Posted: October 27th, 2022

Process Control and Instrumentation Homework for Chemical Engineering

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CBE 43

0

Week 0

3

Reading
โ€ข Ch. 3

Laplace Transforms
Mathematical models of real processes often result in systems of linear ODEs.

The Laplace transform can reduce the time required to find a solution, by
converting ODEs to equations.

Letโ€™s review Laplace transforms!

โ„’ ๐‘“ ๐‘ก

= เถฑ
0

โˆž

๐‘“ ๐‘ก ๐‘’โˆ’๐‘ ๐‘ก๐‘‘๐‘ก = ๐น

๐‘ 

โ„’โˆ’1 ๐น ๐‘  = ๐‘“

๐‘ก

Laplace Domain

๐‘“(๐‘ก) ๐น(๐‘ )

Time Domain

Slide 03.0

1

Topics
โ€ข Laplace

transforms
โ€ข Solving

ODEs
โ€ข Properties

of

โ„’[ ]

โ€ข Example

3.6
โ€ข MATLAB

CBE 430
Week 03

Reading
โ€ข Ch. 3

Laplace Transforms
Constant function:

Unit step function:

Exponential:

Rectangular

Pulse:

โ„’ ๐‘Ž = เถฑ
0

โˆž

๐‘Ž๐‘’โˆ’๐‘ ๐‘ก๐‘‘๐‘ก = โˆ’

๐‘Ž

๐‘ 
๐‘’โˆ’๐‘ ๐‘ก

0
โˆž

=
๐‘Ž

๐‘ 

โ„’ ๐‘“ ๐‘ก

=
1

๐‘ 

Notice, the Laplace transform
doesnโ€™t care what happens at ๐‘ก < 0! So the unit step at ๐‘ก = 0 is the same as the constant with ๐‘Ž = 1.

โ„’ ๐‘“(๐‘ก) = เถฑ
0

โˆž

๐‘’โˆ’๐‘Ž๐‘ก๐‘’โˆ’๐‘ ๐‘ก๐‘‘๐‘ก = เถฑ
0

โˆž

๐‘’โˆ’(๐‘ +๐‘Ž)๐‘ก๐‘‘๐‘ก = โˆ’
๐‘’โˆ’ ๐‘ +๐‘Ž ๐‘ก

๐‘  + ๐‘Ž

0

โˆž
=
1
๐‘  + ๐‘Ž

๐‘“ ๐‘ก = แ‰
0 ๐‘ก < 0 โ„Ž 0 โ‰ค ๐‘ก <

๐‘ก๐‘ค

0 ๐‘ก โ‰ฅ ๐‘ก๐‘ค

โ„’ ๐‘“(๐‘ก) = เถฑ
0
๐‘ก๐‘ค

โ„Ž๐‘’โˆ’๐‘ ๐‘ก๐‘‘๐‘ก = โˆ’
โ„Ž

๐‘ 
๐‘’โˆ’๐‘ ๐‘ก แ‰š

0
๐‘ก๐‘ค

=
โ„Ž

๐‘ 
1 โˆ’ ๐‘’โˆ’๐‘ก๐‘ค๐‘ 

๐‘“ ๐‘ก = ๐‘Ž

๐‘“ ๐‘ก = ๐‘† ๐‘ก = แ‰Š
0 ๐‘ก < 0 1 ๐‘ก โ‰ฅ 0

๐‘“ ๐‘ก = ๐‘’โˆ’๐‘Ž๐‘ก

Slide 03.0

2

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties

of โ„’[ ]
โ€ข Example

3.6
โ€ข MATLAB

CBE 430
Week 03
Reading
โ€ข Ch. 3

How About a Time Delay (Lag).

In chemical and biological processes, one function is often related to another by
a time delay (lag). The simplest example is:

๐‘“๐‘‘ ๐‘ก = ๐‘“ ๐‘ก โˆ’ ๐œƒ ๐‘† ๐‘ก โˆ’ ๐œƒ ๐‘† ensures that ๐‘“๐‘‘ = 0 before the lag
time. This can happen when ๐‘“ and ๐‘“

๐‘‘

measure the same thing (e.g. a
temperature or concentration) at
different points in a , for
example.

What is the Laplace transform of ๐‘“๐‘‘(๐‘ก)?

Slide 03.03

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

CBE 430
Week 03
Reading
โ€ข Ch. 3
How About a Time Delay (Lag).

๐‘“๐‘‘ ๐‘ก = ๐‘“ ๐‘ก โˆ’ ๐œƒ ๐‘† ๐‘ก โˆ’

๐œƒ

โ„’ ๐‘“๐‘‘ ๐‘ก = โ„’ ๐‘“ ๐‘ก โˆ’ ๐œƒ ๐‘† ๐‘ก โˆ’ ๐œƒ

= เถฑ
0
โˆž

๐‘“ ๐‘ก โˆ’ ๐œƒ ๐‘† ๐‘ก โˆ’ ๐œƒ ๐‘’โˆ’๐‘ ๐‘ก

๐‘‘๐‘ก

= เถฑ
0
๐œƒ

๐‘“ ๐‘ก โˆ’ ๐œƒ 0 ๐‘’โˆ’๐‘ ๐‘ก๐‘‘๐‘ก + เถฑ
๐œƒ

โˆž

๐‘“ ๐‘ก โˆ’ ๐œƒ ๐‘’โˆ’๐‘ ๐‘ก๐‘‘๐‘ก

= เถฑ
๐œƒ

โˆž

๐‘“ ๐‘ก โˆ’ ๐œƒ ๐‘’โˆ’๐‘  ๐‘กโˆ’๐œƒ ๐‘’โˆ’๐‘ ๐œƒ ๐‘‘ ๐‘ก โˆ’ ๐œƒ

= ๐‘’โˆ’๐‘ ๐œƒเถฑ
0

โˆž

๐‘“ ๐‘กโˆ— ๐‘’โˆ’๐‘ ๐‘ก
โˆ—
๐‘‘๐‘กโˆ—

โ„’[๐‘“๐‘‘ ๐‘ก ] = ๐‘’
โˆ’๐‘ ๐œƒ๐น ๐‘ 

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.04

CBE 430
Week 03
Reading
โ€ข Ch. 3

Now you try one:

Use the property:

and the Euler formula for the cosine:

To find

โ„’ ๐‘ฅ ๐‘ก + ๐‘ฆ ๐‘ก = โ„’ ๐‘ฅ ๐‘ก + โ„’ ๐‘ฆ ๐‘ก

=
1
2
1

๐‘  โˆ’ ๐‘—๐œ”
+

1

๐‘  + ๐‘—๐œ”

cos ๐œ”๐‘ก =
๐‘’๐‘—๐œ”๐‘ก + ๐‘’โˆ’๐‘—๐œ”๐‘ก

2
; ๐‘— โ‰ก โˆ’1

โ„’[cos ๐œ”๐‘ก ] =
1

2
โ„’ ๐‘’๐‘—๐œ”๐‘ก +

1

2
โ„’ ๐‘’โˆ’๐‘—๐œ”๐‘ก

=
1
2

๐‘  + ๐‘—๐œ” + ๐‘  โˆ’ ๐‘—๐œ”

๐‘ 2 +๐œ”2

=

๐‘ 
๐‘ 2 +๐œ”2

โ„’[cos ๐œ”๐‘ก ]

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.05

CBE 430
Week 03
Reading
โ€ข Ch. 3

What about this one?

๐‘ข = ๐‘ก ๐‘‘๐‘ฃ = ๐‘’โˆ’ ๐‘ +๐‘Ž ๐‘ก๐‘‘๐‘ก

๐‘‘๐‘ข = ๐‘‘๐‘ก ๐‘ฃ

= โˆ’
1

๐‘  + ๐‘Ž
๐‘’โˆ’ ๐‘ +๐‘Ž ๐‘ก

โ„’ ๐‘ก๐‘’๐‘Ž๐‘ก = โˆ’
๐‘ก๐‘’โˆ’ ๐‘ +๐‘Ž ๐‘ก

๐‘  + ๐‘Ž
๐‘ก=0

๐‘ก=โˆž

+เถฑ
0

โˆž ๐‘’โˆ’ ๐‘ +๐‘Ž ๐‘ก

๐‘  + ๐‘Ž
๐‘‘๐‘ก

= 0 โˆ’ 0 โˆ’
๐‘’โˆ’ ๐‘ +๐‘Ž ๐‘ก

๐‘  + ๐‘Ž 2

๐‘ก=0

๐‘ก=โˆž
=
1
๐‘  + ๐‘Ž 2

โ„’ ๐‘ก๐‘’โˆ’๐‘Ž๐‘ก

Integration by parts! เถฑ
0

โˆž

๐‘ข๐‘‘๐‘ฃ = ๐‘ข๐‘ฃ โˆ’เถฑ
0

โˆž

๐‘ฃ๐‘‘๐‘ข

= เถฑ
0
โˆž

๐‘ก๐‘’โˆ’๐‘Ž๐‘ก๐‘’โˆ’๐‘ ๐‘ก๐‘‘๐‘ก = เถฑ
0

โˆž

๐‘ก๐‘’โˆ’ ๐‘Ž+๐‘  ๐‘ก๐‘‘๐‘ก

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.06

CBE 430
Week 03
Reading
โ€ข Ch. 3

One more important exampleโ€ฆ

The impulse (with area ๐‘Ž): ๐‘“ ๐‘ก = ๐‘Ž๐›ฟ ๐‘ก

โ„’ ๐‘“(๐‘ก) = lim

๐‘ก๐‘คโ†’0

๐‘Ž

๐‘ก๐‘ค๐‘ 
1 โˆ’ ๐‘’โˆ’๐‘ก๐‘ค๐‘ 

Where ๐›ฟ ๐‘ก is the function. ๐›ฟ ๐‘ก has an area of 1, a
height of infinity, and a width of 0. This can be modeled as a pulse
of โ„Ž = 1/๐‘ก๐‘ค as ๐‘ก๐‘ค โ†’ 0. Using the result for the pulseโ€ฆ

Apply lโ€™Hรดpitalโ€™s rule (differentiate by ๐‘ก๐‘ค):

โ„’ ๐‘Ž๐›ฟ ๐‘ก = lim
๐‘ก๐‘คโ†’0

๐‘Ž 1 โˆ’ ๐‘’โˆ’๐‘ก๐‘ค๐‘ 

๐‘ก๐‘ค๐‘ 
= lim

๐‘ก๐‘คโ†’0

๐‘Ž๐‘ ๐‘’โˆ’๐‘ก๐‘ค๐‘ 

๐‘ 
= ๐‘Ž

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.07

CBE 430
Week 03
Reading
โ€ข Ch. 3

Why is this useful?

Finding Laplace transforms is fun and easy, but thatโ€™s not
why we use them. In fact, we usually donโ€™t bother finding
them at allโ€ฆ we in a table (like Table 3.1
on pp. 40-41 of Seborg).

Laplace transforms are useful because they
help us to find solutions to differential
equations.

โ€ฆhereโ€™s howโ€ฆ โ†’

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.08

CBE 430
Week 03
Reading
โ€ข Ch. 3

Time-Derivatives
What is the Laplace transform of the time-derivative?

How about the second
time derivative?

For the ๐‘›๐‘กโ„Ž derivative:

โ„’

๐‘‘๐‘“

๐‘‘๐‘ก
= เถฑ

0

โˆž๐‘‘๐‘“

๐‘‘๐‘ก
๐‘’โˆ’๐‘ ๐‘ก๐‘‘๐‘ก = ๐‘“๐‘’โˆ’๐‘ ๐‘ก แ‰š

0

โˆž
โˆ’เถฑ

0
โˆž

๐‘“ ๐‘ก
๐‘‘ ๐‘’โˆ’๐‘ ๐‘ก

๐‘‘๐‘ก
๐‘‘๐‘ก

= โˆ’๐‘“ 0 + เถฑ
0

โˆž

๐‘“ ๐‘ก ๐‘ ๐‘’โˆ’๐‘ ๐‘ก๐‘‘๐‘ก

= โˆ’๐‘“ 0 + ๐‘ โ„’ ๐‘“ ๐‘ก

โ„’
๐‘‘2๐‘“

๐‘‘๐‘ก2
= โ„’

๐‘‘๐œ™

๐‘‘๐‘ก
; where ๐œ™ =

๐‘‘๐‘“
๐‘‘๐‘ก

= โˆ’๐œ™ 0 + ๐‘  ๐‘ ๐น ๐‘  โˆ’ ๐‘“ 0

= โˆ’๐‘“ 0 + ๐‘ ๐น ๐‘ 

= ๐‘ 2๐น ๐‘  โˆ’ แ‰ค
๐‘‘๐‘“

๐‘‘๐‘ก
๐‘ก=0

โˆ’ ๐‘ ๐‘“ 0

This is awesome!
Now we can use
Laplace transforms
to replace first
derivatives!

๐‘‘๐‘› ๐‘“ ๐‘ก

๐‘‘๐‘ก๐‘›
= ๐‘ ๐‘›๐น ๐‘  โˆ’ เท

๐‘š=0

๐‘›โˆ’1

๐‘ ๐‘š เธญ
๐‘‘๐‘›โˆ’1โˆ’๐‘š๐‘“ ๐‘ก

๐‘‘๐‘ก๐‘›โˆ’1โˆ’๐‘š
๐‘ก=0

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.09

CBE 430
Week 03
Reading
โ€ข Ch. 3

A First-Order ODE Example

5

๐‘‘๐‘ฆ

๐‘‘๐‘ก
+ 4๐‘ฆ = 2; ๐‘ฆ 0 = 1

5โ„’
๐‘‘๐‘ฆ

๐‘‘๐‘ก
+ 4โ„’ ๐‘ฆ = โ„’ 2

5 ๐‘ ๐‘Œ ๐‘  โˆ’ ๐‘ฆ 0 + 4

๐‘Œ ๐‘  =
2

๐‘ 

5๐‘ ๐‘Œ ๐‘  โˆ’ 5 1 + 4๐‘Œ ๐‘  =
2

๐‘ 

5๐‘  + 4 ๐‘Œ ๐‘  =
2

๐‘ 
+ 5

๐‘Œ ๐‘  =
2

๐‘ 

5๐‘  +

4

+

5
5๐‘  + 4
โ„’[ ]

Solve
for ๐‘Œ ๐‘ 

Now invert using Table
3.1. Notice that entries 5
and 6, and 9 and 10 are
equivalent alternative
representations.

๐‘Œ ๐‘  =
2
5
1

๐‘  + 0

๐‘  +
4
5

+
1

๐‘  +
4
5

๐‘ฆ ๐‘ก = โˆ’0.5 ๐‘’โˆ’0.8๐‘ก โˆ’ 1 + ๐‘’โˆ’0.8๐‘ก

= 0.5 1 โˆ’ ๐‘’โˆ’0.8๐‘ก

Transform
equation

โ„’โˆ’1[ ]

Notice the gets
used here!

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.10

CBE 430
Week 03
Reading
โ€ข Ch. 3

A Higher-Order ODE Example

๐‘‘3๐‘ฆ

๐‘‘๐‘ก3
+ 6

๐‘‘2๐‘ฆ

๐‘‘๐‘ก2
+ 11

๐‘‘๐‘ฆ

๐‘‘๐‘ก
+ 6๐‘ฆ = 1;

๐‘Œ ๐‘  =
1

๐‘ (๐‘ 3 + 6๐‘ 2 + 11๐‘  + 6)

เธญ
๐‘‘2๐‘ฆ

๐‘‘๐‘ก2
0

= 0; แ‰ค
๐‘‘๐‘ฆ

๐‘‘๐‘ก
0

= 0; ๐‘ฆ 0 = 0

How many ICโ€™s do we need?

โ„’
๐‘‘3๐‘ฆ

๐‘‘๐‘ก3
+ 6โ„’

๐‘‘2๐‘ฆ

๐‘‘๐‘ก2
+ 11โ„’

๐‘‘๐‘ฆ

๐‘‘๐‘ก
+ 6โ„’ ๐‘ฆ = โ„’ 1

๐‘ 3๐‘Œ ๐‘  + 6๐‘ 2๐‘Œ ๐‘  + 11๐‘ ๐‘Œ ๐‘  + 6๐‘Œ ๐‘  =
1

๐‘ 

Find the (4) roots of the (4th-order)
polynomial in the denominator, so it
can be factored!
Roots are ๐‘  = 0, -1, -2, and -3. ๐‘Œ ๐‘  =

1

๐‘ (๐‘  + 1)(๐‘  + 2)(๐‘  + 3)

Perform .
๐‘Œ ๐‘  =

๐›ผ1
๐‘ 
+

๐›ผ2

๐‘  + 1

+
๐›ผ3

๐‘  + 2
+

๐›ผ4

๐‘  + 3

๐›ผ1 =
1

6
; ๐›ผ2 = โˆ’

1

2
; ๐›ผ3 =

1

2
; ๐›ผ4 = โˆ’

1

6
Perform
Heaviside
expansion.

๐‘ฆ ๐‘ก =
1

6
โˆ’
1

2
๐‘’โˆ’๐‘ก +

1

2
๐‘’โˆ’2๐‘ก โˆ’

1

6
๐‘’โˆ’3๐‘ก

When did we
use the ICโ€™s?!?!

โ„’[ ]
โ„’โˆ’1[ ]
Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.11

CBE 430
Week 03
Reading
โ€ข Ch. 3

What was that Heaviside thing?

๐‘Œ ๐‘  =

๐‘ ๐‘ 

๐ท ๐‘ 

=

1

๐‘ (๐‘  + 1)(๐‘  + 2)(๐‘  + 3)
=
๐›ผ1
๐‘ 
+

๐›ผ2
๐‘  + 1
+
๐›ผ3
๐‘  + 2
+
๐›ผ4
๐‘  + 3

Find the coefficients ๐›ผ๐‘– using .

1. Multiply the whole equations by one of the terms in the denominator (for
example, the second one).

2. Choose ๐‘  so that all the remaining ๐›ผ๐‘– are . In this case, we
set ๐‘  = โˆ’1. Compute ๐›ผ2.

๐‘  + 1

๐‘  ๐‘  + 1 ๐‘  + 2 ๐‘  + 3
= ๐›ผ2 + ๐‘  + 1

๐›ผ1
๐‘ 
+

๐›ผ3
๐‘  + 2

+
๐›ผ4

๐‘  + 3
1

โˆ’1 โˆ’1 + 2 [ โˆ’1 + 3]
= ๐›ผ2 =

1

โˆ’1 ร— 1 ร—

2
= โˆ’

1
2

In generalโ€ฆ ๐›ผ๐‘– = แ‰ค๐‘  + ๐‘

๐‘–

๐‘ ๐‘ 

๐ท ๐‘ 
๐‘ =โˆ’๐‘๐‘–

What about
repeated roots?…

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.12

CBE 430
Week 03
Reading
โ€ข Ch. 3

An example with repeated roots
๐‘‘2๐‘ฆ

๐‘‘๐‘ก2
+ 4

๐‘‘๐‘ฆ

๐‘‘๐‘ก
+ 4๐‘ฆ = 1; เธญ

๐‘‘2๐‘ฆ
๐‘‘๐‘ก2
0
= 0; แ‰ค
๐‘‘๐‘ฆ
๐‘‘๐‘ก
0

= 1; ๐‘ฆ 0 = 0

๐‘ 2๐‘Œ ๐‘  โˆ’ ๐‘ ๐‘ฆ 0 โˆ’ แ‰ค
๐‘‘๐‘ฆ

๐‘‘๐‘ก
0

+ 4 ๐‘ ๐‘Œ ๐‘  โˆ’ ๐‘ฆ 0 + 4๐‘Œ ๐‘  =
1

๐‘ 

๐‘ 2๐‘Œ ๐‘  โˆ’ 1 + 4๐‘ ๐‘Œ ๐‘  + 4๐‘Œ ๐‘  =
1

๐‘ 

๐‘ 2 + 4๐‘  + 4 ๐‘Œ ๐‘  =
1

๐‘ 
+ 1 =

๐‘  + 1
๐‘ 

๐‘Œ ๐‘  =
๐‘  + 1

๐‘  ๐‘ 2 + 4๐‘  + 4

Use the Heaviside method
to find ๐›ผ1 and ๐›ผ3

๐‘Œ ๐‘  =
๐‘  + 1

๐‘  ๐‘ 2 + 4๐‘  + 4
=
๐›ผ1
๐‘ 
+

๐›ผ2
๐‘  + 2

+
๐›ผ3

๐‘  + 2 2

Canโ€™t use the Heaviside
method for ๐›ผ2 because
multiplication by ๐‘  + 2 ,
and setting
doesnโ€™t eliminate the ๐›ผ3
term; it blows up.

๐›ผ1 =
0 + 1

02 + 4 0 + 4
=
1

4
; ๐›ผ3 =

โˆ’2 + 1

โˆ’2
=
1

2

Then just choose another ๐‘ 
to find ๐›ผ2; choose ๐‘  = โˆ’1

๐›ผ2 =
๐‘  + 1

๐‘ (๐‘ 2 + 4๐‘  + 4)
โˆ’

1

4๐‘ 
โˆ’

1

2 ๐‘  + 2 2
; ๐›ผ2 =

1

4
โˆ’
1

2
= โˆ’
1
4

The denominator is a 3rd-order
polynomial, so there must be 3 constants.

ICโ€™s used here!

โ„’[ ]
Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.13

CBE 430
Week 03
Reading
โ€ข Ch. 3

Or, you can take the derivative
๐‘Œ ๐‘  =

๐‘  + 1
๐‘  ๐‘ 2 + 4๐‘  + 4
=
๐›ผ1
๐‘ 
+
๐›ผ2
๐‘  + 2
+
๐›ผ3
๐‘  + 2 2

For multiple repeated roots, or roots repeated several times, the derivative
method helps ensure you have to solve for all of the
unknown ๐›ผ๐‘–. The general formula that you can use in this case is:

๐›ผ๐‘– =
1

๐‘› โˆ’ ๐‘– !
lim
๐‘ โ†’โˆ’๐‘

๐‘‘๐‘›โˆ’๐‘–

๐‘‘๐‘ ๐‘›โˆ’๐‘–
๐‘  + ๐‘ ๐‘›

๐‘ ๐‘ 
๐ท ๐‘ 

๐›ผ3 = แ‰ค๐‘  + 2
2
๐‘ ๐‘ 

๐ท ๐‘ 
๐‘ =โˆ’2

; ๐›ผ2 = เธญ
๐‘‘

๐‘‘๐‘ 
๐‘  + 2 2

๐‘ ๐‘ 

๐ท

๐‘ 
s=โˆ’2

; ๐›ผ1 = แ‰ค๐‘ 
๐‘ ๐‘ 

๐ท ๐‘ 
๐‘ =0

๐›ผ3 =
1

2
; ๐›ผ2 = แ‰ค

๐‘‘

๐‘‘๐‘ 
1 +

1
๐‘ 
s=โˆ’2
= โˆ’
1

โˆ’2 2
= โˆ’

1

4
; ๐›ผ1 =

1
4

๐‘Œ ๐‘  =
๐›ผ1

(๐‘  + ๐‘)
+

๐›ผ2
๐‘  + ๐‘ 2

+โ‹ฏ
๐›ผ๐‘›

๐‘  + ๐‘ ๐‘›
+โ‹ฏ =เท

๐‘–

๐‘›
๐›ผ๐‘–

๐‘  + ๐‘ ๐‘–
+โ‹ฏ

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.14

CBE 430
Week 03
Reading
โ€ข Ch. 3

What about complex roots?

Heaviside expansion will

give complex ๐›ผ1,2 = โˆ’
1

3
๐‘—

๐‘Œ ๐‘  =
1

๐‘ 2 + ๐‘  + 2.5

=

๐›ผ1
๐‘  + ๐‘1

+
๐›ผ2

๐‘  + ๐‘2

๐‘Œ ๐‘  =
๐‘ ๐‘ 

๐ท ๐‘ 
; ๐ท ๐‘  = ๐‘Ž๐‘ 2 + ๐‘๐‘  + ๐‘; ๐‘2 < 4๐‘Ž๐‘ โ†’ ๐‘๐‘œ๐‘š๐‘๐‘™๐‘’๐‘ฅ ๐‘Ÿ๐‘œ๐‘œ๐‘ก๐‘ 

๐‘1, ๐‘2 =
โˆ’1 ยฑ 12 โˆ’ 4 1 2.5

2 1
= โˆ’

1

2
ยฑ

โˆ’9

2
= โˆ’
1

2
ยฑ
3

2
๐‘—

Find ๐‘1 and ๐‘2 from the roots of ๐ท ๐‘ 

Instead, lets put ๐‘Œ(๐‘ ) in the form of one of the entries in the Table 3.1
on p. 42. โ€“ How about number 17?

๐‘Œ ๐‘  = ๐‘
๐œ”

๐‘  + ๐‘ 2 + ๐œ”2
=

1
๐‘ 2 + ๐‘  + 2.5

This term has to give both

๐‘ 2 and ๐‘ . So ๐‘ =
1

2
.

Now the denominator is:

๐ท ๐‘  = ๐‘  +
1

2
2

+ ๐œ”2 = ๐‘ 2 + s +
1

4
+ ๐œ”2 = ๐‘ 2 + s + 2.5

๐œ”2 = 2.25; ๐œ” =

1.5

๐œ”๐‘ = 1; ๐‘ =
2

3
๐‘Œ ๐‘  =
2
3
1.5

๐‘  +
1
2

2

+ 1.52

๐‘ฆ ๐‘ก =
2

3
๐‘’โˆ’0.5๐‘กsin(1.5๐‘ก)

Find ๐‘:

Complex roots will always give in the time domain.

Find ๐œ”:
โ„’โˆ’1[ ]

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.15

CBE 430
Week 03
Reading
โ€ข Ch. 3

Some Properties of Laplace Transforms
1. The transform and reverse transform are both linear operations, allowing you to use superposition:

2. The theorem:

3. The initial value theorem:

4. The theorem:

5. Convolution:

6. When ๐น ๐‘  has positive real denominator roots (๐‘๐‘– < 0), ๐‘“(๐‘ก) will grow exponentially.

7. When ๐น(๐‘ ) has complex denominator roots (๐‘๐‘–
2< 0), ๐‘“ ๐‘ก will oscillate .

8. When ๐น(๐‘ ) has an exponential in ๐‘ , ๐‘“(๐‘ก) has a .

โ„’ ๐‘ฅ ๐‘ก + ๐‘ฆ ๐‘ก = โ„’ ๐‘ฅ ๐‘ก + โ„’ ๐‘ฆ ๐‘ก

lim
๐‘กโ†’โˆž

๐‘ฆ(๐‘ก) = lim
๐‘ โ†’0

[๐‘ ๐‘Œ ๐‘  ]

๐น ๐‘  = ๐บ ๐‘  ๐ป ๐‘ 

๐‘“ ๐‘ก = โ„’โˆ’1 ๐บ ๐‘  ๐ป ๐‘  = เถฑ
0

๐‘ก

๐‘” ๐œ โ„Ž ๐‘ก โˆ’ ๐œ ๐‘‘๐œ = เถฑ
0

๐‘ก

๐‘” ๐‘ก โˆ’ ๐œ โ„Ž ๐œ ๐‘‘๐œ

lim
๐‘กโ†’0

๐‘ฆ(๐‘ก) = lim
๐‘ โ†’โˆž

[๐‘ ๐‘Œ ๐‘  ]

โ„’ ๐‘“ ๐‘ก โˆ’ ๐‘ก0 ๐‘† ๐‘ก โˆ’ ๐‘ก0 = ๐‘’
โˆ’๐‘ก0๐‘ ๐น(๐‘ )

Great for checking answers!

Weโ€™ll make use of these a lot.

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.16

CBE 430
Week 03
Reading
โ€ข Ch. 3

Letโ€™s do Example 3.6

Check out how the CSTR works. (Are there
recirculation zones, dead zones, or other non-ideal
mixing?)

Do an impulse test with a tracer, compare the
concentration coming out, ๐‘2(๐‘ก), to the predicted
concentration assuming .

Compare the theoretical solutions of the responses
to (perfect) impulse, and (physically realizable)
rectangular pulse test.

Find:
a) The magnitude of an impulse that would model

the total amount of material added with the
pulse shown at the right.

b) The impulse and pulse responses of the reactant
concentration leaving the reactor, ๐‘.

๐‘‰ = 4 ๐‘š3

๐‘ž = 2 ๐‘š3 ๐‘š๐‘–๐‘›๐‘ข๐‘ก๐‘’โˆ’1

๐‘๐‘– = 1 ๐‘˜๐‘”๐‘š๐‘œ๐‘™ ๐‘š
โˆ’3

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.17

CBE 430
Week 03
Reading
โ€ข Ch. 3

Example 3.6(a)

Find:
a) The magnitude of an impulse that would

model the total amount of material added
with the pulse shown at the right.

๐‘Ž๐›ฟ ๐‘ก ; ๐ด๐‘–๐‘š๐‘๐‘ข๐‘™๐‘ ๐‘’ = ๐‘Ž

Define the ๐œ’๐‘– = ๐‘๐‘– โˆ’ เดฅ๐‘๐‘– and ๐œ’ = ๐‘ โˆ’ าง๐‘
Where เดฅ๐‘๐‘– and าง๐‘ are the respective values.

๐œ’๐‘–,๐‘๐‘ข๐‘™๐‘ ๐‘’ = เต
6 โˆ’ 1 ๐‘˜๐‘”๐‘š๐‘œ๐‘™ ๐‘šโˆ’3 0 โ‰ค ๐‘ก < 0.25

0 ๐‘˜๐‘”๐‘š๐‘œ๐‘™ ๐‘šโˆ’3 ๐‘ก < 0; 0.25 โ‰ค ๐‘ก

๐ด๐‘๐‘ข๐‘™๐‘ ๐‘’ = 1.25 ๐‘˜๐‘”๐‘š๐‘œ๐‘™ ๐‘š๐‘–๐‘›๐‘ข๐‘ก๐‘’๐‘  ๐‘š
โˆ’3

โˆด ๐œ’๐‘–,๐‘–๐‘š๐‘๐‘ข๐‘™๐‘ ๐‘’ = 1.25 ๐›ฟ(๐‘ก)

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.18

CBE 430
Week 03
Reading
โ€ข Ch. 3

Example 3.6(b)

Find:
b) The impulse and pulse responses of the reactant concentration

leaving the reactor, ๐‘. (Letโ€™s find ๐œ’, instead.)

๐‘‰
๐‘‘๐‘

๐‘‘๐‘ก
= ๐‘ž ๐‘๐‘– โˆ’ ๐‘

Write a species mole balance on the tracer:

๐‘‰
๐‘‘ ๐‘ โˆ’ าง๐‘

๐‘‘๐‘ก
= ๐‘ž ๐‘๐‘– โˆ’ เดฅ๐‘๐‘– โˆ’ ๐‘ โˆ’ าง๐‘

Therefore, at steady
state: เดฅ๐‘๐‘– = าง๐‘

Convert this mole balance to a differential
equation for the ๐œ’.

๐‘‘๐œ’

๐‘‘๐‘ก
=
๐‘ž

๐‘‰
๐œ’๐‘– โˆ’ ๐œ’ =

1

๐œ
๐œ’๐‘– โˆ’ ๐œ’

๐œ’ 0 = 0

Now replace the inlet
concentration (perturbation)
with ๐œ’๐‘–,๐‘๐‘ข๐‘™๐‘ ๐‘’ and ๐œ’๐‘–,๐‘–๐‘š๐‘๐‘ข๐‘™๐‘ ๐‘’.

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.19

๐‘‰
๐‘‘ าง๐‘

๐‘‘๐‘ก
= ๐‘ž เดฅ๐‘๐‘– โˆ’ าง๐‘ = 0

CBE 430
Week 03
Reading
โ€ข Ch. 3

Example 3.6(b) continued

Impulse:

๐‘‘๐œ’

๐‘‘๐‘ก
=
1

๐œ
(๐œ’๐‘– โˆ’ ๐œ’)

๐‘‘๐œ’
๐‘‘๐‘ก
=
1

๐œ
๐œ’๐‘–,๐‘–๐‘š๐‘๐‘ข๐‘™๐‘ ๐‘’ โˆ’ ๐œ’

=
1

๐œ
๐‘Ž๐›ฟ ๐‘ก โˆ’ ๐œ’

๐‘ ๐‘‹ ๐‘  โˆ’ ๐œ’ 0 =
1

๐œ
๐‘Ž โˆ’

๐‘‹ ๐‘ 

๐‘‹ ๐‘  = ๐‘Ž
1

๐œ๐‘  + 1

๐œ’ ๐‘ก =
๐‘Ž

๐œ
๐‘’โˆ’ ฮค๐‘ก ๐œ

Pulse:
๐‘‘๐œ’
๐‘‘๐‘ก
=
1

๐œ
๐œ’๐‘–,๐‘๐‘ข๐‘™๐‘ ๐‘’ โˆ’ ๐œ’ ๐‘ ๐‘‹ ๐‘  โˆ’ ๐œ’ 0 =

5

๐œ๐‘ 
1 โˆ’ ๐‘’โˆ’0.25๐‘  โˆ’

๐‘‹ ๐‘ 

๐œ1

๐‘‹ ๐‘  = 5
1

๐‘  ๐œ๐‘  + 1

โˆ’ ๐‘’โˆ’0.25๐‘ 

1
๐‘  ๐œ๐‘  + 1

๐œ’ ๐‘ก = 5 1 โˆ’ ๐‘’โˆ’ ฮค๐‘ก ๐œ โˆ’ 5 1 โˆ’ ๐‘’โˆ’ ๐‘กโˆ’0.25 /๐œ ๐‘†(๐‘ก โˆ’ 0.25)

This is a time lag!

Transform this term,
then change ๐‘ก to ๐‘ก โˆ’
0.25 and multiply by
the step function.๐œ’ ๐‘ก = แ‰

5 1 โˆ’ ๐‘’โˆ’ ฮค๐‘ก ๐œ ; ๐‘ก < 0.25

5 ๐‘’โˆ’ ฮค๐‘กโˆ’0.25 ๐œ โˆ’ ๐‘’โˆ’ ฮค๐‘ก ๐œ ; ๐‘ก โ‰ฅ 0.25

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.20

CBE 430
Week 03
Reading
โ€ข Ch. 3
Example 3.6(b) continued
๐‘‘๐œ’
๐‘‘๐‘ก
=
1
๐œ
(๐œ’๐‘– โˆ’ ๐œ’)
๐œ’ ๐‘ก =
๐‘Ž

๐œ1
๐‘’โˆ’ ฮค๐‘ก ๐œ

๐œ’ ๐‘ก = แ‰
5 1 โˆ’ ๐‘’โˆ’ ฮค๐‘ก ๐œ ; ๐‘ก < 0.25

5 ๐‘’โˆ’ ฮค๐‘กโˆ’0.25 ๐œ โˆ’ ๐‘’โˆ’ ฮค๐‘ก ๐œ ; ๐‘ก โ‰ฅ 0.25
Impulse:
Pulse:

0.6

0.5

0.4

0.3

0.2

0.1

0.0

๏ฃ

1086420

Time, t (min)

impulse
pulse

๐‘Ž = 1.25

๐œ =
๐‘‰

๐‘ž
= 2 ๐‘š๐‘–๐‘›๐‘ข๐‘ก๐‘’๐‘ 

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.21

CBE 430
Week 03
Reading
โ€ข Ch. 3

Laplace transforms using MATLAB

MATLAB can really help with
Laplace transforms!

Use symbolic math toolbox
and commands

laplace

ilaplace

Consider from slide 04.06 โ€ฆ

>>syms t f(t)

>>f =

t*exp(-t)

f =

t*exp(-t)

>> F = laplace(f)

F =

1/(s + 1)^2

>> f = ilaplace(F)

f =

t*exp(-t)
Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.22

CBE 430
Week 03
Reading
โ€ข Ch. 3
Laplace transforms using MATLAB

How about for ODEs?
Consider from slide 03.10:

>>syms t y(t)

>>dydt = diff(y,t)

dydt(t) =

diff(y(t), t)

>>laplace(5*dydt + 4*y==2)

ans =

5*s*laplace(y(t), t, s) – 5*y(0) + 4*laplace(y(t), t, s) == 2/s

Interpret this:

5
๐‘‘๐‘ฆ

๐‘‘๐‘ก
+ 4๐‘ฆ = 2; ๐‘ฆ 0 = 1

Define t and y(t) as symbolic variables. Make
sure to tell MATLAB that y is a function of t.

Next, create a symbolic variable for
the . (You can also define
higher order derivatives with diff.)

Take the Laplace transform of the defining
equation! Remember to use ==.

5๐‘ ๐‘Œ ๐‘  โˆ’ 5๐‘ฆ 0 + 4๐‘Œ ๐‘  =
2

๐‘ 
Compare to slide 03.10

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.23

CBE 430
Week 03
Reading
โ€ข Ch. 3
Laplace transforms using MATLAB

Take care defining constants and variables!

>>syms t y1(t) a

>>Y1 = laplace(y1)

Y1 =

laplace(y(t), t, s)

>>A = laplace(a)

A = 1/s^2

Matlab didnโ€™t know what you wanted the independent variable to be, so it a. This
is the Laplace transform of .

To take the Laplace transform of a constant, a:

>>syms a, t, s

>>laplace(a, t, s)

ans =

a/s

Expected behavior. What will happen for a?

???
What is this ?!!

y1 is a function, a is just another variable.

This tells MATLAB everything it needs to know:
โ€ข Take the Laplace transform of a;
โ€ข Use t as the independent variable;
โ€ข Use s is the Laplace-domain independent variable.
(Otherwise, MATLAB has to guess what you want!)

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.24

CBE 430
Week 03
Reading
โ€ข Ch. 3

Summary
1. Laplace transforms convert time-domain functions to equivalent functions in a Laplace

domain. The independent variable ๐‘  is .

2. This is useful for solving differential equations, because derivatives are converted to
algebraic equations when converting from the time domain to the Laplace domain.

3. Conversion to the Laplace domain decomposes a function helps to find characteristic ,
frequencies, and .

4. Some features of the time-domain solution can be easily identified in the Laplace domain (e.g.
an exponential decay, a sinusoidal oscillation, a time lag). Weโ€™ll get more practice at identifying
these. โ†’With experience you can tell a lot about time-domain dynamics without inverting the
Laplace-domain equations!

5. You can use MATLAB laplace and ilaplace

Topics
โ€ข Laplace
transforms
โ€ข Solving
ODEs
โ€ข Properties
of โ„’[ ]
โ€ข Example
3.6
โ€ข MATLAB

Slide 03.25

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