document4

1Digital Signal Processing A.S.Kayhan

DIGITAL SIGNAL

PROCESSING

Part 3

Digital Signal Processing A.S.Kayhan

IIR Filters:Infinite impulse response (IIR) filters have rational transfer functions as

N

k

kk

M

k

kk

za

zbzH

0

0)(

Some IIR filter types are: Butterworth, Chebyshev, Elliptical. IIR filters have to be stable. IIR filters may not have linear-phase.


Filter Specifications (Low-pass):

)(log20 10

Design analog filter transform to Digital filter


Analog Butterworth Filters:Approximate ideal LPF with magnitude-squared response

c

cKH,0

,)(

2

by the following rational function:

0,11

)(22

1

221

212

nn

n

nn b

bb

aaKH

|H()|2 must be even function of , and denominator degree must be higher than degree of numerator (lowpass).


.1,,2,1, niba ii

.1)( 22

nnb

KH

Then, we have

For maximal flatness at the origin, =0, The first 2n-1 derivatives of |H()|2 must be zero. This requires

Maximal flatness also at , =, requires

0,11

)(22

1

221

212

nn

n

nn b

bb

bbKH

.1,,2,1,0 niba ii


.

1

1)( 2

2

n

o

H

Then, we have the Butterworth response

Let |H(=0)|2 = 1. Then

At the half-power frequency

n

onn

on

bb

H

2

2

2 1

1

1

2

1)(

.11

)(2

2

nnb

H


Consider the filter specifications:

.dB)(log20 10 H

.dB1log10

2

10

n

o

n

o

2

10/ 110 no 2/110/ 110

Using max and pn

o

p

2

10/ 110 max

Using min and sn

o

s

2

10/ 110 min

To find n:


Dividing these equations

n

p

s

2

10/

10/

110

110max

min

Taking logarithm, we get the filter order as:

p

s

nlog2

110

110log 10/

10/

max

min


then, set the denominator to zero

We can normalize the frequency so that o = 1. Then

To find the filter poles, we let =s/j,

nnn sjssHsH

22 )1(1

1

)/(1

1)()(

.11

)(2

2

nH

1)1(0)1(1 22 nnnn ss

If n is even, then

.12,...,1,0,

.12,...,1,0,1

)2

21(

)2(2

nkes

nkes

n

kj

k

kjn


If n is odd, then

.12,...,1,0,

.12,...,1,0,1 22

nkes

nkes

n

kj

k

kjn

Note that all the poles lie on a circle with radius 1, because the frequency is normalized. (If we want to design analog filter we need to correct this). Also, choose the poles in the left-half plane to get a stable filter.


Example: Design specs. for a Butterworth LP filter:At f = 2000 Hz, max= 3dBAt f = 3000 Hz, min= 10dBThen, the filter order n isn 2.7154 n = 3.Poles are at

.5,...,1,0,3 kesk

j

k

.

))23

21

())(23

21

()(1(

1

)1)(1(

1)(

2

jsjss

ssssH


Analog Chebyshev Filters:We can generalize Butterworth response as:

.)(11

)(1

1)(

22

2

nn F

H

Fn() is a function of . Now, consider:

))(coscos()cos(

)(cos)cos(1

1

xnny

xx

n

This is the Chebyshev polynomial of order n. When

))(coshcosh()(,1 1 xnxCx n

When x 1 , Cn(x) 0.


Cn(x) has zeros in -1< x < 1 :

-1 0 1-1

0

1

x

C1(x)

-1 0 1-1

0

1

x

C2(x)

-1 0 1-1

0

1

x

C3(x)

-1 0 1-1

0

1

x

C4(x)

-1 0 1-1

0

1

x

C5(x)

.)(,1)(

),()(2)(

1

11

xxCxC

xCxCxxC

o

nnn


Use Cn(.) in :

)(1

1)(

22

2

nCH

))(coscos()(,1

))(coshcosh()(,11

1

nC

nC

n

n

2 1 , is known as the ripple factor.


Behaviour at =0:

even isn if,)1(

1)0(,1)(

odd isn if,1)0(,0)(

2

22

22

HC

HC

n

n

Behaviour at =1:

)1(

1)1(

n allfor ,1)(

2

2

2

H

C n


110)(1log10 10max/22 nCThe attenuation is

When .dB01.31)(22 nC

This defines the half-power frequency (3dB cut-off) hp.Then,

).1()).1

(cosh1

cosh(

)1

(cosh1

)(cosh

))(coshcosh(1

)(

1

11

1

hphp

hp

hphpn

n

n

nC


The specifications for a Chebyshev filter are max, min, s(p=1). Bandpass is between 0 1 rad/s.To find n:

)(110

)(1log102210/

22min

minsn

sn

C

C

With 110 10max/2

110

110))(coshcosh(

10/

10/1

max

min

sn

Finally,

)(cosh

110110cosh

1

10/

10/1

max

min

s

n


To find the filter poles, we let =s/j,

)/(1

1)()(

22 jsCsHsH

n

then

1

)/(0)/(1 22 jjsCjsC nn

Letjsjvuw /)cos()cos(

wjs )/(cos 1

With

)sinh()sin()cosh()cos()/(

)cos())/(coscos()/( 1

nvnujnvnujsC

nwjsnjsC

n

n

)cosh(2

)cos(,2

)cos( xee

jxee

xxxjxjx

10


With

1

)/( jjsC n then

1

)sinh()sin(

0)cosh()cos(

nvnu

nvnu

cosh(nv) can never be zero, therefore cos(nu) must be zero. It is possible for

.12,,1,0),12(2

,2

5,

2

3,

2

nkkn

u

nnnu

k

k


For these values of u, sin(nu) = 1, then

.)1

(sinh1 1 an

v k

Remember that

jvuwjs )/(cos 1then

jakn

jwjs kk 122cos)cos(

The poles are :12,,1,0, nkjs kkk

)cosh()2

12cos(

)sinh()2

12sin(

an

k

an

k

k

k

Choose left half poles.

11


Discrete Time Filter Design(IIR):There are 3 approches:1-Sampling (Impulse invariance)2-Bilinear transformation3-Optimal procedures

Impulse invariance method:We can sample the impulse response hc(t) of an analog filter with desired specifications as (Td is sampling interval):

.][ dcd nThTnh then

).2

()( kTT

HHdk d

c


If dc TH /,0)( then

.),()(

d

c THH

Analog and digital frequencies have a linear relation: . dT

Consider the transfer function of a system expressed as

,)(1

N

k k

kc ss

AsH

Then, the impulse response is

.0,)(1

teAthN

k

tskc

k

12


The impulse response of the discrete time filter is

.)(

)(

1

1

nueATnh

nueATnThTnh

N

k

nTskd

N

k

nTskddcd

dk

dk

Transfer (or system) function of the discrete time filter is

.1

)(1

1

N

kTs

kd

ze

ATzH

dk

Poles are .dk Tsk ezss

If Hc(s) is stable (k < 0), then H(z) is also stable (|z| < 1).


Example: Transfer function of analog filter is

.

)23

21

(

12

121

)23

21

(

12

121

1

1

)1)(1(

1)(

2

js

j

js

j

s

ssssH c

then

.

1

)12

121

(

1

)12

121

(

1)(

)2

3

2

1(1)2

3

2

1(1

1

jT

d

jT

d

Td

dd

d

ez

jT

ez

jT

ez

TzH

13


Bilinear Transformation:Transformation needed to convert an analog filter to a discrete time filter must have following properties:1- j axis of the s-plane must be mapped onto the unit circle of the z-plane,2- stable analog filters must be tranformed into stable discrete time filters (left hand plane of the s-plane must be mapped into inside the unit circle of the z-plane).Following BT satisfies these conditions:

./1

/1

1

11

1

Ks

Ksz

z

zKs


Let jrezjs ,

then

.)/()/1(

)/()/1(22

222

KK

KKr

Therefore

)u.c. theinside(1)planeLH(0 r

)u.c. the(1)axis (0 rj

)u.c. theoutside(1)planeRH(0 r

14



Now, let jezjs ,

then.

)(

)(

1

12/2/2/

2/2/2/

jjj

jjj

j

j

eee

eeeK

e

eKj

).2/tan()2/cos(

)2/sin(

jKKjj

or).2/tan( K

15


Observations:1- According to Taylor series expansion of tan(), for small

.2/24/2/ 3 KK 2- For high frequencies, the relation is nonlinear causing a distortion called warping effect. Therefore, BT is usually used for the design of LPF to avoid this.


Discrete Butterworth Filter Design:

We convert a frequency normalized analog Butterworth filterto a discrete filter using the BT.

Normalized half power frequecy HP=1 is mapped into the discrete half power frequency HP. To do that, we set K to

).2/cot()2/tan(/)1( HPHPHPbK

16


Then, we use ).5.0tan(/)5.0tan()5.0tan( HPbK

in nnH 2

2

1

1)(

And get the discrete Butterworth Low Pass Filter as:

n

HP

nH 22

)5.0tan()5.0tan(

1

1)(

Using the design specifications, we find the filter order as

)5.0tan()5.0tan(log2

110110log 10/

10/

max

min

p

s

n


The half-power freq. is

npHP 2/110/

1

110

)5.0tan(tan2

max

).5.0cot( HPbK

Example: Consider the second order analog filter

)12(

1)(

22

sssH

Applying the BT, we get (Kb = 1)

2929.0)1716.0(

)1()(

2

2

2

z

zzH

17


Remarks:1- Given the specs.:

a) find the filter order nb) obtain the analog filter H(s)(using LHP poles)c) calculate HP and Kbd) use biliear transformation to find H(z).

2- H(z) is BIBO stable, because H(s) is stable.3- Applying the BT to high order filters may be tedious. Therefore, first express H(s) as product or some of first and second order functions, then apply the BT.


Example: Design specs. for a LP digital Butterworth filter:At f = 0 Hz, 1= 18dBAt f = 2250 Hz, 2= 21dBAt f = 2500 Hz, 3= 27dBSampling freq. fsmp= 9000 Hz(ej0)= 18dBmax = 2 1=3dBmin = 3 1=9dB smp

ss

smp

pp f

f

f

f 2,2

Filter order n 5.52,n = 6, Kb = 1 and a gain G for = 18dB

)02.0.0)(17.0)(59.0(

)1(0037.0)(

222

6

6

zzz

zzH

18


Discrete Chebyshev Filter Design:

We find the constant K by transforming (normalized passband freq.) P=1 into the discrete passband frequency P. We get

)5.0tan(/)5.0tan()5.0tan(/1 PPcK

And get the discrete Chebyshev Low Pass Filter as:

))5.0tan(/)5.0(tan(1

1)(

22

2

pnCH


Using C1(1)=1, we let max, p

110

1log10

10/

2max

min

Also using min, sWe find the filter order

)]5.0tan(/)5.0[tan(cosh

110110cosh

1

10/

10/1

max

min

ps

n

19


Example: Design specs. for a LP digital Chebyshev filter:At f = 0 Hz, 1= 0dBAt f = 2250 Hz, max= 3dBAt f = 2500 Hz, min= 10dBSampling freq. fsmp= 9000 Hz

Filter order n = 3, Kc = 1

)54.0)(72.015.0(

)1(09.0)(

2

3

3

zzz

zzH

smp

ss

smp

pp f

f

f

f 2,2


Example: Design specs. for a LP digital Chebyshev filter:At = 2/5 rad, max= 1dBAt = /2 rad, min= 9dB

Filter order n = 3

)472.0)(619.0505.0(

)1(736.0)(

2

3

3

zzz

zzH

Example: Design specs. for a LP digital Butterworth filter:At = /2 rad, max= 3dB = 0 rad, = 0dBAt = 5/9 rad, min= 10dB

Filter order n = 7, Kb = 1

)052.0)(232.0)(636.0(

)1(01656.0)(

222

7

7

zzzz

zzH

20


Frequency Transformations:

The objective is to obtain transfer functions of other types of dicrete filters from already available prototype Low Pass Filters using tranformation functions g(z) as

))(()( zgHzH LP

Example: Design specs. for a LP digital Chebyshev filter:At f = 0 Hz, = 0dBAt f = 2250 Hz, 2= 21dBAt f = 2500 Hz, 3= 27dBSampling freq. fsmp= 9000 Hz


39.0802.0691.0

)1(09.0)(

23

3

zzz

zzH LPF

21


Now, we want a HPF with cutt-off at 3.6kHz, we use following transformation

)5095.01(

)5095.0(1

11

z

zz

6884.0102.2361.2

)133(0066.0)(

23

23

zzz

zzzzH HPF


Design of FIR Filters by Windowing:

Ideal frequency response and corresponding impulse response functions are .),( nheH djd

Generally hd[n] is infinitely long. To obtain a causal practical FIR filter, we can truncate it as

otherwise.,0

0, Mnnhnh d

In a general form we can write it as nwnhnh d

Where w[n] is window function.

22


In frequency domain, we have

.)()(2

1)(

deWeHeH jjdj


Some commonly used window functions are Rectangular, Bartlett, Hamming, Hanning, Blackman, Kaiser.

Some important factors in choosing windows are: Main lobe width and Peak side lobe level. For rectangular:

23


Main lobe width and transition region; Peak side lobe level and oscillations of filter are related.


24



25


Discrete Time Fourier Series (DTFS):

Consider a periodic sequence with period N:

.][11

0

2~~

N

k

nkN

jekX

Nnx

Which can be represented by a Fourier series as:

.][21

0

~~ nkN

jN

n

enxkX

The DTFS coefficients are obtained as:

.~~

Nnxnx

nx~


Let NjN eW

2

The DTFS analysis and synthesis equations are:

.][1

0

~~nk

N

N

n

WnxkX

.][11

0

~~

N

k

nkNWkXN

nx

Example: Consider the periodic impulse train:

r

rNnnx ][~

The DTFS

. allfor 1][1

0

~

kWnkX nkNN

n

26


Similarly, ][~~

lkXnxW nlN

Properties: 1- Linearity: It is linear.2- Shift of a sequence: If ][

~~

kXnx

then ][~~

kXWmnx kmN

3- Periodic convolution: Consider two periodic sequences with period N

][~

1

~

1 kXnx ][~

2

~

2 kXnx

Then,

][][][~

2

~

1

~

3 kXkXkX .][][1

0

~

2

~

13

~

N

m

mxmnxnx


Example: Consider periodic convolution of two sequences:

27


Sampling the Fourier Transform:

Consider signal x[n] with DTFT X() and assume by sampling X() , we get

kN

XXkXk

N

2|)(][ 2

~

][~

kX could be the sequence of DTFS coefficients of a periodic signal which may be obtained as

.][11

0

~~

N

k

nkNWkXN

nx


Substituting,

k

Nm

mj XkXemxX 2

~

|)(,

.*~

rr

rNnxrNnnxnx then,

28


If N is too small, then aliasing occurs in the time domain.

If there is no aliasing, we can recover x[n].


Discrete Fourier Tranform (DFT):

DFT is obtained by taking samples of the DTFT. Remember DTFT is defined as:

n

njj enxeX ][)(

We take samples of X(ej) at uniform intervals as:

.1,1,0,)(][ 2

NkeXkXk

N

j

We define: N

j

N eW2

29


Then the DFT is defined as (analysis equation):

N

n

knNWnxkX

0

][][

for k=0,1,...,N-1. The inverse DFT is defined as (synthesis equation):

N

n

knNWkXN

nx0

][1

][

for n=0,1,...,N-1.


Example:N=5

30


N=10


Properties:1- Linearity: DFT is a linear operation.2- Circular shift:

].[10,2

kXeNnmnxkm

Nj

N

31


3- Circular convolution:

kXkXkX

mxmnxnxnxnxN

mN

213

1

021213 ][][

Example:


Example: N=L

32


N=2L


4- Multiplication(Modulation):

.213213 kXkXkXnxnxnx

Linear Convolution Using the DFT:

Since there are efficient algorithms to take DFT, like FFT, we can use it instead of direct convolution as:

1. Compute N point DFTs, 2. Multiply them to get 3. Compute the inverse DFT, to get:

. and 21 kXkX .213 kXkXkX

nxnxnx 213

33


Linear Convolution of Finite Length Signals:Consider two sequences x1[n] of length L and x2[n] of

length P, and x3[n] = x1[n]* x2[n].Observe that x3[n] will be of length (L+P-1).

Circular Convolution as Linear Convolution with Aliasing :

Consider again x1[n] of length L and x2[n] of length P, and x3[n] = x1[n]* x2[n].

)()()( 213 jjj eXeXeX


Taking N samples .213 kXkXkX

Now, taking the inverse DFT, we have

otherwise.,0

10,33

NnrNnxnx

rp

nxnxnx p 213

This circular convolution is identical to the linear convolution corresponding to X1(ej) X2(ej), if N (L+P-1).

34


Example:


Example:

35



LTI Systems Using the DFT :Consider x[n] of length L and h[n] of length P, and y[n] =

x[n]* h[n] will be of length (L+P-1).To obtain the result of linear convolution using the DFT,

we must use N ( (L+P-1))-point DFTs. For that x[n] and h[n] must be augmented with zeros (zero-padding).

Overlap-Add Method :Consider h[n] of length P, and length of x[n] is much

grater than P.We can write x[n] as combination of length L segments:

,otherwise,0

10,

LnrLnx

nx r

r

r rLnxnx

36


Since the system is LTI then

nhnxnyrLnyny rrr

r *,

Each output segment yr[n] can be obtained using (L+P-1) point DFT.

Example:


37


Overlap-Save Method :Consider again h[n] of length P, and length of x[n] is much

grater than P.In this method L-point circular convolution (or DFT) is

used. Since the first (P-1) points will be incorrect, input segments must overlap.

We can write each L sample segment of x[n] as :

10,)1()1( LnPPLrnxnx r

11,

,)1()1(

LnPnyny

PPLrnyny

rpr

rr


Example:

38


Efficient Computation of the DFT :Remember the definition of the DFT and inverse DFT

1

0

][][N

n

knNWkXnx

1

0

][][N

n

knNWnxkX

N2 comlex multiplications and N(N-1) additions or 4N2 real mult. and 4N(N-2) real additions necessary:

N

n

knN

knN

knN

knN

Wnx

Wnxj

Wnx

Wnx

kX0

}Re{}][Im{

}Im{}][Re{

}Im{}][Im{

}Re{}][Re{

][


To improve the efficiency, we can use some properties as:

nNkN

knN

nNkN

knN

knN

nNkN

WWW

WWW)()(

*)(

)2

)()1

}.Re{}][Re{}][Re{}Re{}][Re{}Re{}][Re{ )(

knN

nNkN

knN

WnNxnx

WnNxWnx

Example:

39


Fast Fourier Transform(FFT):Fast Fourier Transform algorithms are used to implement DFT efficiently to reduce the number of multiplication and addition operations.Two main algorithms are:Decimation in time and Decimation in frequency. Both these algorithms require that N=2m.The number of operations using either of these will require (N/2)log2N complex multiplications and Nlog2Nadditions.Direct implementation of DFT requires N2 complex multiplications and additions. For N=1024= 210

N2=1048576 but Nlog2N=10240.


Decimation in time FFT Algorithm:Assume N=2m

1,,1,0,][][1

0

NkWnxkXN

n

knN

Which can be written as

oddn

knN

evenn

knN WnxWnxkX ][][][

With n = 2r (even) and n = 2r+1 (odd)

.

]12[)(]2[

)(]12[)(]2[][

12/

02/

12/

02/

12/

0

212/

0

2

kHWkG

WrxWWrx

WrxWWrxkX

kN

N

rN

kN

N

rN

N

r

rkN

kN

N

r

rkN

rkrk

40


.][ kHWkGkX kN

Requires N+2(N/2)2 multiplications


Since G[k] and H[k] requires 2(m-1)-point DFTs, each one can be similarly decomposed until we reach 2-point DFTs.

41



Butterfly:(in place comp.)

Bit reversed order

42


Decimation in frequency FFT Algorithm:Again assume N=2m

1,,1,0,][][1

0

NkWnxkXN

n

knN

then)12/(,,1,0,][]2[

1

0

2

NrWnxrXN

n

rnN

1

2/

212/

0

2 ][][]2[N

Nn

rnN

N

n

rnN WnxWnxrX

12/

0

)2/(212/

0

2 ]2/[][]2[N

n

NnrN

N

n

rnN WNnxWnxrX

12/

0

12/

02/2/ ][])2/[][(]2[

N

n

N

n

rnN

rnN WngWNnxnxrX


Similarly for

.][

])2/[][(]12[

12/

02/

12/

02/

N

n

rnN

nN

N

n

rnN

nN

WWnh

WWNnxnxrX

)12/(0 Nr

43


Procedure is repeated until 2-point DFTs


Computation of Inverse DFT:

1,,1,0,][1

][0

NnWkXN

nxN

n

knN

1,,1,0,][][0

NkWnxkXN

n

knN

][DFT1][1][ *0

** kXN

WkXN

nxN

n

knN

** ][DFT1][ kXN

nx

44


End of Part 3

document4

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