Download - Optimal FIR Filters

7: Optimal FIR filters

7: Optimal FIR filters Optimal Filters Alternation Theorem Chebyshev Polynomials Maximal Error Locations Remez ExchangeAlgorithm Determine Polynomial Example Design FIR Pros and Cons Summary MATLAB routines

DSP and Digital Filters (2014-5490) Optimal FIR: 7 1 / 11

Optimal Filters



We restrict ourselves to zero-phase filters of odd length M + 1, symmetricaround h[0], i.e. h[n] = h[n].

Optimal Filters




H() = H(ej) =M

2

M2

h[n]ejn

Optimal Filters




H() = H(ej) =M

2

M2

h[n]ejn= h[0] + 2M

2

1 h[n] cosn

Optimal Filters




H() = H(ej) =M

2

M2

h[n]ejn= h[0] + 2M

2

1 h[n] cosn

H() is real but not necessarily positive (unlikeH(ej)).

Optimal Filters




H() = H(ej) =M

2

M2

h[n]ejn= h[0] + 2M

2

1 h[n] cosn


Weighted error: e() = s()(H() d()

)where d() is the target.

Optimal Filters




H() = H(ej) =M

2

M2

h[n]ejn= h[0] + 2M

2

1 h[n] cosn




Choose s() to control the error variation with .

Optimal Filters




H() = H(ej) =M

2

M2

h[n]ejn= h[0] + 2M

2

1 h[n] cosn




Choose s() to control the error variation with .Example: lowpass filter

Optimal Filters




H() = H(ej) =M

2

M2

h[n]ejn= h[0] + 2M

2

1 h[n] cosn





d() =

{1 0 1

0 2

Optimal Filters




H() = H(ej) =M

2

M2

h[n]ejn= h[0] + 2M

2

1 h[n] cosn





d() =

{1 0 1

0 2

s() =

{1 0 1

1 2

Optimal Filters




H() = H(ej) =M

2

M2

h[n]ejn= h[0] + 2M

2

1 h[n] cosn





d() =

{1 0 1

0 2

s() =

{1 0 1

1 2

e() = 1 when H() lies at the edge of the specification.

Optimal Filters




H() = H(ej) =M

2

M2

h[n]ejn= h[0] + 2M

2

1 h[n] cosn





d() =

{1 0 1

0 2

s() =

{1 0 1

1 2

e() = 1 when H() lies at the edge of the specification.

Minimax criterion: h[n] = argminh[n]max |e()|: minimize max error

Alternation Theorem



Want to find the best fit line: with the smallest maximal error.

2 4 6 82

4

6

8

Alternation Theorem




Best fit line always attains themaximal error three times withalternate signs

2 4 6 82

4

6

8

Alternation Theorem





2 4 6 82

4

6

8

Proof:Assume the first maximal deviation from the line is negative as shown.

Alternation Theorem





2 4 6 82

4

6

8

Proof:Assume the first maximal deviation from the line is negative as shown.There must be an equally large positive deviation; or else just move the linedownwards to reduce the maximal deviation.

Alternation Theorem





2 4 6 82

4

6

8

Proof:Assume the first maximal deviation from the line is negative as shown.There must be an equally large positive deviation; or else just move the linedownwards to reduce the maximal deviation.This must be followed by another maximal negative deviation; or else youcan rotate the line and reduce the deviations.

Alternation Theorem





2 4 6 82

4

6

8


Alternation Theorem:A polynomial fit of degree n to a bounded set of points is minimax if andonly if it attains its maximal error at n+ 2 points with alternating signs.

Alternation Theorem





2 4 6 82

4

6

8


Alternation Theorem:A polynomial fit of degree n to a bounded set of points is minimax if andonly if it attains its maximal error at n+ 2 points with alternating signs.There may be additional maximal error points.

Alternation Theorem





2 4 6 82

4

6

8


Alternation Theorem:A polynomial fit of degree n to a bounded set of points is minimax if andonly if it attains its maximal error at n+ 2 points with alternating signs.There may be additional maximal error points.Fitting to a continuous function is the same as to an infinite number ofpoints.

Chebyshev Polynomials



H() = H(ej) = h[0] + 2M

2

1 h[n] cosn




H() = H(ej) = h[0] + 2M

2

1 h[n] cosn

But

cos 2 = 2 cos2 1




H() = H(ej) = h[0] + 2M

2

1 h[n] cosn

But

cos 2 = 2 cos2 1cos 3 = 4 cos3 3 cos




H() = H(ej) = h[0] + 2M

2

1 h[n] cosn

But cosn = Tn(cos): Chebyshev polynomial of 1st kind

cos 2 = 2 cos2 1 = T2(cos)cos 3 = 4 cos3 3 cos = T3(cos)




H() = H(ej) = h[0] + 2M

2

1 h[n] cosn



Recurrence Relation:Tn+1(x) = 2xTn(x) Tn1(x) with T0(x) = 1, T1(x) = x




H() = H(ej) = h[0] + 2M

2

1 h[n] cosn




Proof: cos (n + ) + cos (n ) = 2 cos cosn




H() = H(ej) = h[0] + 2M

2

1 h[n] cosn





So H() is an M2 order polynomial in cos: alternation theorem applies.




H() = H(ej) = h[0] + 2M

2

1 h[n] cosn






Example: Low-pass filter of length 5 (M = 4)

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

|

g

|

M=4




H() = H(ej) = h[0] + 2M

2

1 h[n] cosn







0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

|

g

|

M=4

0 0.5 1 1.5 2 2.5 3

-15

-10

-5

0

|

g

|

(

d

B

)

M=4




H() = H(ej) = h[0] + 2M

2

1 h[n] cosn







0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

|

g

|

M=4

0 0.5 1 1.5 2 2.5 3

-15

-10

-5

0

|

g

|

(

d

B

)

M=4

-1 -0.5 0 0.5 1

0

0.5

1

cos()

g

M=2

Maximal Error Locations



Maximal error locations occur either at bandedges or when dH

d= 0

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

|

H

|

M=18





d= 0

H() = h[0] + 2M

2

1 h[n] cosn

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

|

H

|

M=18





d= 0

H() = h[0] + 2M

2

1 h[n] cosn= P (cos)

where P (x) is a polynomial of order M2 . 0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

|

H

|

M=18





d= 0

H() = h[0] + 2M

2



0.2

0.4

0.6

0.8

1

1.2

|

H

|

M=18

dHd

= P (cos) sin





d= 0

H() = h[0] + 2M

2



0.2

0.4

0.6

0.8

1

1.2

|

H

|

M=18

dHd

= P (cos) sin

= 0 at = 0, and at most M2 1 zeros of polynomial P(x).





d= 0

H() = h[0] + 2M

2



0.2

0.4

0.6

0.8

1

1.2

|

H

|

M=18

dHd

= P (cos) sin


With two bands, we have at most M2 + 3 maximal error frequencies.





d= 0

H() = h[0] + 2M

2



0.2

0.4

0.6

0.8

1

1.2

|

H

|

M=18

dHd

= P (cos) sin


With two bands, we have at most M2 + 3 maximal error frequencies.We require M2 + 2 of alternating signs for the optimal fit.





d= 0

H() = h[0] + 2M

2



0.2

0.4

0.6

0.8

1

1.2

|

H

|

M=18

dHd

= P (cos) sin



Only three possibilities exist (try them all):(a) = 0 + two band edges + M2 1 zeros of P (x).(b) = + two band edges + M2 1 zeros of P (x).





d= 0

H() = h[0] + 2M

2



0.2

0.4

0.6

0.8

1

1.2

|

H

|

M=18

dHd

= P (cos) sin



Only three possibilities exist (try them all):(a) = 0 + two band edges + M2 1 zeros of P (x).(b) = + two band edges + M2 1 zeros of P (x).(c) = {0 and } + two band edges + M2 2 zeros of P (x).

Remez Exchange Algorithm



1. Guess the positions of the M2 + 2 maximal error frequencies and givealternating signs to the errors (e.g. choose evenly spaced ).





2. Determine the error magnitude, , and the M2 + 1 coefficients ofthe polynomial that passes through the maximal error locations.

0 0.5 1 1.5 2 2.5 3

-0.5

0

0.5

1

1.5

|

g

|

M = 4Iteration 1






3. Find the local maxima of the error function by evaluatinge() = s()

(H() d()

)on a dense set of .

0 0.5 1 1.5 2 2.5 3

-0.5

0

0.5

1

1.5

|

g

|

M = 4Iteration 1







(H() d()


4. Update the maximal error frequencies to be an alternating subset ofthe local maxima + band edges + {0 and/or }.

0 0.5 1 1.5 2 2.5 3

-0.5

0

0.5

1

1.5

|

g

|

M = 4Iteration 1







(H() d()



If maximum error is > , go back to step 2.

0 0.5 1 1.5 2 2.5 3

-0.5

0

0.5

1

1.5

|

g

|

M = 4Iteration 1







(H() d()



If maximum error is > , go back to step 2.

0 0.5 1 1.5 2 2.5 3

-0.5

0

0.5

1

1.5

|

g

|

M = 4Iteration 1

0 0.5 1 1.5 2 2.5 3

0

0.5

1

|

g

|

M = 4Iteration 2







(H() d()



If maximum error is > , go back to step 2. (typically 15 iterations)

0 0.5 1 1.5 2 2.5 3

-0.5

0

0.5

1

1.5

|

g

|

M = 4Iteration 1

0 0.5 1 1.5 2 2.5 3

0

0.5

1

|

g

|

M = 4Iteration 2







(H() d()



If maximum error is > , go back to step 2. (typically 15 iterations)

0 0.5 1 1.5 2 2.5 3

-0.5

0

0.5

1

1.5

|

g

|

M = 4Iteration 1

0 0.5 1 1.5 2 2.5 3

0

0.5

1

|

g

|

M = 4Iteration 2

0 0.5 1 1.5 2 2.5 3

0

0.5

1

|

g

|

M = 4Iteration 3







(H() d()



If maximum error is > , go back to step 2. (typically 15 iterations)5. Evaluate H() on M + 1 evenly spaced and do an IDFT to get h[n].

0 0.5 1 1.5 2 2.5 3

-0.5

0

0.5

1

1.5

|

g

|

M = 4Iteration 1

0 0.5 1 1.5 2 2.5 3

0

0.5

1

|

g

|

M = 4Iteration 2

0 0.5 1 1.5 2 2.5 3

0

0.5

1

|

g

|

M = 4Iteration 3

Determine Polynomial



For each extremal frequency, i for 1 i M2 + 2

d(i) = H(i) +(1)is(i)





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2

n=1 h[n] cosni +(1)is(i)





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2


Method 1:Solve M2 + 2 equations in

M2 + 2 unknowns for h[n] + .





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2


Method 1:Solve M2 + 2 equations in


In step 3, evaluate H() = h[0] + 2M

2

n=1 h[n] cosni





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2


Method 1: (Computation time M3)Solve M2 + 2 equations in



2

n=1 h[n] cosni





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2





2

n=1 h[n] cosni

Method 2: Dont calculate h[n] explicitlyMultiply equations by ci =

j 6=i

1cosicosj

and add:M2+2

i=1 ci

(h[0] + 2

M2

n=1 h[n] cosn +(1)is(i)

)=M

2+2

i=1 cid(i)





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2





2

n=1 h[n] cosni


j 6=i

1cosicosj

and add:M2+2

i=1 ci

(h[0] + 2

M2


)=M

2+2

i=1 cid(i)

All terms involving h[n] sum to zero leavingM2+2

i=1(1)icis(i)

=M

2+2

i=1 cid(i)





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2





2

n=1 h[n] cosni


j 6=i

1cosicosj

and add:M2+2

i=1 ci

(h[0] + 2

M2


)=M

2+2

i=1 cid(i)


i=1(1)icis(i)

=M

2+2

i=1 cid(i)

Solve for





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2





2

n=1 h[n] cosni


j 6=i

1cosicosj

and add:M2+2

i=1 ci

(h[0] + 2

M2


)=M

2+2

i=1 cid(i)


i=1(1)icis(i)

=M

2+2

i=1 cid(i)

Solve for then calculate the H(i)





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2





2

n=1 h[n] cosni


j 6=i

1cosicosj

and add:M2+2

i=1 ci

(h[0] + 2

M2


)=M

2+2

i=1 cid(i)


i=1(1)icis(i)

=M

2+2

i=1 cid(i)

Solve for then calculate the H(i) then use Lagrange interpolation:H() = P (cos) =

M2+2

i=1 H(i)

j 6=icoscosjcosicosj





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2





2

n=1 h[n] cosni


j 6=i

1cosicosj

and add:M2+2

i=1 ci

(h[0] + 2

M2


)=M

2+2

i=1 cid(i)


i=1(1)icis(i)

=M

2+2

i=1 cid(i)


M2+2

i=1 H(i)

j 6=icoscosjcosicosj(

M2 + 1

)-polynomial going through all the H(i) [actually order M2 ]





d(i) = H(i) +(1)is(i)

= h[0] + 2M

2





2

n=1 h[n] cosni

Method 2: Dont calculate h[n] explicitly (Computation time M2)Multiply equations by ci =

j 6=i

1cosicosj

and add:M2+2

i=1 ci

(h[0] + 2

M2


)=M

2+2

i=1 cid(i)


i=1(1)icis(i)

=M

2+2

i=1 cid(i)


M2+2

i=1 H(i)

j 6=icoscosjcosicosj(

M2 + 1

)-polynomial going through all the H(i) [actually order M2 ]

Example Design



Filter Specifications:Bandpass = [0.5, 1],

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

Example Design



Filter Specifications:Bandpass = [0.5, 1], Transition widths: = 0.2

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

Example Design



Filter Specifications:Bandpass = [0.5, 1], Transition widths: = 0.2Stopband Attenuation: 25 dB and 15 dB

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

Example Design



Filter Specifications:Bandpass = [0.5, 1], Transition widths: = 0.2Stopband Attenuation: 25 dB and 15 dBPassband Ripple: 0.3 dB

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

Example Design




Determine gain tolerances for each band:25 dB = 0.056, 0.3 dB = 1 0.034, 15 dB = 0.178

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

Example Design





Predicted order: M = 36

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

Example Design





Predicted order: M = 36M2 + 2 extremal frequencies are distributed between the bands

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

Example Design






Filter meets specs ,

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

Example Design






Filter meets specs ,; clearer on a decibel scale

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

0 0.5 1 1.5 2 2.5 3-30

-25

-20

-15

-10

-5

0

|

H

|

(

d

B

)

M=36

Example Design






Filter meets specs ,; clearer on a decibel scaleMost zeros are on the unit circle + three reciprocal pairs

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

0 0.5 1 1.5 2 2.5 3-30

-25

-20

-15

-10

-5

0

|

H

|

(

d

B

)

M=36

-1 0 1 2

-1

-0.5

0

0.5

1

Example Design






Filter meets specs ,; clearer on a decibel scaleMost zeros are on the unit circle + three reciprocal pairs

Reciprocal pairs give a linear phase shift

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

0 0.5 1 1.5 2 2.5 3-30

-25

-20

-15

-10

-5

0

|

H

|

(

d

B

)

M=36

-1 0 1 2

-1

-0.5

0

0.5

1

FIR Pros and Cons



Can have linear phase no envelope distortion, all frequencies have the same delay , symmetric or antisymmetric: h[n] = h[n]n or h[n]n antisymmetric filters have H(ej0) = H(ej) = 0 symmetry means you only need M2 + 1 multiplications

to implement the filter.

FIR Pros and Cons





Always stable ,

FIR Pros and Cons





Always stable ,

Low coefficient sensitivity ,

FIR Pros and Cons





Always stable ,


Optimal design method fast and robust ,

FIR Pros and Cons





Always stable ,



Normally needs higher order than an IIR filter / Filter order M dBatten3.5 where is the most rapid transition

FIR Pros and Cons





Always stable ,



Normally needs higher order than an IIR filter / Filter order M dBatten3.5 where is the most rapid transition

Filtering complexity M fs dBatten3.5 fs =dBatten3.5f

2s

f2s for a given specification in unscaled units.

Summary



Optimal Filters: minimax error criterion

Summary




use weight function, s(), to allow different errorsin different frequency bands

Summary





symmetric filter has zeros on unit circle or in reciprocal pairs

Summary





symmetric filter has zeros on unit circle or in reciprocal pairs Response of symmetric filter is a polynomial in cos

Summary





symmetric filter has zeros on unit circle or in reciprocal pairs Response of symmetric filter is a polynomial in cos Alternation Theorem: M2 + 2 maximal errors with alternating signs

Summary






Remez Exchange Algorithm (also known as Parks-McLellan Algorithm)

Summary






Remez Exchange Algorithm (also known as Parks-McLellan Algorithm) multiple constant-gain bands separated by transition regions

Summary






Remez Exchange Algorithm (also known as Parks-McLellan Algorithm) multiple constant-gain bands separated by transition regions very robust, works for filters with M > 1000

Summary






Remez Exchange Algorithm (also known as Parks-McLellan Algorithm) multiple constant-gain bands separated by transition regions very robust, works for filters with M > 1000 Efficient: computation M2

Summary






Remez Exchange Algorithm (also known as Parks-McLellan Algorithm) multiple constant-gain bands separated by transition regions very robust, works for filters with M > 1000 Efficient: computation M2 can go mad in the transition regions

Summary







Modified version works on arbitrary gain function

Summary








Does not always converge

Summary








Does not always converge

For further details see Mitra: 10.

MATLAB routines



firpm optimal FIR filter designfirpmord estimate require order for firpmcfirpm arbitrary-response filter designremez [obsolete] optimal FIR filter design

7: Optimal FIR filtersOptimal FiltersAlternation TheoremChebyshev PolynomialsMaximal Error LocationsRemez Exchange AlgorithmDetermine PolynomialExample DesignFIR Pros and ConsSummaryMATLAB routines

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