A square-wave signal of amplitude and period takes the value for and for

PHYSICS · Multi-part · Fermi University Physics · Oscillations, Waves & Acoustics · Coupled oscillators, normal modes and Fourier analysis · Coupled oscillators, normal modes and Fourier analysis

Modes, Packets and Media, Ch The continuum limit, Fourier series and the spectrum of a measured signal

Problem

A square-wave signal f(t)f(t) of amplitude V0=2.00 VV_0 = 2.00\,\mathrm V and period T=1.00 msT=1.00\,\mathrm{ms} takes the value +V0+V_0 for 0<t<T/20 \lt t \lt T/2 and −V0-V_0 for T/2<t<TT/2 \lt t \lt T, repeating indefinitely. Its Fourier series is f(t)=4V0π∑n=1,3,5,…∞1nsin⁡(2πntT).f(t) = \frac{4V_0}{\pi}\sum_{n=1,3,5,\ldots}^{\infty}\frac{1}{n}\sin\left(\frac{2\pi n t}{T}\right). ((a)) Compute the mean square ⟨f2⟩=1T∫0Tf(t)2 dt\langle f^2\rangle = \frac1T\int_0^T f(t)^2\,\mathrm dt directly from the two constant values ff takes. ((b)) For a signal written as a sine series f(t)=∑nbnsin⁡(2πnt/T)f(t)=\sum_n b_n\sin(2\pi n t/T), Parseval's theorem for a periodic signal reads 1T∫0Tf2 dt=12∑nbn2\frac1T\int_0^T f^2\,\mathrm dt=\frac12\sum_n b_n^2. Using this and the standard sum ∑n odd1/n2=π2/8\sum_{n\ \mathrm{odd}}1/n^2=\pi^2/8, show that the Fourier series reproduces the mean square found in part (a), and state the fraction of the total mean square carried by the fundamental alone.

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