AC Analysis
Table of Contents
1. Phasors
We know that Euler’s identity is \(e^{j\theta} = \cos\theta + j\sin\theta\). Oftentimes, we wish to analyze signals of the form \(A\cos(\omega t + \phi)\). This is just the real part of \(Ae^{j(\omega t + \phi)}\):
\begin{align} \text{Re}\{ Ae^{j(\omega t + \phi)}\} &= \text{Re}\{A\cos(\omega t + \phi) + Aj\sin(\omega t + \phi) \notag \\ \text{Re}\{ Ae^{j\omega t}e^{j\omega}\} &= A\cos(\omega t + \phi) \notag \end{align}Then, \(A\) is the amplitude, \(\phi\) gives the phase, and \(e^{j\omega t}\) provides the time component. The phasor is defined as \(Ae^{j\phi}\), or just:
\begin{align} \boxed{A\angle\phi = Ae^{j\phi}} \end{align}When we do phasor analysis, we often use it with signals that have the same frequency, so we can leave that part out. Additionally, we often use the notation \(\tilde{X}\) to represent that \(X\) is in the phasor domain.
2. Impedance
In the phasor domain, the impedance \(\tilde{Z}\) of a circuit element is defined as the ratio of the phasor voltage across it to the phasor current entering through its positive terminal:
\begin{align} \boxed{\tilde{Z} = \frac{\tilde{V}}{\tilde{I}}} \end{align}For resistors, impedance is the same as resistance. However, resistance can only be used for DC, while impedance can be extended to AC analysis.
Similar to how conductance is the inverse of resistance, we define admittance as the inverse of impedance:
\begin{align} \boxed{\tilde{Y} = \frac{\tilde{I}}{\tilde{V}} = \frac{1}{\tilde{Z}}} \end{align}Additionally, the same circuit transformations that we used for resistance can also be used for impedance.
2.1. Impedance of Resistors
By Ohm’s Law, we know that for a resistor:
\begin{align} v_R = Ri_r \notag \end{align}If \(i_R\) is a sinusoidal functoin of time, then the same is true for \(v_R\). We can relate the time domain to the phasor domain as follows:
\begin{align} v_R &= \text{Re}\{\tilde{V_R}e^{j\omega t}\} \notag \\ i_R &= \text{Re}\{\tilde{I_R}e^{j\omega t}\} \notag \end{align}Substituting, we get:
\begin{align} \text{Re}\{\tilde{V_R}e^{j\omega t}\} &= R \cdot \text{Re}\{\tilde{I_R}e^{j\omega t}\} \notag \\ \text{Re}\{\tilde{V_R}e^{j\omega t}\} - \text{Re}\{R\tilde{I_R}e^{j\omega t}\} &= 0 \notag \\ \text{Re}\{\left(\tilde{V_R}-R\tilde{I_R}\right)e^{j\omega t}\} &= 0 \notag \end{align}Since \(e^{j\omega t}\) is usually nonzero, we know that \(\tilde{V_R} - R\tilde{I_R}\) must be zero:
\begin{align} \tilde{V_R}-R\tilde{I_R} &= 0 \notag \\ \tilde{V_R} &= R\tilde{I_R} \notag \end{align}We see that in the phasor domain, Ohm’s Law holds. Then, the impedance is just:
\begin{align} \boxed{\tilde{Z_R} = R} \end{align}2.2. Impedance of Capacitors
We know that for capacitors:
\begin{align} i_C = C\frac{\text{d}v_c}{\text{d}t} \notag \end{align}Then, going to the phasor domain:
\begin{align} \text{Re}\{\tilde{I}_ce^{j\omega t}\} &= C\frac{d}{dt}\text{Re}\{\tilde{V}_ce^{j\omega t}\} \notag \\ \text{Re}\{\tilde{I}_ce^{j\omega t}\} &= \text{Re}\left\{C\tilde{V}_c\frac{d}{dt}e^{j\omega t}\right\} \notag \\ \text{Re}\{\tilde{I}_ce^{j\omega t}\} &= \text{Re}\{j\omega C\tilde{V}_ce^{j\omega t}\} \notag \\ \text{Re}\left\{\left(\tilde{I}_c-j\omega C\tilde{V}_c\right)e^{j\omega t}\right\} &= 0 \notag \end{align}Therefore, we see that:
\begin{align} \tilde{I_C} = j\omega C\tilde{V_C} \notag \end{align}Then, the impedance of a capacitor is:
\begin{align} \boxed{\tilde{Z}_C = \frac{1}{j\omega C}} \end{align}Realize that if we have DC, then \(\omega =0\) and \(\tilde{Z_C}\) goes to infinity. In other words, the capacitor blocks DC current. However if we have AC, then when \(\omega = \infty\), \(\tilde{Z_C}\) goes to zero, so the capacitor lets AC current through.
2.3. Impedance of Inductors
For inductors, we know that:
\begin{align} V_L = L\frac{\text{d}i}{\text{d}t} \notag \end{align}Converting to the phasor domain, we have:
\begin{align} \text{Re}\{\tilde{V}_Le^{j\omega t}\} &= L\frac{\text{d}}{\text{d}t}\text{Re}\{\tilde{I}_Le^{j\omega t}\} \notag \\ &= \text{Re}\{\frac{\text{d}}{\text{d}t}L\tilde{I}_Le^{j\omega t}\} \notag \\ &= \text{Re}\{j\omega L\tilde{I}_Le^{j\omega t}\} \notag \end{align}Rearranging, we get:
\begin{align} \text{Re}\{e^{j\omega t}\left(\tilde{V}_L-j\omega L\tilde{I}_L\right)\} = 0 \notag \end{align}Since \(e^{j\omega t}\) is nonzero, we have:
\begin{align} \tilde{V}_L &= j\omega L\tilde{I}_L \notag \\ \frac{\tilde{V}_L}{\tilde{I}_L} &= j\omega L \notag \end{align}Thus, the impedance of an inductor is:
\begin{align} \boxed{\tilde{Z}_L = j\omega L} \end{align}3. Phasor Domain Analysis
It turns out that KVL and KCL both work in the phasor domain:
\begin{align} \tilde{V_1} + \tilde{V_2} + \cdots + \tilde{V_n} &= 0 \\ \tilde{I_1} + \tilde{I_2} + \cdots + \tilde{I_n} &= 0 \end{align}This implies that all other techniques based on KVL and KCL are also valid. In general, we can do phasor domain analysis by following these steps:
- Adopt a cosine reference.
- Transform to the phasor domain.
- Cast equations in phasor form.
- Solve for unknown variables in the phasor domain.
- Transform solutions back to the time domain.
Example: AC analysis
Consider the following circuit, where \(v_s(t) = 12\sin(\omega t - 45^{\circ})\):
Turning the voltage into cosine, we have \(v_s(t) = 12\cos(\omega t - 135^{\circ})\). Then, using KVL in phasor form, we have:
\begin{align} \tilde{V}_S - \tilde{Z}_R\tilde{I} - \tilde{Z}_C\tilde{I} &= 0 \notag \\ \tilde{V}_S - R\tilde{I} - \frac{1}{j\omega C}\tilde{I} &= 0 \notag \\ \left(R + \frac{1}{j\omega C}\right)\tilde{I} &= \tilde{V}_S \notag \end{align}Then, isolating \(I\):
\begin{align} \tilde{I} &= \frac{\tilde{V}_S}{R+ \frac{1}{j\omega C}} \notag \\ \tilde{I} &= \frac{\tilde{V}_Sj\omega C}{1+j\omega RC} \notag \end{align}Substituting, we get:
\begin{align} \tilde{I} = 6e^{-j105^{\circ}} \notag \end{align}Thus, the real part and our answer is:
\begin{align} \boxed{6\cos(\omega t - 105^{\circ})} \notag \end{align}