EI331 Signals and Systems Lecture 27 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 27 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University May 30, 2019

Contents 1. Evaluation of Definite Integrals Using Residues 2. Z -transform 3. Properties of Z -transform 4. Analysis of DT LTI Systems by Z -transform 1/35

� ∞ −∞ R ( x ) dx R ( x ) = N ( x ) D ( x ) is a rational function of x , where N , D are polynomials with deg D ≥ deg N + 2 , and R has no singularity on the real axis. y 1. Pick a r large enough s.t. the upper half disk centered at 0 contains all C r the singularities z 1 , . . . , z K , of R ( z ) in z 2 the upper half plane (we don’t care z 1 z 3 about those in the lower half plane) r x − r � r K � � 2. R ( x ) dx + R ( z ) dz = j 2 π Res( R , z k ) − r C r k = 1 � 3. lim R ( z ) dz = 0 by the condition deg D ≥ deg N + 2 r →∞ C r � ∞ K � 4. R ( x ) dx = j 2 π Res( R , z k ) −∞ k = 1 2/35

� ∞ −∞ R ( x ) dx y Lemma. Suppose f ( z ) is continuous on D = { z : R < | z | < ∞ , θ 1 ≤ arg z ≤ θ 2 } , γ r where 0 ≤ θ 1 < θ 2 ≤ 2 π . Let γ r be the θ 2 θ 1 arc z ( t ) = re j θ , r > R , θ ∈ [ θ 1 , θ 2 ] . If lim sup D ∋ z →∞ zf ( z ) = 0 , then r x − r � − R R lim f ( z ) dz = 0 r →∞ γ r Proof. Since lim sup D ∋ z →∞ zf ( z ) = 0 , given any ǫ > 0 , there exists an R ǫ ǫ s.t. | zf ( z ) | < θ 2 − θ 1 for z ∈ D and | z | > R 0 . For r > R 0 , � � � � � ǫ � � f ( z ) dz � ≤ | f ( z ) | ds ≤ ( θ 2 − θ 1 ) rds = ǫ � � � γ r γ r γ r NB. Item 3 of the previous slide follows from the lemma with θ 1 = 0 and θ 2 = π . 3/35

Example � ∞ x 2 dx Evaluate I = ( x 2 + a 2 )( x 2 + b 2 ) , where a , b > 0 −∞ z 2 Solution. R ( z ) = ( z 2 + a 2 )( z 2 + b 2 ) has four simple poles at z = ± ja , ± jb . Two of them, ja , jb , are in the upper half plane. The residues are a Res( R , ja ) = lim z → ja ( z − ja ) R ( z ) = 2 j ( a 2 − b 2 ) − b Res( R , jb ) = lim z → jb ( z − jab ) R ( z ) = 2 j ( a 2 − b 2 ) Thus π I = j 2 π [Res( R , ja ) + Res( R , jb )] = a + b 4/35

� ∞ −∞ R ( x ) e jax dx ( a > 0 ) R ( x ) = N ( x ) D ( x ) is a rational function of x , where deg D ≥ deg N + 1 , and R has no singularity on the real axis. y 1. Pick a r large enough s.t. the upper C r half disk centered at 0 contains all z 2 the singularities z 1 , . . . , z K , of R ( z ) in z 1 z 3 the upper half plane r x − r � r K � � R ( x ) e jax dx + R ( z ) e jaz dz = j 2 π Res[ R ( z ) e jaz , z k ] 2. − r C r k = 1 � R ( z ) e jaz dz = 0 by the condition deg D ≥ deg N + 1 and 3. lim r →∞ C r y Jordan’s Lemma − r r � ∞ K x � R ( x ) e jax dx = j 2 π Res[ R ( z ) e jaz , z k ] z ′ 4. z ′ 1 3 z ′ −∞ k = 1 2 C ′ r NB. If a < 0 , use the lower half disk instead. 5/35

Jordan’s Lemma y Lemma. Suppose f ( z ) is continuous on D = { z : R < | z | < ∞ , Im z ≥ 0 } . If C r D ∋ z →∞ f ( z ) = 0 , and a > 0 , then lim � f ( z ) e jaz dz = 0 lim r →∞ C r r x where C r is z ( t ) = re j θ , r > R , θ ∈ [ 0 , π ] . − r − R R � Proof. Let I r = C r f ( z ) e jaz dz and M r = max z ∈ C r | f ( z ) | . Then � π � π/ 2 � e − ar sin θ rd θ = 2 M r e − ar sin θ rd θ | f ( z ) e jaz | ds ≤ M r | I r | ≤ γ r 0 0 Using sin θ ≥ 2 π θ for θ ∈ [ 0 , π 2 ] and lim r →∞ M r = 0 , � π/ 2 π θ rd θ = π M r e − ar 2 a ( 1 − e − ar ) → 0 , as r → ∞ | I r | ≤ 2 M r 0 NB. If a < 0 , the lemma still holds if we replace D and C r by { z : R < | z | < ∞ , Im z ≤ 0 } and z ( t ) = re − j θ , r > R , θ ∈ [ 0 , π ] . 6/35

Example � ∞ x sin x Evaluate I = x 2 + a 2 dx , where a > 0 0 z Solution. R ( z ) = z 2 + a 2 has two simple poles at z = ± ja . The pole ja is in the upper half plane, z → ja ( z − ja ) R ( z ) e jz = e − a Res[ R ( z ) e jz , ja ] = lim 2 Thus � ∞ xe jx x 2 + a 2 dx = j 2 π Res[ R ( z ) e jz , ja ] = j π e − a −∞ Since x sin x x 2 + a 2 is even, � ∞ � ∞ xe jx x sin x x 2 + a 2 dx = π I = 1 x 2 + a 2 dx = 1 2 e − a 2 Im 2 −∞ −∞ 7/35

Example 1 Find the inverse CTFT of X ( j ω ) = a + j ω , where a > 0 � ∞ Solution. x ( t ) = 1 1 a + j ω e j ω t d ω 2 π −∞ 1 R ( z ) = a + jz has a simple poles at z = ja in the upper half plane. y For t > 0 , C r � ∞ x ( t ) = 1 ja X R ( ω ) e j ω t d ω = j Res[ R ( z ) e jzt , ja ] 2 π −∞ r x − r z → ja ( z − ja ) R ( z ) e jzt = e − at = j lim y For t < 0 , R ( z ) is analytic in the lower half plane, ja X so − r r � ∞ x ( t ) = 1 R ( ω ) e j ω t d ω = 0 x 2 π −∞ C ′ Therefore, x ( t ) = e − at u ( t ) . r 8/35

Contents 1. Evaluation of Definite Integrals Using Residues 2. Z -transform 3. Properties of Z -transform 4. Analysis of DT LTI Systems by Z -transform 9/35

Z -transform Recall the response of a DT LTI system to the input x [ n ] = z n is y [ n ] = ( x ∗ h )[ n ] = H ( z ) z n where h is the impulse response of the system and ∞ � h [ k ] z − k H ( z ) = k = −∞ The system function H ( z ) is called the z -transform of h . In general, the z -transform of a DT signal x [ n ] is ∞ � x [ n ] z − n X ( z ) = n = −∞ also denoted by Z X = Z { x } , or x [ n ] ← − − → X ( z ) 10/35

Z -transform Note ∞ ∞ x [ n ] z − n = � � c n z n X ( z ) = n = −∞ n = −∞ is a Laurent series at z = 0 whose coefficient for z n is c n = x [ − n ] . As a Laurent series, the z -transform converges on an annulus centered at z = 0 , called its region of convergence (ROC) Relation with DTFT ∞ z = re j ω = { x [ n ] r − n } e − j ω n = F { x [ n ] r − n } � ⇒ X ( re j ω ) = n = −∞ If the ROC includes the unit circle, setting r = 1 yields � z = e j ω = X ( e j ω ) = F { x } ( e j ω ) X ( z ) � 11/35

Example Im For x [ n ] = a n u [ n ] , ∞ 1 z a n z − n = � X ( z ) = 1 − az − 1 = z − a , X Re n = 0 a 1 with ROC | z | > | a | . If | a | < 1 , the ROC contains the unit circle, Im 1 F { x } ( e j ω ) = X ( e j ω ) = 1 − ae − j ω If | a | > 1 , the DTFT does not exist. X Re a 1 If | a | = 1 , the DTFT exists as a distribution, F { x } ( e j ω ) � = X ( z ) � � z = e j ω 12/35

Example Im For x [ n ] = − a n u [ − n − 1 ] , − 1 1 z a n z − n = � X ( z ) = − 1 − az − 1 = z − a , X Re a n = −∞ 1 with ROC | z | < | a | . If | a | > 1 , the ROC contains the unit circle, Im 1 F { x } ( e j ω ) = X ( e j ω ) = 1 − ae − j ω If | a | < 1 , the DTFT does not exist. X Re a 1 If | a | = 1 , the DTFT exists as a distribution, F { x } ( e j ω ) � = X ( z ) � � z = e j ω 13/35

Importance of ROC Im z Z x 1 [ n ] = a n u [ n ] ← − − → X 1 ( z ) = z − a , | z | > | a | X Re a 1 z Z x 2 [ n ] = − a n u [ − n − 1 ] ← − − → X 2 ( z ) = z − a , | z | < | a | Im Different signals can have the same X ( z ) but different ROCs, consistent with the X Re fact a function has different Laurent a 1 series in different annuli of analyticity. Always specify ROC for z -transforms! 14/35

Example � 1 � n u [ n ] − 6 � 1 � n u [ n ] , For x [ n ] = 7 3 2 ∞ � n � n � � � 1 � 1 � z − n X ( z ) = 7 − 6 3 2 n = 0 7 6 Im = 3 z − 1 − 1 − 1 1 − 1 2 z − 1 1 − 3 2 z − 1 = ( 1 − 1 3 z − 1 )( 1 − 1 2 z − 1 ) z ( z − 3 2 ) Re = X X 1 ( z − 1 3 )( z − 1 1 1 3 2 ) 3 2 2 with ROC | z | > 1 2 . | x [ n ] | = 1 � R = lim sup n 2 n →∞ 15/35

Example � n sin � π � 1 � 1 4 � n u [ n ] − 1 � 1 4 � n u [ n ] , � u [ n ] = 1 3 e j π 3 e − j π For x [ n ] = 4 n 3 2 j 2 j X ( z ) = 1 1 4 z − 1 − 1 1 1 − 1 3 e j π 1 − 1 3 e − j π 2 j 2 j 4 z − 1 1 Im 2 z √ 3 = ( z − 1 3 e j π 4 )( z − 1 3 e − j π 4 ) with ROC | z | > 1 3 . X Re 1 1 X 3 e j π 3 e − j π Simple poles at z = 1 4 and z = 1 3 4 A simple zero at z = 0 1 2 ζ By ζ = 1 √ z , X ( ζ ) = 3 4 ζ ) , so X ( z ) also has a simple 3 e j π 3 e − j π ( 1 − 1 4 ζ )( 1 − 1 zero at ∞ . 16/35

Rational Transforms A rational transform X has the following form X ( z ) = N ( z ) D ( z ) where N , D are polynomials that are coprime, i.e. they have no common factors of degree ≥ 1 . By the Fundamental Theorem of Algebra, � n k = 1 ( z − z k ) X ( z ) = A � m k = 1 ( z − p k ) with the convention � 0 k = 1 · = 1 . • z 1 , . . . , z n are the finite zeros of X • p 1 , . . . , p m are the finite poles of X • If n > m , X has a pole of order n − m at ∞ • If n < m , X has a zero of order m − n at ∞ 17/35

Rational Transforms A rational function X is determined by its zeros and poles in C , including their orders, up to a multiplicative constant factor. A rational z -transform is determined by its pole-zero plot and ROC, up to a multiplicative constant factor. Im Example. ( z + 2 )( z − e j π 6 ) X ( z ) = A 2 e j 5 π ( z − 3 2 )( z − 1 4 )( z − 2 e − j π 4 ) e j π 2 e j 5 π 6 1 4 X X Re Four possibilities for ROC 3 2 • | z | < 1 / 2 • 1 / 2 < | z | < 3 / 2 X − 2 2 e − j π • 3 / 2 < | z | < 2 4 • 2 < | z | < ∞ 18/35