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Proof Using Trigonometry

We want to show it is always possible to solve

$\displaystyle A\cos(\omega t + \phi) = A_1\cos(\omega t + \phi_1) + A_2\cos(\omega t + \phi_2) + \cdots + A_N\cos(\omega t + \phi_N) \protect$ (A.2)

for $ A$ and $ \phi$ , given $ A_i, \phi_i$ for $ i=1,\ldots,N$ . For each component sinusoid, we can write
$\displaystyle A_i\cos(\omega t + \phi_i)$ $\displaystyle =$ $\displaystyle A_i\cos(\omega t)\cos(\phi_i) - A_i\sin(\omega t)\sin(\phi_i)$  
  $\displaystyle =$ $\displaystyle \left[A_i\cos(\phi_i)\right]\cos(\omega t) - \left[A_i\sin(\phi_i)\right]\sin(\omega t)$ (A.3)

Applying this expansion to Eq.(A.2) yields

\begin{eqnarray*} \left[A\cos(\phi)\right]\cos(\omega t) &-&\left[A\sin(\phi)\right]\sin(\omega t)\\ &=& \left[\sum_{i=1}^N A_i\cos(\phi_i)\right]\cos(\omega t) - \left[\sum_{i=1}^N A_i\sin(\phi_i)\right]\sin(\omega t). \end{eqnarray*}

Equating coefficients gives

$\displaystyle A\cos(\phi)$ $\displaystyle =$ $\displaystyle \sum_{i=1}^N A_i\cos(\phi_i) \isdefs x$  
$\displaystyle A\sin(\phi)$ $\displaystyle =$ $\displaystyle \sum_{i=1}^N A_i\sin(\phi_i) \isdefs y. \protect$ (A.4)

where $ x$ and $ y$ are known. We now have two equations in two unknowns which are readily solved by (1) squaring and adding both sides to eliminate $ \phi$ , and (2) forming a ratio of both sides of Eq.(A.4) to eliminate $ A$ . This gives

\begin{eqnarray*} A &=& \pm\sqrt{x^2+y^2}\\ \phi &=& \tan^{-1}\left(\frac{y}{x}\right) \end{eqnarray*}

for any values of $ A_i$ and $ \phi_i$ . Since $ \tan^{-1}(y/x) = \tan^{-1}(-y/-x)$ , we have $ \phi\in[-\pi/2,\pi/2)$ . To impose $ A\ge 0$ and $ \phi\in[-\pi,\pi)$ , a four-quadrant arctangent$ (y,x)$ must be used, normally written atan2(y,x) in computer languages.

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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition)
Copyright © 2024-09-03 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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