计算机网络(computer networks)2.1.1 节 Fourier 分析，没看懂

In the early 19th century, the French mathematician Jean-Baptiste Fourier proved that any reasonably behaved periodic function, g(t) with period T, can be constructed as the sum of a (possibly infinite) number of sines and cosines:

g(t) = (1/2)c + Σ(an sin(2πnft), n=1->∞) + Σ(bn cos(2πnft), n=1->∞) (2-1)

where f = 1/T is the fundamental frequency, an and bn are the sine and cosine amplitudes of the nth harmonics (terms), and c is a constant. Such a decomposition is called a Fourier series. From the Fourier series, the function can be reconstructed. That is, if the period, T, is known and the amplitudes are given, the original function of time can be found by performing the sums of Eq. (2-1).

A data signal that has a finite duration, which all of them do, can be handled by just imagining that it repeats the entire pattern over and over forever (i.e., the interval from T to 2T is the same as from 0 to T, etc.).

The an amplitudes can be computed for any given g(t) by multiplying both sides of Eq. (2-1) by sin(2πkft) and then integrating from 0 to T. Since

∫(sin(2πkft)sin(2πnft)dt, t=0->T) = { 0 for k ≠ n T/2 for k = n }

only one term of the summation survives: an. The bn summation vanishes completely. Similarly, by multiplying Eq. (2-1) by cos(2πkft) and integrating between 0 and T, we can derive bn. By just integrating both sides of the equation as it stands, we can find c. The results of performing these operations are as follows:

an = (2/T)∫(g(t)sin(2πnft)dt, t=0->T) bn = (2/T)∫(g(t)cos(2πnft)dt, t=0->T) c = (2/T)∫(g(t)dt, t=0->T)

公式(2-1)两边乘上 sin(2πkft)，再对 t 从 0 到 T 积分，到底是怎么算出 an 的，公式(2-1)中还有常量 c，是怎么消掉的，原来还有 1 到无穷的求和，怎么也没了