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01/09/2013 10 Applications of Laplace interactive mathematics Learn math by playing with it! home sitemap math interactives math blog about feedback Search site Chapter Contents Laplace Transforms 1a The Unit Step Function - Definition Oliver Heaviside 1b The Unit Step Function - Products Laplace Transform Definition Table of Laplace Transformations Properties of Laplace Transform Transform of Unit Step Functions Transform of Periodic Functions Transforms of Integrals Inverse of the Laplace Transform Using Inverse Laplace to Solve DEs Integro-Differential Equations and Systems of DEs 10 Applications of Laplace Transform Laplace Transforms Problem Solver Comments, Questions? www.intmath.com/laplace-transformation/10-applications.php 1/30 01/09/2013 10 Applications of Laplace Summer holidays I feel summer holidays are: Too long About the right length Not long enough Vote! 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Online Algebra Solver » Share Share this page From the math blog Who's Counting - interesting real-life math applications www.intmath.com/laplace-transformation/10-applications.php 2/30 01/09/2013 10 Applications of Laplace John Allen Paulos discusses why it is important to understand math in the news more » Producing mathematics for the Web I recently got an enquiry from a mathematics teacher who likes my Interactive Mathematics site more » Math graphs on the Web without images Here's one way to plot good looking graphs on the Web - ASCIIsvg more » Earth killer - composite trigonometry CO2 graph Here's a satisfactory attempt to model the Earth's increase in CO2 concentration, using Excel and Scientific Notebook more » Graphs using free math software Do we still need to graph on paper, or should we use free math graphers instead? more » www.intmath.com/laplace-transformation/10-applications.php 3/30 01/09/2013 10 Applications of Laplace 10 Applications of Laplace Transforms Circuit Equations There are two (related) approaches: Derive the circuit (differential) equations in the time domain, then transform these ODEs to the s-domain; Transform the circuit to the s-domain, then derive the circuit equations in the s-domain (using the concept of "impedance") We will use the first approach We will derive the system equations(s) in the t-plane, then transform the equations to the s-plane We will usually then transform back to the t-plane Example Consider the circuit when the switch is closed at t www.intmath.com/laplace-transformation/10-applications.php , = V C (0) = 1.0 V Solve for the current i(t) in the circuit 4/30 01/09/2013 10 Applications of Laplace Answer ∫ i dt + Ri = V C 10 −6 ∫ i dt + 10 i = Multiplying throughout by 10−6 gives: ∫ i dt + 10 −3 i = × 10 −6 Taking Laplace transform: I ( [∫ + s ) + 10 i dt] s −3 × 10 −6 I = s t=0 Now in this example, we are told V c(0) = 1.0 So 1 [∫ V c(0) = idt] C That is: [∫ = C t=0 10 NOTE: [∫ −6 [∫ i dt] = = q0 = t=0 Therefore: [∫ t=0 idt] idt] i dt] = 10 −6 t=0 t=0 www.intmath.com/laplace-transformation/10-applications.php ( + ) + −3 I = 5/30 01/09/2013 10 Applications of Laplace I ( 10 −6 ) + 10 + s −3 × 10 −6 I = s s Collecting I terms and subtracting 10 −6 from both sides: s ( + 10 −3 × 10 −6 )I = s 10 −6 − s × 10 −6 = s s Multiply throughout by s : (1 + 10 −3 s) = × 10 −6 Solve for I : × 10 I = + 10 −6 −3 = (4 × 10 −3 ) s 1000 + s Finding the inverse Laplace transform gives us the current at time t: i = × 10 −3 e −1000t www.intmath.com/laplace-transformation/10-applications.php 6/30 01/09/2013 10 Applications of Laplace Note: Throughout this page these problems are also solved using Scientific Notebook They are TEX files and you need Scientific Notebook or similar, to view them Alternative answer using Scientific Notebook (.tex file) iPage® Web Hosting www.iPage.com Web Hosting, Email & Free Domains! Trusted by Millions Example Solve for i(t) for the circuit, given that V(t) = 10 sin5t V, R = W and L = H www.intmath.com/laplace-transformation/10-applications.php 7/30 01/09/2013 10 Applications of Laplace Answer di Ri + L = V dt di 4i + = 10 sin 5t dt 4I + 2(sI − i(0)) = 10 s Assume i0 = i(0) = + 25 50 4I + 2sI = s + 25 25 (2 + s)I = s So I + 25 25 A = = (s + 2)( s 25 = A(s 2 + 25) Bs + C + s + s + 25 + 25) + (Bs + C )(s + 2) www.intmath.com/laplace-transformation/10-applications.php 8/30 01/09/2013 10 Applications of Laplace We need to solve for A, B and C First, let s = −2 and this gives 25 = 29A Thus A = 25 29 Next, we equate coefficients of s : = A+ B gives B = 25 − 29 Equating coefficients of s : = 2B + C gives C 50 = 29 So 25 I = (s + 2)( s + 25) 25 25 = s − 29 29(s + 2) 25 ( = 29 s 50 s + s 29 + 25 s − + + 25 + + 25 s s ) + 25 So we have 25 (e i = −2t − cos 5t + 29 sin 5t) A Here is the graph of i(t): www.intmath.com/laplace-transformation/10-applications.php 9/30 01/09/2013 10 Applications of Laplace Alternative answer using Scientific Notebook (.tex file) Example In the circuit shown below, the capacitor is uncharged at time t = If the switch is then closed, find the currents i1 and i2, and the charge on C at time t greater than zero Answer www.intmath.com/laplace-transformation/10-applications.php 10/30 01/09/2013 10 Applications of Laplace I 5I + 0.1(sI − i ) + 5000( + q0 s I 5I + 0.1(sI − 2) + 5000( ) = s 10 −3 ) = + s s I 5I + 0.1sI − 0.2 + 5000 + s = s Multiplying by 10s: 50sI + s I − 2s + 5000I + 50 = (s + 50s + 50000)I = 2s − 50 Solving for I and completing the square on the denominator gives us: 2s − 50 I = s + 50s + 50000 2s − 50 = s + 50s + 625 + 49375 2s − 50 = (s + 25) + 49375 s − 25 ≅2 (s + 25) + 222.2 s + 25 = 2( (s + 25) 50 − + 222.2 25 (s + 25) s + 25 = 2( (s + 25) 50 − + 222.2 222.2 www.intmath.com/laplace-transformation/10-applications.php ) + 222.2 222.2 (s + 25) ) + 222.2 16/30 01/09/2013 10 Applications of Laplace So the transient current is: i = 2(e −25t 50 cos 222.2t − e −25t sin 222.2t) 222.2 = e −25t (2 cos 222.2t − 0.45 sin 222.2t) We could transform the trigonometric part of this to a single expression: cos 222.2t − 0.45 sin 222.2t = R cos(222.2t + α) − − − − − − − − − 2 R = √2 + 0.45 = 2.05 0.45 α = arctan = 0.22131 So i = 2.05 e −25t cos(222.2t + 0.22) Alternative answer using Scientific Notebook (.tex file) www.intmath.com/laplace-transformation/10-applications.php 17/30 01/09/2013 10 Applications of Laplace Example The system is quiescent Find the loop current i2(t) Answer Quiescent implies i1, i2 and their derivatives are zero for t = 0, ie i1(0) = i2(0) = i1'(0) = i2'(0) = For loop 1: 10 i + 10(i − i ) = 20 i − 10 i = 10 i − i = i1 = 10 (2 + i ) .(1) For loop 2: di www.intmath.com/laplace-transformation/10-applications.php 20 + 0.1 + 10( − ) = 18/30 01/09/2013 10 Applications of Laplace 20 i + 0.1 di dt + 10(i − i ) = −10 i + 30 i + 0.1 −100 i + 300 i + di = dt di = dt Substituting our result from (1) gives: −100 × 10 (2 + i ) + 300 i + di −10(2 + i ) + 300 i + 250 i + di di = dt = dt = 20 dt Taking Laplace transform: 20 250 I + (sI − i (0)) = 20 (250 + s)I = s s since i2 (0) = 20 I2 = s(s + 250) 20 Let A = s(s + 250) So 20 B + s s + 250 = A(s + 250) + Bs 20 www.intmath.com/laplace-transformation/10-applications.php 20 = 250A ⇒ A = = 19/30 01/09/2013 10 Applications of Laplace 20 20 = 250A ⇒ A = = 250 25 A+ B = ⇒ B = − 25 20 So s(s + 250) ( = 25 ) − s s + 250 Taking Inverse Laplace: So i2 (1 − e = −250t ) 25 Alternative answer using Scientific Notebook (.tex file) Example Consider a series RLC circuit where R = 20 W, L = 0.05 H and C = 10-4 F and is driven by an alternating emf given by E = 100 cos 200t Given that both the circuit current i and the capacitor charge q are zero at time t = 0, find an expression for i(t) in the region t > www.intmath.com/laplace-transformation/10-applications.php 20/30 01/09/2013 10 Applications of Laplace Answer We use the following equation from before: di Ri + L ∫ + dt i dt = E C and obtain: di 20i + 0.05 + dt 10 −4 ∫ i dt = 100 cos 200t After multiplying throughout by 20, we have: di 400i + + 20 × 10 ∫ idt = 2000 cos 200t dt Taking Laplace transform and using the fact that i(0) = 0: I 400I + sI − i(0) + 200000 s = 2000 s s s 2 + 200 400sI + s I + 200000I = 2000 s (s 2 + 200 s 2 + 400s + 200000)I = 2000 s s 2 + 200 2 I = 2000 (s + 400s + 200000)(s 2 + 200 ) Using Scientific Notebook to find the partial fractions: www.intmath.com/laplace-transformation/10-applications.php 2000 21/30 01/09/2013 10 Applications of Laplace s 2000 (s + 400s + 200000)(s 2000 − s = s 2 + 200 ) −400 + s + + 400s + 200000 s + 40000 So s − 2000 I = − s s − 400 + + 400s + 200000 s + 200 s − 2000 = − s 2 s − 400 + + 400s + 40000 + 160000 s − 2000 = − (s + 200) (s + 200) + 400 s + 200 + 400 (s + 200) + 200 2 2200 + (s + 200) s + 200 = − s − 400 + s + 200 = s + 400 2200 + 400 s + 400 400 + s + 200 − s 400 (s + 200) + 200 s + 400 + s + 200 200 − s + 200 So i = −e −200t 11 cos 400t + e −200t sin 400t + cos 200t −2 sin 200t NOTE: Scientific Notebook can all this for us very easily In one step, we have: www.intmath.com/laplace-transformation/10-applications.php ⎧ ⎪ ⎫ ⎪ 22/30 01/09/2013 10 Applications of Laplace ⎧ ⎪ L −1 s ⎫ ⎪ ⎨2000 ⎩ ⎪ = −e −200t ⎬ (s + 400s + 200000)(s 11 cos 400t + e −200t ⎭ + 200 ) ⎪ sin 400t + cos 200t − sin 200t Transient part: −e−200t cos 11 400t + e −200t sin 400t www.intmath.com/laplace-transformation/10-applications.php 23/30 01/09/2013 Steady state part: cos 10 Applications of Laplace 200t − sin 200t Alternative answer using Scientific Notebook (.tex file) Example www.intmath.com/laplace-transformation/10-applications.php R (t) 24/30 01/09/2013 10 Applications of Laplace A rectangular pulse vR (t) is applied to the RC circuit shown Find the response, v(t) Graph of vR (t): Note: v(t) = V for all t < s implies v(0-) = V Answer Now V R (t) = u(t) − u(t − 2) To solve this, we need to work in voltages, not current We start with Ri + ∫ i dt = V C The voltage across a capacitor is given by v = ∫ i dt C It follows that C dv = i dt So for this example we have: dv RC + v = u(t) − u(t − 2) dt www.intmath.com/laplace-transformation/10-applications.php 25/30 ... Circuit Equations There are two (related) approaches: Derive the circuit (differential) equations in the time domain, then transform these ODEs to the s-domain; Transform the circuit to the s-domain,... the circuit equations in the s-domain (using the concept of "impedance") We will use the first approach We will derive the system equations(s) in the t-plane, then transform the equations to the... C First, let s = −2 and this gives 25 = 29A Thus A = 25 29 Next, we equate coefficients of s : = A+ B gives B = 25 − 29 Equating coefficients of s : = 2B + C gives C 50 = 29 So 25 I = (s + 2)(