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ERT 210/4 Process Control & Dynamics. DYNAMIC BEHAVIOR OF PROCESSES : Laplace Transforms. MISS. RAHIMAH BINTI OTHMAN (Email: rahimah@unimap.edu.my). COURSE OUTCOME 1 CO1) 1. Theoretical Models of Chemical Processes 2. Laplace Transform

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## ERT 210/4 Process Control & Dynamics

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**ERT 210/4Process Control & Dynamics**DYNAMIC BEHAVIOR OF PROCESSES : Laplace Transforms MISS. RAHIMAH BINTI OTHMAN (Email: rahimah@unimap.edu.my)**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

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