Electrical – How does one relate an RC circuit’s time domain solution to phasors

capacitorcircuit analysiscircuit-theoryphasor

It seems to be a given for phasor analysis that the voltage across a capacitor can be defined as follows:

$$ V_{C}(t) = V_{A}\cos(\omega t + \phi)\tag{1} $$

Where Vc(t) is the capacitor's voltage at time t, Va is the amplitude of the voltage supply's sinusoid, ω is the angular frequency of the signal, and ϕ is the phase offset.

From there, you can go on to define the voltage as a phasor, for use in steady-state analysis. Unfortunately, I'm having trouble proving to myself that this sinusodial behaviour can be assumed of a capacitor's voltage.

Sinusoidal RC circuit

I know that, for a fixed source, the voltage across the capacitor becomes the same voltage as the supply. In the case of a sinusoidal supply, however, it would seem possible to me that the supply would change faster than the voltage across the capacitor could, causing, at the very least, the voltage across the capacitor to have a different amplitude to the source voltage.

I tried to find the mathematical solution for this circuit by equating the current through the resistor to the current "through" the capacitor:

$$ \frac{V_{A}\cos(\omega t + \phi) – V(t)}{R} = C\frac{\textrm{d}V(t)}{\textrm{d}t} $$

Plugging this into Wolfram Alpha yields the following:

$$ V(t) = c_{1}\exp\left(-\frac{t}{RC}\right) + \frac{A}{(RC\omega)^{2} + 1}\Bigl[RC\omega\sin(\omega t + \phi) + \cos(\omega t + \phi)\Bigr] $$

Which seems to imply a non-source amplitude, as well as the composition of two sinusoidal functions rather than one. Is the calculation I've tried to perform correct? If so, how would one relate this solution to (1)?

Best Answer

Your question gets to the assumptions and fundamental of the theory of impedance.

You start off by writing out the differential equation for the system with a cosine drive. Remember that solutions to differential equations consist of the natural response and the forced response. The forced response depends on the external input to the circuit, which in your case is the applied cosine signal.

Looking at your result from Wolfram Alpha we can identify the natural response as $$K_1 e^{-t/RC}$$. We'll ignore this transient portion for now as the assumption for use of impedances is sinusoidal steady state, which implies that we are far away from the transient response.

The forced response is some sum of a sin and cos, which we do see in your expression. The key part that you are missing is that you can express this sum of sine and cosine as a single cosine expression with a scaling coefficient and phase offset.

$$K_2 sin(wt) + K_3 cos(wt) = K_4 cos(wt + \phi)$$

To calculate out K_4 and \phi, you make use of the trigonometric identity for the sine of a sum of angles. [1]

So when a forced sinusoid is applied to the system, the output is an amplitude scaled and phase offset version of the signal. This is what a Bode plot of an LTI system is showing. It gives the amplitude scaling and the phase offset.

The amount of math required for this differential equations and trigonometric identities analysis is cumbersome. Impedances are a shortcut to reach the same answer by making use of Euler's identity, superposition, and the complex plane. This ends up reducing the differential equations to algebraic equations, which are much easier to solve. A detailed proof of impedances is beyond the scope of this answer. Reference [2] has brief derivation of impedance from the differential equations and trig. Most introductory circuits textbooks should also include a derivation of impedance as part of their introduction to impedance.

[1] https://www.myphysicslab.com/springs/trig-identity-en.html

[2] https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/lecture-notes/09_sss.pdf

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