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Butterworth Filter Butterworth Filter 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 frequency to pass to its output and reject all other signals Filters can be classified according to range of signal frequencies in the passband Lowpass filter Highpass filter Bandpass filter Stopband (bandreject) filter Digital Signal Processing ٢ Slide ١ Filter types Digital Signal Processing ٣ Slide Filter types Filter types according to its frequency response Butterworth filter Chebychev I filter Chebychev II filter Elliptic 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 prototype filter Lowpass prototype filter: it is a lowpass filter with cutoff frequency 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 filter design 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 transfer function of analogue filter is given as rational function of the form 2 m co c1s c2s cms Hs 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 jH * 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 sH 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 1D2N 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 ripple 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 decibels, 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 N1) /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 Dp The denominator polynomial D (p) is calculated by n Dp 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 cosk 1 / 2 d d k 1,2,3,, N k k k1 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 2logs 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: j2k2 /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 j0.866p 1p 0.5 j0.866 D( p) p 2 p 1p 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 ps /1000 3 2 s s s 2 2 1 1000 1000 1000 1.7783.109 H (s) s3 2000s2 2106 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 2logs And the poles are given by 1/ N j(2kN 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.
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