The Laplace Transform

Introduction There are two common approaches to the developing and understanding the Laplace transform It can be viewed as a generalization of the CTFT to include some signals with infinite energy It can be seen as a natural consequence of the fact that an LTI system excited by a complex exponential responds with another complex exponential 5/10/04 M. J. Roberts - All Rights Reserved

Generalization of the CTFT The CTFT expresses a time-domain 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, ( ())= ()= () st L x t X s x t e dt A function and its Laplace transform form a transform pair which is conveniently indicated by the notation, L x() t X() s 5/10/04 M. J. Roberts - All Rights Reserved 3

Generalization of the CTFT The variable, s, is viewed as a generalization of the variable, ω, of the form, s= σ + jω. Then, when the real part of s, σ, is zero, the Laplace transform reduces to the CTFT. Using s= σ + jω the Laplace transform is X ()= () σ+ jω t s x t e dt [ σt ] = F x() te ( ) and can be viewed as the CTFT of, not x(t), but rather x() te σt. 5/10/04 M. J. Roberts - All Rights Reserved 5

Generalization of the CTFT e t The extra factor, σ, is sometimes called a convergence factor because, when chosen properly, it makes the integral converge for some signals when 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, σ, is chosen appropriately, the CTFT integral will converge. 5/10/04 M. J. Roberts - All Rights Reserved 6

Complex Exponential Excitation If a continuous-time LTI system is excited by a complex exponential, x()= t Ae st, the 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, as first shown in Chapter 3, that turns out to be y ()= ( ) st s t { Ae τ h τ e dτ x( t ) 1444 3 Laplace transform of h( t ) So for any particular value of s the complex constant is the Laplace transform of the impulse response. 5/10/04 M. J. Roberts - All Rights Reserved 8

Region of Convergence The causal function, ()= () > αt g t Ae u t, α 1 0 does not have a CTFT, even in the generalized sense. But it does have a Laplace transform which is G 1 ()= () = = α s Ae t u t e st dt A e s α t α σ dt A e t e j ω t dt ( ) 0 0 This integral converges if σ > α and this inequality defines what is known as the region of convergence (ROC) of this Laplace transform. ( ) 5/10/04 M. J. Roberts - All Rights Reserved 9

The ROC of the integral G 1 ()= 0 Region of Convergence ( ) α σ t jωt s A e e dt is the region of the s plane for which σ > α and the integral is G 1 s This function has a pole at s = α and the ROC is the region to the right of that point. A ()= α s 5/10/04 M. J. Roberts - All Rights Reserved 10

Region of Convergence By similar reasoning, the ROC of Laplace transform of the anti-causal function, g αt ()= t Ae u( t) = g ( t), α > 1 0 is the region, σ < α, in the s plane and the integral is G ()= s A s + α This function has a pole at s = -α and the ROC is the region to the left of that point. 5/10/04 M. J. Roberts - All Rights Reserved 11

Region of Convergence The following two Laplace transform pairs illustrate the importance of the region of convergence. e e αt αt u () t, σ > α s + α L 1 u ( t), σ < α s + α L 1 The two time-domain functions are different but the algebraic expressions for their Laplace transforms are the same. Only the ROC s are different. 5/10/04 M. J. Roberts - All Rights Reserved 1

Unilateral Laplace Transform Definition G ()= () st s g t e dt 0 Only the lower limit of the integral has changed A function and its Laplace transform are uniquely related only if the function is causal. The ROC is always the region of the s plane to the right of the pole in the transform with the most positive real part. Any function which grows no faster than an exponential in positive time has a unilateral Laplace transform. 5/10/04 M. J. Roberts - All Rights Reserved 13

Unilateral Laplace Transform The unilateral Laplace transform will be referred to simply as the Laplace transform and the bilateral Laplace transform will be identified specifically. There is an inversion integral which is the same for both forms, σ + j π σ j 1 L -1 ( G() s )= g()= t G() se + j The inversion integral is rarely used in practice. Instead tables and properties are used. st ds 5/10/04 M. J. Roberts - All Rights Reserved 14

Properties Let g(t) and h(t) both be causal functions and let them form the following transform pairs, Linearity L L g() t G() s h() t H() s L αg()+ t βh() t αg s βh s ()+ () Time Shifting L st g( t t ) G () s e, t > Complex-Frequency Shifting 0 0 0 0 ( ) st e 0 L g() t G s s 0 5/10/04 M. J. Roberts - All Rights Reserved 16

Time Differentiation Once Properties Time Scaling L 1 s g( at) G, a a Frequency Scaling 1 t g G as, a a L d dt 5/10/04 M. J. Roberts - All Rights Reserved 17 a a > ( ) > L ( g() t ) sg s g 0 0 () ( 0 ) Time Differentiation Twice d L d ( g() t ) s G() s sg( 0 ) g() t dt dt ( ) t= 0

Properties Complex-Frequency Differentiation L d tg() t G ds Multiplication-Convolution Duality L g() t h() t G s H s Integration () () Initial Value Theorem 0 t ( ) () s σ + j g() t h() t G( w) H( s w) dw jπ σ j L 1 L 1 g( τ) dτ G() s s g( )= lim sg() s 0 + s 5/10/04 M. J. Roberts - All Rights Reserved 18

Final Value Theorem Properties lim g()= t lim sg() s t s 0 This theorem only applies if the limit actually exists. It is possible for the limit, lim sg s, to exist even though the s 0 () limit, limg t does not exist. For example, () t L s x()= t cos( ω 0t) X()= s s + ω s lim sx()= s lim = 0 s 0 s 0 s + ω but lim cos t does not exist. t ( ) ω 0 5/10/04 M. J. Roberts - All Rights Reserved 19 0 0

Partial-Fraction 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 N N 1 bn s + bn 1s + L+ bs+ b D D 1 s + a s + La s+ a D 1 1 0 1 0 The denominator can be factored, putting it into the form, ()= G s N N 1 bn s + bn 1s + L+ bs+ b s p s p L s p 1 0 ( )( ) ( ) 1 D 5/10/04 M. J. Roberts - All Rights Reserved 0

Partial-Fraction Expansion For now, assume that there are no repeated poles and that D > N, making the fraction proper in s. Then it is possible to write the expression in the form, where K1 K KD G()= s + + L+ s p s p s p 1 D N N 1 bn s + bn 1s + Lbs+ b s p s p L s p 1 0 D K1 s p ( )( ) ( ) = 1 1 K KD + + L+ s p s p D The K s can be found be any convenient method. 5/10/04 M. J. Roberts - All Rights Reserved 1

Partial-Fraction Expansion Multiply both sides of the previous expression by s p. s p b s N b s N 1 bs b N + N 1 + L+ + 1 s p s p L s p ( ) 1 0 ( )( ) ( ) 1 D 1 K1 K ( s p1 ) + ( s p1 ) s p1 s p = KD + ( s p1 ) s pd Then, since the expression must be valid for any value of s, let s= p 1. All the terms on the right except one are then zero and N N 1 bn p1 + bn p1 + L+ b1p1+ b0 K1 = p p L p p ( ) ( ) 1 1 All the K s can be found by the same method and the inverse Laplace transform is then found by table look-up. D + L 5/10/04 M. J. Roberts - All Rights Reserved

Partial-Fraction Expansion If the expression has a repeated pole of the form, ()= G s N N 1 b s + b s + L+ bs+ b N N 1 1 0 ( s p ) ( s p ) L( s p ) 1 3 D the partial fraction expansion is of the form, ()= G s K1 K ( s p ) + s 1 11 p 1 K3 KD + + L+ s p s p 3 D K 1 K 11 and can be found using the same method as before. But cannot be found using the same method. 5/10/04 M. J. Roberts - All Rights Reserved 3

Partial-Fraction Expansion Instead can be found by using the more general formula K 11 K qk = 1 ( m k)! d ds m k m k ( ) ( ) m s pq H s, k = 1,,, m s p q where m is the order of the qth pole, which applies to repeated poles of any order (pp. 647-649). 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 (p. 649). 5/10/04 M. J. Roberts - All Rights Reserved 4

Partial-Fraction Expansion ()= H s 10s K K ( s+ )( s+ ) = 1 s + + 4 9 4 s + 9 K K 1 = s+ 4 ( ) = s+ 9 ( ) 10s ( s+ 4) s+ 9 ( ) 10s ( s+ 4) s+ 9 ( ) = 10s s + 9 s= 4 s= 4 = 10s s + 4 s= 9 s= 9 40 = = 8 5 90 = 18 5 = s s H()= s 8 s+ + 18 s+ = 8 7 + 18 + 7 4 9 s+ 4 s 9 ( )( + ) = 10s ( s+ 4) s+ 9 ( ) Check. h ( ) () ()= + t e 4t 9t 8 18e u t 5/10/04 M. J. Roberts - All Rights Reserved 5

()= H s Partial-Fraction Expansion s ()= H s 10s + 13s+ 36 Synthetic Division H s 10s s+ 4 s 9 ( )( + ) s + 13s+ 36 10s 130s + 360 3 16 10 s+ 4 s 9 s 4 s 9 ()= 10 ( )( + ) = + + + h ()= () Improper in s 5/10/04 M. J. Roberts - All Rights Reserved 6 ) 10 10s + 130s+ 360 130s 360 [ ] () 9t 4t t 10δ t 16e 3e u t

Partial-Fraction Expansion ()= H s 10s K K K ( s+ ) ( s+ ) = 1 ( s + ) + 11 4 9 4 s + 4 + s + 9 K K 1 Using Dk, Repeated Pole = s+ 4 ( ) 1 d = ( m k)! ds 10s 40 = = 8 s+ 4 ( s 9) 5 s= 4 ( ) + m k m k [ m ( D) ()] = s s p H s, k 1L,,, m p D K 11 = 1 d 1! ds ( ) 1 1 [ ( s+ 4) H( s) ] = s 4 d ds 10s s + 9 s 4 5/10/04 M. J. Roberts - All Rights Reserved 7

Partial-Fraction Expansion ( s+ 910 ) 10s 18 K11 = = ( s + 9) 5 s= 4 18 18 18 8 K = H()= s 5 ( s + ) + 5 s+ + 5 4 4 s + 9 8s 7+ 18 ( 18 s + 13s+ 36) s + 8s+ 16 H()= s 5 5 s+ 4 s 9 ()= H s ( ) ( + ) 10s ( s+ 4) s+ 9 Check ( ) 4t 18 4t 18 9t h()= t 8te + e e u t 5 5 ( ) () 5/10/04 M. J. Roberts - All Rights Reserved 8

Laplace-Transform Fourier- Transform Equivalence If the region of convergence of the Laplace transform contains the ω axis (where σ is zero) then the Laplace transform, along that axis, is the same as the Fourier transform In such a case the Fourier transform can be found from the Laplace transform by substituting jω for s 5/10/04 M. J. Roberts - All Rights Reserved 9

Solution of Differential Equations The unilateral Laplace transform is particularly well suited for the solution of differential equations with initial conditions because the time differentiation properties of the unilateral Laplace transform call for the initial conditions in a systematic way. For example, the Laplace transform of the differential equation, d d [ x() t ]+ 7 [ x() t ]+ 1x()= t 0 dt dt is d s X() s sx( 0 ) ( x() t ) + dt 7 [ sx() s x( 0 ) 1 s 0 t 0 ]+ X()= = an algebraic equation in s which can be solved by algebraic methods. Then, when X(s) is found, its inverse Laplace transform, x(t), is the solution of the original differential equation. 5/10/04 M. J. Roberts - All Rights Reserved 30

The Bilateral Laplace Transform Applies to non-causal signals and systems Can be found using tables of the unilateral Laplace transform Any signal can be expressed as the sum of three parts, the part before time, t = 0 (the anticausal part), the part at time, t = 0 and the part after time, t = 0 (the causal part). ()= ()+ ()+ () x t x t x t x t ac 0 c 5/10/04 M. J. Roberts - All Rights Reserved 31

The Bilateral Laplace Transform To find a bilateral Laplace transform (pp. 657-660) 1. Find the unilateral Laplace transform of the causal part along with its ROC. Find the unilateral Laplace transform of the time inverse of the anti-causal part along with its ROC 3. Change s to -s in the result of step and in its ROC 4. If there is an impulse at time, t = 0, find its Laplace transform and its ROC, the entire s plane 5. Add the results of steps 1, 3 and 4 and form the ROC from the region in the s plane common to all the ROC s. If such a region does not exist, the bilateral transform does not exist either 5/10/04 M. J. Roberts - All Rights Reserved 33