P1
For the DC Level in WGN detection problem assume that we wish to have P F A = 1 0 − 4 P_{FA}=10^{-4} P F A = 1 0 − 4 P D = 0.99 P_D=0.99 P D = 0 . 9 9 10 log 10 A 2 / σ 2 = − 30 dB 10\log_{10}A^2/\sigma^2=-30\text{dB} 1 0 log 1 0 A 2 / σ 2 = − 3 0 dB
Solution
P F A = 1 0 − 4 , P D = 0.99 , SNR = 10 log 10 A 2 / σ 2 = − 30 P D = Q ( Q − 1 ( P F A ) − N A 2 σ 2 Q − 1 ( P D ) = Q − 1 ( P F A ) − N A 2 σ 2 Q − 1 ( P D ) = Q − 1 ( P F A ) − N ⋅ 1 0 − 3 N = ( Q − 1 ( P F A ) − Q − 1 ( P D ) ) 2 / 1 0 − 3 = 36546.43 P_{FA}=10^{-4},\;P_D=0.99,\;\text{SNR}=10\log_{10}A^2/\sigma^2=-30\\[0.3cm] P_D=Q(Q^{-1}(P_{FA})-\sqrt{\frac{NA^2}{\sigma^2}}\\[0.3cm] Q^{-1}(P_D)=Q^{-1}(P_{FA})-\sqrt{\frac{NA^2}{\sigma^2}}\\[0.3cm] Q^{-1}(P_D)=Q^{-1}(P_{FA})-\sqrt{N\cdot 10^{-3}}\\[0.3cm] N=\left(Q^{-1}(P_{FA})-Q^{-1}(P_D)\right)^2/10^{-3}=36546.43 P F A = 1 0 − 4 , P D = 0 . 9 9 , SNR = 1 0 log 1 0 A 2 / σ 2 = − 3 0 P D = Q ( Q − 1 ( P F A ) − σ 2 N A 2 Q − 1 ( P D ) = Q − 1 ( P F A ) − σ 2 N A 2 Q − 1 ( P D ) = Q − 1 ( P F A ) − N ⋅ 1 0 − 3 N = ( Q − 1 ( P F A ) − Q − 1 ( P D ) ) 2 / 1 0 − 3 = 3 6 5 4 6 . 4 3 P2
Find the matrix prewhitener D D D
C = [ 1 ρ ρ 1 ] C=\left[\begin{matrix}1&\rho\\ \rho&1\end{matrix}\right] C = [ 1 ρ ρ 1 ] Hint. use the eigenanalysis decomposition V T C V = Λ , V^TCV=\Lambda, V T C V = Λ , V = [ V 1 V 2 ] , Λ = diag ( λ 1 , λ 2 ) V=[V_1\;V_2],\Lambda=\text{diag}(\lambda_1,\lambda_2) V = [ V 1 V 2 ] , Λ = diag ( λ 1 , λ 2 ) V T = V − 1 V^T=V^{-1} V T = V − 1
Solution
V T C V = Λ , V T = V − 1 C = V Λ V T C − 1 = V Λ − 1 V T = D T D D T = V T Λ − 1 / 2 , D = Λ − 1 / 2 V → V = [ 1 / 2 1 / 2 1 / 2 − 1 / 2 ] , Λ = [ 1 + ρ 0 0 1 − ρ ] Λ − 1 / 2 = [ 1 / 1 + ρ 0 0 1 / 1 − ρ ] D = [ 1 / 1 + ρ 0 0 1 / 1 − ρ ] ⋅ [ 1 / 2 1 / 2 1 / 2 − 1 / 2 ] = [ 1 / 2 + 2 ρ 1 / 2 + 2 ρ 1 / 2 + 2 ρ − 1 / 2 − 2 ρ ] V^TCV=\Lambda,\;V^T=V^{-1}\\[0.2cm] C=V\Lambda V^T\\[0.3cm] C^{-1}=V\Lambda^{-1}V^T=D^TD\\[0.2cm] D^T=V^T\Lambda^{-1/2},\;D=\Lambda^{-1/2}V\\[0.3cm] \rightarrow V=\left[\begin{matrix}1/\sqrt 2&1/\sqrt 2\\1/\sqrt2& -1/\sqrt 2\end{matrix}\right],\;\Lambda=\left[\begin{matrix}1+\rho&0\\0&1-\rho\end{matrix}\right]\\[0.3cm] \Lambda^{-1/2}=\left[\begin{matrix}1/\sqrt{1+\rho}&0\\0&1/\sqrt{1-\rho}\end{matrix}\right]\\[0.3cm] D=\left[\begin{matrix}1/\sqrt{1+\rho}&0\\0&1/\sqrt{1-\rho}\end{matrix}\right]\cdot \left[\begin{matrix}1/\sqrt2& 1/\sqrt2\\1/\sqrt2 & -1/\sqrt 2\end{matrix}\right]=\left[\begin{matrix}1/\sqrt{2+2\rho}&1/\sqrt{2+2\rho} \\1/\sqrt{2+2\rho}&-1/\sqrt{2-2\rho}\end{matrix}\right] V T C V = Λ , V T = V − 1 C = V Λ V T C − 1 = V Λ − 1 V T = D T D D T = V T Λ − 1 / 2 , D = Λ − 1 / 2 V → V = [ 1 / 2 1 / 2 1 / 2 − 1 / 2 ] , Λ = [ 1 + ρ 0 0 1 − ρ ] Λ − 1 / 2 = [ 1 / 1 + ρ 0 0 1 / 1 − ρ ] D = [ 1 / 1 + ρ 0 0 1 / 1 − ρ ] ⋅ [ 1 / 2 1 / 2 1 / 2 − 1 / 2 ] = [ 1 / 2 + 2 ρ 1 / 2 + 2 ρ 1 / 2 + 2 ρ − 1 / 2 − 2 ρ ] 