330 likes | 670 Views
ERT 210/4 Process Control & Dynamics. DYNAMIC BEHAVIOR OF PROCESSES : Laplace Transforms. MISS. RAHIMAH BINTI OTHMAN (Email: email@example.com). COURSE OUTCOME 1 CO1) 1. Theoretical Models of Chemical Processes 2. Laplace Transform
E N D
ERT 210/4Process Control & Dynamics DYNAMIC BEHAVIOR OF PROCESSES : Laplace Transforms MISS. RAHIMAH BINTI OTHMAN (Email: firstname.lastname@example.org)
COURSE OUTCOME 1 CO1) 1. Theoretical Models of Chemical Processes 2. Laplace Transform EXPLAIN, REPEAT, APPLY and ANALYZE Laplace transforms techniques to solve linear differential equations. 3. Transfer Function Models 4. Dynamic Behavior of First-order and Second-order Processes 5. Dynamic Response Characteristics of More Complicated Processes 6. Development of Empirical Models from Process Data
Laplace Transforms • 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
Definition The Laplace transform of a function, f(t), is defined as 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, .
Inverse Laplace Transform, L-1: By definition, the inverse Laplace transform operator, L-1, converts an s-domain function back to the corresponding time domain function: Important Properties: Both L and L-1 are linear operators. Thus,
where: - x(t) and y(t) are arbitrary functions - a and b are constants - = = Similarly,
Laplace Transforms of Common Functions • Constant Function • Let f(t) = a (a constant). Then from the definition of the Laplace transform in (3-1),
Step Function • The unit step function is widely used in the analysis of process control problems. It is defined as: = Because the step function is a special case of a “constant”, it follows from (3-4) that
Derivatives • 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 initial condition at t = 0 Similarly, for higher order derivatives:
where: - n is an arbitrary positive integer - = Special Case: All Initial Conditions are Zero Suppose Then 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.
Exponential Functions • Consider where b > 0. Then, • Rectangular Pulse Function • It is defined by:
Time, t The Laplace transform of the rectangular pulse is given by: (3-22)
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 • the area under the pulse constant at one. (i.e., let ) • Let, • Then, = Solution of ODEs by Laplace Transforms • Procedure: • Take the L of both sides of the ODE. • Rearrange the resulting algebraic equation in the s domain to solve for the L of the output variable, e.g., Y(s). • Perform a partial fraction expansion. • Use the L-1 to find y(t) from the expression for Y(s).
Table 3.1. Laplace Transforms • See page 54 of the text.
Example 3.1 Solve the ODE, First, take L of both sides of (3-26), Rearrange, Take L-1, (3-35) Simplify, (3-36) From Table 3.1 (No.11),
Partial Fraction Expansions 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: Perform a partial fraction expansion (PFE) where coefficients and have to be determined.
Use Method 3 (page 59) – “Heaviside expansion” In this method multiply both sides of the equation by one of the denominator terms (s + bi) and then set s = - bi, which causes all terms except one to be multiplied by zero. To find : Multiply both sides by s + 1 and let s = -1 To find : Multiply both sides by s + 4 and let s = -4 A General Partial Fraction Expansions (PFE) Consider a general expression,
Here D(s) is an n-th order polynomial with the roots all being real numbers which are distinct so there are no repeated roots. The PFE is: Note:D(s) is called the “characteristic polynomial”. • Special Situations: • Two other types of situations commonly occur when D(s) has: • Complex roots: e.g., • Repeated roots (e.g., ) • For these situations, the PFE has a different form. See SEM • text (pp. 61-64) for details.
Example 3.2 (cont’) Recall that the ODE, , with zero initial conditions resulted in the expression The denominator can be factored as Note: Normally, numerical techniques are required in order to calculate the roots. The PFE for (3-40) is
Solve for coefficients to get (For example, find , by multiplying both sides by s and then setting s = 0.) Substitute numerical values into (3-51): Take L-1 of both sides: From Table 3.1,
General Procedure for Solving Differential Equations Note that; The solution for the differential equation involves use of Laplace transforms as an intermediate step. Step 3 can be bypassed if the transform found in Step 2 matches an entry in Table 3.1. In step 3, other types of situations can occur. Both repeated factors require modifications of the partial expansion procedure.
Example 3.3 (Repeated Factors) For (3-54) Evaluate the unknown coefficient i. Solution To find in (3-54), the “Heaviside rule” cannot be used for multiplication by (s+2), because s = -2 causes the second term on the right side to be bounded, rather than 0 as desired. But, 1 and 2 can be found by using “Heaviside rule” Therefore;
Example 3.4 (Complex Factors) Find the inverse Laplace transform of Solution Using the quadratic formula, the factors s2 + 4s + 5 are found to be (s + 2 + j) and (s + 2 - j), so that; The partial fraction expansion is; (3-79) Therefore; Use (3-79) and Table 3.1 to obtain the corresponding time-domain expression;
Important Properties of Laplace Transforms • Final Value Theorem 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: providing that the limit exists (is finite) for all where Re (s) denotes the real part of complex variable, s.
Example 3.5 Apply the initial value theorems to the transport derived in Example 3.1 (page 57). Solution; Suppose, Then, Initial Value; Final Value; The initial value of 1 corresponds to the initial condition given in Eq.3-26. The final value of 0.5 agrees with the time-domain solution in Eq.3-37. Both the theorems are useful for checking mathematical errors that may occur in the course of obtaining Laplace transform solution.
2. Time Delay • Time delays commonly occur due to; • - the transport time required for a fluid to flow through piping • - fluid flow, time required to do an analysis • (e.g., gas chromatograph) • The delayed signal can be represented as; Also,
Thank you Prepared by, MISS RAHIMAH OTHMAN