Course: CHE417 - Chemical Engineering Laboratory Exercise No.: 6
Process Dynamics and Control
Group No.: 9 and 10
Section: CHE41S3
Group Members:
Date Performed: October 20, 2022
1. Rivamonte, Robert Z.
2. Rivera, John Mitch M.
3. Rosqueta, Noriel James
4. .Sabater, Stephanie
5.
Salonga, Lianne Alyssa B.
6.
Salvador, Nehemiah Gabriel B.
7. Samson, Karen P.
8. Serito, Klarence Nicole T.
9.
Trigal, Reñelle Maxine Z.
10. Vasquez, Chanelo Paol M.
11. Vibar, Rizza Anne M.
Date Submitted: October 21, 2022
Professor: Engr. Crispulo G. Maranan
12. Villota, Hazel Grace M.
Poles and Zeros of a Transfer Function
P1
Results
Single real zero
1
2
−
Pair of real poles
− 3, − 2
Gain constant K
1
2
P2
Results
Transfer function
Differential equation
3(
2
𝑑𝑦
2
𝑑𝑡
+ 2
𝑑𝑦
𝑑𝑡
𝑠+4
2
𝑠 +2𝑠+5
)
+ 5𝑦 = 3
P3
Results
One real pole
Complex conjugate pole pair
− 1
𝑝1, 𝑝2 =− 1 ± 𝑗2
Single real zero
0 − 𝑗2
Gain constant K
3
𝑑𝑢
𝑑𝑡
+ 12𝑢
Transfer function
3𝑠+6
3
(𝑠−(−2))
= 3( (𝑠−(−1))(1−(−1−2𝑠))(𝑠−1) )
2
𝑠 +3𝑠' +7𝑠+5
Differential equation
3
𝑑𝑦
3
𝑑𝑡
2
+ 3
𝑑𝑦
2
𝑑𝑡
+ 7
𝑑𝑦
𝑑𝑡
5𝑦 = 3
𝑑𝑢
𝑑𝑡
+ 6𝑢
P4
Results
Real pole/s
𝑝 =− 0. 3536 ± 0. 3536𝑖
Complex conjugate pole pair
𝑝 =− 0. 3536 ± 0. 3536𝑖
Real zero/s
𝑧 = 0, − 1. 5000
Gain constant K
P5
𝑘=2
Pole-Zero Map
Conclusion
We learned that the values of a system's poles and zeros affect whether or not the system is stable
and how well it functions. We concluded that in order to find the poles and zeros of a chemical process,
one must first understand the process' ordinary differential equation and transfer function, where the
numerator of the transfer function represents the process's zeros and the denominator represents the
process's poles. After all, MATLAB is useful for determining the transfer function and pole map of an
ordinary differential equation.
F1
F2
F3
F4
F5
F6
F7
F8
F9
F10
F11
F12
F13
END of Laboratory Exercise No. 3