Process Control and Instrumentation Homework for Chemical Engineering
KKK8008
Week03Notes.pdf
CBE 430 Week 03
Reading • 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.
Let’s review Laplace transforms!
ℒ 𝑓 𝑡 = න 0
∞
𝑓 𝑡 𝑒−𝑠𝑡𝑑𝑡 = 𝐹 𝑠
ℒ−1 𝐹 𝑠 = 𝑓 𝑡
Laplace Domain
𝑓(𝑡) 𝐹(𝑠)
Time Domain
Slide 03.01
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
CBE 430 Week 03
Reading • Ch. 3
Laplace Transforms Constant function:
Unit step function:
Exponential:
Rectangular Pulse:
ℒ 𝑎 = න 0
∞
𝑎𝑒−𝑠𝑡𝑑𝑡 = − 𝑎
𝑠 𝑒−𝑠𝑡
0
∞
= 𝑎
𝑠
ℒ 𝑓 𝑡 = 1
𝑠
Notice, the Laplace transform doesn’t care what happens at 𝑡 < 0! So the unit step at 𝑡 = 0 is the same as the constant with 𝑎 = 1.
ℒ 𝑓(𝑡) = න 0
∞
𝑒−𝑎𝑡𝑒−𝑠𝑡𝑑𝑡 = න 0
∞
𝑒−(𝑠+𝑎)𝑡𝑑𝑡 = − 𝑒− 𝑠+𝑎 𝑡
𝑠 + 𝑎 0
∞
= 1
𝑠 + 𝑎
𝑓 𝑡 = ቐ 0 𝑡 < 0 ℎ 0 ≤ 𝑡 < 𝑡𝑤 0 𝑡 ≥ 𝑡𝑤
ℒ 𝑓(𝑡) = න 0
𝑡𝑤
ℎ𝑒−𝑠𝑡𝑑𝑡 = − ℎ
𝑠 𝑒−𝑠𝑡 ቚ
0
𝑡𝑤
= ℎ
𝑠 1 − 𝑒−𝑡𝑤𝑠
𝑓 𝑡 = 𝑎
𝑓 𝑡 = 𝑆 𝑡 = ቊ 0 𝑡 < 0 1 𝑡 ≥ 0
𝑓 𝑡 = 𝑒−𝑎𝑡
Slide 03.02
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
CBE 430 Week 03
Reading • 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:
𝑓𝑑 𝑡 = 𝑓 𝑡 − 𝜃 𝑆 𝑡 − 𝜃 𝑆 ensures that 𝑓𝑑 = 0 before the lag time. This can happen when 𝑓 and 𝑓𝑑 measure the same thing (e.g. a temperature or concentration) at different points in a , for example.
What is the Laplace transform of 𝑓𝑑(𝑡)?
Slide 03.03
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
CBE 430 Week 03
Reading • Ch. 3
How About a Time Delay (Lag).
𝑓𝑑 𝑡 = 𝑓 𝑡 − 𝜃 𝑆 𝑡 − 𝜃
ℒ 𝑓𝑑 𝑡 = ℒ 𝑓 𝑡 − 𝜃 𝑆 𝑡 − 𝜃
= න 0
∞
𝑓 𝑡 − 𝜃 𝑆 𝑡 − 𝜃 𝑒−𝑠𝑡𝑑𝑡
= න 0
𝜃
𝑓 𝑡 − 𝜃 0 𝑒−𝑠𝑡𝑑𝑡 + න 𝜃
∞
𝑓 𝑡 − 𝜃 𝑒−𝑠𝑡𝑑𝑡
= න 𝜃
∞
𝑓 𝑡 − 𝜃 𝑒−𝑠 𝑡−𝜃 𝑒−𝑠𝜃 𝑑 𝑡 − 𝜃
= 𝑒−𝑠𝜃න 0
∞
𝑓 𝑡∗ 𝑒−𝑠𝑡 ∗ 𝑑𝑡∗
ℒ[𝑓𝑑 𝑡 ] = 𝑒 −𝑠𝜃𝐹 𝑠
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.04
CBE 430 Week 03
Reading • Ch. 3
Now you try one:
Use the property:
and the Euler formula for the cosine:
To find
ℒ 𝑥 𝑡 + 𝑦 𝑡 = ℒ 𝑥 𝑡 + ℒ 𝑦 𝑡
= 1
2
1
𝑠 − 𝑗𝜔 +
1
𝑠 + 𝑗𝜔
cos 𝜔𝑡 = 𝑒𝑗𝜔𝑡 + 𝑒−𝑗𝜔𝑡
2 ; 𝑗 ≡ −1
ℒ[cos 𝜔𝑡 ] = 1
2 ℒ 𝑒𝑗𝜔𝑡 +
1
2 ℒ 𝑒−𝑗𝜔𝑡
= 1
2
𝑠 + 𝑗𝜔 + 𝑠 − 𝑗𝜔
𝑠2 +𝜔2 =
𝑠
𝑠2 +𝜔2
ℒ[cos 𝜔𝑡 ]
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.05
CBE 430 Week 03
Reading • Ch. 3
What about this one?
𝑢 = 𝑡 𝑑𝑣 = 𝑒− 𝑠+𝑎 𝑡𝑑𝑡
𝑑𝑢 = 𝑑𝑡 𝑣 = − 1
𝑠 + 𝑎 𝑒− 𝑠+𝑎 𝑡
ℒ 𝑡𝑒𝑎𝑡 = − 𝑡𝑒− 𝑠+𝑎 𝑡
𝑠 + 𝑎 𝑡=0
𝑡=∞
+න 0
∞ 𝑒− 𝑠+𝑎 𝑡
𝑠 + 𝑎 𝑑𝑡
= 0 − 0 − 𝑒− 𝑠+𝑎 𝑡
𝑠 + 𝑎 2 𝑡=0
𝑡=∞
= 1
𝑠 + 𝑎 2
ℒ 𝑡𝑒−𝑎𝑡
Integration by parts! න 0
∞
𝑢𝑑𝑣 = 𝑢𝑣 −න 0
∞
𝑣𝑑𝑢
= න 0
∞
𝑡𝑒−𝑎𝑡𝑒−𝑠𝑡𝑑𝑡 = න 0
∞
𝑡𝑒− 𝑎+𝑠 𝑡𝑑𝑡
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.06
CBE 430 Week 03
Reading • Ch. 3
One more important example…
The impulse (with area 𝑎): 𝑓 𝑡 = 𝑎𝛿 𝑡
ℒ 𝑓(𝑡) = lim 𝑡𝑤→0
𝑎
𝑡𝑤𝑠 1 − 𝑒−𝑡𝑤𝑠
Where 𝛿 𝑡 is the function. 𝛿 𝑡 has an area of 1, a height of infinity, and a width of 0. This can be modeled as a pulse of ℎ = 1/𝑡𝑤 as 𝑡𝑤 → 0. Using the result for the pulse…
Apply l’Hôpital’s rule (differentiate by 𝑡𝑤):
ℒ 𝑎𝛿 𝑡 = lim 𝑡𝑤→0
𝑎 1 − 𝑒−𝑡𝑤𝑠
𝑡𝑤𝑠 = lim
𝑡𝑤→0
𝑎𝑠𝑒−𝑡𝑤𝑠
𝑠 = 𝑎
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.07
CBE 430 Week 03
Reading • Ch. 3
Why is this useful?
Finding Laplace transforms is fun and easy, but that’s not why we use them. In fact, we usually don’t bother finding them at all… 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.
…here’s how… →
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.08
CBE 430 Week 03
Reading • Ch. 3
Time-Derivatives What is the Laplace transform of the time-derivative?
How about the second time derivative?
For the 𝑛𝑡ℎ derivative:
ℒ 𝑑𝑓
𝑑𝑡 = න
0
∞𝑑𝑓
𝑑𝑡 𝑒−𝑠𝑡𝑑𝑡 = 𝑓𝑒−𝑠𝑡 ቚ
0
∞ −න
0
∞
𝑓 𝑡 𝑑 𝑒−𝑠𝑡
𝑑𝑡 𝑑𝑡
= −𝑓 0 + න 0
∞
𝑓 𝑡 𝑠𝑒−𝑠𝑡𝑑𝑡
= −𝑓 0 + 𝑠ℒ 𝑓 𝑡
ℒ 𝑑2𝑓
𝑑𝑡2 = ℒ
𝑑𝜙
𝑑𝑡 ; where 𝜙 =
𝑑𝑓
𝑑𝑡
= −𝜙 0 + 𝑠 𝑠𝐹 𝑠 − 𝑓 0
= −𝑓 0 + 𝑠𝐹 𝑠
= 𝑠2𝐹 𝑠 − ቤ 𝑑𝑓
𝑑𝑡 𝑡=0
− 𝑠𝑓 0
This is awesome! Now we can use Laplace transforms to replace first derivatives!
𝑑𝑛 𝑓 𝑡
𝑑𝑡𝑛 = 𝑠𝑛𝐹 𝑠 −
𝑚=0
𝑛−1
𝑠𝑚 อ 𝑑𝑛−1−𝑚𝑓 𝑡
𝑑𝑡𝑛−1−𝑚 𝑡=0
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.09
CBE 430 Week 03
Reading • Ch. 3
A First-Order ODE Example
5 𝑑𝑦
𝑑𝑡 + 4𝑦 = 2; 𝑦 0 = 1
5ℒ 𝑑𝑦
𝑑𝑡 + 4ℒ 𝑦 = ℒ 2
5 𝑠𝑌 𝑠 − 𝑦 0 + 4𝑌 𝑠 = 2
𝑠
5𝑠𝑌 𝑠 − 5 1 + 4𝑌 𝑠 = 2
𝑠
5𝑠 + 4 𝑌 𝑠 = 2
𝑠 + 5
𝑌 𝑠 = 2
𝑠 5𝑠 + 4 +
5
5𝑠 + 4
ℒ[ ]
Solve for 𝑌 𝑠
Now invert using Table 3.1. Notice that entries 5 and 6, and 9 and 10 are equivalent alternative representations.
𝑌 𝑠 = 2
5
1
𝑠 + 0 𝑠 + 4 5
+ 1
𝑠 + 4 5
𝑦 𝑡 = −0.5 𝑒−0.8𝑡 − 1 + 𝑒−0.8𝑡
= 0.5 1 − 𝑒−0.8𝑡
Transform equation
ℒ−1[ ]
Notice the gets used here!
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.10
CBE 430 Week 03
Reading • Ch. 3
A Higher-Order ODE Example
𝑑3𝑦
𝑑𝑡3 + 6
𝑑2𝑦
𝑑𝑡2 + 11
𝑑𝑦
𝑑𝑡 + 6𝑦 = 1;
𝑌 𝑠 = 1
𝑠(𝑠3 + 6𝑠2 + 11𝑠 + 6)
อ 𝑑2𝑦
𝑑𝑡2 0
= 0; ቤ 𝑑𝑦
𝑑𝑡 0
= 0; 𝑦 0 = 0
How many IC’s do we need?
ℒ 𝑑3𝑦
𝑑𝑡3 + 6ℒ
𝑑2𝑦
𝑑𝑡2 + 11ℒ
𝑑𝑦
𝑑𝑡 + 6ℒ 𝑦 = ℒ 1
𝑠3𝑌 𝑠 + 6𝑠2𝑌 𝑠 + 11𝑠𝑌 𝑠 + 6𝑌 𝑠 = 1
𝑠
Find the (4) roots of the (4th-order) polynomial in the denominator, so it can be factored! Roots are 𝑠 = 0, -1, -2, and -3. 𝑌 𝑠 =
1
𝑠(𝑠 + 1)(𝑠 + 2)(𝑠 + 3)
Perform . 𝑌 𝑠 =
𝛼1 𝑠 +
𝛼2 𝑠 + 1
+ 𝛼3
𝑠 + 2 +
𝛼4 𝑠 + 3
𝛼1 = 1
6 ; 𝛼2 = −
1
2 ; 𝛼3 =
1
2 ; 𝛼4 = −
1
6 Perform Heaviside expansion.
𝑦 𝑡 = 1
6 − 1
2 𝑒−𝑡 +
1
2 𝑒−2𝑡 −
1
6 𝑒−3𝑡
When did we use the IC’s?!?!
ℒ[ ]
ℒ−1[ ]
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.11
CBE 430 Week 03
Reading • Ch. 3
What was that Heaviside thing?
𝑌 𝑠 = 𝑁 𝑠
𝐷 𝑠 =
1
𝑠(𝑠 + 1)(𝑠 + 2)(𝑠 + 3) = 𝛼1 𝑠 +
𝛼2 𝑠 + 1
+ 𝛼3
𝑠 + 2 +
𝛼4 𝑠 + 3
Find the coefficients 𝛼𝑖 using .
1. Multiply the whole equations by one of the terms in the denominator (for example, the second one).
2. Choose 𝑠 so that all the remaining 𝛼𝑖 are . In this case, we set 𝑠 = −1. Compute 𝛼2.
𝑠 + 1
𝑠 𝑠 + 1 𝑠 + 2 𝑠 + 3 = 𝛼2 + 𝑠 + 1
𝛼1 𝑠 +
𝛼3 𝑠 + 2
+ 𝛼4
𝑠 + 3
1
−1 −1 + 2 [ −1 + 3] = 𝛼2 =
1
−1 × 1 × 2 = −
1
2
In general… 𝛼𝑖 = ቤ𝑠 + 𝑏𝑖 𝑁 𝑠
𝐷 𝑠 𝑠=−𝑏𝑖
What about repeated roots?...
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.12
CBE 430 Week 03
Reading • Ch. 3
An example with repeated roots 𝑑2𝑦
𝑑𝑡2 + 4
𝑑𝑦
𝑑𝑡 + 4𝑦 = 1; อ
𝑑2𝑦
𝑑𝑡2 0
= 0; ቤ 𝑑𝑦
𝑑𝑡 0
= 1; 𝑦 0 = 0
𝑠2𝑌 𝑠 − 𝑠𝑦 0 − ቤ 𝑑𝑦
𝑑𝑡 0
+ 4 𝑠𝑌 𝑠 − 𝑦 0 + 4𝑌 𝑠 = 1
𝑠
𝑠2𝑌 𝑠 − 1 + 4𝑠𝑌 𝑠 + 4𝑌 𝑠 = 1
𝑠
𝑠2 + 4𝑠 + 4 𝑌 𝑠 = 1
𝑠 + 1 =
𝑠 + 1
𝑠
𝑌 𝑠 = 𝑠 + 1
𝑠 𝑠2 + 4𝑠 + 4
Use the Heaviside method to find 𝛼1 and 𝛼3
𝑌 𝑠 = 𝑠 + 1
𝑠 𝑠2 + 4𝑠 + 4 = 𝛼1 𝑠 +
𝛼2 𝑠 + 2
+ 𝛼3
𝑠 + 2 2
Can’t use the Heaviside method for 𝛼2 because multiplication by 𝑠 + 2 , and setting doesn’t eliminate the 𝛼3 term; it blows up.
𝛼1 = 0 + 1
02 + 4 0 + 4 = 1
4 ; 𝛼3 =
−2 + 1
−2 = 1
2
Then just choose another 𝑠 to find 𝛼2; choose 𝑠 = −1
𝛼2 = 𝑠 + 1
𝑠(𝑠2 + 4𝑠 + 4) −
1
4𝑠 −
1
2 𝑠 + 2 2 ; 𝛼2 =
1
4 − 1
2 = −
1
4
The denominator is a 3rd-order polynomial, so there must be 3 constants.
IC’s used here!
ℒ[ ]
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.13
CBE 430 Week 03
Reading • Ch. 3
Or, you can take the derivative 𝑌 𝑠 =
𝑠 + 1
𝑠 𝑠2 + 4𝑠 + 4 = 𝛼1 𝑠 +
𝛼2 𝑠 + 2
+ 𝛼3
𝑠 + 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 𝛼𝑖. The general formula that you can use in this case is:
𝛼𝑖 = 1
𝑛 − 𝑖 ! lim 𝑠→−𝑏
𝑑𝑛−𝑖
𝑑𝑠𝑛−𝑖 𝑠 + 𝑏 𝑛
𝑁 𝑠
𝐷 𝑠
𝛼3 = ቤ𝑠 + 2 2 𝑁 𝑠
𝐷 𝑠 𝑠=−2
; 𝛼2 = อ 𝑑
𝑑𝑠 𝑠 + 2 2
𝑁 𝑠
𝐷 𝑠 s=−2
; 𝛼1 = ቤ𝑠 𝑁 𝑠
𝐷 𝑠 𝑠=0
𝛼3 = 1
2 ; 𝛼2 = ቤ
𝑑
𝑑𝑠 1 +
1
𝑠 s=−2
= − 1
−2 2 = −
1
4 ; 𝛼1 =
1
4
𝑌 𝑠 = 𝛼1
(𝑠 + 𝑏) +
𝛼2 𝑠 + 𝑏 2
+⋯ 𝛼𝑛
𝑠 + 𝑏 𝑛 +⋯ =
𝑖
𝑛 𝛼𝑖
𝑠 + 𝑏 𝑖 +⋯
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.14
CBE 430 Week 03
Reading • Ch. 3
What about complex roots?
Heaviside expansion will
give complex 𝛼1,2 = − 1
3 𝑗
𝑌 𝑠 = 1
𝑠2 + 𝑠 + 2.5 =
𝛼1 𝑠 + 𝑏1
+ 𝛼2
𝑠 + 𝑏2
𝑌 𝑠 = 𝑁 𝑠
𝐷 𝑠 ; 𝐷 𝑠 = 𝑎𝑠2 + 𝑏𝑠 + 𝑐; 𝑏2 < 4𝑎𝑐 → 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑟𝑜𝑜𝑡𝑠
𝑏1, 𝑏2 = −1 ± 12 − 4 1 2.5
2 1 = −
1
2 ±
−9
2 = −
1
2 ± 3
2 𝑗
Find 𝑏1 and 𝑏2 from the roots of 𝐷 𝑠
Instead, lets put 𝑌(𝑠) in the form of one of the entries in the Table 3.1 on p. 42. – How about number 17?
𝑌 𝑠 = 𝑐 𝜔
𝑠 + 𝑏 2 + 𝜔2 =
1
𝑠2 + 𝑠 + 2.5
This term has to give both
𝑠2 and 𝑠. So 𝑏 = 1
2 .
Now the denominator is:
𝐷 𝑠 = 𝑠 + 1
2
2
+ 𝜔2 = 𝑠2 + s + 1
4 + 𝜔2 = 𝑠2 + s + 2.5
𝜔2 = 2.25; 𝜔 = 1.5
𝜔𝑐 = 1; 𝑐 = 2
3
𝑌 𝑠 = 2
3
1.5
𝑠 + 1 2
2
+ 1.52
𝑦 𝑡 = 2
3 𝑒−0.5𝑡sin(1.5𝑡)
Find 𝑐:
Complex roots will always give in the time domain.
Find 𝜔: ℒ−1[ ]
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.15
CBE 430 Week 03
Reading • 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 𝐹 𝑠 has positive real denominator roots (𝑏𝑖 < 0), 𝑓(𝑡) will grow exponentially.
7. When 𝐹(𝑠) has complex denominator roots (𝑏𝑖 2< 0), 𝑓 𝑡 will oscillate .
8. When 𝐹(𝑠) has an exponential in 𝑠, 𝑓(𝑡) has a .
ℒ 𝑥 𝑡 + 𝑦 𝑡 = ℒ 𝑥 𝑡 + ℒ 𝑦 𝑡
lim 𝑡→∞
𝑦(𝑡) = lim 𝑠→0
[𝑠𝑌 𝑠 ]
𝐹 𝑠 = 𝐺 𝑠 𝐻 𝑠
𝑓 𝑡 = ℒ−1 𝐺 𝑠 𝐻 𝑠 = න 0
𝑡
𝑔 𝜏 ℎ 𝑡 − 𝜏 𝑑𝜏 = න 0
𝑡
𝑔 𝑡 − 𝜏 ℎ 𝜏 𝑑𝜏
lim 𝑡→0
𝑦(𝑡) = lim 𝑠→∞
[𝑠𝑌 𝑠 ]
ℒ 𝑓 𝑡 − 𝑡0 𝑆 𝑡 − 𝑡0 = 𝑒 −𝑡0𝑠𝐹(𝑠)
Great for checking answers!
We’ll make use of these a lot.
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.16
CBE 430 Week 03
Reading • Ch. 3
Let’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, 𝑐2(𝑡), 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, 𝑐.
𝑉 = 4 𝑚3
𝑞 = 2 𝑚3 𝑚𝑖𝑛𝑢𝑡𝑒−1
𝑐𝑖 = 1 𝑘𝑔𝑚𝑜𝑙 𝑚 −3
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.17
CBE 430 Week 03
Reading • 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.
𝑎𝛿 𝑡 ; 𝐴𝑖𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑎
Define the 𝜒𝑖 = 𝑐𝑖 − ഥ𝑐𝑖 and 𝜒 = 𝑐 − ҧ𝑐 Where ഥ𝑐𝑖 and ҧ𝑐 are the respective values.
𝜒𝑖,𝑝𝑢𝑙𝑠𝑒 = ൝ 6 − 1 𝑘𝑔𝑚𝑜𝑙 𝑚−3 0 ≤ 𝑡 < 0.25
0 𝑘𝑔𝑚𝑜𝑙 𝑚−3 𝑡 < 0; 0.25 ≤ 𝑡
𝐴𝑝𝑢𝑙𝑠𝑒 = 1.25 𝑘𝑔𝑚𝑜𝑙 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 𝑚 −3
∴ 𝜒𝑖,𝑖𝑚𝑝𝑢𝑙𝑠𝑒 = 1.25 𝛿(𝑡)
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.18
CBE 430 Week 03
Reading • Ch. 3
Example 3.6(b)
Find: b) The impulse and pulse responses of the reactant concentration
leaving the reactor, 𝑐. (Let’s find 𝜒, instead.)
𝑉 𝑑𝑐
𝑑𝑡 = 𝑞 𝑐𝑖 − 𝑐
Write a species mole balance on the tracer:
𝑉 𝑑 𝑐 − ҧ𝑐
𝑑𝑡 = 𝑞 𝑐𝑖 − ഥ𝑐𝑖 − 𝑐 − ҧ𝑐
Therefore, at steady state: ഥ𝑐𝑖 = ҧ𝑐
Convert this mole balance to a differential equation for the 𝜒.
𝑑𝜒
𝑑𝑡 = 𝑞
𝑉 𝜒𝑖 − 𝜒 =
1
𝜏 𝜒𝑖 − 𝜒
𝜒 0 = 0
Now replace the inlet concentration (perturbation) with 𝜒𝑖,𝑝𝑢𝑙𝑠𝑒 and 𝜒𝑖,𝑖𝑚𝑝𝑢𝑙𝑠𝑒.
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.19
𝑉 𝑑 ҧ𝑐
𝑑𝑡 = 𝑞 ഥ𝑐𝑖 − ҧ𝑐 = 0
CBE 430 Week 03
Reading • Ch. 3
Example 3.6(b) continued
Impulse:
𝑑𝜒
𝑑𝑡 = 1
𝜏 (𝜒𝑖 − 𝜒)
𝑑𝜒
𝑑𝑡 = 1
𝜏 𝜒𝑖,𝑖𝑚𝑝𝑢𝑙𝑠𝑒 − 𝜒
= 1
𝜏 𝑎𝛿 𝑡 − 𝜒
𝑠𝑋 𝑠 − 𝜒 0 = 1
𝜏 𝑎 − 𝑋 𝑠
𝑋 𝑠 = 𝑎 1
𝜏𝑠 + 1 𝜒 𝑡 = 𝑎
𝜏 𝑒− Τ𝑡 𝜏
Pulse:
𝑑𝜒
𝑑𝑡 = 1
𝜏 𝜒𝑖,𝑝𝑢𝑙𝑠𝑒 − 𝜒 𝑠𝑋 𝑠 − 𝜒 0 =
5
𝜏𝑠 1 − 𝑒−0.25𝑠 −
𝑋 𝑠
𝜏1
𝑋 𝑠 = 5 1
𝑠 𝜏𝑠 + 1 − 𝑒−0.25𝑠
1
𝑠 𝜏𝑠 + 1
𝜒 𝑡 = 5 1 − 𝑒− Τ𝑡 𝜏 − 5 1 − 𝑒− 𝑡−0.25 /𝜏 𝑆(𝑡 − 0.25)
This is a time lag!
Transform this term, then change 𝑡 to 𝑡 − 0.25 and multiply by the step function.𝜒 𝑡 = ቐ
5 1 − 𝑒− Τ𝑡 𝜏 ; 𝑡 < 0.25
5 𝑒− Τ𝑡−0.25 𝜏 − 𝑒− Τ𝑡 𝜏 ; 𝑡 ≥ 0.25
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.20
CBE 430 Week 03
Reading • Ch. 3
Example 3.6(b) continued
𝑑𝜒
𝑑𝑡 = 1
𝜏 (𝜒𝑖 − 𝜒)
𝜒 𝑡 = 𝑎
𝜏1 𝑒− Τ𝑡 𝜏
𝜒 𝑡 = ቐ 5 1 − 𝑒− Τ𝑡 𝜏 ; 𝑡 < 0.25
5 𝑒− Τ𝑡−0.25 𝜏 − 𝑒− Τ𝑡 𝜏 ; 𝑡 ≥ 0.25
Impulse:
Pulse:
0.6
0.5
0.4
0.3
0.2
0.1
0.0
1086420
Time, t (min)
impulse pulse
𝑎 = 1.25
𝜏 = 𝑉
𝑞 = 2 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.21
CBE 430 Week 03
Reading • 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 …
>>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 • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.22
CBE 430 Week 03
Reading • 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 𝑑𝑦
𝑑𝑡 + 4𝑦 = 2; 𝑦 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 ==.
5𝑠𝑌 𝑠 − 5𝑦 0 + 4𝑌 𝑠 = 2
𝑠 Compare to slide 03.10
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.23
CBE 430 Week 03
Reading • 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 didn’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: • Take the Laplace transform of a; • Use t as the independent variable; • Use s is the Laplace-domain independent variable. (Otherwise, MATLAB has to guess what you want!)
Topics • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.24
CBE 430 Week 03
Reading • Ch. 3
Summary 1. Laplace transforms convert time-domain functions to equivalent functions in a Laplace
domain. The independent variable 𝑠 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). We’ll get more practice at identifying these. →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 • Laplace
transforms • Solving
ODEs • Properties
of ℒ[ ] • Example
3.6 • MATLAB
Slide 03.25
