考虑在高斯噪声背景中检测高斯信号的问题。设信号模型为 H0:x(t)=n(t) 0≤t≤T H1:x(t)=s(t)+n(t) 0≤t≤T

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考虑在高斯噪声背景中检测高斯信号的问题。设信号模型为

   H0:x(t)=n(t),0≤t≤T

   H1:x(t)=s(t)+n(t),0≤t≤T

 其中,n(t)和s(t)分别是零均值的高斯噪声和高斯信号,其带宽限于|ω|﹤Ω=2πB,功率谱密度分别为N0/2和S0/2。假如以π/Ω为间隔取2BT个样本的方式进行统计信号检测,试求似然比检验系统。

热搜:关于高斯噪声,下列说法错误的是

答案解析

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  • 因为噪声的功率谱密度为
       latex.action?latex=UChuKSjPiSk9XGxlZnRce1xiZWdpbnthcnJheX17Y31cZnJhY3tOX3swfX17Mn0sfM%2BJfO%2B5pM6p%0D%0AXFwwLOWFtuWugyAgXGVuZHthcnJheX0%3D
       所以,其自相关函数为
       latex.action?latex=cl97bn0oz4QpPVxmcmFjezF9ezLPgH1caW50X3st4oiefV574oiefVBfe259KM%2BJKWVee2rPic%2BE%0D%0AfWTPiQ%3D%3D
       latex.action?latex=PVxmcmFjezF9ezLPgH1caW50X3stzql9XnvOqX1cZnJhY3tOX3swfX17Mn1lXntqz4nPhH1kz4k%3D
       latex.action?latex=PVxmcmFje05fezB9fXs0z4B9XGZyYWN7MX17as%2BJfShlXntqzqnPhH0tZV57LWrOqc%2BEfSk%3D
       latex.action?latex=PVxmcmFje05fezB9fXsyz4DPhH1zaW7Oqc%2BEPVxmcmFje05fezB9zql9ezLPgH1cZnJhY3tzaW7O%0D%0Aqc%2BEfXvOqc%2BEfT1OX3swfUJcZnJhY3tzaW7Oqc%2BEfXvOqc%2BEfQ%3D%3D
       类似地,信号的功率谱密度为
      latex.action?latex=UChzKSjPiSk9XGxlZnRce1xiZWdpbnthcnJheX17Y31cZnJhY3tOX3swfX17Mn0sfM%2BJfO%2B5pM6p%0D%0AXFwwLOWFtuWug1xlbmR7YXJyYXl9
       其自相关函数为
      latex.action?latex=cl97c30oz4QpPVxmcmFjezF9ezLPgH1caW50X3stzql9XnvOqX1cZnJhY3tTX3swfX17Mn1lXntq%0D%0Az4nPhH1kz4k9U197MH1CXGZyYWN7c2luzqnPhH17zqnPhH0%3D
       这样,以Ωτ-π,即latex.action?latex=z4Q9XGZyYWN7z4B9e86pfQ%3D%3D为间隔进行采样,所得样本xk(k=1,2,…,N)是互不相关的高斯随机变量,因而也是相互统计独立的。在(0,T)时间内的样本数为
        latex.action?latex=Tj1cZnJhY3tUfXvPhH09XGZyYWN7VM6pfXvPhH09MkJU
       在假设H0下,样本xk(k=1,2,…,2BT)的均值、方差和相应的x=(x1,x2,…,x2BT)T的概率密度函数分别为
       E(xk|H0)=E(nk)=0
        latex.action?latex=VmFyKHhfe2t9fEhfezB9KT1FKG5fe2t9XnsyfSk9cl97bn0oz4QpfF97z4Q9MH09Tl97MH1C
       和
       latex.action?latex=UCh4fEhfezB9KT1cbGVmdCggXGZyYWN7MX17Ms%2BATl97MH1CfSBccmlnaHQpXntCVH1leHBcbGVm%0D%0AdFsgLVxzdW1fe2s9MX1eezJCVH1cZnJhY3t4X3trfV57Mn19ezJOX3swfUJ9IFxyaWdodF0%3D
      在假设H1下,样本xk(k=1,2..,2BT)的均值、方差和相应的x=(x1,x2...,x2BT)T的概率密度函数分别为
      E(xk|H1)=E(sk+nk)=0
      latex.action?latex=VmFyKHhfe2t9fEhfezF9KT1FWyhzX3trfStuX3trfSleezJ9XT1FKFNfe0t9XnsyfSkrRShuX3tr%0D%0AfV57Mn0p
      latex.action?latex=PXJfe3N9KM%2BEKXxfe8%2BEPTB9K3Jfe259KM%2BEKXxfe8%2BEPTB9PVNfezB9QitOX3swfUI9KFNfezB9%0D%0AK05fezB9KUI%3D
      和
       latex.action?latex=UCh4fEhfezF9KT1cbGVmdCggXGZyYWN7MX17Ms%2BAKFNfezB9K05fezB9KUJ9IFxyaWdodClee0JU%0D%0AfWV4cFxsZWZ0WyAtXHN1bV97az0xfV57MkJUfVxmcmFje3hfe2t9XnsyfX17MihTX3swfStOX3sw%0D%0AfSlCfSBccmlnaHRd
      这样,似然比检测门限为η的似然比检验为
      latex.action?latex=zrsoeCk9XGZyYWN7UCh4fEhfezF9KX17UCh4fEhfezB9KX09XGZyYWN7XGxlZnQoIFxmcmFjezF9%0D%0AezLPgChTX3swfStOX3swfSlCfSBccmlnaHQpXntCVH1leHBcbGVmdFsgLVxzdW1fe2s9MX1eezJC%0D%0AVH1cZnJhY3t4X3trfV57Mn19ezIoU197MH0rTl97MH1CKX0gXHJpZ2h0XX17XGxlZnQoIFxmcmFj%0D%0AezF9ezLPgE5fezB9Qn0gXHJpZ2h0KV57QlR9ZXhwXGxlZnRbIC0gXHN1bV97az0xfV57MkJUfVxm%0D%0AcmFje3hfe2t9XnsyfX17Mk5fezB9Qn1ccmlnaHRdfQ%3D%3D
       xhjcCF786D.jpg
      化简整理得判决表达式
      xhjcCFF1B4.jpg
      最佳监测系统如图所示。
     xhjcD0C484.jpg