Real and Complex Sinusoids

Many problems in engineering (Power Engineering ring a bell?) involve inputs and outputs oscillating at the same frequency. The problem is to find the amplitude and phase of the resultinng output.

The solution to y' - ay = A \, \cos \omega t + B \, \sin \omega t is y = M \, \cos \omega t + N \, \sin \omega t. We simplify by turning the cosine and sine terms into a complex exponential.

1 Real Solution

We know that y = M \, \cos \omega t + N \, \sin \omega t is the form of the solution, so we can simply input that solution into y' - ay = A \, \cos \omega t + B \, \sin \omega t. Then perform the differentiation and create a system of equations:

\cos \omega t terms: -a\,M + \omega \, N = A \sin \omega t terms: -\omega\,M - a \, N = A

For two equations, you can use the 2x2 identity matrix to find the solution.

\begin{bmatrix} -a & \omega \\ -\omega & -a \end{bmatrix} \begin{bmatrix} M \\ N \end{bmatrix} = \begin{bmatrix} A \\ B \end{bmatrix} \hspace{1em} \textnormal{gives} \hspace{1em} \begin{bmatrix} M \\ N \end{bmatrix} = \frac{1}{\omega^2 + a^2} \begin{bmatrix} -a & -\omega \\ \omega & -a \end{bmatrix} \begin{bmatrix} A \\ B \end{bmatrix}

Therefore,

M = -\frac{aA + \omega B}{\omega^2 + a^2} \hspace{2em} N = \frac{\omega A - aB}{\omega^2 + a^2} \tag{1}

2 Complex Solution

Here the input is q(t) = R\,e^{i\omega t}. It oscillates with frequency \omega rad/sec. The output y(t) will oscillate with the same frequency \omega. This happens when a is constant.

Plug in Y\,e^{i\omega t} to the differential equation y' - ay = R\, e^{\omega t}:

i\omega Y \, e^{i\omega t} - aY \, e^{i \omega t} = R \, e^{i\omega t}

From which:

Y = \frac{R}{i\omega - a} \hspace{1em} \textnormal{and} \hspace{1em} y = Y\,e^{i\omega t}