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Discrete - Time Signals and Systems Z-Transform-FIR filters Yogananda Isukapalli FIR filters-Review M y[]nbx= å k [n - k]...general difference equation for FIR filters k =0 Filter Order = M: No. of memory blocks required in the filter implementation Filter Length, L = M+1: Total No. of samples required in calculating the output, M from memory (past) and one present sample Filter coefficients {bk}: Completely defines an FIR filter. All the properties of the filter can be understood through the coefficients Z- transform of impulse response h[] n results in 'transfer function', it is also known as 'system function' Hz( )[]= å hnz-n n From the previous lecture recall that Transfer function Yz() Hz()= Xz() Þ Yz( )()= HX z ()z Notice the mathematical simplicity of the above result Convolution becomes a simple multiplicatio n hn[]*« xn []() H z Xz() Calculating the output of a FIR filter using Z- transforms Steps involved 1) Find the Z- transform of input signal x[ n ] ¥ Xz()= å xnz []-n n=-¥ 2) Find the Z- transform of impulse response h[ n ] M Hz()= å h [nz] -n n=0 3) Multiply X( z ) and H(z ) to get Y() z 4) Obtain output y[] n by applying inverse Z- transform to Y ()z --1 yn[]¬¾! ¾® Y( z) Why to operate in transforms? Z-TRANSFORM-DOMAIN M Hz( )= å hnz [ ] -n n=0 {bk } TIME-DOMAIN FREQ-DOMAIN Fig 12.1 M M jwˆ jwˆ - jwˆ k y[n] = b x[n - k] He()= Hz () ˆ H (e ) = bke å k ze= jw å k=0 k=0 \ z= ejwˆ , makes both domains equal x[n] h[n] = d [n] -d [n -1] y[n] x[n] y[n] jwˆ - jwˆ n x[n] jwˆ - jwˆ y[n] H (e ) = åh[n]e H (e ) = 1- e n -n x[n] -1 y[n] Hz()= å hnz [] Hz()= 1- z n Fig 12.2 Same output from all three domains Computational complexity determines the domain Some of the filter properties are better understood in frequency or Z-domains Z and frequency transforms are related Example 1 Calculating the transfer function of the FIR filter with impulse response co- efficients given by, {[]}{2,0,3,0,2}hn = - {[]}{2,0,3,hn = - 0, 2} 4 Hz()= å hnz [] -n n=0 Hz( )=+ h [0] h[1] z----1234 + h [2] z + h [3]zhz + [4] =+2 0zzzz--12-- 3 0 - 3+ 2 - 4 = 2- 3z--2 + 2z 4 Example 2 Calculating the transfer function of the FIR filter described by the difference equation yn[]=+ xn [] 2[ xn -1]-3[ xn -2]-4[ xn -3] hn[]= bk hn[]= { 1,2,-3,-4} 3 H (zn)[= å hz] -n n=0 Hz( )=+ h [0] h [1]zhzhz---123 + [2] + [3] =+1 2z-12--34z--z 3 Example 3 Calculating the output of the FIR filter xn[ ]=+dd [ n -1]- [ n - 2] dd [ n -3]- [ n - 4] hn[ ]=+dd [ n ] 2 [ n -1] + 3 d [ n - 2] + 4 d [ n -3] z -n0 !d[-nn0 ]¬¾® z Xz()= zzzz--1234- - + - - Hz()=+1324z-12 +zz--+ 3 !Yz()= XzHz () () =( zzzz----1234- + - )( 1++24zz - 13z- 2+ - 3) =zz--12+ (- 1++ 2) (1- 2++ 3) z - 3 (- 1+ 2- 3+ 4) z - 4 +(- 2+ 3- 4)zzz---567+ (- 3++ 4) (- 4) =z--12++zzzzzz 2 - 3 + 2 - 4- 3- 56+ -- 4 - 7 Apply the inverse Z- transform --1 -n0 Z ! znn¬¾ ¾® d[-0 ] Yz()=+ z-1 z-2 + 2 z-34567+ 23zzzz----- + - 4 yn[ ]=++dd [ n -1] [ n - 2] 2 d [ n -3] + 2 d [ n - 4]- 3 d [ n -5] +dd[nn - 6]- 4 [ - 7] Example 4 Find the impulse response of the FIR filter xn[ ]= d [ n - 2] yn[ ]=+dd [ n ] 2 [ n -1] + 3 d [ n - 2] + 4 d [ n -3] z -n0 !d[-nn0 ]¬¾® z Xz()= z-2 Yz()=+ 1 2 z--1 ++ 3z 23 4z - Yz() ! Hz()= Xz() 12++zzz-12 3--+ 4 3 =234=+z2 z + + z-1 z-2 h[ n ]=++++dd [ n 2] 2dd [ n 1] 3 [ n ]+ 4 [ n-1]... non causal Example 5 Calculating the output of the FIR filter An 0££ 4 xn[]= 0 elsewhere n 1 2 0££n 3 hn[]= ( ) 0 elsewhere ¥ Xz()= å xnz []-n n=-¥ 4 ==+++++å Az-----n A(1 z1234 z z z ) n=0 ¥ Hz()= å hnz []-n n=-¥ 3 n ==å (1 2) z-n n=0 =+(1 (12) zzz---12+ ( 14) + (1 8)) 3 !Yz()= HzXz () () =(1Az+++++----1234 z z z )(112 +( ) z - 1 +( 14) z - 2 +( 18) z - 3 ) =+Az(1( 3 2) --12 +( 7 4) z +( 15 8) z - 3 +( 15 8) z - 4 + (7 8) z---56+ ( 3 8) zz+ (18) 7 ) yn[] = A(dd[nn ]+ ( 3 2) [- 1]+ ( 7 4) d [ n- 2]+ ( 15 8) d [ n- 3] +( 15 8)ddd [nnn- 4]+ ( 7 8) [- 5]+ ( 3 8) [- 6] + (1 8)d [n - 7]) Cascading Systems Fig 12.3 x[ n ]... input signal to the 1st FIRfilter h[ n ]... impulse response of the 1st FIRfilter w[ n ]... output of the 1st FIR filter and input to the 2 nd FIR filter y[ n ]... output of the 2nd filter, also this is the overall output Fig 12.4 z hn[]= h1 [ n][* h2 n]¬¾® Hz()= H12 () zH( z) Cascaded system can be replaced by a single filter with system function H() z Example 1 The impulse responses in a cascaded system are hn1[]= dd [] n-- [ n 1] hn2[]=+dd [] n [ n- 1] Find the impulse response of an effective system that can replace the cascading arrangement z -n0 !d[-nn0 ]¬¾® z zz hn1122[ ]¬¾®Hz(), h[ n] ¬¾® H () z -1 Hz1()= 1- z -1 Hz2 ()=+ 1 z --11 - 2 H (z) ==HzHz12() ()= ( 1- z)( 1+ z) 1- z hn[]= d [nn]-d[- 2] Example 2 Consider a system described by the difference equations wn[]= 3[] xn-- xn [ 1] yn[]= 2[] wn-- wn [ 1] Find the impulse response of an effective system that can replace the cascading arrangement hn11[]== bk () [3,1]- hn1[]= 3[]dd n-- [ n 1] hn22[]== bk () [2,1]- hn2[]= 2[]dd n-- [ n 1] z -n0 !d[-nn0 ]¬¾® z zz h1122[nhn ]¬¾® Hz(), [ ]¬¾® H () z -1 Hz1()= 3- z -1 Hz2 ()= 2- z --11 Hz()= HzHz12() ()= ( 3-- z)( 2 z ) --12 =+6 - 5zz hn[ ]= 6dd [ n ]-- 5 [ n 1]+ d [ n- 2] The system can now be expressed as one difference equation y[nxnxnxn ]= 6 [ ]-- 5 [ 1]+ [- 2] Example 3 The impulse response of an effective system, hn[ ]= dd [ n ]-- 2 [ n 1]+ 2 d [ n-- 2] d [ n - 3] Split the above filter into two cascaded filters such that the 1st system is described by, wn[]= xn []-- xn [ 1] z -n0 !d[-nn0 ]¬¾® z Hz()= 1- 2 z---123+ 2 z- z hn1[]= dd [] n-- [ n 1] -1 Hz1()= 1- z ! Hz()= H12 () zH () z Hz() Hz2 ()= Hz1() 12- zzz---123+ 2 - = 1- z-1 =1- zz--12+ h2[nnn ]= dd [ ]-- [ 1]+ d [ n- 2] yn[ ]= wn [ ]- wn [-- 1]+ w [n 2]ü ýComplete system wn[]= x []nxn- [1]- þ Fig 12.5 Example 4: Deconvolution Cascading of filters has an important practical application Undoing the effects of the first filter Example: Communication channel Undoing the effects of channel on signal is called equalization Yz()= H1 ()z HzXz2 () () if Y() z= X () z Þ HzHz12() ()= 1 -1 Assume H1 ( z )= 1- z 1 Hz2 ()= 1- z-1 Z-Transform & Unit circle The frequency or wˆ -- domain is a subset of z domain The general expression for Z is, z= re jwˆ , where, ' r ' is the radius of the circle z= ejwˆ , makes both domains equal jwˆ He()= Hz () ˆ ze= jw ˆ ! z= ejw = 1, the unit circle has a unique significance in the z- domain Gray dots, z= {- j,1, j} ìüpp wˆ = íý- ,0, îþ22 General Z- plane z= ejwˆ , special case z- domain= wˆ - domain Fig 12.6 Note that 'wˆ ' is the discrete- time frequency discussed in previous lectures -pwp££¬ˆ ¾® -11 £z £ Periodicity in wpˆ --- domain2 radians, one cycle in z - domain Zeros & Poles Consider the transfer function of an FIR filter Hz()= 1- 2 z---12+ 2 z- z3 Convert the above function as a polynomial in ' z ' (12- zz-1233+ 2--- zz) Hz( ) = z3 z32- 221zz+ - = z3 write the numerator in factored form, z3 - 221(1)()()zz233+ - = zzeze--jjpp -- (1)(zzeze--jjpp33 )( -- ) H ()z = z3 Zeros....The values of z for which H()0 z = In this example, H( z )== 0 for z{1,, ejjpp33 e- } Poles....The values of z for which H() z ® ¥ H( z )® ¥ for z = {0,0,0} note that z3 = 0 results in 3 roots It is also known as a 3rd order pole at z = 0 Significance of zeros in an p j FIR system : 3 ze2 = Except for a constant FIR system is completely characterized by the zeros 3 poles z1 = 1 The difference equation which desribes the relation between x[ n ] and y[n ], can p be found with the knowledge - j 3 ze3 = of zero locations Fig 12.7 Example 1 Find the poles and zeros for the transfer function? Hz( ) = 13- z--12+ 2 z Converting into a polynomial in '' z zz2 - 32+ Hz()= z2 write the numerator in factored form, zz2 - 3+= 2 ( z-- 2)( z 1) zeros==[ z1,][2, z2 1] nd polesp= [,1 p2 ]== [0,0] 2 order pole at z =0 Example 1 continued… Fig 12.8 Notice that both the zeros are on real axis, also notice one pole is not included in the unit circle. There is a double pole at z=0 Example 2 Find the poles and zeros of an FIR system described by the difference equation, yn[]= xn []--x [n 1]+ x [n - 2] Theimpulseresponsefunctionisgivenby hn[ ]= dd [ n ]-- [ n 1]+ d [ n- 2] z -n0 !d[-nn0 ]¬¾® z Hz( )1= - z--12+ z Converting into a polynomial in '' z zz2 - +1 Hz()= z2 write the numerator in factored form, zz233- +=1( ze--jjpp )( ze- ) zeros= [, ejjpp33 e- ], both on unit circle poles==[0,0] 2nd order pole at z =0 p 3 2 poles - p 3 Fig 12.9 Example 3 The zeros and poles of an FIR filter are given, zeros= [, ejjpp4 e- 4 ] poles = [0,0] Find the difference equation of the filter The numerator is obtained through multiplication of factors ()(ze--jjp 442 ze- p ) = z- 2 z+1 denominator through poles, z2 zz2 - 21+ Hz()= z2 = 1- 2zz--12+ Apply inverse Z- transform, hn[]= ddd [] n-- 2[ n 1]+ [ n- 2] The difference equation yn[]= xn []-- 2[ xn 1]+ xn [- 2] Notice that, because of conjugate zeros a sinusoid signal with p 4 frequency p 4 will be nulled out by this FIR filter Fig 12.10 Example 4, Application: Nulling filters The Graphical Design of a Comb filter • In medical applications, the 60Hz frequency of the power supply is often “picked up” by the test equipment (EKG recorder) • Also harmonically related frequencies such as f2 = 2x60 = 120Hz, and f3 = 3x60 = 180 Hz are generated because of non-linear phenomena.
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    2065-27 Advanced Training Course on FPGA Design and VHDL for Hardware Simulation and Synthesis 26 October - 20 November, 2009 Digital Signal Processing Filter Design Massimiliano Nolich DEEI Facolta' di Ingegneria Universita' degli Studi di Trieste via Valerio, 10, 34127 Trieste Italy Filter design 1 Design considerations: a framework |H(f)| 1 C ıp 1 ıp ıs f 0 fp fs Passband Transition Stopband band The design of a digital filter involves five steps: Specification: The characteristics of the filter often have to be specified in the frequency domain. For example, for frequency selective filters (lowpass, highpass, bandpass, etc.) the specification usually involves tolerance limits as shown above. Coefficient calculation: Approximation methods have to be used to calculate the values hŒk for a FIR implementation, or ak, bk for an IIR implementation. Equivalently, this involves finding a filter which has H.z/ satisfying the requirements. Realisation: This involves converting H.z/ into a suitable filter structure. Block or flow diagrams are often used to depict filter structures, and show the computational procedure for implementing the digital filter. 1 Analysis of finite wordlength effects: In practice one should check that the quantisation used in the implementation does not degrade the performance of the filter to a point where it is unusable. Implementation: The filter is implemented in software or hardware. The criteria for selecting the implementation method involve issues such as real-time performance, complexity, processing requirements, and availability of equipment. 2 Finite impulse response (FIR) filter design A FIR filter is characterised by the equations N 1 yŒn D hŒkxŒn k kXD0 N 1 H.z/ D hŒkzk: kXD0 The following are useful properties of FIR filters: They are always stable — the system function contains no poles.
  • Control System Design Methods

    Control System Design Methods

    Christiansen-Sec.19.qxd 06:08:2004 6:43 PM Page 19.1 The Electronics Engineers' Handbook, 5th Edition McGraw-Hill, Section 19, pp. 19.1-19.30, 2005. SECTION 19 CONTROL SYSTEMS Control is used to modify the behavior of a system so it behaves in a specific desirable way over time. For example, we may want the speed of a car on the highway to remain as close as possible to 60 miles per hour in spite of possible hills or adverse wind; or we may want an aircraft to follow a desired altitude, heading, and velocity profile independent of wind gusts; or we may want the temperature and pressure in a reactor vessel in a chemical process plant to be maintained at desired levels. All these are being accomplished today by control methods and the above are examples of what automatic control systems are designed to do, without human intervention. Control is used whenever quantities such as speed, altitude, temperature, or voltage must be made to behave in some desirable way over time. This section provides an introduction to control system design methods. P.A., Z.G. In This Section: CHAPTER 19.1 CONTROL SYSTEM DESIGN 19.3 INTRODUCTION 19.3 Proportional-Integral-Derivative Control 19.3 The Role of Control Theory 19.4 MATHEMATICAL DESCRIPTIONS 19.4 Linear Differential Equations 19.4 State Variable Descriptions 19.5 Transfer Functions 19.7 Frequency Response 19.9 ANALYSIS OF DYNAMICAL BEHAVIOR 19.10 System Response, Modes and Stability 19.10 Response of First and Second Order Systems 19.11 Transient Response Performance Specifications for a Second Order