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Description of Fourier series by complex expression

The following comment shows, how is possible to convert the trigonometric expression of Fourier series into complex formula. The conversion will be carried by means of Euler’s formula.

\text{cos}(k\omega t)=\frac{e^{jk\omega t}+e^{-jk\omega t}}{2}\\ 
j\text{sin}(k\omega t)=\frac{e^{jk\omega t}-e^{-jk\omega t}}{2}

Describe $B_{k}\text{sin}(k\omega t) in the following way: $-jB_{k} \frac{e^{jk\omega t}-e^{-jk\omega t}}{2}=B_{k}\text{sin}(k\omega t)
Write down the trigonometric expression of Fourier series: $f(t)= A_{0}+\sum_{k=1}^{\infty}\left(A_{k}\text{cos}(k\omega t)+B_{k}\text{sin}(k\omega t)\right)

Substitute the above written expressions into the Fourier series formula in order to obtain the Euler’s description way:

f(t)=A_{0}+\sum_{k=1}^{\infty}\left[A_{k} \frac{e^{jk\omega t}+e^{-jk\omega t}}{2}-jB_{k}\frac{e^{jk\omega t}-e^{-jk\omega t}}{2}\right]

f(t)=A_{0}+\sum_{k=1}^{\infty}\left[\frac{A_{k}-jB_{k}}{2}e^{jk\omega t}+\frac{A_{k}+jB_{k}}{2}e^{-jk\omega t}\right]

f(t)=A_{0}+\sum_{k=1}^{\infty}\frac{A_{k}-jB_{k}}{2}e^{jk\omega t}+\sum_{k=-1}^{-\infty}\frac{A_{-k}+jB_{-k}}{2}e^{+jk\omega t}

Apply the following denoting:

\frac{A_{k}-jB_{k}}{2}, & \text{ for } k \geq 0\\ 
\frac{A_{-k}+jB_{-k}}{2}, & \text{ for } k < 0

f(t)= A_{0} + \sum_{k=1}^{\infty}c_{k}e^{jk\omega t}+\sum_{k=-1}^{-\infty}c_{k}e^{jk\omega t}

Use common $\sum for separate summing operation of Fourier series:

f(t)=\sum_{k=-\infty}^{\infty}c_{k}e^{jk\omega t} \text{, where } c_{k}=\frac{1}{T}\int_{0}^{T}f(t)e^{-jk\omega t}dt

Return to the “Characteristic Input Function” chapter

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