Part A: Signal Proceing
Chapter 5: Digital Filter Deign 5. Chooing between FIR and IIR filter 5. Deign Technique 5.3 IIR filter Deign 5.3. Impule Invariant Method 5.3. Bilinear Tranformation 5.3.3 Digital to Digital Tranformation 5.4 Window function 5.5 Deign Method for FIR filter 5.5. Deign of FIR Filter Uing Window 5.5. Deign Procedure 5.6 Frequency Sampling Filter 5.7 Problem heet A5
5.0 Digital Filter Deign The deign of a digital filter i the tak of determining a tranfer function which i a rational function of - (e.g. a 0 a b a b in the cae of a recurive filter (IIR or a polynomial in -, (a 0 +a - +a - +a 3-3 in the cae of a non-recurive filter.
+ - Performance pecification: A typical amplitude characteritic of a low-pa filter i hown below. A ( Pa band L u Tranition band Stop band Note: i the 0dB point f L f u f / Digital Freq ( Analogue Freq (f
The goal of the deign i to determine a tranfer function H( o that it amplitude characteritic H( atifie the condition. H ( for 0 H ( for U L
5. Chooing between FIR and IIR filter. [4] FIR Filter Sytem function contain only ero Non-recurive or recurive tructure are both poible; the bet known i the non-recurive (tranveral tructure. FIR Filter can have an exactly linear phae repone. The implication of thi i that no phae ditortion i introduced into the ignal by the filter. The effect of uing a limited number of bit to implement filter uch a round off noie and quantiation error are much le evere in FIR than in IIR. IIR Filter Contain pole and ero (normally Only recurive tructure i poible; the mot widely ued form i the cacade connection of firt-order and econd order ection. The phae repone of IIR filter are nonlinear, epecially at the band edge. Becaue of quantiation of the filter coefficient, a pole can in principle move from a poition inide the unit circle to a poition outide the unit circle and hence caue intability.
FIR require more coefficient for harp cut-off filter than IIR. Thu for a given amplitude repone pecification, more proceing time and torage will be required for FIR implementation. Complexity i proportional to the length of the impule repone. FIR filter have no analogue counterpart. FIR deign procedure are normally iterative procedure. Deign equation do not exit IIR require fewer coefficient for harp cut off filter than FIR. No direct relation between the complexity and the length of the impule repone (which i infinite by definition Filter with high electivity can be realied with relatively low complexity. Analogue filter can be readily tranformed into equivalent IIR digital filter meeting imilar pecification. IIR filter can be deigned uing deign formulae.
Example: An FIR filter i to be deigned to meet the following frequency repone pecification. Pa-band 0.8-0.33 (normalied Tranition band 0.04 (normalied Stop-band deviation 0.00 Pa-band deviation 0.05 (i Sketch the tolerance cheme for the filter. (ii Expre the filter band edge frequencie in the tandard unit of kh, auming a ampling frequency of 0kH and the top band and pa band deviation in db.
The tolerance diagram for the filter, i hown below: H(f + - 0.4 0.8 0.33 0.37 f =0kH, therefore Pa band:.8-3.3 kh Stop band: 0-.4 kh and 3.7-5kH Stop band attenuation: -0log0(0.00 = -60dB Pa band ripple: 0log0(+0.05 = 0.4 db f 0.5 f (normalied
Example: The following tranfer function repreent two different filter meeting identical amplitude frequency repone pecification. a a a 0 ( b b H Filter Where a 0 = 0.49889; a = 0.974777; a = 0.49889; b = -0.6744878; b = -0.363348;
0 ( k k k h H Filter Where 6 0 0.5789400 5 7 0 0.634840 4 8 0 0.55384370 3 9 0 0.696940 0 0 0.45068750 0 0.5460380 0 0 h h h h h h h h h h h h
Determine and comment on the computational and torage requirement. x[n] Filter -b w[n] -b T w( n x ( n b w( n y( n a + + o w( n a a 0 T a w( n a b w( n a w( n y[n]
x[n] x[n-] T T T h(0 h( h( + Filter y[n] x[n-] h(
Number of Multiplication Number of addition Storage location (coefficient and data FIR (Filter 4 IIR (Filter It i evident that the IIR filter i more economical in both computation and torage requirement than the FIR filter. 5 4 8
5. Deign Technique [4] window The method ued to calculate the filter coefficient (h k for FIR, a k and b k for IIR depend on whether the filter i IIR or FIR type. There are everal method of calculating filter coefficient of which the following are the mot widely ued. Frequency ampling FIR digital filter Optimiation method (e.g Reme Algorithm Y( (N h h h... h N 0 X( Y ( X ( IIR digital filter Impule invariant tranformation Bilinear tranformation Pole-ero placement method a0 a b... a n... b ( N L L
We chooe the method that bet uit our particular application. In mot cae, if the FIR propertie are vital then a good candidate i the optimiation method, where a, if IIR propertie are deirable, then the bilinear method will in mot cae uffice.
5.3 IIR Filter Deign In tranforming an analogue filter to digital filter, we mut obtain either H( or h[n] from the analogue filter deign. In uch tranformation, we generally require that the eential propertie of the analogue frequency repone be preerved in the frequency repone of the reulting digital filter. Thi implie that we want the imaginary axi of the -plane to map into the unit circle of -plane. A econd condition i that a table analogue filter hould be tranformed to a table digital filter. That i if the analogue ytem ha two pole only in the left half -plane, then the digital filter mut have pole inide the unit circle.
5.3. Impule invariant method [] In thi method we tart from an analogue filter of impule repone h a (t and the ytem function H a (. The objective of our deign i to realie an IIR filter with an impule repone h[n] which atifie : h[n] = h a (nt where T-ampling frequency The characteritic property preerved by thi tranformation i that the impule repone of the reulting digital filter i a ampled verion of the impule repone of the analogue filter.
h a (t h[n] Impule repone T t n -/T -/T H a ( H( /T Frequency Repone /T Aliaing /T - - 0 We ee that with thi method there are problem to a greater or leer extent depending on the choice of T.
The ampling frequency affect the frequency repone of the impule invariant dicrete filter. A ufficient high ampling frequency i neceary for the frequency repone to be cloe to that of the equivalent analogue filter. Thu due to aliaing, the frequency repone of the digital filter will not be identical to that of the analogue filter. So how do we find the filter coefficient of the IIR filter in thi deign method?
To obtain the mapping let, H a b, b 0 h a h a ( t ( nt e e bt bnt Uually written a H Sampled equence h n e 0 ( bt e Invere Laplace tranform bnt n 0 n 0 -tranform
It i een that H( i obtained from H a ( by uing the mapping relationhip b e bt j 3/T /T /T, b 0-3/T -plane Each trip map onto the interior of the unit circle T-ampling period = =0 = - In thi kind of mapping, the perimeter i the imaginary axi. Note: Mapping doe not exit for the ero. -plane =
Example: Uing partial fraction thi can be written a 3 ( ( ( H 3 ( H e b bt 4 3 3 3 ( ( e e e e e e e H T T T T T T T
Example: Ue the impule invariant method to deign digital filter from an analogue prototype that ha a ytem function. To deign a filter uing the impule invariant method, expand H( in a partial fraction form: ( ( b a a H ( ( ( ( ( ( ( jb a jb a jb a jb a jb a B jb a A jb a jb a a H
e c ct ( ( ( ( ( ( co( co( ( ( ( e bt e bt e e e e e e e H at at at T jb a T jb a T jb a T jb a T jb a T jb a Subtituting (Impule invariant tranform
a Hence, where H a ( b b at at bt, b e co bt b e at e co and ( a a b at e co( bt at e co( bt e at Note that the ero at =-a i mapped to a ero at =e -at co(bt. Thu, the location of the ero in the dicrete time filter depend on the poition of the pole a well a the ero in the analogue filter.
Example: Uing impule invariant method deign a digital filter to approximate the following normalied analogue tranfer function: H ( [] Aume that the 3dB cut-off frequency of the digital filter i 50H and the ampling frequency i.8 kh
Solution: Before applying the impule invariant method, we need to de-normalie the tranfer function. ( ( c H H ( c c c H 4778 94. 50 c ( p B p A H c c c where
4, j j p p c c c c c c j p j p c c j B j A j p j p c c 666.434( 666.434(
( ( ( ( ( e e e B e A e B e A H T p T p T p T p T p p T ( ( ( ( e e e Be Ae B A H T p T p T p p T T p 0.3530.0308 393.964 ( H 0.3530.0308 393.964 ( j j j e e e H 3 0.3530.0308 393.964 ( 0 H Thi i approximately equal to the ampling frequency
393.964 H ( 0 3.03080.3530 Such a large gain i characteritic of impule invariant filter. To keep the gain down (and to avoid overflow when the filter i implemented. It i common practice to divide the gain by f. Thu the new tranfer function become H 0.3078.0308 0.3530 (
x[n] 0.3078 +.0308-0.3530 - - y[n] If x[n]=[n] then y[n]=h[n].
Impule repone of the analogue filter h(t Impule repone of the digital filter i identical to that of analogue filter h(t = h(n h[n] 0 3 4 5 Note: The ampling frequency affect the frequency repone of the digital filter obtained uing impule invariant tranformation. t n
A ufficient high ample frequency i neceary for the frequency repone to be cloer to that of the equivalent analogue filter (ee below Low degree of aliaing can be achieved by making the ampling frequency high. H( aliaing 0 H(j 4 f / f f f
5.3. Bilinear Tranformation [] The bilinear tranformation yield table digital filter from table analogue filter (the impule invariant technique may not. Alo the bilinear tranformation avoid the problem of aliaing encountered with the ue of the impule invariant tranformation, becaue it map the entire imaginary axi in the -plane on to the unit circle in the -plane.
j -plane -plane Image of the left hand -plane.
e T e T T e T T T T T T T...... (drop out higher order term
The price paid for the avoidance of aliaing i an introduction of ditortion in the frequency axi. Conequently, the deign of digital filter uing the bilinear tranformation i only ueful when the ditortion can be compenated.
Example: We have a ytem function H a ( uch that H a ( b b Applying bilinear tranformation, H b T ( Let b 000 & T 000 b bt bt( ( bt
The modulu of the frequency repone H( and H( are hown below: f -3000-500 H(j H( H( 500 3000 j - -/ / b j b -000-500 500 000 f
There i a very important property of the bilinear tranformation that can be een in the above example. The entire frequency range (- a of the continuou ytem map into the fundamental interval (- of the dicrete ytem, where = 0 correpond to = 0, = to = and = - to = -.
To demontrate that thi mapping ha the property that the imaginary axi in the -plane map onto the unit circle, let = e j and = j. T tan j T j e e T j j j tan T tan T Then ince, Hence, or
We ee that a nonlinear relation exit between and. Thi effect i called Warping and i hown below. At low frequencie T. - tan T T T T
The great advantage of warping i that no aliaing of the frequency characteritic can occur in the tranformation of an analogue filter to a dicrete filter, which we encountered in the impule-invariant method. We mut however check carefully jut how the variou characteritic frequencie of the continuou characteritic frequencie of the dicrete filter. We can illutrate thi with the aid of a diagram (next lide for a band-pa filter.
3 - Digital filter repone 3 Analogue filter repone The effect of warping in the converion of H( H( i een from the above diagram.
In deigning a digital filter by thi method we mut firt pre-wrap the given filter pecification to find Example: the continuou filter to which we are going to apply the bilinear tranformation. The pecification of a deired digital low pa filter i hown below. Sampling frequency: f = 8kH (T =/f = 5 L A pa band up to f L =.6 kh ( T 0. 65 u and a top band, f u, above 3kH ( T 0. 75. u L L L f f f f
+ - 0 H( Digital filter Repone L u.6kh 3kH f /
We mut tart from an analogue filter with L L f l tan (455 f L 4.55kH T u u f h tan( (648 fu 6.48kH T 0 H( 4.55kH 4H 6.48 8kH Analogue filter Repone
Example: Determine, uing bilinear tranformation method, the tranfer function and difference equation for the digital equivalent of the RC filter. The normalied tranfer function for the RC filter i H ( Aume a ampling frequency of 50H and a cut-off frequency of 30H.
Solution: T c c f c f c (30 0. 4 50 f c ' i alway > f c and after warping, before pre-warping Pre-warped frequency (un-warped freq= 30 H f c ' = 34.68H.
The de-normalied analogue filter tranfer function (thi i achieved by replacing with i obtained from H( a c 0.765 0.765 ( ( ' ' ' T T H H c c 0.584 0.408( 0.765 765 ( 0.765( ( ( H H T
Example: It i required to deign a digital filter to approximate the following analogue tranfer function [] Uing the Bi-linear tranformation method obtain the tranfer function, H( of the digital filter auming a 3dB cut-off frequency of 50H and a ampling frequency of.8kh. H (
Solution : ' c T f c = 50H f c 50 5 c ( f 80 64 T 5 c 64 tan 80 tan 987.5009 0.3859rad T T ' 987.5009 f c 57. 656H f = 80H The analogue frequency after pre-warping ' f c f c i alway greater than. f / ec
(0.3857 (0.3857 0.3857 ( T T T H 0.3857 0.3857 0.3857 ( ( T T T T T H H T 0.356.0048 0.0878 ( H Prewarped analogue filter i given by,
' c 0.0878(.0048 0.356 ' c c' All pole 0 gain (db 50H f c Bilinear tranformation digital frequency repone H( 640H f / Analogue frequency repone H(j Pole & ero f (rad/ Note: Same cut-off frequency increaed roll off and attenuation in top band.
Example : y (a An analogue tranfer function can be converted to a digital tranformation uing the bilinear tranformation. Derive thi tranform relationhip uing the following equation. T n y n x n x n H( Analogue Integrator Digital Integrator T-ampling period, x[n] - input, y[n] - output
Solution: ( ( ( ( ( ( H T H X X T Y Y Setting ( = (, T T
(b Convert the analogue filter H( 0. H ( ( 0. 6 into a digital IIR filter by mean of the bilinear tranformation. The digital filter i to have a reonant frequency. 0 H( 0. 0. 6.0 0
The analogue filter ha a reonant frequency The frequency i to be mapped into by electing the value of the parameter T. 0 T tan 0 4 tan tan 4 T T 4 T T in order to have Thu the deired mapping i 0 0 rad / ec 4 0 4 rad/ ec.
reonant frequency 4 0.975 0. 0.006 0.8 0.975 0.0006 0. 0.006 0.8 6.0 4 0. 4 0. 4 ( ( H H 0.987 0.987 j j e p e p Thi filter ha pole at and ero at = - and = 0.95.
Example : H( Convert the imple low pa filter into an equivalent high pa dicrete filter. Aume f = 50H, analogue cut-off f c = 30H. 30 c 0. 4 50 ' c T 0.4 tan( 0.765 T LPF to HPF tranformation H ' ( H ( ' c ' c ' c
0.765( 0.765 ( ( ' T T T H H T 0.584 0.579 ( H
Example : Let u now apply thi approach to the deign of a digital low-pa filter. The amplitude repone pecification i Sampling Frequency : Paband: Stopband: Maximum ripple in paband: Minimum attenuation in topband: f = 8 kh f c = 0 to.3 kh f h.6 kh = 0. db = -33.5 db
Step : From the ampling frequency f and the cut-off frequency f c we can determine c and h : fc fh c.00 and h. 040 f f
Step : Determine the top-band edge frequency f h or h uing c T c tan and h T h tan In filter deign table c i normalied to rad/ec. T h c.7856 c tan.040.7856 tan.939
Step 3 : Determine an appropriate tranfer function. Thu we need an analogue filter with a maximum ripple of 0.dB in the pa-band (0 and a minimum attenuation of -33.5 in the top-band (.94. From the table we chooe a 3rd order elliptic filter with a tranfer function H a 5.888( 0.089.0398( 0.8700 which meet thee pecification..6674
Step 4 : Apply the bilinear tranformation to obtain a digital filter tranfer function: and.7856 T Subtituting for, we get H 0.56(.0478 ( 0.640 ( 0.4748 ( 0.553 (
% Exercie. IIR digital filter deign % We wih to deign an IIR digital filter given ome pecification. % We ue the example given in page 30 of chapter 5. % The tak i to deign a low-pa filter, whoe magnitude repone pecification % are a follow: ampling_frequency = 8000 % f = 8 kh paband_frequency = 300 % fc =.3 kh topband_frequency = 600 % fh =.6 kh paband_ripple = 0. % 0. db topband_ripple = 33.5 % 33.5 db % Step. Evaluate digital frequencie paband_digital_frequency = *pi*paband_frequency/ampling_frequency topband_digital_frequency = *pi*topband_frequency/ampling_frequency % Step. Evaluate analog frequencie paband_analog_frequency = *ampling_frequency*tan(paband_digital_frequency/ topband_analog_frequency = *ampling_frequency*tan(topband_digital_frequency/ % Step 3. Determine the normalied analog low-pa filter [filter_order, cutoff_frequency] = ellipord(... paband_analog_frequency, topband_analog_frequency,... paband_ripple, topband_ripple, '' [numerator, denominator] = ellip (filter_order,... paband_ripple, topband_ripple, cutoff_frequency, ''
% Plot the magnitude repone of the analog low-pa filter frequency = :ampling_frequency; angular_frequency = frequency**pi; [frequency_repone] = freq(numerator, denominator, angular_frequency; magnitude_repone = 0*log0(ab(frequency_repone; figure(, clf plot(frequency, magnitude_repone grid on title ('Magnitude repone of the analog filter' ylabel('magnitude (db' xlabel('analog frequency' hold on emilogx([0 paband_analog_frequency//pi], [-paband_ripple - paband_ripple], 'r' emilogx([paband_analog_frequency paband_analog_frequency]//pi, [-paband_ripple - topband_ripple*], 'r' emilogx([topband_analog_frequency topband_analog_frequency]//pi, [ -topband_ripple], 'r' emilogx([topband_analog_frequency//pi ampling_frequency], [- topband_ripple -topband_ripple], 'r' axi([0 ampling_frequency -topband_ripple* 0] hold off
Magnitude(dB 0-0 -0-30 -40-50 -60 Magnitude repone of the analog filter 0 000 000 3000 4000 5000 6000 7000 8000 Analog frequency
% Step 4. Apply bilinear tranformation to obtain the digital filter tranfer function [digital_numerator, digital_denominator] = bilinear(numerator, denominator, ampling_frequency % Impule invariant method can be elected intead of bilinear tranformation. % Matlab provide a build-in command called `impinvar'. % You may experiment with `impinvar' and compare the outcome with thoe of the bilinear tranformation. % Plotting the magnitude repone digital_frequency = (0:0.0:*pi; frequency_repone = freq(digital_numerator, digital_denominator, digital_frequency; magnitude_repone = 0*log0(ab(frequency_repone; figure(, clf plot(digital_frequency, magnitude_repone grid on title ('Magnitude repone of the digital filter' ylabel('magnitude (db' xlabel('digital frequency' hold on emilogx([0 paband_digital_frequency], [-paband_ripple -paband_ripple], 'r' emilogx([paband_digital_frequency paband_digital_frequency], [-paband_ripple - topband_ripple*], 'r' emilogx([topband_digital_frequency topband_digital_frequency], [ -topband_ripple], 'r' emilogx([topband_digital_frequency pi], [-topband_ripple -topband_ripple], 'r' hold off axi([0 pi -topband_ripple* 0]
% There are everal method to produce the impule repone of the digital IIR filter. % One method i to feed a unit impule function into the filter. impule_repone_iir = filter(digital_numerator, digital_denominator, [ ero(,50]; figure(3, clf tem(impule_repone_iir % Another method i to ue Matlab' built-in command, `imp'. % `imp' alo accept additional input. Ue `help' to find out more. [impule_repone_iir, time] = imp(digital_numerator, digital_denominator; tem(impule_repone_iir % You may try different pecification. Other analog filter type, uch a, chebyhev type % I can alo be ued, imply by replacing `ellipord' and `ellip' with `chebord' and `cheby'.
M agnitude (db 0-0 -0-30 -40-50 -60 Magnitude repone of the digital filter 0 0.5.5.5 3 Digital frequency
0.5 0.4 0.3 0. 0. 0-0. Impule repone of the IIR filter -0. 0 5 0 5 0 5 30
Example : We wih to deign a digital filter which meet the following pecification: Low-pa filter: 0 to 0 kh paband Sampling frequency: f = 00 kh Tranition band: 0 kh to 0 kh Stopband attenuation : -0 db(starting at 0 kh The filter mut be monotonic in the pa and top band, (i.e. no ripple.
Step : The monotonic requirement indicate a Butterworth filter. f c 0,000 c 0. f 00,000 and fh h 0. 4 f
Step : We can calculate the correponding analogue filter frequencie: c T tan 0. f f tan(0..4530 h h tan(0. 0.64980 5 5
Step 3 : The required order of the Butterworth filter can be determined by uing the following equation, enuring at leat 0 db attenuation at H a ( N 0log ( H a h c N.453 0.6498.453 0log 0.6498 0 N.367 N
Thu we chooe N = (even. The pole occur in complex conjugate pair and lie on the left-hand ide of the -plane. 3 j 4 5 p ce 0.64980 ( 0.707j0.707 5 j 4 5 p ce 0.64980 ( 0.707j 0.707 H( p p p 5 9 p 4.30 0.99 0 9 4.3 0
Step 4 : Now apply the bilinear tranformation to obtain H( 5 0 0.0675 0.349 0.0675 H ( (.43 0.43
5.3.3 Digital -To-Digital Tranformation We have een in the lecture note that one method for the deign of analogue filter relied on applying a tranformation to an analogue low-pa filter with a unit bandwidth. It wa hown that we could obtain low-pa, high-pa, band-pa and band-top filter by electing the appropriate tranformation. Similarly, a et of tranformation can be formed that take a low-pa digital filter and turn it into highpa, band-pa, and band-top or another low-pa digital filter.
The tranformation are given below: 0 k Digital to Digital Tranformation 0 log H( p Low-pa to Low-pa in[( p p '/] in[( '/] p p 0 k 0 log H( p
0 k 0 log H( p Low-pa to High-pa 0 k 0 log H( Simplet tranformation i to change the ign of pole and ero of the LP LP tranformation co[( p p'/ ] co[( '/ ] p p p
0 k 0 log H( p co[( u L / ] co[( / ] u k L Lowpa to Band-pa cot(( / u L 0-3 db 0 log H( L k k k k k k k k p tan u
Lowpa to Band-top u L 0 log H( p k 0 0 log H( k k k k k k
5.4 Window Function Some of the mot commonly ued window function are: Rectangular Blackman Hanning 0.8 0.6 0.4 0. Rectangular 0 0 5 0 5 0 0.8 0.6 0.4 0. Blackman 0 0 5 0 5 0 0.8 0.6 0.4 0. Hanning 0 0 5 0 5 0 0.8 0.6 0.4 0. Bartlett 0 0 5 0 5 0 0.8 0.6 0.4 0. Hamming 0 0 5 0 5 0 0.8 0.6 0.4 0. Kaier, beta=4 0 0 5 0 5 0 Bartlett Hamming Kaier, beta=4
w[n] To Analye a truncation proce we model it a a multiplication of the deired equence by finite duration window equence denote by w[n]. Truncation of a equence [n] i equivalent to placing a rectangular time window around [n]. Frame Frame [n] peech ignal
Vocal tract hape change every 5m. When uing a ampling frequency of 8 kh (T = 5 with 00 ample in each frame (.5m, y n n w n In frequency domain, Y( S W ( (Multiplication become convolution in the frequency domain Thu when window i applied, the frequency domain convolution caue ditortion in the pectrum Y(. It can be hown that the rectangular window create ripple in the pectrum Y(. To reduce the ditortion, ue Hamming or Kaier Window.
Rectangular Window : N.0 w[n] n otherwie 0 0 N n w n ( 0... ( ( n w W N N N n n j jn e e e W W j ( (
W N N j e in N ( - -/N W( /N W N in in Main Lobe Side lobe or ripple
- -/N W( /N If N increae the width of the main lobe decreae but the peak amplitude of the ide lobe grow in a manner uch that the area under each lobe i contant while the width of each lobe decreae with N. Main Lobe Side lobe or ripple
5.5 Deign Method for FIR filter [] The mot eential feature of FIR filter i, by definition, the finite length of the impule repone. Another important point i that it can be eem directly from the impule repone of an FIR filter whether we have a linear phae characteritic or not. A filter i aid to have a linear phae repone of it repone atifie one of the following relationhip. ( a ( b a ( where a and b are contant.
( a ( b a ( It can be hown that for condition ( above to be atified the impule repone of the filter mut have poitive ymmetry. h N h n n where N denote the filter length., N a
When the condition given in ( i atified, the impule repone of the filter ha negative ymmetry. h n hn n, a N and b centre of ymmetry (Poitive ymmetry 0 4 6 8 0 N = 3 (odd n
0 0 4 6 8 centre of ymmetry (N even, poitive ymmetry N = (even n
5.5. Deign of FIR filter uing Window The eaiet way to obtain an FIR filter i to imply truncate the impule repone of an IIR filter. If h d [n] repreent the impule repone of a deired IIR filter, then an FIR filter with impule repone h[n] can be obtained a follow: h n h d 0 n N n N otherwie
h n h d 0 n N n In general h[n] can be thought of a being formed by the product h d [n] and a window function w[n] a follow: h n h d n w n H( H ( W ( d N otherwie let it be, for example a rectangular window let it be an ideal low pa filter with cut off frequency 0
H d ( * - 0 0 - - H( - - 0 0 /N
Therefore it i een that the convolution produce a meared verion of the ideal low pa frequency repone H d (. In general, the wider the main lobe of W(, the more preading, where a the narrower the main lobe (larger N the cloer H( come to H d (. In general we are left with a trade off of making N large enough o that mearing i minimied, yet mall enough to allow reaonable implementation.
% Exercie. FIR filter deign % We now conider the FIR filter deign. % The firt method we conider i the windowing method. % Matlab provide a command called `fir' to deign the FIR filter uing the windowing method. % We deign a FIR filter uing the ame pecification a in exercie. filter_order = 500; paband_norm_frequency = paband_frequency/ampling_frequency * *pi / pi topband_norm_frequency = topband_frequency/ampling_frequency * *pi / pi cutoff_frequency = paband_frequency/ampling_frequency * *pi / pi filter_coeff = fir(filter_order, cutoff_frequency; digital_frequency = (0:0.00:*pi; frequency_repone = freq(filter_coeff,, digital_frequency; magnitude_repone = 0*log0(ab(frequency_repone; figure(4, clf plot(digital_frequency/pi, magnitude_repone grid on title ('Magnitude repone of the digital filter' ylabel('magnitude (db' xlabel('normalied digital frequency' hold on emilogx([0 paband_norm_frequency], [-paband_ripple -paband_ripple], 'r' emilogx([paband_norm_frequency paband_norm_frequency], [-paband_ripple -topband_ripple*], 'r' emilogx([topband_norm_frequency topband_norm_frequency], [ -topband_ripple], 'r' emilogx([topband_norm_frequency ], [-topband_ripple -topband_ripple], 'r' hold off axi([0 -topband_ripple* 0] % Oberve that we can nearly meet the required pecification when the filter order i over 500. % By default the `fir' ue a hamming window. % There i a variety of window to chooe from. % You may experiment with different window and oberve it impact on the magnitude repone. % (You may need to reduce the filter order and not to worry about meeting the pecification.
Magnitude (db 0-0 -0-30 -40-50 -60 Magnitude repone of the digital filter 0 0. 0. 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Normalied digital frequency
% We now conider another method of deigning a FIR filter, which i baed on % the Park-McClellan optimal algorithm. % The command i available in Matlab and i called `reme'. % We deign a FIR filter uing the ame pecification a in exercie. filter_order = 3 % The lowet filter order that meet the required pecification. paband_norm_frequency = paband_frequency/ampling_frequency * *pi / pi topband_norm_frequency = topband_frequency/ampling_frequency * *pi / pi paband_delta = 0^(0./0- % Converting from db topband_delta = /0^(33.5/0 % Converting from db frequency_band = [0 paband_norm_frequency topband_norm_frequency ]; % The paband frequency i 0.pi and the topband frequency i 0.4pi. amplitude = [ 0 0]; % Thi amplitude configuration repreent a lowpa filter. weight = [/paband_delta /topband_delta] % Setting error weight at the paband and topband filter_coeff = reme(filter_order, frequency_band, amplitude digital_frequency = (0:0.0:*pi; frequency_repone = freq(filter_coeff,, digital_frequency; magnitude_repone = 0*log0(ab(frequency_repone; figure(5, clf plot(digital_frequency/pi, magnitude_repone grid on title ('Magnitude repone of the digital filter' ylabel('magnitude (db' xlabel('normalied digital frequency' hold on emilogx([0 paband_norm_frequency], [-paband_ripple -paband_ripple], 'r' emilogx([paband_norm_frequency paband_norm_frequency], [-paband_ripple -topband_ripple*], 'r' emilogx([topband_norm_frequency topband_norm_frequency], [ -topband_ripple], 'r' emilogx([topband_norm_frequency ], [-topband_ripple -topband_ripple], 'r' hold off axi([0 -topband_ripple* 0] % Note that given ome deign pecification, we find the filter order by trial-and-error % until the tranition bandwidth, paband and topband ripple atify the required % pecification. % Generally, a FIR filter require higher order than an IIR filter given the ame pecification. % You may try different filter order and oberve it impact on the tranition bandwidth, % paband and topband ripple. % Another method of deigning a FIR filter i to ue the leat-quare error minimiation algorithm. % The Matlab command to do thi i `firl'.
Magnitude (db 0-0 -0-30 -40-50 -60 Magnitude repone of the digital filter 0 0. 0. 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Normalied digital frequency
5.5. Deign Procedure [] An ideal low pa filter with linear phae of lope -β and cut-off ω c can be characteried in the frequency domain by H The correponding impule repone can be obtained by taking the invere Fourier tranform of and eaily hown to be d e ( 0 j h w c w n c h d n c in n n d H d
A caual FIR filter with impule repone h[n] can be n obtained by multiplying by a window beginning at the origin and ending at N- a follow : h d For h[n] to be a linear phae, βmut be elected o that the reulting h[n] i ymmetric. n A i ymmetric about n= βand the window i ymmetric about in c ( n h ( n n n in c ( n N n N. Symmetric about β
Example : (a Determine the impule repone of the lowpa filter whoe frequency repone i given by H( 0 (b To obtain a finite impule repone from rectangular window of length N = 9 i ued. Compute the coefficient of the FIR filter with a linear phae characteritic and with thi finite impule repone. 0 3 3 h d n n h d a
Solution: n n jn e d e H n h n j n j d d 3 3 3 in ( 0 / 3 in( 0 3 n n n n n h d (a
(b For a linear phae filter, n i ymmetric about n = 0 h d ( 9 n N 4 and the window i ymmetric about. n n n h h d h h h h h 0 3 4 0.333 3 3 4 0 3 8
The coefficient are n 3 8 0 0 3 4 3 3 0.333 4 5 3 Symmetry about n = 4 4 3 6 0 7 3 8 8
5.6 Frequency-Sampling Filter [] Although it i implied that all FIR filter are nonrecurive, thi i not the cae. To illutrate the approach, let u conider the following FIR filter having a caual finite-duration unit-ample repone containing N element of contant value. h n g 0 0 N N n 0 otherwie
The correponding filter ytem function i equal to 0 0 0 0 ( 0... ( g N N g N g N g H N N N p p N Comb filter H c ( Reonator H R0 (
Thi analytic form of the ytem function ugget a novel way to implement the above filter, a the twotage cacade tructure hown below. x[n] -N - Comb filter + + /N Reonator - g 0 y[n]
Example: Let N = 8 and g 0 = x[n] -8 0 3 - /8 + /8 /8 0 + H 8 ( -/8 8-8 y[n] 0 5 8 0 5 6 7 8 9 n n Filter implementation uing the comb filter reonator tructure. /8
H c ( i given a H c ( N N c i hown in the next lide for N = 8. ha N ero equally paced over the frequency range 0, Becaue the magnitude repone reemble a comb, thi filter ha become known a a comb filter. j N N e j j N H ( e in c N N H ( ( H C j ( 8 e 8 j 8 in 4 je j 4 H c in 4
/4 H c ( H c ( in 4 4 0 /8 H c ( v for a Comb filter The pole/ero pattern for the comb filter can be obtained uing the tranfer function H c ( N The above equation explicitly indicate that there are N equally paced ero over the unit circle with the firt ero at = 0. N N N N N N k 0 ( e k j N
e e j 3 j 4 e j e j e j 4 Pole/ero pattern for Comb filter N = 8 8 pole
HR 0 ( i the ytem function of a reonator, or a filter that ha pole on the unit circle at frequency = 0 i.e. at =. Since thi pole i not trictly within the unit circle, the filter i not table. When thi reonator i cacaded with the comb filter, however, the ero of the comb filter located at = cancel the reonator pole, making the combination table. The tranfer function of thi two-tage filter i equal to j N N g0 e g j 0 H( e j N e N N in in
in in ( 0 0 N e N g e e N g H N j j N j ( N g H N o in in ( 0 N e N g H N j The tranfer function of thi two-tage filter i equal to
Example : For N=8, H( ha a maximum at =0, equal g 0 and i equal to ero at k. k N g 0 H( Magnitude repone of Comb filter plu reonator (N = 8 - /N Let u conider a econd-order reonator, whoe coefficient are real-valued, and the pole are ituated at the following ero location of the comb filter.
N N 8 pole at the origin e e j 3 j 4 3 4 e j e j e j (Let N = 8 e j 4 4 e j
The ytem function of the econd-order reonator can be written in parallel form a Thi procedure can be generalied to include each econd-order reonator cancelling a pair of ero of the comb filter. Thee reonator are all connected in parallel and thi parallel combination i then connected in cacade with the comb filter to produce the total filter H T ( given by e g e g H N j N j R 0 ( N k N k j k N T e g N H
- -N Y( X( /N 0 g e g e g N j N j e g e g k N j k N k j k 4 4 e g e g N j N j + k N k j g e H k N k j g e H
Example: Let u conider implementing the linear interpolator with the comb and reonator tructure. The impule repone of the linear interpolator i given by h h 0 ; h ; h n 0 otherwie and
Solution: Since the number of element in the unit ample repone i equal to 3, we chooe N = 3. The comb filter ytem function i then H( H( H c j e ( co ( 3 3 H c ( ha three ero located at for k = 0, &. 3 e k j
The ero at = will be cancelled by a real reonator, and ero at will be cancelled by a pair of complex reonator. The gain of the reonator are equal to 3 j e 3 3 3 3 0 * 3 3 4 3 4 3 3 0 0 ( (, ( (, ( e e e e H H g e e H g e H g H g j j j j R R j j j
x[n] -3 - + -/3 + - - - - - + + - -/ + 0 /3 0 y[n] / 0 / 3 4 /3 /6 -/6
Note: If the comb filter i followed by a recurive network that ha a number of pole that coincide exactly with the ame number of ero of the comb filter, we obtain a frequency-ampling filter. The frequencyampling filter can be particularly attractive olution if we want to make a narrow-band filter; the N i large, while the number of recurive ection can be very mall. Thi mean that the filter contain only a few multiplier and adder.
For practical application we have to keep an eye on a number of thing. In theory we tart by auming that a number of pole and ero coincide exactly on the unit circle in the -plane. Thi require that we hould be able to realie the filter coefficient to an accuracy of 00%. However, apart from a few obviou exception uch a -,0 and + thi i never the cae. Thi mean that while we can locate the ero of the combfilter ection in exactly the right poition, we cannot do o far the correponding pole. The bet we can do i to get them in the vicinity. Thi can lead to a very erratic local variation of the frequency repone, or even an untable ytem if one of the pole lie jut outide the unit circle. In practice both ero and pole are therefore deliberately located jut inide the unit circle. For comb filter the ytem N function i then choen a H ( ( a where a i made lightly le than.
Problem Sheet A5 [,] Q. Conider the following analogue ytem with a tranfer function t H( h( t e where a = 04 rad/ec and the ampling period T i 00. (i What i the dc gain of H(? (ii At what frequency i the H( equal to ero? (digital frequency. (iii Calculate the impule repone h[n]. (iv Auming that the impule repone decay to /e of it initial value at n = N ample, how that: N a ln a ln( a
Q. A digital low-pa filter i required to meet the following pecification: Paband ripple db (peak to peak ripple Paband edge : 4kH Stopband attenuation : 40dB Stopband edge : 6kH Sampling rate : 4kH The filter i to be deired by performing a bilinear tranformation on an analogue filter. Determine what order Chebyhev deign mut be ued to meet the pecification in the digital implementation. (Hint: ue the deign formula given in the lecture note {An: n = 6}.
Q3. The block diagram how a digital ocillator [n] which i initialied by an impule a hown in the figure below. The deired unit impule repone i h n A n u n -b -b Z - Z - co( 0 a 0 + + a a a a a 0 ( b b H
(a Auming 0 i the reonant frequency of the digital ocillator and writing appropriate equation find the value of a 0, a, a, b and b. (An: a 0 =A, a = -Aco 0, a =0, b = co= co 0, b =
(b The frequency of ocillation i determined by the coefficient b. Show that f where f i the ampling frequency (6kH and f 0 i the deired frequency of ocillation. f b f 0 4in f Calculate the highet frequency reolution that i obtainable and how that thi obtainable accuracy depend on the deired ocillation frequency. You may aume that b i repreented by a K-bit number (fractional arithmetic and i given by b K 0 and b where K = 6. K f 0. 0 0777H (An:
(c Uing the equation obtained and the value in part (a and (b above, how that the lowet ocillation frequency obtainable i given by f f0 min for 56 f0 f x You may aume that co( x, given mall f 0.
(d For the tructure hown in figure below, write down the appropriate difference equation and hence tate the function of thi tructure. - - + y [n] - co + - in y [n]
Q4. (a A econd-order analogue band pa filter with an -domain tranfer function i given by, bp H( b Where ω p and b p are the centre frequency and bandwidth of the filter; repectively, both expreed in rad/. By applying the bilinear tranformation to equation ( a digital filter with the following tranfer function can be obtained: H a b ( 0 b p a p (. (.
0 4 4 4 8 4 T T b T b T b T T b T b T T b T b a a p p p p p p p p p p Show that the digital filter coefficient are given by
(b A digital filter with a centre frequency of 000 H and a bandwidth of 50 H i required. Auming a ampling frequency of 0kH, compute the digital filter coefficient a 0, a, b & b and how that a 0 = a = 0.044095; b = -.55 and b = 0.9876 (c Comment on the tability of the digital filter H( (ee equation which you have obtained. H a b a 0 ( b (. (An : Stable
Q5. A econd-order reonant ytem i given by, H( p p Q p By applying the impule invariant tranformation to the above equation, a digital filter with the following tranfer function can be obtained. H k b p Bandwidth ( b b p Q p reonant frequency Q-factor Aociated with the pole
Show that the filter coefficient b and b are given by, Alo how that, where p T pt b e co( qt; b e K p T 4qe in( q T p p and q p 4Qp Q p
Q6. (a An FIR filter ha an impule repone hown in the figure below. h[n] 0 3 4 5 n Show that the frequency repone of thi filter i given by 5 H( e j ( in in where i the digital frequency.
(b Draw an alternate implementation of thi filter uing pole and ero configuration. (c Draw an FIR filter tructure that ha the ame impule repone a the figure below. State clearly the value of the FIR filter coefficient. X( - b - + c d e Y( +
Q7. (a Determine the impule repone h d [n] of the bandpa filter whoe frequency repone i given by H d j 3 e ( 4 4 0 otherwie (b To obtain a finite impule repone from h d [n] a Bartlett window of length N = 7 i ued. Compute the coefficient of the FIR filter with thi impule repone. 3
Note: The Bartlett Window function i given by otherwie 0 0 N n N N n N n N n n w B Where N i the length of the window.
An : An: h d [n] n h d h d [0] 0 n 3 ( n in( 3 in( 3 3 3 n 4 n 4 n n 3 0 3 0 3 3 4 after windowing 0 5 6
Q8. (i (a Find the tranfer function for the following: X( (a If (b If H ( (b Draw a pole-ero diagram (c Sketch the magnitude frequency repone. H(. H (. 5 H ( 3 H 3 /5 H ( e j 5 j 5, ketch the magnitude repone 3 ( 3 3 5, ketch the magnitude repone of e 3 5 Y(
Q9. A frequency ampling filter i hown below & N = 3 X( - -N /N a e 0 j N (a Determine a 0, a & a uch that thi filter ha a real impule repone h[n], where a e a j N + Y(
3 3 6 3 ( and 3 0 ( j H H k N k j k N k j a e H a e H ( ;
(b Draw the frequency-ampling filter tructure uing delay element, multiplier and adder. (c Derive a general expreion for H(. (d Give a filter that ha ame frequency repone H(, but i realied an FIR filter.
Summary of Part A Chapter 5 At the end of thi chapter, it i expected that you hould know: The propertie of FIR and IIR filter, o that you can jutify your choice of filter type for a given problem. The linear phae property of ymmetric FIR filter i particularly important. Deign technique, the difference between them, and why each are ued. FIR: windowing and frequency ampling. IIR: impule invariant tranformation (when and why it i ued, bilinear tranformation (when and why it i ued and pole-ero placement. Prewarping and it role in bilinear tranformation baed deign.
Given a digital filter pecification, how to deign an IIR digital filter uing an analog prototype. Digital to digital tranformation, epecially lowpa to lowpa, and lowpa to highpa. Type of window function, the difference between them, and the fundamental trade-off between time and frequency reolution. Student hould undertand the importance of topband attenuation, and the role of the main lobe width in the definition of frequency reolution.
Comb filter a a frequency ampling filter. Comb filter a a particular cae of FIR filter deign. How to draw comb filter tructure, and limitation aociated with the comb filter tructure.