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Fourier Series Coefficients

Jul 22, 20251 min read

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$c_{k} = \frac{1}{T _{0}} \int_{τ}^{τ + T_{0}} f (t) e^{- jk ω_{0} t} d t$

Fourier series coefficients

Properties (assuming

$R (c_{k}) = R (c_{- k})$

$I (c_{k}) = - I (c_{- k})$

$c_{k}^{*} = c_{- k}$

$∣ c_{k} ∣ = ∣ c_{- k} ∣$

$∠ c_{k} = - ∠ c_{k}^{*}$

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$\displaystyle c_{k}=\frac{1}{T_{0}}\int_{\tau}^{\tau+T_{0}} f(t)e^{-jk{\omega}_{0}t} \, \mathrm{d}t$
Properties (assuming
$\displaystyle \mathfrak{R}(c_{k})=\mathfrak{R}(c_{-k})$
$\displaystyle \mathfrak{I}(c_{k})=-\mathfrak{I}(c_{-k})$
$\displaystyle c^{*}_{k}=c_{-k}$
$\displaystyle \lvert c_{k}\rvert=\lvert c_{-k}\rvert$
$\displaystyle \angle c_{k}=-\angle c_{k}^{*}$

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Fourier Series Coefficients

$c_{k} = \frac{1}{T _{0}} \int_{τ}^{τ + T_{0}} f (t) e^{- jk ω_{0} t} d t$

Properties (assuming

$R (c_{k}) = R (c_{- k})$

$I (c_{k}) = - I (c_{- k})$

$c_{k}^{*} = c_{- k}$

$∣ c_{k} ∣ = ∣ c_{- k} ∣$

$∠ c_{k} = - ∠ c_{k}^{*}$

Graph View

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Fourier Series Coefficients

ck​=T0​1​∫ττ+T0​​f(t)e−jkω0​tdt

Properties (assuming

R(ck​)=R(c−k​)

I(ck​)=−I(c−k​)

ck∗​=c−k​

∣ck​∣=∣c−k​∣

∠ck​=−∠ck∗​

Graph View

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$c_{k} = \frac{1}{T _{0}} \int_{τ}^{τ + T_{0}} f (t) e^{- jk ω_{0} t} d t$

$R (c_{k}) = R (c_{- k})$

$I (c_{k}) = - I (c_{- k})$

$c_{k}^{*} = c_{- k}$

$∣ c_{k} ∣ = ∣ c_{- k} ∣$

$∠ c_{k} = - ∠ c_{k}^{*}$