coltech main slides

8/13/2019 ColTech Main Slides

1/98

D
http://find/


2/98

D

H(z) = +z1

1 +z1,

*

1
http://find/


3/98

D

max

p

lim0


4/98

D

2

c

z =

z

2

c

p

=

ej

c c

z p =

complementary

angle to z

= cz cp=

2 z

2 p

= p z = (z p) = H(ej)
http://find/


5/98

D

(0) = 0 () =

( )

1

0 2

2

( )
http://find/


6/98

D

= 1

1 = +

0 2

2

( )

1

( ) changes rapidly

when approaches
http://find/


7/98

D

[ ]y n[ ]x n

All-pass

network 1

All-pass

network 2

+

+
http://find/


8/98

D

0 2

2

( ) 1 2both ( ) & ( )

1 2

pass band

( ) ( ) =

2 ( )

1( )

2 1

stop band

( ) ( ) = 2 1

transision band

( ) ( )
http://find/http://goback/


9/98

D

(low rate)0

3

3

( )

for both high rate

and low rate 2

2

2

3

2

(high rate)

low-rate spectrum is

squeezed by a factor of 2

0

H(z) =HL(z2) =

+z2

1 +z2 H(ej) =HL(ej2)
http://find/


10/98

D

H(z) =HL(z2)

1

1j

1j

j

j

( )L

H z( )2( ) LH z H z=
http://find/


11/98

D

H(ej) = + ejM

1 +ejM

=ej

M2 +ej

M2

ejM

2 +ejM

2

=1j

11+

tan

M2

1 +j 11+ tan M2

2

denomin

ator

numerator

1

1tan

1 2

M

+

1tan

1 2

M

+

0
http://find/


12/98

D

H(z) = +z1

1 +z1

1

[ ]x n [ ]n[ ]v n

direct form II
http://find/


13/98

D

1z

[ ]x n [ ]n[ ]v n

1[ ]q n

2 1[ ] [ 1]q n q n=

1

[ ]q n

1[ ]q n

1

[ ]q n

1

2[ ]q n

( )z
http://find/


14/98

D

H(z) = +z1

1 +z1

1z

[ ]n [ ]y n[ ]v n

1z
http://find/


15/98

D1

z

[ ]x n [ ]y n[ ]v n

1z

1z

[ ]x n [ ]y n

1z

do not need this

1z

[ ]x n

[ ]n

1z

+

The node labeled [ ]

has been removed

v n

Rearranged direct form I,

i.e., a derivative from

direct form I
http://find/


16/98

D

1z

[ ]n

[ ]y n

1z

+

[ ]q n

[ 1]x n

( )Q z

1

( )qY z

Yq(z) =Q(z) z1Yq(z) Hq(z) = Yq(z)Q(z)

= 1

1 +z1
http://find/


17/98

D

1z

[ ]n

[ ]y n

1z

+

[ ]v n[ ]n

1

( )X z

( )Y z

1z

+

( )z

out[ ]n in [ ]n

( )
http://find/


18/98

DH(z) =

(z)

X(z) =

1 z21 +z1

=(1 z1)(1 +z1)

1 +z1

2 2 2 (2 )
http://find/


19/98

D

H(ej)2 = 22 cos(2)1+2+2 cos

0 0.1 0.2 0.3 0.4 0.50

0.5

1

1.5

2

2.5

3

3.5

4

frequency in cycles per sample,i.e.,/(2)

H(ej)

2

= 0.99 = 0.99

= 0.7

= 0

= 0.7
http://find/


20/98


21/98

D

[n]

x[n]

[n]

[n]

= 2x[n]

x[n] [n] x[n] [n] [n]

x[n]

x[n]

v[n] x[n 1]

x[n]

[n]

v[n]

[n] v[n]
http://goforward/http://find/http://goback/


22/98

D

( )X z 1z

( )20H z ( )2

1H z

( )Y z

( )20H z

( )21H z

1z

( )Y z( )X z

z2

z2

z1

j
http://find/


23/98

D0 (e )jH

2 1

(e )jH

( )(e )

jH

+

Hej2 + Hej()2 = 4
http://find/


24/98

D

( )

2

0

z

( )21H z

1z

( )Y z( )X z .5

Hej2 + Hej()2 = 1

( )jH
http://find/


25/98

D0 (e )jH

2

0.5

0.707

1

0p

0s

H

ej

2

= 1 H

ej()

2

H

ej

2 1

Hej 2 2 = 12

= 2

2p2

2s(2+)2
http://find/


26/98

D

p(2 )

2 +

60

= 20 log10s

s = 10

6020 = 103 = 0.001

p = 1

1 2s = 1

1 106 = 5.0000012 107

p =2s

2 =

106

2 = 5.0000000 107

12

50000012 =

2.4

105%
http://find/


27/98

D

0

(e )j

H

2

0.5

0.707

1

0p

0s

2 s

+ 2

s

2s

stop band template

p s p = 2s2
http://find/


28/98

D

s

s

s

s
http://find/


29/98

D12

3

gain 1 at=
http://find/


30/98

D0(e )jH

2

0.5

0.707

1

1

template

3 2

1

2

3

does not meet SPEC

1 2 3

g

, ,

and at

=

s
http://find/


31/98

D

H0(z) = 0+z

1

1 +0z1, H1(z) = 1

1

1z

( )Y z( )X z 0.5

( )20 z

2

0

2

01

z

z

+

+

( )21 z

H(z) =

0+z

2

1 +0z2+z1

0.5 =

z(1 +0z

2) + (0+z2)

(0+z2)z

0.5
http://find/


32/98

DH(z) =

0+z+z2 +0z

3

(0+z2)z 0.5

rej 1r

ej rej 1r

ej

0 0


33/98

D z1 1, z2 r, z3 z2 r 1 r 1

0= 1

1

r

1

r

1r=1

0 13

r=1

z1= 1, z2 = ej, z3=z2 = ej

0=

1

1 2cos , 0

=, 0=1

3

3

,1

3

0 3

0, 0 1

Plot of0= 1/(1 2cos)
http://find/


34/98

D0 0.2 0.4 0.6 0.8 1

2

1.5

1

0.5

0

0.5

1

1.5

2

/

0

0 /( )

This is the region of interest
http://find/


35/98

D

s() = 0.01

z = ej z= ej(8)

0

= 78
http://find/


36/98

D8

0 0.2 0.4 0.6 0.8 160

50

40

30

20

10

0

10

frequency normalized to radians per sample, i.e., /

20log10

H

ej

Trial magnitude response for zeros on the unit circle at = 0.875

= 0.818
http://find/


37/98

D0 0.2 0.4 0.6 0.8 1

60

50

40

30

20

10

0

10


20log10

H

ej


s 0.791

stop band
http://find/


38/98

D0 0.1 0.2 0.3 0.4 0.5

3.5

3

2.5

2

1.5

1

0.5

0


20log1

0

H

e

j


0.405 0.435 0.5
http://find/


39/98

D0 0.1 0.2 0.3 0.4 0.5

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0x 10

3


20log10

H

ej

Magnitude response for zeros at = 0.818

20log10(1 p)

2

= 0.182

2 (0 01)2
http://find/


40/98

Dp =

2s2

=(0.01)2

2 = 5.0 105

20 log(1 p) = 0.42 103

1 p = 100.42103

20 = 102.1105

p = 1 102.1105 = 4.8 105.
http://find/


41/98

D0() = H0(z

2)z=ej1() = z

1H1(z2)

z=ej

H(ej) = 0.5 ej0()

H0(ej2)+ ej1() ejH1(ej2)

H(ej) = 0.5ej0()+1()

2

ej

1()0()2 + ej

1()0()2

H(ej

) = ej0()+1()

2 cos1() 0()2 H(ej)

=

cos

1() 0()

2

H(ej) = 0() +1()

2

0() 1()
http://find/


42/98

D|1() 0()| 2s( )

0() 1()

() =1() s( )

() =0

()

s

(

)

H1(z) H1(z) =z

M H(z)

() = k

z1H1(z2)

s( )

H0(z)
http://find/


43/98

D0() 1() (2k+ 1) 2s()

H(ej) =1()

k+12

s()

H(ej) =0() +

k+

1

2

s()

0

Phase response for zeros on the unit circle at = 0.818
http://find/


44/98

D0 0.2 0.4 0.6 0.8 1

4.5

4

3.5

3

2.5

2

1.5

1

0.5

0


p

hasenormalized

to,i.e.,

Hej

/

zero inH(z) atz = ej0.818

H(ej) 1H ( 2)

( )
http://find/


45/98

Dz1H1(z

2)z=ej

=1() =

0 0.05 0.1 0.15 0.2 0.250.02

0.015

0.01

0.005

0

0.005

0.01

0.015

0.02

frequency normalized to radians persample, i.e.,/

deviationin

phasefrom

1

()=

(unitsradian

s)

Illustration of being nearly linear phase in the pass band

s = 40dB, which is 0.01

natural pass band corner frequency

D
http://find/


46/98

D

N

Kp

H0(z)

H1(z)

p s

2 p= 0.4

(0.4)2

Npaths Kp p

D
http://find/


47/98

D

D

Impulse response of 2-path filter
http://find/


48/98

D0 20 40 60 80 1000.2

0

0.2

0.4

samples

amplitude

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5100

80

60

40

20

0

frequency in cycles per sample, i.e., /(2)

20lo

g10

H

e

j

Frequency response of 2-path filter

D

Roots of 2-path, 4-coefficient filter
http://find/


49/98

D1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1

1

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

D

0

Phase profiles of 2-path filter
http://find/


50/98

D0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

4

3.5

3

2.5

2

1.5

1

0.5


Normalizedph

ase,i.e.,()/(2

)

phase ofH(z)

phase ofH0(z)

phase ofz1H1(z2)

D
http://find/


51/98

D

HLP(z) = H0(z2

) +z1

H1(z2

)2

HHP(z) = H0(z

2) z1H1(z2)2

|HLP(z)|2 + |HHP(z)|2 = H0(z2)2

+ z12H1(z2)22

H0(z2)

H1(z2)

HLP(ej)2 + HHP(ej)2 = 1 HLP(z) HHP(z)

D
http://find/


52/98

D2M

z

( )2 z

( )Y z( ) z

D


53/98

D

2( 1)M

( )21H z

1z

( )Y z( )X z1

z

( )20this becomes H z

( )21this becomes H z

D

( ) 2 z z

=
http://find/


54/98

D4

( )2 z

( )Y z( )X z

H(z) = 1

z4 +1

z1 +1z2

1+z21 +3z2

3+z2

0.35 2 1 3

D
http://find/


55/98

D1

3

0.56245

0.11285

=

=

zeros on unit circle

frequency in cycles/sample

(

)

10

20log

ej

H

D

4

Pole-zero diagram
http://find/


56/98

D4 2 0 2 4

4

3

2

1

0

1

2

3

rogue zeros

These zeros force high attenuation

in the stop band and therebyforce a nearly flat pass band

D

0

Phase response
http://find/


57/98

D0 0.1 0.2 0.3 0.4 0.5

15

10

5


H

ej

(radians)

phase response for z4, i.e.,H0 ej2

2

2

4

4.5

D

1 2
http://find/


58/98

DH(z) =

c1+c0z1 +z2

1 +c0z1 +c1z2

H(z)

1z

1

[ ]n [ ]n1

c

1

z

1z

1

z

0c

0c

1c

+

D

[ ]x n [ ]y n1c

+
http://find/


59/98

D1

z

1

1z

1

1z

0c

0c

1c

+

1z

[ ]x n

[ ]y n

1z

+

1c

0c+

1

1

D

H(z) = c1+c0z

1 +z2

1 1 2
http://find/


60/98

D( )

1 +c0z1 +c1z2

er

1e

j

r

e jr

1e j

r

D
http://find/


61/98

D

D
http://find/


62/98

D

D
http://find/


63/98

D

D
http://find/


64/98

Dzero #1

frequency in cycles/sample

(

)

10

20log

ej

H

change angle (really frequency)

by this amount (unitscycles/sample)

number of zeros

D
http://find/


65/98

D

c0 c1

D

2.5

Pole-zero diagram
http://find/


66/98

D2 1 0 1 2

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

D

0

Phase response
http://find/


67/98

D0 0.1 0.2 0.3 0.4 0.5

25

20

15

10

5


H

ej

(radians)

6.5

D

1

Impulse response


68/98

D0 20 40 60 80 100

0.4

0.2

0

0.2

0.4

0.6

0.8

Samples

Amplitude

D

HG(z) = H

1

( )

, g(z)
http://find/


69/98

DHG(z) H

g(z)

, g(z)

1z

T

1z

+

1z

[ ]n [ ]n0b

1

1b

2b

[ ]x n [ ]y n0b

1b

2b

1z

T

1z

+

GH z H z

D

HG(z)

g

ej

= ej()
http://find/


70/98

Dg

HG ej=Hej () = Hej ()

() = ()

HG ej Hej

()>

()<

()

=

()

= g z1

D

ld

warp 0 0( ) ( ) =
http://find/


71/98

Dused in (e )j

GH

0

2

warp( ) ( ) =

2

0

olds

news

oldz

newz

1

2

location of the zero in (e )j

GH

location of a zero in (e )jH

D

(e )jH
http://find/


72/98

D0 2

oldz

0

(e )jGH

( )old warp news s =

oldnew

2

ss

= newz

D

0.45

0.5

0.95

0 90

0.95

0 90

0.95

0 90

0.95

0 90

0.95

0 90

0.95

0 90

0.95

0 90

0.95

0 90

0.95

0 90

0.95

0 90

0.95

0 90cycles)

Phase response curves for first-order APF with as indicated
http://find/


73/98

D0 0.1 0.2 0.3 0.4 0.5

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

frequency in cycles per sample, i.e., /(2) -fornew filter

Negativeof

phaseanglenormalizedto2,i.e.,

()

2

(unitsc

D

filter


74/98

D( )(e ) (e )s sj j

GH H

=3dB of (e )

2

j

GH

(

)

2

s

3dB

(

)

2

3dB

3dB

(

)

3dB

1

(e

)

(e

)

(

)

0.2

5

2

forh

alf-band

2

j

j

G

H

H

=

=

=

2

s

D


75/98

Dg(z) =

0.3 +z1

1 0.3z1

HG(z)

D
http://find/


76/98

D3dB 2 0.157 =

1dB 2 0.147 =

,1 2 0.222z =

,2 2 0.291z =

2 0.214s

=

0.2

5

D
http://find/


77/98

D

D
http://find/


78/98

D3dB

2 0.157 =

1dB 2 0.147 =

,1 2 0.222

z =

,2 2 0.291

z =

2 0.214s

=

D

T

2 0.4
http://find/


79/98

D

0 0.1 0.2 0.3 0.4 0.50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

0.00

0.95

0.90

0.80

0.60

0.30

0.30

0.60

0.80

0.90

0.95

frequency in cycles per sample, i.e., /(2) -for newfilter

Negativeofphas

eanglenormalizedto2,

i.e.,

()

2

(unitscycles)

Phase response curves for first-order APF with as indicated

(3dB)2

= 0.25

3dB

9 mm

4 mm

D

c
http://find/


80/98

D c

HG

ejc

= H

ejc

= H

ej(c)

.

T =sin cc2 sin

c+c2

c = 2

T =sin

2 4

sin

2 + 4

=tan

2

1tan

2

+ 1

D

T = 0.5 s= 0.4
http://find/


81/98

Ds

s = 0.5 + ej0.4

1 0.5ej0.4 = 0.363 2

H

ej

=Hej ejM

HG

ej

= H

ej()

=H

ej()

ejM()

D

H(z) =HAP(z2) HAP(z) =

0+z1

1+0z1
http://find/


82/98

DHG(z) = H(z)z= 1

g(z)=HAP(z

2

)z= 1g(z)

=

c2+c1z1 +z2

1 +c1z1 +c2z2

c1 =2T(0+ 1)

1 +02T, c2=

0+2T

1 +02T

1z

( )X z

( )Y z

1z

+

2c

1c+

1

z

1z

D

T

[ ]x n

1

+

1z

1
http://find/


83/98

D0

[ ]x n

1z

+

1z 1

z

1z

[ ]y n

1z

T

1z

+

T

1z

1z

T

1z

1z

0+

[ ]y n

1z

[ ]x n

[ ]y n1

z

+

2c

1c

+

1

z

1z

1z

01 2

0

2

02 2

0

2 ( 1)

1

1

T

T

T

T

c

c

+=

+

+=

+
http://find/


84/98

D

(e )jBP

H


85/98

D0 20.707

1

2

3

2

4

3

4

bndwidth2

=

D

(e )jLPH

1

Response of prototype lowpass filter
http://find/


86/98

D0 20.5

0.707

2

2

frequency for prototype response

0 2

( )

2

3

2

frequencyfortransformedfilter

4

2

34 2

1

( ) 2 = +

2Plot of (e ) versus for ( )j

g g z z =

2 ( 2 )(e ) e =e (e ) 2j j j j

g g += = +

35

2

35

2

D

(e )jLPH

0.707

1
http://find/


87/98

D0 2

0.5

2

3

2

35

2

0

(e )j

BPH

0.5

0.707

1

2

4

3

4

D

c+z1

1+cz1 z1

( ) 1c + z

1
http://find/


88/98

Dg(z) = z1 c+z

1 +cz1

0 2

2

0c

=

2Note that 0 has ( )c

g z z = =

3

(e ) ( )jg =

0.4c

0.4c

for

0.4

c

c

for

0.4

c

c

D

H(z) c HG(z) c

ejc = g(ejc) ejc = ejc c+ ejc
http://find/


89/98

De =g(e ) e = e

1 +cejc

c = 0 c = cos cc

c

These distances are not equal

0

(e )jGH

2

c

3

4

0.707

1

D

1.5

c = 0.4

scycles)

Negative phase response curves forz1 c+z11+cz1

with c indicated

negative phase of z1
http://find/


90/98

D0 0.5 1 1.5

0

0.5

1

c = 0.4

c = 0.4

c = 0.4

c2

=cos1(c)

2

negative phase of z1 c+ z1

1 + cz1

negative phase of c+ z

1

1 + cz1

frequency in cycles per sample, i.e., /(2) fornew filter

Negativeofp

haseanglenormalize

dto2,i.e.,

()

2

(units negative phase of z

D
http://find/


91/98

D

HGLP(z) =H

1

g(z)

, g(z) =

T+z1

1 +Tz1

i+z2

1+iz2

c2+c1z1+z21+c1z1+c2z2

c1 =2T(i+ 1)

1 +i2T

c2 = i+

2T

1 +i2T

z1 H1(z) T+z

1

1+Tz1

D

HGLP(z)

HG BP (z) = HG LP

1

g(z) = z1 c+z

1
http://find/


92/98

DHGBP(z) =HGLP

g(z)

, g(z) = z

1 +cz1

i+z2

1 +iz2

d4+d3z1 +d2z

2 +d1z3 +z4

1 +d1z1 +d2z2 +d3z3 +d4z4 z1

T+z1

1 +iz1

z1=z1 c+z11+cz1

= e2+e1z1 +z2

1 +e1z1 +e2z2

D

1z

( )X z

+

4d

d+

1z 1 1z
http://find/


93/98

D ( )Y z1z3d+

1

+

+

1 1

z

d

1d

1

( )X z

( )Y z

1

+

+

2e

1e

1z

1z

++

+

D


94/98

D

D

Lowpass filter


95/98

Dpolynomial order

0.137

2

s

=

D

c1 c2
http://find/


96/98

D

c1 c2

z1

D

Bandpass filter
http://find/


97/98

D

D

d1 d2 d3 d4
http://find/


98/98

D

d1 d2 d3 d4

e1 e2
http://find/

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