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Proof Using Complex Variables

To show by means of phasor analysis that Eq.$ \,$ (A.2) always has a solution, we can express each component sinusoid as

$\displaystyle A_i\cos(\omega t + \phi_i) =$   re$\displaystyle \left\{A_i e^{j(\omega t + \phi_i)}\right\}
$

Equation (A.2) therefore becomes

\begin{eqnarray*}
\mbox{re}\left\{A e^{j(\omega t + \phi)}\right\} &=& \sum_{i=1}^N \mbox{re}\left\{A_i e^{j(\omega t + \phi_i)}\right\}
= \mbox{re}\left\{\sum_{i=1}^N A_i e^{j(\omega t + \phi_i)}\right\}\\
& = & \mbox{re}\left\{\underbrace{\sum_{i=1}^N \left(A_i e^{j\phi_i}\right)}_{\isdef Ae^{j\phi}} e^{j(\omega t)}\right\}
\isdef \mbox{re}\left\{\left(A e^{j\phi}\right)e^{j(\omega t)}\right\}\\
&=& \mbox{re}\left\{A e^{j(\omega t+\phi)}\right\}.
\end{eqnarray*}

Thus, equality holds when we define

$\displaystyle A e^{j\phi} \isdef \sum_{i=1}^N A_i e^{j\phi_i}. \protect$ (A.5)

Since $ A e^{j\phi}$ is just the polar representation of a complex number, there is always some value of $ A\geq 0$ and $ \phi\in[-\pi,\pi)$ such that $ A e^{j\phi}$ equals whatever complex number results on the right-hand side of Eq.$ \,$ (A.5).

As is often the case, we see that the use of Euler's identity and complex analysis gives a simplified algebraic proof which replaces a proof based on trigonometric identities.


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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA