ENGIN 211, Engineering Math. Laplace Transforms
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1 ENGIN 211, Engineering Math Laplace Transforms 1
2 Why Laplace Transform? Laplace transform converts a function in the time domain to its frequency domain. It is a powerful, systematic method in solving differential equations. It converts differential equations into algebraic equations. Initial or boundary conditions are automatically accounted for. In situations where other methods fail because the solution is discontinuous, it can succeed. 2
3 Definition L f t = e st f t dt = F s 0 The parameter s known as the complex frequency is assumed to have positive and large enough real part to ensure that the integral converges. 3
4 Useful Transforms f t = a F s = a s f t = e at F s = 1 s a f t = sin ωt F s = ω s 2 +ω 2 f t = cos ωt F s = s s 2 +ω 2 f t = t n F s = n! s n+1 f t = sinh ωt F s = ω s 2 ω 2 f t = cosh ωt F s = s s 2 ω 2 4
5 The Unit Impulse δ(t) The unit impulse function is defined as δ(t) = du(t) dt where u(t) is the unit step function. Total area underneath δ(t): Sifting property of δ(t): δ(t t 0 )dt f(t)δ(t t 0 )dt δ(t t 0 ) = 1 = f(t 0 ) Laplace transform: L δ(t) = 1, L δ(t t 0 ) = e st 0 5
6 Properties Linearity L f t ± g t = L f t ± L g t But L kf t = kl f t L f t g t L f t L g t Rather the convolution of two function, L f t g t = L f t L g t 6
7 Examples L 2 sin 3t + 4 sinh 3t = s 2 +9 s 2 9 = 18 s2 +3 s 4 81 L 5e 4t + cosh 2t = 5 s 4 + s s 2 4 = 6s2 4s 20 s 4 s 4 4 L t3 + 2t 2 4t + 1 = 3! s ! s s s = 1 s 4 s3 4s 2 + 4s + 6 7
8 Main Theorems Assume L f t = F s : Similarity theorem L f at = 1 a F s a Real Shifting theorem: for a > 0 L f t a = e as F s Complex Shifting theorem: for a > 0 L e at f t = F s + a 8
9 Main Theorems (Cont d) Derivative theorem: L df t dt = sf s f 0 + L d2 f t dt 2 = s 2 F s sf 0 + f 0 + Complex differentiation theorem: L t n f t = 1 n dn F s ds n 9
10 Main Theorems (Cont d) Real integral theorem: L t f τ 0 dτ = F s s Complex integral theorem: L f t t = F σ s f t dσ if lim t 0 t exists t Convolution theorem: if f 1 f 2 = f 1 τ 0 f 2 t τ dτ L f 1 f 2 = F 1 s F 2 s 10
11 Main Theorems (Cont d) Convolution theorem: Initial theorem: f 1 f 2 = L f 1 f 2 t f 1 τ 0 f 2 t τ dτ = F 1 s F 2 s f 0 + Final theorem: = lim s sf s if lim t 0 f t exists f = lim s 0 sf s if lim f t exists t 11
12 Example (Shifting theorem) The real shifting theorem states that if L f t = F s, then L e at f t = F s + a Because L sin 3t = 3 s 2 +9, then L e 2t sin 3t = Because L t 3 = 3! s4, then L t 3 e 4t = 3 s = 3 s 2 + 4s ! s = 6 s
13 Example (Complex Differentiation) Complex differentiation theorem: L t n f t = 1 n dn F s ds n If L cosh 3t = s, then s 2 9 L t 2 cosh 3t = d2 ds 2 s s 2 9 = 2s s s
14 Example (Complex Integral) L f t t Example: find L 1 cos 2t Since lim t 0 t = F σ s 1 cos 2t t =? 2 sin = lim 2 t t 0 t f t dσ if lim t 0 t exists = 0 does exist, we obtain L 1 cos 2t t = ln = σ s 1 σ σ s σ σ = 0 ln dσ = lnσ 1 2 ln σ2 + 4 s s = ln s2 + 4 s s 14
15 Inverse Laplace Transform Given Laplace transform F s in s-domain, find its function f t in t-domain. L 1 F s = f(t) Useful pairs F s = a s F s = 1 s a F s = F s = f t = a f t = ω s 2 f t + ω2 s s 2 f t + ω2 eat = sin ωt = cos ωt F s = F s = F s = n! f t sn+1 = ω s 2 f t ω2 s s 2 f t ω2 tn = sinh ωt = cosh ωt 15
16 Example Find L 1 4s 2 5s+6 s+1 s 2 +4 =? Solution: 4s 2 5s+6 = A + Bs+C s+1 s 2 +4 s+1 s s 2 5s + 6 = A s Bs + C s + 1 s + 1 = 0 A = 3 Equating coefficients of various powers B = 1, and C = 6 L 1 3 s s 6 s = L 1 3 s = 3e t + cos 2t 3 sin 2t s s s
17 Partial Fractions Order in numerator less than denominator Factorize the denominator into prime factors A linear factor s + a gives A s+a A repeated factor s + a 2 gives A s+a + B s+a 2 A quadratic factor s 2 + ps + q gives Ps+Q s 2 +ps+q 17
18 Example (Partial Fractions) s 2 9s + 7 s 2 + s 2 s 2 + 2s + 2 = A s B s 1 + Cs + D s 2 + 2s + 2 A = s 2 9s+7 s 1 s 2 +2s+2 s= 2 = 4 and B = s 2 9s+7 s+2 s 2 +2s+2 s=1 = 2 5 s 2 9s + 7 = A s 1 s 2 + 2s B s + 2 s 2 + 2s Cs + D s 2 + s 2 Equating highest-order s 3 coefficient A + B + C = 0 C = 22 Equating lowest-order s 0 coefficient 2A + 4B 2D = 0 D = 16 L 1 A s B s 1 + Cs + D s 2 + 2s + 2 = L 1 4 s s = L 1 4 s s s s = L 1 4 s s s + 1 s = 4e 2t 2 5 et e t sin t 6 5 e t cos t s s + 8 s
19 Applying to Differential Equations With the use of Derivative theorems: L df t dt = sf s f 0 + L d2 f t dt 2 = s 2 F s sf 0 + f 0 + We can convert differential equations into algebraic equations where all initial conditions are taken into account. Effectively, we are converting problems in t-domain into s-domain. In the end, we need to perform inverse transform to go back to t-domain. 19
20 1 st Order Differential Equation Example: solve dx dt + 4x = 2e 2t + 4e 4t, at t = 0, x = 2. Taking Laplace on both sides to s-domain: sx 2 + 4X = 2 s s + 4 Then X = 2 s s+2 s s+4 2 = 2 s s+2 1 s s+4 2 Inverse Laplace back to t-domain X = 1 s s s x t = L 1 X = e 2t + e 4t + 4te 4t 20
21 2 nd Order Differential Equation Example: x" + 5x + 6x = 24 sin 2t, at t = 0, x = 0 and x = 0 Taking Laplace transform on both sides: Plug in initial conditions: s 2 X sx 0 x sx x(0) + 6X = 24 s 2 + 5s + 6 X = 48 s s So X = 48 = A + B + Cs+D s 2 +5s+6 s 2 +4 s+2 s+3 s 2 +4 where A = And 48 (s+3) s 2 +4 s= 2 = 6, B = 48 = 48 (s+2) s 2 +4 s= 3 13 A s + 3 s B s + 2 s Cs + D s + 2 s + 3 = 48 21
22 2 nd Order Differential Equation (Cont d) Equating the coefficients for s 3 and s 0, we obtain Thus, Taking inverse Laplace: A + B + C = 0, and 12A + 8B + 6D = 48 C = A + B = A 4B, and D = 13 3 x t = L 1 X = L 1 A s B Cs + D + s + 3 s = = 6L 1 1 s L 1 s s 13 L 1 s L 1 s = 6e 2t e 3t cos 2t + 6 sin 2t 13 22
23 System of Differential Equations x" + 2x y = 0 y" + 2y x = 0, with initial conditions x 0 = 4 y 0 = 2 and x 0 = 0 y 0 = 0 Taking Laplace transforms using L f" = s 2 F s sf 0 f 0 In matrix form s 2 X sx 0 x 0 + 2X Y = 0 s 2 Y sy 0 y 0 + 2Y X = 0 or s X Y = 4s X + s Y = 2s s s Using inverse matrix, we obtain X Y = 4s 2s X Y = 1 s s s s 2s = 2s s s s Thus X = 2s 2s2 +5 s = 2s 2s2 +5 s 2 +1 s 2 +3, and Y = 2s s2 +4 s = 2s s2 +4 s 2 +1 s
24 System of Differential Equations (Cont d) Now let s find the inverse Laplace of X, (a) Multiply s and then let s 2 = 1, X = 2s 2s2 + 5 s s = A 1s + B 1 s C 1s + D 1 s A 1 s + B 1 = 2s 2s2 + 5 s s 2 = 1 = 3s Thus, A 1 = 3, and B 1 = 0 (b) Multiply s and then let s 2 = 3, Thus, C 1 = 1, and D 1 = 0 So we have X = 3s s s s 2 +3 C 1 s + D 1 = 2s 2s2 + 5 s s 2 = 3 = s Inverse Laplace: x t = L 1 X = 3 cos t + cos 3t 24
25 System of Differential Equations (Cont d) Next let s find the inverse Laplace of Y, Y = 2s s s s = A 2s + B 2 s C 2s + D 2 s (a) Multiply s and then let s 2 = 1, A 2 s + B 2 = 2s s2 + 4 s s 2 = 1 = 3s Thus, A 2 = 3, and B 2 = 0 (b) Multiply s and then let s 2 = 3, C 2 s + D 2 = 2s s2 + 4 s s 2 = 3 = s Thus C 2 = 1, and D 2 = 0 We have Y = 3s s 2 +1 s s 2 +3 Inverse Laplace: y t = L 1 Y = 3 cos t cos 3t Solution: x = 3 cos t + cos y = 3 cos t cos 3t 3t 25
26 Impulse Response Network Network Network Network 26
27 Convolution and Laplace If we define the impulse response as h(t), then the output y(t) is related to the input x(t) via the convolution integral: Convolution in the time domain is multiplication in the frequency domain: 27
28 Convolution Theorem for Inverse Laplace This theorem can be used to find inverse Laplace transform L 1 1 s 2 s 3 =? F 1 s = 1 s 2, and F 2 s = 1 s 3 Their inverse: f 1 t = t, and f 2 t = e 3t Using Convolution Theorem L 1 1 s 2 s 3 = xe3(t x) dx 0 = e 3t xe 3x dx 0 t t t = e 3t xe 3x dx 0 = 1 9 e3t 3t 1 28
29 Summary Key points: Useful Laplace transform pairs δ-function and its properties Laplace transform of derivatives Inverse Laplace using partial fractions Solving differential equations with Laplace Supplemental: Convolution theorem 29
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