PROCESS CONTROL

1. Process Control
Course II
Lecture 1
Closed Loop system

2. Outline
2- Definition of closed loop system
1- Open loop Vs closed loop system
3 -Variables in open loop and closed loop system
4- Components of closed loop system
5- Transfer function of closed loop
6- Regulator VS. Servo loop

3. Open loop VS. Closed loop system
In simple words
Open system is the system which is not under control. It means that To will be changed if any
change occurs in Ti or Q or m.
As an example ,Let’s take the CSTH(continuous stirred tank heater)
m
Ti
cp
m
To
cp
M
Q
Open loop system
Variables
Input variables output variables
m
Ti
Q
To
Fig.1 CSTH (open loop system)

4. Transfer function of open loop
Ti(s)
𝑘2
𝜏𝑠 + 1
To(s)
m(s)
+
1
+
-
Q(s)
𝑘1
𝜏𝑠 + 1
𝑘3
𝜏𝑠 + 1
Previously we discussed the transfer function of the CSTH and we found that 𝑇𝑜 = 𝑓 ( 𝑇𝑖, 𝑚, 𝑄)
To =
𝑘1
𝜏𝑠 + 1
𝑇𝑖 +
𝑘2
𝜏𝑠 + 1
𝑄 −
𝑘3
𝜏𝑠 + 1
𝑚
The signal flow Block diagram (SFBD)is shown in Fig.2
Fig.2 SFBD for open system.

5. Closed loop system
In simple words
Closed system is the system which is under control. It means that To will remain constant if any
change occurs in Ti or Q or m.
Components of closed system
1- Process
2- measuring element
3Comparator
4- Controller
5- Final Control Element (Control Valve)
m
cp
Ti
Steam in
m
cp
To
M
Q
Tm
Tsp
E
Comparator
Process
Final control element
Temperature
measuring
element
Controller
Control Valve
P
Fig.3 Closed System.

6. Variables in Closed loop System
Variables
Load variables Manipulating
variable
m
Ti Q
Controlled
variable
To

7. Signal Flow Block Diagram (SFBD) for closed system
Control
valve
Process
Controller
Measuring
element
Load 2
∑
∑
+
+
+
-
Ti
Tsp
To
Comparator
Controlled
variable
Tm
Measured
variable
Set point
Load 1
∑
∑
Σ
m
To
Q
E P
+
+

8. Signal flow block diagram always written in terms of transfer function as below
∑
∑
+
+
+
-
Ti
Tsp
To
Comparator
Controlled
variable
Tm
Measured
variable
Set point
∑
∑
Σ
m
To
Q
E P
+
+
𝐺𝐿1
𝐺𝑐
𝐺𝐿2
𝐺𝑃
𝐺𝑣
𝐺𝑚
𝐺𝑐 Controller Transfer
Function
𝐺𝑣 Valve Transfer Function
𝐺𝑃 Process Transfer Function
𝐺𝑚 Measuring element
Transfer function
𝐺𝐿 Load Transfer Function

9. Transfer function of closed loop
The transfer function of a closed system depends mainly upon Load variables and set point
Type equation here.Referring to the previous example of CSTH, we can construct three transfer functions as follows
𝐺1 𝑠 =
𝑇𝑜(𝑠)
𝑇𝑖(𝑠)
, 𝐺2 𝑠 =
𝑇𝑜(𝑠)
𝑚 (𝑠)
, 𝐺3 𝑠 =
𝑇𝑜(𝑠)
𝑇𝑠𝑝(𝑠)

10. ∑
∑
+
+
+
-
Ti (s)
Tsp (s)
To(s)
Comparator
Controlled
variable
Tm
Measured
variable
Set point
∑
∑
Σ
m(s)
To
Q
E P
+
+
𝐺𝐿1
𝐺𝑐
𝐺𝐿2
𝐺𝑃
𝐺𝑣
𝐺𝑚
𝐺1 𝑠 =
𝑇𝑜(𝑠)
𝑇𝑖(𝑠)
=
𝐺𝐿1
1 + 𝐺𝑐𝐺𝑣 𝐺𝑃𝐺𝑚
𝐺2 𝑠 =
𝑇𝑜(𝑠)
𝑚 (𝑠)
=
𝐺𝐿2
1 + 𝐺𝑐𝐺𝑣 𝐺𝑃𝐺𝑚
𝐺3 𝑠 =
𝑇𝑜(𝑠)
𝑇𝑠𝑝(𝑠)
=
𝐺𝑐𝐺𝑣 𝐺𝑃
1 + 𝐺𝑐𝐺𝑣 𝐺𝑃𝐺𝑚

11. Derivation of Transfer function of a closed loop system
∑
∑
+
+
+
-
Ti (s)
Tsp (s)
To(s)
Comparator
Controlled
variable
Tm
Measured
variable
Set point
∑
∑
Σ
m(s)
To
Q
E P
+
+
𝐺𝐿1
𝐺𝑐
𝐺𝐿2
𝐺𝑃
𝐺𝑣
𝐺𝑚
Z2
Z1
Z3
Z4
We try to derive the transfer function
𝐺1 𝑠 =
𝑇𝑜(𝑠)
𝑇𝑖(𝑠)
Note that the only input variable is Ti
So, m and Tsp are both constants
m=0 and Tsp=0
𝑇𝑜 𝑠 = 𝑍1 + 𝑍4 … … (1)
𝑍1 = 𝑍3 + 𝑍2
Z1= 𝑚 𝑠 . 𝐺𝐿2 + 𝑇𝑖(𝑠)𝐺𝐿1
Z1= 𝑇𝑖 𝑠 𝐺𝐿1 … … … … . . (2)
Z1= 𝑚 𝑠 . 𝐺𝐿2 + 𝑇𝑖 𝑠 𝐺𝐿1 𝑏𝑢𝑡 𝑚 𝑠 = 0 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Subs. Eq.(2) in (1) gives
𝑇𝑜 𝑠 = 𝑇𝑖 𝑠 𝐺𝐿1 + 𝑍4 … … (3)

12. ∑
∑
+
+
+
-
Ti (s)
Tsp (s)
To(s)
Comparator
Controlled
variable
Tm
Measured
variable
Set point
∑
∑
Σ
m(s)
To
Q
E P
+
+
𝐺𝐿1
𝐺𝑐
𝐺𝐿2
𝐺𝑃
𝐺𝑣
𝐺𝑚
Z2
Z1
Z3
Z4
𝑍4 = 𝐺𝑃𝐺𝑣𝑝
𝑍4 = 𝐺𝑃𝑄
𝑍4 = 𝐺𝑃𝐺𝑣𝐺𝑐 E
𝑍4 = 𝐺𝑃𝐺𝑣𝐺𝑐 𝑇𝑠𝑝 − 𝑇𝑚
𝑏𝑢𝑡 𝑇𝑠𝑝 = 0
𝑍4 = − 𝐺𝑃𝐺𝑣𝐺𝑐 𝑇𝑚 … … … . 4
Subs. Eq.(4) in (3) gives
𝑇𝑜 𝑠 = 𝑇𝑖 𝑠 𝐺𝐿1 + 𝑍4 … … (3)
𝑇𝑜 𝑠 = 𝑇𝑖 𝑠 𝐺𝐿1 − 𝐺𝑃 𝐺𝑣𝐺𝑐 𝑇𝑚 … … (5)
𝑇𝑜 𝑠 = 𝑇𝑖 𝑠 𝐺𝐿1 − 𝐺𝑃 𝐺𝑣𝐺𝑐𝐺𝑚𝑇𝑜(𝑠)
𝑇𝑜 𝑠 (1 + 𝐺𝑃 𝐺𝑣𝐺𝑐𝐺𝑚) = 𝑇𝑖 𝑠 𝐺𝐿1
𝑇𝑜 𝑠
𝑇𝑖 𝑠
=
𝐺𝐿1
1 + 𝐺𝑃 𝐺𝑣𝐺𝑐𝐺𝑚
𝐺1 𝑠 =
𝑇𝑜 𝑠
𝑇𝑖 𝑠
=
𝐺𝐿1
1 + 𝐺𝑃 𝐺𝑣𝐺𝑐𝐺𝑚
In the same way, we can derive the other two transfer functions.
𝐺2 𝑠 =
𝑇𝑜(𝑠)
𝑚 (𝑠)
=
𝐺𝐿2
1 + 𝐺𝑐𝐺𝑣 𝐺𝑃𝐺𝑚
𝐺3 𝑠 =
𝑇𝑜(𝑠)
𝑇𝑠𝑝(𝑠)
=
𝐺𝑐𝐺𝑣 𝐺𝑃
1 + 𝐺𝑐𝐺𝑣 𝐺𝑃𝐺𝑚
and

13. Example 1: A closed loop system consists of the following items:
- Process: 𝑦 𝑠 = 𝑓 𝑥 𝑠 , 𝑧 𝑠 𝑤ℎ𝑒𝑟𝑒 𝑥 𝑠 𝑖𝑠 𝑙𝑜𝑎𝑑 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑎𝑛𝑑 𝑧 𝑠 𝑖𝑠 𝑚𝑎𝑛𝑖𝑝𝑢𝑙𝑎𝑡𝑖𝑛𝑔 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒.
𝑦 𝑠 =
2
3𝑠+1
𝑥 𝑠 +
4
10𝑠+1
𝑧(𝑥)
- Controller transfer function: 𝐺𝑐 𝑠 = 0.2
- Valve transfer function: 𝐺𝑣 𝑠 = 2
- Measurement transfer function: 𝐺𝑚 𝑠 =
1
2𝑠+1
Construct the signal flow block diagram (SFBD) for the closed system.
2
3𝑠 + 1
∑
∑
+
+
+
-
ysp (s)
y(s)
ym
Set point
X(s)
y
Z
E P
0.2
4
10𝑠 + 1
2
1
2𝑠 + 1
Solution

14. Regulator VS. Servo loop
Regulator Loop: It is a closed loop in which the load is varied and the set point is constant.
Servo Loop: It is a closed loop in which set point is varied and the load is constant.
𝐺𝑙(𝑠)
∑
∑
+
+
+
-
ysp (s)
y(s)
ym
Set point
X(s)
y
Z
E P
𝐺𝑝(𝑠)
𝐺𝑣(𝑠)
𝐺𝑚(𝑠)
𝐺𝑐(𝑠)
Consider the closed loop shown below
For the regulator loop, the transfer function is
𝐺 𝑠 =
𝑦(𝑠)
𝑥(𝑠)
=
𝐺𝑙(𝑠)
1 + 𝐺𝑐(𝑠)𝐺𝑣(𝑠)𝐺𝑝(𝑠)𝐺𝑚(𝑠)
For the Servo loop, the transfer function is
𝐺 𝑠 =
𝑦(𝑠)
𝑦𝑠𝑝(𝑠)
=
𝐺𝑐(𝑠)𝐺𝑣(𝑠)𝐺𝑝(𝑠)
1 + 𝐺𝑐(𝑠)𝐺𝑣(𝑠)𝐺𝑝(𝑠)𝐺𝑚(𝑠)

15. Example 2
A close loop system in which the output variable 𝜃𝑜 is function of both load variable𝜃𝐿 and
the set point 𝜃𝑠𝑝. i.e. 𝜃𝑜 = 𝑓(𝜃𝐿, 𝜃𝑠𝑝). The system has the following elements with their transfer functions:
Process : 𝐺𝑝 𝑠 =
3
10𝑠+1
Load : 𝐺𝐿 𝑠 =
1
10𝑠+1
Measurement : 𝐺𝑚 𝑠 = 1
Control Valve : 𝐺𝑣 𝑠 = 1.5
Controller : 𝐺𝑐 𝑠 = 2
a. Draw the block diagram for this closed loop.
b. Find the transfer function if the system operates as servo.
c. Find the transfer function if the system operates as regulator.
d. Find the response, if a unit step change occurs in the load and sketch it.

16. 1
10𝑆 + 1
1.5
3
10𝑆 + 1
2
1
∑ ∑
+
+
+
_
𝜃L (s)
𝜃 (s)
𝜃sp (s)
Solution
a. The block diagram is shown below
b. Transfer function for Servo problem
𝐺 𝑠 =
𝜃 (s)
𝜃sp
(s)
=
2(1.5)(
3
10𝑠+1)
1+2(1.5)(
3
10𝑠+1)
(1)
=
0.9
𝑠+1

17. c. Transfer function for Regular problem
𝐺 𝑠 =
𝜃 (s)
𝜃L (s)
=
(
1
10𝑠 + 1
)
1 + 2(1.5)(
3
10𝑠 + 1)
(1)
=
1
10(𝑠 + 1)
=
0.1
(𝑠 + 1)
d. For a unit step change in load of value =1, 𝜃𝐿 𝑠 = 1
𝑠
𝜃𝑜 𝑡 = ℒ−1 0.1
𝑠(𝑠+1)
= 0.1(1 − 𝑒−𝑡)
0
0.1
𝜃𝑜(𝑡)
𝑡

18. Example 3
A closed loop system has the following items;
Process : 𝐺𝑝 𝑠 = 3 𝑒−2𝑠
10𝑠+1
, Load : 𝐺𝐿 𝑠 =
𝑒−2𝑠
10𝑠+1
, Measurement: 𝐺𝑚 𝑠 = 1, Valve: 𝐺𝑣 𝑠 = 1.5 , 𝐺𝑐 𝑠 = 2
The output variable (controlled variable) 𝜃𝑜 is function of load variable 𝜃𝐿 and set point 𝜃𝑠𝑝 , 𝜃𝑜 = 𝑓(𝜃𝐿, 𝜃𝑠𝑝).
a. Draw the signal flow block diagram.
b. Determine the system response for a unit step in load variable
Solution
1.5
2
1
∑ ∑
+
+
+
_
𝜃L (s)
𝜃 (s)
𝜃sp (s)
𝑒−2𝑠
10𝑠 + 1
3𝑒−2𝑠
10𝑠 + 1
a.

19. a. The change occurs in load variable, so the system is regulator problem.
𝜃𝑜 𝑠 =
𝐺𝐿 𝑠
1 + 𝐺𝑐 𝑠 . 𝐺𝑣 𝑠 𝐺𝑝(𝑠)𝐺𝑚(𝑠)
𝜃L 𝑠
𝐺 𝑠 =
𝜃𝑜 𝑠
𝜃L 𝑠
=
𝑒−2𝑠
10𝑠 + 1
1 + 2(1.5)(
3𝑒−2𝑠
10𝑠 + 1
)
1.5
2
1
∑ ∑
+
+
+
_
𝜃L (s)
𝜃 (s)
𝜃sp (s)
𝑒−2𝑠
10𝑠 + 1
3𝑒−2𝑠
10𝑠 + 1
using pade approximation 𝑒−𝜏𝐷 =
1−0.5 𝜏𝐷𝑆
1+0.5𝜏𝐷𝑆
=
1−0.5(2)𝑠
1+0.5(2)𝑠
=
1−𝑠
1+𝑠
𝐺 𝑠 =
𝑒−2𝑠
10𝑠 + 1
1 +
2 1.5 3 𝑒−2𝑠 1
10𝑠 + 1
=
1 − 𝑠
1 + 𝑠)(10𝑠 + 1
1 +
9(
1 − 𝑠
1 + 𝑠
10𝑠 + 1

20. 𝐺 𝑠 = 0.1[
−𝑠 + 1
𝑠2 + 0.2𝑠 + 1
]
𝜃L 𝑠 =
1
𝑆
∴ 𝜃𝑜(𝑠) =
1
10
[
−𝑠 + 1
𝑠(𝑠2 + 0.2𝑠 + 1)
]
𝜃𝑜 𝑡 = ℒ−1
1
𝑠
1
10
−𝑠 + 1
𝑠2 + 0.2𝑠 + 1
= 0.1ℒ−1
−𝑠 + 1
𝑠 𝑠2 + 0.2𝑠 + 1
From Laplace inverse table
∴ 𝜃𝑜 𝑡 = ℒ−1
𝑇1𝑠 + 1
𝑠 𝑇2𝑠2 + 2𝛿𝑇𝑠 + 1
= 1 +
1
𝐴
1 − 2𝑇1𝑐 +
𝑇1
𝑇
2
1
2
𝑒−𝑐𝑡 sin 𝐵𝑡 − ∅
∅ = tan−1
𝑇1𝐵
1 − 𝑇1𝑐
− tan−1
𝐴
−𝛿
𝑇1 = −1, 𝑇 = 1, 𝛿 = 0.1
𝐴 = 1 − 𝛿2 = 0.994, 𝐵 =
1−𝛿2
𝑇
= 0.994 , 𝑐 =
𝛿
𝐴
= 0.1
∴ ∅ = 0.736
‫البالس‬ ‫تحويالت‬ ‫جدول‬ ‫من‬

21. ∴ 𝜃𝑜 𝑡 =
1
10
[1 +
1
0.994
1 − 2(−1)(0.1) +
−1
2
2
1
2
𝑒−(0.1)𝑡 sin 0.994𝑡 − 0.736
= 0.1 1 + 1.49𝑒− 0.1 𝑡 sin 0.994𝑡 − 0.736
𝜃𝑜 𝑡 = 0.1 + 0.149 𝑒−0.1𝑡 sin 0.994𝑡 − 0.736
t
ϴo
0.1
0