KING FAHD UNIVERSITY CHEMICAL ENGINEERING COURSE NOTES (Process Control)-Lec9

KING FAHD - 1 Chapter 3 Example 2 Example 2 system at rest(s.s Step 1 Take L.T(note zero initial conditions 2 4 6 11 6 2 2 3 3 at t= dt du)= y)=

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Unformatted text preview: 1 Chapter 3 Example 2: Example 2: system at rest (s.s.) Step 1 Take L.T. (note zero initial conditions) 2 4 6 11 6 2 2 3 3 at t= dt du )= ( y )= ( y )= y( u dt du y dt dy dt y d dt y d = ′ ′ ′ + = + + + 3 2 6 11 6 ( ) 4 2 s Y(s)+ s Y(s)+ sY(s) Y s = sU(s) + U(s) + Chapter 3 To find transient response for u(t) = unit step at t > 0 1. Take Laplace Transform (L.T.) 2. Factor, use partial fraction decomposition 3. Take inverse L.T. 2 Chapter 3 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: ( 29 ( 29 ( 29 5 (3-41) 1 4 s Y s s s + = + + Perform a partial fraction expansion (PFE) ( 29 ( 29 1 2 5 (3-42) 1 4 1 4 s s s s s α α + = + + + + + where coefficients and have to be determined. 1 α 2 α 3 Chapter 3 To find : Multiply both sides by s + 1 and let s = -1 1 α 1 1 5 4 4 3 s s s α =- + ∴ = = +...
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This note was uploaded on 09/22/2009 for the course CHEMICAL CHE 401 taught by Professor Dr.muhammadal-arfaj during the Spring '09 term at King Fahd University of Petroleum & Minerals.

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KING FAHD - 1 Chapter 3 Example 2 Example 2 system at rest(s.s Step 1 Take L.T(note zero initial conditions 2 4 6 11 6 2 2 3 3 at t= dt du)= y)=

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