Derivation: Fourier Series Coefficients

## Fourier Series Coefficients

Derivation of the equation for An. Multiply both sides of the Fourier series equation by cos(2πmf 0t) and integrate over one repetition period, t0.

0 t 0 F(t)cos(2πm f 0 t)dt = A 0 2 0 t 0 cos(2πm f 0 t)dt + n=1 A n 0 t 0 cos(2πn f 0 t)cos(2πm f 0 t)dt + n=1 B n 0 t 0 sin(2πn f 0 t)cos(2πm f 0 t)dt

When nm, all terms on the right side integrate to zero. The proof of this assertion involves the trig identities 2cosαcosβ = cos(α+β) + cos(α-β), etc. When n = m, only the cosine product has a non-zero integral.

A m 0 t 0 cos(2πm f 0 t)dtcos(2πm f 0 t)dt = A m t 0 2

This result can be used to compute each Am.

A m t 0 2 = 0 t 0 F(t)cos(2πm f 0 t)dt A m = 2 t 0 0 t 0 F(t)cos(2πm f 0 t)dt

The zero frequency term. When n = m = 0, the first term on the right of the top equation integrates to A0t/2. When added to the integral of the cosine product, the result is A0t. This result can be used to compute A0.

A 0 = 1 t 0 0 t 0 F(t)cos(0)dt

This result (the correct answer) is half that obtained by the general formula for Am. This explains why, in the Fourier series expansion, the value of A0 is divided by 2.

Derivation of the equation for Bn. Multiply both sides of the Fourier series equation by sin(2πmf 0t) and integrate over one repetition period, t0. The derivation follows that show above for An. That is, all terms will integrate to zero except that with the product of two sines.

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