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.