科恩系列分佈

當Cohen's class分佈中的${\displaystyle \Phi \left(\eta ,\tau \right)=1}$時，Cohen's class分佈會成韋格納分布(Wigner distribution function)${\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }x(t+\tau /2)x^{*}(t-\tau /2)e^{-j2\pi f\tau }\,d\tau }$。

利用韋格納分佈對函數${\displaystyle x(t)=exp(j0.015t^{4}+j0.06t^{3}-j0.3t^{2}+jt)}$作時頻分析的結果可見右圖。

當Cohen's class分佈中的${\displaystyle \phi (t,\tau )={\frac {1}{\left|\tau \right|}}exp(-2\pi \alpha \tau ^{2})\Pi ({\frac {t}{\tau }})}$，且${\displaystyle \Phi \left(\eta ,\tau \right)=sinc(\eta \tau )exp((-2\pi \alpha \tau ^{2})}$時，

其中${\displaystyle \phi (t,\tau )=\int _{-\infty }^{\infty }\Phi (\eta ,\tau )exp(j2\pi \eta t)d\eta }$，Cohen's class分佈會成錐狀分布。

右圖為不同的${\displaystyle \alpha }$值下的錐狀分佈時頻分析圖。

當Cohen's class分佈中的${\displaystyle \Phi \left(\eta ,\tau \right)=exp[-\alpha (\eta \tau )^{2}]}$時，Cohen's class分佈會成喬伊-威廉斯分布。

右圖為不同的${\displaystyle \alpha }$值下的錐狀分佈時頻分析圖。

對${\displaystyle \ \left|\eta \right|>B\ }$或${\displaystyle \ \left|\tau \right|>C}$，若${\displaystyle \Phi (\eta ,\tau )=0}$，則${\displaystyle C_{x}(t,f)=\int _{-C}^{C}\int _{-B}^{B}\int _{-\infty }^{\infty }x(u+{\frac {\tau }{2}})x^{*}(u-{\frac {\tau }{2}})\Phi (\eta ,\tau )e^{-j2\pi u\eta +j2\pi (\eta t-\tau f)}dud\tau d\eta }$

${\displaystyle C_{x}(t,f)=\int _{-C}^{C}\int _{-\infty }^{\infty }x(u+{\frac {\tau }{2}})x^{*}(u-{\frac {\tau }{2}})[\int _{-B}^{B}\Phi (\eta ,\tau )e^{-j2\pi ,\eta (t-u)}d\eta ]e^{-j2\pi \tau ,f}dud\tau }$

${\displaystyle =\int _{-C}^{C}\int _{-\infty }^{\infty }x(u+{\frac {\tau }{2}})x^{*}(u-{\frac {\tau }{2}})\Phi (\tau ,t-u)e^{-j2\pi \tau ,f}dud\tau }$，其中

${\displaystyle \Phi (\tau ,t-u)=\int _{-B}^{B}\Phi (\eta ,\tau )e^{-j2\pi ,\eta (t-u)}d\eta }$，由於${\displaystyle \Phi (\tau ,t-u)}$和輸入無關，可事先算出，因此可簡化成兩個積分式。

${\displaystyle C_{x}(t,f)=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }x(u+{\frac {\tau }{2}})x^{*}(u-{\frac {\tau }{2}})\phi (t-u,\tau )due^{-j2\pi f\tau }d\tau }$，其中

${\displaystyle \phi (t,\tau )=\int _{-\infty }^{\infty }\Phi (\eta ,\tau )exp(j2\pi \eta t)d\eta }$。對${\displaystyle \left|t\right|>b}$或是${\displaystyle \left|\tau \right|>c}$，則

${\displaystyle C_{x}(t,f)=\int _{-c}^{c}\int _{t-b}^{t+b}x(u+{\frac {\tau }{2}})x^{*}(u-{\frac {\tau }{2}})\phi (t-u,\tau )due^{-j2\pi f\tau }d\tau }$，上式為一摺積式。

我們來看一下 ${\displaystyle x(t)}$ 對於模糊函數的影響

(1) 假設 ${\displaystyle x_{1}(t)}$ 是一個高斯函數: ${\displaystyle ae^{-(t-b)^{2}/2c^{2}}}$, 其中 ${\displaystyle a=1,b=0,c={\sqrt {\tfrac {1}{2\alpha }}}}$

那麼我們可以得到 ${\displaystyle x_{1}(t)=e^{-\alpha \pi t^{2}}}$, 代入模糊函數 ${\displaystyle A_{x}\left(\eta ,\tau \right)}$ 中：

${\displaystyle A_{x_{1}}\left(\eta ,\tau \right)=\textstyle \int \limits _{-\infty }^{\infty }\displaystyle e^{-\alpha \pi {(t+{\tfrac {\tau }{2}})}^{2}}\ e^{-\alpha \pi {(t-{\tfrac {\tau }{2}})}^{2}}\ e^{-j2\pi t\eta }\ dt}$

${\displaystyle =\textstyle \int \limits _{-\infty }^{\infty }\displaystyle e^{-\alpha \pi (2t^{2}+{\tfrac {\tau }{2}}^{2})}\ e^{-j2\pi t\eta }\ dt}$

(2) 假設 ${\displaystyle x_{2}(t)}$ 是一個經過 shifting 和 modulation 的高斯函數:

那麼我們可以得到 ${\displaystyle x_{2}(t)=e^{-\alpha \pi (t-t_{0})^{2}+j2\pi f_{0}t}}$, 代入模糊函數 ${\displaystyle A_{x}\left(\eta ,\tau \right)}$ 中：

${\textstyle A_{x_{2}}\left(\eta ,\tau \right)=\textstyle \int \limits _{-\infty }^{\infty }\displaystyle e^{-\alpha \pi {(t+{\tfrac {\tau }{2}}-t_{0})}^{2}+j2\pi f_{0}(t+{\tfrac {\tau }{2}})}\ e^{-\alpha \pi {(t-{\tfrac {\tau }{2}}-t_{0})}^{2}-j2\pi f_{0}(t-{\tfrac {\tau }{2}})}\ e^{-j2\pi t\eta }\ dt}$

${\displaystyle =\textstyle \int \limits _{-\infty }^{\infty }\displaystyle e^{-\alpha \pi (2(t-t_{0})^{2}+{\tau /2}^{2})+j2\pi f_{0}\tau }\ e^{-j2\pi t\eta }\ dt}$

${\displaystyle =\textstyle \int \limits _{-\infty }^{\infty }\displaystyle e^{-\alpha \pi (2t^{2}+{\tfrac {\tau ^{2}}{2}})}\ e^{j2\pi f_{0}\tau }\ e^{-j2\pi t\eta }\ e^{-j2\pi t_{0}\eta }\ dt}$

我們可以看到 ${\displaystyle |A_{x_{1}}\left(\tau ,\eta \right)|=|A_{x_{2}}\left(\tau ,\eta \right)|}$,

因此我們可以得出 time shifting ${\displaystyle t_{0}}$ 和 modulation ${\displaystyle f_{0}}$ 並不會影響 ${\displaystyle |A_{x}\left(\tau ,\eta \right)|}$

積分後，${\displaystyle A_{x}\left(\tau ,\eta \right)={\sqrt {\tfrac {1}{2\alpha }}}e^{-\pi ({\tfrac {\alpha \tau ^{2}}{2}}+{\tfrac {\eta ^{2}}{2\alpha }})}e^{j2\pi (f_{0}\tau -t_{0}\eta )}}$

所以 ${\displaystyle A_{x}\left(\tau ,\eta \right)}$ 在 ${\displaystyle \tau =0,\eta =0}$ 的地方會有最大的 ${\displaystyle |A_{x}\left(\tau ,\eta \right)|}$

上述所列出來的是當 ${\displaystyle x(t)}$ 只有一項而已 (one term only)，如果 ${\displaystyle x(t)}$ 有兩項以上的元素構成 (more than two terms), ${\displaystyle x(t)=x_{1}(t)+x_{2}(t)+\cdot \cdot \cdot +x_{n}(t)}$，依然會有交叉項 (cross-term) 的問題存在。

假設 ${\displaystyle x(t)=x_{1}(t)+x_{2}(t)}$ , 其中

${\displaystyle {\begin{cases}x_{1}(t)=e^{-\alpha _{1}\pi (t-t_{1})^{2}+j2\pi f_{1}t}\\x_{2}(t)=e^{-\alpha _{2}\pi (t-t_{2})^{2}+j2\pi f_{2}t}\end{cases}}}$

將 ${\displaystyle x(t)}$ 代入模糊函數 ${\displaystyle A_{x}\left(\eta ,\tau \right)=\int _{-\infty }^{\infty }x(t+{\tfrac {\tau }{2}})x^{*}(t-{\tfrac {\tau }{2}})e^{-j2\pi t\eta }\,dt}$ 中：

${\displaystyle A_{x}\left(\eta ,\tau \right)=\underbrace {\int _{-\infty }^{\infty }x_{1}(t+{\tfrac {\tau }{2}})x_{1}^{*}(t-{\tfrac {\tau }{2}})e^{-j2\pi t\eta }\,dt} _{A_{x_{1}}(\tau ,\eta )}+\underbrace {\int _{-\infty }^{\infty }x_{2}(t+{\tfrac {\tau }{2}})x_{2}^{*}(t-{\tfrac {\tau }{2}})e^{-j2\pi t\eta }\,dt} _{A_{x_{2}}(\tau ,\eta )}}$

${\displaystyle +\underbrace {\int _{-\infty }^{\infty }x_{1}(t+{\tfrac {\tau }{2}})x_{2}^{*}(t-{\tfrac {\tau }{2}})e^{-j2\pi t\eta }\,dt} _{A_{{x_{1}}{x_{2}}}(\tau ,\eta )}+\underbrace {\int _{-\infty }^{\infty }x_{2}(t+{\tfrac {\tau }{2}})x_{1}^{*}(t-{\tfrac {\tau }{2}})e^{-j2\pi t\eta }\,dt} _{A_{{x_{2}}{x_{1}}}(\tau ,\eta )}}$

其中 ${\displaystyle {\begin{cases}Auto-terms:\quad A_{x_{1}}(\tau ,\eta ),\ A_{x_{2}}(\tau ,\eta )\\Cross-terms:\ A_{{x_{1}}{x_{2}}}(\tau ,\eta ),\ A_{{x_{2}}{x_{1}}}(\tau ,\eta )\end{cases}}}$

${\displaystyle A_{x_{1}}\left(\tau ,\eta \right)={\sqrt {\tfrac {1}{2\alpha _{1}}}}\ e^{-\pi ({\tfrac {\alpha _{1}\tau ^{2}}{2}}+{\tfrac {\eta ^{2}}{2\alpha _{1}}})}\ e^{j2\pi (f_{1}\tau -t_{1}\eta )}}$

${\displaystyle A_{x_{2}}\left(\tau ,\eta \right)={\sqrt {\tfrac {1}{2\alpha _{2}}}}\ e^{-\pi ({\tfrac {\alpha _{2}\tau ^{2}}{2}}+{\tfrac {\eta ^{2}}{2\alpha _{2}}})}\ e^{j2\pi (f_{2}\tau -t_{2}\eta )}}$

(1) ${\displaystyle \alpha _{1}\neq \alpha _{2}}$

${\displaystyle A_{{x_{1}}{x_{2}}}(\tau ,\eta )={\sqrt {\tfrac {1}{(\alpha _{1}+\alpha _{2})}}}\ e^{-\pi ({\tfrac {(\alpha _{1}+\alpha _{2})(\tau -t_{1}+t_{2})^{2}}{4}}\ +\ {\tfrac {[(\alpha _{1}-\alpha _{2})(\tau -t_{1}+t_{2})-j2(\eta -f_{1}+f_{2})]^{2}}{4(\alpha _{1}+\alpha _{2})}})}\ e^{j2\pi [({\tfrac {f_{1}+f_{2}}{2}})\tau -{\tfrac {t_{1}+t_{2}}{2}}\eta +{\tfrac {(f_{1}-f_{2})(t_{1}+t_{2})}{2}}]}}$

${\displaystyle ={\sqrt {\tfrac {1}{2\alpha _{u}}}}\ e^{-\pi ({\tfrac {\alpha _{u}(\tau -t_{d})^{2}}{2}}\ +\ {\tfrac {[\alpha _{d}(\tau -t_{d})-j2(\eta -f_{d})]^{2}}{8\alpha _{u}}})}\ e^{j2\pi (f_{u}\tau -t_{n}\eta +f_{d}t_{u})}}$

${\displaystyle A_{{x_{2}}{x_{1}}}(\tau ,\eta )=A_{{x_{1}}{x_{2}}}^{*}(-\tau ,-\eta )}$

${\displaystyle {\begin{cases}t_{u}={\tfrac {t_{1}+t_{2}}{2}},\ f_{u}={\tfrac {f_{1}+f_{2}}{2}},\ \alpha _{u}={\tfrac {\alpha _{1}+\alpha _{2}}{2}}\\t_{d}=t_{1}-t_{2},\ f_{d}=f_{1}-f_{2},\ \alpha _{d}=\alpha _{1}-\alpha _{2}\end{cases}}}$

(2) ${\displaystyle \alpha _{1}=\alpha _{2}}$

${\displaystyle A_{{x_{1}}{x_{2}}}(\tau ,\eta )={\sqrt {\tfrac {1}{2\alpha _{u}}}}\ e^{-\pi ({\tfrac {\alpha _{u}(\tau -t_{d})^{2}}{2}}\ +\ {\tfrac {(\eta -f_{d})^{2}}{2\alpha _{u}}})}\ e^{j2\pi (f_{u}\tau -t_{n}\eta +f_{d}t_{u})}}$

${\displaystyle A_{{x_{2}}{x_{1}}}(\tau ,\eta )=A_{{x_{1}}{x_{2}}}^{*}(-\tau ,-\eta )}$

因此，我們目前得到 ${\displaystyle A_{x_{1}}\left(\tau ,\eta \right),A_{x_{2}}\left(\tau ,\eta \right)}$ (auto-terms) 和 ${\displaystyle A_{{x_{1}}{x_{2}}}(\tau ,\eta ),A_{{x_{2}}{x_{1}}}(\tau ,\eta )}$ (cross-terms) 的公式，我們再仔細的分析 auto-terms 和 cross-terms 分別發生最大值的位置。

Ambiguity Function 分析圖

首先，先看 Auto-terms：

${\displaystyle |A_{x_{1}}\left(\tau ,\eta \right)|}$最大值發生在 ${\displaystyle \tau =0,\eta =0}$的地方

${\displaystyle |A_{x_{2}}\left(\tau ,\eta \right)|}$最大值發生在 ${\displaystyle \tau =0,\eta =0}$的地方

而 Cross-terms：

${\displaystyle |A_{{x_{1}}{x_{2}}}(\tau ,\eta )|}$最大值發生在 ${\displaystyle \tau =t_{d},\eta =f_{d}}$的地方

${\displaystyle |A_{{x_{2}}{x_{1}}}(\tau ,\eta )|}$最大值發生在 ${\displaystyle \tau =-t_{d},\eta =-f_{d}}$的地方

換句話說，如果我們繪製一個 x軸為 ${\displaystyle \tau }$, y軸為 ${\displaystyle \eta }$ 的座標圖，Auto-terms發生在原點 ${\displaystyle (0,0)}$ 的位置，而 Cross-terms 則是以原點為對稱中心，在第一象限和第三象限的位置，

這也是為什麼可以透過一個低通函數來濾除雜訊，把主成分 Auto-terms 分離出來，避免交叉項的問題。