ALaRI2011_Lecture4

8/3/2019 ALaRI2011_Lecture4

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p. 1DSP-II

Lecture 4:

FIR and IIR Filter Realization

Marc Moonen

Dept. E.E./ESAT, K.U.Leuven

[email protected]

www.esat.kuleuven.be/scd/


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ALARI/DSP Marc Moonen - Version 2011 p. 2Lecture-4 : FIR and IIR FilterRealization

Filter Design/Realization

Step-1 : define filter specs(pass-band, stop-band, optimization criterion,)

Step-2 : derive optimal transfer funcion

FIR

or IIR

Step-3 : filter realization (block scheme/flow graph)

direct form realizations, lattice realizations,

Step-4 : filter implementation (software/hardware)

finite word-length issues,

question: implemented filter = designed filter ?

Lecture-3

Lecture-4


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Lecture 4 : Filter Realizations

FIR Filter Realizations

IIR Filter Realizations

PS: We will assume real-valuedfilter coefficients


4/64



FIR Filter Realization

=Construct (realize) LTI system (with delay elements,

adders and multipliers), such that I/O behavior is given by..

Several possibilities exist

1. Direct form

2. Transposed direct form

3. Lattice (LPC lattice)

4. Lossless lattice

PS: Frequency-domain realization: see part filter banks

][....]1[.][.][ 10 Nkubkubkubky N !


5/64



1. Direct form

u[k]

( ( ( (

u[k-4]u[k-3]u[k-2]u[k-1]

x

bo

+

x

b4

x

b3

+

x

b2

+

x

b1

+

y[k]

][....]1[.][.][ 10 Nkubkubkubky N !


6/64



2. Transposed direct formStarting point is direct form :

`Retiming = select subgraph (shaded)remove one delay element on all inbound arrowsadd one delay element on all outbound arrows

u[k]

( ( ( (

u[k-4]u[k-3]u[k-2]u[k-1]

x

bo

+

x

b4

x

b3

+

x

b2

+

x

b1

+

y[k]


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`Retiming : results in...

u[k]

(

(

u[k-1]

x

bo

+

x

b1

+y[k]

( (

u[k-3]u[k-2]

x

b4

x

b3

+

x

b2

+

(=different software/hardware, same i/o-behavior)


8/64



`Retiming : repeated application results in...

i.e. `transposed direct form

u[k]

xbo

+

y[k]

( (

xb1

+

xb2

+ (

xb3

+ (

xb4

(=different software/hardware, same i/o-behavior)


9/64



3. Lattice formDerived from combined realization of

with `flipped version of H(z)

Reversed (real-valued) coefficient vector results in...

i.e. - same magnitude response

- different phase response

][....]1[.][.][~:)(.)(~

01

1 NkubkubkubkyzHzzH NNN !!

212)(...)(~).(~)(~ [[[ jjj ezezez

zHzHzHzH!!

!!!!

][....]1[.][.][:)( 10 NkubkubkubkyzH N !


10/64



Hence starting point is

b4

u[k] u[k-1]

(

u[k-2]

xb3

+

xb2

+

x

+

x

+

b1

u[k-3]

(

xb1

+

b2x

+

xb4

+

(

y[k]

box

+

u[k-4]

(

xbo

b3x

y[k]~


11/64



With this can be rewritten as(ps: find fix for case bo=0)

? A

? A

2

0

10332

0

20222

0

3011,00

3210

01231

0

0

3210

0123

0

0

1',

1',

1''

]1[]...[.''''

''''

.10

0

.1

1

][]...[.0''''

''''0

1

1

][

][~

O

O

O

O

O

O

O

O

O

O

!

!

!!

-

-

-

!

-

-

!

-

bbb

bbb

bbbbb

Nkukubbbb

bbbbz

Nkukubbbb

bbbb

ky

ky

T

T

? AT

Nkukubbbbb

bbbbb

ky

ky][]...[.

][

][~

43210

01234

-!

-

1if, 00)0(assume

0

40 {! { OO b

b

b

`sloppy notation


12/64



This is equivalent to...

Now repeat procedure for shaded graph

(=same structure as the one we started from)

u[k] u[k-3]

(

u[k-2]

xb3

+

xb2

+

x

+

x

+

bo

u[k-2]

(

xb1

+

b1x

+

bo

(

xb2

xb3

y[k]

+

+

x

x ko

y[k]~

(


13/64



Repeated application results in `lattice formu[k]

y[k]

+

+

x

x

ko

+

+

x

x

k1

( +

+

x

x

k2

+

+

x

x

k3

xbo

y[k]~

(= different software/hardware, same i/o-behavior)

explain

(((


14/64



Lattice form : Also known as `LPC Lattice (`linear predictive coding lattice)

Kis are so-called `reflection coefficients

Every set of bis corresponds to a set of Kis, and vice versa.

Procedure for computing Kis from bis corresponds to thewell-known `Schur-Cohn stability test(from control theory):

problem = for a given polynomial B(z), how do we find out

if all the zeros of B(z) are stable (i.e. lie inside unit circle) ?solution = from bis, compute reflection coefficients Kis

(following procedure on previous slides). Zeros are (proven to be)

stable if and only if all reflection coefficients statisfy |Ki|


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Procedure (page 11) breaks down if |Ki|=1 is encountered.Means at least one root of B(z) lies on or outside the unitcircle (cfr Schur-Cohn). Then have to select other realization(direct form, lossless lattice, ) for B(z).

Lattice form is often applied to `minimum phase filters, i.e.

filter with only stable zeros (roots of B(z) strictly inside unit

circle). Then design procedure never breaks down.

Lattice form not overly relevant at this point, but sets stage

for similar derivations that lead to more relevant realizations

(read on)


16/64



4. Lossless lattice :Derived from combined realization of H(z)

with

which is such that

(*)

PS : interpretation ? (see next slide)

PS : may have to scale H(z) to achieve this (why?) (scaling omitted here)

][.~~

...]1[.~~

][.~~

][~~:)(

~~10 NkubkubkubkyzH N !

][....]1[.][.][:)( 10 NkubkubkubkyzH N !

1)(

~~

).(

~~

)().(11

!

zHzHzHzH


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PS : Interpretation ?When evaluated on the unit circle, formula (*) is

equivalent to (for filters with real-valued coefficients)

i.e. and are `power complementary

(= form a 1-input/2-output `lossless system, see also below)

1)(~~)(2

2 !!

! [[

jj

ezez

zHzH

)(~~zH)(zH

TT

2

)(

~~ [jeH

2

)( [jeH

1


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PS : How is computed ?

Note that if `a is a root ofR(z), then `1/a is also aroot ofR(z). Hence can factorize R(z) as

Note that ais can be selected such that all of them lie inside the

unit circle. Then is a minimum-phase FIR filter.

This is referred to as spectral factorization, =spectral factor.

)(

~~zH

)(

11 )().(1)(~~

).(~~

zR

zHzHzHzH !

! i

ii azazzHzH ))(()(~~

).(~~ 11 !

i

iazzH )()(~~ 1

)(~~zH

)(~~zH


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Starting point is

u[k] u[k-1]

(

u[k-2]

x

+

x

+

x

+

x

+

u[k-3]

(

x

+

x

+

x

+

(

y[k]

x

+

u[k-4]

(

x

b1 b2 b3bo b4x

4~~b0~~b 1

~~b 2

~~b 3

~~b

y[k]~~


20/64



From (*) (page 16), it follows that (prove it)

Hence there exists a theta_0 such that

? A

? AT

T

Nkukubbbb

bbbbz

Nkukubbbb

bbbbkyky

]1[]...[.''''

'~~

'~~

'~~

'~~

.10

0.

cossin

sincos

][]...[.0'''''

~~'

~~'

~~'

~~0

cossinsincos

][][

~~

3210

3210

1

00

00

3210

3210

00

00

-

-

-

!

-

- !

-

UU

UU

UUUU

? AT

Nkukubbbbb

bbbbb

ky

ky ][]...[.

~~

~~

~~

~~

~~

][

][~~

43210

43210

-!

-

-B

-!

4

4

0

04040

~~~~

0.~~.~~bb

bbbbbb

`sloppy notation


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This is equivalent to...

Now shaded graph can again be proven to be power

complementary system (Why ? Intuition? Hint: page 23).

Hence can repeat procedure

u[k] u[k-3](

u[k-2]

x

+

x

+

x

+

x

+

u[k-2](

x

+

x

+

bo b1 b2 b3

(

x

x

y[k]

0cosU

+

+

x

(x

x

x

0sinU

0cosU

y[k]

~~

0sinU


22/64



Repeated application results in `lossless lattice explain

2cosU

+

+

xx

x

x

2sinU

2cosU

2sinU

y[k]

0cosU

+

+

xx

x

x

0sinU

0cosU

y[k]~~

0sinU

(

u[k]

4sinU

4cosU

x

x

1cosU

+

+

xx

x

x

1sinU

1cosU

1sinU

((

3cosU

+

+

xx

x

x

3sinU

3cosU

3sinU

(


23/64


Lossless lattice :

also known as `paraunitary lattice (see also Lecture 6)

each 2-input/2-output section is based on an orthogonaltransformation, which preserves norm/energy/power

i.e. forms a 2-input/2-output `lossless system.

Overall system is realized as cascade of lossless sections,

hence is itself also `lossless (see also next slides)


2

2

2

1

2

2

2

1

2

1

2

1)()()()(.

cossin

sincosOUTOUTININ

IN

IN

OUT

OUT!

-

-

!

-

UU

UU


24/64



State-space descriptionExample 2nd-order system :

- R is `realization matrix (A-B-C-D-matrix) for 1-input/2-output system

- R is orthogonal ! Orthogonality implies losslessness (see next slide)

verify!

2

][

][2

][1

2

][

][~~

]1[2

]1[1

2

1

2

2

11

11

00

00

2

1

hence

100

010

001

.

][][

][

.

cos00

010

001sin00

.

cos0sin0

0100

sin0cos00001

.

cossin00

sincos00

00100001

][

][~~

]1[

]1[

-

!

-

-

!

-

-

-

-

!

-

ku

kx

kx

ky

ky

kx

kx

T

R

RR

kukx

kx

ky

ky

kx

kx

U

U

UU

UU

UU

UU

!!

1cosU

+

+

x

x

x

x

][1 kx

y[k]

0cosU

+

+

x

x

x

x

y[k]~~ u[k]2cosU

x

x((

][2kx


25/64



Orthogonality ofR implies `losslessness :assume initial internal state is x1[0], x2[0] (2nd order example)

input sequence is u[0],u[1],u[2],,u[L]

corresponding output sequences are y[0],y[1], and y[0],y[1],...

then orthogonality implies that

ènergy in initial states + inputs = ènergy in final states + outputs

Equivalent transfer function based property (L=infinity) is

i.e. H(z) is èmbedded in 1-input/2-output

lossless system (see page 17)

][~~...]0[

~~][...]0[]1[]1[

][...]1[]0[]0[]0[

22222

2

2

1

2222

2

2

1

LyyLyyLxLx

Luuuxx

!

1)(~~

)(2

2

!!

! [[

jj

ezez

zHzH

~~~~


26/64



PS : Relevance of lattice realizations : robustnessexample :

o = original transfer function

+ = transfer function after 8-bit

truncation of lattice filter

parameters

- = transfer function after 8-bit

truncation of direct-form

coefficients (bis)

PS : A 2x2 orthogonal transformation (rotation) may be implemented inhardware based on a so-called `CORDIC (`COR for CO-ordinate

Rotation) architecture, which is more efficient (and more robust) than a

multiply-add based implementation


27/64



PS : can be generalized to 1-input M-output lossless systems(will be used in part II on filter banks) (compare to page 22 !)

21,UU u[k]

explain/derive!

x

x

x

y[k]

(

2cosU

+

+

x

x

x

x

2sinU

2cosU

2sinU

1cosU

+

+

x

x

x

x

1cosU

1sinU

1sinU

43,UU 65 ,UU

(

y[k]~

y[k]~~

(

1)(

~

~).(

~

~)(~).(~)().( 111 ! zHzHzHzHzHzH

M

=3


28/64



ConstructLT

I system such that I/O behavior is given by..

Several possibilities exist

1. Direct form

2. Transposed direct form

PS: Parallel and cascade realization

3. Lattice-ladder form

4. Lossless lattice

PS: State space realizations

N

N

N

N

zaza

zbzbb

zA

zBzH

!!...1

...

)(

)()(

1

1

1

10


29/64



1. Direct formStarting point is u[k]

( ( ( (

xbo xb4xb3xb2xb1

+ +++y[k]

( ( ( (

+ +++

x-a4

x-a3

x-a2

x-a1

)(

1

zA

)(zB

N

N

N

N

zaza

zbzbb

zA

zBzH

!!

...1

...

)(

)()(

1

1

1

10


30/64



which is equivalent to...

PS : If all a_i=0 (i.e. H(z) is FIR), then this reduces to a direct form FIR

( ( ( (

xbo

xb4

xb3

xb2

xb1

+ +++y[k]

u[k] + +++

x-a4

x-a3

x-a2

x-a1

Direct form B(z)

)(

1

zA

)(zB


31/64



-State-space description:

-N delay elements (=minimum number =max of numerator anddenominator polynomial order = max(p,q))

-2N+1 multipliers, 2N adders (=minimum number)

? A ][.

][

][

][

][

.....][

][.

0

0

0

1

][

][

][

][

.

0100

0010

0001

]1[

]1[

]1[

]1[

0

4

3

2

1

044033022011

4

3

2

13321

4

3

2

1

kub

kx

kx

kx

kx

babbabbabbabky

ku

kx

kx

kx

kxaaaa

kx

kx

kx

kx

-

!

-

-

-

!

-

u[k]

( ( ( (

xbo

xb4

xb3

xb2

xb1

+ +++

y[k]

+ +++

x-a4

x-a3

x-a2

x-a1

x1[k] x2[k] x3[k] x4[k]


32/64



2. Transposed direct formStarting point is

N

N

N

N

zaza

zbzbb

zA

zBzH

!!...1

...

)(

)()(

1

1

1

10

)(

1

zA

)(zB

+ +++

( ( ( (

x-a4

x-a3

x-a2

x-a1

y[k]

u[k]( ( ( (

x

bo

x

b4

x

b3

x

b2

x

b1

+ +++


33/64



which is equivalent to...

)(

1

zA

)(zB

( ( ( (

x-a4

x-a3

x-a2

x-a1

y[k]

u[k]( ( ( (

xbo

xb4

xb3

xb2

xb1

+ +++


34/64



Transposed direct form is obtained after retiming ...

PS : If all a_i=0 (i.e. H(z) is FIR), then this reduces to a transposed direct form FIR

Transposeddirect form B(z)

( ( ( (

x-a4

x-a3

x-a2

x-a1

y[k]

u[k]

xbo

xb4

xb3

xb2

xb1

+ +++

)(zB

)(

1

zA


35/64



- State-space description

i.e. direct forms state space matrices are `transpose of

each other (which justifies the name `transposed dir.form).

? A ][.

][

][

][

][

.0001][

][.

.

.

.

.

][

][

][

][

.

000

100

010

001

]1[

]1[

]1[

]1[

0

4

3

2

1

044

033

022

011

4

3

2

1

4

3

2

1

4

3

2

1

kub

kx

kx

kx

kx

ky

ku

bab

bab

bab

bab

kx

kx

kx

kx

a

a

a

a

kx

kx

kx

kx

-

!

-

-

-

!

-

u[k]

( ( ( (

x-a4

x-a3

x-a2

x-a1

y[k]

xbo

xb4

xb3

xb2

xb1

+ +++

x1[k] x2[k] x3[k] x4[k]

T

DC

BA

DC

BA

formdirectTransposedformDirect

-

!

-


36/64



PS: Parallel & Cascade Realization

Parallel real. obtained from partial fraction

decomposition, e.g. for simple poles:

similar for the case of multiple poles

each term realized in, e.g., direct form

transmission zeros are realized iff signals from different sectionsexactly cancel out.

=Problem in finite word-length implementation

!

! !!

21

121

1

1

1

110 11)(

)()(

N

i N

i

N

i i

i

zzz

zc

zAzBzH

JIHK

FE

u[k]

y[k]

+

+


37/64



PS: Parallel & Cascade Realization

Cascade realization obtained from

pole-zero factorization of H(z)

e.g. for N even:

similar for N odd

each section realized in, e.g., direct form

second-order sections are called bi-quads

!

!!2/

121

21

0..1

..1.

)(

)()(

N

i ii

ii

zz

zzb

zA

zBzH

HK

FE

u[k]

y[k]


38/64



Cascade realization is non-unique:- multiple ways of pairing poles and zeros

- multiple ways of ordering sections in cascade

``Pairing procedure :

- pairing of little importance in high-precision (e.g. floating point

implementation), but important in fixed-point implementation (with

short word-lengths)

- principle = pair poles and zeros to produce a frequency response for

each section that is as flat as possible (i.e. ratio of max. to min.

magnitude response close to unity)- obtained by pairing each pole to a zero as close to it as possible

- procedure : start with pole pair nearest to the unit circle, and pair this

to nearest complex zeros. remove pole-zero pair, and repeat, etc.

u[k]

y[k]


39/64



3. Lattice-ladder formDerived from combined realization of

with...

- numerator polynomial is denominator polynomial with

reversed coefficient vector(see also page 9)- hence is an `all-pass (=`SISO lossless) filter :

N

N

N

N

zaza

zbzbb

zA

zBzH

!!...1

...

)(

)()(

1

1

1

10

N

N

N

NN

zaza

zzaa

zA

zAzH

!!...1

.1...

)(

)(~

)(~

1

1

1

1

1

)(

)(~

)(~

2

2

2

!!

!

!

!

[

[

[

j

j

j

ez

ez

ez

zA

zAzH

)(~zH


40/64



Starting point is

u[k]

+ +

x xb1

+

b2 x

+

b0x b3 x b4x x

a3x

a2x

a1x

1

( (( (

xa1 xa2 xa3 xa4

+ + ++

+ + ++

y[k]

y[k]~

a4

x[k]

-


41/64



This is equivalent to

+ +

x x

+

x x

x x x x

( (

(

(

+ + +

x[k-1]

x

xu[k]

b0

y[k]

y[k]~

x

+

x

0cosU

4.41

1.43

aa

aaa

4.41

2.42

aa

aaa

4.41

3.41

aa

aaa

1

4.41

1.01

aa

abb

4.41

2.02

aa

abb

4.41

3.03

aa

abb

4.41

4.04

aa

abb

+

x

x x x

+ ++-

4.41

3.41

aa

aaa

4.41

2.42

aa

aaa

4.41

1.43

aa

aaa

0sinU

0sinU

0cosU

04

01

sin ka (

!!U

prove it !

+


42/64



Right-hand part has the same structure as what we started

from, hence procedure can be repeated.

Repeated application leads to lattice-ladder form

u[k]

y[k]

y[k]~ x

x

x

+

x

+

0cosU

+

x

0cosU

0sinU

0sinUbo

( ( ( (

x


43/64



Lattice-Ladder form : Kis are `reflection coefficients

Procedure for computing Kis (=sin(theta_i) !) from ais again

corresponds to `Schur-Cohn stability test(cfr. supra):

all zeros ofA(z) are stable (i.e. lie inside unit circle)

iff all reflection coefficients statisfy |Ki|


44/64



State-space description

Example: 2nd-order system:

- R is `realization matrix (A-B-C-D-matrix) for u[k] -> y[k] (SISO)- R is orthogonal (R^T.R=I), which implies `losslessness , i.e.

(cfr. page 39)

? A

-

!

-

!

-

-

-

!

-

100

010

001

.

][

][

][

.][

and

][

][

][

.

100

0sincos

0cossin

.

sincos0

cossin0

001

][~]1[

]1[

2

1

012

2

1

11

11

00

002

1

RR

ku

kx

kx

ky

ku

kx

kx

ky

kx

kx

T

R

EEE

UU

UU

UU

UU

~

1)(~ 2

!! [jez

zH

!!

verify!


45/64



PS : Note that the all-pass part corresponds toA

(z) (i.e. N angles theta_icorrespond to N coefficients a_i), while the ladder part corresponds to

B(z). If all a_i=0 (i.e. H(z) is FIR), then all theta_i=0, hence the all-pass

part reduces to a delay line, and the lattice-ladder form reduces to a

direct-form FIR.

PS : All-pass part (SISO u[k]->y[k]) is known as `Gray-Markel structure

( ( ( (

x

x

x

+

0cosU

+

x0cosU

0sinU

0sinU

in

out


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4. Lossless-lattice :Derived from combined realization of(possibly rescaled, as on page 16)

with...

- numerator polynomial is C(z) is such that

i.e. and are `power complementary (p.17-18)

N

N

N

N

zaza

zbzbb

zA

zBzH

!!...1

...

)(

)()(

1

1

1

10

N

N

N

N

zaza

zczcc

zA

zCzH

!!...1

....

)(

)()(

~~1

1

1

10

)(~~zH )(zH

)().()().()().(1)(

~~

).(

~~

)().(111

11

!!

zAzAzCzCzBzBzHzHzHzH


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`Lossless-lattice :similar derivation (but more complicated -hence skipped) leads to

( ( (

x

u[k]

y[k]y[k]~

xN

]sin

N]cos

0cos]

+

+

x

x

xx

0cos]

0sin]

0sin]

x

x

x

+

x

0cosU

0cosU

00 sinU!k

0sinU

+


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Orthogonal transformations correspond to (3-input 3-output)`lossless sections

State-space description with orthogonal realization matrix R(try it) implies overall lossless characteristic, i.e.

(cfr. Page 46)

PS : If all a_i=0 (i.e. H(z) is FIR), then all theta_i=0 and then this reducesto FIR lossless lattice

PS : If all phi_i=0, then this reduces to (retimed) Gray-Markel structure

2

3

2

2

2

1

2

3

2

2

2

1

3

2

1

3

2

1

)()()()()()(

.

...

...

...

OUTOUTOUTINININ

IN

IN

IN

OUT

OUT

OUT

!

-

-

!

-

1)(~~

)(2

2

!!

! [[

jj

ezez

zHzH

!!


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PS : can be generalized to 1-input M-output lossless systems

(=combine p.27 & p.47 !)

x

x

x

y[k]

(

2cosU

+

+

x

x

x

x

2sinU

2cosU

2sinU

1cosU

+

+

x

x

x

x

1cosU

1sinU

1sinU

(

y[k]~

y[k]~~

x

x

x

+

x

0cosU

0cosU

00 sinU!k

0sinU

+

u[k]


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State-space Realizations : State-space description:

State-space realization = realization such that all the

elements ofA,B,C,D are the multiplier coefficients in thestructure

Example: (transposed) direct form is NOT a state-space realization, cfr.

elements (bi-bo.ai) in B or C (page 31 & 35)

][.][.][

][.][.]1[

kuDkxCky

kuBkxAkx

!

!

(+

+

CB

y[k]

u[k]

D

A

N


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Example: bi-quad `coupled realization- direct form realization of bi-quad (2nd order IIR) may be unsatisfactory,

e.g. for short word-lengths and poles near z=1 or z=-1 (see lecture-4)

- alternative realization:

state space description :

? A ][][

][.][

][.0

1

][

][.

]1[

]1[

2

1

2

1

2

1

kukx

kxky

kukx

kx

kx

kx

-

!

-

-

-

!

-

PO

LQ

QL

21

21

..1..1)(

!

zzzzzH

ii

iii

HKFE

poles : QL .js

(

(

+

+ +

O

L

Q

P

L

Q

-

y[k]

u[k]


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State-space description/realization is non-unique:any non-singular matrix T transforms A,B,C,D into an alternative state-spacedescription/realization

with

State-space realization with orthogonal realization matrix is

referred to as `orthogonal filter :

Relevance : implementation aspects

][.~

][~.~

][

][.~

][~.~

]1[~

kuDkxCky

kuBkxAkx

!

!

][.][~

10

0..

10

0~~

~~

1

1

kxTkx

T

DC

BAT

DC

BA

!

-

-

-

!

-

IRRRR

DC

BAR

TT !!

-!

..


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Addendum: Filter Implementation

The finite word-length problem : So far assumed all signals/coefficients (=numbers) are

represented exactly and arithmetic operations can be

performed to an infinite precision.

In practice, numbers can be represented only to a finiteprecision, and arithmetic operations are subject to errors

(truncation/rounding/...)

Issues:

- quantization of filter coefficients

- quantization & overflow in arithmetic operations

We considerfixed-point filter implementations, with a

`short word-length.


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Filter Implementation : Coefficient Quantization

In signal processors with a `sufficiently long word-length,e.g. with 16 bits (=4 decimal digits) or 24 bits (=7 decimal

digits) precision, or with floating-point representations and

arithmetic, finite word-length issues are less relevant.

The coefficient quantization problem :

Filter design in Matlab (e.g.) provides filter coefficients to

15 decimal digits (such that filter meets specifications)

For implementation, need to quantize coefficients to the

word length used for the implementation.

As a result, implemented filter may fail to meet

specifications ??


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Coefficient quantization effect on pole locations : example : 2nd-order system (e.g. for cascade realization)

`triangle of stability : denominator polynomial is stable (i.e.

roots inside unit circle) iff coefficients lie inside triangle

proof: apply Schur-Cohn stability test (cfr. supra). Stable iff all |ki|


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example (continued) :with 5 bits per coefficient, all possible `quantized pole positions are...

Low density of `quantized pole locations at z=1, z=-1, hence problem

for narrow-band LP and HP filters (cfr. supra).

- 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5- 1 . 5

- 1

- 0 . 5

0

0 . 5

1

1 . 5

end

end)plot(roots

1:0625.0:1for

2:1250.0:2for

!

!

i

i

H

K


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example (continued) :

possible remedy: `coupled realization

poles are where are realized/quantized

hence `quantized pole locations are (5 bits) :

- 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5- 1 . 5

- 1

- 0 . 5

0

0 . 5

1

1 . 5

QL .js QL,

(

(

+

+ +

O

L

Q

PL

Q

-

y[k]

u[k]


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Coefficient quantization effect on pole locations :

example : higher-order systems (first-order analysis)

-> tightly spaced poles (e.g. for narrow band filters) imply

high sensitivity of pole locations to coefficient quantization

-> hence preference for low-order systems (parallel/cascade)

!{

}

N

k

kk

ml

lm

kNm

mm

N

NN

N

N

N

aapp

ppp

ppp

zazaza

ppp

zazaza

1

21

22

11

21

2

2

1

1

).()(

,...,,:areroots'`quantized

......1:polynomial'`quantized

,...,,:areroots

......1:polynomial

C ff Q


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Coefficient Quantization

PS: Coefficient quantization in lossless lattice realizations,

o = original transfer functiono = original transfer function

+ = transfer function after 8+ = transfer function after 8--bitbit

truncationtruncation of lossless latticeof lossless lattice

filter coefficientsfilter coefficients

-- = transfer function after 8= transfer function after 8--bitbit

truncationtruncation of directof direct--formform

coefficients (bis)coefficients (bis)

In lossless lattice, all coefficients are sines and cosines, hence allIn lossless lattice, all coefficients are sines and cosines, hence all

values betweenvalues between 1 and +1, i.e. `dynamic range and coefficient1 and +1, i.e. `dynamic range and coefficient

quantization error well under control.quantization error well under control.


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Filter Implementation : Arithmetic Operations

Finite word-length effects in arithmetic operations:

In linear filters, have to consider additions & multiplications

Addition:

if, two B-bit numbers are added, the result has (B+1) bits.

Multiplication:

if a B1-bit number is multiplied by a B2-bit number, the

result has (B1+B2-1) bits.

For instance, two B-bit numbers yield a (2B-1)-bit product

Typically (especially so in an IIR (feedback) filter), the result of anaddition/multiplication has to be represented again as a B-bit number

(e.g. B=B). Hence have to remove most significant bits and/or least

significant bits


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Option-1: Most significant bits (MSBs)If the result is known to be upper bounded such that 1 or more MSBsare always redundant, these MSBs can be dropped without loss ofaccuracy. Dropping MSBs leads to better usage of available word-length, hence better SNR.

This implies we have to monitor potential overflow(=dropping MSBsthat are non-redundant), and possibly introduce additional scalingtoavoid overflow.

Option-2 : Least significant bits (LSBs)

Rounding/truncation/ to B bits introduces quantization noise.

The effect of quantization noise is usually analyzed in a statisticalmanner.

Quantization, however, is a deterministic non-linear effect, which maygive rise to limit cycle oscillations.


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zero-input limit cycle oscillations :

Example:

y[k] = -0.625.y[k-1]+u[k]4-bit rounding arithmetic

input u[k]=0, y[0]=3/8

output y[k] = 3/8, -1/4, 1/8, -1/8, 1/8, -1/8, 1/8, -1/8, 1/8,..

=oscillations in the absence of input (u[k]=0)


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Example: y[k] = -0.625.y[k-1]+u[k]4-bit truncation (instead of rounding)

input u[k]=0, y[0]=3/8

output y[k] = 3/8, -1/4, 1/8, 0, 0, 0,.. (no limit cycle!)

Example: y[k] = 0.625.y[k-1]+u[k]

4-bit rounding

input u[k]=0, y[0]=3/8

output y[k] = 3/8, 1/4, 1/8, 1/8, 1/8, 1/8,..

Example: y[k] = 0.625.y[k-1]+u[k]

4-bit truncation

input u[k]=0, y[0]=-3/8

output y[k] = -3/8, -1/4, -1/8, -1/8, -1/8, -1/8,..

Conclusion: weird, weird, weird, !


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Limit cycle oscillations are clearly unwanted (e.g. may beaudible in speech/audio applications)

Limit cycle oscillations can only appear if the filter has

feedback. Hence FIR filters cannot have limit cycle

oscillations.Mathematical analysis is verydifficult.

Truncation i.o. rounding often helps to avoid limit cycles (e.g.

magnitude truncation, where absolute value of quantizer

output is never larger than absolute value of quantizer input

( passive quantizer)).

Some filter structures can be made limit cycle free, e.g.

coupled realization, lattice filters, .

ALaRI2011_Lecture4

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