# Magnitude of Complex voltage vs Real Part

1. Feb 9, 2012

### teroenza

1. The problem statement, all variables and given/known data
I have been asked to plot the "magnitude" of a function for V_out of a circuit. It has an imaginary component. I know I need to plot the physical, real portion. But I do not understand if this means the magnitude (modulus) of this complex function, or just the real part. The function has two terms, one of which is real, the other imaginary. By plotting the "real part" I mean, just plotting the term that does not contain the imaginary unit (j).

In terms of the complex plane. Should I be plotting the magnitude of the line, or just it's projection onto the real axis?

Thank you

2. Relevant equations

a+bi=z
sqrt(a^2+b^2)=Magnitude(z)

3. The attempt at a solution

2. Feb 9, 2012

### cepheid

Staff Emeritus
Most probably they want the real part, not the modulus. The reasoning is as follows. For some AC voltage$$v(t) = V_0 \cos(\omega t + \phi)$$we typically define a complex voltage: $$V(t) = V_0e^{i(\omega t + \phi)}$$such that$$\Re[V(t)] = v(t)$$where $\Re$ denotes the real part. You can verify for yourself that this is true using the Euler equation for complex exponentials. The reason for doing this is because complex exponentials are much easier to work with than sinusoids, and a lot of algebraic manipulations become much simpler. Furthermore, a vector in the complex plane is also a natural way of thinking of an AC signal. To see this, rewrite the equation as$$V(t) = \widetilde{V}e^{i\omega t}$$where$$\widetilde{V} = V_0e^{i\phi}$$So this complex amplitude, which consists of two real numbers (a magnitude and a phase) is just like a sinusoid, which has an amplitude and a phase. Those two numbers together contain all the information you need to know (the amplitude and the phase) to fully represent that AC signal (assuming that the frequency is a given). This complex voltage $\widetilde{V}$ is called a "phasor" (from "phased vector"), and engineers also sometimes use the notation $\widetilde{V} = V_0 \angle \phi$ to represent phasors.

3. Feb 11, 2012

### teroenza

You were correct, that is what they desired. Thank you very much.