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 1
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 2
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 3
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 4
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 5
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) 6
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) 7
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. 8
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 9
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) 10
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 = 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 See page 54 of the text. 12
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, 1 5s 2 yt () + = L s( 5s+ 4) From Table 3.1, 0.8t () = + e yt 0.5 0.5 (3-37) 13
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 14
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) 15
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 s α i + 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. 16
Example 3.2 (continued) Recall that the ODE, y + + 6 y+ 11y + 6y = 1, with zero initial conditions resulted in the expression Y s = 1 ( 3 + 6 2 + 11 + 6) s s s s The denominator can be factored as ( 3 2 + + + ) = ( + )( + )( + ) (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) 17
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) 18
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. 19
Example: 2. Time Delay Suppose, Then, Y s 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 20