[系統理論] 連續時間週期訊號的 Fourier Series Representation (4)-Parseval's relation for periodic signal

令 $x(t)$ 與 $y(t)$ 為具有週期為 $T$ 的連續時間週期訊號，且存在 Fourier Series Representation 如下
\[\begin{array}{l}
x\left( t \right) = \sum\limits_{k =  - \infty }^\infty  {{a_k}} {e^{jk{\omega _0}t}}\\
y\left( t \right) = \sum\limits_{k =  - \infty }^\infty  {{b_k}} {e^{jk{\omega _0}t}}
\end{array}
\]

我們首先證明下面的結果：

Fact: 給定兩時域訊號相乘 $\Rightarrow$ 頻域訊號 convolution
定義兩訊號乘積 $z(t) := x(t) y(t) = \sum\limits_{k =  - \infty }^\infty  {{c_k}} {e^{jk{\omega _0}t}}$，則 其乘積的 Fourier Series Coefficient 為離散 convolution
\[
c_k = \sum_{n=-\infty}^{\infty}a_n b_{k-n}
\]Proof:
觀察 $z(t) = x(t)y(t)$ 如下
\[\begin{array}{l}
x\left( t \right)y\left( t \right) = \sum\limits_{n =  - \infty }^\infty  {{a_n}} {e^{jn{\omega _0}t}}\sum\limits_{k =  - \infty }^\infty  {{b_k}} {e^{jk{\omega _0}t}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \sum\limits_{k =  - \infty }^\infty  {\left( {{b_k}\sum\limits_{n =  - \infty }^\infty  {{a_n}} {e^{jn{\omega _0}t}}} \right){e^{jk{\omega _0}t}}}
\end{array}\]又注意到 $b_k$ 為 $y(t)$ 的 Fourier Series Coefficient，故我們有
\[{b_k} = \frac{1}{T}\int_T^{} {y\left( t \right){e^{ - jk{\omega _0}t}}} dt
\] 帶入上式可得
\[\begin{array}{l}
x\left( t \right)y\left( t \right) = \sum\limits_{k =  - \infty }^\infty  {\left( {\frac{1}{T}\int_T^{} {y\left( t \right){e^{ - jk{\omega _0}t}}} dt\sum\limits_{n =  - \infty }^\infty  {{a_n}} {e^{jn{\omega _0}t}}} \right){e^{jk{\omega _0}t}}} \\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \sum\limits_{k =  - \infty }^\infty  {\left( {\sum\limits_{n =  - \infty }^\infty  {{a_n}} \underbrace {\frac{1}{T}\int_T^{} {y\left( t \right){e^{ - j\left( {k - n} \right){\omega _0}t}}dt} }_{ = {b_{k - n}}}} \right){e^{jk{\omega _0}t}}} \\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \sum\limits_{k =  - \infty }^\infty  {\left( {\sum\limits_{n =  - \infty }^\infty  {{a_n}} {b_{k - n}}} \right){e^{jk{\omega _0}t}}}
\end{array}\]亦即
\[{c_k} = \sum\limits_{n =  - \infty }^\infty  {{a_n}} {b_{k - n}} \ \ \ \ \square
\]
Theorem: Parseval's Relation for Periodic Signals
現在我們令上述 $y(t) := x^*(t)$ (其中 ${}^*$ 表 complex conjugate)。則下列結果成立
\[
\frac{1}{T} \int_0^T |x(t)|^2 dt =\sum_{k=-\infty}^{\infty}|a_k|^2
\]

Comment: 上式說明了一個週期訊號 $x(t)$ 的 單一週期的 total energy 除以 週期 $T$ (或者 直接說 average power) 等同於 其對應的 Fourier Series coefficient 的平方 $|a_k|^2$ 做無窮級數。 (此平方項 $|a_k|^2$又稱 k-th harmonic components)

Proof:
觀察 ${\left| {x\left( t \right)} \right|^2} = x\left( t \right){x^*}\left( t \right)$，故我們有
\[\frac{1}{T}\int_0^T {{{\left| {x\left( t \right)} \right|}^2}dt}  = \frac{1}{T}\int_0^T {x\left( t \right){x^*}\left( t \right)dt} \]接著帶入 $x(t)$ (對應的 Fourier Series coefficient 為 $a_k$) 與 $x^*(t)$ ( $x^*(t)$ 對應的 Fourier Series 為 $a_{-k}^*$)。
\[\begin{array}{*{20}{l}}
{\frac{1}{T}\int_0^T {{{\left| {x\left( t \right)} \right|}^2}dt}  = \frac{1}{T}\int_0^T {\sum\limits_{n =  - \infty }^\infty  {{a_n}} {e^{jn{\omega _0}t}}\sum\limits_{k =  - \infty }^\infty  {a_{ - k}^*} {e^{ - jk{\omega _0}t}}dt} }\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{T}\int_0^T {\sum\limits_{k =  - \infty }^\infty  {\sum\limits_{n =  - \infty }^\infty  {a_{ - k}^*{a_n}{e^{ - jk{\omega _0}t}}} {e^{jn{\omega _0}t}}} dt} }\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{T}\sum\limits_{k =  - \infty }^\infty  {a_{ - k}^*\underbrace {\sum\limits_{n =  - \infty }^\infty  {{a_n}\int_0^T {{e^{ - jk{\omega _0}t}}{e^{jn{\omega _0}t}}dt} } }_{ = {a_k}T}} }\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{T}\sum\limits_{k =  - \infty }^\infty  {a_{ - k}^*{a_k}T}  = \sum\limits_{k =  - \infty }^\infty  {{{\left| {{a_k}} \right|}^2}} } \ \ \ \ \square
\end{array}
\]

ref: A.V. Oppenheim, A. S. Willsky, S. H. Nawab, Signals and Systems