Lecture 6: Properties of ROC and Inverse z-transform

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1 EE58 Digital Signal Processing University of Washington Autumn Dept. of Electrical Engineering Lecture 6: Properties of ROC and Inverse z-transform Oct 7, Prof: J. Bilmes TA: Mingzhou ng 6. Properties of ROC (continued) More on Property 5 x[n] is a right sided sequence and x[n] = for n < N. Let r such that x[n] r n n= = x[n] r n < n=n Now see if Consider r r and let r = cr for c. {z : z = r } ROC {z : z > r } ROC x[n] r n n=n = x[n] (cr ) n n=n = x[n] c n r n n=n c N + n= x[n] r n + c N n=n x[n] c n r n n= x[n] r n =A x[n] r n < ( {z : z = r } ROC) n=n where A = ( r r ) N. which means x[n] r n < n=n {z : z > r } ROC Property 7. If x[n] is two sided sequence, and if the circle z = r is in ROC, then the ROC will consist of a ring in the z-plane which includes z = r. Why? x[n] = x L [n] + x R [n] 6-

2 6- where x L [n] is a left sided sequence and x R [n] is a right sided sequence. ROC for x R [n] is bounded inside by outmost pole of X R (z). ROC for x L [n] is bounded outside by innermost pole of X L (z). Add the two together. Either the two ROC s overlap, which is a ring shaped ROC for x[n], or there is no ROC for x[n]. Property 8. ROC must be a connected region. Note from Property 7, any sequence is the sum of a left sided sequence and a right sided sequence. if intersection exists, it must be formed by d R < z < d L Hence, poles determine possible types of ROC s. When there are two poles d and d, and d < d, there are three possible ROC candidates, i.e. ROC ={z : z < d } ROC ={z : d < z < d } ROC 3 ={z : d < z } 3 log( H(z) ) Re z 3 Im z Figure 6.: A z-transform H(z) =.z +.88z.8z +.6z with poles around. ± j.698 and zeros around. ± j. 6. Inverse z-transform z-transform is n= x[n]z n Let z = re jω n= x[n](re jω ) n = n= x[n]r n e jωn = X(re jω ) = FT {x[n]r n }

3 6-3 Then FT {X(re jω )} = x[n]r n x[n] =r n FT {X(re jω )} =r n π X(re jω )e jωn dω π π = π X(re jω )(re jω ) n dω π π Since z = re jω, that is, dz = jre jω dω = jzdω dω = j z dz Therefore, x[n] = π X(z)z n dz π j π But r can be anything inside the ROC. The inverse z-transform equation can be written as x[n] = X(z)z n dz (6.) π j c The contour integral c is performed at a counter-clockwise closed circular contour centered at the origin with radius r, for all r such that {z : z = r} ROC Another way to derive this. Theorem 6. (Cauchy Integral Theorem). z k dz = π j c { k = k (6.) Proof. See Churchill & Brown (98), Introduction to Complex Variables. Consider X(z)z n dz = π j c π j c = = k= k= =x[n] k= x[k]z k+n dz x[k] z k+n dz π j c x[k]δ[n k] ( Cauchy Integral Theorem) Note: the result is valid for both positive and negative values of n. Note: for z = e jω, this reduces to inverse FT, since X(z)z n dz π j c since dz = jre jω dω. How to calculate this in general? z=e jω = π X(e jω )e jωn e jω je jω dω π j π

4 6- Theorem 6. (Cauchy Residue Theorem). Proof. See Churchill & Brown (98). x[n] = X(z)z n dz = π j [residues of X(z)z n at poles inside c] (6.3) c Finding residues (to be defined shortly) is often difficult. But if X(z)z n is a rational function of z, X(z)z n = If X(z)z n has s poles at z = d and Ψ(z) has no poles at z = d, then Residue [X(z)z n at z = d ] = Ψ(z) (z d ) s (6.) (s )! [ d s ] Ψ(z) dz s z=d (6.5) if s = (st order pole), then Residue [X(z)z n at z = d ] = Ψ(d ) Find the inverse z-transform of x[n] = π j c az z > a z n dz = az π j c z n z a dz The radius of contour must be greater than a since the contour has to be in ROC. We will consider the contour integral for n and n < separately.. n There is only one pole at z = a, i.e., Ψ(z) = z n and Ψ(a) = a n. Therefore since s = and no derivative is necessary.. n < z n has only one pole at z = a z a x[n] = a n n There are multiple order poles at z =, with the order depending on n. There is one pole at z = a. When n =, there are a st order pole at z = and a st order pole at z = a. residues sum to, i.e., Residue [X(z)z n n= at z = a] = a ( Ψ(z) = z n ) Residue [X(z)z n n= at z = ] = a ( Ψ(z) = z a ) x[ ] = When n =, there are two poles at z = and one pole at z = a. X(z)z n = z z a = Ψ (z) (z a) = Ψ (z) (z ) with Ψ (z) = z and Ψ (z) = (z a)

5 6-5 (Pole at z = a, s = ) (Pole at z =, s = ) Residue [X(z)z n n= at z = a] = Ψ (a) = a Residue [X(z)z n n= at z = ] = d dz Ψ (z) = d z= dz z= (z a) = (z a) z= = a Residues cancel out again, so We can continue this for x[ 3],x[ ], x[ ] = Easier Ways to Compute Inverse z-transform Inspection. Use z-transform table. if x[n] is a right sided sequence. so a n Z u[n] az z > a 3 z Z ( ) n u[n] = x[n] 3 a z + a z a < z < a x[n] = a n u[n] an u[ n ] since ROCs of each of the terms in X(z) must overlap for the existence of X(z). Partial Fraction Expansion. (There are some nice forms of z-transform that we can get their inverse relatively easily.) When We can factor the denominator and numerator and get P(z) Q(z) = M k= b kz k N k= a kz k (6.6) b a M k= ( c kz ) N k= ( d kz ) (6.7). N > M and the poles are all first order poles (i.e., all the poles are unique). N k= A k d k z (6.8) Note: since A k = ( d k z )X(z) z=dk (6.9) ( d k z ) A k + N l k,l= A l ( d k z ) d l z

6 6-6 and 3 z + 8 z = ( z )( z ) = A z + A z A = ( z )X(z) z= = A = ( z )X(z) z= = Z z x[n] = z ( ) n u[n] ( ) n u[n]. N M and the poles are all first order poles (i.e., all the poles are unique). M k= b kz k M N N k= a kz k = B r z r + r= N k= A k d k z (6.) The first term is obtainable by long division (example later). The second term is obtainable by the same procedure used when N > M. In general where max(m N,) r= max(m N,) = r= B r z r + M k= b kz k N k= a kz k (M < N) B r z r + N u S i i= k= A ik d i z N u # of unique poles in denominator S i = order of i-th pole B r is obtained as before (6.) A ik = (s i k)!( d i ) s i k [ d s i k dw s i k ( ( di w) s i X(w ) )] w=di We can verify this by multiplying both sides of X(z) by ( d i z ) s i and following the above operations. Power Series Expansion. then z z + z x[n] = δ[n + ] δ[n + ] δ[n] + δ[n ] (6.) Power series. log( + az ) = n= ( ) n+ a n z n n Z x[n] = { ( ) n+ a n n n n Long Division. +.z +.z.z = +.6z.5z +.z 3 +

7 6-7 Then x[n] = n < x[] = x[] =.6 x[] =.5 x[3] =..