EI331 Signals and Systems Lecture 31 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 31 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University June 13, 2019

Contents 1. Unilateral Laplace Transform 2. Review 1/25

Unilateral Laplace Transform Recall the (bilateral) Laplace transform of a CT signal x � ∞ x ( t ) e − st dt X ( s ) = −∞ The unilateral Laplace transform of x is � ∞ x ( t ) e − st dt X ( s ) = 0 − also denoted UL X = UL { x } , x ( t ) ← − − → X ( s ) ROC of X is always a right half-plane NB. Information about x ( t ) at t < 0 is lost in X . NB. L { x } = UL { x } iff x ( t ) is causal, i.e. x ( t ) = 0 for t < 0 2/25

Examples The calculation of unilateral Laplace transforms is almost the same as for bilateral Laplace transforms. Example. x 1 ( t ) = e − at u ( t ) , 1 X 1 ( s ) = X 1 ( s ) = Re s > − Re a s + a , Example. x 2 ( t ) = e − a ( t + 1 ) u ( t + 1 ) = x 1 ( t + 1 ) , e s X 2 ( s ) = e s X 1 ( s ) = bilateral Re s > − Re a s + a , � ∞ e − a ( t + 1 ) e − st dt = e − a unilateral X 2 ( s ) = s + a , Re s > − Re a 0 NB. X 2 � = X 2 , since x 2 is noncausal. NB. X 2 � = e s X 1 , time-shift property fails for unilateral transform. 3/25

Examples The calculation of unilateral Laplace transforms is almost the same as for bilateral Laplace transforms. Example. x 3 ( t ) = e − a | t | with Re a > 0 − 2 a bilateral X 3 ( t ) = s 2 − a 2 , − Re a < Re s < Re a 1 unilateral X 3 ( s ) = s + a , Re s > − Re a NB. X 3 � = X 1 , since x 3 � = x 1 NB. X 3 = X 1 , since x 3 and x 1 differ only for t < 0 Example. x ( t ) = δ ( t ) + 2 δ ′ ( t ) + e t u ( t ) 1 X ( s ) = X ( s ) = 1 + 2 s + Re s > 1 s − 1 , NB. The lower limit of integration is 0 − so δ and δ ′ contribute to X 4/25

Examples The calculation of inverse unilateral Laplace transforms is the same as for bilateral Laplace transforms, but we can only recover x ( t ) for t ≥ 0 ! Example. Given unilateral Laplace transform 1 1 1 X ( s ) = ( s + 1 )( s + 2 ) = s + 1 − s + 2 the only possibility for ROC is Re s > − 1 . Thus x ( t ) = e − t − e − 2 t , t ≥ 0 X provides no information about x ( t ) for t < 0 Example. X ( s ) = s 2 − 3 1 s + 2 = − 2 + s + s + 2 . The ROC must be Re s > − 2 , and x ( t ) = − 2 δ ( t ) + δ ′ ( t ) + e − 2 t , t ≥ 0 5/25

Properties of Unilateral Laplace Transform Many properties are the same as for bilateral Laplace transform Property Signal ULT – x ( t ) X ( s ) – y ( t ) Y ( s ) Linearity ax ( t ) + by ( t ) a X ( s ) + b Y ( s ) e s 0 t x ( t ) Shifting in s -domain X ( s − s 0 ) 1 a X ( s Time scaling x ( at ) , a > 0 a ) x ∗ ( t ) X ∗ ( s ∗ ) Conjugation d − tx ( t ) ds X ( s ) Differentiation in s domain 6/25

Convolution Property of Unilateral Laplace Transform If x ( t ) = y ( t ) = 0 for t < 0 , then UL ( x ∗ y )( t ) ← − − → X ( s ) Y ( s ) , ROAC ⊃ ROAC X ∩ ROAC Y Proof. Note ( x ∗ y )( t ) = 0 for t < 0 . UL { x ∗ y } = L { x ∗ y } = L { x } L { y } = UL { x } UL { y } Caution. The assumption x ( t ) = y ( t ) = 0 for t < 0 is important! Example. x ( t ) = δ ( t ) − δ ( t − 1 ) , y ( t ) = δ ( t + 1 ) , X ( s ) = 1 − e − s , Y ( s ) = 0 , Note ( x ∗ y )( t ) = δ ( t + 1 ) − δ ( t ) , and UL { x ∗ y } = − 1 � = X ( s ) Y ( s ) 7/25

Integration in Time Domain If UL x ( t ) ← − − → X ( s ) , with ROAC = R then � t → 1 UL x ( τ ) d τ ← − − s X ( s ) , with ROAC ⊃ R ∩ { Re s > 0 } 0 − Proof. Apply the convolution property to ˜ x ∗ u , where � x ( t ) , t ≥ 0 ˜ x ( t ) = 0 , t < 0 Example. x ( t ) = δ ( t ) , X ( s ) = 1 . � t → 1 UL x ( τ ) d τ = u ( t ) ← − − s , Re s > 0 0 − 8/25

Differentiation in Time Domain If UL x ( t ) ← − − → X ( s ) , with ROC = R t → + ∞ x ( t ) e − st = 0 for s ∈ R 0 , then and lim d UL dtx ( t ) ← − − → s X ( s ) − x ( 0 − ) , with ROC ⊃ R ∩ R 0 NB. R 0 ⊃ ROAC of X ( s ) (also true for bilateral Laplace transform) Proof. Use integration by parts � ∞ � ∞ x ′ ( t ) e − st dt = x ( t ) e − st � � ∞ x ( t ) e − st dt = − x ( 0 − ) + s X ( s ) t = 0 − + s 0 − 0 − � ∞ 0 + x ( t ) e − st dt , we would NB. Had we used the definition X ( s ) = UL have d dt x ( t ) ← − − → s X ( s ) − x ( 0 + ) 9/25

Differentiation in Time Domain Example. x ( t ) = e − t , X ( s ) = 1 s + 1 with ROAC Re s > − 1 . By the differentiation property, s 1 UL x ′ ( t ) ← − − → s + 1 − 1 = − ROC ⊃ { Re s > − 1 } s + 1 , By direct calculation, 1 UL x ′ ( t ) = − e − t ← − − → = − s + 1 , ROC = ROAC = { Re s > − 1 } Example. x ( t ) = e − t u ( t ) , X ( s ) = 1 s + 1 with ROAC Re s > − 1 . By the differentiation property, s UL x ′ ( t ) ← − − → ROC ⊃ { Re s > − 1 } s + 1 , By direct calculation, x ′ ( t ) = δ ( t ) − e − t u ( t ) , and 1 s UL { x ′ } = − s + 1 + 1 = s + 1 , ROC = ROAC = { Re s > − 1 } 10/25

Differentiation in Time Domain Under appropriate conditions, e.g. x ( k ) ( t ) = O ( e γ t ) , we can extend the differentiation property to higher derivatives, n − 1 � UL x ( n ) ( t ) → s n X ( s ) − s n − 1 − k x ( k ) ( 0 − ) ← − − k = 0 Since solutions to constant coefficient ODEs is of the form n � p i ( t ) e a i t x ( t ) = k = 1 where p i are polynomials, the above condition is satisfied. Unilateral Laplace transform is useful for solving ODEs with nonzero initial condtions. 11/25

Initial Value Theorem Theorem. If the unilateral Laplace transform X ( s ) of x ( t ) is a proper rational function, then x ( 0 + ) = lim s →∞ s X ( s ) Proof. X ( s ) has partial fraction expansion, r N i A i , k i � � X ( s ) = ( s + a i ) k i i = 1 k i = 1 r N i A i , k i t k i − 1 � � ( k i − 1 )! e − a i t , x ( t ) = t ≥ 0 i = 1 k i = 1 r r N i A i , k i s � � � x ( 0 + ) = A i , 1 = lim ( s + a i ) k i = lim s →∞ s X ( s ) s →∞ i = 1 i = 1 k i = 1 Example. x ( t ) = 2 e − t + e − 2 t , X ( s ) = 3 s + 5 s 2 + 3 s + 2 , x ( 0 + ) = lim s →∞ s X ( s ) = 3 . 12/25

Initial Value Theorem If X ( s ) is rational but not proper, then n � a k s k + X 1 ( s ) X ( s ) = k = 0 where X 1 ( s ) is a proper rational function. Taking inverse transform, n � a k δ ( k ) ( t ) + x 1 ( t ) x ( t ) = k = 0 so x ( 0 + ) = x 1 ( 0 + ) = lim s →∞ s X 1 ( s ) Example. x ( t ) = δ ( t ) + 2 e − t , X ( s ) = 1 + s + 1 = s + 3 2 s + 1 , 2 s x ( 0 + ) = 2 = lim s + 1 � = lim s →∞ s X ( s ) s →∞ 13/25

Example Suppose a CT LTI system has the following properties. 1. The system is causal 2. The system function H ( s ) is a proper rational function and has only two poles, at s = − 2 and s = − 4 . 3. For input x ( t ) = 1 , the output is y ( t ) = 0 4. The impulse response satisfies h ( 0 + ) = 4 Determine the system function H ( s ) . Solution. By 1 and 2 as + b H ( s ) = ( s + 2 )( s + 4 ) , Re s > − 2 By 3, H ( 0 ) = 0 , so b = 0 . By 4, lim s →∞ sH ( s ) = 4 , so a = 4 , and 4 s H ( s ) = ( s + 2 )( s + 4 ) , Re s > − 2 14/25

Final Value Theorem Theorem. Suppose x ( t ) does not contain δ or its derivatives for t ≥ 0 . If X ( s ) converges for real s > 0 and the x ( t ) has a finite limit as t → + ∞ , then t → + ∞ x ( t ) = lim lim s → 0 + sX ( s ) Proof. Since A � lim t → + ∞ x ( t ) exists, x ( t ) is bounded on ( 0 , + ∞ ) , i.e. | x ( t ) | ≤ M . For any ǫ > 0 , ∃ T s.t. | x ( t ) − A | < ǫ for t ≥ T . For s > 0 , � ∞ [ x ( t ) − A ] e − st dt s X ( s ) − A = s 0 � ∞ �� T � | x ( t ) − A | e − st dt = I 1 + I 2 | s X ( s ) − A | ≤ s + 0 T � ∞ � ∞ T e − st dt ≤ ǫ s 0 e − st dt = ǫ . where I 1 ≤ sT ( M + A ) and I 2 ≤ ǫ s s → 0 + | s X ( s ) − A | ≤ ǫ = lim ⇒ lim s → 0 + | s X ( s ) − A | = 0 = ⇒ lim s → 0 + s X ( s ) = A 15/25

Final Value Theorem Example. x ( t ) = 2 + 3 e − t + e − 2 t , X ( s ) = 2 3 1 s + s + 1 + s + 2 , Re s > 0 Check s → 0 + s X ( s ) = 2 = lim lim t → + ∞ x ( t ) The Final Values Theorem can be extended to allow finitely many δ and its derivatives in x ( t ) on the positive real axis. We can rewrite x ( t ) as n n � � UL a i δ ( k i ) ( t − t i ) a i s k i e − st i x ( t ) = x 1 ( t ) + ← − − → X ( s ) = X 1 ( s ) + i = 1 i = 1 t → + ∞ x ( t ) = lim lim t → + ∞ x 1 ( t ) = lim s → 0 + s X 1 ( s ) = lim s → 0 + s X ( s ) Example. x ( t ) = δ ( t ) + 2 , X ( s ) = 1 s + 1 , x (+ ∞ ) = 2 = lim s → 0 + s X ( s ) . 16/25

Linear Constant-coefficient ODE Consider ODE N M d k y d k x � � dt k = a k b k dt k k = 0 k = 0 with initial condition y ( k ) ( 0 − ) , k = 0 , 1 , . . . , N − 1 , and causal input, i.e. x ( t ) = 0 for t < 0 . Take unilateral Laplace transform of both sides � � N k − 1 M � � � s k Y ( s ) − s k − 1 − ℓ y ( ℓ ) ( 0 − ) b k s k X ( s ) a k = k = 0 ℓ = 0 k = 0 so � M � N � k − 1 ℓ = 0 s k − 1 − ℓ y ( ℓ ) ( 0 − ) k = 0 b k s k k = 0 a k Y ( s ) = k = 0 a k s k X ( s ) + � N � N k = 0 a k s k � �� � � �� � zero-state response zero-input response 17/25