EE6110 Adaptive Signal Processing Problem Set 2 Solutions

EE6110 Adaptive Signal Processing Problem Set 2 Solutions

1.

Xˆ = R_XYR⁻_y¹Y R_XY = E

S0

S1

Y₀^∗ Y₁^∗

=

E(S⁰(S⁰+V0)^∗) E(S⁰(S¹+ 0.5S⁰+V1)^∗) E(S¹(S⁰+V0)^∗) E(S¹(S¹+ 0.5S⁰+V1)^∗)

=

1 0.5 0 1

RY = E Y0

Y1

Y₀^∗ Y₁^∗

=

E((S⁰+V0)(S⁰+V0)^∗) E((S⁰+V0)(S¹+ 0.5S⁰+V1)^∗) E((S¹+ 0.5S⁰+V1)(S⁰+V0)^∗) E((S¹+ 0.5S⁰+V1)(S¹+ 0.5S⁰+V1)^∗)

=

2 0.5 0.5 2.25

Xˆ =

1 0.5 0 1

2 0.5 0.5 2.25

−1

Y

= 4 7

2 ¹₂

−

1

2 2

Y

Sˆ0 = 8

17Y0+ 2 17Y1

Sˆ1 = − 2

17Y0+ 8 17Y1.

1

2. Case i: ∆ = 0

Sˆ_i = α0Y_i+α1Y_i−1+α2Y_i−2 ,k0^HY where,k0^H = [α⁰, α1, α2]

Y =



 Y_i Y_i−1

Y_i−2





Y_i = S_i+ 0.5Si−1+V_i R_Y = E Y Y^H

=





R_Y(0) R_Y(1) R_Y(2) R^∗_Y(1) R_Y(0) R_Y(1) R^∗_Y(2) R^∗_Y(1) R_Y(0)



where,R_Y(k) = E Y_iY_i^∗

−k

R_SiY = E S_iY^H

=

E(SiY_i^∗) E S_iY_i−^∗1

E S_iY_i−^∗2

E(SiY_i^∗) = E(Si(Si+ 0.5Si−1+V_i)^∗) = 1 E S_iY_i^∗

−1

= E(S_i(S_i−1+ 0.5S_i−2 +V_i−1)^∗) = 0 E(SiY_i^∗) = E(Si(Si−2+ 0.5Si−3 +Vi−2)^∗) = 0

R_Y(0) = E(YiY_i^∗)

= E((Si+ 0.5Si−1+V_i)Y_i^∗)

= E(S_iY_i^∗) + 0.5E(Si−1Y_i^∗) +E(ViY_i^∗)

= 1 + 0.25 + 1

= 2.25 R_Y(1) = E Y_iY_i^∗

−1

= E (Si+ 0.5Si−1+V_i)Y_i^∗₋1

= E S_iY_i^∗₋1

+ 0.5E Si−1Y_i^∗₋1

+E V_iY_i^∗₋1

= 0 + 0.5 + 0

= 0.5 R_Y(0) = E(YiY_i^∗)

= E((S_i+ 0.5S_i−1+V_i)Y_i^∗)

= E SiY_i^∗₋2

+ 0.5E Si−1Y_i^∗₋2

+E ViY_i^∗₋2

= 0

k^H0 = R_XYR⁻_Y¹

=

1 0 0





2.25 0.5 0 0.5 2.25 0.5

0 0.5 2.25





−1

=

0.4688 −0.1096 0.0244

2

Case (ii) ∆ = 1

Sˆ_i−1 = α0Y_i+α1Y_i−1 +α2Y_i−2

R_Si−1Y =

E(Si−1Y_i^∗) E S_i−1Y_i^∗

−1

E S_i−1Y_i^∗

−2

=

0.5 1 0

k^H0 =

α0 α1 α2

= R_Si−1YR⁻_Y¹

=

0.5 0 1 R⁻_Y¹

The model and equation for general L and ∆ is in Section 5.4 of [Sayed].

3.

X = X_C+Z Assume X_C and Z are zero-mean

Xˆ = K0(Y −E(Y)) whereK0R_Y =R_XY RXY = E

XY^H

= E((XC +Z)(Y −E(Y)))

= E(XC(Y^H)−E(Y))

= R_XcY

R_Y = E (Y −E(Y))(Y −E(Y))^H Therefore, ˆX_C =K0(Y −E(Y)) = ˆX.

3