[DetnEst] Assignment 7

KBC·2024년 12월 12일
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Detection and Estimation

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P1

Let Y=X2Y=X^2, and let XX be a uniform random variable over (1,1)(-1, 1)

  1. Find the linear minimum mean squared estimator of YY in terms of XX
  2. Find its mean squared error

Solution

i) mse(Y^)=E[(YY^)2]=E[(YaXb)2]mse(Y^)a=E[2(YaXb)(X)]=E[2aX2]=0=2aE[X2]=0a=0mse(Y^)b=E[2(YaXb)(1)]=E[2Y+2b]=0b=E[Y]=E[X2]=1112x2dx=13Y^=13\text{i) mse}(\hat Y)=E[(Y-\hat Y)^2]=E[(Y-aX-b)^2]\\[0.2cm] \frac{\partial\text{mse}(\hat Y)}{\partial a}=E\left[2(Y-aX-b)\cdot (-X)\right]=E[2aX^2]=0\\[0.2cm] =2aE[X^2]=0\rightarrow a=0\\[0.3cm] \frac{\partial \text{mse}(\hat Y)}{\partial b}=E[2(Y-aX-b)(-1)]=E[-2Y+2b]=0\\[0.2cm] \rightarrow b =E[Y]=E[X^2]=\int^1_{-1}\frac{1}{2}x^2dx=\frac{1}{3} \\[0.3cm] \therefore \hat Y=\frac{1}{3}
ii) mse(Y^)=E[(Y13)2]=E[Y223Y+19]=E[X4]23E[X2]+19=1112x4dx2313+19=1529+19=845\text{ii) mse}(\hat Y)=E\left[\left(Y-\frac{1}{3}\right)^2\right]=E\left[Y^2-\frac{2}{3}Y+\frac{1}{9}\right]\\[0.2cm] =E[X^4]-\frac{2}{3}E[X^2]+\frac{1}{9}\\[0.3cm] =\int^1_{-1}\frac{1}{2}x^4dx-\frac{2}{3}\cdot\frac{1}{3}+\frac{1}{9}=\frac{1}{5}-\frac{2}{9}+\frac{1}{9}=\frac{8}{45}

P2

Let us consider the estimation of the value of an (unknown) constant s=s[n]s=s[n], nn given measurements

x[n]=s[n]+w[n]x[n]=s[n]+w[n]

that are corrupted (but uncorrelated) with a zero mean white noise w[n]w[n] that has variance σw2\sigma^2_w

  1. Write the estimation problem as a Kalman filter problem and compute the Kalman gain K[n]K[n] and the variance of the estimation error M[nn]M[n|n]
    (You are asked to find close forms of K[n]K[n] and M[nn]M[n|n] as functions of n,σw2,M[11]=e02n,\sigma^2_w,M[-1|-1]=e^2_0
  2. Now suppose that we do not have no a priori information about ss
    (i.e. s^[11]=0\hat s[-1|-1]=0 and e02e^2_0\rightarrow \infty)
    Show that the Kalman filter simply becomes the sample mean
    s^=1Nn=0N1x[n]\hat s=\frac{1}{N}\sum^{N-1}_{n=0}x[n]

Solution

i) Since s[n]=s,  x[n]=s+w[n],  s[n]=as[n1]+u[n]s=as+u[n]a=1,  u[n]=0N(0,0)M[nn1]=a2M[n1n1]+σa2,  σa2=0=M[n1n1]=e02 where M[11]=e02,  s^[11]=0M[01]=e02K[0]=e02σn2+e02M[00]=(σn2σn2+e02)e02=M[10]K[1]=σn2e02en2+e02σn2+σn2+e02σn2+e02=e02σn2+2e02M[11]=σn2+e02σn2+e02σn2e02σn2+e02=σn2e02σn2+2e02=M[21]K[2]=σn2+e02σn2+2e02σn2+σn2e02σn2+2e02=e02σn2+3e02If) M[n1n1]=M[nn1]=σn2e02σn2+ne02,K[n]=e02σn2+(n+1)e02Then) M[nn]=σn2e02σn2+(n+1)e02=M[n+1n]K[n+1]=σn2e02σn2+(n+1)e02σn2+σn2e02σn2+(n+1)e02=e02σn2+(n+2)e02\text{i) Since }s[n]=s,\;x[n]=s+w[n],\;s[n]=as[n-1]+u[n]\\[0.2cm] \rightarrow s=as+u[n]\\[0.2cm] \rightarrow a=1,\;u[n]=0\sim\mathcal{N}(0, 0)\\[0.2cm] \rightarrow M[n|n-1]=a^2M[n-1|n-1]+\sigma^2_a,\;\sigma^2_a=0\\[0.2cm] =M[n-1|n-1]=e^2_0\text{ where }M[-1|-1]=e^2_0,\;\hat s[-1|-1]=0\\[0.2cm] \rightarrow M[0|-1]=e^2_0\\[0.2cm] K[0]=\frac{e^2_0}{\sigma^2_n+e^2_0}\rightarrow M[0|0]=\left(\frac{\sigma^2_n}{\sigma^2_n+e^2_0}\right)e^2_0=M[1|0]\\[0.3cm] K[1]=\frac{\frac{\sigma^2_ne^2_0}{e^2_n+e^2_0}}{\sigma^2_n+\frac{\sigma^2_n+e^2_0}{\sigma^2_n+e^2_0}}=\frac{e^2_0}{\sigma^2_n+2e^2_0}\\[0.3cm] M[1|1]=\frac{\sigma^2_n+e^2_0}{\sigma^2_n+e^2_0}\cdot\frac{\sigma^2_ne^2_0}{\sigma^2_n+e^2_0}=\frac{\sigma^2_ne^2_0}{\sigma^2_n+2e^2_0}=M[2|1]\\[0.3cm] K[2]=\frac{\frac{\sigma^2_n+e^2_0}{\sigma^2_n+2e^2_0}}{\sigma^2_n+\frac{\sigma^2_ne^2_0}{\sigma^2_n+2e^2_0}}=\frac{e^2_0}{\sigma^2_n+3e^2_0}\\[0.4cm] \text{If) }M[n-1|n-1]=M[n|n-1]=\frac{\sigma^2_ne^2_0}{\sigma^2_n+ne^2_0},\\[0.2cm] K[n]=\frac{e^2_0}{\sigma^2_n+(n+1)e^2_0}\\[0.2cm] \text{Then) }M[n|n]=\frac{\sigma^2_ne^2_0}{\sigma^2_n+(n+1)e^2_0}=M[n+1|n]\\[0.3cm] K[n+1]=\frac{\frac{\sigma^2_ne^2_0}{\sigma^2_n+(n+1)e^2_0}}{\sigma^2_n+\frac{\sigma^2_ne^2_0}{\sigma^2_n+(n+1)e^2_0}}=\frac{e^2_0}{\sigma^2_n+(n+2)e^2_0}
  • 다음 문제
    s^[nn1]=as^[n1n1]=s^[n1n1]e02,K[n]=1n+1,M[nn]=σn2n+1s^[nn]=s^[n1n1]+K[n](x[n]s^[n1n1])=s^[n1n1]+x[n]n+1s^[n1n1]n+1=nn+1s^[n1n1]+x[n]n+1()\hat s[n|n-1]=a\hat s[n-1|n-1]=\hat s[n-1|n-1]\\[0.2cm] e^2_0\rightarrow \infty,\quad K[n]=\frac{1}{n+1},\quad M[n|n]=\frac{\sigma^2_n}{n+1}\\[0.3cm] \hat s[n|n]=\hat s[n-1|n-1]+K[n](x[n]-\hat s[n-1|n-1])\\[0.3cm] =\hat s[n-1|n-1]+\frac{x[n]}{n+1}-\frac{\hat s[n-1|n-1]}{n+1}\\[0.3cm] =\frac{n}{n+1}\hat s[n-1|n-1]+\frac{x[n]}{n+1}\cdots \color{red}(*)
where s^[11]=0s^[00]=x[0]s^[11]=12x[0]+12x[1]()if s^[n1n1]=1ni=1N1x[i]s^[nn]=nn+1(1ni=0N1x[i])+x[n+1]n+1=1n+1i=0nx[i]\text{where }\hat s[-1|-1]=0\rightarrow \hat s[0|0]=x[0]\\[0.2cm] \hat s[1|1]=\frac{1}{2}x[0]+\frac{1}{2}x[1]\\[0.3cm] \textcolor{red}{(*)} \rightarrow \text{if }\hat s[n-1|n-1]=\frac{1}{n}\sum^{N-1}_{i=1}x[i]\\[0.3cm] \hat s[n|n]=\frac{n}{n+1}\cdot \left(\frac{1}{n}\sum^{N-1}_{i=0}x[i]\right)+\frac{x[n+1]}{n+1}\\[0.3cm] =\frac{1}{n+1}\sum^n_{i=0}x[i]

P3

For the scalar state-scalar observation Kalman filter assume that σn2=0\sigma^2_n=0 for all nn so that we observe s[n]s[n] directly. Find the innovation wequence. Is it white?

Solution

σn2=0x[n]=s[n]+w[n]=as[n1]+u[n]K[n]=M[nn1]M[nn1]=1s^[nn]=s^[nn1]+x[n]s^[nn1]=x[n]Then, x[n]=x[n]s^[nn1]=s[n]s^[nn1]=s[n]as^[n1n1]=s[n]as[n1]=u[n]White\sigma^2_n=0\rightarrow x[n]=s[n]+w[n]=as[n-1]+u[n]\\[0.2cm] \rightarrow K[n]=\frac{M[n|n-1]}{M[n|n-1]}=1\\[0.3cm] \hat s[n|n]=\hat s[n|n-1]+x[n]-\hat s[n|n-1]=x[n]\\[0.2cm] \text{Then, }x[n]=x[n]-\hat s[n|n-1]\\[0.2cm] =s[n]-\hat s[n|n-1]\\[0.2cm] =s[n]-a\hat s[n-1|n-1]\\[0.2cm] =s[n]-as[n-1]=u[n]\\[0.2cm] \therefore \text{White}
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