BELC3663 – CONTROL AND INSTRUMENTATION
SEMESTER 2 - 2023/2024
ANSWER - ASSIGNMENT 1
Objective:
To design cascaded controllers via root locus techniques and Ziegler-Nichols Method
using calculation and then verify by simulation MATLAB.
Instruction for students:
1. This is a GROUP assignment. (2-3 students per group)
2. In the task, each group need to design a controller or compensator via the classical design
technique (root locus technique) and Ziegler-Nichols Method using calculation. Show
all the calculation in the report.
3. Verify your answer in 2 via MATLAB simulation.
4. Discuss the results.
1
Question:
An adaptive suspension vehicle (ASV) will be a mobility system for use in very rough terrain. It
uses a legged locomotion principle rather than wheels or tracks. An adaptive suspension vehicle
that used a legged locomotion principle. The closed-loop control of the system can be
represented as shown in Figure 1. Do the following:
1. Design a K Gain using Root Locus technique to yield percentage of overshoot (%OS)
≤ 10 % using calculation. (Use the simulation for uncompensated dominant poles
location). Then simulate using in MATLAB.
2. Design a PID using Root Locus technique to yield requirement in Table 1. Then simulate
using in MATLAB.
Table 1: Design requirement
Settling time Reduced by a factor of 2
Percentage of Overshoot (%OS) Same in 1
Steady-state error 0
3. Design a PID using Ziegler-Nichols Method. Then simulate using in MATLAB.
4. Write a report that shows all calculation in detail. Print screen the simulation results and
Complete Table 1 to present your findings. Please discuss the results.
R(s) +¿ +¿ Y (s)
10
GC (s) 3 2
s + 14 s +43 s+28
−¿ −¿
0.2
Figure 1
2
Table 1: Performance analysis of the uncompensated and compensated systems
Uncompensated Uncompensated Compensated systems Compensated systems via
Simulation Simulation
System System via Root Locus method Ziegler-Nichols Method
(Root Locus) (Ziegler-Nichols)
(Calculation) (Simulation) (Theoretical) (Theoretical)
Transfer
function
4.5226 43.172
10
=4.317
3.58(s+0.1)(s+5.14)
s
33.65 1+ (1
0.479 s
+0.12 s ) 2.6(s+ 0.85)(s+5.14 )
s
Dominant poles −1.7+ j2.02 −1.7+ j2.02 −3.4+ j 4.23 −2.5 4 + j 3.25
Proportional
Gain, K P
4.5226 4.317 3 .58 ( 0.1+5.14 ) =18.76 33.65 2.6 ( 0. 85+5.14 )=15.57
Integral Gain, 33.65
KI - - 3 .58(0.1 ×5.14 )=1.84 =70.25 2.6 (0. 85× 5.14)=11.36
0.479
Derivative
Gain, K D
- - 3 .58 33.65 ×0.12=4.038 2.6
%OS 8% 6.8% 8% 9.45%
Damping ratio 0.6266 0.645 0.6266 0.616
4
Settling time =2.35 2.38 1.175 1.3
1.7
π π
=1.56 =0.74
Peak Time 1.68 4.23 0.864
2.02
Steady state 0.398
error
1−0.593=0.407 0 0 0 0
3
Design PID controller via Root Locus Method
R(s) +¿ +¿ Y (s)
10
GC (s) 3 2
s + 14 s +43 s+28
−¿ −¿
0.2
R(s) +¿ Y (s)
10
GC (s) 3 2
s + 14 s +43 s+28+10 (0.2)
−¿
R(s) +¿ 10 Y (s)
GC (s) 3 2
s + 14 s +43 s+30
−¿
Poles location : −1 ,−3 ,−10
Design Gain Controller (Uncompensated)
Calculation
Determine the sd (un comp) - choose %OS = 8% (from simulation sd (uncomp)=−1.7+ j2.02 )
Design K
Find K
K= | (s +1)(s +3)(s +10)
10 | s=−2.25+ j 3.41
K=4.5226
4
Find steady-state error
4.5226(1 0)
K p=lim KG ( s )=¿ =1.5 075 ¿
s →0 (1 ×3 ×10)
1 1
e ss = = =0.399
1+ K p 1+1. 51
5
Simulation
6
Design PID Controller (Compensated) via Root Locus
Calculation
Determine the sd (comp) =¿ - based on design requirement.
T s (uncomp) 2.35
T s (comp )= = =1.175
2 2
Using T s (comp ), we can find
4
ζ ω n= =3.4
1.175
Given %OS=8 % , ∴ ζ=0.577
3.4
ω n= =5.43
0. 6266
So, ω d=ωn √ 1−ζ 2=5.43 √ 1−0.62662=4.23
Therefore,
sd (comp) =−3.4+ j 4.23
Calculate the angle of θ zpd
−3.4+ j 4.23
.
θ3
θ zpd θ1 θ2
O X X X
−Z pd=¿ −10 −3 −1
7
−1 4.23
θ1=tan =32.66 °
1 0−3.4
−1 4.23
θ2=180 ° −tan =95.4 °
3.4−3
−1 4.23
θ 4=180 °−tan =119.57 °
3.4−1
Apply the formula
∑ θ z −¿ ∑ θ p=−180° ¿
θ zpd =−180 ° + ( 247.63 ° )=67.63 °
Find the location of z PD
−1 4.23
θ zpd =tan =67.63 °
z pd−3.4
Z pd =5.14
Design PI
K (s +0.1)
G PI (s)=
s
Determine K for PID
K=
| 1
G ( s) H ( s ) | |
s=sd
¿
( s+1)(s +3)(s +10)(s)
|
10( s+5.14)( s+ 0.1) s=−3.4 + j 4.23
K=3.58
Write the PID transfer function
3.58 (s +0.1)(s +5. 14)
G PID (s)=
s
8
3.57 (1 0 )( 0.1 ) ( 5.14 )
K p=lim KG ( s )=¿ =∞ ¿
s →0 0 ( 1 ×3 ×10 )
1 1
e ss = = =0
1+ K p 1+ ∞
Simulation: PID Controller (Compensated) via Root Locus
Table 1: Design requirement (Simulation)
Settling time (In simulation, +2 0 % is acceptable)
1.175 +0.235 = 1.41
Percentage of Overshoot (%OS) ≤ 10 %
Steady-state error 0
9
10
Design PID Controller (Compensated) via Ziegler-Nichols
Calculation
Find the closed-loop transfer function
10 1M
T 1 ( s) = 3 2
s +14 s + 43 s +30
10 K 2M
T ( s )= 3 2
s +14 s + 43 s +30+10 K
Construct RH table
s
3
1 43
s
2
14 30+10 K
s
1
602−30−10 K
14
s
0
30+10 K
602−30−10 K
>0
14
572
K<
10
K <57.2
∴ K critical=57.2
Find theω c 0
2
1 4 s +3 0+10 K =0
2
1 4 s +3 0+10 (57.2)=0
s=± j6.56
∴ ωc 0=6.56 rad/s
11
Calculate
K ult =57.2
2π 2π
T ult = = =0. 958
ωc 0 6.56
Use the table to find
K ult 57.2
K p= = =33.65
1.7 1.7
T ult 0.958
T i= = =0. 479
2 2
T ult
T d= =0.12
8
(
G PID ( s )=K p 1+
1
Tis ) (
+T d s =33.65 1+
1
0. 479 s
+0. 12 s )
12
Simulation: PID Controller (Compensated) via Ziegler-Nichols
13
