ELEG 305: Digital Signal Processing Lecture 4: Inverse z Transforms & z Domain Analysis Kenneth E. Barner Department of Electrical and Computer Engineering University of Delaware Fall 008 K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 / 4 Outline Review of Previous Lecture Lecture Objectives 3 Inverse z Transform Distinct Poles Case Multiple Order Poles Case 4 Analysis of LTI Systems in the z Domain Response of Systems with Rational System Functions Causality and Stability Stability of Second Order Systems K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 / 4
Review of Previous Lecture Review of Previous Lecture The Rational z Transform conversion from difference equations, the system function (H(z)), poles and zeroes, the relation between pole location and time domain behavior The Inverse z Transform contour integration and power series expansion (function of z or z ) K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 3 / 4 Lecture Objectives Lecture Objectives Objective Evaluate the inverse z transform using partial fraction expansion and table lookup; Analysis of LTI systems in the z domain (transient vs. steady state response and stability); Stability & response of second order systems Reading Chapter 3 (3.4-3.5); Next lecture, the remainder of Chapter 3 (3.6) the one sided z transformation K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 4 / 4
Commonly Used Inverse z Transform Methods Direct evaluation of x(n) = X(z)z n dz πj by contour integration Power series expansion X(z) = n= x(n)z n 3 Partial fraction expansion and table lookup K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 5 / 4 Inverse z Transform Approach Express X(z) as a linear sum of components that have simple inverse transformations (that are known or can be looked up) If we can express X(z) as where it is known that Then, X(z) =α X (z)+α X (z)+ + α k X k (z) x k (n) =Z {X k (z)} x(n) =α x (n)+α x (n)+ + α k x k (n) K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 6 / 4
Systems of interest: Rational functions that are proper Proper = numerator poly. order < denominator poly. order If function not proper, use division to write it as a polynomial and proper rational function Example Determine the inverse transformation of X(z) = + z + z 3 z + z X(z) is not a proper function Reverse the order of the polynomials Divide until remainder is proper z 3 z + z + z + K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 7 / 4 Inverse z Transform + 5z X(z) = + 3 z + z Note: The rational function is now proper. Next, factor the denominator 3 z + z = ( ) ( z z ) + 5z X(z) = + 3 z + z + 5z = + ( z )( z ) A = + ( z ) + A ( z ) }{{} partial fraction expansion K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 8 / 4
Approach: Solve for A and A by isolating the terms X(z) = + 5z + ( z )( z ) ( ) = A + ( z ) + A ( z ) ( ) A To isolate A : () multiply ( ) by the denominator of and () ( z ) evaluate at the associated root, z = ( X(z) ) z= z = ( ) z + A + A ( z ) ( z ) = 0 + A + 0 = A z= K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 9 / 4 Inverse z Transform Repeat the process for ( ), + 5z X(z) = + ( z )( z ) ( X(z) ) z= z = ( ) + z 5z + ( z ) z= + 0 = 0 + = 9 Combining the results for ( ) and ( ) A = 9 Next: Repeat the process for A, i.e., multiply ( ) and ( ) by ( z ) and evaluate at z =. K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 0 / 4
For ( ), ( X(z) z ) ( z= = z ) + A ( z ) + A z z= = 0 + 0 + A = A and ( ) ( X(z) z ) z= = ( z ) + 5z + z z= = 0 + + 5 = 8 Combining the results for ( ) and ( ) A = 8 Final Result: Combining results into final partial fraction expression: 9 X(z) = + + 8 z z K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 / 4 Inverse z Transform Next, determine x(n) depends on X(z) and ROC 9 X(z) = + + 8 z z Case : ROC: z > causal signal { } { } x(n) = Z {} 9Z + 8Z z z ( ) n = δ(n) 9 u(n)+8u(n) K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 / 4
9 X(z) = + + 8 z z Case : ROC: z < anti causal signal ( ) n x(n) =δ(n)+9 u( n ) 8u( n ) Case 3: ROC: < z < two sided signal 9 8 X(z) = + + }{{ z }} {{ z } causal anti-causal x(n) =δ(n) 9 ( ) n u(n) 8u( n ) K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 3 / 4 Inverse z Transform General Partial Fraction Procedure Express X(z) in proper rational function form X(z) = + M i= b iz i + N i= a (M < N) iz i Factor the denominator X(z) = + M i= b iz i N ( i= pi z i) where the p i s are the roots of + N i= a iz i 3 Express as a sum of first order terms (assumes distinct poles) X(z) = N A i p i= i z i 4 Determine A i =( p i z i )X(z) z=pi (assumes distinct poles) K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 4 / 4
Multiple Order Poles Case Note: For poles with multiplicity m >, m terms are needed Required Terms: For a pole ( pz ) m, required partial fraction expansion terms are A pz + A ( pz ) + + A m ( pz ) m where A i = { d m i [ (m i)!( p) m i dz m i ( pz) m X(z )] } z=p K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 5 / 4 Inverse z Transform Example Determine the inverse z transform of Expand into partial fraction form X(z) = B = X(z) = ( z )( z ) B + A z z + A ( z ) ( ) z X(z) = z= A = ( z ) X(z) = z= = ( ( ) ) = ( ) = K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 6 / 4
For A, we must use the derivative method: { d m i [ A i = (m i)!( p) m i dz m i ( pz) m X(z )] } z=p Note that X(z )=, p =, and m i = =. Thus, ( z)( z) d [ A =!( ) ( z) X(z )] = d dz z= dz ( z) z= = ( z) = ( ) = z= Combining the three partial fraction components, = X(z) = + z z + ( z ) K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 7 / 4 Inverse z Transform Alternate Method for nd Order Poles B and A are easy to determine thus we know X(z) = Problem: Need to find A z + A z + ( z ) Observation: Only one unknown need one equation to solve Solution: Evaluate X(z) at value pole & solve for A, e.g. choose z = K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 8 / 4
Inverse Transformation of Partial Fraction Expansion Partial fraction expansion yields: X(z) = A p z + + A N p N z Inverse transform each component (and sum results) { } { Z A k Ak p p k z = k nu(n), ROC: z > p k A k pk nu( n ), ROC: z < p k Note: IfX(z) contains complex conjugate poles (ROC: z > p ) A pz + A p z [Ap n + A (p ) n ]u(n) = A r n cos(βn + α)u(n) where α and β are the phase components of A and p, i.e., A = A e jα and p = re jβ K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 9 / 4 Analysis of LTI Systems in the z Domain System Response: Suppose, X(z) = N(z) and Q(z) Response of Systems with Rational System Functions H(z) = B(z) A(z) Y (z) = H(z)X(z) = B(z)N(z) N A(z)Q(z) = A k L p k= k z + Q k q k= k z where p k are the poles of H(z) and q k are the poles of X(z) N L = y(n) = A k (p k ) n u(n) + Q k (q k ) n u(n) k= k= }{{}}{{} natural response forced response Observations: If poles of H(z) are inside the unit circle, then the natural response 0asn : transient If poles of X(z) are on the unit circle, the forced response is periodic: steady state K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 0 / 4
Analysis of LTI Systems in the z Domain Causality and Stability Causality and Stability Recall n h(n) < = BIBO stable Approach: Assume BIBO stability & evaluate H(z) on the unit circle H(z) = h(n)z n n h(n)z n n h(n) z n [ z n = ] n = n h(n) < [by BIBO assumption] H(z) BIBO stable ROC contains the unit circle (converse also true) Result: a LTI system is BIBO stable iff the ROC includes the unit circle Result: a causal LTI system is BIBO stable iff all poles are inside the unit circle K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 / 4 Analysis of LTI Systems in the z Domain Stability of Second Order Systems Consider the stability of a second order system y(n) = a y(n ) a y(n )+b 0 x(n) H(z) = Y (z) X(z) = b 0 + a z + a z p, p = a ± a 4a 4 Note: Three stable cases exist: Case : Real distinct (a > 4a ) Case : Real and equal (a = 4a ) Case 3: Complex Conjugates (a < 4a ) K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 / 4
Analysis of LTI Systems in the z Domain Stability of Second Order Systems Case : Time domain signal difference of exponentials h(n) = b ( ) 0 p n+ p p p n+ u(n) Case : Time domain signal ramp, exponential product h(n) =b 0 (n + )p n u(n) Case 3: Time domain signal oscillating exponential h(n) = b 0r n sin((n + )ω 0 )u(n) sin ω 0 K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 3 / 4 Lecture Summary Lecture Summary The Inverse z Transform partial fraction expansion; consideration of distinct and multiple order poles Analysis of LTI systems in the z domain response of systems with rational system functions (transient vs. steady state response), causality and stability (BIBO stable pole location restrictions), and stability & response of second order systems Next Lecture one sided z transformation (systems with non zero initial conditions) K. E. Barner (Univ. of Delaware) ELEG 305: Digital Signal Processing Fall 008 4 / 4