PROCESS CONTROL

1. Process Control
Course II
Lecture 5
The Controllers
Part II
1
By
Prof. Alaa Kareem Mohammed

2. 2
Example 1
Consider the following closed system:
0
𝑋𝐿(s)
𝑄(s) 1
2𝑠 + 1
3
2𝑠 + 1
𝑦(s)
+
+
0
𝑦𝑠𝑝(s)
−
+
0.5 2
3
a. For a unit step change in XL , find the time constant 𝜏 of the closed loop, the final
value of response 𝑦(∞) and the offset. Sketch the response.
b. Repeat (a) for value of Kc = 1 and Kc = 2.
c. Conclude the effect of the controller gain Kc on the time constant 𝜏 and the offset.
d. Repeat (a and b) for unit step change in set point.(Homework)
‫البيتي‬ ‫الواجب‬ ‫حل‬

3. 3
c. For the three values of controller gain Kc, the values of 𝜏 and offset are given in Table 2
𝐊𝐜 𝛕 𝐨𝐟𝐟𝐬𝐞𝐭
0.5 0.5 0.75
1 0.285 0.428
2 0.15 0.28
Table 2
From Table 2, we can conclude that:
∴ 𝝉 ∝
𝟏
𝑲𝒄
[𝐼𝑡 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑒𝑓𝑓𝑒𝑐𝑡 𝑡𝑜 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝐾𝑐 for faster response]
𝒐𝒇𝒇𝒔𝒆𝒕 ∝
𝟏
𝑲𝒄
[𝐼𝑡 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑒𝑓𝑓𝑒𝑐𝑡 𝑡𝑜 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝐾𝑐 for lower value of offset]

4. 4
𝑡
𝐾𝑐 = 0.5
𝑦(𝑡)
0
𝐾𝑐 = 1
𝐾𝑐 = 2
𝑜𝑓𝑓𝑠𝑒𝑡1
𝑜𝑓𝑓𝑠𝑒𝑡2
𝑜𝑓𝑓𝑠𝑒𝑡3
0.75
0.428
0.28

5. 5
d. Unit step change in set point 𝑦𝑠𝑝that means the loop is Servo.
Kc= 0.5
𝐺 𝑠 =
𝑦 𝑠
𝑦𝑠𝑝 𝑠
=
0.5 ∗ 2 ∗
1
2𝑠 + 1
1 + 0.5 ∗ 2 ∗
1
2𝑠 + 1
∗ 3
=
0.25
0.5𝑠 + 1
𝐓𝐢𝐦𝐞 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 τ = 0.5
𝑦𝑠𝑝 𝑠 =
1
𝑠
𝑦(𝑡) = ℒ−1
𝑦𝑠𝑝 𝑠 ∙ 𝐺 𝑠
𝑦 𝑡 = ℒ−1
1
𝑠
∗
0.25
0.5𝑠 + 1
𝑦 𝑡 = 0.25(1 − 𝑒−2𝑡
)
Final value of the response:
𝑦 ∞ = lim
𝑠→0
𝑠 ∗ 𝑦(𝑠) = lim
𝑠→0
𝑠 ∗ [
1
𝑠
∗
0.25
0.5𝑠 + 1
]
∴ 𝑦 ∞ = 0.25
0
𝑋𝐿(s)
𝑄(s) 1
2𝑠 + 1
3
2𝑠 + 1
𝑦(s)
+
+
0
𝑦𝑠𝑝(s)
−
+
0.5 2
3

6. 6
Offset = ∆𝑦𝑠𝑝 − 𝑦 ∞ = 1 − 0.25 = 0.75
Offset= 0.75
∴ 𝑭𝒐𝒓 𝒌𝒄 = 𝟎. 𝟓 , 𝝉 = 𝟎. 𝟓 , 𝒐𝒇𝒇𝒔𝒆𝒕 = 𝟎. 𝟕𝟓
0.25
𝑡
Offset =0.75
∆ 𝑠𝑒𝑡 𝑝𝑜𝑖𝑛𝑡 1
0
F𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
This is the offset
in Servo loop

7. 7
For 𝑲𝒄 = 𝟏
𝐺 𝑠 =
𝑦 𝑠
𝑋𝐿 𝑠
=
1 ∗ 2 ∗
1
2𝑠 + 1
1 + 1 ∗ 2 ∗
1
2𝑠 + 1
∗ 3
=
0.285
0.285𝑠 + 1
𝐓𝐢𝐦𝐞 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 τ = 0.285
𝑦𝑠𝑝 𝑠 =
1
𝑠
𝑦(𝑡) = ℒ−1
𝑦𝑠𝑝 𝑠 ∙ 𝐺 𝑠
𝑦 𝑡 = ℒ−1
1
𝑠
∗
0.285
0.285𝑠 + 1
𝑦 𝑡 = 0.285(1 − 𝑒−𝑡/0.285
)
Final value of the response:
𝑦 ∞ = lim
𝑠→0
𝑠 ∗ 𝑦 𝑠 = lim
𝑠→0
𝑠 ∗ [
1
𝑠
∗
0.285
0.285𝑠 + 1
]
𝑦 ∞ = 0.285
Offset= ∆𝑦𝑠𝑝 − 𝑦 ∞ = 1 − 0.285 = 0.715
∴ 𝑜𝑓𝑓𝑠𝑒𝑡 = 0.715
∴ 𝑭𝒐𝒓 𝑲𝒄 = 𝟏 , 𝝉 = 𝟎. 𝟐𝟖𝟓 , 𝒐𝒇𝒇𝒔𝒆𝒕 = 𝟎. 𝟕𝟏𝟓
0.285
𝑡
Offset = 0.715
∆ 𝑠𝑒𝑡 𝑝𝑜𝑖𝑛𝑡 1
0
F𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒

8. 8
For 𝒌𝒄 = 𝟐
𝐺 𝑠 =
𝑦 𝑠
𝑋𝐿 𝑠
=
2 ∗ 2 ∗
1
2𝑠 + 1
1 + 2 ∗ 2 ∗
1
2𝑠 + 1
∗ 3
=
0.3
0.15𝑠 + 1
𝐓𝐢𝐦𝐞 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 τ = 0.15
𝑦𝑠𝑝 𝑠 =
1
𝑠
𝑦(𝑡) = ℒ−1 𝑦𝑠𝑝 𝑠 ∙ 𝐺 𝑠
𝑦 𝑡 = ℒ−1
1
𝑠
∗
0.3
0.15𝑠 + 1
𝑦 𝑡 = 0.3 (1 − 𝑒−𝑡/0.15)
Final value of the response:
𝑦 ∞ = lim
𝑠→0
𝑠 ∗ 𝑦 𝑠 = lim
𝑠→0
𝑠 ∗ [
1
𝑠
∗
0.3
0.15𝑠 + 1
]
𝑦 ∞ = 0.3
Offset= ∆𝑦𝑠𝑝 − 𝑦 ∞ = 1 − 0.3 = 0.7
∴ 𝑜𝑓𝑓𝑠𝑒𝑡 = 0.7
∴ 𝑭𝒐𝒓 𝑲𝒄 = 𝟐 , 𝝉 = 𝟎. 𝟏𝟓 , 𝒐𝒇𝒇𝒔𝒆𝒕 = 𝟎. 𝟕
0.15
𝑡
Offset = 0.7
∆ 𝑠𝑒𝑡 𝑝𝑜𝑖𝑛𝑡 1
0
𝑓𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒

9. 9
K𝒄 𝝉 𝒐𝒇𝒇𝒔𝒆𝒕
0.5 0.5 0.75
1 0.285 0.715
2 0.15 0.7
Also we notice that:
𝜏 ∝
1
𝐾𝑐
𝑂𝑓𝑓𝑠𝑒𝑡 ∝
1
𝐾𝑐

10. 10
𝑀
m, cp, T
TT
m, cp, 𝑇𝑖
𝑄
Tc
𝑇𝑠𝑝
Example 2
Consider the controlled heating tank that shown in Figure below. Given the following information:
Proportional controller 𝐺𝑐 𝑠 = 0.5
Measurement 𝐺𝑚 𝑠 = 3
Control valve 𝐺𝑣 𝑠 = 2
𝑚𝑐𝑝 = 10 𝑘𝑔/𝑘
𝑀𝑐𝑝 = 20 𝑘𝑔/𝑘

11. 11
a. Find the final value of response (T) and the offset of the system for a unit step change in 𝑇𝑖.
b. Repeat (a) for values of 𝐾𝑐 = 1, 𝐾𝑐 = 2.
c. What is the relation between 𝐾𝑐 and offset and time constant of the system.
d. Find the final value of response and the offset of the system for a unit step change in
Set point 𝑇𝑠𝑝. (Homework) [ Ans.: 𝑻 ∞ = 𝟎. 𝟎𝟕𝟔𝟗, 𝒐𝒇𝒇𝒔𝒆𝒕 = 𝟎. 𝟗𝟐𝟑]
e. Repeat (d) for values of 𝐾𝑐 = 1, 𝐾𝑐 = 2. (Homework)[Ans.: for kc=1, 𝑻 ∞ = 𝟎. 𝟏𝟐𝟓, 𝒐𝒇𝒇𝒔𝒆𝒕 = 𝟎. 𝟖𝟕𝟓
Solution
𝑇 𝑠 = 𝑓 𝑇𝑖 𝑠 , 𝑄 𝑠
Heat balance
𝑚 𝑐𝑝 𝑇𝑖 + 𝑄 = 𝑚 𝑐𝑝 𝑇 + 𝑀 𝑐𝑝
𝑑𝑇
𝑑𝑡
𝑀𝑐𝑝
𝑚 𝑐𝑝
𝑑𝑇
𝑑𝑡
+ 𝑇 =
1
𝑚𝑐𝑝
𝑄 + 𝑇𝑖
for kc=2, 𝑻 ∞ = 𝟎. 𝟏𝟖𝟏𝟖, 𝒐𝒇𝒇𝒔𝒆𝒕 = 𝟎. 𝟖𝟏𝟖𝟐

12. 12
𝑀𝑐𝑝
𝑚 𝑐𝑝
𝑑𝑇
𝑑𝑡
+ 𝑇 =
1
𝑚𝑐𝑝
𝑄 + 𝑇𝑖
𝜏
𝑑𝑇
𝑑𝑡
+ 𝑇 =
1
𝑚
𝑄 + 𝑇𝑖
𝜏𝑠 + 1 𝑇 𝑠 = 𝑘1𝑄 𝑠 + 𝑇𝑖 𝑠
𝑇 𝑠 =
𝑘1
𝜏𝑠 + 1
𝑄 𝑠 +
1
𝜏𝑠 + 1
𝑇𝑖 𝑠
Where 𝜏 =
20
10
= 2 , 𝑘1 =
1
10
= 0.1
∴ 𝑇 𝑠 =
0.1
2𝑠 + 1
𝑄 𝑠 +
1
2𝑠 + 1
𝑇𝑖 𝑠
𝐺𝑃
𝐺𝐿

13. 13
The signal flow block diagram (SFBD) of the closed system is given in Figure below
0
𝑇𝑖(s)
0.1
2𝑠 + 1
1
2𝑠 + 1
𝑇(s)
+
+
𝟎
𝑇𝑠𝑝(s)
−
+
0.5
3
2
a. Unit step change in 𝑇𝑖that means the loop is regulator.
𝐾𝑐 = 0.5
𝑇 𝑠
𝑇𝑖 𝑠
=
1
2𝑠 + 1
1 + 0.5 ∗ 2 ∗
0.1
2𝑠 + 1
∗ 3
=
0.769
1.538𝑠 + 1

14. 14
𝑇𝑖 𝑠 =
1
𝑠
𝑇(𝑡) = ℒ−1
𝑇𝑖 𝑠 ∙ 𝐺 𝑠
𝑇 𝑡 = ℒ−1
1
𝑠
∗
0.769
1.538𝑠 + 1
𝑇 𝑡 = 0.769 (1 − 𝑒−𝑡/1.538)
𝑂𝑓𝑓𝑠𝑒𝑡 = 𝑇 ∞ = lim
𝑠 →0
𝑠 ∗ 𝑇 𝑠 = lim
𝑠 →0
𝑠 ∗ [
1
𝑠
∗
0.769
1.538𝑠 + 1
]
∴ 𝑜𝑓𝑓𝑠𝑒𝑡 = 0.769
𝐹𝑜𝑟 𝐾𝑐 = 0.5 , 𝜏 = 1.538 , 𝑜𝑓𝑓𝑠𝑒𝑡 = 0.769
𝑇(𝑡)
𝑡
offset
0.769
0

15. 15
𝑭𝒐𝒓 𝑲𝒄 = 𝟏
𝑇 𝑠
𝑇𝑖 𝑠
=
1
2𝑠 + 1
1 + 1 ∗ 2 ∗
0.1
2𝑠 + 1
∗ 3
=
0.625
1.25𝑠 + 1
𝑇𝑖 𝑠 =
1
𝑠
𝑇(𝑡) = ℒ−1
𝑇𝑖 𝑠 ∙ 𝐺 𝑠
𝑇 𝑡 = ℒ−1
1
𝑠
∗
0.625
1.25𝑠 + 1
𝑇 𝑡 = 0.625 (1 − 𝑒−𝑡/1.25
)
𝑂𝑓𝑓𝑠𝑒𝑡 = 𝑇 ∞ = lim
𝑠 →0
𝑠 ∗ 𝑇 𝑠 = lim
𝑠 →0
𝑠 ∗ [
1
𝑠
∗
0.625
1.25𝑠 + 1
]
𝑂𝑓𝑓𝑠𝑒𝑡 = 0.625
∴ 𝑭𝒐𝒓 𝑲𝒄 = 𝟏 , 𝝉 = 𝟏. 𝟐𝟓 , 𝒐𝒇𝒇𝒔𝒆𝒕 = 𝟎. 𝟔𝟐𝟓
𝑇(𝑡)
𝑡
offset
0.625
0

16. 16
𝑭𝒐𝒓 𝑲𝒄 = 𝟐
𝑇 𝑠
𝑇𝑖 𝑠
=
1
2𝑠 + 1
1 + 2 ∗ 2 ∗
0.1
2𝑠 + 1
∗ 3
=
0.454
0.91𝑠 + 1
𝑇𝑖 𝑠 =
1
𝑠
𝑇(𝑡) = ℒ−1 𝑇𝑖 𝑠 ∙ 𝐺 𝑠
𝑇 𝑡 = ℒ−1
1
𝑠
∗
0.454
0.91𝑠 + 1
𝑇 𝑡 = 0.454 (1 − 𝑒−𝑡/0.91)
𝑂𝑓𝑓𝑠𝑒𝑡 = 𝑇 ∞ = lim
𝑠 →0
𝑠 ∗ 𝑇 𝑠 = lim
𝑠 →0
𝑠 ∗ [
1
𝑠
∗
0.454
0.91𝑠 + 1
]
𝑂𝑓𝑓𝑠𝑒𝑡 = 0.454
∴ 𝑭𝒐𝒓 𝑲𝒄 = 𝟐 , 𝝉 = 𝟎. 𝟗𝟏 , 𝒐𝒇𝒇𝒔𝒆𝒕 = 𝟎. 𝟒𝟓𝟒
𝑇(𝑡)
𝑡
offset
0.454
0

17. 17
c. Now we can conclude the following
𝒌𝒄 𝝉 𝑶𝒇𝒇𝒔𝒆𝒕
0.5 1.538 0.769
1 1.25 0.625
2 0.91 0.454
We notice that:
 Increasing 𝐾𝑐will decrease the time constant of the closed
system (positive effect).
 Increasing 𝐾𝑐will decrease the value of offset (positive
effect).
∴ 𝜏 ∝
1
𝐾𝑐
𝑎𝑛𝑑 𝑂𝑓𝑓𝑠𝑒𝑡 ∝
1
𝐾𝑐
𝑡
0.769
0.454
0.625
𝑇
0
𝐾𝑐 = 𝑜. 5
𝐾𝑐 = 1
𝐾𝑐 = 2

18. 18
𝑄𝑜
LT
𝑄𝑖
𝐻𝑠𝑝
LC
𝐻
Consider the liquid-level tank system shown in Figure below. Given the following data
Proportional controller 𝐺𝑐 𝑠 = 4
Measurement 𝐺𝑚 𝑠 = 2
Control valve 𝐺𝑣 𝑠 = 1
𝐴 = 1 𝑚2
Example 3
a. Draw the signal flow block diagram SFBD.
b. Find the transfer function for regulator loop.
c. For unit step change in 𝑄𝑜, calculate the response and the
offset, and sketch it.

19. 19
Solution
Material balance
In = out +Accumulation
𝑄𝑖 = 𝑄0 + 𝐴
𝑑𝐻
𝑑𝑡
𝐴
𝑑𝐻
𝑑𝑡
= 𝑄𝑖 − 𝑄0
1 𝑠 𝐻 𝑠 = 𝑄𝑖 𝑠 − 𝑄𝑜 𝑠
𝐻 𝑠 =
1
𝑠
𝑄𝑖 𝑠 −
1
𝑠
𝑄𝑜 𝑠
b. Regular loop
𝐻 𝑠
𝑄𝑜 𝑠
=
−
1
𝑠
1 +
4 ∗ 2
𝑠
=
−
1
𝑠
𝑠 + 8
𝑠
=
−0 ∙ 125
0 ∙ 125 𝑠 + 1
0
𝑄𝑜(s)
1
𝑆
−
1
𝑆
𝐻(s)
+
0
𝐻𝑠𝑝(s)
−
+
4
2
1
+

20. 20
c. 𝑄𝑖 𝑠 =
1
𝑠
𝐻 𝑠 = 𝑄𝑖 𝑠 ∗ (𝐺(𝑠)
𝐻 𝑡 = ℒ−1
1
𝑠
∗
−0 ∙ 125
0 ∙ 125 𝑠 + 1
= − 0 ∙ 125(1 − 𝑒−8𝑡)
𝐻 ∞ = −0 ∙ 125
Offset = 𝐻 ∞ = - 0.125
−0.125
offset= = - 0.125
0
H

21. 21
Example 4
Two liquid level tanks are connected in series as shown in Figure below.
𝑞𝑜
𝑞𝑖
𝐻𝑠𝑝(𝑠)
𝐻1
𝑞2
𝐴1
𝐴2 𝐻2
𝑅2
𝑅1
LT
LC
𝑞1
Given the following data
Proportional controller 𝐺𝑐 𝑠 = 𝐾𝑐
Measurement 𝐺𝑚 𝑠 = 0.5
Control valve 𝐺𝑣 𝑠 = 0.5
𝐴1 = 𝐴2 = 2 𝑚2
𝑅1 = 𝑅2 = 0.5 (s/m2)
a. For open loop, find 𝐻2 𝑠 = 𝑓(𝑞𝑖 𝑠 , 𝑞2 𝑠 )
b. Sketch the SFBD for closed system.
c. For unit step change in 𝑞2, calculate the offset for 𝐾𝑐 = 10.
d. Repeat (c) for 𝐾𝑐 = 20 and 𝐾𝑐 = 40.
e. What is the effect of 𝐾𝑐 on values of time constant T ,
damping factor 𝛿 and the offset.

22. 22
Solution
a. Material balance on tank 1
In = out +Accumulation
𝑞𝑖 = 𝑞1 + 𝐴1
𝑑𝐻1
𝑑𝑡
𝑞𝑖 =
𝐻1
𝑅1
+ 𝐴1
𝑑𝐻1
𝑑𝑡
𝜏1
𝑑𝐻1
𝑑𝑡
+ 𝐻1 = 𝑅1𝑞𝑖
𝜏1𝑠 + 1 𝐻1 𝑠 = 𝑅1𝑞𝑖 𝑠
𝐻1 𝑠 =
𝑅1
𝜏1𝑠 + 1
𝑞𝑖 𝑠 =
0 ∙ 5
𝑠 + 1
𝑞𝑖 𝑠 … … … … (∗)
Material balance on tank 2
𝑞1 + 𝑞2 = 𝑞𝑜 + 𝐴2
𝑑𝐻2
𝑑𝑡

23. 23
𝐻1
𝑅1
+ 𝑞2 =
𝐻2
𝑅2
+ 𝐴2
𝑑𝐻2
𝑑𝑡
𝑅2𝐴2
𝑑𝐻2
𝑑𝑡
+ 𝐻2 =
𝐻1 𝑅2
𝑅1
+ 𝑞2𝑅2
𝜏2
𝑑𝐻2
𝑑𝑡
+ 𝐻2 =
𝐻1 𝑅2
𝑅1
+ 𝑞2𝑅2
(𝜏2𝑠 + 1) 𝐻2 𝑠 =
𝑅2
𝑅1
𝐻1 𝑠 + 𝑞2(𝑠)𝑅2
𝜏2 = 𝑅2𝐴2 = 1 ,
𝑅2
𝑅1
= 1
𝑠 + 1 𝐻2 𝑠 = 𝐻1 𝑠 + 0 ∙ 5 𝑞2(𝑠)
𝐻2 𝑠 =
1
𝑠 + 1
𝐻1 𝑠 +
0 ∙ 5
𝑠 + 1
𝑞2 𝑠 … … … … (∗∗)
Substitute Eq. (*) in (**) gives
𝐻2 𝑠 =
1
𝑠 + 1
0 ∙ 5
𝑠 + 1
𝑞𝑖 𝑠 +
0 ∙ 5
𝑠 + 1
𝑞2(𝑠)
𝐻2 𝑠 =
0 ∙ 5
(𝑠 + 1)2
𝑞𝑖 𝑠 +
0 ∙ 5
𝑠 + 1
𝑞2(𝑠)
𝐺𝑃
𝐺𝐿
𝑞𝑜
𝑞𝑖
𝐻𝑠𝑝(𝑠)
𝐻1
𝑞2
𝐴1
𝐴2 𝐻2
𝑅2
𝑅1
LT
LC
𝑞1

24. 24
c. For unit step change in load
𝐾𝑐 = 10
𝐺 𝑠 =
𝐻2 𝑠
)
𝑞2(𝑠
=
0 ∙ 5
𝑠 + 1
1 +
10 ∗ 0 ∙ 5 ∗ 0 ∙ 5 ∗ 0.5
𝑠 + 1 2
=
0 ∙ 5
𝑠 + 1
)
𝑠 + 1 2 + 0 ∙ 125 (10
𝑠 + 1 2
=
0.22 𝑠 + 1
0.44𝑠2 + 0.88𝑠 + 1
∴ 𝐺(𝑠) =
0.22 𝑠 + 1
0.44𝑠2 + 0.88𝑠 + 1
0
𝑞2(s) 0.5
𝑆 + 1
𝐻2(s)
+
0
𝐻𝑠𝑝(s)
−
+
0.5
0.5
0.5
𝑆 + 1 2
𝐾𝑐
+

25. 25
𝑇2 = 0.44, 𝑇 = 0.44 = 0.66
2𝛿𝑇 = 0.88 , 𝛿 = 0.66
𝑜𝑓𝑓𝑠𝑒𝑡 = 𝐻2 ∞ = lim
𝑠 →0
𝑠 ∗ 𝐻 𝑠
= lim
𝑠 →0
𝑠 ∗ [
1
𝑠
∗
0.22 𝑠 + 1
0.44𝑠2 + 0.88𝑠 + 1
]
𝑜𝑓𝑓𝑠𝑒𝑡 = 0.22
0
Offset = 0.22
0.22
𝐻2
𝑡

26. 26
d. For 𝑲𝒄 = 𝟐𝟎
𝐻2 𝑠
𝑞2(𝑠)
=
0 ∙ 5
𝑠 + 1
1 +
20 ∗ 0 ∙ 5 ∗ 0 ∙ 5 ∗ 0.5
(𝑠 + 1)2
𝐺 𝑠 =
0.142 𝑠 + 1
0.285𝑠2 + 0.571𝑠 + 1
𝑇2 = 0.285, 𝑇 = 0.285 = 0.533
2𝛿𝑇 = 0.571 , 𝛿 = 0.535
𝑜𝑓𝑓𝑠𝑒𝑡 = 𝐻2 ∞ = lim
𝑠 →0
𝑠 ∗ 𝐻 𝑠
= lim
𝑠 →0
𝑠 ∗ [
1
𝑠
∗
0.142 𝑠 + 1
0.285𝑠2 + 0.571𝑠 + 1
]
𝑜𝑓𝑓𝑠𝑒𝑡 = 0.142
0
Offset = 0.142
0.142
𝐻2
𝑡

27. 27
For 𝒌𝒄 = 𝟒𝟎
𝐻2 𝑠
𝑞2(𝑠)
=
0 ∙ 5
𝑠 + 1
1 +
0 ∙ 5 (40) ∗ 0 ∙ 5 ∗ 0 ∙ 5
(𝑠 + 1)2
G(s) =
0.083 𝑠 + 1
0.166𝑠2 + 0.33𝑠 + 1
𝑇2
= 0.166, 𝑇 = 0.166 = 0.407
2𝛿𝑇 = 0.33 , 𝛿 = 0.41
𝑜𝑓𝑓𝑠𝑒𝑡 = 𝐻2 ∞ = lim
𝑠 →0
𝑠 ∗ 𝐻 𝑠
= lim
𝑠 →0
𝑠 ∗ [
1
𝑠
∗
0.083 𝑠 + 1
0.166𝑠2 + 0.33𝑠 + 1
]
𝑜𝑓𝑓𝑠𝑒𝑡 = 0.083
0
Offset = 0.083
0.083
𝐻2
𝑡

28. 28
e. We can summarize the obtained results in Table below.
Kc T δ offset
10 0.66 0.66 0.22
20 0.533 0.535 0.142
40 0.407 0.41 0.083
From the Table , we can conclude the following
𝑇 ∝
1
𝐾𝑐
, 𝛿 ∝
1
𝐾𝑐
, 𝑜𝑓𝑓𝑠𝑒𝑡 ∝
1
𝐾𝑐
So increasing the value of 𝑲𝒄 will lead to
i. Decreasing the value of time constant T. This is a positive effect, since the response of the system will be faster.
ii. Decreasing the value of damping coefficient 𝛿 . This is a negative effect because of decreasing the value of 𝛿 will increase
the value of overshoot and hence decreasing the stability of the system
iii. Decreasing the value of offset and this is a positive effect.

29. 29
Advantages of Proportional Controller
Now let us discuss some advantages of the proportional controller.
1. It is cheap and easy to manufacture.
2. Slow response of the over damped system can be made faster with the help of these controllers.
Disadvantages of Proportional Controller
There are some serious disadvantages of these controllers and these are:
1. Due to the presence of these controllers, we get some offsets in the system.
2. Proportional controllers also increase the maximum overshoot of the system and hence make the
system less stable.

30. 30