Spring 2012

© Ammar Abu-Hudrouss -Islamic ١ University Gaza

Filter types

What are the function of Filters ?

A filter is a system that allow certain to pass to its output and reject all other signals

Filters can be classified according to range of signal in the

Lowpass filter Highpass filter Bandpass filter Stopband (bandreject) filter

Digital ٢ Slide

١ Filter types

Digital Signal Processing ٣ Slide

Filter types

Filter types according to its

Butterworth filter Chebychev I filter Chebychev II filter

Digital Signal Processing ٤ Slide

٢ Butterworth filter

Ideal lowpass filter is shown in the figure

The passband is normalised to one. Tolerance in passband and stopband are allowed to enable the construction of the filter. Digital Signal Processing ٥ Slide

Lowpass Lowpass prototype filter: it is a lowpass filter with p =1.

The frequency scale is normalized by p. We use  = / p.

Lowpass filter

Highpass filter Lowpass Frequency prototype Transformation filter Bandpass filter

Bandreject filter Digital Signal Processing ٦ Slide

٣ Lowpass prototype filter Notation In analogue we will use

s to denote complex frequency

 to denote analogue frequency

p to denote complex frequency at lowpass prototype frequencies.

 to denote analogue frequency at the lowpass prototype frequencies.

Digital Signal Processing ٧ Slide

Magnitude Approximation of Analog Filters

 The of is given as rational function of the form

2 m co  c1s  c2s  cms Hs  2 n m  n d0  d1s  d2s  dn s

 The Fourier transform is given by

2 m m co  jc1  c2   j cm H  H (s) s j  2 n n d0  jd1  d2   j dn

H  H ( j) e j 

Digital Signal Processing ٨ Slide

٤ Magnitude Approximation of Analog Filters

 Analogue filter is usually expressed in term of

H  j 2  H  jH *  j H  j  2 H  j

 Example  Consider the transfer function of analogue filter, find s 1 H s  s2  2s  2

2 s 1  s 1 H  j  H sH  s  s j s 2  2s  2 s 2  2s  2 s j 2 2  1 H  j  4  4

Digital Signal Processing ٩ Slide

Butterworth filter

2 H  j will have only even powers of , or

2 4 2m 2 Co  C2  C4  C2m H ()  2 4 2n 1 D2  D4  D2n

In order to approximate the ideal filter

1) The magnitude at  = 0 is normalized to one 2) The magnitude monotonically decreases from this value to zero as ∞. 3) The maximum number of its derivatives evaluated at  = 0 are zeros. 2 1 H ()  This can be satisfied if 2N 1 D2N 

Digital Signal Processing ١٠ Slide

٥ Butterworth filter

The following specification is usually given for a lowpass Butterworth filter is

1) The magnitude of H0 at  = 0

2) The bandwidth p.

3) The magnitude (-Ap dB) at the bandwidth p.

4) The stopband frequency s.

5) The magnitude (-As dB) at the stopband frequency s .

6) The transfer function is given by

2 H0 H()  2N 1D2N 

Digital Signal Processing ١١ Slide

Butterworth filter

To achieve the equivalent lowpass prototype filter 1) We scale the cutoff frequency to one using transformation  = / p. 2) We scale the max. magnitude to 1 to one by dividing the magnitude by H0 . The transfer function become

2 1 H()  ' 2N 1 D2N 

2 We denotes D’2N as  where  is the factor, then

2 1 H()  1  2 2N

Digital Signal Processing ١٢ Slide

٦ Butterworth filter

2 If the magnitude at the bandwidth  = p = 1 is given as (1 - p) or −Ap ,

the value of  2 is computed by

2 10log H ()  Ap p 1 1 10log  A 1  2 p

 2 100.1Ap 1

2 If we choose Ap = -3dB   = 1. this is the most common case and gives

2 1 H ()  1 2 N

Digital Signal Processing ١٣ Slide

Butterworth filter

If we use the complex frequency representation

2 2 1 H( p)  H  p / j  N 1  p 2 

The poles of this function occurs at

e 2 jk / 2 N k  1,2,..., 2N , n even p  k  2k 1 j / 2 N e k  1,2,..., 2N , n odd Or in general

j(2k N 1) / 2N pk  e k  1,2,...,2N Poles occurs in complex conjugates Poles which are located in the LHP are the poles of H (s)

j(2k N1) /2N pk  e k 1,2,..., N

Digital Signal Processing ١٤ Slide

٧ Butterworth filter

When we found the N poles we can construct the filter transfer function as 1 H p  Dp

The denominator polynomial D (p) is calculated by

n Dp   p  pk  k 1

Digital Signal Processing ١٥ Slide

Butterworth filter

Another method to calculate D (p ) using

D( p)  p  pk  k

2 n D( p) 1 d1 p  d2 p  dn p cosk 1 / 2 d  d k  1,2,3,, N k  k  k1 sin 2N 

The coefficients dk is calculated recursively where d0 = 1

Digital Signal Processing ١٦ Slide

٨ Butterworth filter

The minimum attenuation as dB is usually given at certain frequency s. The order of the filter can be calculated from the filter equation

2 10log H (s )  As 10log 1 2 N  s  H() dB log10As /10 1 N    2logs 

Digital Signal Processing  (rad/sec) s ١٧ Slide

Design Steps of Butterworth Filter

1. Convert the filter specifications to their equivalents in the lowpass prototype frequency.

2. From Ap determine the ripple factor .

3. From As determine the filter order, N. 4. Determine the left-hand poles, using the equations given. 5. Construct the lowpass prototype filter transfer function. 6. Use the frequency transformation to convert the LP prototype filter to the given specifications.

Digital Signal Processing ١٨ Slide

٩ Butterworth filter

Example: Design a lowpass Butterworth filter with a maximum gain of 5 dB and a cutoff frequency of 1000 rad/s at which the gain is at least 2 dB and a stopband frequency of 5000 rad/s at which the magnitude is required to be less than −25dB.

Solution:

p = 1000 rad/s , s = 5000 rad/s, By normalization,

p = p/ p = 1 rad/s,

s = s / p = 5 rad/s,

And the stopband attenuation As = 25+ 5 =30 dB The filter order is calculated by log(10 As /10 1) N   2.146  3 2log(5)

Digital Signal Processing ١٩ Slide

Butterworth filter

The pole positions are:

j2k2 /6 pk  e k 1,2,3 j2 /3 j j4 /3 pk  e ,e ,e  0.5 j0.866,1,0.5 j0.866 D( p)  p  0.5  j0.866p 1p  0.5  j0.866 D( p)  p 2  p 1p 1 D( p)  p3  2 p 2  2 p 1 Hence the transfer function of the normalized prototype filter of third order is 1 H ( p)  p3  2 p2  2 p 1

Digital Signal Processing ٢٠ Slide

١٠ Butterworth filter

To restore the magnitude, we multiply be H0 H H ( p)  0 p3  2 p2  2 p 1

20logH0 = 5dB which leads H0 = 1.7783 To restore the frequency we replace p by s/1000

1.7783 H (s)  H p ps /1000  3 2  s   s   s     2   2  1 1000  1000  1000 

1.7783.109 H (s)  s3  2000s2  2106 s 109

Digital Signal Processing ٢١ Slide

Butterworth filter

If the passband edge is defined for Ap  3 dB (i.e.   1).

The design equation needs to be modified. The formula for calculating the order will become

A /10 log 10As /10 1 10 p 1 N     2logs  And the poles are given by

1/ N j(2kN 1) / 2N pk   e k 1,2,...,N

Home Study: Repeat the previous example if Ap = 0.5 dB

Digital Signal Processing ٢٢ Slide

١١