Example Laplace Transform For Solving Differential Equations

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Initial & Final Value Theorems Lecture 7How to find the initial and final values of a function x(t) if we know itsLaplace Transform X(s)? (t 0 , and t )Initial Value TheoremMore onLaplace Transform Conditions: lim x(t ) x(0 ) lim sX ( s)(Lathi 4.3 – 4.4)t 0Final Value TheoremPeter CheungDepartment of Electrical & Electronic EngineeringImperial College London s Conditions: lim x(t ) x( ) lim sX (s)t s 0 URL: www.ee.imperial.ac.uk/pcheung/teaching/ee2 signalsE-mail: p.cheung@imperial.ac.ukPYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 1PYKC 8-Feb-11Lecture 7 Slide 2E2.5 Signals & Linear SystemsLaplace Transform for Solving Differential EquationsFind the initial and final values of y(t) if Y(s) is given by:Y ( s) Laplace transforms of x(t) anddx/dt exist.sX(s) poles are all on the LeftPlane or origin.L4.2-5 p370Example Laplace transforms of x(t) anddx/dt exist.X(s) numerator power (M) isless than denominator power(N), i.e. M N. Remember the time-differentiation property of Laplace Transform10(2s 3)s(s 2 2s 5)dk y s kY (s)dt k Exploit this to solve differential equation as algebraic equations:initial value: y (0 ) lim sY ( s)s 10(2s 3) 0s ( s 2 2 s 5) limfinal value:x (t )y ( ) lim sY ( s)x (t )L4.2-5 p371E2.5 Signals & Linear Systemsy (t )X ( s ) frequency-domain Y ( s )s 010(2s 3) lim 2 6s 0 ( s 2 s 5)PYKC 8-Feb-11time-domainanalysissolve differentialequationsLecture 7 Slide 3PYKC 8-Feb-11Lanalysissolve algebraicequationsE2.5 Signals & Linear SystemsL-1y (t )L4.3 p371Lecture 7 Slide 4

Example (1) Example (2)Time DomainSolve the following second-order linear differential equation:d2ydydx 5 6 y(t ) x(t )dt 2dtdt 4 tGiven that y (0 ) 2, y (0 ) 1 and input x (t ) e u (t ).Time DomainLaplace (Frequency) DomainLaplace (Frequency) DomainL4.3 p371PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 5PYKC 8-Feb-11Zero-input & Zero-state Responses Lecture 7 Slide 6E2.5 Signals & Linear SystemsLaplace Tranform and Transfer FunctionLet’s think about where the terms come from: Let’s express input x(t) as a linear combination of exponentials est:Kx(t ) X ( si )e siti 1Initial conditiontermInput term H(s) can be regarded as the system’s response to each of the exponentialKcomponents, in such a way that the output y(t) is:sty (t ) X ( si ) H ( si )e i Therefore, we getx (t )LX (s)X (s)i 1Y ( s) H ( s) X ( s)H(s)system responseto X(s)est isH(s)X(s)estY ( s) H ( s) X ( s)Transfer FunctionH(s)L4.3 p373PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 7L-1y (t )Y ( s) H ( s) X ( s)L4.3 p377PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 8

Transfer Function Examplesx (t )X (s)Delay by T secH ( s) Y ( s)X ( s) e sTDifferentiatord/dtx (t )X (s)H ( s) sInitial conditions in systems (1)Y (s) X (s)e sT In circuits, initial conditions may not be zero. For example,capacitors may be charged; inductors may have an initial current.How should these be represented in the Laplace (frequency)domain?Consider a capacitor C with an initial voltage v(0-):Now take Laplace transform on both sides: Rearrange this to give: y (t ) x(t T )ShiftingProperty y(t ) dx dtY ( s) sX ( s)DifferentiationPropertyvoltage acrosscharged capacitortx (t )Integratory (t ) x(τ )dτ0X (s)H ( s) PYKC 8-Feb-111sY ( s) 1X ( s)sE2.5 Signals & Linear Systemsvoltage acrosscapacitor with nochargeIntegrationPropertyL4.3 p380Lecture 7 Slide 9L4.4 p387PYKC 8-Feb-11Initial conditions in systems (2) Similarly, consider an inductor L with an initial current i(0-):Consider a capacitor C with an initial voltage v(0-): Now take Laplace transform on both sides: effect of the initialcharge voltagesourceE2.5 Signals & Linear SystemsLecture 7 Slide 10Solving Transient Behaviour in circuits – Example 1(1) The switch in the circuit here is in closed position for a long time before t 0, whenit is opened instantaneously. Find the current y1(t) and y2(t) for t 0.y1(t) Rearrange this to give: voltage acrossinductorvoltage acrossinductor with no initialcurrentFirst determine the initial condition at t 0.effect of the initialcurrent voltagesourceL4.4 p388PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 11L4.4 p389PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 12

Example 1 (2) From this we can rewrite as in matrix form: We need to solve for Y1(s) and Y2(s). We do this by applying Cramer’s rule, which is:Given Az c, where A is a square matrix, z and c are column vectors,det( Ai )the vector z can be solve by:zi det( A)where Ai is the matrix A with its ith column replaced by column vector c. Example 1(3)We readily obtain:and therefore: 1 1 s 5det( A) det 1 51 4 s5det 2 6 s 5 2 24( s 2) 24 48Y1 ( s) det( A)s 2 7 s 12 s 3 s 4 Inverse Laplace gives us: Similarly we obtain: Therefore:L4.4 p389PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 13Solving Transient Behaviour in circuits – Example 2(1) Find the transfer function H(s) relating the output vo(t) to the input voltage vi(t) forthe Sallen and Key filter shown below. Assume that initial condition is zero.1 1 25 ( s 7 s 12) 6 s 10s 5 2 L4.4 p389PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 14Solving Transient Behaviour in circuits – Example 2(2)Step 3: Sum current atnode aStep 1: Form equivalent circuitStep 2: Pick “variables” – nodalvoltages at a and bStep 4: Sum current atnode bStep 5: Put in matrixformL4.4 p394PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 15PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 16

Solving Transient Behaviour in circuits – Example 2(3)Relating this lecture to other courses Step 6: Apply Cramer’srule You have done much of the circuit analysis in your first year, but Laplacetransform provides much more elegant method in find solutions to BOTHtransient and steady state condition of circuits.You have done Sallen-and-Key filter in your 2nd year analogue circuitscourse. Here we derive the transfer function from first principle, usingonly tools you know about.The treatment provided in this lecture also enhances what you have beenlearning in your 2nd year control course.Step 7: Derive H(s)PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 17PYKC 8-Feb-11E2.5 Signals & Linear SystemsLecture 7 Slide 18

Laplace Transform for Solving Differential Equations Remember the time-differentiation property of Laplace Transform Exploit this to solve differential equation as algebraic equations: () k k k dy sY s dt time-domain analysis solve differential equations xt() yt() frequency-domain analysis solve algeb

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