The Laplace Transform


 Darrell Thornton
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1 The Laplace Transform
2 Generalizing the Fourier Transform The CTFT expresses a timedomain signal as a linear combination of complex sinusoids of the form e jωt. In the generalization of the CTFT to the Laplace transform, the complex sinusoids become complex exponentials of the form e st where s can have any complex value. Replacing the complex sinusoids with complex exponentials leads to this definition of the Laplace transform. L( x( t) ) = X s L x t = x t X s e st dt 1/13/11 M. J. Roberts  All Rights Reserved 2
3 Generalizing the Fourier Transform The variable s is viewed as a generalization of the variable ω of the form s = σ + jω. Then, when σ, the real part of s, is zero, the Laplace transform reduces to the CTFT. Using s = σ + jω the Laplace transform is = x t X s e ( σ + jω )t dt = F x( t)e σt which is the Fourier transform of x( t)e σt 1/13/11 M. J. Roberts  All Rights Reserved 3
4 Generalizing the Fourier Transform The extra factor e σt is sometimes called a convergence factor because, when chosen properly, it makes the integral converge for some signals for which it would not otherwise converge. For example, strictly speaking, the signal Au t does not have a CTFT because the integral does not converge. But if it is multiplied by the convergence factor, and the real part of s is chosen appropriately, the CTFT integral will converge. Au( t)e jωt dt Ae σt u( t)e jωt dt 0 = A e jωt dt Does not converge = A e ( σ + jω )t dt Converges (if σ > 0) 0 1/13/11 M. J. Roberts  All Rights Reserved 4
5 Complex Exponential Excitation If a continuoustime LTI system is excited by a complex exponential x t = Ae st, where A and s can each be any complex number, the system response is also a complex exponential of the same functional form except multiplied by a complex constant. The response is the convolution of the excitation with the impulse response and that is = h τ y t x( t τ )dτ = h τ The quantity H s of h( t). = h τ Ae s t τ dτ = Ae st h( τ )e sτ dτ 1/13/11 M. J. Roberts  All Rights Reserved 5 x( t) e sτ dτ is called the Laplace transform
6 Complex Exponential Excitation Let x t = ( 6 + j3) e 3 j 2 A ( s ) t = ( )e ( 3 j 2)t and let h( t) = e 4t u( t). Then H( s) = 1, σ > 4 and, s + 4 in this case, s = 3 j2 = σ + jω with σ = 3 > 4 and ω = 2. y( t) = x( t)h( s) = 6 + j3 3 j2 + 4 e ( 3 j 2)t = ( )e ( 3 j 2)t. The response is the same functional form as the excitation but multiplied by a different complex constant. This only happens when the excitation is a complex exponential and that is what makes complex exponentials unique. 1/13/11 M. J. Roberts  All Rights Reserved 6
7 PierreSimon Laplace 3/23/17493/2/1827 1/13/11 M. J. Roberts  All Rights Reserved 7
8 The Transfer Function Let x( t) be the excitation and let y( t) be the response of a system with impulse response h( t). The Laplace transform of y( t) is Y s = y t Y s e st dt = h( t) x t = h τ Y s = h τ x( t τ )dτ e st dt dτ x( t τ )e st dt e st dt 1/13/11 M. J. Roberts  All Rights Reserved 8
9 The Transfer Function Y s = h τ dτ Let λ = t τ dλ = dt. Then Y s = h τ x( t τ )e st dt dτ x( s( λ+τ ) λ)e dλ = h( τ )e sτ dτ x( λ)e sλ dλ H( s) X( s) Y( s) = H( s)x( s) H( s) is called the transfer function. y( t) = x( t) h( t) L Y( s) = H( s)x( s) 1/13/11 M. J. Roberts  All Rights Reserved 9
10 The Transfer Function Let x( t) = u( t) and let h( t) = e 4t u( t). Find y( t). y( t) = x( t) h( t) = x τ y( t) = y t h( t τ )dτ = u τ ( 4 t τ )e u ( t τ )dτ t t 4( t τ ) e 4t 1 1 e dτ = e 4t e 4τ dτ = e 4t e 4t =, t > , t < 0 ( 1 e 4t )u( t) = 1/ 4 X( s) = 1/ s, H( s) = 1 s + 4 Y ( s ) = 1 s 1 s + 4 = 1/ 4 s x( t) = ( 1/ 4) 1 e 4t u t 1/ 4 s + 4 1/13/11 M. J. Roberts  All Rights Reserved 10
11 CascadeConnected Systems If two systems are cascade connected the transfer function of the overall system is the product of the transfer functions of the two individual systems. 1/13/11 M. J. Roberts  All Rights Reserved 11
12 Direct Form II Realization A very common form of transfer function is a ratio of two polynomials in s, = Y( s) X( s) = N b k s k=0 H s N k=0 k a k s k = b N s N + b N 1 s N b 1 s + b 0 a N s N + a N 1 s N a 1 s + a 0 1/13/11 M. J. Roberts  All Rights Reserved 12
13 The transfer function can be conceived as the product of two transfer functions, and Direct Form II Realization H 1 H 2 ( s) = Y s 1 X s ( s) = Y s s Y 1 = 1 a N s N + a N 1 s N a 1 s + a 0 = b s N N + b N 1 s N b 1 s + b 0 1/13/11 M. J. Roberts  All Rights Reserved 13
14 From we get or X s H 1 ( s) = Y s 1 X s = 1 a N s N + a N 1 s N a 1 s + a 0 X( s) = a N s N + a N 1 s N a 1 s + a 0 Y 1 ( s) = a N s N Y 1 ( s) + a N 1 s N 1 Y 1 ( s) + + a 1 s Y 1 ( s) + a 0 Y 1 ( s) Rearranging s N Y 1 Direct Form II Realization ( s) = 1 a N { X( s) a N 1 s N 1 Y 1 ( s) + + a 1 s Y 1 ( s) + a 0 Y 1 ( s) } 1/13/11 M. J. Roberts  All Rights Reserved 14
15 Direct Form II Realization 1/13/11 M. J. Roberts  All Rights Reserved 15
16 Direct Form II Realization 1/13/11 M. J. Roberts  All Rights Reserved 16
17 Direct Form II Realization A system is defined by y H( s) = ( t) + 3 y t s 5 s 2 + 3s y( t) = x ( t) 5x t. 1/13/11 M. J. Roberts  All Rights Reserved 17
18 Inverse Laplace Transform There is an inversion integral y( t) = 1 j2π σ + j Y( s)e st ds, s = σ + jω σ j for finding y( t) from Y( s), but it is rarely used in practice. Usually inverse Laplace transforms are found by using tables of standard functions and the properties of the Laplace transform. 1/13/11 M. J. Roberts  All Rights Reserved 18
19 If x t Existence of the Laplace Transform Time Limited Signals = 0 for t < t 0 and t > t 1 it is a time limited signal. If is also bounded for all t, the Laplace transform integral x t converges and the Laplace transform exists for all s. 1/13/11 M. J. Roberts  All Rights Reserved 19
20 Existence of the Laplace Transform Let x( t) = rect ( t) = u( t +1/ 2) u( t 1/ 2). X s = rect t e st dt = e st dt = e s/2 e s/2 s 1/2 1/2 = es/2 e s/2 s, All s 1/13/11 M. J. Roberts  All Rights Reserved 20
21 Existence of the Laplace Transform Right and LeftSided Signals RightSided LeftSided 1/13/11 M. J. Roberts  All Rights Reserved 21
22 Existence of the Laplace Transform Right and LeftSided Exponentials = e αt u t t 0 x t RightSided, α x t LeftSided = eβt u t 0 t, β 1/13/11 M. J. Roberts  All Rights Reserved 22
23 Existence of the Laplace RightSided Exponential x( t) = e αt u( t t 0 ), α = e αt e st dt X s If Re s t 0 Transform = e ( α σ )t e jωt dt = σ > α the asymptotic behavior of e ( α σ )t e jωt as t is to approach zero and the Laplace transform integral converges. t 0 1/13/11 M. J. Roberts  All Rights Reserved 23
24 Existence of the Laplace LeftSided Exponential x( t) = e βt u t 0 t = e βt e st dt X s, β Transform t 0 = e ( β σ )t e jωt dt t 0 If σ < β the asymptotic behavior of e ( β σ )t e jωt as t is to approach zero and the Laplace transform integral converges. 1/13/11 M. J. Roberts  All Rights Reserved 24
25 Existence of the Laplace Transform The two conditions σ > α and σ < β define the region of convergence (ROC) for the Laplace transform of right and leftsided signals. 1/13/11 M. J. Roberts  All Rights Reserved 25
26 Existence of the Laplace Transform Any rightsided signal that grows no faster than an exponential in positive time and any leftsided signal that grows no faster than an exponential in negative time has a Laplace transform. = x r ( t) + x l ( t) where x r ( t) is the rightsided part and ( t) is the leftsided part and if x r ( t) < K r e αt and x l ( t) < K l e βt If x t x l and α and β are as small as possible, then the Laplacetransform integral converges and the Laplace transform exists for α < σ < β. Therefore if α < β the ROC is the region α < β. If α > β, there is no ROC and the Laplace transform does not exist. 1/13/11 M. J. Roberts  All Rights Reserved 26
27 Laplace Transform Pairs The Laplace transform of g 1 G 1 ( s) = Ae αt u( t)e st dt ( t) = Ae αt u( t) is 0 = A e ( s α )t dt = A e ( α σ )t e jωt dt = A s α This function has a pole at s = α and the ROC is the region to the right of that point. The Laplace transform of g 2 ( t) = Ae βt u t G 2 0 is ( s) = Ae βt u( t)e st dt = A e ( β s)t dt = A e ( β σ )t e jωt dt = A s β 0 This function has a pole at s = β and the ROC is the region to the left of that point. 0 1/13/11 M. J. Roberts  All Rights Reserved 27
28 Region of Convergence The following two Laplace transform pairs illustrate the importance of the region of convergence. e αt u t L 1 s + α, σ > α e αt L u( t) 1 s + α, σ < α The two timedomain functions are different but the algebraic expressions for their Laplace transforms are the same. Only the ROC s are different. 1/13/11 M. J. Roberts  All Rights Reserved 28
29 Region of Convergence Some of the most common Laplace transform pairs (There is more extensive table in the book.) ramp t u( t) L = t u( t) L e αt u( t) L L δ ( t) 1, All σ u( t) L = t u( t) L, σ > α e αt u( t) L 1/ s, σ > 0 1/ s 2, σ > 0 ramp t 1/ s + α 1/ s, σ < 0 1/ s 2, σ < 0 1/ s + α, σ < α e αt sin( ω 0 t)u t e αt cos( ω 0 t)u t L L ω 0 ( s + α ) ω 0 s + α ( s + α ) ω 0, σ > α e αt sin( ω 0 t)u t, σ > α e αt cos( ω 0 t)u t L L ω 0 ( s + α ) ω 0 s + α ( s + α ) ω 0, σ < α, σ < α
30 Laplace Transform Example Find the Laplace transform of = e t u( t) + e 2t u( t) x t e t u t e 2t u t L 1 s +1, σ > 1 L 1 s 2, σ < 2 e t u( t) + e 2t L u( t) 1 s +1 1 s 2, 1 < σ < 2 1/13/11 M. J. Roberts  All Rights Reserved 30
31 Laplace Transform Example Find the inverse Laplace transform of The ROC tells us that rightsided signal and that a leftsided signal. X( s) = 4 s s 6, 3 < σ < 6 x t 4 must inverse transform into a s must inverse transform into s 6 = 4e 3t u( t) +10e 6t u( t) 1/13/11 M. J. Roberts  All Rights Reserved 31
32 Laplace Transform Example Find the inverse Laplace transform of X( s) = 4 s s 6, σ > 6 The ROC tells us that both terms must inverse transform into a rightsided signal. = 4e 3t u( t) 10e 6t u( t) x t 1/13/11 M. J. Roberts  All Rights Reserved 32
33 Laplace Transform Example Find the inverse Laplace transform of X( s) = 4 s s 6, σ < 3 The ROC tells us that both terms must inverse transform into a leftsided signal. = 4e 3t u( t) +10e 6t u( t) x t 1/13/11 M. J. Roberts  All Rights Reserved 33
34 MATLAB System Objects A MATLAB system object is a special kind of variable in MATLAB that contains all the information about a system. It can be created with the tf command whose syntax is sys = tf(num,den) where num is a vector of numerator coefficients of powers of s, den is a vector of denominator coefficients of powers of s, both in descending order and sys is the system object. 1/13/11 M. J. Roberts  All Rights Reserved 34
35 MATLAB System Objects For example, the transfer function H 1 ( s) = can be created by the commands s s 5 + 4s 4 + 7s 3 +15s s + 75»num = [1 0 4] ; den = [ ] ;»H1 = tf(num,den) ;»H1 Transfer function: s ^ s ^ s ^ s ^ s ^ s /13/11 M. J. Roberts  All Rights Reserved 35
36 PartialFraction Expansion The inverse Laplace transform can always be found (in principle at least) by using the inversion integral. But that is rare in engineering practice. The most common type of Laplacetransform expression is a ratio of polynomials in s, G( s) = b s M M + b M 1 s M b 1 s + b 0 s N + a N 1 s N 1 + a 1 s + a 0 The denominator can be factored, putting it into the form, G s = b M s M + b M 1 s M b 1 s + b 0 ( s p 1 )( s p 2 ) s p N 1/13/11 M. J. Roberts  All Rights Reserved 36
37 PartialFraction Expansion For now, assume that there are no repeated poles and that N > M, making the fraction proper in s. Then it is possible to write the expression in the partial fraction form, where G( s) = K 1 + K s p 1 s p 2 K N s p N b M s M + b M 1 s M 1 + b 1 s + b 0 = K 1 + K ( s p 1 )( s p 2 ) ( s p N ) s p 1 s p 2 The K s can be found be any convenient method. K N s p N 1/13/11 M. J. Roberts  All Rights Reserved 37
38 PartialFraction Expansion Multiply both sides by s p 1 ( s p 1 ) b s M M + b M 1 s M b 1 s + b 0 ( s p 1 )( s p 2 ) ( s p N ) = K 1 + ( s p 1 ) + ( s p 1 ) K 2 + s p 2 K N s p N K 1 = b p M M M 1 + b M p b 1 p 1 + b 0 ( p 1 p 2 ) ( p 1 p N ) All the K s can be found by the same method and the inverse Laplace transform is then found by table lookup. 1/13/11 M. J. Roberts  All Rights Reserved 38
39 H( s) = PartialFraction Expansion 10s s + 4 K 1 = s + 4 ( s + 9) = K 1 K 2 = s + 9 H( s) = 8 s s s ( s + 4) s + 9 s K 2 s + 9, σ > 4 10s ( s + 4) s + 9 h( t) = ( 8e 4t +18e 9t )u( t) = s= 4 = s= 9 10s s s s + 4 8s s + 72 = s + 4 ( s + 9) = 40 s= 4 5 = 8 = 90 s= 9 5 = 18 = 10s s + 4 ( s + 9). Check. 1/13/11 M. J. Roberts  All Rights Reserved 39
40 PartialFraction Expansion If the expression has a repeated pole of the form, = b M s M + b M 1 s M b 1 s + b 0 G s ( s p 1 ) 2 ( s p 3 ) ( s p N ) the partial fraction expansion is of the form, G( s) = K 12 K N ( s p 1 ) + K 11 + K s p 1 s p 3 s p N and K 12 can be found using the same method as before. But K 11 cannot be found using the same method. 1/13/11 M. J. Roberts  All Rights Reserved 40
41 PartialFraction Expansion Instead K 11 can be found by using the more general formula K qk = 1 ( m k)! d m k ds m k ( s p ) m q H s s pq, k = 1,2,, m where m is the order of the qth pole, which applies to repeated poles of any order. If the expression is not a proper fraction in s the partialfraction method will not work. But it is always possible to synthetically divide the numerator by the denominator until the remainder is a proper fraction and then apply partialfraction expansion. 1/13/11 M. J. Roberts  All Rights Reserved 41
42 PartialFraction Expansion H( s) = 10s ( s + 4) 2 s + 9 Repeated Pole K 12 = s + 4 Using K qk = K 11 = = K 12 ( s + 4) + K s ( s + 4) 2 ( s + 9) 1 d m k ( m k)! ds m k 1 ( 2 1)! d 2 1 s + 4 ds 2 1 s= 4 ( s p ) m q H s 2 H( s) s K 2 s + 9, σ > 4 = 40 5 = 8 s pq = d s 4 ds, k = 1,2,, m 10s s + 9 s 4 1/13/11 M. J. Roberts  All Rights Reserved 42
43 PartialFraction Expansion 1/13/11 M. J. Roberts  All Rights Reserved 43
44 H( s) = H( s) = PartialFraction Expansion 10s 2 s + 4 ( s + 9) 10s 2 s 2 +13s + 36, σ > 4, σ > 4 Improper in s Synthetic Division s 2 +13s s 2 H( s) = s s + 4 ( s + 9) h( t) = 10δ ( t) 162e 9t 32e 4t 10 10s s = 10 s s + 9 u t 130s 360, σ > 4 1/13/11 M. J. Roberts  All Rights Reserved 44
45 Inverse Laplace Transform Example G( s) = G( s) = G( s) = 3 / 2 g t Method 1 s ( s 3) s 2 4s + 5 s ( s 3) s 2 + j ( s 2 j) / 4 / 4 s j s 2 + j = 3 2 e3t j 3 j s 2 j, σ < 2 4 e ( 2 j )t + 3 j 4 e 2+ j t, σ < 2, σ < 2 u( t) 1/13/11 M. J. Roberts  All Rights Reserved 45
46 Inverse Laplace Transform Example = 3 2 e3t j g t 4 e ( 2 j )t + 3 j 4 e 2+ j t u( t) This looks like a function of time that is complexvalued. But, with the use of some trigonometric identities it can be put into the form = ( 3 / 2) e 2t cos( t) + 1/ 3 g t which has only real values. { sin( t) e3t }u t 1/13/11 M. J. Roberts  All Rights Reserved 46
47 Inverse Laplace Transform Example G( s) = G( s) = G( s) = 3 / 2 Method 2 s ( s 3) s 2 4s + 5 s ( s 3) s 2 + j ( s 2 j) / 4 / 4 s j s 2 + j 3 j s 2 j, σ < 2, σ < 2, σ < 2 Getting a common denominator and simplifying G( s) = 3 / 2 s s 10 s 2 4s + 5 = 3 / 2 s s 5 / 3 ( s 2) 2 + 1, σ < 2 1/13/11 M. J. Roberts  All Rights Reserved 47
48 Inverse Laplace Transform Example Method 2 G( s) = 3 / 2 s s 5 / 3 ( s 2) 2 +1, σ < 2 The denominator of the second term has the form of the Laplace transform of a damped cosine or damped sine but the numerator is not yet in the correct form. But by adding and subtracting the correct expression from that term and factoring we can put it into the form G( s) = 3 / 2 s s 2 ( s 2) / 3 s , σ < 2 1/13/11 M. J. Roberts  All Rights Reserved 48
49 Inverse Laplace Transform Example Method 2 G( s) = 3 / 2 s 3 3 s 2 2 ( s 2) / 3 ( s 2) 2, σ < 2 +1 This can now be directly inverse Laplace transformed into g( t) = ( 3 / 2) { e 2t cos( t) + ( 1/ 3)sin( t) e3t }u( t) which is the same as the previous result. 1/13/11 M. J. Roberts  All Rights Reserved 49
50 Inverse Laplace Transform Example Method 3 When we have a pair of poles p 2 and p 3 that are complex conjugates we can convert the form G( s) = A s 3 + K 2 + K 3 s p 2 s p 3 K 3 p 2 K 2 p 3 into the form G( s) = A s 3 + s K 2 + K 3 = A s 2 ( p 1 + p 2 )s + p 1 p 2 s 3 + Bs + C s 2 ( p 1 + p 2 )s + p 1 p 2 In this example we can find the constants A, B and C by realizing that G( s) = s ( s 3) ( s 2 4s + 5) A s 3 + Bs + C s 2 4s + 5, σ < 2 is not just an equation, it is an identity. That means it must be an equality for any value of s. 1/13/11 M. J. Roberts  All Rights Reserved 50
51 Inverse Laplace Transform Example Method 3 A can be found as before to be 3 / 2. Letting s = 0, the identity becomes 0 3 / C and C = 5 / 2. Then, letting 5 s = 1, and solving we get B = 3 / 2. Now G( s) = 3 / 2 s 3 + ( 3 / 2)s + 5 / 2, σ < 2 s 2 4s + 5 or G( s) = 3 / 2 s 3 3 s 5 / 3 2 s 2 4s + 5, σ < 2 This is the same as a result in Method 2 and the rest of the solution is also the same. The advantage of this method is that all the numbers are real. 1/13/11 M. J. Roberts  All Rights Reserved 51
52 Use of MATLAB in Partial Fraction Expansion MATLAB has a function residue that can be very helpful in partial fraction expansion. Its syntax is [r,p,k] = residue(b,a) where b is a vector of coefficients of descending powers of s in the numerator of the expression and a is a vector of coefficients of descending powers of s in the denominator of the expression, r is a vector of residues, p is a vector of finite pole locations and k is a vector of socalled direct terms which result when the degree of the numerator is equal to or greater than the degree of the denominator. For our purposes, residues are simply the numerators in the partialfraction expansion. 1/13/11 M. J. Roberts  All Rights Reserved 52
53 Let g t and h t Laplace Transform Properties L and h( t) form the transform pairs, g( t) G( s) L H( s) with ROC's, ROC G and ROC H respectively. Linearity L α g( t) + β h( t) α G s + β H( s) ROC ROC G ROC H L Time Shifting g( t t 0 ) G( s)e st 0 sdomain Shift ROC = ROC G L e s0t g t G s s 0 ROC = ROC G shifted by s 0, (s is in ROC if s s 0 is in ROC G ) 1/13/11 M. J. Roberts  All Rights Reserved 53
54 Laplace Transform Properties Time Scaling Time Differentiation sdomain Differentiation L g at ( 1/ a )G s / a ROC = ROC G scaled by a (s is in ROC if s / a is in ROC G ) d dt g( t) L ROC ROC G t g t L sg( s) d ds G( s) ROC = ROC G 1/13/11 M. J. Roberts  All Rights Reserved 54
55 Laplace Transform Properties Convolution in Time Time Integration g( t) h t L G( s)h( s) ROC ROC G ROC H t L g( τ )dτ G( s) / s ROC ROC G σ > 0 If g( t) = 0, t < 0 and there are no impulses or higherorder singularities at t = 0 then Initial Value Theorem: g( 0 + ) = lim sg s s Final Value Theorem: lim g t t = lim s 0 sg s if lim t g t exists 1/13/11 M. J. Roberts  All Rights Reserved 55
56 Laplace Transform Properties Final Value Theorem = lim limg t sg s t s 0 This theorem only applies if the limit lim g t t It is possible for the limit lim sg s s 0 limit lim g t t does not exist. For example x( t) = cos ω 0 t L actually exists. to exist even though the X( s) = s lim s X s s 0 2 = lim s 0 s 2 + ω = but lim cos ω 0 t t does not exist. s s 2 + ω 0 2 1/13/11 M. J. Roberts  All Rights Reserved 56
57 Laplace Transform Properties Final Value Theorem if all the lie in the open left half of the s plane. Be sure must could have a single The final value theorem applies to a function G s poles of sg s to notice that this does not say that all the poles of G s lie in the open left half of the s plane. G s pole at s = 0 and the final value theorem would still apply. 1/13/11 M. J. Roberts  All Rights Reserved 57
58 Use of Laplace Transform Properties = u( t) u( t a) and L L L Find the Laplace transforms of x t x( 2t) = u( 2t) u( 2t a). From the table u t Then, using the timeshifting property u t a Using the linearity property u( t) u t a Using the timescaling property L u( 2t) u( 2t a) e as s s s/2 = 1 / s, σ > 0. e as / s, σ > 0. ( 1 e as ) / s, σ > 0. 1 e as/2 s, σ > 0 1/13/11 M. J. Roberts  All Rights Reserved 58
59 Use of Laplace Transform Properties Use the sdomain differentiation property and L u t 1 / s, σ > 0 to find the inverse Laplace transform of 1 / s 2. The sdomain differentiation property is t g t L d ( ds G( s) ). Then t u t L d ds 1 s = 1. Then using the linearity property 2 s t u t L 1 s. 2 1/13/11 M. J. Roberts  All Rights Reserved 59
60 The Unilateral Laplace Transform In most practical signal and system analysis using the Laplace transform a modified form of the transform, called the unilateral Laplace transform, is used. The unilateral Laplace transform is defined by G s = g t 0 e st dt. The only difference between this version and the previous definition is the change of the lower integration limit from to 0. With this definition, all the Laplace transforms of causal functions are the same as before with the same ROC, the region of the s plane to the right of all the finite poles. 1/13/11 M. J. Roberts  All Rights Reserved 60
61 The Unilateral Laplace Transform The unilateral Laplace transform integral excludes negative time. If a function has nonzero behavior in negative time its unilateral and bilateral transforms will be different. Also functions with the same positive time behavior but different negative time behavior will have the same unilateral Laplace transform. Therefore, to avoid ambiguity and confusion, the unilateral Laplace transform should only be used in analysis of causal signals and systems. This is a limitation but in most practical analysis this limitation is not significant and the unilateral Laplace transform actually has advantages. 1/13/11 M. J. Roberts  All Rights Reserved 61
62 The Unilateral Laplace Transform The main advantage of the unilateral Laplace transform is that the ROC is simpler than for the bilateral Laplace transform and, in most practical analysis, involved consideration of the ROC is unnecessary. The inverse Laplace transform is unchanged. It is g( t) = 1 j2π σ + j σ j G( s)e + st ds 1/13/11 M. J. Roberts  All Rights Reserved 62
63 The Unilateral Laplace Transform Some of the properties of the unilateral Laplace transform are different from the bilateral Laplace transform. TimeShifting L g( t t 0 ) G( s)e 0, t 0 > 0 Time Scaling L g( at) ( 1 / a )G s / a First Time Derivative Nth Time Derivative Time Integration d N dt N t 0 d dt g t L ( g( t) L ) sg( s) g( 0 ) s N G s L g( τ )dτ G( s) / s, a > 0 N d n 1 s N n n=1 dt n 1 ( g( t) ) t =0 1/13/11 M. J. Roberts  All Rights Reserved 63
64 The Unilateral Laplace Transform The time shifting property applies only for shifts to the right because a shift to the left could cause a signal to become noncausal. For the same reason scaling in time must only be done with positive scaling coefficients so that time is not reversed producing an anticausal function. The derivative property must now take into account the initial value of the function at time t = 0 and the integral property applies only to functional behavior after time t = 0. Since the unilateral and bilateral Laplace transforms are the same for causal functions, the bilateral table of transform pairs can be used for causal functions. 1/13/11 M. J. Roberts  All Rights Reserved 64
65 The Unilateral Laplace Transform The Laplace transform was developed for the solution of differential equations and the unilateral form is especially well suited for solving differential equations with initial conditions. For example, d 2 dt 2 x t + 7 d dt with initial conditions x 0 x t = 2 and d dt + 12 x( t) = 0 ( x( t) ) t =0 = 4. Laplace transforming both sides of the equation, using the new derivative property for unilateral Laplace transforms, s 2 X( s) s x( 0 ) d dt ( x( t) ) t =0 + 7 s X( s) x X s = 0 1/13/11 M. J. Roberts  All Rights Reserved 65
66 The Unilateral Laplace Transform Solving for X( s) or X( s) = x t =2 =2 = 4 s x( 0 ) + 7x( 0 ) + d ( x( t) ) dt t =0 s 2 + 7s s + 10 s 2 + 7s + 12 = 4 s The inverse transform yields s + 4. This solution solves the differential equation X( s) = u t = 4e 3t 2e 4t with the given initial conditions. 1/13/11 M. J. Roberts  All Rights Reserved 66
67 PoleZero Diagrams and Frequency Response If the transfer function of a stable system is H(s), the frequency response is H(jω). The most common type of transfer function is of the form, H s Therefore H(jω) is H jω = A s z 1 s z 2 ( s p 1 ) s p 2 = A jω z 1 s z M s p N jω z 2 ( jω p 1 ) jω p 2 jω z M jω p N 1/13/11 M. J. Roberts  All Rights Reserved 67
68 Let H( s) = 3s s + 3. H( jω ) = 3 jω jω + 3 The numerator jω and the denominator jω + 3 can be conceived as vectors in the s plane. H( jω ) = 3 PoleZero Diagrams and Frequency Response jω jω + 3 H jω = 3 =0 + jω ( jω + 3 ) 1/13/11 M. J. Roberts  All Rights Reserved 68
69 PoleZero Diagrams and Frequency Response lim ω 0 H( jω ) = lim 3 ω 0 jω jω + 3 = 0 lim ω 0 + H( jω ) = lim 3 ω 0 + jω jω + 3 = 0 lim H( jω ) = lim 3 ω ω jω jω + 3 = 3 lim H( jω ) = lim 3 ω + ω + jω jω + 3 = 3 1/13/11 M. J. Roberts  All Rights Reserved 69
70 PoleZero Diagrams and Frequency Response lim H ( jω ) = π ω = π 2 lim H ( jω ) = π + ω = π 2 lim H jω ω = π 2 π 2 = 0 lim H jω ω + = π 2 π 2 = 0 1/13/11 M. J. Roberts  All Rights Reserved 70
71 PoleZero Diagrams and Frequency Response 1/13/11 M. J. Roberts  All Rights Reserved 71
72 PoleZero Diagrams and Frequency Response 1/13/11 M. J. Roberts  All Rights Reserved 72
73 PoleZero Diagrams and Frequency Response 1/13/11 M. J. Roberts  All Rights Reserved 73
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