EI331 Signals and Systems Lecture 29 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 29 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University June 6, 2019

Contents 1. Laplace Transform 2. Region of Convergence 3. Properties of Laplace Transform 1/35

Laplace Transform Recall the response of a CT LTI system to the input x ( t ) = e st is y ( t ) = ( x ∗ h )( t ) = H ( s ) e st where h is the impulse response of the system and � ∞ h ( t ) e − st dt H ( s ) = −∞ The system function H ( s ) is called the Laplace transform of h . In general, the Laplace transform of a CT signal x ( t ) is � ∞ � T 2 x ( t ) e − st dt = lim x ( t ) e − st dt X ( s ) = T 1 →∞ −∞ − T 1 T 2 →∞ also denoted by L X = L { x } , or x ( t ) ← − − → X ( s ) 2/35

Laplace Transform The set of s for which the integral defining Laplace transform � ∞ � T 2 x ( t ) e − st dt = lim x ( t ) e − st dt X ( s ) = T 1 →∞ −∞ − T 1 T 2 →∞ converges is called its region of convergence (ROC) Relation with CTFT For s = σ + j ω , � ∞ { x ( t ) e − σ t } e − j ω t dt = F { x ( t ) e − σ t } X ( σ + j ω ) = −∞ If the ROC includes the imaginary axis, setting σ = 0 yields � X ( s ) s = j ω = X ( j ω ) = F { x } ( j ω ) � 3/35

Example Im For x ( t ) = e − at u ( t ) , − a X � ∞ 1 − e − ( s + a ) T 1 Re e − at e − st dt = lim X ( s ) = = s + a , s + a T →∞ 0 with ROC given by Re s > − Re a . If Re a > 0 , the ROC contains the imaginary axis, 1 Im � F { x } ( j ω ) = X ( s ) z = e j ω = � j ω + a X − a If Re a < 0 , the CTFT does not exist. Re If Re a = 0 , the CTFT exists only as a distribution, 1 ⇒ F { x } ( e j ω ) = � a = j ω 0 = j ( ω + ω 0 )+ πδ ( ω + ω 0 ) � = X ( s ) � s = j ω 4/35

Example Im For x ( t ) = − e − at u ( − t ) , X − a � 0 1 − e ( s + a ) T 1 Re e − at e − st dt = lim X ( s ) = − = s + a , s + a T →∞ −∞ with ROC given by Re s < − Re a . If Re a < 0 , the ROC contains the imaginary axis, 1 Im � F { x } ( j ω ) = X ( s ) z = e j ω = � j ω + a − a X If Re a > 0 , the CTFT does not exist. Re If Re a = 0 , the CTFT exists only as a distribution, 1 ⇒ F { x } ( e j ω ) = � a = j ω 0 = j ( ω + ω 0 )+ πδ ( ω + ω 0 ) � = X ( s ) � s = j ω 5/35

Importance of ROC Im 1 L x 1 ( t ) = e − at u ( t ) ← − − → X 1 ( s ) = s + a , − a X Re ROC Re s > − Re a Im 1 L x 2 ( t ) = − e − at u ( − t ) ← − − → X 2 ( s ) = s + a , − a X Re ROC Re s < − Re a Different signals can have the same X ( s ) but different ROCs Always specify ROC for Laplace transforms! 6/35

Example Im For x ( t ) = 3 e − 2 t u ( t ) − 2 e − t u ( t ) , � ∞ � 3 e − 2 t u ( t ) − 2 e − t u ( t ) � e − st dt X ( s ) = 0 3 2 X X Re = s + 2 − − 2 − 1 1 s + 1 s − 1 = ( s + 2 )( s + 1 ) with ROC Re s > − 1 . Two simple poles at s = − 2 and s = − 1 A simple zero at s = 1 Also a simple zero at ∞ . 7/35

Example For x ( t ) = e − 2 t u ( t ) + e − t cos( 3 t ) u ( t ) = e − 2 t + 1 2 e − ( 1 − 3 j ) t + 1 2 e − ( 1 + 3 j ) t , s + 2 + 1 1 s + ( 1 − 3 j ) + 1 1 1 X ( s ) = 2 2 s + ( 1 + 3 j ) 2 s 2 + 5 s + 12 = ( s 2 + 2 s + 10 )( s + 1 ) Im √ √ ( s + 5 + j 71 )( s + 5 − j 71 ) − 1 + 3 j X 4 4 = ( s + 2 )( s + 1 − 3 j )( s + 1 + 3 j ) with ROC Re s > − 1 . Re X − 2 Simple poles at s = − 2 and s = − 1 ± 3 j √ Simple zeros at s = − 5 ± j 71 4 − 1 − 3 j X Also a simple zero at ∞ . 8/35

Rational Transforms A rational transform X has the following form X ( s ) = N ( s ) D ( s ) 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 ( s − z k ) X ( s ) = A � m k = 1 ( s − 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 ∞ 9/35

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 Laplace transform is determined by its pole-zero plot and ROC, up to a multiplicative constant factor. Im Example. ( s − 1 ) 2 X ( s ) = A ( s + 1 )( s − 2 ) We will see there are three X X 2 Re possibilities for the ROC − 1 1 • Re s < − 1 • − 1 < Re s < 2 • Re s > 2 10/35

Contents 1. Laplace Transform 2. Region of Convergence 3. Properties of Laplace Transform 11/35

Convergence of Laplace Transform Assume x ( t ) is integrable on any finite interval [ T 1 , T 2 ] . The convergence of the Laplace transform � ∞ � T 2 x ( t ) e − st dt = lim x ( t ) e − st dt X ( s ) = T 1 →∞ −∞ − T 1 T 2 →∞ is equivalent to the convergence of the following two integrals � ∞ � T 2 x ( t ) e − st dt = lim x ( t ) e − st dt ( ⋆ ) T 2 →∞ 0 0 � 0 � 0 x ( t ) e − st dt = lim x ( t ) e − st dt T 1 →∞ −∞ − T 1 The ROC for the Laplace transform is the intersection of the ROCs for the above two one-sided integrals. NB. The integral in ( ⋆ ) is the unilateral Laplace transform of x . 12/35

Convergence of Unilateral Laplace Transform Theorem. If the integral in ( ⋆ ) converges for s = s 0 = σ 0 + j ω 0 , then it converges for any s = σ + j ω with σ > σ 0 . Proof. Let � t x ( τ ) e − s 0 τ d τ y ( t ) = 0 t →∞ y ( t ) exists and hence M � sup By assumption, lim | y ( t ) | < ∞ . t ≥ 0 Integration by parts yields � T � T x ( t ) e − st dt = y ′ ( t ) e − ( s − s 0 ) t dt 0 0 � T = e − ( s − s 0 ) T y ( T ) + ( s − s 0 ) y ( t ) e − ( s − s 0 ) t dt 0 As T → ∞ , e − ( s − s 0 ) T y ( T ) → 0 . 13/35

Convergence of Unilateral Laplace Transform Theorem. If the integral in ( ⋆ ) converges for s = s 0 = σ 0 + j ω 0 , then it converges for any s with Re s > σ 0 . Proof (cont’d). Let s = σ + j ω . � T � T | y ( t ) e − ( s − s 0 ) t | dt = | y ( t ) | e − ( σ − σ 0 ) t dt 0 0 � T M Me − ( σ − σ 0 ) t dt ≤ ≤ σ − σ 0 0 � ∞ 0 y ( t ) e − ( s − s 0 ) t dt converges absolutely. so the integral Therefore, ( ⋆ ) converges and � ∞ � ∞ x ( t ) e − st dt = ( s − s 0 ) y ( t ) e − ( s − s 0 ) t dt 0 0 14/35

ROC of Unilateral Laplace Transform Three possibilities for the convergence of the integral ( ⋆ ), (a). it converges for every s ∈ C (b). it diverges for every s ∈ C (c). it converges for Re s > σ c ∈ R and diverges for Re s < σ c In case (c), σ c ∈ R is called the abscissa of convergence, and the line Re s = σ c is called the axis of convergence The ROC 1 is always a right half-plane Im ROC = { s ∈ C : Re s > σ c } Re σ c We also write σ c = −∞ in case (a), and σ c = + ∞ in case (b). 1 More precisely, the interior of the ROC. 15/35

ROC of Unilateral Laplace Transform � ∞ 0 e t 2 e − st dt diverges for every s ∈ C , i.e. σ c = + ∞ Example. Proof. Let s = σ ∈ R . Note � ∞ e t 2 e − σ t dt = + ∞ 0 Since σ is arbitrary, the previous theorem implies the integral diverges for any s ∈ C . � ∞ 0 e − t 2 e − st dt converges for every s ∈ C , i.e. σ c = −∞ Example. Proof. Let s = σ ∈ R . Note � ∞ � ∞ � ∞ e − t 2 dt = √ π e σ 2 / 4 e − t 2 e − σ t dt = e σ 2 / 4 e − ( t + σ/ 2 ) 2 dt ≤ e σ 2 / 4 −∞ 0 0 The integral converges absolutely for every s ∈ C . � ∞ 0 e − st dt has σ c = 0 Example. 16/35

Absolute Convergence of Unilateral Laplace Transform Theorem. If the integral in ( ⋆ ) converges absolutely for s = s 0 = σ 0 + j ω 0 , then it converges absolutely and uniformly for s = σ + j ω with σ ≥ σ 0 . Proof. � ∞ � ∞ � ∞ | x ( t ) e − st | dt = | x ( t ) | e − σ t dt ≤ | x ( t ) | e − σ 0 t dt < ∞ 0 0 0 As in the case of convergence, the region of Im absolute convergence (ROAC) is always a right half-plane Re ROAC = { s ∈ C : Re s > σ a } σ a where σ a ∈ ¯ R is the abscissa of absolute convergence, and the line Re s = σ a is the axis of absolute convergence. 17/35

ROAC of Unilateral Laplace Transform The axis of convergence and the axis of absolute convergence need not coincide ! � ∞ 0 e kt sin( e kt ) e − st dt Example. Let k > 0 . To see σ a = k , note converges absolutely for Re s > k , since for s = σ + j ω , | e kt sin( e kt ) e − st | ≤ e − ( σ − k ) t ∈ L 1 [ 0 , ∞ ) It is not absolutely convergent for s = k , since � ∞ � ∞ | sin( e kt ) | dt = 1 | sin u | du = ∞ . Im k u 0 1 Let s = σ ∈ R . Re � ∞ � ∞ sin u 0 k e kt sin( e kt ) e − σ t dt = 1 u σ/ k du k 0 1 Dirichlet’s test implies σ c = 0 . NB. We will use ROC, although in most cases ROC = ROAC. 18/35

ROC of (Bilateral) Laplace Transform The ROC of � ∞ x ( t ) e − st dt 0 is a right half-plane Re s > σ 1 . The ROC of � 0 � ∞ x ( t ) e − st dt = x ( − t ) e − ( − s ) t dt −∞ 0 is a left half-plane, Re s < σ 2 . The ROC of � ∞ � ∞ � 0 x ( t ) e − st dt = x ( t ) e − st dt + x ( t ) e − st dt X ( s ) = −∞ −∞ 0 is a strip σ 1 < Re s < σ 2 , which is nonempty iff σ 1 < σ 2 19/35