Designing of a Controller for Hydraulic Suspension Test Facility

7/28/2019 Designing of a Controller for Hydraulic Suspension Test Facility

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CONSULTANCY PROPOSAL ON DESIGNING OF ACONTROLLER FOR HAUDRAULLIC SUSPENSION

TEST FACILITY

To : Dr Director,

Proper Shafts Ltd.

GROUP T16

REF : XF-16-T16

Prepared by:

Consulted by:

Muhammad Naim Bin Abdul Talib 13729 (Chief Eng.)

Muhamad Hanif Muqsit Bin Azhar 13185

Muhammad Hariz Bin Mohd. Noor 13954

Luqman Hakim Bin Zulkefli 13861

Ahmad Yazid Azizy Bin Mohamad Idris 13630

Suraya Nabilah Binti Zaini 13917

2012

SUPER SUSPENSION LTD.

Technology Beyond Boundary

7/15/2012


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MAB 3023 / MBB 2094 Group Project 1

Form B

Group Assessment Form

Team Name: Super Suspension Ltd. (Group T16)

Overall Group Mark:

Student Name &

Signature

Peer Assessment Mod.

Factor

Final

Mark1 2 3 4 5 6

1. Muhammad Naim BAbdul Talib

2. Muhammad Hariz B

Mohd Noor

3. Muhamad Hanif

Muqsit B Azhar

4. Luqman Hakim B

Zulkifli

5. Ahmad Yazid Azizy B

Mohamad Idris

6. Suraya Nabilah Bt

Zaini


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TASK A1: EXPLANATION OF THE GIVEN TRANSFER FUNTIONS

a) Show the Transfer Function GA(S) Relating Displacement, X, to Oil Flow from the Valve, Qin, Is

Third Order:

---------------------------- (1)

--------------------------- (2)

------------------------- (3)Substitute equation(2) Into equation (3), thus we get;

------------------------ (4)Substitute equation(4) Into equation (1), thus we get;

------------ multiplywith a

]

-------- multiply with

, thus we get;


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b)Prove that

Neglect oil leakage; ---------------------------------- (1)

--------------------------- (2)

------------------------- (3)

Substitute equation (2) Into equation (3), thus we get;

------------------------ (4)Substitute equation(4) Into equation (1), thus we get;

--------------- multiply with

, thus it is prove that;


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c) Show that the overall forward path transfer function neglecting the oil leakage is

Based on previous equation we know that;

So , G(s) will be equal to ;

TASK A2: BLOCK DIAGRAM

a) Block Diagram Model of the Testing Facility System.

1) Valve Component

2) Actuator Component


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3) 1/4 Car Component


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b) Simulink Model


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TASK B: CRITICAL ANALYSIS OF THE OUTPUT RESPONSE

a) Analysing the Forward Path Transfer Function and the Type of Response.

These are the specification details from Proper Shaft Ltd. :

1) The Electro-Hydraulic Actuator:

Gain, Av = 1.138 x 10-2

Time constant, Tv = 6.2 sec

2) The Hydraulic Actuator:

Cross sectional area, a = 8 x 10-2

m2

Volume, V = 3.2 x 10-2

m3

Bulk modulus, B = 1.5 M N/m2

3) The displacement transducer:

Ah = 1 volts/m

The car model has the following specifications:

Mass, m = 300 kg

Damper coefficient, c = 2000 N s/m

Spring constant, k = 2700 N/m

Forward Path Transfer Function:

G(s) =

=

We are assuming that the characteristic equation below;

behave like order system.By putting the value give by Proper Shaft Ltd. Thus, we get :

= 300 + 2000s + 302700We are using MATLAB Software to find the roots of this equation. Thus, the roots are;


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This is the S-Plane for the characteristic equation;

Based on the S-plane we can see that we have 2 complex poles, hence It is underdamped response.

Explanation:

Pole: 2 Complex poles at -3.3333Natural Response: Damped sinusoid with an exponential envelope whose time constant equal to

reciprocal of the poles real part. The radian frequency is equal to the imaginary part of the poles. In

this case the solution can be generally written as C(t) = A .General Second Order System;

G(s)=

Comparing the characteristic equation

We are assuming the characteristic equation behave like 2nd

order system.

Equating the left and right component to get:

-3.3333

31.5894

-31.5894


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2 =

Since we noticed that the damping ratio is between 0 and 1, it is proven that the transientresponse is underdamped.

To clarify the type of response, we have used MATLAB Software to analyze the response by putting

the specification details into the Simulink Model that we have made in the PART A2.

This the how the response looks like by using MATLAB Software;

Figure 1 : Transient Response Diagram (Open Loop)


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Performance measures for underdamped second-order system:

Rise time,

Setting time,

Peak time,

Percentage %OS:

= exp = 0.7% 0%Since we are approximating there is no overshoot, we need to introduce overshoot 0 < % OS < 10%

in order to avoid any breakdown to the suspension. This will be explained details in PART C.

b) Critical Comment of the observed response of displacement, x.

We can see that the response of displacement, x is underdamped response. An underdamped

response is one that oscillates within a decaying envelope. The more underdamped the system, the

more oscillations and longer it takes to reach steady-state. Based on the calculation made before,

the damping ratio is


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))(1(

.

)(2

2

V

BakcsmsTvss

V

aBAv

sG

)30000027002000300)(2.61(

426752

ssss

)3027002000300)(2.6(

4267522

ssss

23423 18767401240018603027002000300

42675

ssssss

ssss 3027001878740127001860

42675234

TASK C: CONTROLLER DESIGN

a) Using A Simple Gain Controller, Gc(S)=K Which Gives A %OS Of Less Than 10 %

Below is the forward path transfer function, G(s) with value based on the specification details given

by Proper Shaft Ltd.;

+ pure gain system

-

Kp

1

G(s)


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,)(1

)()('

sGHG

sGGsG

c

c

,)( pc KsG 1)( sH

)375270023268740127001860

426751(

)375270023268740127001860

42675(

234

234

ssss

Kssss

K

p

p

p

p

Kssss

K

426753027001878740127001860

42675

234

pKssss 426753027001878740127001860234

The overall transfer function, G(s) is;

Characteristic Equation

Time-Domain Analysis and Design Method

By using Final Value Theory ;

= = = 0Based on the Final Value theory we can say that, the system has zero steady state error.


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pKssss 426753027001878740127001860234

Below is the table to determine either the system has oscillatory response or not;

Kp S1 S2 S3 S4

0.5 -3.3334 +31.5892i -3.3334 -31.5892i -0.0806 + 0.0698i -0.0806 - 0.0698i

1.0 -3.3334 +31.5890i -3.3334 -31.5890i -0.0806 + 0.1275i -0.0806 - 0.1275i

1.5 -3.3334 +31.5889i -3.3334 -31.5889i -0.0805 + 0.1662i -0.0805 - 0.1662i2.0 -3.3335 +31.5887i -3.3335 -31.5887i -0.0805 + 0.1975i -0.0805 - 0.1975i

2.5 -3.3335 +31.5885i -3.3335 -31.5885i -0.0805 + 0.2244i -0.0805 - 0.2244i

3.0 -3.3336 +31.5883i -3.3336 -31.5883i -0.0804 + 0.2485i -0.0804 - 0.2485i

3.5 -3.3336 +31.5881i -3.3336 -31.5881i -0.0804 + 0.2704i -0.0804 - 0.2704i

4.0 -3.3336 +31.5880i -3.3336 -31.5880i -0.0804 + 0.2907i -0.0804 - 0.2907i

4.5 -3.3337 +31.5878i -3.3337 -31.5878i -0.0803 + 0.3097i -0.0803 - 0.3097i

5.0 -3.3337 +31.5876i -3.3337 -31.5876i -0.0803 + 0.3275i -0.0803 - 0.3275i

5.5 -3.3337 +31.5874i -3.3337 -31.5874i -0.0802 + 0.3444i -0.0802 - 0.3444i

6.0 -3.3338 +31.5873i -3.3338 -31.5873i -0.0802 + 0.3606i -0.0802 - 0.3606i

Figure 2: Table to determine oscillatory of the response

From the table above we can see that as the forward path gain is increased the controller

proportional action, the system become more oscillatory since imaginary component of the root S2,

S3, S4 are increasing.

In order to determine the stability of the system, we need to use Routh-Hurwitz Criterion.

Below are the steps to determine the systems stability;

Characteristics equation as polynomial

= s4

+ s3+ 1010s

2+ s +

Kp

S4

1 1010 22.94Kp

S3

635/93 5045/31 0

S2

986.17 22.94Kp 0

S1

162.74-0.1588Kp 0

To find the range of Kp we have to equate;

22.94Kp>0 and 162.74-0.1588Kp>0

-0.1588Kp>-162.74

Kp < 1024.8

Thus , we have known that the range of Kp in the controller of this system must be less than 1024.8

in order the system to stable. If the Kp is greater than 1024.8 the system will no longer stable.


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Frequency-Domain Analysis and Design Method

Below is the damping ratio calculated by the following equation;

Note: we have chosen %OS to be 8% which is less than 10% as required.

= 0.6266

Find ;

= = = 60.877Locate the phase plot where the = 60.877

= -180 + 60.877 = -119.123 = -119.123 -119

By using MATLAB, get the bode plote in order to determine the frequency, w.

Below is the bode plot created using the MATLAB;

Figure 3: Bode Plot


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Based on the bode plot we get the frequency , =0.0904 rad/swhen the phase margin = 119. At -180, the magnitude is -50dB which is below the 0dB line, therefore the system is very stable.Now locate the magnitude at=0.0904 rad/s from the bode plot, we get magnitude is equal to -2.61dB. Thus, from that we can get the value of Kp ;

Kp=1.35

b) The Effect of the Designed Controller to the Output Response of the Closed-Loop System

Based on the Time-Domain and Frequency-domain method we have used before, the most suitable

gain, Kp for the controller is 1.35. In figure shows how is the designed controller affect the output

response of the system.

Figure 4: The Output Response with the Controller

From the figure above, we can see that the %OS is less than 10% which is about 8%OS. Thus, the

controller is valid to the requirement from Proper Shaft Ltd.


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C) How the Changes in Parameter Values of the Car Could Affect the Designed Controller?

In order to determine the effect of changing parameter values of the car to the controller, we have

to analyse the output response one by one parameters.

The parameter of the car part are:

c = damping ratio

k= spring constant

m = mass

Now, we analyse the output response with respect to changing in parameter value of the car.

Test 1- Mass is varied meanwhile damping ratio,c and spring constant,k are fixed.

With m= 100 kg the output response with step input ;

Figure 5: Output response with changing mass = 100kg


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With m= 900 kg the output response with step input is :

m = 700 kg


With m= 2500 kg the output response with step input is :



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Test 2- Damping ratio is varied meanwhile mass,m and spring constant,k are fixed.

With c= 500 Ns/m the output response with step input is :

Figure 8: Output response with changing damping ratio= 500Ns/m

With c= 5000 Ns/m the output response with step input is :

Figure 9: Output response with changing damping ratio= 5000Ns/m


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Test 3- Spring constant,k is varied meanwhile mass,m and damping ratio,c are fixed.

With k= 1500 N/m the output response with step input is :

Figure 10: Output response with changing spring constant= 1500N/m

With k= 5000 N/m the output response with step input is :

Figure 11: Output response with changing spring constant= 5000N/m

Since all the output responses are same with changing in parameter value, we can conclude that the

controller that we have designed with Kp=1.35 is so stable. The changing in parameter values of the

car would not give any effect to the output response of the system. Thus, this controller can beused by Proper Shaft Ltd. to the other vehicles.


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INVOICE

Super Suspension Ltd.Technology Beyond Boundary INVOICE #1

DATE: JULY 15, 2012

TO: Dr Director,

Proper Shafts Ltd.,

[email protected]

Ref : xf-16-T16

SALESPERSON JOB PAYMENT TERMS DUE DATE

Mr Muhammad Naim Chief Engineer Due on receipt July 30,2012

NO. DESCRIPTION UNIT PRICE LINE TOTAL

1

Providing consultancy service in a project entitled Design

of a Controller for a Hydraullic Suspension Test Facility.

Note: 6 Consultants have been selected to provide the

serviceRM 3500 RM3500

SUBTOTAL

SALES TAX

TOTAL RM3500.00

Make all checks payable to Super Suspension Ltd.

THANK YOU FOR YOUR BUSINESS!

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Designing of a Controller for Hydraulic Suspension Test Facility