Fourier Series — Question 3

Question 3 – Solution

Strategy. Rectification combines an odd fundamental with an even function of half the period.

Step 1: Derive the expansion. Since $h=(\sin x+|\sin x|)/2$, its mean is $1/\pi$. Direct integration on $(0,\pi)$ gives $b_1=1/2$ and $b_n=0$ for $n\ne 1$. For $k\geq 1$, product-to-sum integration gives $$\frac 1\pi\int_0^\pi\sin x\cos(2kx)\,dx =-\frac 2{\pi(4k^2-1)}.$$ Odd cosine coefficients vanish, including $a_1$. Thus $$\boxed{h(x)=\frac 1\pi+\frac 12\sin x -\frac 2\pi\sum_{k=1}^{\infty}\frac{\cos(2kx)}{4k^2-1}.}$$ The coefficients are absolutely summable; the standard piecewise smooth Fourier theorem identifies the sum with this continuous target, including its corners.

Step 2: Explain the selection rules. The function $|\sin x|$ is even, so contributes only cosines. It is also $\pi$-periodic, eliminating odd cosine frequencies in a $2\pi$ expansion. The remaining odd part is exactly $\sin x/2$, so it contributes only $b_1$. The full rectified waveform is neither even nor odd.

Step 3: Account for the average power. Direct integration gives $(2\pi)^{-1}\int h^2=1/4$. Orthogonality separates the nonnegative contributions: $$\boxed{P_{\mathrm{constant}}=\frac 1{\pi^2},\quad P_{\mathrm{fundamental}}=\frac 18,\quad P_{\mathrm{higher}}=\frac 18-\frac 1{\pi^2}.}$$ They sum to $1/4$. The higher-harmonic contribution is positive; the mean squared is only the constant contribution, not the total average squared output.

Step 4: Prove the fundamental periods. The maxima of $h$ occur exactly at $\pi/2+2k\pi$. Any positive period must take one maximum to another, hence be a positive multiple of $2\pi$; $2\pi$ works. The maxima of $|\sin x|$ occur at $\pi/2+k\pi$, so its least positive period is $\pi$. Thus the full output has fundamental period $2\pi$ despite its even-frequency cosine terms: its nonzero sine fundamental matters.