首先回憶給定平滑有界之 週期訊號 $x(t)$ 且假設此訊號具有基本頻率 $f_0$ (亦即此訊號具有基本週期 $T_0 :=1/f_0$ ),那麼在 Fourier analysis 中我們說此訊號可以表示為
\[
x(t) = \sum_{k = -\infty}^\infty a_k e^{j 2 \pi f_0 k t }
\]且其對應的 Fourier Series 可由下列積分求得
\[
a_k := \frac{1}{T_0} \int_0^{T_0} x(t) e^{-j 2 \pi f_0 k t} dt \;\;\;\;\; (*)
\]
但事實上由於 $x(t)$ 為週期訊號,上述積分範圍可以為 任意單位週期長度 而不需拘泥於 $[0,T_0]$,一般而言,最常見到的另一種 Fourier Series coefficient 形式為 : \[{a_k} = \frac{1}{{{T_0}}}\int_{ - {T_0}/2}^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} \]事實上此式子與 $(*)$ 等價。 我們將此寫作以下 FACT:
===================
FACT:
給定平滑有界 週期訊號 $x(t)$ 表示為
\[
x(t) = \sum_{k = -\infty}^\infty a_k e^{j 2 \pi f_0 k t }
\]則其 Fourier Series 係數亦可表為
\[{a_k} = \frac{1}{{{T_0}}}\int_{ - {T_0}/2}^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} \]===================
Proof:
我們僅需證明
\[ \frac{1}{{{T_0}}}\int_{ - {T_0}/2}^{{T_0}/2} {x\left( t \right){e^{ - j2\pi {f_0}kt}}dt} = \frac{1}{{{T_0}}}\int_0^{{T_0}} {x\left( t \right){e^{ - j2\pi {f_0}kt}}dt} \]故現在觀察左式,利用積分的線性性質:
\begin{align*}
{a_k} &:= \frac{1}{{{T_0}}}\int_{ - {T_0}/2}^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt}\\
& = \frac{1}{{{T_0}}}\left( {\int_{ - {T_0}/2}^0 {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \;\;\;\; (\star)
\end{align*}現在對第一項積分做變數變換,令 $u : = t + {T_0}$ 其中 $T_0:=1/f_0$ 則前述 $(\star)$ 中的第一項積分可改寫為
\begin{align*}
{a_k} &= \frac{1}{{{T_0}}}\left( {\int_{ - {T_0}/2}^0 {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \hfill \\
&= \frac{1}{{{T_0}}}\left( {\int_{{T_0}/2}^{{T_0}} {x\left( {u - {T_0}} \right){e^{ - j{2 \pi f_0}k\left( {u - {T_0}} \right)}}du} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \hfill \\
&= \frac{1}{{{T_0}}}\left( {\int_{{T_0}/2}^{{T_0}} {x\left( u \right){e^{ - j{2 \pi f_0}k\left( u \right)}}\underbrace {{e^{j{2 \pi f_0}k\left( {{T_0}} \right)}}}_{ = 1}du} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \hfill \\
&= \frac{1}{{{T_0}}}\left( {\int_{{T_0}/2}^{{T_0}} {x\left( u \right){e^{ - j{2 \pi f_0}k\left( u \right)}}du} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \hfill \\
&= \frac{1}{{{T_0}}}\left( {\int_{{T_0}/2}^{{T_0}} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \hfill \\
&= \frac{1}{{{T_0}}}\int_0^{{T_0}} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} .
\end{align*}
讀者可注意到上述第三條等式成立之原因為 $x(t)$ 為週期訊號故 $x(t -T_0) = x(t)$$\square$
\[
x(t) = \sum_{k = -\infty}^\infty a_k e^{j 2 \pi f_0 k t }
\]且其對應的 Fourier Series 可由下列積分求得
\[
a_k := \frac{1}{T_0} \int_0^{T_0} x(t) e^{-j 2 \pi f_0 k t} dt \;\;\;\;\; (*)
\]
但事實上由於 $x(t)$ 為週期訊號,上述積分範圍可以為 任意單位週期長度 而不需拘泥於 $[0,T_0]$,一般而言,最常見到的另一種 Fourier Series coefficient 形式為 : \[{a_k} = \frac{1}{{{T_0}}}\int_{ - {T_0}/2}^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} \]事實上此式子與 $(*)$ 等價。 我們將此寫作以下 FACT:
===================
FACT:
給定平滑有界 週期訊號 $x(t)$ 表示為
\[
x(t) = \sum_{k = -\infty}^\infty a_k e^{j 2 \pi f_0 k t }
\]則其 Fourier Series 係數亦可表為
\[{a_k} = \frac{1}{{{T_0}}}\int_{ - {T_0}/2}^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} \]===================
Proof:
我們僅需證明
\[ \frac{1}{{{T_0}}}\int_{ - {T_0}/2}^{{T_0}/2} {x\left( t \right){e^{ - j2\pi {f_0}kt}}dt} = \frac{1}{{{T_0}}}\int_0^{{T_0}} {x\left( t \right){e^{ - j2\pi {f_0}kt}}dt} \]故現在觀察左式,利用積分的線性性質:
\begin{align*}
{a_k} &:= \frac{1}{{{T_0}}}\int_{ - {T_0}/2}^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt}\\
& = \frac{1}{{{T_0}}}\left( {\int_{ - {T_0}/2}^0 {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \;\;\;\; (\star)
\end{align*}現在對第一項積分做變數變換,令 $u : = t + {T_0}$ 其中 $T_0:=1/f_0$ 則前述 $(\star)$ 中的第一項積分可改寫為
\begin{align*}
{a_k} &= \frac{1}{{{T_0}}}\left( {\int_{ - {T_0}/2}^0 {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \hfill \\
&= \frac{1}{{{T_0}}}\left( {\int_{{T_0}/2}^{{T_0}} {x\left( {u - {T_0}} \right){e^{ - j{2 \pi f_0}k\left( {u - {T_0}} \right)}}du} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \hfill \\
&= \frac{1}{{{T_0}}}\left( {\int_{{T_0}/2}^{{T_0}} {x\left( u \right){e^{ - j{2 \pi f_0}k\left( u \right)}}\underbrace {{e^{j{2 \pi f_0}k\left( {{T_0}} \right)}}}_{ = 1}du} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \hfill \\
&= \frac{1}{{{T_0}}}\left( {\int_{{T_0}/2}^{{T_0}} {x\left( u \right){e^{ - j{2 \pi f_0}k\left( u \right)}}du} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \hfill \\
&= \frac{1}{{{T_0}}}\left( {\int_{{T_0}/2}^{{T_0}} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} + \int_0^{{T_0}/2} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} } \right) \hfill \\
&= \frac{1}{{{T_0}}}\int_0^{{T_0}} {x\left( t \right){e^{ - j{2 \pi f_0}kt}}dt} .
\end{align*}
讀者可注意到上述第三條等式成立之原因為 $x(t)$ 為週期訊號故 $x(t -T_0) = x(t)$$\square$
留言
張貼留言