Pierre Simon Laplace Born: 23 March 1749 in Beaumont-en-Auge, Normandy, France Died: 5 March 1827 in Paris, France Laplace Transforms Dr. M. A. A. Shoukat Choudhury 1
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 Dr. M. A. A. Shoukat Choudhury 2
Definition The Laplace transform of a function, f(t), is defined as [ ] () Fs () L ft () f te dt (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 0 st Dr. M. A. A. Shoukat Choudhury 3
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: 1 L F( s) f t = Important Properties: Both L and L -1 are linear operators. Thus, () + = + = ax ( s) + by ( s) (3-3) L axt byt al xt bl yt Dr. M. A. A. Shoukat Choudhury 4
where: - x(t) and y(t) are arbitrary functions Similarly, - a and b are constants - X ( s) = L x( t) and Y( s) = L y( t) L + = + 1 ax s by s ax t b y t Dr. M. A. A. Shoukat Choudhury 5
Laplace Transforms of Common Functions 1. Constant Function Let f(t) = a (a constant). Then from the definition of the Laplace transform in (3-1), L st a st a a a = ae dt = e = 0 = (3-4) s s s 0 0 Dr. M. A. A. Shoukat Choudhury 6
2. Step Function The unit step function is widely used in the analysis of process control problems. It is defined as: () S t = 0 fort < 0 1 fort 0 (3-5) Because the step function is a special case of a constant, it follows from (3-4) that () L S t = 1 s (3-6) Dr. M. A. A. Shoukat Choudhury 7
3. 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 df L = sf s dt f 0 (3-9) initial condition at t = 0 Similarly, for higher order derivatives: L n d f n n 1 n 2 () 1 s F s s f 0 s f ( 0) n = dt ( n ) 2 1 ( n )... sf 0 f 0 (3-14) Dr. M. A. A. Shoukat Choudhury 8
where: - n is an arbitrary positive integer - ( k f ) ( 0) d k f k dt = Special Case: All Initial Conditions are Zero Suppose t 0 1 1 ( n ) f 0 = f 0 =... = f 0. L n d f n = dt = n s F s 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. Dr. M. A. A. Shoukat Choudhury 9
4. Exponential Functions Consider bt = e f t where b > 0. Then, bt bt st ( b + s) t L e = e e dt = e dt 0 0 1 ( b+ s) t 1 = e = b+ s 0 s+ b 5. Rectangular Pulse Function It is defined by: (3-16) 0 for t < 0 f () t = h for 0 t < tw (3-20) 0 for t tw Dr. M. A. A. Shoukat Choudhury 10
h f () t t w Time, t The Laplace transform of the rectangular pulse is given by h s ( ts = e ) F s w 1 (3-22) Dr. M. A. A. Shoukat Choudhury 11
6. Impulse Function (or Dirac Delta Function) The impulse function is obtained by taking the limit of the rectangular pulse as its width, t w, goes to zero but holding 1 the area under the pulse constant at one. (i.e., let h = ) t Let, = impulse function w δ ( t) Then, ( t) 1 L δ = Dr. M. A. A. Shoukat Choudhury 12
Table 3.1. Laplace Transforms See page 54 of the text. Dr. M. A. A. Shoukat Choudhury 13
Laplace Transform Table Dr. M. A. A. Shoukat Choudhury 14
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). Dr. M. A. A. Shoukat Choudhury 15
Example 3.1 Solve the ODE, dy 5 4y 2 y( 0) 1 (3-26) dt + = = First, take L of both sides of (3-26), 2 5( sy ( s) 1) + 4Y ( s) = s Rearrange, 5s + 2 Y( s) = (3-34) s 5s+ 4 Take L -1, From Table 3.1, () yt 1 5s + 2 = L s( 5s+ 4) 0.8t () = + e yt 0.5 0.5 (3-37) Dr. M. A. A. Shoukat Choudhury 16
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: Y s = s + 5 ( s+ 1)( s+ 4) Perform a partial fraction expansion (PFE) s + 5 α1 α = + 2 s+ 1 s+ 4 s+ 1 s+ 4 (3-41) (3-42) where coefficients and α have to be determined. α1 2 Dr. M. A. A. Shoukat Choudhury 17
To find α 1 : Multiply both sides by s + 1 and let s = -1 s + 5 4 α1 = = s + 4 3 s= 1 To find α 2 : Multiply both sides by s + 4 and let s = -4 s + 5 1 α2 = = s + 1 3 A General PFE s= 4 Consider a general expression, Y s N s = = D s n π i= 1 N s ( s+ b ) i (3-46a) Dr. M. A. A. Shoukat Choudhury 18
Here D(s) is an n-th order polynomial with the roots s = b i all being real numbers which are distinct so there are no repeated roots. The PFE is: Y s N s = = n n π = i i= 1 ( s+ b ) i 1 αi s+ b Note: D(s) is called the characteristic polynomial. Special Situations: i (3-46b) Two other types of situations commonly occur when D(s) has: i) Complex roots: e.g., bi = 3± 4j j = 1 ii) Repeated roots (e.g., b = b = ) 1 2 3 For these situations, the PFE has a different form. See SEM text (pp. 61-64) for details. Dr. M. A. A. Shoukat Choudhury 19
Example 3.2 The ODE, &&& y ++ 6&& y+ 11y& + 6y = 1 resulted in the expression Y s = 1 ( 3 + 6 2 + 11 + 6) s s s s The denominator can be factored as ( 3 2 + + + ) = ( + )( + )( + ), with zero initial conditions (3-40) s s 6s 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 α1 α2 α3 α = = + + + 4 s s+ 1 s+ 2 s+ 3 s s+ 1 s+ 2 s+ 3 (3-51) Dr. M. A. A. Shoukat Choudhury 20
Solve for coefficients to get 1 1 1 1 α1 =, α2 =, α3 =, α4 = 6 2 2 6 (For example, find α, 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: L 1 1 1/6 1 1/2 1 1/2 1 1/6 Y s = L + s L s+ 1 L s+ 2 L s+ 3 From Table 3.1, 1 1 1 1 yt e e e 6 2 2 6 t 2t 3t () = + (3-52) Dr. M. A. A. Shoukat Choudhury 21
Important Properties of Laplace Transforms 1. 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: lim t = lim sy ( s) yt s 0 providing that the limit exists (is finite) for all Re( s) 0, where Re (s) denotes the real part of complex variable, s. Dr. M. A. A. Shoukat Choudhury 22
Example: 2. Time Delay Suppose, Y s = Then, y 5s + 2 s ( 5s+ 4) y( t) (3-34) 5s + 2 = lim = lim 0.5 t = s 0 5 s + 4 Time delays occur due to fluid flow, time required to do an analysis (e.g., gas chromatograph). The delayed signal can be represented as Also, yt ( θ) θ= time delay θs ( θ) = L yt e Y s Dr. M. A. A. Shoukat Choudhury 23