#### Transcript Chapter 3 (10-5

Laplace Transforms Chapter 3 • Important analytical method for solving linear ordinary differential equations. - Application to nonlinear ODEs? Must linearize first. • Laplace transforms play a key role in important process control concepts and techniques. - Examples: • Transfer functions • Frequency response • Control system design • Stability analysis 1 Definition The Laplace transform of a function, f(t), is defined as Chapter 3 F ( s) L f (t ) f t e st dt 0 (3-1) where F(s) is the symbol for the Laplace transform, L is the Laplace transform operator, and f(t) is some function of time, t. Note: The L operator transforms a time domain function f(t) into an s domain function, F(s). s is a complex variable: s = a + bj, j 1 2 Inverse Laplace Transform, L-1: Chapter 3 By definition, the inverse Laplace transform operator, L-1, converts an s-domain function back to the corresponding time domain function: f t L1 F s Important Properties: Both L and L-1 are linear operators. Thus, L ax t by t aL x t bL y t aX s bY s (3-3) 3 where: - x(t) and y(t) are arbitrary functions Chapter 3 - a and b are constants - X s L x t and Y s L y t Similarly, L1 aX s bY s ax t b y t 4 Laplace Transforms of Common Functions Chapter 3 1. Constant Function Let f(t) = a (a constant). Then from the definition of the Laplace transform in (3-1), L a ae 0 st a st dt e s 0 a a 0 s s (3-4) 5 2. Step Function Chapter 3 The unit step function is widely used in the analysis of process control problems. It is defined as: S t 0 for t 0 1 for t 0 (3-5) Because the step function is a special case of a “constant”, it follows from (3-4) that 1 L S t s (3-6) 6 3. Derivatives Chapter 3 This is a very important transform because derivatives appear in the ODEs we wish to solve. In the text (p.53), it is shown that df L sF s f 0 dt (3-9) initial condition at t = 0 Similarly, for higher order derivatives: dn f L n dt n n 1 n 2 1 s F s s f 0 s f 0 n2 n 1 ... sf 0 f 0 (3-14) 7 where: - n is an arbitrary positive integer Chapter 3 - f k 0 dk f dt k t 0 Special Case: All Initial Conditions are Zero Suppose 1 n1 f 0 f 0 ... f 0 . Then dn f L n dt n s F s In process control problems, we usually assume zero initial conditions. Reason: This corresponds to the nominal steady state when “deviation variables” are used, as shown in Ch. 4. 8 4. Exponential Functions Consider f t ebt where b > 0. Then, Chapter 3 b s t L ebt ebt e st dt e dt 0 0 1 b s t 1 e 0 bs sb (3-16) 5. Rectangular Pulse Function It is defined by: 0 for t 0 f t h for 0 t t w 0 for t t w (3-20) 9 h Chapter 3 f t tw Time, t The Laplace transform of the rectangular pulse is given by F s h 1 e t w s s (3-22) 10 Chapter 3 6. Impulse Function (or Dirac Delta Function) The impulse function is obtained by taking the limit of the rectangular pulse as its width, tw, goes to zero but holding 1 the area under the pulse constant at one. (i.e., let h ) tw Let, t impulse function Then, L t 1 Solution of ODEs by Laplace Transforms Procedure: 1. Take the L of both sides of the ODE. 2. Rearrange the resulting algebraic equation in the s domain to solve for the L of the output variable, e.g., Y(s). 3. Perform a partial fraction expansion. 4. Use the L-1 to find y(t) from the expression for Y(s). 11 Table 3.1. Laplace Transforms Chapter 3 See page 54 of the text. 12 Example 3.1 Chapter 3 Solve the ODE, dy 5 4y 2 y 0 1 dt First, take L of both sides of (3-26), 2 5 sY s 1 4Y s s Rearrange, 5s 2 Y s s 5s 4 Take L-1, (3-26) (3-34) 1 5s 2 y t L s 5s 4 From Table 3.1, y t 0.5 0.5e0.8t (3-37) 13 Partial Fraction Expansions Chapter 3 Basic idea: Expand a complex expression for Y(s) into simpler terms, each of which appears in the Laplace Transform table. Then you can take the L-1 of both sides of the equation to obtain y(t). Example: Y s s5 s 1 s 4 Perform a partial fraction expansion (PFE) 1 2 s5 s 1 s 4 s 1 s 4 (3-41) (3-42) where coefficients 1 and 2 have to be determined. 14 To find 1 : Multiply both sides by s + 1 and let s = -1 s5 1 s4 4 s 1 3 Chapter 3 To find 2 : Multiply both sides by s + 4 and let s = -4 s5 2 s 1 s 4 1 3 A General PFE Consider a general expression, Y s N s Ds N s n s bi (3-46a) i 1 15 Here D(s) is an n-th order polynomial with the roots s bi all being real numbers which are distinct so there are no repeated roots. The PFE is: Chapter 3 Y s N s n s bi n i 1 i (3-46b) s bi i 1 Note: D(s) is called the “characteristic polynomial”. Special Situations: Two other types of situations commonly occur when D(s) has: i) Complex roots: e.g., bi 3 4 j ii) Repeated roots (e.g., b1 b2 3 ) j 1 For these situations, the PFE has a different form. See SEM text (pp. 61-64) for details. 16 Example 3.2 (continued) Chapter 3 Recall that the ODE, y 6 y 11y 6 y 1, with zero initial conditions resulted in the expression Y s 1 s s 6s 11s 6 3 2 (3-40) The denominator can be factored as s s3 6s 2 11s 6 s s 1 s 2 s 3 (3-50) Note: Normally, numerical techniques are required in order to calculate the roots. The PFE for (3-40) is Y s 1 1 2 3 4 (3-51) s s 1 s 2 s 3 s s 1 s 2 s 3 17 Solve for coefficients to get 1 1 1 1 1 , 2 , 3 , 4 6 2 2 6 Chapter 3 (For example, find , by multiplying both sides by s and then setting s = 0.) Substitute numerical values into (3-51): 1/ 6 1/ 2 1/ 2 1/ 6 Y ( s) s s 1 s 2 s 3 Take L-1 of both sides: 1 1 1/ 6 1 1/ 2 1 1/ 2 1 1/ 6 L Y s L L L L s s 1 s 2 s 3 From Table 3.1, y t 1 1 t 1 2t 1 3t e e e 6 2 2 6 (3-52) 18 Important Properties of Laplace Transforms 1. Final Value Theorem Chapter 3 It can be used to find the steady-state value of a closed loop system (providing that a steady-state value exists. Statement of FVT: lim y t lim sY s t s0 providing that the limit exists (is finite) for all Re s 0, where Re (s) denotes the real part of complex variable, s. 19 Example: Suppose, Chapter 3 Y s 5s 2 s 5s 4 (3-34) Then, 5s 2 y lim y t lim 0.5 t s 0 5s 4 2. Time Delay Time delays occur due to fluid flow, time required to do an analysis (e.g., gas chromatograph). The delayed signal can be represented as y t θ θ time delay Also, L y t θ eθsY s 20