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TIME-FREQUENCY ANALYSIS Example 1.3

信号処理

TIME-FREQUENCY ANALYSISの例題を解き進める．

Example 1.3

\begin{eqnarray*} s(t) = (\alpha / \pi)^{1/4} \mathrm{e}^{-\alpha t^2 /2 + j\beta t^2 /2 + j \omega_0 t} \end{eqnarray*}
のBandwidth \sigma_\omegaを求めよ．



解答

解答方針

{\sigma_{\omega}}^2 = \langle \omega^2 \rangle - \langle \omega \rangle^2であり，

今回は1.4 SIMPLE CALCULATION TRICKSで紹介されている
\langle \omega \rangle = \int \omega |S(\omega)|^2d\omega=\int s^* (t) \frac{1}{j}\frac{d}{dt}s(t)dt
\begin{eqnarray*} \langle \omega^2 \rangle = \int \omega^2 |S(\omega)|^2d\omega&=&\int s^* (t) \left( \frac{1}{j}\frac{d}{dt} \right) ^2 s(t)dt\\ &=&-\int s^*(t)\frac{d^2}{dt^2}s(t)dt\\ &=&\int \left|\frac{d}{dt}s(t) \right|^2dt \end{eqnarray*}
の結果を用いて計算する．

\langle \omega \rangleの導出

\begin{eqnarray*} \frac{1}{j}\frac{d}{dt}s(t)&=&\frac{1}{j} (\alpha / \pi)^{1/4} \left( -\alpha t^2 /2 + j\beta t^2 /2 + j \omega_0 t \right)' \mathrm{e}^{-\alpha t^2 /2 + j\beta t^2 /2 + j \omega_0 t}\\ &=& \frac{1}{j} (\alpha / \pi)^{1/4} \left( -\alpha t + j \beta t + j\omega_0 \right) \mathrm{e}^{-\alpha t^2 /2 + j\beta t^2 /2 + j \omega_0 t}\\ &=& (\alpha / \pi)^{1/4} \left( j\alpha t + \beta t + \omega_0 \right) \mathrm{e}^{-\alpha t^2 /2 + j\beta t^2 /2 + j \omega_0 t}\\ \end{eqnarray*}

s^*(t) = (\alpha / \pi)^{1/4} \mathrm{e}^{-\alpha t^2 /2 - j\beta t^2 /2 - j \omega_0 t}

\begin{eqnarray*} \langle \omega \rangle &=&\int s^* (t) \frac{1}{j}\frac{d}{dt}s(t)dt\\ &=& \int (\alpha / \pi)^{1/4} \mathrm{e}^{-\alpha t^2 /2 - j\beta t^2 /2 - j \omega_0 t} (\alpha / \pi)^{1/4} \left( j\alpha t + \beta t + \omega_0 \right) \mathrm{e}^{-\alpha t^2 /2 + j\beta t^2 /2 + j \omega_0 t}dt\\ &=& \sqrt{\frac{\alpha}{\pi}} \int \left( j\alpha t + \beta t + \omega_0 \right) \mathrm{e}^{-\alpha t^2}dt\\ &=& \sqrt{\frac{\alpha}{\pi}} \int \left( j\alpha t + \beta t \right) \mathrm{e}^{-\alpha t^2}dt + \omega_0\sqrt{\frac{\alpha}{\pi}} \int \mathrm{e}^{-\alpha t^2}dt\\ &=& \sqrt{\frac{\alpha}{\pi}} 0 + \omega_0 \sqrt{\frac{\alpha}{\pi}} \sqrt{\frac{\pi}{\alpha}}\\ &=& \omega_0 \end{eqnarray*}



\langle \omega^2 \rangleの導出

\begin{eqnarray*} \langle \omega^2 \rangle &=&\int \left|\frac{d}{dt}s(t) \right|^2dt\\ &=& \int \left| (\alpha / \pi)^{1/4} \left( -\alpha t + j \beta t + j\omega_0 \right) \mathrm{e}^{-\alpha t^2 /2 + j\beta t^2 /2 + j \omega_0 t} \right|^2 dt\\ &=& \sqrt{\frac{\alpha}{ \pi}} \int \left| -\alpha t + j \beta t + j\omega_0 \right|^2 \mathrm{e}^{-\alpha t^2}dt\\ &=& \sqrt{\frac{\alpha}{ \pi}} \int \left( \left( \alpha t \right)^2 + \left( \beta t + \omega_0 \right)^2 \right) \mathrm{e}^{-\alpha t^2}dt\\ &=& \sqrt{\frac{\alpha}{ \pi}} \int \left( \alpha^2 t ^2 + \beta ^2 t^2 + 2 \beta \omega_0 t + \omega_0 ^2 \right) \mathrm{e}^{-\alpha t^2}dt\\ &=& \sqrt{\frac{\alpha}{ \pi}} \int \left( \left( \alpha^2 + \beta^2\right) t ^2 + 2 \beta \omega_0 t + \omega_0 ^2 \right) \mathrm{e}^{-\alpha t^2}dt\\ &=& \sqrt{\frac{\alpha}{ \pi}} \left( \left( \alpha^2 + \beta^2 \right) \int t^2 \mathrm{e}^{-\alpha t^2}dt + 2 \beta \omega_0 \int t \mathrm{e}^{-\alpha t^2}dt+ \omega_0 ^2 \int \mathrm{e}^{-\alpha t^2}dt \right) \cdots (*)\\ &=& \sqrt{\frac{\alpha}{ \pi}} \left( \left( \alpha^2 + \beta^2 \right) \frac{1}{2 \alpha} \sqrt{\frac{\pi}{\alpha}} + 2 \beta \omega_0 0+ \omega_0 ^2 \sqrt{\frac{\pi}{\alpha}} \right) \cdots (**)\\ &=& \frac{\alpha^2 + \beta^2}{2 \alpha} + \omega_0 ^2 \end{eqnarray*}

(*)→(**)の変形に関しては，
ガウス積分の公式から\int t^2 \mathrm{e}^{-\alpha t^2}dt = \frac{1}{2 \alpha} \sqrt{\frac{\pi}{\alpha}}，\int \mathrm{e}^{-\alpha t^2}dt= \sqrt{\frac{\pi}{\alpha}}を用いた．
また， \int t^2 \mathrm{e}^{-\alpha t^2}dtは\int (偶関数) dt = 0である．

まとめ

\begin{eqnarray*} {\sigma_{\omega}}^2 &=& \langle \omega^2 \rangle - \langle \omega \rangle^2\\ &=& \frac{\alpha^2 + \beta^2}{2 \alpha} + \omega_0 ^2 - \omega_0 ^2\\ &=& \frac{\alpha^2 + \beta^2}{2 \alpha} \end{eqnarray*}