3. The Wiener Filter

3.1 The Wiener-Hopf Equation

The Wiener filter theory is characterized by:

1. The assumption that both signal and noise are random processed with known spectral characteristics or, equivalently, known auto- and cross-correlation functions. 2. The criterion for best performance is minimum mean-square error. (This is partially to make the problem mathematically tractable, but it is also a good physical criterion in many applications.) 3. A solution based on scalar methods that leads to the optimal filter weighting function (or transfer function in the stationary case).

z(nnn )=+ s( ) v( ) hi() or H ( z ) s(ˆ n )

JWSS LTI WSS

RRRRsvzszs, , , or SzSzSzS (),(),(), v z sz () z known 2 J=− E{}[] sn() snˆ ()

Fig. 3.1-1 Wiener Filter Problem

1 We now consider the filter optimization problem that Wiener first solved in the 1940s. Referring to Fig. 3.1-1, we assume the following:

1. The filter input is an additive combination of signal and noise, both of which are jointly wide- sense stationary (JWSS) with known auto- and cross-correlation functions (or corresponding spectral functions). 2. The filter is linear and time-invariant. No further assumption is made as to its form. 3. The output is also wide-sense stationary. 4. The performance criterion is minimum mean-square error.

The estimate s(ˆ n ) of a signal s(n) is given by the convolution representation

∞∞ s()()z()()z()()z(ˆ nhnn=∗=∑∑ hnii − = hini −), (3.1-1) ii=−∞ =−∞ where z(i) is the measurement and hn() is the impulse response of the . Let H denote the region of support of hn(), defined by

H =≠{nhn:() 0}.

Then, Eq. (3.1-1) can be rewritten as

2 s(ˆ nhin )=∑ ( )z(−i ). (3.1-2) i∈H Let the mean-square estimation error (MSE) J be

2 JE=−{[s( n ) s(ˆ n )] }

⎪⎪⎧⎫⎡⎤⎡⎤ =−En⎨⎬⎢⎥s( )∑∑ hinin ( )z( −− )⎢⎥ s( ) hjnj ( )z( − ) (3.1-3) ⎩⎭⎪⎪⎣⎦ij∈∈HH⎣⎦ =−E{}s2 ( n ) 2∑∑ hiE (){} s()z( n n −+ i )∑ hih () ( jE ) z( n −− i )z( { n j ). } ii∈∈HHj∈H

To minimize the MSE, take the partial derivatives of J with respect to hi(), for each hi()≠ 0. Then, set the result equal to zero

∂J =−2Enni{} s( )z( − ) + 2∑ hjEninj ( ) z(− )z( { − ) } ∂hi() j∈H (3.1-4) = 0.

Solving Eq. (3.1-4), we find

∑ hjE( ){ z( n− i )z( n−= j )} E{ s( n )z( n − i )} , i ∈H (3.1-5) j∈H

3 which we may express in the form of

∑ hjR()zs ( j− i )= Rz (), i i∈H , (3.1-6) j∈H

where Rkz () is the function of z(n) and Rksz () is the crosscorrelation function of s(n) and z(n).

Eq. (3.1-6) is the discrete-time Wiener-Hopf equation. It is the basis for the derivation of the Wiener filter.

3.2 The FIR Wiener Filter

Suppose the filter has an impulse response hn() with support H, where H is a finite set, e.g., H = {0,1,2,…,N-1}. Then the impulse response hn() has a finite duration, and this type of filter is called a finite impulse response (FIR) filter.

For the FIR filter, the Wiener-Hopf equation is written

N −1 ∑hjR()zs ( j− i )=≤≤ Rz (),0 i i N− 1. (3.2-1) j=0

4 This may be written in matrix form

⎡⎤RRzz(0) (1) " RN z (− 1) ⎡⎤ h (0) ⎡ Rsz (0) ⎤ ⎢⎥RR(1) (0) " RN (− 2) ⎢⎥ h (1) ⎢ R (1) ⎥ ⎢⎥zz z ⎢⎥= ⎢ sz ⎥, (3.2-2) ⎢⎥##%# ⎢⎥ # ⎢ # ⎥ ⎢⎥⎢⎥⎢ ⎥ ⎣⎦RNzz(−− 1) RN ( 2)" R z (0) ⎣⎦hN(1)− ⎣RNsz (1)− ⎦

where we have used the fact that Rkzz()= R (− k). Denote Eq. (3.2-2) in a convenient form

Rhz= rsz (3.2-3)

where the autocorrelation function Rz is symmetric and positive-definite.

Eq. (3.2-3) is solved for h,

−1 hR= zrsz. (3.2-4)

This is the FIR Wiener filter of order N −1. Note that Rz is a and Eq. (3.2-4) may be solved using a computationally-efficient method such as the Levinson-Durbin algorithm.

5 Mean-Square Error (MSE)

We may write the MSE of the FIR Wiener filter

⎧⎫⎡⎤N −1 MSE=−−En⎨⎬ s( )⎢⎥ s( n )∑ hini ( )z( ) ⎩⎭⎣⎦i=0 (3.2-5) N −1 =−Enn{} s( )s( )∑ hiEnni ( ) s( )z( { − ) . } i=0

To simplify Eq. (3.2-5), rewrite Eq. (3.1-5) in the following form

⎧⎫⎪⎪⎡⎤ Enin⎨⎬z(−− )s()⎢⎥∑ hjnj ()z( −= ) 0 ⎩⎭⎪⎪⎣⎦j∈H Enin{} z(−−= )[] s( ) s(ˆ n ) 0 (3.2-6) Enin{} z(− )s( )=∈ 0, iH.

This relation is known as the orthogonality principle for LTI . Applying this relation to Eq. (3.2-5) gives

6 MSE= Enn{ s( )s( )} N−1 =−En{} s2 ( )∑ hiEnni ( ){} s( )z( − ) i=0 N −1 =−RhiRiss (0)∑ ( )z ( ) i=0 (3.2-7) T =−Rhrss (0)z .

Observe that if no filtering is performed and we simply use s(ˆ nz )= (n ), the MSE is

2 MSEno filter =−=−+EnznR{[] s() ()} s (0)2 Rsz (0) R z (0). (3.2-8)

We can use Eq. (3.2-8) to measure the effectiveness of the Wiener filter. The reduction in MSE due to Wiener filtering is given by

⎛⎞MSEno filter reduction in MSE= 10log10 ⎜⎟dB . (3.2-9) ⎝⎠MSEfilter

We can inspect the MSE to determine whether the filter performance is acceptable. If the MSE is too high, we may wish to use a larger value of N.

7 Example 3.2-1

k Suppose we have a signal s(n ) with autocorrelation function Rks ( )= 0.95 . The signal is 2 observed in the presence of additive white noise with variance σ v = 2 . Hence Rkv ()= 2(δ k). s(n ) and v(n ) are uncorrelated, zero-mean, JWSS random processes. We would like to design the optimum LTI second-order filter h.

Since the filter is the second-order, we set N = 3. The matrix equation Eq. (3.2-2) to be solved is

⎡⎤RRRzzz(0) (1) (2) ⎡⎤h(0) ⎡Rsz (0)⎤ ⎢⎥RR(1) (0) R (1)⎢⎥ h (1)= ⎢ R (1) ⎥. (3.2-10) ⎢⎥zz z⎢⎥⎢ sz⎥ ⎢ ⎥ ⎣⎦⎢⎥RRRzzz(2) (1) (0) ⎣⎦⎢⎥h(2) ⎣Rsz (2)⎦

For an additive-noise measurement model

z(nn )=+ s( ) v(n ),

the autocorrelation functions Rkz () and Rksz () may be given as follows

8 Rkz ()=− E{ z()z( n n k )} =+En{}[][] s( ) v( n ) s( nknk −+− ) v( ) (3.2-11) =−+E{}[] s( n )s( nk ) E{}[] v( n )v( nk− )

=+Rksv ( ) Rk ( ), and

Rksz ()=−+− E{ s()s( n[ nk ) v( nk )]} =−+E{}[] s( n )s( nk ) E{}[] s( n )v( nk− ) (3.2-12)

= Rks ( ).

According to Eqs. (3.2-11) and (3.2-12), we have

k Rz (kk )=+ 0.95 2δ ( ) k Rksz ( )= 0.95 .

Then Eq. (3.2-10) becomes

9 ⎡⎤3 0.95 0.9025⎡h (0)⎤⎡ 1 ⎤ ⎢⎥0.95 3 0.95⎢h (1)⎥= ⎢ 0.95 ⎥. (3.2-13) ⎢⎥⎢⎥⎢⎥ ⎣⎦⎢⎥0.9025 0.95 3⎣⎢h (2)⎦⎥⎣⎢ 0.9025⎦⎥

The solution of Eq. (3.2-13) is

T h = [0.2203 0.1919 0.1738] .

The MSE of this estimator can be computed via Eq. (3.2-7),

T MSE=−=Rhrss (0)z 0.4405. (3.2-13)

For comparison, we remark that without any filtering,

MSEno filter= RRRR s (0)−+= 2sz (0) z (0)v (0)= 2. (3.2-14) The FIR Wiener filter has reduced the MSE by

⎛⎞MSEno filter reduction in MSE= 10log10 ⎜⎟dB ⎝⎠MSEfilter ⎛⎞2 = 10log10 ⎜⎟ (3.2-15) ⎝⎠0.4405 ≈ 6.5dB .

10 3.3 The Noncausal Wiener Filter

Rewrite the Wiener-Hopf equation, Eq. (3.1-6),

∑ hjR()zs ( j− i )=∈ Rz (), i i H j∈H and let H = Z, where Z is the set of integers: Z =−−{",2,1,0,1,2,"}. Then, the Wiener-Hopf equation becomes

∞ ∑ hjR( )zs ( j− i )= Rz ( i ), for all i∈Z . (3.3-1) j=−∞

Take the z-transform of Eq. (3.3-1)

H ()zSzsz () z= S () z so that

Sz() Hz()= sz . (3.3-2) Szz () Eq. (3.3-2) is the solution for the noncausal Wiener filter.

11 Mean-Square Error (MSE)

As with the FIR Wiener filter of Section 3.2, we can derive an expression for the MSE. The MSE similar to Eq. (3.2-7) can be written as

∞ MSE=−RhiRs (0)∑ ()sz ()i. (3.3-3) i=−∞

Example 3.3-1

Let us apply a noncausal Wiener filter to the filtering problem of Example 3.2-1. In that example we found that

k k Rz (kk )=+ 0.95 2δ ( ), and Rksz ( )= 0.95 . (3.3-4)

Take z-transforms of Eq. (3.3-4)

1− (0.95)2 Sz()=+2 z (1−− 0.95zz−1 )(1 0.95 )

2.3955(1−− 0.7931zz−1 )(1 0.7931 ) 1 =<, 0.7931 z < (1−− 0.95zz−1 )(1 0.95 ) 0.7931

12 and

1− (0.95)2 1 Sz()=<, 0.95 z<. sz (1−− 0.95zz−1 )(1 0.95 ) 0.95

The noncausal Wiener filter, Eq. (3.3-2), is given by

Sz() 0.0975 Hz()==sz Sz( ) 2.3955(1−− 0.7931z−1 )(1 0.7931 z ) z 0.1097(1− (0.7931)2 ) 1 =<, 0.95z < . (1−− 0.7931zz−1 )(1 0.7931 ) 0.95

The impulse response of the filter is

hn( )= 0.1097(0.7931) n .

The mean-square errors associated with the noncausal Wiener filter is according to Eq. (3.3-3),

13 ∞ MSE=−RhnRnss (0)∑ ()z () n=−∞ ∞ =− (0.95)0 ∑ 0.1097(0.7931)nn (0.95) n=−∞ ∞−∞ ⎡ nn −−n n⎤ =− 1 0.1097⎢∑∑ (0.7931) (0.95)+ (0.7931) (0.95) ⎥ ⎣ nn==01−⎦ ∞ ⎡⎤nn =− 1 0.1097⎢⎥ 2∑ (0.7931) (0.95)− 1 ⎣⎦n=0 ⎡⎤2 =− 1 0.1097 ⎢⎥−=1 0.2195. ⎣⎦1− (0.7931)(0.95)

Since the MSE with no filter is 2 according to Eq. (3.2-14), the improvement of the noncausal filter over no filtering is

⎛⎞MSEno filter reduction in MSE= 10log10 ⎜⎟dB ⎝⎠MSEfilter ⎛⎞2 = 10log10 ⎜⎟ ⎝⎠0.2195 ≈ 9.6dB .

14 Compared to the second-order filter in Eq. (3.2-15) in Ex. (3.2-1), the noncausal Wiener filter reduces the estimation error by an additional 3 dB .

3.4 The Causal Wiener Filter

Rewrite the Wiener-Hopf equation, Eq. (3.1-6),

∑ hjR()zs ( j− i )=∈ Rz (), i i H j∈H and let H = N, where N is the set of integers: N = {0,1, 2," }. Then, the Wiener-Hopf equation becomes

∞ ∑hjR()zs ( j− i )= Rz (),for i all i∈N. (3.4-1) j=0

Eq. (3.4-1) can not be simplified by taking the z-transform because of its causality restriction. To solve the Wiener-Hopf equation, the spectral factorization and the causal-part extraction are necessary.

Theorem 3.4-1 (Spectral Factorization)

Let x(n ) be a real-valued, zero-mean, WSS random process with power density spectrum Szx (),

15 where Szx () is rational in z and has no poles on the unit circle. Then Szx () can be factored into the product

+− Szxx()= S() zSzx (), (3.4-2) where

+ − Sx ()z and Szx () are rational in z, + if zi is a pole of Szx (), then zi <1, + if zi is a zero of Szx (), then zi ≤1, − if zi is a pole of Szx (), then zi >1, − if zi is a zero of Szx (), then zi ≥1, and +−−1 Szxx()= Sz( ).

Example 3.4-1

Let s(n) be a random process described by the difference equation

s(nn ) =−1.1s( 1)− 0.24s(n −++ 2) 2w( nn ) 3w(− 1) , (3.4-3)

n where w(n) is a zero-mean, WSS random process with autocorrelation function Rnw ( )= 5(0.6) .

16 We want to find the spectral factorization of Szs ().

Take the z-transform of Eq. (3.4-3)

Sz()=− 1.1 z−−12 Sz () 0.24 z Sz () + 2 Wz ()+3z−1 Wz ().

The system transfer function is

Sz() 2++ 3z−−112(1 1.5z ) Hz()== = . Wz( ) 1−+ 1.1 z−−12 0.24 z (1−0.3z − 1 )(1− 0.8z − 1 )

Using the input-output power spectrum relation, the power spectrum of s(n ) is given by

−1 Szsw()= HzHz () ( ) S () z 2(1++ 1.5zz−1 ) 2(1 1.5 ) 3.2 (3.4-4) = , (1−− 0.3zz−−11 )(1 0.8 ) (1 − 0.3 zzzz )(1− 0.8 ) (1 −− 0.6−1 )(1 0.6 ) where it is used that

1− (0.6)2 Sz()= 5 . w (1−− 0.6zz−1 )(1 0.6 )

17 + Collect the terms in Eq. (3.4-4) that have poles or zeros inside the unit circle to form Szs (),

23.2(11.5)+ z Sz+ ()= . (3.4-5) s (1−−− 0.3zz−−11 )(1 0.8 )(1 0.6z−1 )

− Collect the terms in Eq. (3.4-4) that have poles or zeros outside the unit circle to form Szs (),

23.2(11.5)+ z−1 Sz− ()= . (3.4-6) s (1−−− 0.3zz )(1 0.8 )(1 0.6z )

From Eqs. (3.4-5) and (3.4-6), it is noticed that

−+−1 Szss()= Sz ( ).

The Causal-Part Extraction

Consider the impulse response hn() of a real-valued LTI system with rational z-transform H ()z . In the time domain we can split hn() into its causal and anticausal parts such that

hn( )=+ causal part of hn ( ) anticausal part of hn ( ), (3.4-7) where

18

causal part of hn ( )≡ [ hn ( )] = hn ( )1( n ), + anticausal part of hn()≡ hn ()=−− hn ()1( n 1). []−

By the linearity of the z-transform, Eq. (3.4-7) becomes,

H ()zHzHz=+ () () , (3.4-8) [ ]+ [ ]− where

[Hz()] = Z{ hn ()1(), n} + (3.4-9) Hz()=− hn ()1( n− 1). []− Z{}

We now develop a method for determining H ()z and H ()z . By long division in z, z−1 , or [ ]+ [ ]− both for a rational H (z), we can always convert it into the form

⎛⎞N −1 bz−n LM−N⎜⎟∑ n nn− ⎜n=0 ⎟ Hz()=+∑∑ c−nn z cz +N (3.4-10) nn==10⎜⎟−n   azn Pz() Pz() ⎜⎟∑ AC⎝⎠n =0 Qz()

19

−1 H (z) consists of two polynomial terms, PA ()z and PC ()z , and a proper fraction in z , Qz() .

PA ()z corresponds to a purely anticausal sequence, and PC (z) to a purely causal sequence.

Now we examine Qz(). Let K ≤ N be the number of distinct poles of Qz(), and denote these poles and their associated degrees by pk and mk , respectively, kK=1," , . Then Qz() may be written as

N −1 −n ∑bzn ⎛⎞1 n=0 Qz()= ⎜⎟K . ⎝⎠a0 −1 mk ∏(1− pzk ) k=1

Via partial fractions, we can write this expression as

KKmk q Qz()=≡km, Q( z). (3.4-11) ∑∑−1 m ∑ k km==11(1− pzk ) k=1

Suppose that H ()z be stable. Then its ROC includes the unit circle and the ROC of each Qzk () must include the unit circle. Let N A be the total order of the poles of H (z) outside the unit circle, and NC be the total order of the poles of H (z) inside the unit circle with the exception

20 of the origin. Then NNAC+ = N and we may write

N A −1 −n K ∑ λn z n=0 QzAk()==∑ Q()z K , k=1 (1 − pz−1) mk pk >1 ∑ k k=1 pk >1 and

NC −1 −n K ∑ μn z n=0 QzCk()==∑ Qz () K . k=1 (1− pz−1 ) mk 01<

QzA () is the anticausal, proper, rational portion of H ()z , and QzC () is the causal, proper, rational portion of H (z). Then for a stable rational H (z), we have

H ()zPzQ=+ () ()z, [ ]+ CC which is the sum of a purely causal sequence and all terms in the partial fraction expansion of

21 H (z) that have poles inside the unit circle.

Example 3.4-2

We want the causal part of

6zz21−+− 51 128 109 z− + 197 z−−−2 − 232 z3 + 80 z4 Hz()= , 217− zzz−−−123+− 40 16

⎧⎫1 with ROC=<<⎨⎬zz : 4 . ⎩⎭2

Long division in z gives

22 3z2 +4 2−+ 17zzzzz−−−1232 40 − 16 6 −+−+ 51 128 109 z−1 197 z − 2 − 232 zz −− 34 + 80 6zz21−+− 51 120 48 z− ______

8−+ 61zzzz−−12 197 − 232 − 34 + 80 − 8−+ 68zz−−12 160 − 64 z − 3 ______7zzzz−−12+−+ 37 168 −− 34 80

7zzzz−12+−+ 3716880−−− 34 Hz()=++ 3 z2 4 . 217− zzz−−−123+− 40 16

Then long division in z−1 produces

23 −−5z−1 2 −+16zzz−−−321 40 −+ 17 2 80 z − 4 − 168 zzz −−− 321 + 37 + 7 80zzzz−−−−4321−+− 200 85 10 ______32zzz−−−321−+ 48 17 32zzz−−−321− 80+− 34 4 ______32zz−−21−+ 17 4

417−+zz−−12 32 Hz()=++−−+ 3 z21 4 (2)5z− . 217− zz−123+− 40−− 16z

Let Qz() be the rational term and find its partial fractions

17 −−12 −−12216−+zz 417−+zz 32 Qz()==2 −−−1231 217−+zzz 40 − 16 (1−−zz−−112 )(1 4 ) 2 11 = + . −12 1 (1− 4z ) 1− z−1 2

24

Hence,

11 Hz()=+−+ 3 z21 2 5 z− + . −12 1 (1− 4z ) 1− z−1 2

So the causal part of H ()z is

1 []Hz()=− 2 5 z−1 + , + 1 1− z−1 2

n ⎛⎞1 which corresponds to the sequence 2()δδnn−−+ 5( 1)⎜⎟ 1(n). Also the anticausal part of ⎝⎠2 H (z) is

1 []Hz()=+ 3 z2 , − (1− 4z−12 ) which corresponds to 3(δ nnn+ 2)+−⎡ 4nn 1( −− 1)⎤⎡ ∗− 4 1( −− 1) ⎤. ⎣()⎦⎣ ()⎦

25

The Causal Wiener Filter

Assume that Szz () is rational in z and has no poles or zeros on the unit circle. The equation that we must solve to obtain the causal Wiener filter is as is given in Eq. (3.4-1)

∞ ∑hjR()zs ( j− i )=≥ Rz (), i i 0. j=0

To solve the Wiener-Hopf equation, define

∞ ′ hi()=− Rsz () i∑ hjR ( )z ( j − i ), for all i∈Z. (3.4-12) j=0

Since h is the causal Wiener filter, hi′() can be written

⎧0, i ≥ 0 ⎪ hi′()= ⎨ ∞ (3.4-13) Ri()− hjRji ( ) (− ), i< 0. ⎪ sz ∑ z ⎩ j=−∞

Now we can take the z-transform of both sides of Eq. (3.4-12) and obtain

26 Hz′()=− Ssz () z HzSz ()z () +− =−Szsz () HzSzSz ()z () z ().

− Dividing by Szz (), we have

Hz′() Szsz () + −−=−H ()zSz ( z). Szzz() Sz ()

Extract the causal part of both sides of this equation and find

⎡⎤⎡⎤Hz′() Sz() sz ⎡ + ⎤ ⎢⎥⎢⎥−−=−H ()zSz () z . (3.4-14) Sz() Sz () ⎣ ⎦+ ⎣⎦⎣⎦zz++

From Eq. (3.4-13), hi′() is purely anticausal. Therefore H′(z) contains only poles outside the

− 1 unit circle. Since the zeros of Szz () lie outside the unit circle, the poles of − also lie Szz () H′()z outside the unit circle. We conclude that all the poles of − are outside the unit circle, and Szz () thus

27 ⎡⎤Hz′() ⎢⎥− = 0. Sz() ⎣⎦z +

+ H (z) is causal by definition, so its poles lie within the unit circle. The poles of Szz () are also + within the unit circle. Hence all the poles of H ()zSz () z are inside the unit circle, and

⎡⎤H ()zS++ () z= HzS () ( z). ⎣⎦zz+

Szsz () We can not conclude anything about − . Therefore, Eq. (3.4-14) becomes Szz ()

⎡⎤Szsz () + 0(=−⎢⎥− H zS)z ( z), Sz() ⎣⎦z + which produces the causal Wiener filter, described by its system function H (z)

1 ⎡ Szsz ()⎤ Hz()= +−⎢ ⎥ . (3.4-15) SzSz() () zz⎣ ⎦+

Example 3.4-3

28

Consider the same situation of Examples 3.2-1 and 3.3-1 in which s(n ) and v(n ) are uncorrelated, zero-mean, JWSS random processes with

1− (0.95)2 Sz()= , s (1−− 0.95zz−1 )(1 0.95 ) and

Szv ()= 2.

We observe z(nn )= s( )+ v(n ) and estimate s(n ).

From Example 3.3-1, we have

2.3955(1−− 0.7931zz−1 )(1 0.7931 ) Sz()= . z (1−− 0.95zz−1 )(1 0.95 )

Perform a spectral factorization on Szz () and obtain

(1− 0.7931z−1 ) Sz+ ( )= 1.5477 , z (1− 0.95z−1 )

29 and

(1− 0.7931z ) Sz− ( )= 1.5477 . z (1− 0.95z )

Szsz()= Sz s () because s(n ) and v(n ) are uncorrelated. Therefore, we have

Szsz () 0.0630 −−= 1 Szz ( ) (1−− 0.95 z )(1 0.7931 z ) −0.0794z−1 = (1−− 0.95zz−−11 )(1 1.2608 ) 0.2555 0.2555 =− 1− 0.95zz−−11 1− 1.2608 and

⎡⎤Szsz () 0.2555 ⎢⎥−−= 1 . Sz() 1− 0.95 z ⎣⎦z +

The causal Wiener filter is then

30 1 ⎡⎤Szsz () Hz()= +−⎢⎥ SzSz() () zz⎣⎦+ 1− 0.95z−1 0.2555 0.1651 =•=, 1.5477(1−−− 0.7931zz−11 ) 1 0.95−− 1 0.7931 z 1 and

hn( )= 0.1651(0.7931)n 1(n ).

The MSE associated with this filter is computed using Eq. (3.3-3)

∞ ii MSEC =− 1 0.1651∑ (0.7931) (0.95)= 0.3302. i=0

Compared to no filtering (MSE = 2), the causal Wiener filter reduces the MSE by 7.8 dB . (6.5 dB in FIR filter and 9.6 dB in noncausal filter.)

31