EKT430/4
DIGITAL SIGNAL
PROCESSING
2007/2008
CHAPTER 7
ANALOG FILTER
Analog Low-Pass Filters: a review
In this section, we shall take a review of
performance 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 stopband.
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 splane.
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 + k2 = a2.
or
k + jk = a
Coefficients of the Butterworth
polynomials for various number of poles at
c =1.
N
ao
a1
a2
a3
a4
a5
a6
a7
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
a8
See the
symmetry of
coefficients
1
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.
 = (1/n) sinh-1 (1/)
Minor
axis
Pass band ch. of 6th order Butterworth
and Tchebyshev
1.1
1.05
1
magnitude in dB
6
4
2
p
0.95
1
0.9
5
3
0.85
3.8
0.8
5
0.75
0.7
0
2
4
6
frequency in rad/sec
8
10
12
Tchebyshev Filters
is also an all pole filter.
Poles lie on ellipse and follows following equation:
k
sinh
2
k
2
 1
cosh
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 + jk = -sin sinh + j cos cosh,
while  = (1/n) sinh-1(1/)
[Ambarder p 415]
Or (1/) = sinh (n)
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:
H( 0 ) 
1 if n is odd
1
if n is even
r
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.
Calculations of  and n from ripple…
say -1.1dB





We know that
20 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 + jk = -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 + o2)
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= ejn for integer k.
and -1 = ej(2k+1) for k0
S2n = ej(2k+1) ejn =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
Should be
p= 10, Gp = -2dB
integer
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.
…..
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 +1
Or, 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:
1
Amplitude of a normalized
Tchebyshev
filter is given
H( j ) 
2
2
by
1 :  C
n
The nth order Tchebyshev polynomial Cn() is
alternatively given by
1
Cn (  )  cos ncos (  )
1
Cn (  )  cosh ncosh (  )
Use of first one is preferred used when || <1.
Cn can also be represented by polynomial.
Designing Tchebyshev LP filter:
The following properties of Tchebyshev
Polynomial helps calculate higher order
polynomials iteratively..
Cn() = 2Cn-1() -Cn-2() for n>2, and
C0() =1
and
C1() = .
Tchebyshev polynomials
Order n
00
Cn()
01
02
03
04
1

22 -1
43 -3
84 - 82+1
05
06
07
165-203+5
326-484+182-1
647-1125+563-7
08
09
10
1288-2566+1604-322+ 1
2569 - 5767 + 4325 - 1203 + 9
51210-12808+11206 - 4004+502 -1
Tchebyshev……

To normalize with pass-band frequency we
substitute s by , s/p to get
G
s
cosh ( n  cosh )
1
s
p

10
10
1
R
10
10
1
The number of poles n can now be calculated as:
Tchebyshev Algorithm contd….
G
s
1
n
cosh
1
s
 cosh
p
1
10
10
1
R
10
10
1
Location of poles are given by the equation
sk 
( 2 k 1)  
sin
sinh 
2n
( 2 k 1)  
jcos
cosh
2n
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 band
gain 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
1
Butterworth
Tchebyshev
Inverse Tchebyshev
Elliptical
magnitude in dB
0.8
0.6
0.4
0.2
0
0
10
20
30
40
frequency in rad/sec
50
60
Phase-frequency characteristic
50
Butterworth
Tchebyshev
Inverse Tchebyshev
Elliptical
0
Phase in degrees
-50
-100
-150
-200
-250
-300
-350
0
10
20
30
40
frequency in rad/sec
50
60
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 + o2)
A note of BP and BR filters
These filters observes the geometric mean.
o 2 =  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,
o22 =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)/ Bs
For a given value of s=j x in RHS, we get
normalized frequency n  (x2- o2)/Bx
For B= 14, 02=576 : n  (x2- 576)/14x
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 = 2.57
We get normalized LP Frequencies
[-1.429, -1, 1, 2.57]
For least transit band,
s/ p = min[1/2; 4/3] = 1.429
The 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  Bx/(o2- x2 )
For B= 32, 02=768 : n  32x/(768 -x2)
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.
2
4
2
 f0 = f4f1= 6x 10 =(center frequency)
 Controlling LP to BS transformation is:
 fn = Bfx/(f02-fx2)
4
2
 fn = 500xfx/(6x10 – fx )
 f1= 500x100/(6x104-1x104)
=1
 f2= 500x200/(6x104-4x104)
=5
 f3=500x400 /(6x104-16x104) = -2
4
4
 f4=500x600 /(6x10 -36x10 ) = -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 1000s/(s2+ 2.3687 x 106);
The resulting polynomial would be: HBS(s)= N(s)/D(s)
where
N(s) = (s2+ 02)4
D(s) = (s4+a1s3+b1s2+c1s+d1) (s4+a2s3+b2s2+c2s+d2)
Butterworth filter
Where
02= 2.3687 x 106
a1 = 2.4045 x 103
b1= 1.4607 x 107
c1= 5.6955 x 109
d1= 5.6108 x 1012
a2= 5.8049
b2=1.4607
c2= 1.3750
d2= 5.6108
and location of poles:
-180  620j
-1022 3523j
-667  512j
- 2235 1714j
x 103
x 107
x 1010
x 1012
Designing Tchebyshev_1
The control equations is where
Given:Gs =-20 dB, R=-1.1 dB and
Calculated s/p= 2
1
n
cosh
1
s
p
 cosh
1
G
s
10
10
1
R
10
10
1/cosh-1(2) =0.7593
cosh-1( )=cosh-1(18.53) = 3.612
n= 2.74  3 for better transit characteristic
1
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 1000s/(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
02 = 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]