Initial & Final Value Theorems. Lecture 7. More on Laplace Transform (Lathi ) Example. Laplace Transform for Solving Differential Equations

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1 Initial & Final Value Theorem ecture 7 More on aplace Tranform (athi ) How to find the initial and final value of a function x(t) if we now it aplace Tranform? (t 0 +, and t ) Initial Value Theorem Condition: aplace tranform of x(t) and lim xt ( ) = x(0 + ) = lim X( ) t 0 dx/dt exit. numerator power (M) i le than denominator power (N), i.e. M<N. Peter Cheung Department of Electrical & Electronic Engineering Imperial College ondon UR: p.cheung@imperial.ac.u Final Value Theorem Condition: aplace tranform of x(t) and lim xt ( ) = x( ) = lim X( ) t 0 dx/dt exit. pole are all on the eft Plane or origin p370 ecture 7 Slide ecture 7 Slide Example aplace Tranform for Solving Differential Equation Find the initial and final value of y(t) if Y() i given by: Remember the time-differentiation property of aplace Tranform 0( Y () = ( + + d y dt Y() initial value: y(0 + ) = lim Y( ) 0( = lim = 0 ( + + Exploit thi to olve differential equation a algebraic equation: time-domain analyi olve differential equation final value: y( ) = lim Y( ) 0 0( = lim = 6 0 ( p37 frequency-domain analyi olve algebraic equation Y() p37 ecture 7 Slide 3 ecture 7 Slide 4

2 Example () Example () Solve the following econd-order linear differential equation: d y dy dx = + dt dt dt Given that y (0 ) =, y (0 ) =and input x (t ) = e 4t u (t ). Time Domain aplace (Frequency) Domain Time Domain aplace (Frequency) Domain 4.3 p37 ecture 7 Slide 5 ecture 7 Slide 6 Zero-input & Zero-tate Repone aplace Tranform and Tranfer Function et thin about where the term come from: Initial condition term Input term et expre input x(t) a a linear combination of exponential e t : K t i = X( i ) e H() can be regarded a the ytem repone to each of the exponential K component, in uch a way that the output y(t) i: t i yt ( ) = X( i) H( i) e Therefore, we get i= Y () = H () X () i= H() ytem repone to e t i H()e t Y () = HX () () p373 Tranfer Function Y () = HX () () H() 4.3 p377 ecture 7 Slide 7 ecture 7 Slide 8

3 Tranfer Function Example Initial condition in ytem () Delay by T ec = xt ( T) H () = Y () = e T Y () = Xe () T Differentiator y() t = dx dt d/dt H () = Integrator H () Y () = X () t = x( τ) dτ 0 Y () = X () Shifting Differentiation Integration = 4.3 p380 ecture 7 Slide 9 In circuit, initial condition may not be zero. For example, capacitor may be charged; inductor may have an initial current. How hould thee be repreented in the aplace (frequency) domain? Conider a capacitor C with an initial voltage v(0 - ): Now tae aplace tranform on both ide: Rearrange thi to give: voltage acro charged capacitor voltage acro capacitor with no charge effect of the initial charge = voltage ource 4.4 p387 ecture 7 Slide 0 Initial condition in ytem () Solving Tranient Behaviour in circuit Example () Similarly, conider an inductor with an initial current i(0 - ): Conider a capacitor C with an initial voltage v(0 - ): The witch in the circuit here i in cloed poition for a long time before t=0, when it i opened intantaneouly. Find the current y(t) and y(t) for t>0. Now tae aplace tranform on both ide: Rearrange thi to give: y (t) Firt determine the initial condition at t = 0. voltage acro inductor voltage acro inductor with no initial current effect of the initial current = voltage ource 4.4 p p389 ecture 7 Slide ecture 7 Slide

4 Example () Example (3) From thi we can rewrite a in matrix form: We need to olve for Y () and Y (). We do thi by applying Cramer rule, which i: Given Az = c, where A i a quare matrix, z and c are column vector, the vector z can be olve by: det( Ai ) zi = det( A) where A i i the matrix A with it i th column replaced by column vector c. We readily obtain: det( A) = det = ( + 7+ ) and therefore: 5 det ( + ) 4 48 Y () = = = + det( A) Invere aplace give u: Similarly we obtain: 4.4 p389 Therefore: 4.4 p389 ecture 7 Slide 3 ecture 7 Slide 4 Solving Tranient Behaviour in circuit Example () Solving Tranient Behaviour in circuit Example () Find the tranfer function H() relating the output v o (t) to the input voltage v i (t) for the Sallen and Key filter hown below. Aume that initial condition i zero. Step 3: Sum current at node a Step : Form equivalent circuit Step : Pic variable nodal voltage at a and b Step 4: Sum current at node b Step 5: Put in matrix form 4.4 p394 ecture 7 Slide 5 ecture 7 Slide 6

5 Solving Tranient Behaviour in circuit Example (3) Relating thi lecture to other coure Step 6: Apply Cramer rule You have done much of the circuit analyi in your firt year, but aplace tranform provide much more elegant method in find olution to BOTH tranient and teady tate condition of circuit. You have done Sallen-and-Key filter in your nd year analogue circuit coure. Here we derive the tranfer function from firt principle, uing only tool you now about. The treatment provided in thi lecture alo enhance what you have been learning in your nd year control coure. Step 7: Derive H() ecture 7 Slide 7 ecture 7 Slide 8