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A Sum of Sinusoids at the
Same Frequency is Another
Sinusoid at that Frequency

It is an important and fundamental fact that a sum of sinusoids at the same frequency, but different phase and amplitude, can always be expressed as a single sinusoid at that frequency with some resultant phase and amplitude. An important implication, for example, is that

$\textstyle \parbox{0.8\textwidth}{sinusoids are eigenfunctions of linear time-invariant
(LTI) systems.}$
That is, if a sinusoid is input to an LTI system, the output will be a sinusoid at the same frequency, but possibly altered in amplitude and phase. This follows because the output of every LTI system can be expressed as a linear combination of delayed copies of the input signal. In this section, we derive this important result for the general case of $ N$ sinusoids at the same frequency.

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

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

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$. The results are

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

which has a unique solution for any values of $ A_i$ and $ \phi_i$.

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

\mbox{re}\left\{A e^{j(\omega t + \phi)}\right\} &=& \sum_{i=1...
&=& \mbox{re}\left\{A e^{j(\omega t+\phi)}\right\}.

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.

Phasor Analysis: Factoring a Complex Sinusoid into Phasor Times Carrier

The heart of the preceding proof was the algebraic manipulation

$\displaystyle \sum_{i=1}^N A_i e^{j(\omega t + \phi_i)} = e^{j\omega t} \sum_{i=1}^N A_i e^{j\phi_i}.

The carrier term $ e^{j\omega t}$ ``factors out'' of the sum. Inside the sum, each sinusoid is represented by a complex constant $ A_i e^{j\phi_i}$, known as the phasor associated with that sinusoid.

For an arbitrary sinusoid having amplitude $ A$, phase $ \phi$, and radian frequency $ \omega$, we have

$\displaystyle A\cos(\omega t + \phi) =$   re$\displaystyle \left\{(A e^{j\phi}) e^{j\omega t}\right\}.

Thus, a sinusoid is determined by its frequency $ \omega$ (which specifies the carrier term) and its phasor $ {\cal A}\isdef A e^{j\phi}$, a complex constant. Phasor analysis is discussed further in [84].

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