Tutorial 4 Laplace Transforms 1

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Tutorial 4: Laplace Transforms 1 Refresher… • Application of forward Laplace transform transforms from time domain (t) to complex frequency domain (s) • We employ a mapping process, we do not alter the mathematical meaning of the system • After mapping, system will be algebraic – easier to manipulate and solve • The mapping process involves the application of the derivative law 𝑑𝑑𝑑𝑑 ( 𝑡𝑡 ) 𝑑𝑑𝑡𝑡 → 𝑠𝑠𝑠𝑠 ( 𝑠𝑠 ) − 𝑑𝑑 (0) • Derivative law for first order systems… • Derivative law for second order systems… 𝑑𝑑 2 𝑑𝑑 ( 𝑡𝑡 ) 𝑑𝑑𝑡𝑡 2 → 𝑠𝑠 2 𝑠𝑠 ( 𝑠𝑠 ) − 𝑠𝑠𝑑𝑑 (0) − 𝑑𝑑 ′ (0)

The output y(t) of a dynamic system is required to reach an amplitude of at least 4 three seconds after the system has been activated. The following equation has been used to model the dynamic system. Mathematically confirm the suitability of the system model assuming a step input were to be applied at t=0s. Assume no initial conditions are acting upon the system. ) ( 12 ) ( 6 t x t y dt dy = + What are you being asked to do? As always, we analyse what we know… • The system is dynamic – it is changing with time • The output of the system is rising over time and must reach an amplitude of at least 4 after three seconds • We must apply a step input, therefore we must calculate a step response • Step response is a time domain test – difficult with a differential equation • Perform a forward Laplace transform to transform to s-domain and algebraic format • Perform an inverse Laplace transform using partial fractions & look-up tables to yield time domain equations • Perform step response to determine system output at 3 seconds

) ( 12 ) ( 6 t x t y dt dy = + 1. Forward Laplace Transform… s s Y s sY 12 ) ( ) ( 6 = + From look-up tables, a step input in the s-domain is 𝑎𝑎 𝑠𝑠 Problem? Two instances of output Y(s) – factorise! s 12 1) Y(s).(6s = + ) 6 1 ( 2 ) 1 6 ( 12 ) (       + = + = s s s s s Y ) 6 1 ( 2 ) (       + = s s s Y 2. Reverse Laplace Transform • Cross-multiplication • Cover-up theorem • Partial fractions ) 6 1 ( ) (       + + = s B s A s Y 12 6 1 2 | ) 6 1 ( 2 0 =       =       + = = s s A 12 6 1 2 | ) ( 2 ) 6 1 ( − =       − = = − = s s B From look-up tables we have au(t) + e -at which yields... 72 . 4 12 12 ) ( , 3 12 12 3 * 6 1 6 1 = − = = −       −       − e t y s t with e t

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Inverse Laplace Example Find the inverse Laplace transform… 2𝑠𝑠 − 8 𝑠𝑠 2 − 8𝑠𝑠 + 15 Problem? Cannot decompose the system – factorise! 2𝑠𝑠 − 8 𝑠𝑠 2 − 8𝑠𝑠 + 15 = 2𝑠𝑠 − 8 ( 𝑠𝑠 − 3)( 𝑠𝑠 − 5) ≡ 𝐴𝐴 𝑠𝑠 − 3 + 𝐵𝐵 𝑠𝑠 − 5 To find A set s=3… � 𝐴𝐴 � = 2x3 − 8 𝑠𝑠 − 5 𝑠𝑠=3 = − 2 𝑠𝑠 − 5 𝑠𝑠=3 = − 2 − 2 = 1 To find B, set s=5… � B � = 2x5 − 8 𝑠𝑠 − 3 𝑠𝑠=5 = 2 𝑠𝑠 − 3 𝑠𝑠=5 = 2 2 = 1 ∴ 2𝑠𝑠 − 8 ( 𝑠𝑠 − 3)( 𝑠𝑠 − 5) = 1 𝑠𝑠 − 3 + 1 𝑠𝑠 − 5 From look-up tables… 1 𝑠𝑠 − 𝑎𝑎 = 𝑒𝑒 𝑎𝑎𝑎𝑎 Thus the inverse Laplace transform is… 2𝑠𝑠 − 8 𝑠𝑠 2 − 8𝑠𝑠 + 15 = 𝑒𝑒 3𝑎𝑎 + 𝑒𝑒 5𝑎𝑎