Fourier Series Property

Creator

Creator

Created

Created

2023 Oct 9 12:38

Editor

Editor

Edited

Edited

2023 Oct 10 5:30

Refs

Refs

a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0t} dt \newline x(t) = \sum_{k=-\infty}^\infty a_k e^{jk\omega_0t}

Even Odd

If x is real → $a_{-k} = a_k^\ast$ → $|a_{-k}| = |a_k|$ (absolute value is $\sqrt{a^2 + b^2}$ when $a +bj$ )

If x is real and even → $a_k = a_{-k} = a_k^\ast$ →
Fourier coefficient are real and even

If x is real and odd → $a_k = -a_{-k} = -a_k^\ast$ →
Fourier coefficient are purely imaginary and odd

If $x(t)$ is Real

$a_k = a_{-k}^\ast$

2, 3 property is proved using ignoring the complex part of the representation

Polar form of the
Fourier coefficient is $a_k = A_ke^{j\theta_k}$ , then the trigonometric Fourier representation of $x(t)$ is

x(t) = a_0 + 2\sum_{k=1}^\infty A_k cos(k\omega_0 t + \theta_k)

notion image

Let Cartesian form of the Fourier coefficients is like
Fourier coefficient then trigonometric representation of $x(t)$ is

x(t) = a_0 + 2\sum_{k=1}^\infty[B_kcos(k\omega_0t) - C_ksin(k\omega_0t)]

notion image

If x is real → $a_{-k} = a_k^\ast$ → $|a_{-k}| = |a_k|$ (absolute value is $\sqrt{a^2 + b^2}$ when $a +bj$ )

If x is real and even → $a_k = a_{-k} = a_k^\ast$ →
Fourier coefficient are real and even

If x is real and odd → $a_k = -a_{-k} = -a_k^\ast$ →
Fourier coefficient are purely imaginary and odd

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