5. DIGITAL FILTERS STRUCTURES

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LABORATORY 5

DIGITAL FILTERS STRUCTURES

Infinite Impulse Response filters structures (for more theoretical details, please see the lecture)

Fie funcţia de transfer

0

1

1

Mk

k

k

Nk

k

k

b zB z

H zA z

a z

(0.1)

În domeniul timp, filtrul poate fi caracterizat prin ecuaţia cu diferenţe finite:

1

[ ] [ ] [ ]M N

k k

kk 0

y n = b x n k a y n k

(0.2)

Această ecuaţie permite calculul unui eşantion al ieşirii pe baza a M

eşantioane ale intrării şi a 1N eşantioane anterioare ale ieşirii. Modul în care este

calculată ieşirea (întâi partea nerecursivă şi apoi partea recursivă etc.) poate fi

reprezentat grafic sub forma unei structuri cu unităţi de înmulţire, adunare şi

elemente de întârziere.

5. DIGITAL FILTERS STRUCTURES

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5.1.1. Forma directă

Pornind de la ecuaţia cu diferenţe finite (0.2) operaţiile prin care se

calculează ieşirea filtrului pot fi reprezentate în structuri de tipul forma directă 1

sau forma directă 2 ca în figurile 5.1 şi 5.2 prin simpla identificare a coeficienţilor

ka şi kb .

Figura 5.1. Direct form 1

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1z

0b [ ]y n

1z

1b

1z

2b

Nb

[ ]x n

1a

2a

Na

Figura 5.2 Direct form 2

5.1.2. The cascade form

Funcţia de transfer poate fi factorizată sub forma:

1 * 1 1 2

0, 0, 1, 2,

1 * 1 1 21 1 11, 2,

(1 )(1 )( )

(1 )(1 ) 1

P P Pk k k k k k

k

k= k= k=k k k k

b z z z z b b z b zH z = = H z

p z p z a z a z

(0.3)

unde kz sunt zerourile iar kp sunt polii funcţiei de transfer grupaţi în perechi

complex conjugate pentru obţinerea funcţiilor de transfer de ordin 2 cu coeficienţi

reali ( )kH z . Dacă funcţia de transfer are zerouri sau poli reali atunci se poate face

o descompunere şi cu polinoame de ordin 1. Rezultă o realizare în cascadă,

reprezentată în figura 5.3.

)(1 zH )(2 zH )(zH P

[ ]x n1[ ]y n 2[ ]y n [ ] [ ]Py n y n

Figura 5.3. Realizarea în cascadă

Fiecare din filtrele componente, având o funcţie de transfer de ordinul 2,

poate fi realizată în una din formele directe 1 sau 2.

5. DIGITAL FILTERS STRUCTURES

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The MATLAB functions zp2sos, şi tf2sos allow to determine the 2nd

order transfer functions (Second Order Sections) for the cascade form.

Syntax:

[SOS,G]=tf2sos(B,A)

it returns a SOS matrix which contains the coefficients of the 2nd order sections

for the factorization of the transfer function ( )H z . B represents the vector of the

coefficients of the numerator ( )B z and A represents the vector of the

coefficients of the denominator ( )A z .

SOS is a matrix of the following form: SOS = [ b01 b11 b21 1 a11 a21

b02 b12 b22 1 a12 a22

...

b0L b1L b2L 1 a1L a2L ]

where each line of the matrix contains the coefficients of a 2nd order structure.

1 2

0, 1, 2,

1 2

1, 2,1

k k k

k

k k

b b z b zH z =

a z a z

G is a scalar which represents the global gain of the system. If G is not specified,

it is included in the first section.

[SOS,G]=zp2sos(Z,P,K)

it returns a SOS matrix which contains the coefficients of the 2nd order sections

for the factorization of the transfer function ( )H z . Z represents the vector of

zeros, P represents the vector of poles and K represents the gain of the pole-

zeros decomposition. The poles and zeros must be given in complex conjugated

pairs.

G is a scalar which represents the global gain of the system. If G is not specified,

it is included in the first section.

5.1.3. The parallel form

În acest caz se porneşte de la descompunerea în fracţii simple a funcţiei

( )H z . Presupunând coeficienţii funcţiei de transfer ,k ka b R , fracţiile simple (cu

numitor de gradul 1) au în general coeficienţi complecşi. Grupând însă fracţiile

5. DIGITAL FILTERS STRUCTURES

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corespunzătoare perechilor de poli complex conjugaţi, rezultă funcţiile ( )pH z de

gradul 2, cu coeficienţi reali, , ,,k p k pa b R . Funcţia de transfer poate fi scrisă sub

forma:

1

0, 1,

1 21 11, 2,1

P Pp p

p

p pp p

b b zH z C C H z

a z a z

(0.4)

Se obţine schema din figura 5.4.

1[ ]y n

1( )H z

2 ( )H z

)(zH p

2[ ]y n

[ ]py n

[ ]y n

[ ]x n

C

Figura 5.4. The parallel form

The MATLAB function residuez achieves the decomposition in simple

fractions of the transfer function ( )

( )( )

B zH z

A z . It can be applied in 2 ways:

[R,P,K]=residuez(B,A)

it decomposes in simple fractions the function ( )

( )( )

B zH z

A z defined by the

coefficients vectors B and A as follows:

111 21 1

1

( )

( ) 1 1

N

N

B z r rk k z

A z p z p z

R and P are column vectors containing the residuez and poles, respectively.

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K isa line vector for the free terms (if the order of the numerator is greater than

the one of the denominator).

If jp is a multiple pole of order m, the decomposition in simple fractions will

contain terms of the form:

1 1

21 1 11 1 1

j j j m

m

j j j

r r r

p z p z p z

[B,A]=residuez(R,P,K)

it makes the transfer function ( )H z from the decomposition in simple fractions.

In this way, in order to obtain the decomposition in the parallel structure with

real coefficients, the following method can be applied (only if the poles are ordered

in complex conjugated pairs).

[r,p,k]=residuez(b,a)

[b1,a1]=residuez(r(1:2),p(1:2),[])

[b2,a2]=residuez(r(3:4),p(3:4),[])

etc...

5.1.4. The lattice form

Sinteza funcţiei de transfer a unui filtru RII:

( ) ( )

( )( ) ( )

B z Y zH z

A z X z (0.5)

în forma latice se face completând structura latice recursivă cu o structură în scară

(figura 5.5).

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1z 1z1z

[ ]x n

Nk 1k

Nk 1k1Nk

1 Nk

Nc1Nc 1c 0c

[ ]y n

2Nc

2[ ]y n

1[ ]y n

Figura 5.5. Lattice-ladder form

În figură apar pe lângă nodurile de intrare [ ]x n şi de ieşire [ ]y n încă două

noduri 1[ ]y n şi 2[ ]y n . Se demonstrează că funcţiile parţiale de transfer asociate

acestor două ieşiri sunt:

11

( ) 1( )

( ) ( )

Y zH z

X z A z (0.6)

1

22

( ) ( ) ( )( )

( ) ( ) ( )

NY z z A z A zH z

X z A z A z

(0.7)

Din ecuaţia (0.6) rezultă că în cazul unei funcţii de transfer numai cu poli

structura latice nu are şi structura în scară, iar ieşirea filtrului va fi 1[ ]y n .

În ecuaţia (0.7) s-a notat cu ( )A z polinomul reciproc al numitorului ( )A z

obţinut din polinomul ( )A z prin inversarea ordinii coeficienţilor.

1

1 0( ) N

N NA z a a z a z

(0.8)

Evident funcţia 2 ( )H z este o funcţie trece-tot şi aceasta poate fi sintetizată în

forma latice fără structura în scară, ieşirea filtrului fiind 2[ ]y n .

Sinteza structurii latice pentru o funcţie de transfer RII se face pornind de la

coeficienţii 0 1, ,..., Na a a ai numitorului ( )A z şi se obţin coeficienţii de reflexie

1,..., Nk k după algoritmul recursiv:

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,

,

, ,

1, 2

1 1,1

for = 0 :1:

end

for = : -1: 2

for = 1:1: -1

1

end

end

N j j

i i i

i j i i i j

i j

i

j N

a a

i N

k a

j i

a k aa

k

k a

Calculul coeficienţilor scară 0,..., Nc c se face, de asemenea, recursiv folosind

şi coeficienţii 0 1, ,..., Nb b b ai numărătorului ( )B z .

,

1

1, 2, ,0

N N

N

l l i i i l

i l

c b

l N N

c b c a

for

end

The MATLAB function tf2latc computes the coefficients of the lattice

structure.

Syntax:

[K,C]=tf2latc(B,A)

it returns the vector K which contains the reflection coefficients ik and the

vector C which contains the coefficients of the ladder structure ic for an IIR

filter with the coefficients of the numerator in vector B and the coefficients of

the denominator in vector A, normalized to 0a .

K=tf2latc(B)

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it returns the vector K which contains the reflection coefficients ik for a FIR

filter with the coefficients of the transfer function in vector B, normalized to 0b .

K=tf2latc(1,A)

it returns the vector K which contains the reflection coefficients ik for an IIR

filter only with poles, with the coefficients of the denominator in vector A,

normalized to 0a .

Remark. If from the computations one of the reflection coefficients equals 1, then the

function tf2latc will generate an error (because the term 21 ik which appears

in the denominator is 0).

The transfer function can also be obtained starting from the reflection coefficients

and the coefficients of the ladder structure, using the function latc2tf. For

details type help latc2tf.

Example:

Let an IIR filter have the following transfer function: 1 1 1 2

1 3 4

( ) (0.5 )(1 0.5 )(1 1.2 1.8 )( )

( ) 1 0.5 1.2 0.5

B z z z z zH z

A z z z z

Synthetize the structures for the cascade, parallel and lattice forms.

// The polynomial at the denominator is: a=[1 0.5 0 1.2 0.5];

// The polynomial at the numerator is obtained from the zeros of the transfer

function. z1=[-2; -0.5];

z2=roots([1 -1.2 1.8]);

// Because the first free term is 0.5, the polynomial must be multiplied by 0.5. b=0.5*poly([z1; z2])

b =

0.5000 0.2500 -0.5000 2.7500 1.5000

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// The cascade form [SOS,G]=tf2sos(b,a)

SOS =

1.0000 2.5000 1.0000 1.0000 1.5206 0.4564

1.0000 -1.2000 1.8000 1.0000 -1.0206 1.0955

G =

0.5

The cascade decomposition:

1 2 1 2

1 2 1 2 1 2

1 2.5 1 1.2 1.8( ) ( ) 0.5

1 1.52 0.456 1 1.02 1.095

z z z zH z G H z H z =

z z z z

// The parallel form [r,p,k]=residuez(b,a)

r =

-0.6541

-0.6329 + 0.1741i

-0.6329 - 0.1741i

-0.5800

p =

-1.1091

0.5103 + 0.9138i

0.5103 - 0.9138i

-0.4115

k =

3.0000

It can be noticed that there are complex poles and residuez. Because we want a

structure with real coefficients, we will group the pair of complex conjugated

values into a 2nd order transfer function with real coefficients:

[b2,a2]=residuez(r(2:3),p(2:3),[])

b2 =

-1.2659 0.3278

a2 =

1.0000 -1.0206 1.0955

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The parallel decomposition:

1 2 4

1 1

21 4

1

1 1 2 1

( )( )

( )1 1

0.65 1.26 0.32 0.583

1 1.1 1 1.02 1.09 1 0.41

r B z rH z k

A zp z p z

z

z z z z

// The coefficients of the lattice form are obtained directly: [k,c]=tf2latc(b,a)

k =

0.3061

-0.2794

1.2667

0.5000

c =

-2.3898

-1.4985

-0.2333

2.0000

1.5000

Remarks:

The exact values of the coefficients obtained above have a much greater

number of decimal digits than the ones displayed in MATLAB in short format.

In practice, because of the representation of numbers on a finite number of bits,

differences (truncations) of the coefficients’ values with respect to the

simulated ones.

In the previous example it can be seen that the system is unstable, either by

noticing that one of the poles has a modulus greater than 1 or by using the

Schür-Cohn test for the reflection coefficients k and noticing that the coefficient

3k is greater than 1.

E1. Exercises: Given the IIR systems with the following transfer functions:

1. 1 2 3 4

1 2 3 4

2 3.6 1.3 0.4 0.2( )

1 2.8 3.15 1.75 0.425

z z z zH z =

z z z z

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2. 1 2 3 3

2 4 6

(1 2 3 4 )(1 )( )

1 0.8 0.66 0.35

z z z zH z

z z z

3. 1 2 3 4

1( )

1 1.27 1.19 1.18 0.4H z

z z z z

4. 1 2 3

1 2 3

0.4 0.7 0.175( )

1 0.175 0.7 0.4

z z zH z

z z z

5. 0.95 / 4 0.95 / 4 1.05 / 4 1.05 / 4

/ 4 / 4 / 4 / 4

( 0.9 )( 0.9 )( 0.9 )( 0.9 )( )

( 0.95 )( 0.95 )( 0.9 )( 0.9 )

j j j j

j j j j

z e z e z e z eH z

z e z e z e z e

Synthetize and draw the structures for the direct 1 and 2, cascade, parallel, lattice.