Question 2.7. Show that for all periodic physical signals that have finite power, the coefficients of the Fourier series expansion $ x_n $ tends to 0 as $ n \to \infty $
I have computed $ |x_n|^2 $ by multiplying $ x_n $ with $ x_n^*$. However, as per my calculations, $ |x_n|^2 = P_x $ i.e. power of $ x(t) $, which is a constant and independent of $n$. I know that this is fundamentally incorrect because $ |x_n| $ should vary with $n$. If anyone can guide me about where I went wrong that would be very helpful. Thanks and regards.
MY SOLUTION:
Please refer to the solution from the half of the first page.
Answer
A finite power periodic signal will have the following property:
$$ P_x = \frac{1}{T_0} \int_{0}^{T_0} |x(t)|^2 dt < \infty \tag{1} $$
where $P_x$ is the power averaged over a period of the signal. Then from Parseval's theorem, the total power can also be shown to be (for a real signal I'm assuming):
$$P_x = |a_0|^2+2 \sum_{n=1}^{\infty} |a_n|^2 < \infty \tag{2}$$
where the $a_n$'s are the continuous-time Fourier series coefficients for $x(t)$.
From the theory of power series (calculus), it's known that in order for (2) to converge (sum being less than infinity) a necessary condition is
$$ \lim_{n \to \infty} |a_n|^2 = 0 \tag{3}$$
(3) implies therefore that, for finite power periodic signal's CTFS coefficients $a_n$ also goes to zero as $n$ goes to infinity.
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