CHE430 02 03 - notes on process control PDF

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Laplace Transforms I Solving linear ODEs using Laplace transforms

Solving a linear nth order ODE along with its given initial conditions is called a general initial value problem. As you know from ESC 250 background, there are many methods to find the solution, which rely on calculus. You may have mastered these methods and are fully competent in performing the requisite calculus, it is still a tedious and difficult approach. In the figure above, the red arrow from the problem statement ODE and initial conditions to the ultimate solution y(t), indicates this path as a direct method for solution. Throughout the solution procedure we maintain that time is the independent variable, so there is a certain familiarity to this domain. The procedure certainly involves calculus and we acknowledge the difficulties associated with calculus. Therefore, the left half of this figure is the time domain approach to solving ODEs - possible but generally difficult. As a clever and less painful alternative, we consider the path shown in black arrows - a roundabout way of finding the same solution. It is instructive to keep in mind the analogy of the use of logarithms for multiplication. This alternate solution path is in the Laplace domain, the right half of the flowchart above, which is implemented in four steps (Step 3 is not necessary in all cases): Step 1: Transform the problem description into the Laplace domain. This is analogous to taking logarithms. It is usually as simple as inspecting each term of the given ODE and looking up the corresponding Laplace transform from a table. There is no guarantee that every term you encounter has a transform readily included in these tables, in which case the transformation is your responsibility. However, reference books have quite extensive collections of Laplace transforms. Moreover, MatLab, Mathematica and Mathcad can do the job for your as well. 1

Step 2: Rearrange the transformed problem statement into a Laplace domain solution Y (s). I say “rearrange” because all it requires in this step is a few algebraic manipulations - moving terms between the lefthand side and the righthand side and finally isolating Y (s). Generally, this solution ends up with the following structure: Y (s) =

N(s) D ( s)

where N(s) is the numerator polynomial is s and D(s) is the denominator polynomial in s. For instance: Y (s) =

1 N(s) = D ( s) s2 + s + 1

where N(s) is constant or 0th degree polynomial and D(s) is 2nd degree or quadratic polynomial. At the end of this step the solution is essentially done. Y (s) is the solution, but it does not have a physical interpretation until it is translated back into the time domain. Let us take the dashed arrow path in the flowchart to Step 4 directly here (bypassing Step 3). Step 4: If you are lucky, you may find that the Y (s) determined in Step 3 is listed in a table of Laplace transforms and you can immediately look up the corresponding time function y(t), “you have arrived at your destination!” What if this best case scenario is not possible, then follow through Step 3 and come back to Step 4. Step 3: More likely scenario is you do not find Y (s) in a table. This requires you to perform some well known algebraic manipulations (known as partial fractions expansion) to change the structure of Y (s) into something more recognizable. This usually involves breaking Y (s) into a summation series of pieces such that each smaller piece can be looked up from the table separately. Even here, you have help available from MatLab, Mathematica and to do the above mentioned algebraic manipulations. In rare cases where Y (s) has no resemblense to anything listed in the table, you are responsible for inverting the transform.

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Laplace Transforms I Example ODE I 1st order ODE 5

dy + 4y = 2; dt

y (0) = 1

I Standard form ODE 1.25

dy + y = 0. 5; dt

y (0) = 1

I Laplace domain solution Y (s )

=

5s + 2 s (5s + 4)

I Time domain solution y (t ) = 0.5 + 0.5e−t /1.25

Let us practice solving linear ODEs using Laplace transforms with a couple of examples. The first one is a 1st order ODE with a given initial condition. We can imagine this as a dynamic model of the position y for a car in motion, starting from an initial position of y(t = 0) = 1. A solution to the differential equation 5

dy + 4y = 2; dt

y(0) = 1

provides us with the trajectory of motion for the car as time progresses, y(t) = 0.5 + 0.5e−t/1.25 If you check at t = 0, y = 1 from the solution and as t → ∞, the exponential term dies out and the car comes to rest at y = 0.5, its steady state. The car moves from 1 to 0.5 on an exponentially decreasing curve. As a habitual exercise, I suggest you rearrange the given ODE into the standard format. It makes no difference for this example, but the standard form is crucial in the next chapter. The standard form puts (1) the terms of the output variable (position here) on the left hand side, (2) in decreasing order of derivatives, (3) with 1 as the coefficient for the output and (4) everything else on the righthand side: 1.25

dy + y = 0.5; dt

y(0) = 1

Coming back, we would like to find the solution using Laplace transforms. We will follow the flowchart from the previous slide: 3

Step 1: Take the Laplace transforms of each term on both sides of the ODE separately and reassemble the equation. Start with the terms on the lefthand side of the ODE

L [4y] = 4Y (s) Obviously we do not know what the transform is because we do not know what y(t) is yet, we are just saying that we will call the transform Y (s). Next comes the derivative term,   dy L 5 = 5[sY (s) − y(0)] = 5sY (s) − 5 dt Recall the way transforms of derivatives are defined. Lastly on the righthand side,

L [2] =

2 s

Reassemble the transformed pieces of the differential equation in t-domain into an algebraic equation in s-domain, 5sY (s) − 5 + 4Y (s) =

2 s

Step 2: Rearrange the transformation from Step 1 by moving terms on either side of the equation with the ultimate goal of explicitly isolating Y (s) on the lefthand side, Y (s) =

5s + 2 s(5s + 4)

This is the solution we are looking for in the Laplace domain. As it is, we do not know how to use this solution to describe the position trajectory of the car because it is an abstract solution. It must be transformed back into time domain for that purpose. A slight massaging of the Y (s) will make things easier for the inverse transformation. Follow this little trickery Y (s) =

5s + 2 s + 2/5 s + b3 = = s(5s + 4) (s + 0)(s + 4/5) (s + b2 )(s + b1 )

where b1 = 4/5 = 0.8, b2 = 0, b3 = 2/5 = 0.4. This final form of Y (s) is exactly listed in the righthand column of Table 3.1 in the textbook on entry 11, and gives us the corresponding time function in the lefthand column of the table. No need for Step 3, go directly to Step 4. Step 4: Simply note down the structure of the time function and substitute the appropriate numbers to present it to your liking, b3 − b1 −b1t b3 − b2 −b2t e + e b2 − b1 b1 − b2 = 0.5 + 0.5e−0.8t

y(t) =

No calculus!

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Laplace Transforms I Example ODE I 3rd order ODE d 3y dt 3

+6

d 2y dt 2

+ 11

dy + 6y = 1; dt

y (0) = y 0 (0) = y 00 (0) = 0

I Standard form ODE d 2y 11 dy 1 1 d 3y ; + + + y = 3 6 6 dt 6 dt dt 2

y (0) = y 0 (0) = y 00 (0) = 0

I Laplace domain solution Y (s )

1

=

s (s 3 + 6s 2 + 11s + 6)

I Time domain solution y (t ) =

1 6

1

1

1

2

2

6

− e−t + e−2t − e−3t

The second example is a 3rd order ODE with its initial conditions. A 3rd order ODE can be in principle split into a system of three coupled 1st order equations, each with its own initial condition. Recall the two 1rt order ODEs for the stirred tank heater with an electric heating element - those two equations can be combined to create a single 2nd order equation (the textbook shows it, but we skipped it). The 3rd order ODE has three initial conditions. Imagine the dynamics of the position y of a particle moving in space as represented by this 3rd order ODE, the initial conditions are position y(t = 0), velocity y′ (t = 0) and acceleration y′′ (t = 0), all of which are given as zero in this example. d3y d2y dy y(0) = y′ (0) = y′′ (0) = 0 + 6 + 11 + 6y = 1; dt dt 3 dt 2 If the same problem was posed as an equivalent set of three coupled 1st order ODEs, there would be one for position, one for velocity and one for acceleration. In anycase, the solution is 1 1 −t 1 −2t 1 −3t − e + e − e 2 6 6 2 As a side exercise, put the ODE in the standard format (check the four requirements for standardization) y(t) =

1 d 3 y d 2 y 11 dy + 2+ +y = 6 dt 3 6 dt dt

1 ; y(0) = y′ (0) = y′′ (0) = 0 6 Let us move on to the application of Laplace transforms to solve this 3rd order ODE. 5

Step 1: Take the transform of each term on both sides of the ODE separately

L [6y] = 6Y (s)  dy = 11[sY (s) − y(0)] = 11sY (s) L 11 dt  2  d y L 6 2 = 6[s2Y (s) − sy(0) − y′ (0)] = 6s2Y (s) dt  3  d y = s3Y (s) − s2 y(0 − sy′ (0) − y′′ (0) = s3Y (s) L dt 3 1 L [1] = s So far we have seen the transforms of 1st order derivatives only. Here is the pattern for obtaining the transformations of higher order derivatives. We are also lucky to have all the initial conditions as zeros, hence, the transforms of the three derivatives look clean: sY (s) for 1st order, s2Y (s) for 2nd order and s3Y (s) for 3rd order derivatives. 

(Make a note to remember this observation of zero initial conditions keeping the transforms of derivatives clean looking. We will revisit this very soon in the next chapter. This is a desirable situation - meaning, we would like to somehow work with zero initial conditions by using simple tricks even when the initial conditions are not zero.) Reassemble the transformed equation, s3Y (s) + 6s2Y (s) + 11sY (s) + 6Y (s) =

1 s

Step 2: Rearrange the algebraic equation to explicitly isolate Y (s) on the lefthand side, Y (s) =

1 s(s3 + 6s2 + 11s + 6)

This is the Laplace domain solution. This time we are not lucky enough to find it in any table for looking up the inverse Laplace transformation. Therefore, we must go to Step 3 to change its structure into something more recognizable. Step 3: A detailed reexamination of the method of “partial fractions expansion” is beyond the scope of our current focus. You are expected to be knowledgeable of it. It is useful and essential as well as tedious and messy. I will leave you to practice it from Section 3.3 in the book or any introductory algebra book. In addition, software such as MatLab, Mathematica and Mathcad can eliminate much of the tedious work by doing this step for you, you do need to buy the appropriate symbolic processing tool boxes that are not sold as part of the standard student versions. The results of a partial fractions expansion changes the look of Y (s) into preferable one, Y (s) =

1 s(s3 + 6s2 + 11s + 6)

=

1/2 1/6 1/6 1/2 + − − s+0 s+1 s+2 s+3

The structure of each of these four separate terms is exactly the same. This structure shows up in the righthand side column of Table 3.1 in entry 5 for a simple and easy inverse transform. Step 4: Finally, we can assemble the solution as the following the time function, y(t) =

1 1 −t 1 −2t 1 −3t − e + e − e 6 6 2 2 6

Reading Exercise: Sections 3.2-3.3 in chapter 3.

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