Phasors

Section 3.7 Phasors

In this section, we will introduce phasors, a graphical method of representing complex electronics signals. Take a complex signal, e.g.

\begin{equation*} \tilde{V}(t) = \tilde{V}_0 e^{i\omega t} = V_0 e^{i\phi}e^{i\omega t} = V_0 e^{i\left(\omega t + \phi\right)}\text{.} \end{equation*}

For time \(t=0\text{,}\) we can represent this graphically as a vector at an angle \(\phi\) as shown in Figure 3.7.1.(a). This phasor will rotate about a circle with radius \(V_0\) as time advances, pointing at an angle \(\omega t + \phi\) at some later time \(t\) as shown in Figure 3.7.1.(b).

If we have two signals that we would like to compare

\begin{align*} \tilde{V}_\text{out} \amp = V_{\text{out}_0} e^{i\left(\omega t + \phi_2\right)} \\ \tilde{V}_\text{in} \amp = V_{\text{in}_0} e^{i\left(\omega t + \phi_1\right)} \end{align*}

we can see in Figure 3.7.2 that both vectors rotate at the same rate, leaving the phase \(\phi=\phi_2-\phi_1\) between them unchanged as time advances.

Phasors can help us quickly use graphs to determine the amplitudes and phases of signals relative to some other reference signal. As an example, let’s examine the circuit in Figure 3.7.3.(a).

The phasor diagram in Figure 3.7.3.(b) was generated by completing the following steps:

Draw your phasor diagram for the instant when one of your signals (voltage or current) is purely real, meaning its phasor is aligned along the real axis. Choose this signal based on how easily it can be related to other phasors. In this case, we’ll look at an instant when \(\tilde{I}\) is purely real.
Use Ohm’s law to find expressions for voltage changes across components. In this case, we used \(\tilde{V}_C=\tilde{I} Z_C=-(i/\omega C) \tilde{I}\text{.}\)
Plot \(\tilde{V}\) changes across all components. In this example, \(\tilde{I}\) is purely real which means that \(\tilde{V}_C\) is aligned with the negative imaginary axis.

Thus, we find that the phase of the current \(\tilde{I}\) through a capacitor leads the phase of the capacitor voltage \(\tilde{V}_C\) by \(90^\circ\) and the amplitude of the voltage is \(V_{C_0}=I_0/\omega C\text{.}\)

We can proceed with a similar analysis for an inductor. Examining the circuit in Figure 3.7.4.(a).

and generating the phasor diagram in Figure 3.7.4.(b), we find that the phase of the current \(\tilde{I}\) through an inductor trails the phase of the inductor voltage \(\tilde{V}_L\) by \(90^\circ\) and the amplitude of the voltage is \(V_{L_0}=\omega L I_0\text{.}\)

THINGS TO CHECK: WHICH EQUATIONS SHOULD HAVE TILDE VARIABLES AND WHICH SHOULD NOT.

Example 3.7.5. Phasor analysis of the RC High-pass filter.

Let’s revisit the RC high-pass filter circuit in Figure 3.6.1 and analyze the circuit behavior using phasors. Find the voltage gain \(G_V\) and the phase \(\phi=\phi_\text{out}-\phi_\text{in}\text{.}\)

Answer.

\begin{equation*} G_V = \frac{V_{\text{out}_0}}{V_{\text{in}_0}} = \frac{1}{\sqrt{\left(\frac{1}{\omega R C}\right)^2 + R^2}} = \frac{1}{\sqrt{1+\left(f_0/f\right)^2}} \end{equation*}

\begin{equation*} \left| \phi \right| = \tan^{-1} \left(\frac{1}{\omega R C}\right) = \tan^{-1} \left(f_0/f\right) \end{equation*}

\begin{equation*} f_0=1/2\pi R C \end{equation*}

Solution.

To construct the phasor diagram in Figure 3.7.7, we draw \(\tilde{I}\) along the real axis. Then, \(\tilde{V}_\text{out}=\tilde{V}_\text{R}=\tilde{I}R\) is also purely real. We can use Ohm’s law to draw \(\tilde{V}_C=i\tilde{I}/\omega C\) along the negative imaginary axis. Then, since Kirchhoff’s Voltage Law requires \(\tilde{V}_\text{in} - \tilde{V}_C - \tilde{V}_\text{out}=0\text{,}\) we use vector addition to draw \(\tilde{V}_\text{in}=\tilde{V}_C + \tilde{V}_\text{out}\text{.}\)

Based on our phasor diagram, we can already conclude the following:

\(\tilde{V}_\text{out}\) leads \(\tilde{V}_\text{in}\text{.}\)
At high frequencies, \(\tilde{V}_\text{out}\rightarrow \tilde{V}_\text{in}\) and \(\phi\rightarrow 0\) because \(\tilde{V}_C\ll \tilde{V}_\text{out}\text{.}\)
At low frequencies, \(\tilde{V}_\text{out}\rightarrow 0\) and \(\phi\rightarrow 90^\circ\) because \(\tilde{V}_C\gg \tilde{V}_\text{out}\text{.}\)

We can use the phasor diagram to derive more precise relationships. Using the Pythagorean theorem, we find that

\begin{equation*} V_{\text{in}_0} = \sqrt{I_0^2 \left(\frac{1}{\omega C}\right)^2 + I_0^2 R^2} = I_0\sqrt{\left(\frac{1}{\omega C}\right)^2 + R^2}\text{.} \end{equation*}

Also,

\begin{equation*} V_{\text{out}_0} = I_0 R\text{.} \end{equation*}

Combining these results, we find that

\begin{align*} G_V = \frac{V_{\text{out}_0}}{V_{\text{in}_0}} \amp = \frac{R}{\sqrt{\left(\frac{1}{\omega C}\right)^2 + R^2}}= \frac{1}{\sqrt{\left(\frac{1}{\omega R C}\right)^2 + R^2}}\\ \amp = \frac{1}{\sqrt{1+\left(f_0/f\right)^2}} \end{align*}

where \(f_0=1/2\pi R C\) as before. Likewise, we can find the phase between our signals using

\begin{equation*} \left| \phi \right| = \tan^{-1}\left(\frac{V_{C_0}}{V_{R_0}}\right) = \tan^{-1} \left(\frac{1}{\omega R C}\right) = \tan^{-1} \left(f_0/f\right)\text{.} \end{equation*}

These results are identical to what we had calculated previously using Kirchhoff’s laws.

Example 3.7.8. Phasor analysis of a resonant RLC circuit.

Find \(G_V\) and \(\phi=\phi_\text{out}-\phi_\text{in}\) for the resonant RLC circuit shown in Figure 3.7.9.

Answer.

The voltage gain is

\begin{align*} G_v = \frac{V_{\text{out}_0}}{V_{\text{in}_0}} \amp = \frac{R}{\sqrt{R^2 + \left(\omega L -\frac{1}{\omega C}\right)^2}} \\ \amp = \frac{1}{\sqrt{1+\frac{L^2}{\omega^2 R^2}\left(\omega^2 - \omega_0^2\right)^2}} \text{.} \end{align*}

The phase is

\begin{align*} \left|\phi\right| \amp = \tan^{-1}\left(\frac{\omega L - \frac{1}{\omega C}}{R}\right) \\ \amp = \tan^{-1}\left(\frac{L}{\omega R}\left(\omega^2-\omega_0^2\right)\right) \end{align*}

and the resonant frequency is

\begin{equation*} \omega_0 = \frac{1}{\sqrt{LC}} \end{equation*}

Solution.

Based on our phasor diagram, we can already conclude the following:

At high frequencies, \(\tilde{V}_\text{out}\rightarrow 0\) and \(\phi\rightarrow -90^\circ\) because \(\tilde{V}_C\ll \tilde{V}_\text{out}\) and \(\tilde{V}_L \gg \tilde{V}_\text{out}\text{.}\)
At low frequencies, \(\tilde{V}_\text{out}\rightarrow 0\) and \(\phi\rightarrow 90^\circ\) because \(\tilde{V}_C\gg \tilde{V}_\text{out}\) and \(\tilde{V}_L\ll \tilde{V}_\text{out}\text{.}\)
There exists a frequency \(\omega=\omega_0\) where \(\tilde{V}_C + \tilde{V}_L = 0\text{.}\) At this frequency, \(\tilde{V}_\text{out} = \tilde{V}_\text{in}\) and \(\phi = 0^\circ\text{.}\) We call \(\omega_0\) the resonant frequency for this circuit.

We can be more quantitative with results gleaned from the phasor diagram. First, our resonance frequency can be determined by setting \(\tilde{V}_C + \tilde{V}_L=0\text{,}\) giving us

\begin{equation*} \omega_0 L = \frac{1}{\omega_0 C} \end{equation*}

which can be rearranged into

\begin{equation*} \omega_0 = \frac{1}{\sqrt{LC}}\text{.} \end{equation*}

We can also find expressions for voltage gain and phase using trigonometry. Since the loop law gives

\begin{align*} 0 \amp = \tilde{V}_\text{in} - \tilde{V}_L - \tilde{V}_C - \tilde{V}_\text{out}\\ \Rightarrow \tilde{V}_\text{in} \amp = \tilde{V}_L + \tilde{V}_C + \tilde{V}_\text{out}\\ \amp = i\left(\omega L - \frac{1}{\omega C}\right) I_0 + I_0 R \text{,} \end{align*}

we can use Pythagorean’s theorem to find \(V_{\text{in}_0}\)

\begin{align*} V_{\text{in}_0} \amp = \sqrt{\left(\omega L - \frac{1}{\omega C}\right)^2 I_0^2 + I_0^2 R^2}\\ \amp = I_0 \sqrt{\left(\omega L - \frac{1}{\omega C}\right)^2 + R^2}\text{.} \end{align*}

Solving this equation for \(I_0\) in terms of \(V_{\text{in}_0}\) and inserting into \(V_{\text{out}_0}=I_0 R\) gives

The phase is found using

\begin{align*} \left|\phi\right| \amp = \tan^{-1}\left(\frac{\omega L - \frac{1}{\omega C}}{R}\right) \\ \amp = \tan^{-1}\left(\frac{L}{\omega R}\left(\omega^2-\omega_0^2\right)\right) \text{.} \end{align*}

These results are plotted in Figure 3.7.11, demonstrating the same qualitative behaviors that we had deduced at the outset.

Example 3.7.12. Phasor analysis of a parallel RLC circuit.

Examine Figure 3.7.13. Determine the current being drawn from the voltage source and the frequency at which the source current’s magnitude is minimized.

Answer.

\begin{equation*} I_{\text{in}_0} = V_{\text{in}_0} \sqrt{ \frac{1}{R^2} + \left(\omega C - \frac{1}{\omega L}\right)^2} \end{equation*}

\begin{equation*} \phi = \tan^{-1}\left(\frac{1/R^2}{\omega C - 1/\omega L}\right) = \tan^{-1}\left(\frac{1}{R^2\left(\omega C - 1/\omega L\right)}\right) \end{equation*}

Solution.

In order to create the phasor diagram, we draw it at a time at which one convenient current phasor is aligned purely along the real axis. We will choose a time at which \(\tilde{I}_R\) is purely real so that \(\tilde{V}|_\text{in}\) is also purely real. We can then use Ohm’s law to determine currents across the inductor and capacitor and draw them on the phasor diagram because the voltage across each of the three components is \(V_\text{in}\text{.}\)

Now, the junction law can be used to find \(\tilde{I}_\text{in}\text{:}\)

\begin{equation*} \tilde{I}_\text{in} = \tilde{I}_R + \tilde{I}_L + \tilde{I}_C\text{.} \end{equation*}

Using Pythagorean’s theorem,

\begin{align*} I_{in_0}^2 \amp = I_{R_0}^2 + \left(I_{C_0}-I_{L_0}\right)^2\\ \amp = \frac{V_{\text{in}_0}}{R^2}+ \left(\omega C V_{\text{in}_0} - \frac{V_{\text{in}_0}}{\omega L} \right)^2\\ \amp = V_{\text{in}_0}^2\left[\frac{1}{R^2} + \left(\omega C - \frac{1}{\omega L}\right)^2\right] \end{align*}

so that

\begin{equation*} I_{\text{in}_0} = V_{\text{in}_0} \sqrt{ \frac{1}{R^2} + \left(\omega C - \frac{1}{\omega L}\right)^2} \end{equation*}

and

\begin{equation*} \phi = \tan^{-1}\left(\frac{1/R^2}{\omega C - 1/\omega L}\right) = \tan^{-1}\left(\frac{1}{R^2\left(\omega C - 1/\omega L\right)}\right)\text{.} \end{equation*}

In the limits \(\omega\rightarrow 0\) and \(\omega\rightarrow\infty\text{,}\) \(\tilde{I}_\text{in}\rightarrow \infty\) since the inductor or capacitor short out in these limits, respectively. There are intermediate frequencies at which \(\tilde{I}\) is finite. We can find the frequency \(\omega_0\) at which source current is minimized by solving for \(\omega\) when \(\text{d}\tilde{I}_\text{in}/\text{d}V=0\text{.}\) Current is also minimized when \(\text{d}\tilde{I}_\text{in}^2/\text{d}V=0\text{.}\) So, we’ll examine

\begin{equation*} 0 = \frac{\text{d}}{\text{d}V} \tilde{I}_\text{in}^2 = 2\tilde{V}_\text{in}^2\left(\frac{1}{\omega L} - \omega C\right)\left(-\frac{1}{\omega^2 L} - C\right)\text{.} \end{equation*}

Now, it is impossible for the second term in parentheses to equal zero. So, source current is minimized when

\begin{equation*} \frac{1}{\omega_0 L} = \omega_0 C \end{equation*}

\begin{equation*} \omega_0 = \frac{1}{\sqrt{LC}}\text{.} \end{equation*}

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