匹配濾波器

假設 g(t)：傳送訊號

w(t)：可加性高斯白雜訊

x(t) = g(t) + w(t)

h(t)：未知波形

y(t)：解調結果

${\displaystyle 1.x(t)=g(t)+w(t)}$

${\displaystyle 2.y(t)=[g(t)+w(t)]\ast h(t)}$　

${\displaystyle =g(t)\ast h(t)+w(t)\ast h(t)}$

${\displaystyle =G(t)+N(t)}$

${\displaystyle 3.SNR=|G(T)|^{2}/E[N^{2}(T)]|}$

SNR = 信號瞬間功率 / Noise平均功率

信號瞬間功率

${\displaystyle |G(T)|^{2}=\int _{-\infty }^{\infty }H(f)G(f)e^{j2\pi fT}\,df}$

雜訊平均功率

${\displaystyle E[N^{2}(T)]={\frac {N_{0}}{2}}\int _{-\infty }^{\infty }|H(f)|^{2}\,df}$

${\displaystyle SNR={\frac {\int _{-\infty }^{\infty }H(f)G(f)e^{j2\pi fT}\,df}{{\frac {N_{0}}{2}}\int _{-\infty }^{\infty }|H(f)|^{2}\,df}}}$

${\displaystyle \leq {\frac {\int _{-\infty }^{\infty }|H(f)|^{2}e^{j2\pi fT}\,df\int _{-\infty }^{\infty }|G(f)e^{j2\pi fT}|^{2}\,df}{{\frac {N_{0}}{2}}\int _{-\infty }^{\infty }|H(f)|^{2}\,df}}}$

${\displaystyle ={\frac {2}{N_{0}}}\int _{-\infty }^{\infty }|G(f)|^{2}\,df}$

4. 當

${\displaystyle H_{opt}(f)=k[G(f)e^{j2\pi fT}]^{*}}$ , ${\displaystyle SNR_{max}={\frac {2}{N_{0}}}\int _{-\infty }^{\infty }|G(f)|^{2}\,df}$

所以

${\displaystyle h_{opt}(t)=k\int _{-\infty }^{\infty }G(-f)e^{-j2\pi fT}e^{j2\pi ft}\,df}$

${\displaystyle =k\int _{-\infty }^{\infty }G(z)e^{-j2\pi f(T-t)}\,dz}$

${\displaystyle =kg(T-t)}$

(備註)Cauchy-Schwartz inequality:

若

${\displaystyle \int _{-\infty }^{\infty }|A(x)|^{2}\,dx<\infty }$ 且 ${\displaystyle \int _{-\infty }^{\infty }|B(x)|^{2}\,dx<\infty }$

則

${\displaystyle |\int _{-\infty }^{\infty }A(x)B(x)\,dx|^{2}\leq \int _{-\infty }^{\infty }|A(x)|^{2}\,dx\int _{-\infty }^{\infty }|B(x)|^{2}\,dx}$

當 ${\displaystyle A=kB^{*}}$時，等號成立。

${\displaystyle \ x=s+v,\,}$

${\displaystyle \ R_{v}=E\{vv^{\mathrm {H} }\}.\,}$

${\displaystyle \mathrm {SNR} ={\frac {|y_{s}|^{2}}{E\{|y_{v}|^{2}\}}}.}$

${\displaystyle \ |y_{s}|^{2}={y_{s}}^{\mathrm {H} }y_{s}=h^{\mathrm {H} }ss^{\mathrm {H} }h.\,}$

${\displaystyle \ E\{|y_{v}|^{2}\}=E\{{y_{v}}^{\mathrm {H} }y_{v}\}=E\{h^{\mathrm {H} }vv^{\mathrm {H} }h\}=h^{\mathrm {H} }R_{v}h.\,}$

${\displaystyle \mathrm {SNR} ={\frac {h^{\mathrm {H} }ss^{\mathrm {H} }h}{h^{\mathrm {H} }R_{v}h}}.}$

如果我們限制分母為1, 最大化 ${\displaystyle \mathrm {SNR} }$ 的問題可以被簡化為最大化分子.

於是可以使用 拉格朗乘數

${\displaystyle \ h^{\mathrm {H} }R_{v}h=1}$

${\displaystyle \ {\mathcal {L}}=h^{\mathrm {H} }ss^{\mathrm {H} }h+\lambda (1-h^{\mathrm {H} }R_{v}h)}$

${\displaystyle \ \nabla _{h^{*}}{\mathcal {L}}=ss^{\mathrm {H} }h-\lambda R_{v}h=0}$

${\displaystyle \ (ss^{\mathrm {H} })h=\lambda R_{v}h}$

${\displaystyle \ h^{\mathrm {H} }(ss^{\mathrm {H} })h=\lambda h^{\mathrm {H} }R_{v}h.}$

因為 ${\displaystyle ss^{\mathrm {H} }}$ 是一維, 他只有一個非零特徵值. 此特徵值=

${\displaystyle \ \lambda _{\max }=s^{\mathrm {H} }R_{v}^{-1}s,}$

${\displaystyle \ h={\frac {1}{\sqrt {s^{\mathrm {H} }R_{v}^{-1}s}}}R_{v}^{-1}s.}$

若欲偵測一特定信號 h[n]，我們可以將h[n]時域反向並取共軛，當做濾波器。

一維信號

${\displaystyle y[n]=x[n]*h^{*}[-n]=\sum _{\tau =\tau _{1}}^{\tau _{2}}x[n-\tau ]h^{*}[-\tau ]=\sum _{\tau =\tau _{1}}^{\tau _{2}}x[n+\tau ]h^{*}[\tau ]}$

x[n] :數入信號 ，h[n]：欲偵測的特定信號，且假設當${\displaystyle \tau _{1}\leq n\leq \tau _{2}}$ 時， h[n]≠0

二維信號

${\displaystyle y[m,n]=x[m,n]*h^{*}[-m,-n]=\sum _{\tau =\tau _{1}}^{\tau _{2}}\sum _{\rho =\rho _{1}}^{\rho _{2}}x[m+\tau ,n+\rho ]h^{*}[\tau ,\rho ]}$

假設當${\displaystyle \tau _{1}\leq m\leq \tau _{2},\rho _{1}\leq m\leq \rho _{2}}$時， h[m,n]≠0

模擬結果：

未標準化而造成的計算誤差 y[n] = x[n]*h*[-n ]

但由於卷積是線性的，當信號能量大，算出來的值也會跟著變大而有誤差，因此我們需要標準化。

標準化公式

一維信號

當 ${\displaystyle \sum _{s=n+\tau _{1}}^{n+\tau _{2}}|x[s]|^{2}}$ ≠0

${\displaystyle y[n]=}$${\displaystyle {\sum _{\tau =\tau _{1}}^{\tau _{2}}x[n+\tau ]h^{*}[\tau ]} \over {\sqrt {\sum _{s=n+\tau _{1}}^{n+\tau _{2}}|x[s]|^{2}\sum _{s=\tau _{1}}^{\tau _{2}}|h[s]|^{2}}}}$

當 ${\displaystyle \sum _{s=n+\tau _{1}}^{n+\tau _{2}}|x[s]|^{2}}$ =0

${\displaystyle y[n]=0}$

二維信號

當 ${\displaystyle \sum _{s=m+\tau _{1}}^{m+\tau _{2}}\sum _{v=n+\rho _{1}}^{n+\rho _{2}}|x[s,v]|^{2}}$ ≠0

${\displaystyle y[m,n]=}$${\displaystyle {\sum _{\tau =\tau _{1}}^{\tau _{2}}\sum _{\rho =\rho _{1}}^{\rho _{2}}x[m+\tau ,n+\rho ]h^{*}[\tau ,\rho ]} \over {\sqrt {\sum _{s=m+\tau _{1}}^{m+\tau _{2}}\sum _{v=n+\rho _{1}}^{n+\rho _{2}}|x[s,v]|^{2}\sum _{s=\tau _{1}}^{\tau _{2}}\sum _{v=\rho _{1}}^{\rho _{2}}|h[s,v]|^{2}}}}$

當 ${\displaystyle \sum _{s=m+\tau _{1}}^{m+\tau _{2}}\sum _{v=n+\rho _{1}}^{n+\rho _{2}}|x[s,v]|^{2}}$ = 0

${\displaystyle y[m,n]=0}$

標準化後的模擬結果：

標準化後可減少計算誤差

1. Haykin,S. / Moher,M. （中文）.
2. Jian-Jiun Ding, Advanced Digital Signal Processing, the Department of Electrical Engineering, National Taiwan University (NTU), Taipei, Taiwan, 2015.

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