# Using Division Method to Convert the K-Input MIMO System to SISOs System Combined with Optimal Algorithm Application to Control of a Flexible Link System for the Oscillation (K =1,2)

Authors: Nguyen Cong Danh
Journal: Journal of Applied Mathematics
 (2021)

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## Abstract

The flexible link system (FLS) was a highly nonlinear model, multivariable and absolutely unstable dynamic system. In practice, it is common to integrate multiple subsystems into the main system with dynamically turned k-input signals (k =1, 2,…, N) to diversity the functionality of the main system. The flexibility of the division method to convert k-input MIMO system to SISOs system combined with the optimal algorithm creates a powerful tool that can be applied to many different MIMO nonlinear systems with high success rates. The optimal controllers can be created in the future for the flexible system is implemented on an experiment system using Arduino UNO micro-controller KIT. This paper describes division method to convert the k-input MIMO system to SISOs system, after that combined with the optimal algorithm to control for the flexible link system. Specifically, the author will conduct oscillating component analysis of a system with k input pairs (k =1, 2) so that the author can better understand the nature of sub-components as they interact with the system.


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## Body

## 1. Introduction

The flexible link system is a topic that attracts the attention of professionals in forming flexible structures for an automation system, from a simple model to a model with a relatively complex structure. The structural flexibility when using flexible links gives the system many advantages compared to the previous traditional model: the system volume will be less, the operation will be faster, and more flexibility, due to less system volume, the power consumption of the system is reduced. The linear quadratic optimal boundary control of the flexible robotic arm is a new proposition based on the reference of the paper [1]. I am currently researching for using this technique [1] for the other system: Ac-robot. I can apply the techniques in article [2] to flexible robotic arms. The use of techniques in this article [3] in control systems is a promising research topic. The application of distributed parameter models for the robot’s mobile feet based on the reference of the article [4] is a new idea. A new proposal is the use of modern controllers for the design of systems in the paper [5]. The use of the optimal controller in [6, 7] is a new proposition. The application of the fuzzy algorithm in [8, 9] is a promising idea. Path Planning [10] of a planetary robot is an important research work for the field of astronomy. Theoretical and experimental study of DLCC [11] for a flexible link system of any model is welcome. The use of achievements [12] for a soft body is an interesting topic. Maximum Allowable Load [13] of a flexible robotic arm is a promising topic.This article focuses on the MIMO system with k-input pairs in the flexible link system, the author can apply a modern control method: The method of using the optimal controller to control the flexible link system. The application of a novel control method with a MIMO system can be applied to many complex MIMO system models. The state-of-the-art controllers will be created in the future, which will flexibly change the oscillation characteristics. The author will compare the efficiency of the above controller for the k-input system with k =1, 2. This can be checked to show effectiveness of the real system. Through this article, the reader can find out the effectiveness of the above control method for a system. The author will also compare the pros and cons of a system using an optimal controller versus a system without any controller. Examining sub-components in a system with k inputs (k =1, 2) is essential for the maintenance of a system. Sub-components can have more or less impact on a system depending on how much they affect the server. This paper will give an overview through the evaluation of simulation results. From there, the author can confirm their impact on a host system. Previous studies have not been able to show the existence of components that make up a system. Besides, the operation of sub-components in a host system has not been investigated meticulously in previous studies. Sub-components here can be variables that describe an individual form of activity in the constitutive of the system’s motions. Specifically in this article are the levels of flexible link properties, the degree of oscillation around the specified position,....Therefore, controlling the behavior of a sub-component in a large system is an important task. Readers can understand that the components involved in the operation of the system work in harmony with each other or not? Do they interact with each other? The answers to these questions will be presented through the analytical sections of this paper. The difference of this paper compared to other papers is the detailed description of the operation of one of the elements constituting the system with 1 input pair, 2 input pairs... These elements are very diverse in terms of their structure as well as their functions. Each system has many distinct elements. Exploring these elements gives the reader a detailed look at the functions of these categories. This can give designers more incentive to create new components to improve the quality of a system’s features.


## 2. Related Theory

### 2.1. Structure of the FLS

A block diagram showing the controller based on the Arduino UNO Kit is provided in Figure1. The FLS system consists of a flexible metal (LINK) rod attached to one end of the DC deceleration motor, the other (TIP) with an acceleration sensor to measure the horizontal oscillation. Motor rotates the metal horizontally, due to the flexible metal rod, so the TIP will vibrate a lot. A 720ppr encoder was used to measure the position of the current of the LINK is θ and the acceleration sensor measures the deflection of the angle α of the TIP (Figure 2). The control system uses the Arduino UNO Kit, the AVR ATMEGA328 micro-controller reads the value from the encoder, which it receives the value from the accelerometer sensor. It implements the optimal control algorithm and supplies PWM signals to the motor. The sampling frequency of the system is 4 ms.Figure 1
A block diagram of FLS control system.Figure 2
A Spatial model of FLS [14].Table1 shows the simulation results described below.Table 1
Physical specifications of the real system [14].
SymbolDescriptionValueUnitBeqEquivalent viscous damping coefficient0.004Nm/(rad/s)ηgGearbox efficiency0.9KtMotor torque constant0.0134N-m/AKmBack-emf constant0.0134V-s/radJeqMoment of inertia of the rotor of the motor2.08x10-3Kg.m2RmArmature resistance of motor1.9ΩηmMotor efficiency0.8JLEquivalent moment of inertia at the LINK0.0038Kg.m2mMass of LINK0.115KgLLength to LINK’s center of mass0.25mKsStiffness of LINK1.3Nm/radBLEquivalent viscous damping coefficient of LINK0.00004Nm/(rad/s)


### 2.2. A Mathematical Model of FLS

FLS [14, 15] can be converted into blocks for analysis, such as Figure 3.Figure 3
The Analysis of block diagram of FLS system [14].The input of the FLS system is the voltage, which supplies for (Vm) voltage to the motor, the two outputs are controlled by the position of the motor’s rotation and the deviation of the angle of the LINK (Figure 3).Using Lagrange equation to determine the equations of motion of the system through the total kinetic energy and the total potential energy of the elements in the FLS system in motion.The potential energy of FLS system is the elastic energy when the LINK vibrates:(1)V=12KSα2The kinetic energy of a FLS system consists of two components, the kinetic energy component rotated by the motor and the kinetic energy component when the LINK is deviated:(2)T=12Jeqθ̇2+12JLθ̇+α̇2Lagrange’s the equation:(3)L=T−V=12Jeqθ̇2+12JLθ̇+α̇2−12KSα2The differential components of the Lagrange equation followsθ:(4)∂L∂θ̇=Jeqθ̇+JLθ̇+α̇ddt∂L∂θ̇=Jeqθ¨+JLθ¨+α¨∂L∂θ=0The differential components of the Lagrange’s equation followα:(5)∂L∂α̇=JLθ̇+α̇ddt∂L∂α̇=JLθ¨+α¨∂L∂α=−KSαForce of the balance [27] of (4), (5) follows Lagrange equation:(6)Jeq+JLθ¨+JLα¨+Beqθ̇=τJLθ¨+JLα¨+BLα.+KSα=0τ=ηgKgηmKtVm−KgKmθ̇RmThe (BL) the damping coefficient of the LINK is very small, which it can be considered as a ~0, the system of equations (6) becomes:(7)θ¨=−BeqJeqθ̇+KSJeqα−ηgKg2ηmKtKmJeqRmθ̇+ηgKgηmKtJeqRmVmα¨=BeqJeqθ̇−KS1Jeq+1JLα+ηgKg2ηmKtKmJeqRmθ̇−ηgKgηmKtJeqRmVmI set the state and output variables for the FLS system (k =1) as follows:(8)x1=θ,x2=α,x3=θ̇,x4=α̇y1=x1,y2=x2I combine (7) and (8) to obtain the system of state equations describing the FLS system:(9)ẋ1ẋ2ẋ3ẋ4=001000010KsJeq−BeqRm−ηgKg2ηmKtKmJeqRm00−Ks1Jeq+1JLBeqRm+ηgKg2ηmKtKmJeqRm0x1x2x3x4y1y2=10000100x1x2+00ηgKgηmKtJeqRm−ηgKgηmKtJeqRmVmI replaced the physical parameters from Table1 into equation (9), the system of state equation of the FLS system has been designed:(10)ẋ1ẋ2ẋ3ẋ4=001000010623.7−7.200−965.37.20x1x2x3x4+00479.8−479.8Vmy1y2=10000100x1x2The transfer function of the FLS system is:(11)Gs=479.8s2+2.196×10‐10s+1.639×105s4+7.2s3+965.3s2+2460sI set state and output variables for the FLS system with 2 input pairs (k =2) as follows:(12)x1=θ,x2=α,x3=θ̇,x4=α̇,x5=θ1,x6=α1,x7=θ̇1,x8=α̇1y1=x1,y2=x2,y3=x5,y4=x6I combine (7) and (8), (12) to obtain the system of state equations describing the FLS system with 2 input pairs:(13)ẋ1ẋ2ẋ3ẋ4ẋ5ẋ6ẋ7ẋ8=00100000000100000KsJeq−BeqRm−ηgKg2ηmKtKmJeqRm000000−Ks1Jeq+1JLBeqRm+ηgKg2ηmKtKmJeqRm00000000000100000000100000KsJeq−BeqRm−ηgKg2ηmKtKmJeqRm000000−Ks1Jeq+1JLBeqRm+ηgKg2ηmKtKmJeqRm0x1x2x3x4x5x6x7x8y1y2y3y4=10000000010000000000100000000100x1x2x3x4x5x6x7x8+00ηgKgηmKtJeqRm−ηgKgηmKtJeqRm00ηgKgηmKtJeqRm−ηgKgηmKtJeqRmVmI replaced the physical parameters from Table1 into the equation (9), the system of state equations of the FLS system has been designed:(14)ẋ1ẋ2ẋ3ẋ4ẋ5ẋ6ẋ7ẋ8=00100000000100000623.7−7.2000000−965.37.200000000000100000000100000623.7−7.2000000−965.37.20x1x2x3x4x5x6x7x8+00479.8−479.800479.8−479.8Vmy1y2y3y4=10000000010000000000100000000100x1x2x3x4x5x6x7x8The transfer function of the FLS system with 2 input pairs (k =2) is:(15)G1s=479.8s2+1.613×10‐10s+1.639×105s4+7.2s3+965.3s2+2460sG2s=‐479.8s‐2.213×10‐12s3+7.2s2+965.3s+2460G3s=479.8s2+2.196×10−10s+1.639×105s4+7.2s3+965.3s2+2460sG4s=‐479.8s+1.107×10‐12s3+7.2s2+965.3s+2460


### 2.3. Method of Splitting MIMO System into SISOs System

Obviously FLS system is multivariate, the oscillator variable isα and the LINK’s position variable is θ. The control voltage which supplies the motor is Vm voltage. FLS can be separated into two SISO systems. The input of the first SISO system is θ, the output of the first SISO system is the voltage Vm1 which it supplies to the total set. The input of the second SISOs system is α, the output of the second SISO system is VM2 provides to the total set. The KPOS and KOSC coefficients are used to set priorities for the system controller. 2-inputs of the second SISO system are α,α1, 2-outputs of the second SISO system are Vm3,Vm4 provides to the total set. The MIMO system for the FLS with 1 input pair is shown in Figure 4. The MIMO system for the FLS with 2 input pairs is shown in Figure 5.Figure 4
The diagram the optimal controller splitting MIMO into SISOs for FLS system with 1 input pairs (k =1).Figure 5
The diagram the optimal controller splitting MIMO into SISOs for FLS with 2 input pairs (k =2).


## 2.1. Structure of the FLS

A block diagram showing the controller based on the Arduino UNO Kit is provided in Figure1. The FLS system consists of a flexible metal (LINK) rod attached to one end of the DC deceleration motor, the other (TIP) with an acceleration sensor to measure the horizontal oscillation. Motor rotates the metal horizontally, due to the flexible metal rod, so the TIP will vibrate a lot. A 720ppr encoder was used to measure the position of the current of the LINK is θ and the acceleration sensor measures the deflection of the angle α of the TIP (Figure 2). The control system uses the Arduino UNO Kit, the AVR ATMEGA328 micro-controller reads the value from the encoder, which it receives the value from the accelerometer sensor. It implements the optimal control algorithm and supplies PWM signals to the motor. The sampling frequency of the system is 4 ms.Figure 1
A block diagram of FLS control system.Figure 2
A Spatial model of FLS [14].Table1 shows the simulation results described below.Table 1
Physical specifications of the real system [14].
SymbolDescriptionValueUnitBeqEquivalent viscous damping coefficient0.004Nm/(rad/s)ηgGearbox efficiency0.9KtMotor torque constant0.0134N-m/AKmBack-emf constant0.0134V-s/radJeqMoment of inertia of the rotor of the motor2.08x10-3Kg.m2RmArmature resistance of motor1.9ΩηmMotor efficiency0.8JLEquivalent moment of inertia at the LINK0.0038Kg.m2mMass of LINK0.115KgLLength to LINK’s center of mass0.25mKsStiffness of LINK1.3Nm/radBLEquivalent viscous damping coefficient of LINK0.00004Nm/(rad/s)


## 2.2. A Mathematical Model of FLS

FLS [14, 15] can be converted into blocks for analysis, such as Figure 3.Figure 3
The Analysis of block diagram of FLS system [14].The input of the FLS system is the voltage, which supplies for (Vm) voltage to the motor, the two outputs are controlled by the position of the motor’s rotation and the deviation of the angle of the LINK (Figure 3).Using Lagrange equation to determine the equations of motion of the system through the total kinetic energy and the total potential energy of the elements in the FLS system in motion.The potential energy of FLS system is the elastic energy when the LINK vibrates:(1)V=12KSα2The kinetic energy of a FLS system consists of two components, the kinetic energy component rotated by the motor and the kinetic energy component when the LINK is deviated:(2)T=12Jeqθ̇2+12JLθ̇+α̇2Lagrange’s the equation:(3)L=T−V=12Jeqθ̇2+12JLθ̇+α̇2−12KSα2The differential components of the Lagrange equation followsθ:(4)∂L∂θ̇=Jeqθ̇+JLθ̇+α̇ddt∂L∂θ̇=Jeqθ¨+JLθ¨+α¨∂L∂θ=0The differential components of the Lagrange’s equation followα:(5)∂L∂α̇=JLθ̇+α̇ddt∂L∂α̇=JLθ¨+α¨∂L∂α=−KSαForce of the balance [27] of (4), (5) follows Lagrange equation:(6)Jeq+JLθ¨+JLα¨+Beqθ̇=τJLθ¨+JLα¨+BLα.+KSα=0τ=ηgKgηmKtVm−KgKmθ̇RmThe (BL) the damping coefficient of the LINK is very small, which it can be considered as a ~0, the system of equations (6) becomes:(7)θ¨=−BeqJeqθ̇+KSJeqα−ηgKg2ηmKtKmJeqRmθ̇+ηgKgηmKtJeqRmVmα¨=BeqJeqθ̇−KS1Jeq+1JLα+ηgKg2ηmKtKmJeqRmθ̇−ηgKgηmKtJeqRmVmI set the state and output variables for the FLS system (k =1) as follows:(8)x1=θ,x2=α,x3=θ̇,x4=α̇y1=x1,y2=x2I combine (7) and (8) to obtain the system of state equations describing the FLS system:(9)ẋ1ẋ2ẋ3ẋ4=001000010KsJeq−BeqRm−ηgKg2ηmKtKmJeqRm00−Ks1Jeq+1JLBeqRm+ηgKg2ηmKtKmJeqRm0x1x2x3x4y1y2=10000100x1x2+00ηgKgηmKtJeqRm−ηgKgηmKtJeqRmVmI replaced the physical parameters from Table1 into equation (9), the system of state equation of the FLS system has been designed:(10)ẋ1ẋ2ẋ3ẋ4=001000010623.7−7.200−965.37.20x1x2x3x4+00479.8−479.8Vmy1y2=10000100x1x2The transfer function of the FLS system is:(11)Gs=479.8s2+2.196×10‐10s+1.639×105s4+7.2s3+965.3s2+2460sI set state and output variables for the FLS system with 2 input pairs (k =2) as follows:(12)x1=θ,x2=α,x3=θ̇,x4=α̇,x5=θ1,x6=α1,x7=θ̇1,x8=α̇1y1=x1,y2=x2,y3=x5,y4=x6I combine (7) and (8), (12) to obtain the system of state equations describing the FLS system with 2 input pairs:(13)ẋ1ẋ2ẋ3ẋ4ẋ5ẋ6ẋ7ẋ8=00100000000100000KsJeq−BeqRm−ηgKg2ηmKtKmJeqRm000000−Ks1Jeq+1JLBeqRm+ηgKg2ηmKtKmJeqRm00000000000100000000100000KsJeq−BeqRm−ηgKg2ηmKtKmJeqRm000000−Ks1Jeq+1JLBeqRm+ηgKg2ηmKtKmJeqRm0x1x2x3x4x5x6x7x8y1y2y3y4=10000000010000000000100000000100x1x2x3x4x5x6x7x8+00ηgKgηmKtJeqRm−ηgKgηmKtJeqRm00ηgKgηmKtJeqRm−ηgKgηmKtJeqRmVmI replaced the physical parameters from Table1 into the equation (9), the system of state equations of the FLS system has been designed:(14)ẋ1ẋ2ẋ3ẋ4ẋ5ẋ6ẋ7ẋ8=00100000000100000623.7−7.2000000−965.37.200000000000100000000100000623.7−7.2000000−965.37.20x1x2x3x4x5x6x7x8+00479.8−479.800479.8−479.8Vmy1y2y3y4=10000000010000000000100000000100x1x2x3x4x5x6x7x8The transfer function of the FLS system with 2 input pairs (k =2) is:(15)G1s=479.8s2+1.613×10‐10s+1.639×105s4+7.2s3+965.3s2+2460sG2s=‐479.8s‐2.213×10‐12s3+7.2s2+965.3s+2460G3s=479.8s2+2.196×10−10s+1.639×105s4+7.2s3+965.3s2+2460sG4s=‐479.8s+1.107×10‐12s3+7.2s2+965.3s+2460


## 2.3. Method of Splitting MIMO System into SISOs System

Obviously FLS system is multivariate, the oscillator variable isα and the LINK’s position variable is θ. The control voltage which supplies the motor is Vm voltage. FLS can be separated into two SISO systems. The input of the first SISO system is θ, the output of the first SISO system is the voltage Vm1 which it supplies to the total set. The input of the second SISOs system is α, the output of the second SISO system is VM2 provides to the total set. The KPOS and KOSC coefficients are used to set priorities for the system controller. 2-inputs of the second SISO system are α,α1, 2-outputs of the second SISO system are Vm3,Vm4 provides to the total set. The MIMO system for the FLS with 1 input pair is shown in Figure 4. The MIMO system for the FLS with 2 input pairs is shown in Figure 5.Figure 4
The diagram the optimal controller splitting MIMO into SISOs for FLS system with 1 input pairs (k =1).Figure 5
The diagram the optimal controller splitting MIMO into SISOs for FLS with 2 input pairs (k =2).


## 3. Designing the Optimal Controller for Fls with 1 Input Pair (K =1)

### 3.1. Designing the Equation of State for the Oscillation of the FLS with 1 Input Pair

The author considered the system of state equations when only the variable name wasα as follows:(16)ẋ1ẋ2ẋ3ẋ4=001000010623.7000−965.300x1x2x3x4+00479.8052−479.8052Vmy1y2=10000100x1x2The author applied the algorithm according to the controller design of the SISOs system for the oscillation corresponding to the MIMO system with the state equation presented above.


### 3.2. Designing the Optimal Controller Splitting MIMO into SISOs for the FLS with 1 Input Pair

A control system designed in the best working mode is always in the optimal state according to a certain quality (the standard extreme value is reached).Whether or not it is stable depends on the required quality, and understanding of the object and its impacts, based on the working condition of the control system ... In the presentation, The author designed the controller to operate in the optimal state according to the J quality criteria function as required by the problem, that traditional model without using the optimal controller has not been achieved.The diagram of the optimal controller splitting MIMO into SISOs for the FLS is shown below (Figure4).(17)A=001000010623.7−7.200−965.37.20;B=00479.8052−479.8052C=10000100


### 3.3. Designing the Optimal Controller for the Oscillation of FLS with 1 Input Pair

The diagram of the Optimal control for the oscillation of FLS is shown below (Figure6).(18)Aα1=001000010623.7000−965.300;Bα1=00479.8−479.8Cα1=10000100Figure 6
The diagram of the Optimal control for the oscillation.The transfer functionGα1s:(19)Gα11s=479.8s2+1.79×10−11s+1.639×105s4+965.3s2Gα12s=−479.8s2+965.3The author consider the system to have the external impactu≠0:(20)ẋ=Aα1x+Bα1uThe author need to find the matrixKα1 of the optimal control vector: ut=−Kα1∗xt satisfy the quality index value of Jα1 and Jα1 must reach the minimum value:(21)Jα1=∫0∞xTQα1x+uTRα1udtwhere Qα1 is a positive deterministic matrix, Rα1 is a positive deterministic matrix.The matrix ofKα1 is determined from Riccati’s equation of the form:(22)Kα1=Rα1−1Bα1TPThe state feedback control structure is shown below (Figure7).Figure 7
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(23)ut=−Kα1t=−Rα1−1Bα1TPxtThe value of the matrix is P, P must be satisfied the equation:(24)PAα1+Aα1TP+Qα1−PBα1Rα1−1Bα1TP=ṖThe equation (24) is known as Riccati’s equation.The author choose the values of the matrixQα1 and the values of the matrix Rα1 as follows:(25)Qα1=1000010000100001,Rα1=1.The author calculate the value of the matrixKα1 through Matlab software:(26)Kα1=lqrAα1,Bα1,Qα1,Rα1=1.0−15.81421.0509−0.3879


### 3.4. Designing the Optimal Controller for FLS with 2 -Input Pairs (Figure5)

#### 3.4.1. Designing the Optimal Controller for the Oscillation of FLS with 2 Input Pairs (Figure8)

Figure 8
The diagram of the Optimal control for the oscillation.The author considered the system of state equations when only the variable names wereα;α1 as follows:(27)ẋ1ẋ2ẋ3ẋ4ẋ5ẋ6ẋ7ẋ8=00100000000100000KsJeq0000000−Ks1Jeq+1JL000000000000100000000100000KsJeq0000000−Ks1Jeq+1JL00x1x2x3x4x5x6x7x8y1y2y3y4=10000000010000000000100000000100x1x2x3x4x5x6x7x8+00ηgKgηmKtJeqRm−ηgKgηmKtJeqRm00ηgKgηmKtJeqRm−ηgKgηmKtJeqRmVmẋ1ẋ2ẋ3ẋ4ẋ5ẋ6ẋ7ẋ8=00100000000100000623.70000000−965.3000000000000100000000100000623.70000000−965.300x1x2x3x4x5x6x7x8y1y2y3y4=10000000010000000000100000000100x1x2x3x4x5x6x7x8+00479.8−479.800479.8−479.8VmThe author applied the algorithm according to the controller design of the SISOs system for the oscillation corresponding to the MIMO system with the state equation presented above.The transfer functionGα2s:(28)Gα2s=479.8s2+1.79×10−11s+1.639×105s4+965.3×s2Aα2=0−30.17003200001000010;Bα2=64000;Cα2=00.23432.913×10−1580.03The author consider the system to have the external impactu≠0:(29)ẋ=Aα2x+Bα2uI need to find the matrixKα2 of the optimal control vector: ut=−Kα2∗xt satisfy the quality index value of Jα2 and Jα2 must reach the minimum value:(30)Jα2=∫0∞xTQα2x+uTRα2udtwhere Qα2 is a positive deterministic matrix, Rα2 is a positive deterministic matrix.The matrix ofKα2 is determined from Riccati’s equation of the form:(31)Kα2=Rα2−1Bα2TPThe state feedback control structure is shown below (Figure9).Figure 9
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(32)ut=−Kα2t=−Rα2−1Bα2TPxtThe value of the matrix is P, P must be satisfied the equation:(33)PAα2+Aα2TP+Qα2−PBα2Rα2−1Bα2TP=ṖThe equation (33) is known as Riccati’s equation.The author choose the values of the matrixQα2 and the values of the matrix Rα2 as follows:(34)Qα2=1000010000100001;Rα2=1.The author calculate the value of the matrixKα2 through Matlab software:(35)Kα2=lqrAα2,Bα2,Qα2,Rα2Gα3s=−479.8s2+965.3Aα3=0−30.17320;Bα3=40;Cα3=0−3.748The author consider the system to have the external impactu≠0:(36)ẋ=Aα3x+Bα3uThe author need to find the matrixKα3 of the optimal control vector: ut=−Kα3∗xt satisfy the quality index value of Jα3 and Jα3 must reach the minimum value:(37)Jα3=∫0∞xTQα3x+uTRα3udtwhere Qα3 is a positive deterministic matrix, Rα3 is a positive deterministic matrix.The matrix ofKα3 is determined from Riccati’s equation of the form:(38)Kα3=Rα3−1Bα3TPThe state feedback control structure is shown below (Figure10).Figure 10
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(39)ut=−Kα3t=−Rα3−1Bα3TPxtThe value of the matrix is P, P must be satisfied the equation:(40)PAα3+Aα3TP+Qα3−PBα3Rα3−1Bα3TP=ṖThe equation (40) is known as Riccati’s equation.I choose the values of the matrixQα3 and the values of the matrix Rα3 as follows:(41)Qα3=1001;Rα3=1.Then I calculate the value of the matrixKα3 through Matlab software:(42)Kα3=lqrAα3,Bα3,Qα3,Rα3=1.43390.066Gα4s=479.8s2+5.966×10−12s+1.639×105s4+965.3s2Aα4=0−30.17003200001000010;Bα4=64000;Cα4=00.23432.913×10−1580.03I consider the system to have the external impactu≠0:(43)ẋ=Aα4x+Bα4uI need to find the matrixKα4 of the optimal control vector: ut=−Kα4∗xt satisfy the quality index value of Jα4 and Jα4 must reach the minimum value:(44)Jα4=∫0∞xTQα4x+uTRα4udtwhere Qα4 is a positive deterministic matrix, Rα4 is a positive deterministic matrix.The matrix ofKα4 is determined from Riccati’s equation of the form:(45)Kα4=Rα4−1Bα4TPThe state feedback control structure is shown below (Figure11).Figure 11
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(46)ut=−Kα4t=−Rα4−1Bα4TPxtThe value of the matrix is P, P must be satisfied the equation:(47)PAα4+Aα4TP+Qα4−PBα4Rα4−1Bα4TP=ṖThe equation (47) is known as Riccati’s equation.I choose the values of the matrixQα4 and the values of the matrix Rα4 as follows:(48)Qα4=1000010000100001,Rα4=1.Then I calculate the value of the matrixKα4 through Matlab software:(49)Kα4=lqrAα4,Bα4,Qα4,Rα4=1.30350.69911.82791.0000Figures12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 are shown below.Figure 12
The model without using any algorithmsGα11s.Figure 13
The model without using any algorithmsGα12s.Figure 14
The model without using any algorithmsGα2s.Figure 15
The model without using any algorithmsGα3s.Figure 16
The model without using any algorithmsGα4s.Figure 17
Applying the optimal LQR control method for the systemGα1s.Figure 18
The system with the optimal controllerGα1s.Figure 19
Applying the optimal LQR control method for the systemGα3s.Figure 20
The system with the optimal controllerGα3s.Figure 21
Applying the optimal LQR control method for the systemGα4s,Gα2s.Figure 22
The system with the optimal controllerGα4s,Gα2s.


## 3.1. Designing the Equation of State for the Oscillation of the FLS with 1 Input Pair

The author considered the system of state equations when only the variable name wasα as follows:(16)ẋ1ẋ2ẋ3ẋ4=001000010623.7000−965.300x1x2x3x4+00479.8052−479.8052Vmy1y2=10000100x1x2The author applied the algorithm according to the controller design of the SISOs system for the oscillation corresponding to the MIMO system with the state equation presented above.


## 3.2. Designing the Optimal Controller Splitting MIMO into SISOs for the FLS with 1 Input Pair

A control system designed in the best working mode is always in the optimal state according to a certain quality (the standard extreme value is reached).Whether or not it is stable depends on the required quality, and understanding of the object and its impacts, based on the working condition of the control system ... In the presentation, The author designed the controller to operate in the optimal state according to the J quality criteria function as required by the problem, that traditional model without using the optimal controller has not been achieved.The diagram of the optimal controller splitting MIMO into SISOs for the FLS is shown below (Figure4).(17)A=001000010623.7−7.200−965.37.20;B=00479.8052−479.8052C=10000100


## 3.3. Designing the Optimal Controller for the Oscillation of FLS with 1 Input Pair

The diagram of the Optimal control for the oscillation of FLS is shown below (Figure6).(18)Aα1=001000010623.7000−965.300;Bα1=00479.8−479.8Cα1=10000100Figure 6
The diagram of the Optimal control for the oscillation.The transfer functionGα1s:(19)Gα11s=479.8s2+1.79×10−11s+1.639×105s4+965.3s2Gα12s=−479.8s2+965.3The author consider the system to have the external impactu≠0:(20)ẋ=Aα1x+Bα1uThe author need to find the matrixKα1 of the optimal control vector: ut=−Kα1∗xt satisfy the quality index value of Jα1 and Jα1 must reach the minimum value:(21)Jα1=∫0∞xTQα1x+uTRα1udtwhere Qα1 is a positive deterministic matrix, Rα1 is a positive deterministic matrix.The matrix ofKα1 is determined from Riccati’s equation of the form:(22)Kα1=Rα1−1Bα1TPThe state feedback control structure is shown below (Figure7).Figure 7
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(23)ut=−Kα1t=−Rα1−1Bα1TPxtThe value of the matrix is P, P must be satisfied the equation:(24)PAα1+Aα1TP+Qα1−PBα1Rα1−1Bα1TP=ṖThe equation (24) is known as Riccati’s equation.The author choose the values of the matrixQα1 and the values of the matrix Rα1 as follows:(25)Qα1=1000010000100001,Rα1=1.The author calculate the value of the matrixKα1 through Matlab software:(26)Kα1=lqrAα1,Bα1,Qα1,Rα1=1.0−15.81421.0509−0.3879


## 3.4. Designing the Optimal Controller for FLS with 2 -Input Pairs (Figure5)

### 3.4.1. Designing the Optimal Controller for the Oscillation of FLS with 2 Input Pairs (Figure8)

Figure 8
The diagram of the Optimal control for the oscillation.The author considered the system of state equations when only the variable names wereα;α1 as follows:(27)ẋ1ẋ2ẋ3ẋ4ẋ5ẋ6ẋ7ẋ8=00100000000100000KsJeq0000000−Ks1Jeq+1JL000000000000100000000100000KsJeq0000000−Ks1Jeq+1JL00x1x2x3x4x5x6x7x8y1y2y3y4=10000000010000000000100000000100x1x2x3x4x5x6x7x8+00ηgKgηmKtJeqRm−ηgKgηmKtJeqRm00ηgKgηmKtJeqRm−ηgKgηmKtJeqRmVmẋ1ẋ2ẋ3ẋ4ẋ5ẋ6ẋ7ẋ8=00100000000100000623.70000000−965.3000000000000100000000100000623.70000000−965.300x1x2x3x4x5x6x7x8y1y2y3y4=10000000010000000000100000000100x1x2x3x4x5x6x7x8+00479.8−479.800479.8−479.8VmThe author applied the algorithm according to the controller design of the SISOs system for the oscillation corresponding to the MIMO system with the state equation presented above.The transfer functionGα2s:(28)Gα2s=479.8s2+1.79×10−11s+1.639×105s4+965.3×s2Aα2=0−30.17003200001000010;Bα2=64000;Cα2=00.23432.913×10−1580.03The author consider the system to have the external impactu≠0:(29)ẋ=Aα2x+Bα2uI need to find the matrixKα2 of the optimal control vector: ut=−Kα2∗xt satisfy the quality index value of Jα2 and Jα2 must reach the minimum value:(30)Jα2=∫0∞xTQα2x+uTRα2udtwhere Qα2 is a positive deterministic matrix, Rα2 is a positive deterministic matrix.The matrix ofKα2 is determined from Riccati’s equation of the form:(31)Kα2=Rα2−1Bα2TPThe state feedback control structure is shown below (Figure9).Figure 9
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(32)ut=−Kα2t=−Rα2−1Bα2TPxtThe value of the matrix is P, P must be satisfied the equation:(33)PAα2+Aα2TP+Qα2−PBα2Rα2−1Bα2TP=ṖThe equation (33) is known as Riccati’s equation.The author choose the values of the matrixQα2 and the values of the matrix Rα2 as follows:(34)Qα2=1000010000100001;Rα2=1.The author calculate the value of the matrixKα2 through Matlab software:(35)Kα2=lqrAα2,Bα2,Qα2,Rα2Gα3s=−479.8s2+965.3Aα3=0−30.17320;Bα3=40;Cα3=0−3.748The author consider the system to have the external impactu≠0:(36)ẋ=Aα3x+Bα3uThe author need to find the matrixKα3 of the optimal control vector: ut=−Kα3∗xt satisfy the quality index value of Jα3 and Jα3 must reach the minimum value:(37)Jα3=∫0∞xTQα3x+uTRα3udtwhere Qα3 is a positive deterministic matrix, Rα3 is a positive deterministic matrix.The matrix ofKα3 is determined from Riccati’s equation of the form:(38)Kα3=Rα3−1Bα3TPThe state feedback control structure is shown below (Figure10).Figure 10
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(39)ut=−Kα3t=−Rα3−1Bα3TPxtThe value of the matrix is P, P must be satisfied the equation:(40)PAα3+Aα3TP+Qα3−PBα3Rα3−1Bα3TP=ṖThe equation (40) is known as Riccati’s equation.I choose the values of the matrixQα3 and the values of the matrix Rα3 as follows:(41)Qα3=1001;Rα3=1.Then I calculate the value of the matrixKα3 through Matlab software:(42)Kα3=lqrAα3,Bα3,Qα3,Rα3=1.43390.066Gα4s=479.8s2+5.966×10−12s+1.639×105s4+965.3s2Aα4=0−30.17003200001000010;Bα4=64000;Cα4=00.23432.913×10−1580.03I consider the system to have the external impactu≠0:(43)ẋ=Aα4x+Bα4uI need to find the matrixKα4 of the optimal control vector: ut=−Kα4∗xt satisfy the quality index value of Jα4 and Jα4 must reach the minimum value:(44)Jα4=∫0∞xTQα4x+uTRα4udtwhere Qα4 is a positive deterministic matrix, Rα4 is a positive deterministic matrix.The matrix ofKα4 is determined from Riccati’s equation of the form:(45)Kα4=Rα4−1Bα4TPThe state feedback control structure is shown below (Figure11).Figure 11
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(46)ut=−Kα4t=−Rα4−1Bα4TPxtThe value of the matrix is P, P must be satisfied the equation:(47)PAα4+Aα4TP+Qα4−PBα4Rα4−1Bα4TP=ṖThe equation (47) is known as Riccati’s equation.I choose the values of the matrixQα4 and the values of the matrix Rα4 as follows:(48)Qα4=1000010000100001,Rα4=1.Then I calculate the value of the matrixKα4 through Matlab software:(49)Kα4=lqrAα4,Bα4,Qα4,Rα4=1.30350.69911.82791.0000Figures12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 are shown below.Figure 12
The model without using any algorithmsGα11s.Figure 13
The model without using any algorithmsGα12s.Figure 14
The model without using any algorithmsGα2s.Figure 15
The model without using any algorithmsGα3s.Figure 16
The model without using any algorithmsGα4s.Figure 17
Applying the optimal LQR control method for the systemGα1s.Figure 18
The system with the optimal controllerGα1s.Figure 19
Applying the optimal LQR control method for the systemGα3s.Figure 20
The system with the optimal controllerGα3s.Figure 21
Applying the optimal LQR control method for the systemGα4s,Gα2s.Figure 22
The system with the optimal controllerGα4s,Gα2s.


## 3.4.1. Designing the Optimal Controller for the Oscillation of FLS with 2 Input Pairs (Figure8)

Figure 8
The diagram of the Optimal control for the oscillation.The author considered the system of state equations when only the variable names wereα;α1 as follows:(27)ẋ1ẋ2ẋ3ẋ4ẋ5ẋ6ẋ7ẋ8=00100000000100000KsJeq0000000−Ks1Jeq+1JL000000000000100000000100000KsJeq0000000−Ks1Jeq+1JL00x1x2x3x4x5x6x7x8y1y2y3y4=10000000010000000000100000000100x1x2x3x4x5x6x7x8+00ηgKgηmKtJeqRm−ηgKgηmKtJeqRm00ηgKgηmKtJeqRm−ηgKgηmKtJeqRmVmẋ1ẋ2ẋ3ẋ4ẋ5ẋ6ẋ7ẋ8=00100000000100000623.70000000−965.3000000000000100000000100000623.70000000−965.300x1x2x3x4x5x6x7x8y1y2y3y4=10000000010000000000100000000100x1x2x3x4x5x6x7x8+00479.8−479.800479.8−479.8VmThe author applied the algorithm according to the controller design of the SISOs system for the oscillation corresponding to the MIMO system with the state equation presented above.The transfer functionGα2s:(28)Gα2s=479.8s2+1.79×10−11s+1.639×105s4+965.3×s2Aα2=0−30.17003200001000010;Bα2=64000;Cα2=00.23432.913×10−1580.03The author consider the system to have the external impactu≠0:(29)ẋ=Aα2x+Bα2uI need to find the matrixKα2 of the optimal control vector: ut=−Kα2∗xt satisfy the quality index value of Jα2 and Jα2 must reach the minimum value:(30)Jα2=∫0∞xTQα2x+uTRα2udtwhere Qα2 is a positive deterministic matrix, Rα2 is a positive deterministic matrix.The matrix ofKα2 is determined from Riccati’s equation of the form:(31)Kα2=Rα2−1Bα2TPThe state feedback control structure is shown below (Figure9).Figure 9
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(32)ut=−Kα2t=−Rα2−1Bα2TPxtThe value of the matrix is P, P must be satisfied the equation:(33)PAα2+Aα2TP+Qα2−PBα2Rα2−1Bα2TP=ṖThe equation (33) is known as Riccati’s equation.The author choose the values of the matrixQα2 and the values of the matrix Rα2 as follows:(34)Qα2=1000010000100001;Rα2=1.The author calculate the value of the matrixKα2 through Matlab software:(35)Kα2=lqrAα2,Bα2,Qα2,Rα2Gα3s=−479.8s2+965.3Aα3=0−30.17320;Bα3=40;Cα3=0−3.748The author consider the system to have the external impactu≠0:(36)ẋ=Aα3x+Bα3uThe author need to find the matrixKα3 of the optimal control vector: ut=−Kα3∗xt satisfy the quality index value of Jα3 and Jα3 must reach the minimum value:(37)Jα3=∫0∞xTQα3x+uTRα3udtwhere Qα3 is a positive deterministic matrix, Rα3 is a positive deterministic matrix.The matrix ofKα3 is determined from Riccati’s equation of the form:(38)Kα3=Rα3−1Bα3TPThe state feedback control structure is shown below (Figure10).Figure 10
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(39)ut=−Kα3t=−Rα3−1Bα3TPxtThe value of the matrix is P, P must be satisfied the equation:(40)PAα3+Aα3TP+Qα3−PBα3Rα3−1Bα3TP=ṖThe equation (40) is known as Riccati’s equation.I choose the values of the matrixQα3 and the values of the matrix Rα3 as follows:(41)Qα3=1001;Rα3=1.Then I calculate the value of the matrixKα3 through Matlab software:(42)Kα3=lqrAα3,Bα3,Qα3,Rα3=1.43390.066Gα4s=479.8s2+5.966×10−12s+1.639×105s4+965.3s2Aα4=0−30.17003200001000010;Bα4=64000;Cα4=00.23432.913×10−1580.03I consider the system to have the external impactu≠0:(43)ẋ=Aα4x+Bα4uI need to find the matrixKα4 of the optimal control vector: ut=−Kα4∗xt satisfy the quality index value of Jα4 and Jα4 must reach the minimum value:(44)Jα4=∫0∞xTQα4x+uTRα4udtwhere Qα4 is a positive deterministic matrix, Rα4 is a positive deterministic matrix.The matrix ofKα4 is determined from Riccati’s equation of the form:(45)Kα4=Rα4−1Bα4TPThe state feedback control structure is shown below (Figure11).Figure 11
The state feedback control structure.Thus, the optimal control law for an optimal control problem with the quality criteria is a linear equation and it has the form:(46)ut=−Kα4t=−Rα4−1Bα4TPxtThe value of the matrix is P, P must be satisfied the equation:(47)PAα4+Aα4TP+Qα4−PBα4Rα4−1Bα4TP=ṖThe equation (47) is known as Riccati’s equation.I choose the values of the matrixQα4 and the values of the matrix Rα4 as follows:(48)Qα4=1000010000100001,Rα4=1.Then I calculate the value of the matrixKα4 through Matlab software:(49)Kα4=lqrAα4,Bα4,Qα4,Rα4=1.30350.69911.82791.0000Figures12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 are shown below.Figure 12
The model without using any algorithmsGα11s.Figure 13
The model without using any algorithmsGα12s.Figure 14
The model without using any algorithmsGα2s.Figure 15
The model without using any algorithmsGα3s.Figure 16
The model without using any algorithmsGα4s.Figure 17
Applying the optimal LQR control method for the systemGα1s.Figure 18
The system with the optimal controllerGα1s.Figure 19
Applying the optimal LQR control method for the systemGα3s.Figure 20
The system with the optimal controllerGα3s.Figure 21
Applying the optimal LQR control method for the systemGα4s,Gα2s.Figure 22
The system with the optimal controllerGα4s,Gα2s.


## 4. Simulation Results and Discussions

Simulation results are shown Figures23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33.Figure 23
The simulation result without using any algorithmsGα11s.Figure 24
The simulation result without using any algorithmsGα12s.Figure 25
The simulation result without using any algorithmsGα2s.Figure 26
The simulation result without using any algorithmsGα3s.Figure 27
The simulation result without using any algorithmsGα4s.Figure 28
The simulation result using optimal controlGα1s. Initial condition: x1 = 0.04, x2 = 0.03, x3 = 0.03, x4 = 0.03.Figure 29
The simulation result using optimal controlGα1s. Initial condition: x1 = 0.01, x2 = 0.01, x3 = 0.01, x4 = 0.01.Figure 30
The simulation result using optimal controlGα3s. Initial condition: x1 = 0.01, x2 = 0.01.Figure 31
The simulation result using optimal controlGα3s. Initial condition: x1 = 0.04, x2 = 0.03.Figure 32
The simulation result using optimal controlGα4s,Gα2s. Initial condition: x1 = 0.01, x2 = 0.01, x3 = 0.01, x4 = 0.01.Figure 33
The simulation result using optimal controlGα4s,Gα2s. Initial condition: x1 = 0.04, x2 = 0.03, x3 = 0.03, x4 = 0.03.My comments:Figures23, 24, 25, 26, 27 show that the amplitude of oscillation is large.From Figures28, 29, 30, 31, 32, 33 of the simulation results show me with the optimal controller, it will help stabilize the output signal of the system. However, if the input signal is large (Figures 28, 31, 33), then the oscillation range of the signal amplitude at the output is high before it is stabilized. The result in Figures 29, 30, 32 is better than in Figures 28, 31, 33. The simulation results accurately reflect the problem: the optimal controller performs better when there is no external force applied to the system, for example: noise levels of other signals ..The optimal controller used in this case is more efficient than the simulation result without using any algorithms.From Figures28, 29 and Figures 32, 33 it is shown that the system with 2 input pairs has slightly higher amplitude of oscillation than the system with one input pair in using the same controller. This shows that the more complex the system, the larger the amplitude of the vibration compared to the simple structure. Furthermore, this system with complex structure takes faster to return to steady state (2 sec) this the system with simpler structure (4 sec).


## 5. Conclusions

In this paper, it is proposed to investigate the operation of a flexible link system with 2 input pairs based on the simulation results to help the system achieve the desired functions. Besides, the simulation helps me to determine the values of the parameters to have the basis to control the operation of the system to suit the requirements. This also shows the flexibility in adjusting the parameters of modern controllers as it is applied to the above model. Through this survey, the research achievements on modern control theory for systems with relatively complex structure like the above system will be applied in the future. This method is useful for checking the operational status of all the existing components in a system. In the case of possible problems such as a failure of elements inside a server, this survey is a reference for quality control as well as the functions of these elements. For complex structured systems, the author can use control methods for each of the transfer functions in the host system. From there, the author took insight into the problem and the author was able to evaluate the host system’s properties most accurately. In the future, the author can apply modern control algorithms like neural control in above model. Specifically, a neural control method is applied: a model with 1 input pair, a model with 2 input pairs...After that, the author evaluate the effect of the neural network for the above models. With the above controller applied in this paper, its efficiency has been tested for most of the subsystems in a system. This article is a premise for me to study for other complex models based on the above implemented contents. Another advantage of using the above controller for this model is that the time to steady state for the system is faster for the complex structures (2 input pairs) than for the simpler structure (1 input pair). The time to return to steady state is negligible (~0.1 sec -2 sec) for a system with 2 input pairs. This rarely happens for a system with a complex structure.



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*Source: 1880822-2021-08-26.xml*