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# Analogue Filters

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```									                                                                Lecture 6: Analogue filters

Analogue Filters

To design a recursive filter, we begin with a good analogue filter and approximate it
with a digital filter. It is the best way at hand. This requires us to design an analogue
filter first.

These filters are described by a constant coefficient differential equation – output is
the weighted sum of the derivatives of the input and output signals, defined as

N
d k x (t ) M d k y (t )
y t    ak             bk                               (1)
k 0   dt k      k 1 dt k
e.g.

y t   x (t )   with a0=1, ak=0 for k=1, ...,N, and bk=0 for k=1, ...,M.
dx (t )
y t             with a0=0, a1=1, ak=0 for k=2, ...,N, and bk=0 for k=1, ...,M.
dt

Let  e st is an eigenfunction for the analogue filter likewise e jk is an eigenfunction
st
of a digital filter. e is an eigenfunction that can be shown as follows

 
d st
dt
e  s  e st

 
d 2 st
dt 2
e  s 2  e st
(2)


 
d k st
dt k
e  s k  e st

Eigen                                                         Eigen
function                                                       value

Analogue                     H ( s )e st
e st                           filter

Figure 1: Eigen function and eigen value

Eigenfunction and eigenvalue play an important role. Eigen value H(s) is called the
transfer function of the analogue filter

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Lecture 6: Analogue filters

y(t )  H ( s)est where x(t )  e
st
(3)

Replacing y(t) in the general equation (9), we get
N
d k e st  M d k H ( s )e st 
H ( s )e   ak
st
  bk
k 0   dt k       k 1  dt k
N                                  M                                  (4)
H ( s )e   ak s e  H ( s ) bk s e
st             k     st                         k   st

k 0                                k 1
Solving equation (12) for H(s) yields

 N a s k   e st

 k 0 k 
H ( s)             
(5)
 M           
1   bk s k   e st
 k 1
             
The transfer function of the recursive analogue filter is as follows

N
 ak s k
H ( s)         k 0
M
(6)
1   bk s      k

k 1
Since H(s) is a ratio of polynomials, it can be represented by the coefficients of the
polynomials ak and bk.

a 0  a1 s  a 2 s 2    a k s k
H ( s) 
1  b1 s  b2 s 2    bk s k 
(7)

H(s) can also be represented by poles, zeros, and gains.

H ( s)  G
s  z1 s  z2 s  z N 
s  p1 s  p2 s  pM                                       (8)

Where G is the gain, N is the number of zeros and M is the number of poles of the
filter. The poles and zeros can be plotted in complex s-palne shown in Figure 5.

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Lecture 6: Analogue filters

Im{s}

zero
pole


Re{s}

Figure 2: Pole-zero diagram

Example 1: An analogue filter is described by the following differential equation
dx (t )    dy (t )
y (t )  x (t )            2
dt         dt
Find the transfer function of the analogue filter.

Example 2: An analogue filter is described by the transfer function
1 s
H ( s) 
1  2s
Find the pole-zero-gain form of the of the transfer function of the filter and plot the
pole-zero diagram.

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Lecture 6: Analogue filters

Stability of analogue filter
An analogue filter is stable if and only if its impulse response goes to zero, i.e.

Lt h(t )  0                                        (9)
t 

If the impulse response increases with time, the filter is unstable.

In general, the computation of impulse response is very tedious. It requires the
solution of the filter’s differential equation or the evaluation of the inverse Laplace
transform of the transfer function.

Fortunately, to test the stability, we don’t need to compute filter’s impulse response. It
is possible to determine the stability by the position of the poles of the filter’s transfer
function in the s-plane.

The filter is stable if the poles lie in the left half of the s-plane. If the poles lie in the
right half of the s-plane, filters are unstable.

Im{s}                       s-plane

unstable
region
stable

region

Re{s}

Figure 3: Stability from Pole-zero diagram

Example 3: An analogue filter is described by the differential equation
y (t )  a0 x (t )  b1 y (t )

Show the stability of the filter using pole-zero plot.

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Lecture 6: Analogue filters

We know the stability for a single pole filter and this idea can be extended to
realisation of more complicated filters by expressing an arbitrary filter as a collection
of single-pole filters. Lets consider an analogue filter with M poles. The transfer
function can be expressed in terms of poles, zeros and gain

H ( s)  G
s  z1 s  z2 s  z N                       (10)
s  p1 s  p2 s  pM 
Where N is the number of zeros and M is the number of poles.

These single-pole filters can be realised in two ways:

   Parallel combination

By cascade combination it is difficult to visualise the relationship between individual
impulse response and response of the original filter. For this reason, we turn to
parallel combination of single-pole filters.

Pole-zero gain representation of the transfer function can be expressed as the parallel
combination of M-pole filters by performing a partial fraction expansion

g1        g2              gM
H ( s)                                                              (11)
s  p1  s  p2        s  pM 

g1
s  p1

g2
x(t)                                                        y(t)
s  p2


gM
s  pM
Figure 4: Parallel combination of single-poles filters.

In this representation, the impulse response of the original filter is the sum of the
impulse responses of single-pole filters. If any of these filter is unstable, the original
filter will also be unstable.

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Lecture 6: Analogue filters

In other words, M-pole filter is stable, only if each of its single-pole filter is stable.

Therefore, the m-pole filter is stable iff (if and only if) all of its poles lie in the left
half of s-plane.

Example 4: A low-pass analogue filter has three poles
p1 = -5
p2 = -10
p3 = -5

Calculate the gains of the single-pole filters using partial fraction expansion. Give a
parallel realisation of the filter.

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Lecture 6: Analogue filters

Butterworth filter design

The general form of the analogue filter is

d k x (t ) M d k y (t )
N
y t    ak             bk                                                      (12)
k 0   dt k      k 1 dt k
st
Where the complex exponential e is applied to an analogue filter. The output
st
becomes H ( s)e where H (s ) is the eigenvalue or the transfer function. The
transfer function of the general form becomes

N
 ak s k
H ( s)            k 0
M                                             (13)
1   bk s        k

k 1

The frequency response is the filter’s transfer function at s  j . Therefore, the
frequency response of the filter can be rewritten as

N
d k jt M    dk
H ( )e   jt
  a k k e   bk k H ( )e jt
k 0 dt      k 1 dt
N                                         M
H ( )e   jt
e      jt
 a k ( j )       k
e   jt
H ( ) bk ( j ) k
(14)
k 0                                      k 1
N
 a k ( j ) k
H ( )         k 0
M
1   bk ( j ) k
k 1

Butterworth filter is very popular because of its passband and stopband are without
ripples. But it has the widest transition band and cutoff frequency. The Butterworth
filter has two parameters

   Order of the filter  order of the denominator of H (s ) that corresponds to
ˆ
the number of poles of the filter’s transfer function
   Cutoff frequency denoted as  c

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Lecture 6: Analogue filters

The Butterworth filter is defined, in general, by the squared transfer function

H B ( s)  H B ( s) H B ( s)
2                                   *
(15)

If the order of the filter is   m , then the transfer function square becomes

1
H B ( s) 
2
2m
(16)
 s 
 j 
1     
 c

We know that the frequency response is the filter’s transfer function at s  j

1
H B ( ) 
2
2m
 j               
1 
 j               

 c                

1
H B ( ) 
2
2m
(17)
         
1 
         

 c        

Poles of the squared transfer function are values of                s for which

2m
 s 
 j 
1                     0
 c
2m
 s 
 j   1
      
 c
s
 2 m ( 1)
j c

1

s  ( 1)    2m
 j c                                               (18)

Now, (1) and j can be expressed in exponential form as

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Lecture 6: Analogue filters

 1  e j ( 2 k 1)
                                                            (19)
j
je               2

                  
NOTE: cos ,3,5,...  1 , sin  ,3,5,...  0 sin
1                     1                                                     1, and cos       0
2                  2

Therefore, the poles become

1

s  ( 1)        2m
 j c

s  e                      
1           j
j ( 2 k 1)    2m      e       2
c
( 2 k 1)              
j
se      j       2m         e       2
c
( 2 k 1) 

se      j       2m      2       c

( 2 k  m 1)
se   j          2m           c                                                    (20)

The Butterworth filter’s complex pole             s has the magnitude and angle as

| s |  c
2k  m  1                                                                 (21)
s                  
2m

They are distributed around a circle of radius                       c centred at origin at a distance of
360
from each other.
2m

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Lecture 6: Analogue filters

Im{s}
360/2m

                        

|          Re{s}
|c|
                       

Figure 5: Butterworth filter’s poles

Butterworth filter design can be summarised in a three-step design procedure

Step 1: Select the order   m and cutoff frequency  c of Butterworth filter
Step 2: Plug the values into the equation

( 2 k  m 1)
se   j          2m        c
360
and compute the 2m poles. The poles are equally spaced at                         around the circle
2m

Step 3: Select   m stable poles (those are in the left half of s-plane)

Example 5: Find the poles of a third-order Butterworth low-pass filter with a cutoff