SP05 Dig Filters 1 Classif Analysis1 Vs100609

8/2/2019 SP05 Dig Filters 1 Classif Analysis1 Vs100609

1/93

Signal Processing05 digitalfilters(1)seriesofexpertlectures IrR.deWild

ASTRON,32010

1


2/93

learning

goals

2

At the end of this

lecture,

-

from

a given

transfer function

or

a given

impulse

response of a linear

system,

you

can

draw a functional

block

diagram (or

a signal

flow

diagram);

-

from

a given

block

diagram (or

a signal

flow

diagram) of a linear

system,

you

can

derive

the transfer function

and the impulse

response;

-

you

can

convert

a block

diagram from

its

direct form

into

its

transposed

form;

-

you

know

how

to adapt/modify

(in a qualitative

manner) the frequency

response of

a linear

system, by

moving

in the z-plane

the locations

of poles

and zeros;

-

you

can

classify

a digital filter according

to its

impulse

response, its

transfer function,

its

frequency

response and/or

its

functional

block

diagram;

-

you

can

identify

and analyze

comb

filters, all-pass

filters, FIR-filters

and IIR-filters;


3/93

Contents

3

digitalfilters

classification

digitalfilters

analysis

(1)

digitalfilters analysis (2):

Digital Signal Processing

Monson H. Hayes, McGraw-Hill publ.

par. 2.3; 8 (except

8.6)

Signals

& Systems

H.P. Hsu, McGraw-Hill publ. none

see SP06


4/93

systembehaviour four points ofview

)(

]1[))1((

1 zXz

nxTnx s

[ ])(arg)()( fHjefHfH =

)(

]1[))1((

2 fXe

nxTnx

sTfj

s

4

impulse

response

discrete-time: t = n.Ts

y[n] = h[n]

*

x[n]

(h[n]: implicit

expression)

transfer function

Y(z) = H(z)

. X(z)

H(z) = Y(z) / X(z)

frequency

response

H(f)

H(2 .f.Ts

)

complex z-plane

-

poles

& zeros

-

stability

criterium

(1/Ts

)-periodic

f-axis

-

Bode-diagram

(amplitude & phase)

-

stability

criterium

Z-Transform DTFT (Discrete-Time

Fourier

Transform)

peel-off

unit-circle

sTfj

ez

=

2

functional

structure:

block

diagram (schematic)

signal

flow

diagram (directed

graph)

)2( sss TfjTTp eez + ==


5/93

systembehaviour analysis &synthesis

5

Filter Analysis

starting

from

a given

-

impulse

response

-

transfer function

-

functional

structure,

arrive

at the resulting

frequency

response

What

is a filter?

type of signal

processing system

frequency-specific

treatment

of

signal-amplitude

and signal-phase

Filter Synthesis

starting

from

the required

frequency

response,

arrive

at the optimally

fitting

-

impulse

response

-

transfer function

-

functional

structure

= subject ofSP06= subject ofSP05


6/93

Contents

6

digitalfilters

classification

digitalfilters

analysis

(1)

digitalfilters analysis (2): see SP06


7/93

digitalfilters

classification

7

according

to impulse

response

according to transferfunction

according

to systemstructure

according to frequency response


8/93

impulse

response issues

8

relationship:

y[n]=h[n]

*x[n]

linear

difference

equation

inn(t=nTs

)

FIR (Finite Impulse Response)

IIR

(Infinite

Impulse

Response)


9/93

impulse response overview

9

linear

digital system: y[n] = h[n] * x[n]

=

+++++=

==

M

m

m

N

m

m

M

N

mnyamnxbny

Mnyanyanya

Nnxbnxbnxbnxbny

10

21

210

][][][

][]2[]1[

][]2[]1[][][

L

L

impulse

response h[h] linear

difference

equation

in n

Z-Transform

transfer function

H(z) bi-linear

function

in z : H(z) = T(z) / N(z) ,

with

T(z) = degree-N

polynomial

in z

N(z) = degree-M polynomial in z

x[n] [ny[n] h[n]


10/93

digitalfilters

classification

10

according

to impulse

response


according

to systemstructure



11/93

transferfunction issues

11

relationship:Y(z)=H(z).X(z) H(z) =T(z)/N(z)bi

linear function,with T(z)andN(z)polynomials inz

MA(Moving

Average)

H(z)=T(z) zplane:only zeros H(f) amplitudecharacteristic:dips

AR(AutoRegressive)

H(z)=1/N(z) zplane:only poles H(f) amplitudecharacteristic:spikes

ARMA(AutoRegressive

Moving

Average)

H(z)=T(z)/N(z) zplane:zeros &poles


12/93

transferfunction overview

12

linear

digital system: H(z) = T(z) / N(z)

polynomials

in z:

T(z) factorization

N(z) factorization

( )

( )

=

=

=

=

==

=1

0

1

0

1

0)(M

m

m

N

m

m

M

m

m

m

N

m

m

m

pz

zz

za

zb

zH L

complex z-plane:

-

N zeros: z0

,,zM-1

-

M poles: p0

,,pN-1

stability: all poles

within unit circle!

j.Im(z)

Re(z)

10

j

X O

O

X

X O

O

zero

pole


13/93

digitalfilters

classification

13

according

to impulse

response


according

to systemstructure



14/93

systemstructure

issues

14

representation (visualisation):block

diagram

(block

schematic)

signal

flow

diagram(directed

graph)

canonical

structures

directform I&IItransposed

form

I&II

frequencysampled

structure

not

explicitly

treated

MA transversal form FIRonly

outside scope ladder

structure,

lattice structure


15/93

systemstructure overview

15

[derived

from

h[n]

or

from

H(z):]

N(z)

= degree-M

polynomial

in z

AR (Auto Regressive)

part:

feed-back paths poles

T(z)

= degree-N

polynomial

in z

MA (Moving

Average)

part:

feed-forward paths zeros

H(z) =

T(z) / N(z)


16/93

digitalfilters

classification

16

according

to impulse

response


according

to systemstructure



17/93

frequency response issues

17

relationship:

Y(f)=H(f).X(f) H(f) =T(f)/N(f)

Bodediagram: amplitude &phase characteristicsLowPass

&

HighPass

BandPass

&

BandStop

MultiBand

AllPass

(i.e. phaseshift

only)


18/93

frequency response overview

ss

s

zjTfj

Tf

TzTfz

z

ezezfHzH s

2

1

2

1]arg[:periodic2]arg[

circleunit1

:)()(]arg[2

+


19/93

Contents

19

digitalfilters

classification

digitalfilters

analysis

(1)

digitalfilters analysis (2): see SP06


20/93

digitalfilters

analysis

(1)

20

functional systemstructure family tree

directform

I&II

transposed form I&II

lowpass

filterexamples

overview

according

tofrequency

response

transferfunction

&complexzplane

transfer

function

&

impulse

response

transferfunction

&frequency

response

transferfunction

&block

diagram


21/93

block diagramversussignal flow diagram

21

see

Digital Signal

Processing

(M. Hayes): p.288 (CH 8.2)

signal

flow

diagram (graph)

directed

branch

sink

node

source

node

block

diagram (schematic)

functional

block

(sub-system)

functional

interconnection

signal

delay

signal

scaling

signal

addition

input signal

output signal

In this

lecture, we adhere

to block

diagrams

(block

schematics) !


22/93

block schematic(1) Direct

Form I

22

Ts

z

-1

y[n]

Y(z)

x[n]

X(z)b0

Ts

z-1

Ts

z-1

Ts

z-1

Ts

z

-1

Ts

z-1

Ts

z-1

b2

b1

b4

b3

-a1

-a3

-a2

]3[]2[]1[]4[]3[]2[]1[][][ 32143210 ++++= nyanyanyanxbnxbnxbnxbnxbny


23/93

block schematic(2a) Direct

Form II

23

Direct Form

-

I Direct Form - II : step 1 swap T(z) and 1/N(z)

Ts

z-1

y[n]

Y(z)

x[n]

X(z)b0

Ts

z-1

Ts

z-1

Ts

z-1

Ts

z-1

Ts

z-1

Ts

z-1

b2

b1

b4

b3

-a1

-a3

-a2

( ) ( )

( ) ( )]4[]3[]2[]1[][]3[]2[]1[]3[]2[]1[]4[]3[]2[]1[][][

43210321

32143210

+++++=

=+++++=

nxbnxbnxbnxbnxbnyanyanya

nyanyanyanxbnxbnxbnxbnxbny


24/93

block schematic(2b) Direct

Form II

24

Direct Form

-

I Direct Form - II : step 2 combine delay-elements

Ts

z-1

y[n]

Y(z)

x[n]

X(z)b0

Ts

z-1

Ts

z-1

Ts

z-1

b2

b1

b4

b3

-a1

-a3

-a2


25/93

block schematic(3a) Transposed Form I

)}}}({)({)({

)}}}}({)({)({)({)()(

3

1

2

1

1

1

4

1

3

1

2

1

1

1

0

zYazzYazzYaz

zXbzzXbzzXbzzXbzzXbzY

+++

++++=

)()()(

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

3

3

2

2

1

1

4

4

3

3

2

2

1

10

zYzazYzazYza

zXzbzXzbzXzbzXzbzXbzY

++++=

25

Direct Form

-

I Transposed Form - I

rewrite

the Z-domain

input-output

relationship

]3[]2[]1[

]4[]3[]2[]1[][][

321

43210

++++=

nyanyanya

nxbnxbnxbnxbnxbny

Z-Transform


26/93

block schematic(3b) Transposed

Form

I

26

y[n]

Y(z)

x[n]

X(z)b0

Ts

z

-1

b2

b1

b4

b3

-a1

-a3

-a2

Ts

z-1

Ts

z-1

Ts

z-1

Ts

z

-1

Ts

z-1

Ts

z-1


27/93

block schematic(4) Transposed Form II

27

Transposed

Form

-

I

Transposed Form - II:

combine delay-elements

y[n]

Y(z)

x[n]

X(z)b0

Ts

z-1

b2

b1

b4

b3

-a1

-a3

-a2

Ts

z-1

Ts

z

-1

Ts

z-1


28/93

block schematic(5a) FrequencySampled Form

28

=

=

+=

=

==1

0

][][)(Nn

n

nn

n

n znhznhzHFIR-type

filter:

h[n] = h(nTs

) Ts

-sampled

&

NTs

-finite

substitute

Inverse-DFT-expression

of h[n]

into

FIR-expression

of H(z):

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

==

1

0 1

2

1

0 1

2

2

1

0 1

2

2

1

0

1

0

2

1

0

1

0

21

0

1

][1

1

1][

1

1

1][

1][

1

][

1

][)(

Nk

kk

Nj

NNk

kk

Nj

Nkj

Nk

k

k

N

j

NNk

Nj

Nk

k

Nn

n

nnk

Nj

Nn

n

nNk

k

nkN

jNn

n

n

ze

kH

N

z

ze

zekH

N

ze

zekH

NzekH

N

zekHNznhzH

parallelserial


29/93

block schematic(5b) FrequencySampled Form

=

=

=

1

0

220

00 ][)()(Nn

n

nTfjfj senheHfH

=

=

=

=

==

1

0

2

0

1

0

20 0

0

][][)(Nn

n

nkN

jNn

n

nTNT

kj

s

s

enhkHenTh

NT

kH

ss

29

Discrete-Time

Fourier

Transform: h[n] = h(nTs

) Ts

-sampled

(& finite)

(DTFT)

H(f) (1/Ts

)-periodic

& f-continuous

frequency-sampled

approximation

of one

period

of H(f)

Discrete Fourier

Transform: h[n] = h(nTs

) Ts

-sampled

& (NTs

)-limited

(~ period)

(DFT)

H[k] = H(k/NTs

) (1/Ts

)-periodic

& (1/NTs

)-sampled

=

=

=

=

+

=

=

1

0

2

0

1

0

2

0

00

][1

][1

)(Nk

k

nkN

jNk

k

TnNT

kj

s

s ekHN

nheNT

kH

NTnh

ss

Inverse Discrete Fourier

Transform

(apply

duality

property):

see DA05


30/93

block schematic(5c) FrequencySampled Form

30

x[n]

X(z)

1 / N

Ts

z-1

H[0]

y[n]

Y(z)

Ts

z-1

1

H[1]

Ts

z-1

exp

( j2

/ N )

H[N-1]

Ts

z-1exp ( j2(N-1)/ N )

H[k]

Ts

z-1

exp ( j2.k/ N )

N x


31/93

digitalfilters

analysis

(1)

31

functional systemstructure family tree

directform

I&II

transposed form I&II

lowpass

filterexamples

overview

according

tofrequency

response

transferfunction

&complexzplane

transfer

function

&

impulse

response

transferfunction

&frequency

response

transferfunction

&block

diagram


32/93

lowpass filterexamples:overview (1)( )

( )

=

=

=

=

==

=1

0

1

0

1

0)(M

m

m

N

m mMN

M

m

m

m

N

m

m

m

pz

zz

z

za

zb

zH L

32

transfer function complex z-planeH(z) = T(z) / N(z)

M zeros: z0

,,zM-1N poles: p0

,,pN-1


33/93

lowpass filterexamples:overview (2)

33

T(z)order N(z)order filtertype(impulse

response/zplane

/frequency

response)

N=M M=N comb

filter:

type1

N=M

notches,

equidistant

on

unit

circle

|z|

=1

type2

M=Npeaks,equidistant

on

unitcircle

|z|=1

N=M M=N allpass

filter:

N=Mpolezero

pairs:geometric

inversew.r.t.unitcircle

0 M IIRfilter:

Mpoles,within

unitcircle|z|=1

M=1 1pole: real &pos./neg.;[complex outside scope]

M=2 2poles:2xreal ;1x2foldreal;complexconjugate pair

N 0 FIRfilter:

N

zeros;

N

=even/odd

h[n]=geometric

sequence

(H(z):nonlinear

phase)

h[n]=even/odd

function H(z): linear phase 4FIRtypes


34/93

lowpass

filterexamples

comb

filters

allpass filters

FIRfilters

IIRfilters

evaluation


35/93

lowpass filterexamples comb filter intuitive

35

block

diagram

Ts

z-1

1

Ts

z-1

Ts

z-1

- 1

x[n]

X(z)

N delay-units time-shift: = N.Ts [s]

y[n]

Y(z)

harmonic

response

constructive

combining

at output, if

x[n N] and x[n] are in counter-phase

( i.e. for

= , 3, 5,

[rad] )

destructive

combining

at output, if

x[n N] and x[n] are in phase

( i.e. for

= 2, 2, 2,

[rad] )

x[n -N]

X(z) . z-N

l f l l


36/93

lowpass filterexamples comb filter 2types

( )( )N

NNNNNN

z

zzzzzzzH

=++++== )1(12211 111)( L

36

linear


comb

filter

type 1

1 pole

N-fold

at z = 0

N zeros

equidistant

on

circle

|z| =

( < 1)

N notches (dips) equidistant on unit circle |z| = 1

1 -

impulse

response

2 -

transfer function

3 -

block

schematic

comb

filter

type 2

1 zero

N-fold

at z = 0

N poles

equidistant

on

circle

|z| =

( < 1)

N peaks (spikes) equidistant on unit circle |z| = 1

( )( ) NNN

NNNNz

z

zzzzzzH

=

++++=

=

)1(12211 11

1

1

1)(

L

M. Hayes: pp.331

exercise

type

1 : see

next

slides

type 2 : do-it-yourself!

l f l l


37/93

lowpass filterexamples comb filter type1(1a)

37

( )( )N

NNNNNN

z

zzzzzzzH

=++++== )1(12211 111)( L

transfer function

complex Z-plane

comb

filter

type 1

1 pole

N-fold

at z = 0

N zeros

equidistant

on

circle

|z| =

( < 1)

N notches (dips) equidistant on unit circle |z| = 1

NOTE

evaluation slides comb & IIR = FIR

l fil l


38/93

lowpass filterexamples comb filter type1(1b)N

NN

z

zzH

=)(

38

transfer function

complex Z-plane

][1

2 radN+

n=2

0

n=1

n=N-1

O O

O

n=3

O

Re(z)

jj.Im(z)

1

n=0(=N)OXN x

l fil l


39/93

lowpass filterexamples comb filter type1(2)

sNTfjN efH= 21)(

NNzzH

= 1)(

][][][ Nnnnh N =

39

transfer function

impulse

response

frequency

response

t = nTs

[s]0

NTs

N

.(t NTs

)

(t)

NjnjNNT

NT

nj

N

ss

NnjN

NT

NT

nj

N

ss

eee

NT

nH

NT

nf

eeNT

nHNT

nf

ss

s

s

+===

++=

===

=

+

111:for

111:for

22

21

21

2

2

21

maxima

minima

l fil l


40/93

lowpass filterexamples comb filter type1(3)sNTfjN efH

= 21)(

40

transfer function

frequency

response

f [Hz]

1 / Ts1 / NTs0

= 1

0 <

< 1

e.g. N = 5

1 + N

1 -

N

exercise

Applying

the duality

property

of the Fourier

transform,

which

case do you

get

when

you

swap t-

and f-domain?

l filt l


41/93

lowpass filterexamples comb filter type1(4)

41

NNzzH

= 1)(

transfer function

block

diagram

Ts

z-1

1

Ts

z-1Ts

z-1

N

x[n]

X(z)

y[n]

Y(z)

x[n -N]

X(z) . z-N

N

delay-units


42/93

lowpass

filterexamples

comb

filters

allpass filters

FIRfilters

IIRfilters

evaluation

l filt l


43/93

lowpass filterexamples allpass filter (1a)

( ) ( )

( )

( ) ( )

( ) 1,,1,0;

:zeros

1,,1,0;

:poles

2

1

2

2

2

==

==

==

==

+

+

+

+

Nneerz

eererz

Nneerz

eererz

nN

jj

njNjNjN

nN

jj

njNjNjN

L

L

( )( ) NNj

NjN

NN

NN

zer

erz

z

zzH

=

=

11

)()(

43

transfer function

complex Z-plane

N zeros

equidistant

on

circle

|z| =

r < 1

1 -

impulse

response2 -

transfer function

3 - block schematic

N x pole-zero pair: geometric inverse (with respect to unit circle)

N poles

equidistant

on

circle

|z| = r-1

> 1

jj

erer

==

l filt l


44/93

lowpass filterexamples allpass filter (1b)( )( ) NN

NN

NNj

NjN

NN

NN

z

z

zer

erz

z

zzH

=

=

=

)(1

11

)()(

44

][1

2rad

N+

n=2

r

0

n=1

n=N-1

O

O

O

n=3

O

Re(z)

X

j

j.Im(z)

1/r

n=0(=N)O

X

XXX

transfer function

complex Z-plane

1

jj erer ==

l filt l


45/93

lowpass filterexamples allpass filter (2)

][)(][][][ nxNnxNnyny NN =

NN

NN

z

zzH

=

1

)()(

NN

N

NN

N

NN

NN

zz

z

z

zzH

=

=

1

)(

11

)()(

45

transfer function

Impulse

response

][][

][)(][][

2nuNnu

nuNnunh

NN

NNN

=

=

implicit

formulation

of impulse

response ( in terms

of x[n] and y[n] )

explicit

formulation

of impulse

response (= h[n] )

l filt l


46/93

lowpass filterexamples allpass filter (3)( )( ) NNj

NjN

NN

NN

zer

erz

z

zzH

=

=

11

)()(

46

exercises

1

prove that

the amplitude

characteristic

is flat

( i.e. H(f) = 1 for

all f );

2

compute

the phase

characteristic

for

f = 0 ,

1/4Ts

, 1/2Ts

, 3/4Ts

, 1/Ts ;

transfer function

frequency

response

( )

( )s

s

s

s

NTfjNj

NjNTfj

NTfjN

NNTfj

eer

ere

e

efH

=

=

2

2

2

2

11

)()(

(peel-off

the unit circle)

low pass filter examples


47/93

lowpass filterexamples allpass filter (4)

47

NN

NN

z

zzH

=

1

)()(

transfer function

block diagram

Ts

z-1

1

Ts

z-1

Ts

z-1

(*)N

x[n]

X(z)

y[n]

Y(z)

x[n -N]

X(z).z-N

N

delay-units

Ts

z-1

N

N

delay-units

y[n -N]

Y(z).z-N


48/93

lowpass

filterexamples

comb

filters

allpass filters

FIRfilters

IIRfilters

evaluation



49/93

lowpass filterexamples FIR geom.seq.&lin.phase

=

=

+=

=

==Nn

n

nn

n

n znhznhzH0

][][)(

)(1

11)(

11

1

)1(1221

0

=

=++++==

++

++

=

=

zz

z

z

zzzzzzH

N

NNNNNN

Nn

n

nnL

49

linear

phase:

restrictions

on

h[n] 4 types

h[n] sampled

rectangular

pulse

| H(f) | periodic-sinc

(sin(Nx

/ Nsinx

)

arg{H(f)}

linear

function

of f

finite

geometric

sequence:

h[n] = n

, n = 0,1,,N ; 0 <

< 1

=

=

=Nn

n

nzzH

0

)(

special case: uniform weighting

of N successive

samples

N

1h[n]

1

exercise

ForN=4

and N=5, derive

z-plane, impulse

response,

frequency

response, and

block

schematic.

see SP04



50/93

lowpass filterexamples FIR geometric sequence (1a)

50

)(1

11)(

11

1

)1(1

221

=

=++++=

++

++

zz

z

z

zzzzzH N

NNNN

NNL

nN

j

ez 12

+=

transfer function

complex Z-plane

1 -

impulse

response

2 -

transfer function

3 -

block

schematicNOTE

evaluation slides comb & IIR = FIR

N+1 zeros:

1 pole:1 pole

(N-fold):

N zeros:

1 pole

(N-fold):

z = 0

z =

(n=0,1,2,,N)

nNjez 12

+=

z = 0

(n=1,2,,N)

zero-pole

cancellation

for

z=

finite

geometric

sequence: h[n] = n

, n = 0,1,,N see SP04



51/93

lowpass filterexamples FIR geometric sequence (1b)

51

][1

2rad

N+

n=2

0

n=1

n=N

O

O

O

n=3

O

Re(z)X .

N x

j.

j.Im(z)

transfer function

complex Z-plane)()(11

=

++

zz

zzH

N

NN

j

1



52/93

lowpass filterexamples FIR geometric sequence (2)][]2[]1[][][

2

NnnnnnhN

++++= L

sss NTfjNTfjTfj eeefH ++++= 222221)( L

NNzzzzH

++++= L2211)(

52

transfer function

impulse

response

frequency

response

t = nTs

(t) .(t Ts

) N.(t NTs

)

0 NTsTs

maxima & minima ???

finite

geometric

sequence

h[n] = n

n = 0,1,,N

0 <

< 1



53/93

lowpass filterexamples FIR geometric sequence (3a)sss NTfjNTfjTfj

eeefH

++++=

22222

1)(L

NNzzzzH

++++= L2211)(

53

transfer function

frequency

response

|H(f)|

f [Hz]1/NTs+1/2Ts-1/2Ts 0

H(z): N zeros

on

circle

|z|=

H(f): N notches

(dips)

on

unit circle

|z|=1

N notches

oddNif,for

evenNiffor1

1

21

21

221

1

ss

ss

NTN

NTN

NTN

T

N

f

f

+

+

=

==

=

+

minglobal

(1-N+1)/(1-) = global

max

for

f = 0

0 <

< 1



54/93

54

transfer function

frequency

response

lowpass filterexamples FIR geometric sequence (3b)0 <

< 1

min

for

f = 1/2Ts

:

(1-8)/(1+)

=

1

1

:0for

8

fmax

7 notchesexample:

N = 7 (odd)



55/93

55

transfer function

frequency

response

)1(1

1)0(

:0for1

1

1

1

1

1

1

)1(

:1)N,1,2,(nfor2-N

1

1

1

1

1

1

)1(

:N),1,2,(nfor1-N

21

11

2

1

11

2

21

)1(2

)1()1(21

21

21

1

2

1

1

2

21

)1(2

)1()1(

21

21

21

NN

Njn

Nj

N

Njn

Nj

jnjN

TTN

nj

TNTN

n

jN

s

s

nNj

N

nNj

njN

TTN

nj

TNTN

nj

N

s

s

H

f

eeee

ee

e

e

TN

nH

NT

nf

ee

e

e

e

TN

nH

NTnf

ss

ss

ss

ss

++++=

=

=

+=

=

=

+

+

=+

=

=

=

=

+

==

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

L

L

L

maximumglobal

maximalocal

(notches)minima

s

s

Tfj

TNfjN

e

efH

++

=

2

)1(21

1

1)(

lowpass filterexamples FIR geometric sequence (3c)1

turn resp. N+1

turns

on

circles

with

radii

resp. N+1

( circlein complex z-plane

with

radius

)

lowpass filter examples


56/93

lowpass filterexamples FIR geometric sequence (4a)

56

1

)1(1221

1

1

1)(

++

=++++= z

z

zzzzH

NNNN

L

Ts

z-1

y[n]

Y(z)

1

Ts

z-1

Ts

z-1

N2

transversal

structure

(tapped

delay

line; MA only)

x[n]X(z)

transfer function

block

diagram

x[n -N]X(z) . z-N

[N delay-units]



57/93

lowpass filterexamples FIR geometric sequence (4b)

57

1

)1(1221

1

1

1)(

++

=++++= z

z

zzzzH

NNNN

L

Ts

z-1

y[n]

Y(z)

1

Ts

z-1

Ts

z-1

-

N+1 -

cascade of

MA structure

and

AR structure

x[n]

X(z)

[N+1

delay-units]

Ts

z-1

transfer function

block

diagram

correct:

IF

MA-structure

THEN

FIR-behaviour

incorrect: IF FIR-behaviourTHEN MA-structure

x[n (N+1)]

X(z).z-(N+1)

y[n -1]

Y(z).z-1



58/93

lowpass filterexamples FIR linear phase (0a)

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

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

:proof

)()()()(odd)(

)()()()(even)(

:therevisiting

fHfHjfHfHjfHfH

fHjfHfHfhfhth

fHfHththth

fHfHththth

oddevenoddeven

oddevenoddeven

==+=

+=+=

==

==

propertiessymmetryFourier

=

=

+=

=

==Nn

n

nn

n

nznhznhzH

0

][][)(

58

H(f)

with

linear

phase which restrictions to h[n] ???

M.Hayes: pp.189+190

H.P.Hsu: pp.333+334

put restrictions

on

the (anti)symmetry

of

h[n]

!!!

1 -

impulse

response

2 -

transfer function

3 -

block

schematic



59/93

lowpass filterexamples FIR linear phase (0b)

fNTffNTfff

eefHefHfHfH

nNhnhnhh

ss

NTfjfjfj s

~)(2)()(

)()()()(

][][][)symmetric"("even[n]

2)()(

====

==

59

So: h[n] (anti-)symmetric arg{H(f)} ~ f

fNTffNTfff

eeefHefHfHfH

nNhnhnhh

ss

NTfjfjjfj s

~)(2)()(

)()()()(

][][][)symmetric"-anti("odd[n]

2

2)()(

==

==

==



60/93

low pass filterexamples FIR

linear

phase

(0c)

60

Only

digital

(FIR-)filters can

realize

an

exact linear

phase

!

4linearphase

FIRfilter

types

Neven

centre:n=

N/2

Nodd

centre:n=

(N

1)/2

,n=

(N+1)/2

h[n]even:

h[Nn]= h[n]

h[(N/2)m]= h[(N/2)+m]

h[0] = h[N]

h[1] = h[N 1]

....

h[N/2] = h[N/2]trivial

h[(N1)/2m]= h[(N+1)/2 +m]

h[0] = h[N]

h[1] = h[N 1]

....

h[(N1)/2] = h[(N+1)/2]

h[n]odd:

h[Nn]=

h[n]

h[(N/2)m]= h[(N/2)+m]

h[0] =

h[N]

h[1] =

h[N 1]

....

h[(N/2)1] =

h[(N/2)+1]

h[N/2] =

h[N/2] =0

h[(N1)/2m]= h[(N+1)/2 +m]

h[0] =

h[N]

h[1] =

h[N 1]

....

h[(N1)/21]=

h[(N+1)/2+1]

h[(N1)/2] =

h[(N+1)/2]



61/93


linear

phase

(1a)

N

nn

zNnhznhznhnh

znhzHznhzH

=

=

+

++++=

===

][]2[]1[][

][)(][)(

21L

Nn

0n

th :FIRorderN

61

H(z) = degree-N

polynomial

in z equation H(z)=0 has N solutions:

H(z) = 0 z = z0, z1, , zN-1

transfer function

complex Z-plane

=

=

=

=

=

=

===1

000

)(][][)(Nl

l

l

NNn

n

nNNNn

n

nzzzznhzznhzH

location

of the N zeros

z0

,z1

, , zN-1

in the complex z-plane:

-

not

necessarily

equidistant

-

not

necessarily

on

(unit) circle;

-

appearing

as

complex conjugate

pairs (w.r.t. real

axis)

and/or

as reciprocal

pairs (w.r.t. unit circle);-

appearing

as conjugate

reciprocal

quadruples;

N+1 terms N factors



62/93


linear

phase

(1b)

NzNnhznhznhnhzH ++++= ][]2[]1[][)( 21 L

62

transfer function

complex Z-plane

][0 rad

r0

O

O

O

r0

O

Re(z)X .

N x

j.r

j.Im(z)

j

1

compare with geometric-sequence FIR!

O

O

N

zeros

1/r0

][0 rad



63/93


linear

phase

(2a)

63

transfer function

impulse

response

N evenN odd

h[n]

even

h[n]

odd



64/93


linear

phase

(2b)

64

transfer function

impulse

response

frequency

response

h[n]even

h[n]

odd

N odd N even



65/93


linear

phase

(3a)

65

amplitude:

1

main

lobe,

N-2

side

lobes

of unequal

width;

phase: linear

with

f ( incl. phase-jumps

of 2

[rad] )

transfer function

frequency

response

=

=

=Nn

n

nTfj senhfH0

..2][)(

h[n] =1

, for

n=0,1,,N

= 0 , for other n

N oddN even

example



66/93


linear

phase

(3b)

66

example: N = 6 z-plane: 3 x 2 zeroscomplex conjugate

pairs (w.r.t. real

axis)

reciprocal

pairs (w.r.t. unit circle)

transfer function

frequency

response

=

=

=Nn

n

nTfj senhfH0

..2][)(



67/93


linear

phase

(4)

67

=

=

=Nn

n

nznhzH

0

][)(

= MA-part of

- direct form I & II- transposed form I & II

transfer function

block

diagram

Ts

z-1

y[n]

Y(z)

h[n2]

Ts

z-1

Ts

z-1

transversal

structure

(tapped

delay

line; MA only)

x[n]

X(z)

x[n -N]

X(z) . z-N

[N delay-units]

h[n1] h[nN]h[n]

compare with geometric-sequence FIR!

l fil l


68/93

lowpass

filterexamples

comb

filters

allpass filters

FIRfilters

IIRfilters

evaluation



69/93

low pass filterexamples IIR order1&order2(1)

=

=+++=

z

z

zzzzH

1

221

1

11)( L

69

linear


IIR filter: T(z) = 1, N(z) = order-M

polynomial

in z

M poles, within unit circle

M = 1

1 pole: real, pos./neg. ; [complex]M = 2

2 poles: 2 x real

; complex conjugate

pair; xxx

M = 1

M = 2)1)(1(

11)(

1

2

1

1

2

2

1

10

=

++=

zpzpzazaazH

M.H. Hayes, Dig. Signal

Processing:

CH.2 pp. 79 (S.P. 2.10)

M.H. Hayes, Dig. Signal

Processing:

CH.8 pp. 185+186CH.2 pp. 79 (S.P. 2.10)

lowpass filter

examples


70/93

IIR order1&order2(2)

70

IIR-filter:

order 1 1 pole-

1 x real

& positive

-

1 x real

& negative

order 2 2 poles-

2 x real

-

1 x real

& 2-fold

-

2 complex conjugate

transfer function

impulse

response

complex Z-plane



71/93

low pass filterexamples IIR order1(1a)

==+++=

z

z

zzzzH 1221

1

1

1)(L

71

see SP04 transfer functioncomplex Z-plane

NOTE evaluation slides comb & IIR = FIR

infinite

geometric

sequence: h[n] = n

, n = 0,1,,N

1 -

impulse

response

2 -

transfer function

3 -

block

schematic

1 zero: z = 0

1 pole: z = real

stability -1 < < +1

lowpass filterexamples


72/93

p p IIR order1(1b)

=

=

z

z

zzH

11

1)(

72

0

stability: |z| = r < 1

ORe(z)X

X = r . exp[j2.f0

.Ts

]

1

1||

:circleunit2

=

=

z

ez sTfj

j

single real

pole:

single complex pole:

transfer function

complex Z-plane

sTr

=


73/93

p p IIR order1(1c)

=

=

z

z

zzH

11

1)(

73

( )( ) ( ) ( )( ) ( ) ( ){ }( ) ( ) ( ){ } LL

L

L

=+++++

+++++=

=++++=

==

====

33221

33221

3332221

1

3sin2sinsin0

3cos2coscos1

1

1

1)(

)2argandwith(

zrzrzr

zrzrzr

zerzerzer

zerzH

Trer

jjj

j

s

j

j

f0

single complex

pole

z = |H(f)|: not even; single notch at f0 = /(2Ts) [Hz]

h[n]: complex (I-part and Q-part)

= S.S.B.-system outside scope of Theme 8

transfer function

complex Z-plane



74/93

p p IIR order1(2a)L+++=

]2[]1[][][

2

nnnnh

74

real: |H(f)| not even

0 <

< 1 h[n] monotonously decreasing

exp(2.f0

.Ts

) = +1 f0 = 0 [d.c.]

-1 <

< 0

h[n] alternatingly decreasingexp(j2.f0

.Ts

) = -1 f0 = 1/(2Ts) [Nyquist rate]

complex: |H(f)| not even

= r.exp(j), 0 < r < 1 h[n] complex: oscillating & decreasing

exp(j2.f0

.Ts

) = exp(j)

f0 = /(2Ts) = single notch

1,2

= r.exp(j), 0 < r < 1 h[h] real: oscillating & decreasing IIR, order 2

cos(2.f0

.Ts

) = cos() f0 = /(2Ts) = complexconjugate pair

of notches

transfer function

impulse

response

1

221

1

11)(

=+++=

zzzzH

L



75/93

p p IIR order1(2b)

]1[][][]2[]1[][][2

+=+++= nynxnynxnxnxny L

75

transfer function

impulse

response

( )1

221

1

)()(1)(

=+++=

z

zXzXzzzY

L

< 1 stable

> 1 instable



76/93

p p IIR order1(2c)

76

transfer function

impulse

response

frequency

response

( )1

221

1

)()(1)(

=+++=

z

zXzXzzzY

L



77/93

p p IIR order1(3a)

=

=

=+++=

z

z

z

zzzH

1

221

1

1

1)( L

=

=

=+++=

s

s

s

ss

Tfj

Tfj

Tfj

TfjTfj

e

e

e

eefH

2

2

2

2222

1

1

1)( L

77

transfer function

frequency

response

construction rule:

H(z) = bi-linear function of z ifzon circle, thenH(z)on circle

sTfj

ez

=2



78/93

p p IIR order1(3b)

78

transfer function

frequency

response



79/93

p p IIR order1(4)1

221

1

11)(

=+++=

zzzzH

L

79

Ts

z-1

y[n]

Y(z)

x[n]

X(z)

-

transfer function

block

schematic

y[n -N]

Y(z).z-NAR structure



80/93

p p IIR order2(1a)

2

2

11

0

20

2

11

2,1 222

4

:poles a

aa

a

aaaa

p

=

=80

)()()()(1

1

)1)(1(11)(

2121

2

2

2

21

1

21

1

2

1

1

2

2

1

10

ppzppz

z

zppzpp

zpzpzazaazH

++=

++=

=

=++

=

a0

, a1

, a2

: all real

a0

1

a1

, a2 : bounded

by

stability

criteria

transfer function

complex Z-plane

1 -

impulse

response

2 -

transfer function

3 -

block

schematic



81/93

IIR order2(1b)

221)cos2(1

1)(

+=

zrzrzH

cos2;

22222

12

12

1122222

211

2

21211

21

21

2

21

21

12

21

1212

2

1

212

2

1122

2

1112

2

1

=+=====

+

=

=


82/93

IIR order2(1c)

21

2

aa

r == 21

2

aa

r ==

82

sTf

=

2

0Re(z)X

1

j

j.Im(z)

Re(z)

j.Im(z)

j

1X

X

X

0

case A ( B):two real poles

case C ( B):pair ofcomplexconjugate poles

transfer function

complex Z-plane)()()(

2121

2

2

21

2

0

2

ppzppz

z

azaza

zzH

++=

++=

p1p1

p2

p2



83/93

IIR order2(2)( )]2[]1[][

1)(

210

= nxanxanxa

ny

83

transfer function

impulse

response

2

2

1

10

)()( ++

=zazaa

zXzY

a0 1

a1 cos

a2 a

[rad]

= 2.f0

.Ts1/f0

.= 2.Ts

/



84/93

IIR order2(3)

84

)()()(

2121

2

2

21

2

0

2

ppzppz

z

azaza

zzH

++=

++= transfer function

frequency

response



85/93

IIR order2(4)

85

y[n]

Y(z)

-a1

-a2

Ts

z-1

Ts

z

-1

x[n]

X(z)

Ts

z

-1

Ts

z-1

-a1

-a2

x[n]

X(z)

y[n]

Y(z)

AR-structure, direct form AR-structure,

transposed

form

2r.cos

r2

exercise: how to make this filter behave like an oscillator?

transfer function

block

schematic

1;1

)( 022

1

10

=++

=

azazaa

zH



86/93

lowpass

filterexamples

comb

filters

allpass filters

FIRfilters

IIRfilters evaluation

evaluation


87/93

comparison:IIRversusFIR(1)

87

filteringperformance:passbandstopband

range

transitionband rolloffphase linearity stability &settling time effectoffeedbackon signal error accumulation

sensitivity

tocoefficientvariations

filterorder(number ofpoles) size ofblock schematic(amount ofcircuitry)

evaluation

comparison: IIR versus FIR (2)


88/93

comparison:IIRversusFIR(2)

88

evaluation

comb filter & IIR filter = FIR filter (1)


89/93

{ }{ }L

LL

L

++++++=

=++++=

=

==

++

)1(1

)1(1

)1(1221

1

1

1

11

NNNN

NNNN

NN

NN

zz

zz

zzz

zz

(z)H(z)H(z)H IIR1comb1FIRN

NN z= 1(z)Hcomb1

L+++=

=

221

11

11 zz

z

(z)HIIR1

combfilter

&IIRfilter

=FIRfilter

(1)

evaluation



90/93

the cascade of a type 1

comb-filter

of order N with

an

IIR-filter

of order 1

yields

a finite

geometric

sequence

FIR-filter

of order N

{ }{ }L

LL

L

+++

+++++==++++=

==

+

)]1([][

][)]1([]1[][)]1([]2[]1[][

][][][

1

1

12

11

NnNn

NnNnnnNnnnn

nhnhnh

NN

NN

N

IIRcombFIRN

][][][1 NnnnhN

comb =

L

+++= ]2[]1[][][2

1 nnnnhIIR

order-N

FIR

= order-1 IIR

minus N-delayed

order-1 IIR

combfilter

&IIRfilter

=FIRfilter

(2)

evaluation



91/93

t = nTs0 NTs

(t)

combfilter

&IIRfilter

=FIRfilter

(3)

0 NTs

N.(t NTs

)

0

NTs

N.(t NTs

)

(t)

t = nTs

t = nTs

IIR + delayed IIR(parallel system) comb * IIR(cascaded system)

N.(t NTs

)

.(t Ts

)

t = nTs

(t).(t Ts

)

*

t = nTs

(t) .(t Ts

) N.(t NTs

)

0 NTsTs

evaluation



92/93

Ts

z-1

y[n]

Y(z)1

Ts

z-1

Ts

z-1

-

N

-

x[n]

X(z)Ts

z-1

Ts

z-1

y[n]

Y(z)

1

Ts

z-1

Ts

z-1

N-1

2

x[n]

X(z)

combfilter

&IIRfilter

=FIRfilter

(4)

=

x[n -N]

X(z).z-N

y[n -1]

Y(z).z-1

x[n -(N-1)]

X(z).z-(N-1)

exercises


93/93

exercises

see

reference

book: Solved

Problems

SP05 Dig Filters 1 Classif Analysis1 Vs100609

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