EKT430/4DIGITAL SIGNAL

PROCESSING2007/2008

CHAPTER 7ANALOG FILTER

Analog Low-Pass Filters: a review

In this section, we shall take a review ofperformance of polynomial based analog low

pass filters, such as

Butterworth filter,Tchebyshev filter,Inverse Tchebyshev filter.Elliptical filters or Cauer filter, Bessel’s filter.

& transformability from LPF to lowpass, band-pass,

high-pass and band-reject configurations;with frequency scaling.

A preview to Low Pass Analog Filters

Low pass (LP) analog filters are all-pole filters.

The pass band extends from DC to usually -3dB or, half power frequency.A range beyond pass band is

[transit + stop band].Transit band lie between pass and stop-band.The roll-off rate in transit band is @ -nx20 dB/dec. Here n represents number of poles in excess of zeros.

A preview to Low Pass Analog Filters

HP and LP filter have one pass and one stop band.

Band-reject (BR) has 2 pass band and 1 stop band. Band-pass (BP) has 2 stop-bands and 1 pass band.An steeper transit-band is always associated with: (a) abrupt non-linear phase characteristic(b) abrupt peak at the end of pass-band.

Properties of Butterworth Filters

The pass-band is mathematically maximal-flat.All the poles lie on a left hand semi-circle in s-plane. They are equi-distant.Odd pole always lie on negative real axis. Remaining poles are complex-conjugate.The roll off rate of transit characteristic is @–20n dB/dec, n is the number of poles.

Properties of Butterworth Filters

Stop band response asymptotically goes to zero.Location of poles being known,

they are easy to design.follows the equation of a circle

k2 + k

2 = a2.

or k + jk = a

Coefficients of the Butterworth polynomials for various number of poles at c =1.

N ao a1 a2 a3 a4 a5 a6 a7 a8

1 1 1

2 1 1.4142 1

3 1 2 2 1

4 1 2.6131 3.4142 2.6131 1

5 1 3.2361 5.2361 5.2361 3.2361 1

6 1 3.8637 7.4641 9.1416 7.4641 3.8637 1

7 1 4.494 10.098 14.592 14.592 10.098 4.494 1

8 1 5.1258 13.137 21.846 25.688 21.864 13.137 5.1258 1

See the symmetry of coefficients

Poles, angles, Polynomial and Factor form

For N=3: Angles: [ 0, 60]D(s)=1+2s+2s2 +s3 = (1+s)(1+s+s2 )

For N=4: Angles: [22.5, 67.5]D(s) = 1+2.613s+3.414s 2+2.613 s3 +s4 = (1+0.76536s +s2)(1+1.84776s+s 2)

For N=5: Angles: [0, 36, 72]D(s) =1+3.236s+5.236s 2+ 5.236s3 + 3.236s4+ s5

= (1+s)(1+0.618s+s2) (1+1.618s+s2)

For N=6: Angles: [15, 45, 75]D(s)=1+3.864s+7.464s 2+9.141s3 + 7.464s4+3.864 s5+ s6

= (1+0.5176s+s2 ) (1+1.414s+s2 ) (1+1.9318s+s2 )

Pole location on ellipse from a circle:

Butterworth to Tchebyshev.

Major

axis

Minor axis

= (1/n) sinh-1 (1/)

Pass band ch. of 6th order Butterworth and Tchebyshev

0 2 4 6 8 10 120.7

0.75

0.8

0.85

0.9

0.95

1

1.05

1.1

mag

nitu

de in

dB

frequency in rad/sec

1

2

3

4

5

6

5

3.8

p

Tchebyshev Filters

is also an all pole filter. Poles lie on ellipse and follows following equation:

Here sinh and cosh represent the radius of minor and major axes respectively. Since is the angle with respect to y axis,

-sin + jcos are the Butterworth poles, yielding Tchebyshev_1 poles at pk = k + jk = -sin sinh + j cos cosh, while = (1/n) sinh-1(1/) [Ambarder p 415] Or (1/) = sinh (n)

k

sinh

2 k

cosh

2

1

Tchebyshev Filter contd…

For the same order of filter, the poles are obtainable from Butterworth circle by horizontally shifting them on to the ellipse having minor axis sinh and mazor axis cosh . In case of Tchebyshev, ripples are in pass-

band while stop band is flat. For an ‘n’ pole filter, there are n numbers of

maxima and minima of equal heights. Distances between maxima and minima

decrease toward “corner” frequencies. It is also called equi-ripple filter.

Tchebyshev filter contd……

For the normalized value of maxima =1, denoting r = (1+2) = 1/(1-p),

p is the peak to peak ripple factor the minima has the value 1/r resulting in:

A trade off has to be maintained between the pass band ripple and stop band gain:

since any decrease in the pass-band ripple, the gain at stop band increases resulting in poor stop-band

performance. These filters have superior transit-band characteristic.

H 0( ) 1 nif is odd

1

rnif is even

Calculations of and n from ripple…say -1.1dB

We know that20 log10(r )= 10 log (r2) = 1.1

thus r2= 1.288=1 + 2

2 = 0.288 hence 1/ = 1.8626

.n = sinh-1(1/ ) = 1.3804 Where n is the number of poles. Thus = 1.3804/n

Calculation of poles….

Let n = 3 = 1.3804/3 = 0.46. Hence: sinh(0.46)= 0.4764 & cosh(0.46)= 1.1077. Butterworth poles for n=3 are:

[-1, cos(/3) j sin(/3)] = [ -1 -0.5 j 0.866] Multiply real parts by ‘sinh’ and imaginary, by ‘cosh’ we

get: [-0.4764 -0.2382 j 0.9593] These are the locations of poles in Tchebyshev_1filter. Work out the denominator polynomial. Numerator is a

constant and can be calculated from H(0)=1. Soln. Hlp(s) =0.4656/(s+0.4766)(s2+0.4766s+0.977) Try the formulae pk = k + jk = -sin sinh + cos cosh.

Transformation of analog filter from LP LP; LP HP:

Low Pass to Low Pass: The transformation of cut-off frequency from 1rad/sec to p radians per second

can be carried out by the transformation: s (s/p)

Low Pass to High Pass: A low pass prototype can be converted

into a high pass filter of cut-off frequency c by following

transformation. s (c/s)

Transformation of analog filter.....LP BP; LP BR

Let B rad/sec be the –3dB bandwidth, and o rad/sec be the center frequency,

the transformation to operate on Low Pass filter

to get band-pass polynomial is:s (s2 + o

2)/ Bs

and to get band-stop polynomial is:s Bs/(s2 + o

2)

Inverse Tchebyshev Filter

This filter has flat pass band with ripple in stop band.

It’s TF can be derived from that of Tchebyshev filter ::

Step-1: Employ s(1/s) transformation to convert the low-

pass filter into a high pass filter; HT (1/s).

The ripples of HT (1/s) are in pass band of this high pass filter.

Inverse Tchebyshev Filter

Step-2: Subtract the so obtained high pass filter from HT(0).

Thus HIT (s) = [HT (0) – HT(1/s)].

Thus this filter has zeros in the transfer function.

All other characteristics of Tchebyshev filter are maintained.

Thus:The transit band characteristics are same,

Peak to Peak height of the stop-band ripple is decided by the minor to major axis ratio.

Elliptical or, Cauer filter.

It is an extension of Tchebyshev filter.

Here pass-band as well as stop-band

have ripples. Transit band characteristics are at best.

This filter has a zero

at a frequency close to stop band.

[details not to discuss]

BESSEL’S FILTER

All above filters care only for amplitude characteristics.

Their phase characteristics become poor as we increase the order of filter.

Philosophy of this filter is based on Bessel’s Polynomial.

This filter takes care of both amplitude and phase characteristics. (details not to discuss)

Butterworth low-pass polynomial

The Butterworth Polynomial for n pole filter is given by:

H(s)H(-s) = [1+(s/j)2n]-1

The poles lie at s2N = (-1)(j)2n

s2n = (-1)(j)2n Since j = ej/2 & hence j2n= ejn for

integer k.and -1 = ej(2k+1) for k0

S2n = ej(2k+1) ejn =ej(2k+n+1)

sk = e j(2k+n+1)/2n)

= cos (2k+n+1)/2n +j sin (2k+n+1)/2n for k 0.

Poles are separated by angle /N.

Butterworth filter

For odd number of poles, one pole has to lie on the real axis

while remaining all other are complex conjugate in the polynomial:

H(s) = 1/{(s+a1)(s+a2)….(s+an)}.At most one pole will lie on real axis rest

are complex conjugates.The frequency scaling is carried out by

substituting s by s/c, where c is the -3 dB frequency in rad./sec.

Determination of order n of the filter:

Given that H(s)H(-s) = [1+(s/j)2n]-1,Gx = 20 log10 |H(j)| = -10log10[1 + (x/c)2n]

Where s =x, c = cut-off frequency, n is order of filter.

Letting Pass band gain Gp dB occur at p rad/sec

and stop band gain Gs dB occur at s rad/sec.

We get expressions as: Gp = -10log10[1 + (p/c)2n] (p/c)2n = 10-Gp/10 -1

Gs = -10log10[1 + (s/c)2n] (s/c)2n = 10-Gs/10 –1

Hence: (s/p)2n = (10-Gs/10 –1)/ (10-Gp/10 –1)

Conclusion

From here we conclude:

(a) the order of the filter n log10 {(10-Gs/10 –1)/ (10-Gp/10 –1)} /

2log10(s/p).

(b) c, the cut off frequency

= p/{10-Gp/10 –1}1/2n

ALTERNATIVELLY AS

= s/{10-Gs/10 –1}1/2n

Ex.9.01:Designing a maximally flat LP filter

specifications:Pass band gain Gp= -2 dB for 0 rad/sec, and

stop band gain Gs = -20 dB for 20 rad/sec.

Soln:(a) Calculating number of poles:

Given p= 10, Gp = -2dB

and s = 20, Gs = -20 dB.

Substituting the values in the equation, n = log {(10-Gs/10 –1)/ (10-Gp/10 –1)} / 2log(s/p).

we get: n = 3.7 4. …..

Should be integer

design contd..…

(b) since n is taken 4 in place of 3.7, there will be two values of

c, the -3dB cut off frequency: = p/{10-Gp/10 –1}1/2n

and

= s/{10-Gs/10 –1}1/2n

pass band consideration: For n=4, p = 10, we get

-3dB Cut-off frequency c = 10.693 rad/sec.Stop-band consideration: For n=4, s = 20, we get

-3dB Cut-off frequency c = 11.261 rad/sec.

Contd:

We can choose any frequency between 10.693 and 11.261 rad/sec but least width of transit ch. obtained when lowest value is chosen.

So we choose : c = 10.693 rad/sec. Use the prototype denominator

polynomial:D(s)=S4+2.6131S3+3.4142S2 + 2.6163S +1Or, can calculate factored polynomial from

circle at angles 22.5 and 67.5. It is: D(s) = (s2+1.842s+1) (s2+0.771s+1)

design continued…..

(c) The proto-filter function for n =4 is: H(s) = 1/{s4 + 2.6131s3 + 3.4142 s2+2.6131s +1}

Or, H(s)-1 = (s2+1.842s+1) (s2+0.771s+1) Frequency scaling, replace s by s’=(s/10.693),

H(s’) = 13073.7/{s4+27.942s3 +390.4s2

+3194.88s+13073.7} Cascade realization:

H(s) = H1(s) H2(s) =

13073.7/{(s2+8.1844s+114.34) (s2+19.578s+114.34)} Parallel realization:

H(s) = H3(s) + H4(s) =

95.5858/( s2 +19.578s+114.34) + 136.7746/( s2 +8.1844s+114.34)

design contd……..

Each one of these forms can be converted in z-domain format using

impulse-invariance or, bilinear transformation.

Alternatively, one can be converted in desired format and

other realizations can be derived from there.

Designing Tchebyshev LP filter:

Amplitude of a normalized Tchebyshev filter is given by :

The nth order Tchebyshev polynomial Cn() is alternatively given by

Use of first one is preferred used when || <1. Cn can also be represented by polynomial.

H j( )1

1 2 Cn2

Cn

( ) cos ncos 1 ( )

Cn

( ) cosh ncosh 1 ( )

Designing Tchebyshev LP filter:

The following properties of Tchebyshev Polynomial helps calculate higher order polynomials iteratively..

Cn() = 2Cn-1() -Cn-2() for n>2, and C0() =1 and

C1() = .

Tchebyshev polynomials

Order n Cn()

00 1

01

02 22 -1

03 43 -3

04 84 - 82+1

05 165-203+5

06 326-484+182-1

07 647-1125+563-7

08 1288-2566+1604-322+ 1

09 2569 - 5767 + 4325 - 1203 + 9

10 51210-12808+11206 - 4004+502 -1

Tchebyshev……

To normalize with pass-band frequency we substitute s by , s/p to get

cosh n cosh( ) 1s

p

10

Gs

10 1

10

R

10 1

The number of poles n can now be calculated as:

Tchebyshev Algorithm contd….

n1

cosh 1s

p

cosh 1 10

Gs

10 1

10

R

10 1

Location of poles are given by the equation

sk

sin2 k 1( ) 2 n

sinh jcos2 k 1( ) 2 n

cosh

Where = [sinh-1(1/)]/n and k= [0,1, 2,…..,n].

Lathi, ’Signal Processing & Linear Systems’, Oxford,1998, pp. 505-524.

Ex. 9.02: Designing a Tchebyshev LP filter.

Design to meet the parameters:Pass band gain Gp= -2 dB for 0 rad/sec; stop

bandgain Gs = -20 dB for 16.5 rad/sec.Soln:(a) Calculations for no. of Poles:(b) Here pass band gain= peak to peak ripple. n = {1/cosh-1(16.5/10) cosh-1 [{102 –1}/{100.2-1}]1/2

= 2.999, say 3.0(c) Calculation of and x.

(i) In the equation 2 = 10R/10 –1. For R = 2 dB, = 0.7647.

(ii) Putting the values of n and in = [sinh-1(1/)]/n=(1/3)sinh-1(1/0.7647) = 0.3610.Contd..

Design of Tchebyshev LP filter

(d) Calculation of location of poles: (Tchebyshev polynomial can also be used)

s1= -0.3689, s2,s3 = -0.1844 j0.9231 (e) Calculation of normalized Transfer function: that is to satisfy; H(0) = 1,

H(s) = 0.3269/[s3 + 0.7378s2 + 1.0222s + 0.3269]

(f) Calculation of transfer function for the filter: Since the pass band frequency p = 10 rad/sec,

replace s s/10 in the normalized transfer function equation.

We get: H(s) = 326.9/[s3 + 7.378s2 + 102.22s + 326.9]

Elliptical or, Cauer filter.

It is an extension of Tchebyshev filter.

Here pass-band as well as stop-band have ripples.

Transit band characteristics are at best.

This filter has a zero

at a frequency close to stop band.

[details not to discuss]

Magnitude-frequency response

0 10 20 30 40 50 600

0.2

0.4

0.6

0.8

1m

ag

nitu

de

in d

B

frequency in rad/sec

ButterworthTchebyshevInverse TchebyshevElliptical

Phase-frequency characteristic

0 10 20 30 40 50 60-350

-300

-250

-200

-150

-100

-50

0

50

Phase

in d

eg

ree

s

frequency in rad/sec

ButterworthTchebyshevInverse TchebyshevElliptical

Transformation of analog filter.....LP BP; LP BR

Let B rad/sec be the –dB bandwidth at center frequency of o rad/sec,

the polynomial transformation for Low Pass to band-pass is:

s (s2 + o 2)/ Bs

and Low Pass to Band stop is: s Bs/(s2 + o

2)

A note of BP and BR filters

These filters observes the geometric mean.

o2 = l h

where l, h are the two frequency points obtained by intersection of any

line parallel to -axis on the filter characteristic.

If case the edges given in specifications do not satisfy the above equation, then they need to be modified to satisfy the

above requirement.

Ex. 9.3:

(a) Design a low pass filter with fp = 200 Hz and fs = 500 Hz.

(b) Design a high pass filter with p = 500 ; s =200 rad/s

(c) Design a bandpass filter with band edges [16, 18, 32, 48]. (d) Design a bandstop filter with band edges [16, 18, 32, 48]. Soln:(a) Choose a low pass proto-type filter with {s/ p}

= 2.5 and operate with following transformation: s (s/p) = s/[200x2].

Soln. 9.3 contd:

(b) The ratio of stop band to pass band characteristic of high-pass filter is

{p/ s} = 2.5.

The needed characteristic of a low pass filter is

{s/ p} = 2.5.

Choose the low-pass characteristic that meet above requirements, operate transformation as

s (2.5x200/s).

Soln. 9.3 contd…

Given that BP filter has edges [1, 2, 3, 4 ] = [16, 18, 32, 48].

Since bandwidth chosen is B = 3 - 2 = 32-18 = 14 rad/sec.

yielding o1 2 = 2 3 = 32 x 18 = 576.

According to the hypothesis, o2

2 =1 x 4. = 768 > o12

Hence we need to modify one or, both of the two edge frequencies.To compute design within transition band,

4 =o2/1 Or 4= 36 rad/sec.

Soln. 9.3 contd…

The modified edges are [16, 18, 32, 36]. Thus 3 - 2 = 14 rad/sec

and 4 - 1 = 20 rad/sec

choose the low pass filter with characteristics Ratio of stop band width/pass bandwidth =

20/14=1.4286 for the specified Gs and Gp

(Gs and Gp ignored here).

Design a LPF to workout the order of TF and polynomial to be used. use the transformation:

S (s2 + 576)/14s

Soln 9.3…

Consider a band-stop filter with band edges:

[1, 2, 3, 4 ] = [16, 18, 32, 48].

The bandwidth B = 4 - 1 = 32 rad/sec

It makes o2 = 768.

But 2 3 = 576. This necessitates to modify

3 =o2/2 = 42.667.

This has been done to limit the design in the

given band.

Soln 9.3…

The modified edges are: [1, 2, 3, 4 ] = [16, 18, 42.667,

48].Thus 4 -1 =32 & 3 - 2 = 24.667 rad/sec.We choose a low pass filter with

Stop bandwidth/pass bandwidth = 32/(24.667) = 1.2973 for given (Gs and Gp, ignored)and transformation:

s 32s/(s2 + 768).

Direct determination of normalized LPF characteristcs from Band Pass One

The LP BP transformation is:

s (s2 + o 2)/ BsFor a given value of s=j x in RHS, we get normalized frequency n (x

2- o2)/Bx

For B= 14, 02=576 : n (x

2- 576)/14x

To get normalized LPF characteristics, n= [1 2 3 4], put for x= [16,18 32 48];

Hence 1 = (162 – 576)/14x16 = -1.429

Direct determination of LPF ch. from BP

2 = (182 – 576)/14x18 = -1

3 = (322 – 576)/14x32 = 1

4 = (482 – 576)/14x48 = 2.57We get normalized LP Frequencies

[-1.429, -1, 1, 2.57] For least transit band, s/ p = min[1/2; 4/3] = 1.429The Low Pass equivalent frequency normalized specifications are [1 1.429] corresponding to [Gp Gs]. The values are same as before.

Direct determination of normalized LPF characteristics from Band Stop One

The LP to BS transformation is:

s Bs/(s2 + 02)

For a given value of s=j x in RHS, we get normalized frequency n Bx/(o

2- x2 )

For B= 32, 02=768 : n 32x/(768 -x

2)

To get normalized LPF freequencies, n= [1 2 3 4], put for x= [16,18 32 48];

Hence 1 = 32x 16/(768 – 162) = 1

Direct determination of LPF ch. Band Stop

2 = 32x 18/(768 – 182) = 1.297 3 = 32x 32/(768 – 322) = -4.0 4 = 32x 48/(768 – 482) = -1 The normalized frequencies are

[1, 1.297, -4, -1] For least transit band s/P = min(2/1 3/2) =

1.297 The Low Pass equivalent frequency normalized

specifications are [1 1.297] for the given [Gp Gs]

Same as derived earlier.

Example_9.4

A band stop filter to satisfy i. Butterworth Criteria and ii. Tchebyshev Criteria is required to meet following

specifications. to meet the following specifications.

1. Stop band 100 to 600 Hz (assumed at -3dB)

2. At and between 200 and 400 Hz, the magnitude should be at least -20 dB.

3. Maximum Gain (at the zero) =1.4. Pass band ripple 1.1 dB.5. Sampling frequency 2000 Hz.

Conversion to normalized LPF specs.

fn= [f1 f2 f3 f4 ]; fx = [100 200 400 600] Hz -3 dB bandwidth B= f4-f1= 600-100=500 Hz. f0

2 = f4f1= 6x 104 =(center frequency)2

Controlling LP to BS transformation is: fn = Bfx/(f0

2-fx2)

fn = 500xfx/(6x104 – fx2)

f1= 500x100/(6x104-1x104) = 1 f2= 500x200/(6x104-4x104) = 5 f3=500x400 /(6x104-16x104) = -2 f4=500x600 /(6x104-36x104) = -1 (fs/fp) = min(5, 2) = 2.

1. Butterworth filter.

Number of poles of LP filter = log10 {(10-Gs/10 –1)/ (10-Gp/10 –1)} / 2log10 (s/p)

= log10{(102-1)/(100.3-1)/2 log10(2) = 3.318 4. (for better transition characteristic)

The Denominator of LP normalized TF = (s2+0.765s+1)(s2+1.8484s+1) After transforming s 1000s/(s2+ 2.3687 x 106); The resulting polynomial would be: HBS(s)= N(s)/D(s) where N(s) = (s2+ 0

2)4

D(s) = (s4+a1s3+b1s2+c1s+d1) (s4+a2s3+b2s2+c2s+d2)

Butterworth filter

Where 0

2= 2.3687 x 106

a1 = 2.4045 x 103 a2= 5.8049 x 103

b1= 1.4607 x 107 b2=1.4607 x 107

c1= 5.6955 x 109 c2= 1.3750 x 1010

d1= 5.6108 x 1012 d2= 5.6108 x 1012

and location of poles:-180 620j -1022 3523j-667 512j - 2235 1714j

Designing Tchebyshev_1

n1

cosh 1s

p

cosh 1 10

Gs

10 1

10

R

10 1

1/cosh-1(2) =0.7593cosh-1( )=cosh-1(18.53) = 3.612n= 2.74 3 for better transit characteristic

The control equations is where Given:Gs =-20 dB, R=-1.1 dB and

Calculated s/p= 2

Tchebyshev…

Normalized Tchebyshev LP TF for R=1.1 dB Hlp(s) =0.4656/(s+0.4766)(s2+0.4766s+0.977) Transforming to BS by s 1000s/(s2+ 2.3687 x 106); HBP(s) = N(s)/D(s) where

N(s) = (s2+ 02)3 & D(s) = (s2+e1s+f1)

(s4+a3s3+b3s2+c3s+d3) Where 0

2 = 2.3687 x 106

e1 = 6.5923 x 103 f1 = 02

a3 = 1.5322 x 103 b3 = 1.4838 x107

c3 = 3.6294 x 109 d3 = 5.6108 x 1012

and poles are at:

[-381.38, -6210.9, -109.8 620j -656.3 3704.4j]