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      <text>23.6.4.2 Parseval’s theorem in FS coeffs</text>
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      <text><![CDATA[Parseval's theorem can also be written in terms of the Fourier coefficients \(a_n\), \(b_n\) of the
trigonometric Fourier series.
Recall that 
\[
 \begin{array}{lll}
c_0=\dfrac{a_0}{2}\qquad c_n={\frac{a_n- ib_n}{2}}\quad n=1,2,3,\ldots\qquad\qquad c_n={\frac{a_n+ ib_n}{2}}\quad
n=-1,-2,-3,\ldots
\end{array}
\]
so
\[ |c_n|^2=\frac{a_n^2+b_n^2}{4}\qquad\quad n=\pm 1,\pm 2,\pm 3,\ldots \]
so
\[  \sum_{n=-\infty}^\infty |c_n|^2=\frac{a_0^2}{4}+2\sum_{n=1}^\infty \frac{a_n^2+b_n^2}{4} \]
and hence Parseval's theorem becomes
\[ \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f^2(t)dt
=\frac{a_0^2}{4}+\frac{1}{2}\sum_{n=1}^\infty (a_n^2+b_n^2) \]
<p></p>
The engineering interpretation of this theorem is as follows. Suppose \(f(t)\) denotes an electrical
signal (current or voltage), then from elementary circuit theory \(f^2(t)\) is the instantaneous power (in a 1 ohm resistor) so that
\[ \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f^2(t)\,dt \]
is the energy dissipated in the resistor during one period.
<p></p>
Now a sinusoid wave of the form
\[ A\cos \omega t\qquad (\text{ or}\ A\sin\omega t) \]
has a mean square value \(\dfrac{A^2}{2}\) so a purely sinusoidal signal would dissipate a
power \(\dfrac{A^2}{2}\) in a 1 ohm resistor. Hence Parseval's theorem states that the average power dissipated over 1 period equals the sum of the powers of the constant (or d.c.) components and of all the
sinusoidal (or alternating) components.]]></text>
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