INVERSE Z TRANSFORM 8 INVERSE Z TRANSFORM 113 8 INVERSE Z TRANSFORM The process by which a Z transform of a time –series kx , namely X(z), is returned to the time domain is called the inverse Z[.]
8 INVERSE Z-TRANSFORM INVERSE Z-TRANSFORM The process by which a Z-transform of a time –series xk , namely X(z), is returned to the time domain is called the inverse Z-transform The inverse Z-transform is defined by: xk Z 1 Xz Computer study M-file iztrans.m is used to find inverse Z-transform Example 8.1 » syms z k z X( z ) » x=z/(z^2+5*z+6); z 5z » iztrans(x) xk 2 U s k 3 U s k k k ans = (-2)^k-(-3)^k Four practican techniques can be used to implement an inverse transform They are: Long divisions Partial fractions Residue theorem Difference equations 8.1 Inverse Z-transforms via long division For causal sequences, the z-transform X(z) can be expended into a power serises in z 1 For a rational X(z), a convenient way to determine the power serise is an expansion by long division b z n b n 1z n 1 b1 z b X( z ) n n c c1z 1 c z 2 z a n 1z n 1 a 1z a where c , c1 , c are power serise coefficients z Example 8.2 X( z ) z 1.414z 1 z z-1.414+z-1 1.414-z-1 1.414-2z-1+1.414z-2 z-1-1.414z-2 z-1-1.414z-2+z-3 -z-3 -z-3+1.414z-4-z-5 z2-1.414z+1 z-1+1.414z-2+z-3-z-5 X[k]=[k-1]+1.414[k-2]+ [k-3]+0-[k-5] 113 INVERSE Z-TRANSFORM Computer study The inverse of a rational z-transform can also be readıly calculated using MATLAB The function impz can be utilized for this purpose Three versions of this function are as follows: [h,t]=impz(num,den) [h,t]=impz(num,den, L) [h,t]=impz(num,den, L, FT) Where the input data consists of the vector num and den containing the coefficients of the numerator and the denominator polynomials of the z-transform given in the descending powers of z, the output impulse response vector h, and the time index vector t The first form, the length L of h is determined automatically by the computer with t=0:L-1, whereas in the remaining two forms it is supplied by the user through the input data L In the last form, the sampling interval is The default value of FT is The following two FT examples show application h, t impznum, den file to and plot power X ( z) Example 8.3 » num=[1 0]; » den=[1 -1.414 +1]; » L=8; » [x,k]=impz(num,den,L) x= 1.0000 1.4140 0.9994 -0.0009 -1.0006 -1.4140 -0.9988 0.0017 k= » stem(k,x,’fill’,’k’) z z 1.414 z Power series coefficiens for X ( z) z z 1.414 z 1.5 0.5 -0.5 -1 -1.5 k 114 INVERSE Z-TRANSFORM X ( z) z z 1 z 3z Example 8.4 » num=[1 -1]; » den= [2 1]; » L=10; » [x,t]=impz(num,den,L) x= 0.5000 -0.2500 -0.3750 0.6875 -0.8438 0.9219 -0.9609 0.9805 -0.9902 0.9951 » stem(k,x,’fill’,’k’) z z 1 Power series coefficients for X ( z ) 2 z 3z 1 0.8 0.6 0.4 0.2 -0.2 -0.4 -0.6 -0.8 -1 8.2 The Inverse Z-Transform Using Partial Fractions We now derive the expression for the inverse z-transform and outline the two methods for its computation Recall that, for z re j , the z-transform G(z) given by the equation is merely the Fourier transform of the modified sequence gk r k Accordingly, by the inverse Fourier transform, we have: k gk r G (re j )e jk d (8.2) 2 By making the change of variable z re j , the above equation can be converted into a contour integral given by : gk residues of G(z)z k 1at the poles inside C (8.3) Note that theequation mentioned above needs to be equated at all values of k which can be quite complicated in most cases A rational G(z) can be expressed as: P( z) G (z) D( z ) 115 INVERSE Z-TRANSFORM where P(z) and D(z) are the polynomials in z 1 If the degree M of the numerator polynomial P(z) is grester than or equal to the degree N of the denominator polynomial D(z), we can divide P(z) by D(z) and re-express G(z) as: M N P ( z) G ( z ) z k D( z ) i 0 where the degree of the polynomial P1 (z) is less that that of D(z) The rational function P1(z) is called a proper fraction D( z ) The expression of Eq (8.2) can be computed in a number of ways Consider the following cases: Case 1: G(z) is a proper fraction with simple poles Let the poles of G(z) be at z p k , k=1,2,3,……,N, where pk are distinct A partial-fraction expansion of G(z) then is of the form : N az (8.4) G (z) i k 1 z p i where the constants l in the above expression, called the residues, are given by: G (z) (8.5) a i (z p i ) z z pi Each term of the sum on the right-hand side of Eq.(8.4) has an ROC given by z pk, therefore, the inverse transform g[k] of G(z) is given by N gk a i p i U s k k k 1 Note that the above approach with slight modifications can also be used to determine the inverse z-transform of a noncausal sequence with a rational z-transform Example 8.5 Let the z-transform of a causal sequence g[k] be given by : z(z 2.0) (z 0.2)(z 0.6) az a 2z G (z) z 0.2 z 0.6 z 2.0 2.2 a (z 0.2) 2.75 (z 0.2)(z 0.6) z 0.2 0.8 G (z) a (z 0.6) (z 2.0) (z 0.2)(z 0.6) z 0.6 1.4 1.75 0.8 gk 2.75(0.2) k 1.75(0.6) k U s k Example 8.6 Using MATLAB determine the partial fraction expansion of X(z): X( z ) 3z 12 2z 3.5z 1.5 116 INVERSE Z-TRANSFORM num=[3 0 12];den=[2 -3.5 -1.5]; » [r,p,k]=residuez(num,den) r= 1.9219 3.7891 - 0.3013i 3.7891 + 0.3013i p= 1.9477 -0.0989 - 0.6126i 3.6 2.625 -0.0989 + 0.6126i 2.47 X( z) k= z 1.5 z z 0.5 -8 » r=[ 1.9219 3.7891 - 0.3013i + 0.3013i]; 1.5z 3.7891 6z (z) -3 0.6126i -0.0989 + » p=[ 1.9477 X -0.0989 z 1.75z 0.75 0.6126i]; » [num,den]=residuez(r,p,k) num = 1.5001 -0.0000 0.0008 5.9999 den = 1.0000 -1.7499 -0.0000 -0.7500 Multiplying the numerator and the denominator by G (z) 3z 12 2z 3.5z 1.5 Case G(z) has multiple poles, for example, if the pole at z is of multiplicity r and the remaining N-r poles are simple and at z p k , k 1,2,3, N r, then the general partial-fraction expansion of G(z) takes the form G (z) MN a k 0 Nr k z k k i r akz z a ri z p k i 1 (1 z 1 ) i where the constant ari (no longer called the residues for i1) are computed using the formula: d r i G (z) a ri ( z ) r , i=1,2,3,… ,r r i (r i)! d(z) z z Example 8.7 a z a z z2 X(z) 21 22 X(z) ; (z 1) (z 1) (z 1) a a 22 X( z ) 21 ; z (z 1) (z 1) a0 zX(z) 0; z d z 1 X(z) d z 1 dz z dz z 1 z 1 a 22 a 21 117 (z 1) X(z) 1 z z 1 INVERSE Z-TRANSFORM Which results in the following time-series xk 1 k u s k z Example 8.8 G (z) (z 0.5)(z 1) G ( z) G (z) a (z 0.5) 4; a 22 (z 1) 2; z z 0.5 z z 1 a 21 d G (z) ( z ) 4 dz z z 1 0.5 z z z 4 2 z 0.5 z 1 (z 1) Consider the following three cases: 1) z1 g[k] 4(0.5) k Uk 4Uk 2kUk 2) z U s k U s k 1 G ( z) 1.0 z1 0.5 g[k] 4(0.5) k U s k 1 4U s k 1 2kU s k 1 3) z1 g[k] 4(0.5) k U s k 4U s k 1 2kU s k 1 1.0 z 0.5 1.0 Example 8.9 3z 5z 3z (z 1) (z 0.5) a z a z a z X(z) 11 21 22 z 0.5 z (z 1) X( z) z1 where a 11z /( z 0.5) an exponential, a 21z /( z 1) a step function, and a 22z /( z 1) a ramp function What is desired, however, is the partial fraction expansion of X(z)/z, where: a 11 a a 22 X( z) 21 z z 0.5 z (z 1) where a1 (z 0.5)X(z) 3z 5z 3z z z(z 1) z 0.5 a 22 (z 1) X(z) 3z 5z 3z z z(z / 2) z 1 5 z 0.5 2 z 1 9z 10 2(3z 5z 3)(z / 4) d (z 1) X(z) 2 dz z z ( z / ) ( z ( z / )) z 1 z 1 which results in a 21 118 INVERSE Z-TRANSFORM xk 5(0.5) k 2k u s k Example 8.10 18z Solve using Matlab: H( z ) 18z 3z 4z 0.24 0.4 0.36 H( z ) 1 1 z 0.5 0.33z (1 0.33z ) » num=[18]; den=[18 -4 -1]; » [r,p,k]=residuez(num,den) r= 0.2400 0.4000 0.3600 p= -0.3333 -0.3333 0.5000 k =[] » [num,den]=residuez(r,p,k) num = 1.0000 0.0000 0.0000 den = 1.0000 0.1667 -0.2222 -0.0556 Using the numerator and the denominator coefficients we have: X( z ) z3 z 0.1667z 0.2222z 0.0556 It can be seen that the coefficients will be same as in the equation of the question if we multiply each coefficient by 18 Example 8.11 Find the inverse Z-transform of X( z) (z 1) (z 0.5)(z 0.5) a1 a 21 a 22 X( z) a z z (z 0.5) (z 0.5) (z 0.5) a0 zX(z) z 0 z 119 INVERSE Z-TRANSFORM (z 0.5)X(z) a1 z a 22 (z 0.5)(z 1) z 0.5 (z 0.5) (z 0.5) X(z) z z 0.5 (z 1) (z 0.5) z 0.5 z 0.5 27 d (z 0.5) X(z) d (z 1) 27 z 0.5 z 0.5 dz z dz (z 0.5) X( z ) z 27 z 27 z 8 z (z 0.5) (z 0.5) (z 0.5) a 21 27 27 1 xk 8k (0.5) k (0.5) k k (0.5) k U s k 4 4 » num=[1 3 1]; den=poly([0 -0.5 0.5 0.5]) den = 1.0000 -0.5000 -0.2500 0.1250 » [r,p,k]=residue(num,den) r= -0.2500 -6.7500 6.7500 8.0000 p= -0.5000 0.5000 0.5000 k = [] Case X(z) has a complex pole Example 8.12 The second-order X(z) (3z 1.5z) /( z cos( )z / 4) has non repeated complex roots The partial expansion of X(z) is defined by: z z a2 ( z ) (z * ) a1 a2 X( z) z ( z ) ( z * ) X( z ) a where =0.433 j0.25 and a1 ( z ) X( z) (3z 1.5z) z z(z * ) z z 120 INVERSE Z-TRANSFORM a2 ( z * ) X( z ) (3z 1.5z) * a1 z z(z ) z * z * Also note that x k Z 1 k uk z and x k Z 1 ( * ) k uk * z Where =0.433013-j0.25=0.5exp(-j/6) Therefore, z z X( z ) a a 1* z 0.5 exp( j / 6) z 0.5 exp( j / 6) which corresponds to a time-series, for k0 k k 1 1 xk 1.55 exp( j / 12) exp( jk / 6) 1.55 exp( j / 12) exp( jk / 6) 2 2 k 1 1.55 (exp( jk / j / 12) exp( jk / / 12)) 2 k 1 3.1 cos(k / / 12) 2 which is seen to be a causal phase-shifted cosine wave with an exponentially descending envelope Also observe that x0 3.1cos( / 12) , which can be verified using the initial value theorem 8.3 Difference Equations Long division can be intensive and tedious computational process If a computer-based signal processing is desired, the use of difference equation is generally more efficient Assume that the Z-transform of a time series is xk is X(z), where M X(z) bi z i aiz i i 0 N i 0 Recall that the Zk and Zk n z n Therefore, it follows that: a xk a xk 1 a M 1 xk (M 1) a M xk M b k b1k 1 b N 1k ( N 1) b N k N 121 INVERSE Z-TRANSFORM The response xk can be simulated by implementing the difference equation Example 8.13 Consider causal 3z 5z 3z X( z ) (z 1)(z 0.5) from example 11 X( z ) 3z 5z 3z 5z 1 3z 2 5z 1 3z 2 z 2.5z 2z 0.5 (1 z 1 ) (1 0.5z 1 ) 2.5z 1 2z 2 0.5z 3 Which produces a time-serises xk (5(0.5) k 2k)u s k Then xk , for k , can be simulated using xk 2.5xk 1 2xk 2 0.5xk 3 3k 5k 1 3k 2 122