Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond - - PowerPoint PPT Presentation

Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond

Eric Dubois

School of Electrical Engineering and Computer Science University of Ottawa

October 2012

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 1 / 12

Problem Setup

The desired signal is passed through a noisy communication channel: The desired signal is distorted by a filter with transfer function H2(z) = 1 1 − c2z−1 |z| > |c2|, c2 = .05 + √ .8025 ≈ .9458 v2(n) is zero-mean white noise uncorrelated with d. Sv2(z) = σ2

2 = 0.1

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 2 / 12

The Desired Signal

The desired signal is assumed to be the realization of an AR(1) process The synthesis filter is given by H1(z) = 1 1 − c1z−1 |z| > |c1|, c1 = .05 − √ .8025 ≈ −.8458 The white noise input is given by Sv1(z) = σ2

1 = 0.27

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 3 / 12

Autocorrelation Function of the Desired Signal

Recall that rd(n) = rv1(n) ∗ h1(n) ∗ h1(−n). Taking z-transforms Sd(z) = Sv1(z)H1(z)H∗

1(1/z∗)

= Sv1(z)H1(z)H1(z−1) (since h1 is real) = σ2

1

(1 − c1z−1)(1 − c1z) |c1| < |z| < 1 |c1|

• .8458<|z|<1.1823

rd(n) is the inverse z-transform of Sd(z). For n ≥ 0 rd(n) = res[Sd(z)zn−1, z = c1] = res

• σ2

1zn

(z − c1)(1 − c1z), z = c1

• = σ2

1cn 1

1 − c2

1

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 4 / 12

Autocorrelation Function of the Desired Signal (2)

Since the process is real, rd(−n) = rd(n), and inserting the value of c1 rd(n) = 0.9488(−0.8458)|n| σ2

d = 0.9488

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 5 / 12

Autocorrelation Function of the Observed Signal

Since v2(n) and d(n) (and so x(n)) are uncorrelated ru(n) = rx(n) + rv2(n) Su(z) = Sx(z) + Sv2(z) = Sd(z)H2(z)H2(z−1) + 0.1 = σ2

1

(1 − c1z−1)(1 − c1z)(1 − c2z−1)(1 − c2z) + 0.1 The region of convergence is |c2| < |z| < 1 |c2| 0.9458 < |z| < 1.0573 ru(n) is the inverse z-transform of Su(z), using your preferred method (partial fraction, residue theorem, etc.).

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 6 / 12

Autocorrelation Function of the Observed Signal (2)

For n ≥ 0 ru(n) = res[Sx(z)zn−1, z = c1] + res[Sx(z)zn−1, z = c2] + 0.1δ(n) = res

• σ2

1zn+1

(z − c1)(1 − c1z)(z − c2)(1 − c2z), z = c1

• +

res

• σ2

1zn+1

(z − c1)(1 − c1z)(z − c2)(1 − c2z), z = c2

• + 0.1δ(n)

= σ2

1cn+1 1

(1 − c2

1)(c1 − c2)(1 − c1c2) +

σ2

1cn+1 2

(c2 − c1)(1 − c1c2)(1 − c2

2)

+ 0.1δ(n) Thus, using symmetry of ru(n), and the values of the parameters, ru(n) = 0.2488(−0.8458)|n| + 0.7512(0.9458)|n| + 0.1δ(n)

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 7 / 12

Cross-correlation Between Observed and Desired Signals

Since v2(n) and d(n) are uncorrelated rdu(n) = rdx(n) = E[D(m)

∞

• ℓ=0

h2(ℓ)D(m − n − ℓ)] =

∞

• ℓ=0

h2(ℓ)rd(n + ℓ) It follows that Sdu(z) = Sd(z)H2(z−1) = σ2

1

(1 − c1z−1)(1 − c1z)(1 − c2z), |c1| < |z| < 1 |c2|

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 8 / 12

Cross-correlation Between Observed and Desired Signals

Using our preferred inverse z-transform method, for n ≥ 0, rdu(n) = res[Sdu(z)zn−1, z = c1] = res

• σ2

1zn

(z − c1)(1 − c1z)(1 − c2z), z = c1

• =

σ2

1cn 1

(1 − c2

1)(1 − c1c2)

= 0.5271(−0.8458)n With these results, we can build the matrices RM and pM to solve for the FIR Wiener filter of any order M.

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 9 / 12

Correlation matrix R5 and correlation vector p5

R5 =       1.1000 0.5000 0.8500 0.4850 0.7285 0.5000 1.1000 0.5000 0.8500 0.4850 0.8500 0.5000 1.1000 0.5000 0.8500 0.4850 0.8500 0.5000 1.1000 0.5000 0.7285 0.4850 0.8500 0.5000 1.1000       pT

5 =

• 0.5271

−0.4458 0.3771 −0.3190 0.2698

• Eric Dubois (EECS)

Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 10 / 12

Wiener Filters for M from 1 to 7

w0 = 0.4792 Jmin = 0.6962 w0 = 0.8362 -0.7854 Jmin = 0.1579 w0 = 0.7378 -0.8033 0.1378 Jmin = 0.1498 w0 = 0.7438 -0.6992 0.1527 -0.1470 Jmin = 0.1405 w0 = 0.7404 -0.6999 0.1370 -0.1495 0.0256 Jmin = 0.1402 w0 = 0.7407 -0.6962 0.1376 -0.1324 0.0282 -0.0279 Jmin = 0.1399 w0 = 0.7405 -0.6962 0.1370 -0.1325 0.0253 -0.0283 0.0047 Jmin = 0.1399

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 11 / 12

Eigenvalue-eigenvector decomposition of the correlation matrix R5

Q =       0.4597 0.3277 −0.7019 −0.0854 −0.4257 0.4175 −0.5706 −0.0854 0.7019 0.0116 0.4782 0.3662 0.0000 −0.0000 0.7983 0.4175 −0.5706 0.0854 −0.7019 0.0116 0.4597 0.3277 0.7019 0.0854 −0.4257       Λ =       3.6073 1.0633 0.3733 0.2482 0.2079       QTQ = I QTRQ = Λ

Eric Dubois (EECS) Wiener Filter Example Haykin 4e Ch. 2 Problems 11, 12 and beyond October 2012 12 / 12