Assignment 4 (Solutions) NPTEL MOOC (Bayesian/ MMSE Estimation for MIMO/OFDM Wireless Communications)

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1 Assignment 4 Solutions NPTEL MOOC Bayesian/ MMSE Estimation for MIMO/OFDM Wireless Communications The system model can be written as, y hx + The MSE of the MMSE estimate ĥ of the aboe mentioned system model is gien by, E{ ĥ h } r hh r hy R yy r yh σ h σ hx H σ hxx H + σ I xσ h σ4 h x x + σ + σ / x Gien data: µ h + j, /, N 4, σ 3 db 0 log σ 3 σ + j + j x j j, y j + j, + j j x + j + j + j + + j Substituting all the alues in equation, we get E{ ĥ h } 9 + /4 /

2 Ans a Refer to the notes of week 3 for this question MMSE estimate of the complex fading coefficient h is gien by, ĥ ĥr + jĥi From the solution of problem, the MSEs of the real, imaginary parts of ĥ can be obtained as, Ansc MSE of the real part of ĥ MSE of the imaginary part of ĥ E{ ĥr h R } E{ ĥi h I } E{ ĥ h } 8 + σ / x 3 Let h R denotes the real part of the true parameter h and ĥr be the real part of the estimate ĥ Further, ĥr h R gies the estimation error in the real part of the estimate Also, from the solutions to problem and we can say, h R N ĥ R, 8 Therefore, hr ĥr is distributed as a zero-mean Gaussian with ariance /8 Hence, ĥr h R N 0, 8 Further, ĥr h R 8 is a zero-mean unit-ariance Gaussian RV Probability that the real part of the MMSE estimate ĥ lies within a radius / of the unknown parameter h can be calculated as follows, Pr ĥr h R ĥr h R Pr ĥr h R Pr { } ĥr h R ĥr Pr h R + Pr 8 ĥr h R Pr Q Q Further, since the errors in the real and imaginary parts are independent as they are Gaussian, the probability that both the real and imaginary parts of the MMSE estimate ĥ lie within a radius of / from the 8

3 real and imaginary parts of the unknown parameter h respectiely is Q 9 Ans b 4 To estimate the unknown parameter h, we hae each obseration as yk h + k, for k N, where k N 0, σk, h N µ h, By stacking N such obserations, we obtain obseration ector as y h +, where mean of the noise ector is E{} 0 and its coariance matrix is denoted by C E{ T } So, the mean of the obseration ector is denoted by µ y E{y} E{h + } µ h and the obseration coariance matrix can be calculated as R yy E{y µ y y µ y T } E{h µ h + h µ h + T } E{h µ h } T + E{ T } + E{h µ h T } + E{h µ h } T σ h T + C Similarly, R hy E{h µ h y µ y T } E{h µ h h µ h + T } E{h µ h } T + E{h µ h T } σ h T The MMSE estimate of the unknown parameter h is gien by, ĥ R hy R yy y µ y + µ h Substituting the alues of the coariance matrices in the aboe expression, we obtain ĥ σ h T σ h T + C y µ h + µ h 3

4 Simplifying the aboe expression using Woodbury matrix identity, we get T T + C T C C + T C T C σ h T C σ h T C T C T C + + T C T C + T C T C + T C T C After substituting the aboe expression in equation, we obtain ĥ + T C T C y µ h + µ h 3 Simplifying T C, we get Similarly, σ T C [ ] T C y µ h 0 σ σ k k 0 0 σ N N σ k k Now, equation 3 can be written as, yk µh Ans d ĥ µ h + k σ k + k σ k 5 The system model can be written as, y h + yk µh k k yk + µ σk h + σk The MSE of the MMSE estimate ĥ of the unknown Gaussian parameter h is gien by, E{ĥ h } r hh r hy R yy r yh 4 4

5 From the solution of problem 4, R yy, r hy can be written as R yy T + C, r hy T Substituting the alues, equation 4 can be written as Ans a E{ĥ h } T T + C + T C σ T C σ h h k σ h σ k + k + k N σ h + σ k σ k k σ k 6 The LMMSE estimate is identical to the MMSE estimate for a Gaussian parameter Ans c 7 As deried in the lectures notes of week 4, the LMMSE estimate ĥ of the unknown parameter h, which is not necessarily Gaussian is gien by ĥ r hy R yy y Ans d 8 In this scenario, we hae N 4 pilot ectors, each corresponding to M transmit antennas The length of each pilot ector is Hence, M Ans b 9 For a multi-antenna channel estimation scenario with N 4 pilot ectors x [ ] [ ], x, x3 σ h [ ] [ ], x4 5

6 The corresponding pilot matrix X is gien by, x T X x T x3 T x4 T Ans d 0 From the solution of problem 9, the pilot matrix X can be written as, X [ ] c c Columns c and c satisfy the orthogonality property, ie c T c 0 Hence, the pilot matrix X has orthogonal columns Ans c 6