Harmonic Addition Theorem
To convert an equation of the form
\begin{equation}
f(\theta)=a\cos\theta+b\sin\theta
\end{equation}
to the form
\begin{equation}
f(\theta)=c\cos(\theta+\delta),
\end{equation}
This can be done by expanding (2) using the trigonometric addition formulas to obtain
\begin{equation}
f(\theta) = c\cos\theta\cos\delta-c\sin\theta\sin\delta.
\end{equation}
Now equate the coefficients of (1) and (3)
\begin{eqnarray}
a &=& c\cos\delta\\
b &=& -c\sin\delta,
\end{eqnarray}so
\begin{equation}
\tan\delta = - {b\over a}
\end{equation}and
\begin{equation}
a^2+b^2 = c^2,
\end{equation}giving
\begin{eqnarray}
\delta &=& \tan^{-1}\left({- {b\over a}}\right)\\
c &=& \sqrt{a^2+b^2}.
\end{eqnarray}
Given two general sinusoidal functions with frequency $\omega$:
\begin{eqnarray}
\psi_1 &=& A_1\sin(\omega t+\delta_1)\\
\psi_2 &=& A_2\sin(\omega t+\delta_2),
\end{eqnarray}
their sum $\psi$ can be expressed as a sinusoidal function with frequency $\omega$
\begin{eqnarray}
\psi &\equiv& \psi_1+\psi_2= A_1[\sin(\omega t)\cos\delta_1+\sin\delta_1\cos(\omega t)]+A_2[\sin(\omega t)\cos\delta_2+\sin\delta_2\cos(\omega t)]\nonumber\\
&=& [A_1\cos\delta_1+A_2\cos\delta_2]\sin(\omega t)+[A_1\sin\delta_1+A_2\sin\delta_2]\cos(\omega t).
\end{eqnarray}
Now, define
\begin{equation}
A\cos\delta \equiv A_1\cos\delta_1+A_2\cos\delta_2
\end{equation}
\begin{equation}
A\sin\delta \equiv A_1\sin\delta_1+A_2\sin\delta_2.
\end{equation}
Then (12) becomes
\begin{equation}
A\cos\delta \sin(\omega t)+A\sin\delta\cos(\omega t) = A\sin(\omega t+\delta).
\end{equation}
Square and add (13) and (14)
\begin{equation}
A^2 = {A_1}^2+{A_2}^2+2A_1A_2\cos(\delta_2-\delta_1).
\end{equation}
\begin{equation}
\tan\delta = {A_1\sin\delta_1+A_2\sin\delta_2 \over A_1\cos\delta_1+A_2\cos\delta_2},
\end{equation}so
\begin{equation}
\psi = A\sin(\omega t+\delta),
\end{equation}where $A$ and $\delta$ are defined by (16) and (17).
This procedure can be generalized to a sum of $n$ harmonic waves, giving
\begin{equation}
\psi = \sum_{i=1}^n A_i\cos (\omega t+\delta_i)= A\cos (\omega t+\delta),
\end{equation}
where
\begin{eqnarray}
A^2 &\equiv& \sum_{i=1}^n\sum_{j=1}^n A_iA_j\cos(\delta_i-\delta_j)\\
&=& \sum_{i=1}^n {A_i}^2 + 2\sum_{i=1}^n \sum_{j>i}^n A_iA_j\cos(\delta_i-\delta_j)
\end{eqnarray}
and
\begin{equation}
\tan\delta = {\sum_{i=1}^n A_i\sin \delta_i\over \sum_{i=1}^n A_i\cos \delta_i}.
\end{equation}MathWorld | 
icesheep
发表于 2014-7-13 19:52
hbghlyj
发表于 2023-5-6 03:40
