modeling system

Eatern International University

School of Engineering

REPORT 1

MODELING SYSTEM

SUBJECT: AUTOMATIC CONTROL

INSTRUCTION: HUỲNH CÔNG HẢO

STUDENT: TỪ VƯƠNG BẢO NGỌC

CLASS: 11ACE3101

IRN’s: 113 110 0003

DAY: 05/03/2014

Report 1: Modeling System

2 Subject: Automatic Control Instructor: Huỳnh Công Hảo

REPORT 01:

MODELING SYSTEM

I. 2nd Problem:

Modeling below sytem:

2rd Problem

1.1 Forces in Sytem:

- External Force T=T(t)+T0(M). T0(M) is the Torque created by Motor to

balance the Force before the cabin moving.

- Gravity of Cabin Mc.

- Gravity of Counter Weight M.

Because the T0 is unchanged when the cabin is moving. Moreover, if we

consider T0, this will be Differential Equation with 2 variables. Therefore, we

can ignore the T0 while cabin moving.

1.2 Analyze Forces in Sytem:

𝑇(𝑡) + 𝑇0(𝑀) + 𝑀𝐶 . 𝑔 − 𝑀. 𝑔 = (𝑀𝑐 + 𝑀). 𝑥′′(𝑡)

Initially, all force are balanced: 𝑇0(𝑀) + 𝑀𝐶 . 𝑔 − 𝑀. 𝑔 = 0. Therefore:

𝑇(𝑡) = (𝑀𝑐 + 𝑀). 𝑥′′(𝑡)

1.3 Using Laplace Transform:

𝑇(𝑠) = (𝑀𝑐 + 𝑀). 𝑠2. 𝑋(𝑠)

𝑇(𝑠) = (𝑀𝑐 + 𝑀). 𝑠. 𝑉(𝑠)

Transfer Function:

𝐻(𝑠) =𝑋(𝑠)

𝑇(𝑠)=

1

(𝑀𝑐 + 𝑀). 𝑠2; 𝐺(𝑠) =

𝑉(𝑠)

𝑇(𝑠)=

1

(𝑀𝑐 + 𝑀). 𝑠



1.4 Using Simulink for Distance Simulations:

M+Mc=30 M+Mc=50

M+Mc=70

Ramp

Step

Pulse

Ramp

Step

Pulse

Ramp

Step

Pulse



1.5 Using Simulink for Velocity Simulations:

M+Mc=30 M+Mc=50

M+Mc=70

Ramp

Step

Pulse

Ramp

Step

Pulse

Ramp

Step

Pulse



1.6 Evaluation:

From the graphs, we can conclude:

- With the same forces, the bigger volume of cabin and counter weight, the

slower the speed and the shorter distance it reach.

- With Ramp Respone: the speed-time graph is a parabol, increasing faster over

the time (due to the increasing of the force).

- With Step Respone: the speed-time graph is a line, increasing the same value

over the time (due to the unchanged force).

- With Pulse Respone: the speed-time graph is not stable, when there a a pulse,

speed increasing while in the off duty of pulse, the speed is unchanged.



II. 3rd & 4th Problem:


3rd Problem 4th Problem


In 3rd Problem, there are:

- External Force f(t).

- Elastic Force k*x.

- Friction Force of Damper B*v.

In 4th Problem, there are:

- External Force f(t).

- Elastic Force 2*k*x. Because of 2 springs paralleled.

- Friction Force of Dampers 2*B*v. Because of 2 Dampers paralleled.

So, we can see there are the same functions for this 2 systems, the difference is

the strength of Forces.


𝑓(𝑡) − 𝑘. 𝑥(𝑡) − 𝑏. 𝑥′(𝑡) = 𝑀𝑥′′(𝑡)

2.3 Using Laplace Transform:

𝐹(𝑠) − 𝑘. 𝑋(𝑠) − 𝑏. 𝑠. 𝑋(𝑠) = 𝑀. 𝑠2. 𝑋(𝑠)

𝐹(𝑠) − 𝑘.1

𝑠. 𝑉(𝑠) − 𝑏. 𝑉(𝑠) = 𝑀. 𝑠. 𝑉(𝑠)

Transfer Function:

𝐻(𝑠) =𝑋(𝑠)

𝐹(𝑠)=

1

𝑀. 𝑠2 + 𝑏. 𝑠 + 𝑘 ; 𝐺(𝑠) =

𝑉(𝑠)

𝐹(𝑠)=

𝑠

𝑀. 𝑠2 + 𝑏. 𝑠 + 𝑘



2.4 Using Simulink for Distance Simulations:

M=2; b=0.4; k=1 M=2; b=1; k=1

M=2; b=1; k=2 M=4; b=1; k=1

Ramp

Step

Pulse

Ramp

Step

Pulse

Ramp

Step

Pulse

Ramp

Step

Pulse



2.5 Using Simulink for Velocity Simulations:

M=2; b=0.4; k=1 M=2; b=1; k=1

M=2; b=1; k=2 M=4; b=1; k=1

Ramp

Step

Pulse

Ramp

Step

Pulse

Ramp

Step

Pulse

Ramp

Step

Pulse



2.6 Evaluation:

From the graphs, we can conclude the effects of variables:

- The heavier the object is (M), the less fluctuation.

- The bigger b value, the less fluctuation. Dampers consume the power while

moving.

- The bigger k value, the more fluctuation. Springs absorb the power then

reflect totally back.

From the graphs, we can conclude the effects of Forces:

- In Step and Pulse Respone, the object is fluctuated around the Y Point. Y

Point is the Point that the Sum of All Force affected system is 0.

- While in Ramp Respone, the object is also fluctuated but the trend is upward.

(the External Force is stronger gradually, so springs and dampers can’t stand

against it. As result, the object is move far away from 0 point).



III. 1st Problem:


1st Problem


- Height of 1st tank: 𝒉𝟏 = 𝑸(𝒕)/(𝒂. 𝒃)

- Water velocity in the pipe connecting 1st tank to 2nd tank:

Appying Torricelli’s law: 𝒗(𝒕) = √𝟐. 𝒈. 𝒉𝟏(𝒕)

- Water flowing from 1st tank to 2nd tank: 𝑸𝟐(𝒕) = 𝒗(𝒕). 𝑺𝑷

SP is the section of the pipe.

- Height of 2nd tank: 𝒉𝟐 = 𝑸𝟐(𝒕)/𝑺𝟐

S2 is the 2nd tank bottom area.


Applying these abow functions

ℎ2(𝑡) =2. 𝑆𝑝

2. 𝑔

𝑆𝑡. 𝑎. 𝑏. √𝑄(𝑡)

modeling system

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