" reactance. – Alfred Centauri Oct 05 '13 at 11:49
  • @AlfredCentauri Thanks for the catch! I made the edit. – Shabab Oct 05 '13 at 14:57
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A square wave has a rise time and fall time that is zero and the current taken at those instances is infinite. Reactance is therefore fairly meaningless. For a resistor, it's resistance is constant at any frequency or \$\dfrac {dv}{dt}\$ but not so for a capacitor

Andy aka
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