EI331 Signals and Systems Lecture 30 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University June 11, 2019
Contents 1. Properties of Laplace Transforms 2. Inverse Laplace Transform 3. Laplace Transform of Singularity Functions 4. Analysis of CT LTI Systems by Laplace Transform 5. Block Diagram Representations 1/42
Convolution Property If L x ( t ) ← − − → X ( s ) with ROAC = R 1 L y ( t ) ← − − → Y ( s ) with ROAC = R 2 then L ( x ∗ y )( t ) ← − − → X ( s ) Y ( s ) with ROAC ⊃ R 1 ∩ R 2 A more precise statement is the following. � ∞ � ∞ x ( t ) e − st dt and Y ( s ) = y ( t ) e − st dt Theorem. If both X ( s ) = −∞ −∞ converges absolutely at some s = s 0 , then the Laplace transform of z = x ∗ y converges absolutely at s = s 0 , and � ∞ z ( t ) e − s 0 t dt X ( s 0 ) Y ( s 0 ) = Z ( s 0 ) = −∞ 2/42
Convolution Property Proof. � ∞ �� ∞ � y ( τ ) e − s ( v + τ ) dv X ( s ) Y ( s ) = x ( v ) d τ −∞ −∞ � ∞ �� ∞ � y ( t − v ) e − sv dv = x ( v ) ( t = v + τ ) dt −∞ −∞ � ∞ �� ∞ � e − sv dv = x ( v ) y ( t − v ) dt ( Fubini’s Theorem ) −∞ −∞ NB. The ROAC of L { x ∗ y } may be larger than the common ROAC of L { x } and L { y } . s + 1 1 Example. X 1 ( s ) = ( s + 2 ) 2 has ROAC Re s > − 2 , X 2 ( s ) = s + 1 has 1 ROAC Re s > − 1 , but X ( s ) = X 1 ( s ) X 2 ( s ) = ( s + 2 ) 2 with ROAC Re s > − 2 , due to pole-zero cancellation at s = − 1 . 3/42
Differentiation in Time Domain If L x ( t ) ← − − → X ( s ) , with ROC = R t →±∞ x ( t ) e − st = 0 for s ∈ R 0 , then and lim d L dtx ( t ) ← − − → sX ( s ) , with ROC ⊃ R ∩ R 0 Proof. Integration by parts yields � ∞ � ∞ � ∞ ∞ x ′ ( t ) e − st dt = x ( t ) e − st � x ( t ) e − st dt = s x ( t ) e − st dt −∞ + s � � −∞ −∞ −∞ NB. ROC may enlarge or shrink L Example. x ( t ) = ( 1 − e − t ) u ( t ) 1 ← − − → s ( s + 1 ) with ROC = ROAC L Re s > 0 , and x ′ ( t ) = e − t u ( t ) 1 ← − − → s + 1 with ROC = ROAC Re s > − 1 . The ROC of L { x ′ } is larger that that of L { x } . 4/42
Differentiation in Time Domain Example. Consider x ( t ) = e kt sin( e kt ) with k > 0 . • For s = σ ∈ R , u = e kt yields (cf. slide 18) � ∞ � ∞ � ∞ � 1 sin u sin u sin u x ( t ) e − st dt = 1 u σ/ k du = 1 u σ/ k du + 1 u σ/ k du k k k −∞ 0 0 1 ◮ � ∞ sin u u σ/ k du has ROAC Re s > k and ROC Re s > 0 1 � 1 ◮ As u ↓ 0 , sin u sin u u σ/ k ∼ u 1 − σ/ k , so u σ/ k du has ROAC Re s < 2 k 0 ◮ Thus L { x } has ROC k < Re s < 2 k and ROC 0 < Re s < 2 k . • x ′ ( t ) = ke kt sin( e kt ) + ke 2 kt cos( e kt ) . For s = σ ∈ R , � ∞ � ∞ sin u + u cos u x ′ ( t ) e − st dt = du u σ/ k −∞ 0 L { x ′ } has empty ROAC, and ROC k < Re s < 2 k t →±∞ x ( t ) e − st = 0 fails for s with 0 < Re s < k • Note lim 5/42
Differentiation in Time Domain If x ( t ) = O ( e at ) as t → + ∞ and x ( t ) = O ( e bt ) as t → −∞ , then L x ( t ) ← − − → X ( s ) , with ROAC containing a < Re s < b and d L dtx ( t ) ← − − → sX ( s ) , with ROC containing a < Re s < b NB. In general, from the absolute convergence of L { x } at s = s 0 we can only conclude the convergence of L { x ′ } at s = s 0 . NB. We mostly deal with x ( t ) of the form � m k = 0 p k ( t ) e α k t u ( ± t + β k ) , where p k are polynomials. After introducing Laplace transform for singularity functions, we have for such functions, d n L → s n X ( s ) , dt n x ( t ) ← − − with ROC containing a < Re s < b 6/42
Differentiation in s -domain L If x ( t ) ← − − → X ( s ) , ROAC σ 1 < Re s < σ 2 then → d L − tx ( t ) ← − − dsX ( s ) , ROAC σ 1 < Re s < σ 2 “Proof”. Differentiating under integral sign (can be justified) � ∞ � ∞ � ∞ d x ( t ) d x ( t ) e − st dt = dse − st dt = − tx ( t ) e − st dt ds −∞ −∞ −∞ Example. 1 L e − at u ( t ) ← − − → s + a , Re s > − a � n � n ! − d 1 L t n e − at u ( t ) ← − − → s + a = ( s + a ) n + 1 , Re s > − a ds n ! L − t n e − at u ( − t ) Similarly, ← − − → Re s < − a ( s + a ) n + 1 , 7/42
Integration in Time Domain If L x ( t ) ← − − → X ( s ) , ROAC = R then � t → 1 L x ( τ ) d τ ← − − s X ( s ) , ROAC ⊃ R ∩ { Re s > 0 } −∞ Proof. Follows from convolution property and the following � t x ( τ ) d τ = ( x ∗ u )( t ) −∞ and → 1 L u ( t ) ← − − with ROAC = { Re s > 0 } s NB. This property holds also for Laplace transforms of singularity functions and will be useful computing some Laplace transforms. 8/42
Contents 1. Properties of Laplace Transforms 2. Inverse Laplace Transform 3. Laplace Transform of Singularity Functions 4. Analysis of CT LTI Systems by Laplace Transform 5. Block Diagram Representations 9/42
Inverse Laplace Transform Recall Laplace transform is related to CTFT by � ∞ { x ( t ) e − σ t } e − j ω t dt = F { x ( t ) e − σ t } X ( σ + j ω ) = −∞ Take the inverse Fourier transform for s = σ + j ω ∈ ROC, � ∞ x ( t ) e − σ t = 1 X ( σ + j ω ) e j ω t d ω 2 π −∞ so � ∞ x ( t ) = 1 X ( σ + j ω ) e ( σ + j ω ) t d ω 2 π −∞ or � σ + j ∞ � σ + jA 1 1 X ( s ) e st ds = lim X ( s ) e st ds x ( t ) = j 2 π j 2 π A →∞ σ − j ∞ σ − jA 10/42
Inverse Transform by Partial Fraction Expansion For rational Laplace transform, the inverse transform can be found by partial fraction expansion. Recall a proper rational function has the following partial fraction expansion r N i A i , k i � � R ( s ) = ( s + a i ) k i i = 1 k i = 1 Also recall t n − 1 1 L ( n − 1 )! e − at u ( t ) ← − − → ( s + a ) n , Re s > − a t n − 1 1 L ( n − 1 )! e − at u ( − t ) − ← − − → ( s + a ) n , Re s < − a By linearity, L − 1 { R } is a linear combination of terms of the above form, where signs are chosen according to the ROC. 11/42
Example 1 Consider a rational Laplace transform X ( s ) = ( s + 1 )( s + 2 ) 2 s + 1 + − 1 1 − 1 X ( s ) = s + 2 + ( s + 2 ) 2 Two poles at s = − 1 and s = − 2 . 1. If ROC is Re s > − 1 x ( t ) = e − t u ( t ) − ( 1 + t ) e − 2 t u ( t ) 2. If ROC is Re s < − 2 x ( t ) = − e − t u ( − t ) + ( 1 + t ) e − 2 t u ( − t ) 3. If ROC is − 2 < Re s < − 1 x ( t ) = − e − t u ( − t ) − ( 1 + t ) e − 2 t u ( t ) 12/42
Inverse Transform by Contour Integration Suppose X ( s ) has finitely many finite isolated singularities s 1 , . . . , s n , and ROC is −∞ ≤ σ 1 < Re s < σ 2 ≤ + ∞ . y The inverse transform is � σ + jr σ + jr 1 X ( s ) e st ds x ( t ) = lim j 2 π r →∞ σ − jr C r X s 3 where σ 1 < σ < σ 2 . s 1 X For t > 0 , choose a large semicircle C r X x that encloses all s k with Re s ≤ σ 1 . s 2 σ 1 σ 2 If X ( s ) satisfies 1 (a shifted and rotated X s 4 version of) Jordan’s Lemma (slide 6, Lecture 27), Residue Theorem implies σ − jr � Res[ X ( s ) e st , s k ] , x ( t ) = t > 0 k : s k ≤ σ 1 1 satisfied by proper rational functions 13/42
Inverse Transform by Contour Integration Suppose X ( s ) has finitely many finite isolated singularities s 1 , . . . , s n , and ROC is −∞ ≤ σ 1 < Re s < σ 2 ≤ + ∞ . The inverse transform is y � σ + jr 1 X ( s ) e st ds x ( t ) = lim σ + jr j 2 π r →∞ C r σ − jr X s 3 where σ 1 < σ < σ 2 . s 1 X For t < 0 , choose a large semicircle C r that encloses all s k with Re s ≥ σ 2 . X x σ 1 σ 2 s 2 If X ( s ) satisfies (a shifted and rotated X s 4 version of) Jordan’s Lemma, Residue Theorem implies σ − jr � Res[ X ( s ) e st , s k ] , x ( t ) = − t < 0 k : s k ≥ σ 2 Caution. Negative sign due to negative orientation of contour. 14/42
Example 1 Consider a rational Laplace transform X ( s ) = ( s + 1 )( s + 2 ) 2 e st � s = − 1 = e − t Res[ X ( s ) e st , − 1 ] = � ( s + 2 ) 2 � � d e st � = − ( 1 + t ) e − 2 t Res[ X ( s ) e st , − 2 ] = ds s + 1 s = − 2 1. If ROC is Re s > − 1 , then σ 1 = − 1 , σ 2 = + ∞ . For t > 0 , Res[ X ( s ) e st , s k ] = e − t − ( 1 + t ) e − 2 t � x ( t ) = k : Re s k ≤− 1 For t < 0 , � Res[ X ( s ) e st , s k ] = 0 x ( t ) = − k : Re s k ≥ + ∞ 15/42
Example (cont’d) 1 Consider a rational Laplace transform X ( s ) = ( s + 1 )( s + 2 ) 2 e st � s = − 1 = e − t Res[ X ( s ) e st , − 1 ] = � ( s + 2 ) 2 � � d e st � = − ( 1 + t ) e − 2 t Res[ X ( s ) e st , − 2 ] = ds s + 1 s = − 2 2. If ROC is Re s < − 2 , then σ 1 = −∞ , σ 2 = − 2 . For t > 0 , � Res[ X ( s ) e st , s k ] = 0 x ( t ) = k : Re s k ≤−∞ For t < 0 , Res[ X ( s ) e st , s k ] = − e − t + ( 1 + t ) e − 2 t � x ( t ) = − k : Re s k ≥− 2 16/42
Recommend
More recommend
