Notes on phasors
What is a phasor?
A phasor is literally just a number in the complex plane. It is used as a convenient shorthand way of representing sinusoidal signals. For e.g. a signal \(v(t) = v_0 \cos(wt+\phi)\) where \(v_0 \in \mathbb{R}\), can be represented by a phasor \( \tilde{v} = v_0 \exp(j\phi)\). Note that the phasor is not a function of time. To go back to the original signal, we simply multiply the phasor by \(\exp(jwt)\) and take the real part, i.e. \(v(t) = Re(\tilde{v}\exp(jwt)) = Re( v_0 \exp(j\phi) \exp(jwt) ) = v_0 \cos(wt+\phi)\).
How does it help?
We take the example of a simple circuit to illustrate its use. Imagine a series R-L circuit connected to an AC source of the form \(v(t) = v_0 \cos(wt+\phi)\), as before, and the problem is to find an expression for the current, \(i(t)\), flowing in the circuit.
-
Let’s solve it the usual way first, without phasors. KVL gives us \(v(t) = R i(t) + L di(t)/dt\), which is a first order inhomogeneous differntial equation. There are many ways of solving this, for e.g. using the integrating factor approach. Rewriting in the standard form, this becomes \( i(t)' + \frac{R}{L} i(t) = \frac{v_0}{L} \cos(wt+\phi)\), and the integrating factor (IF) is \(I(t)=\exp(\int \frac{R}{L} dt)=\exp(\frac{R}{L}t)\). By multiplying the IF on both sides we get: \( \exp(\frac{R}{L}t) (i(t)' + \frac{R}{L} i(t)) = \frac{v_0}{L} \cos(wt+\phi) \exp(\frac{R}{L}t)\). Recognizing the LHS as a single differential, \( \frac{d}{dt} \{\exp(\frac{R}{L}t) i(t)\} = \frac{v_0}{L} \cos(wt+\phi) \exp(\frac{R}{L}t)\), gives us a solution as: \( \exp(\frac{R}{L}t) i(t) = \int\frac{v_0}{L} \cos(wt+\phi) \exp(\frac{R}{L}t) + K\), where \(K\) is some constant.
This simplifies to \(i(t) = \frac{v_0}{\sqrt{R^2+(wL)^2}} \cos(wt+\phi-\theta) + k\exp(-\frac{R}{L}t) \), where \(\tan(\theta) = \frac{wL}{R}\), and the constant \(k\) is determined from initial conditions.
-
Let’s now solve this using phasors. We start by converting the voltage \(v(t)\) into its phasor form, i.e. \( \tilde{v} = v_0 \exp(j\phi)\). Next we write the KVL equation for the circuit, but in complex impedance form using the phasors. For those unfamiliar with complex impedances, the idea is to use the theory of Fourier transforms (FT) to simplify current-voltage relations. For e.g. we know that for a capacitor, the time domain relation is \( i = C dv/dt\). Taking a FT gives us \(\tilde{i} = jwC \tilde{v}\). Similarly for an inductor, we have \( v = L di/dt\), which translates to \( \tilde{v} = jwL \tilde{i}\). Note that these current-voltage relationships are now linear and thus easier to handle.
Getting back to our R-L circuit, we get \( \tilde{v} = \tilde{i} R + \tilde{i} (jwL)\), which gives \(\tilde{i} = \tilde{v}/(R+jwL)\). Recalling the procedure to go to the original signal from the phasor, we write \( i(t) = Re(\tilde{i}\exp(jwt))\). Substituting gives us \(i(t) = Re[v_0\exp(j\phi)\frac{1}{\sqrt{R^2+(wL)^2}}\exp(-j\theta)\exp(jwt)] = \frac{v_0}{\sqrt{R^2+(wL)^2}} \cos(wt+\phi-\theta) \), where \(\theta\) is as before.
What can we conclude?
-
The two approaches gave identical solutions, except that the phasor approach missed a transient current decay term of the \(\exp(-\frac{R}{L}t)\) form. This is a limitation of the phasor approach. The phasor approach is like a `lazy’ Fourier transform approach, since it misses the transients. If you solved the circuit properly via Fourier transforms you will not miss the transients.
-
However, if we were only interested in the steady state solution, the phasor approach is much simpler and also faster in giving us the correct solution!
-
As the circuits get more complicated, the advantages of the phasor approach become more and more apparent.
Back to the course page