在信号处理中,匹配滤波器可以用来解调基频带脉冲信号,基频带脉冲信号意指信号内容为同一波形信号乘上一个常数,在每个周期出现,每个周期中代表着或多或少的信息量。匹配滤波器解调出来的结果其SNR (Signal Noise Ratio)为最大的,匹配滤波器需要事先知道 此条目包含过多行话或专业术语,可能需要简化或提出进一步解释。 (2014年6月24日) 此条目没有列出任何参考或来源。 (2014年6月24日) 1.传送的信号 2.信号的同步 才能解调出传送的信号。 此外,匹配滤波器也可用于模式识别 、相似度测试(similarity measure)。 最高SNR证明 假设 g(t):传送信号 w(t):可加性高斯白噪声 x(t) = g(t) + w(t) h(t):未知波形 y(t):解调结果 1. x ( t ) = g ( t ) + w ( t ) {\displaystyle 1.x(t)=g(t)+w(t)} 2. y ( t ) = [ g ( t ) + w ( t ) ] ∗ h ( t ) {\displaystyle 2.y(t)=[g(t)+w(t)]\ast h(t)} = g ( t ) ∗ h ( t ) + w ( t ) ∗ h ( t ) {\displaystyle =g(t)\ast h(t)+w(t)\ast h(t)} = G ( t ) + N ( t ) {\displaystyle =G(t)+N(t)} 3. S N R = | G ( T ) | 2 / E [ N 2 ( T ) ] | {\displaystyle 3.SNR=|G(T)|^{2}/E[N^{2}(T)]|} SNR = 信号瞬间功率 / 噪声平均功率 信号瞬间功率 | G ( T ) | 2 = ∫ − ∞ ∞ H ( f ) G ( f ) e j 2 π f T d f {\displaystyle |G(T)|^{2}=\int _{-\infty }^{\infty }H(f)G(f)e^{j2\pi fT}\,df} 噪声平均功率 E [ N 2 ( T ) ] = N 0 2 ∫ − ∞ ∞ | H ( f ) | 2 d f {\displaystyle E[N^{2}(T)]={\frac {N_{0}}{2}}\int _{-\infty }^{\infty }|H(f)|^{2}\,df} S N R = ∫ − ∞ ∞ H ( f ) G ( f ) e j 2 π f T d f N 0 2 ∫ − ∞ ∞ | H ( f ) | 2 d f {\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}}} ≤ ∫ − ∞ ∞ | H ( f ) | 2 e j 2 π f T d f ∫ − ∞ ∞ | G ( f ) e j 2 π f T | 2 d f N 0 2 ∫ − ∞ ∞ | H ( f ) | 2 d f {\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}}} = 2 N 0 ∫ − ∞ ∞ | G ( f ) | 2 d f {\displaystyle ={\frac {2}{N_{0}}}\int _{-\infty }^{\infty }|G(f)|^{2}\,df} 4. 当 H o p t ( f ) = k [ G ( f ) e j 2 π f T ] ∗ {\displaystyle H_{opt}(f)=k[G(f)e^{j2\pi fT}]^{*}} , S N R m a x = 2 N 0 ∫ − ∞ ∞ | G ( f ) | 2 d f {\displaystyle SNR_{max}={\frac {2}{N_{0}}}\int _{-\infty }^{\infty }|G(f)|^{2}\,df} 所以 h o p t ( t ) = k ∫ − ∞ ∞ G ( − f ) e − j 2 π f T e j 2 π f t d f {\displaystyle h_{opt}(t)=k\int _{-\infty }^{\infty }G(-f)e^{-j2\pi fT}e^{j2\pi ft}\,df} = k ∫ − ∞ ∞ G ( z ) e − j 2 π f ( T − t ) d z {\displaystyle =k\int _{-\infty }^{\infty }G(z)e^{-j2\pi f(T-t)}\,dz} = k g ( T − t ) {\displaystyle =kg(T-t)} (备注) 柯西-施瓦茨不等式 若 ∫ − ∞ ∞ | A ( x ) | 2 d x < ∞ {\displaystyle \int _{-\infty }^{\infty }|A(x)|^{2}\,dx<\infty } 且 ∫ − ∞ ∞ | B ( x ) | 2 d x < ∞ {\displaystyle \int _{-\infty }^{\infty }|B(x)|^{2}\,dx<\infty } 则 | ∫ − ∞ ∞ A ( x ) B ( x ) d x | 2 ≤ ∫ − ∞ ∞ | A ( x ) | 2 d x ∫ − ∞ ∞ | B ( x ) | 2 d x {\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} 当 A = k B ∗ {\displaystyle A=kB^{*}} 时,等号成立。 Remove ads匹配滤波器频率响应 x = s + v , {\displaystyle \ x=s+v,\,} R v = E { v v H } . {\displaystyle \ R_{v}=E\{vv^{\mathrm {H} }\}.\,} S N R = | y s | 2 E { | y v | 2 } . {\displaystyle \mathrm {SNR} ={\frac {|y_{s}|^{2}}{E\{|y_{v}|^{2}\}}}.} | y s | 2 = y s H y s = h H s s H h . {\displaystyle \ |y_{s}|^{2}={y_{s}}^{\mathrm {H} }y_{s}=h^{\mathrm {H} }ss^{\mathrm {H} }h.\,} E { | y v | 2 } = E { y v H y v } = E { h H v v H h } = h H R v 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.\,} S N R = h H s s H h h H R v h . {\displaystyle \mathrm {SNR} ={\frac {h^{\mathrm {H} }ss^{\mathrm {H} }h}{h^{\mathrm {H} }R_{v}h}}.} 如果我们限制分母为1, 最大化 S N R {\displaystyle \mathrm {SNR} } 的问题可以被简化为最大化分子. 于是可以使用 拉格朗乘数 h H R v h = 1 {\displaystyle \ h^{\mathrm {H} }R_{v}h=1} L = h H s s H h + λ ( 1 − h H R v h ) {\displaystyle \ {\mathcal {L}}=h^{\mathrm {H} }ss^{\mathrm {H} }h+\lambda (1-h^{\mathrm {H} }R_{v}h)} ∇ h ∗ L = s s H h − λ R v h = 0 {\displaystyle \ \nabla _{h^{*}}{\mathcal {L}}=ss^{\mathrm {H} }h-\lambda R_{v}h=0} ( s s H ) h = λ R v h {\displaystyle \ (ss^{\mathrm {H} })h=\lambda R_{v}h} h H ( s s H ) h = λ h H R v h . {\displaystyle \ h^{\mathrm {H} }(ss^{\mathrm {H} })h=\lambda h^{\mathrm {H} }R_{v}h.} 因为 s s H {\displaystyle ss^{\mathrm {H} }} 是一维, 他只有一个非零特征值. 此特征值= λ max = s H R v − 1 s , {\displaystyle \ \lambda _{\max }=s^{\mathrm {H} }R_{v}^{-1}s,} h = 1 s H R v − 1 s R v − 1 s . {\displaystyle \ h={\frac {1}{\sqrt {s^{\mathrm {H} }R_{v}^{-1}s}}}R_{v}^{-1}s.} Remove ads匹配滤波器模式辨识 若欲侦测一特定信号 h[n],我们可以将h[n]时域反向并取共轭,当做滤波器。 一维信号 y [ n ] = x [ n ] ∗ h ∗ [ − n ] = ∑ τ = τ 1 τ 2 x [ n − τ ] h ∗ [ − τ ] = ∑ τ = τ 1 τ 2 x [ n + τ ] h ∗ [ τ ] {\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] :数入信号 ,h[n]:欲侦测的特定信号,且假设当 τ 1 ≤ n ≤ τ 2 {\displaystyle \tau _{1}\leq n\leq \tau _{2}} 时, h[n]≠0 二维信号 y [ m , n ] = x [ m , n ] ∗ h ∗ [ − m , − n ] = ∑ τ = τ 1 τ 2 ∑ ρ = ρ 1 ρ 2 x [ m + τ , n + ρ ] h ∗ [ τ , ρ ] {\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 ]} 假设当 τ 1 ≤ m ≤ τ 2 , ρ 1 ≤ m ≤ ρ 2 {\displaystyle \tau _{1}\leq m\leq \tau _{2},\rho _{1}\leq m\leq \rho _{2}} 时, h[m,n]≠0 模拟结果: 未标准化而造成的计算误差 y[n] = x[n]*h*[-n] 但由于卷积是线性的,当信号能量大,算出来的值也会跟着变大而有误差,因此我们需要标准化。 标准化公式 一维信号 当 ∑ s = n + τ 1 n + τ 2 | x [ s ] | 2 {\displaystyle \sum _{s=n+\tau _{1}}^{n+\tau _{2}}|x[s]|^{2}} ≠0 y [ n ] = {\displaystyle y[n]=} ∑ τ = τ 1 τ 2 x [ n + τ ] h ∗ [ τ ] ∑ s = n + τ 1 n + τ 2 | x [ s ] | 2 ∑ s = τ 1 τ 2 | h [ s ] | 2 {\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}}}} 当 ∑ s = n + τ 1 n + τ 2 | x [ s ] | 2 {\displaystyle \sum _{s=n+\tau _{1}}^{n+\tau _{2}}|x[s]|^{2}} =0 y [ n ] = 0 {\displaystyle y[n]=0} 二维信号 当 ∑ s = m + τ 1 m + τ 2 ∑ v = n + ρ 1 n + ρ 2 | x [ s , v ] | 2 {\displaystyle \sum _{s=m+\tau _{1}}^{m+\tau _{2}}\sum _{v=n+\rho _{1}}^{n+\rho _{2}}|x[s,v]|^{2}} ≠0 y [ m , n ] = {\displaystyle y[m,n]=} ∑ τ = τ 1 τ 2 ∑ ρ = ρ 1 ρ 2 x [ m + τ , n + ρ ] h ∗ [ τ , ρ ] ∑ s = m + τ 1 m + τ 2 ∑ v = n + ρ 1 n + ρ 2 | x [ s , v ] | 2 ∑ s = τ 1 τ 2 ∑ v = ρ 1 ρ 2 | h [ s , v ] | 2 {\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}}}} 当 ∑ s = m + τ 1 m + τ 2 ∑ v = n + ρ 1 n + ρ 2 | x [ s , v ] | 2 {\displaystyle \sum _{s=m+\tau _{1}}^{m+\tau _{2}}\sum _{v=n+\rho _{1}}^{n+\rho _{2}}|x[s,v]|^{2}} = 0 y [ m , n ] = 0 {\displaystyle y[m,n]=0} 标准化后的模拟结果: 标准化后可减少计算误差 Remove ads参考文献 Haykin,S. / Moher,M. Haykin: Communication Systems 5/E (中文). Jian-Jiun Ding, Advanced Digital Signal Processing, the Department of Electrical Engineering, National Taiwan University (NTU), Taipei, Taiwan, 2015. 参见 霍夫变换 拉东变换 周期图法 迭代稀疏渐近最小方差算法 Loading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.Remove ads