Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 18

One period advances the fundamental phase by $2\pi$, so the angular frequency is

The sine and cosine functions are orthogonal over a complete period. For positive integers $m,n$,while every mixed sine-cosine integral and every integral of a nonconstant single harmonic is zero. These identities follow from the product-to-sum formulas and the integer-period endpoint values.

Multiply the Fourier series by $\cos (m\omega t)$ and integrate. All terms except $a_m{\cos }^2(m\omega t)$ vanish, giving $T{a}_m/2$. Repeating with sine isolates $b_m$. ThusIntegrating without a harmonic also gives $a_0=(2/T){\int }_0^T{V}_1(t)dt$. The factor $1/2$ on the constant term in the chosen convention is essential.

Put $s=\omega t$, so the nonzero half-cycle is $0\le s\le \pi$. The constant coefficient isThe sine coefficients satisfyby sine orthogonality. For the cosine coefficients, the product-to-sum identity gives, when $n\ne 1$,The case $n=1$ is zero directly. HenceThe first five nonzero terms, counting the constant as the zero-frequency term, areThe general coefficient formula specifies every remaining term. This is the Fourier series of a half-wave rectified sinusoid; the waveform is continuous at the joins, so the series gives its actual value there.

The given identity immediately yields $V_2=V_1-V_0\sin \omega t$. Thus only the fundamental sine coefficient changes: $V_0/2$ becomes $-V_0/2$. No new integral is needed. The first five nonzero terms areThis also follows from translation of a Fourier series: a half-period delay multiplies the $n$th harmonic by $(-1{)}^n$, leaving the constant and even harmonics unchanged and reversing the only odd harmonic present.