代做ENGE2002 Dynamics, Instrumentation and Control 2024/25代做留学生Matlab程序

Faculty of Computing, Engineering & Media (CEM)

Coursework Brief 2024/25

Module name:

Dynamics, Instrumentation and Control

Module code:

ENGE2002

Title of the Assessment:

DC Motor Speed and Position Control & PLC

This coursework item is: (delete as appropriate)

Summative

 

This summative coursework will be marked anonymously: (delete as appropriate)

Yes

 

The learning outcomes that are assessed by this coursework are:

1.  A student will understand and be able to analyse analog and digital feedback systems

2.  A student will be able to choose the control system which is fit for the purpose

3.  A student will be able to tune a controller for analog and digital feedback system using computer aided design software

1.  Objectives

•    Design a PI controller for a DC motor speed control.

•    Design a PD controller for a DC motor position control.

•    Examine the behaviour of the PID closed-loop system when it is subjected to a disturbance.

There are six sessions (2 Practical ) to complete these tasks followed up with a submission of a report documenting your work.

A  report  documenting  your  work  must  be  submitted  according to  deadline specified on LZ . The details of report content and submission   together  with the marking rubric are detailed at the end of the tasks.

2.  Theoretical background

a.  Speed control

The speed of the  DC  motor  is  controlled  in  our  system  by  a  proportional- integral  (PI)  controller  with  the  so-called,  `set-point'  weighting.  The  block diagram of the closed-loop system is shown in Figure 1.


Figure 1. DC motor PI speed control block diagram

kp  is the  proportional gain, ki  is the integral gain, s/1 is a transfer function of integration, vd  denotes disturbance,r is the reference signal,u is the control signal,y is the output signal and is equal to the rotational speed measurement ω , Td  denotes disturbance torque, vd  denotes disturbance voltage, and vm denotes the input voltage to the DC motor.Jeq  represents the motor moment of inertia.

Dynamics of the DC motor, i.e. the relationship between its speed and voltage in time is represented by the first-order transfer function shown in Equation (1).  As  you  may  remember,  this  equation  was  introduced  in  the  previous laboratory session.


The symbol τ in  Equation  (1)  is a time constant whilst K is  referred to as a steady-state gain.

The (closed loop) transfer function of the system inFigure 1 can be derived through a series of block diagram simplifications or by implementing Mason's Gain Rule. Derivation of transfer functions for block diagrams of such levels of complexity is beyond the scope of our lectures so, for now, please take it for granted that our closed-loop transfer function Gω ,r  from the speed reference, r , to the angular motor speed output, ωm , is described with the equation below.


b.  Position control

The position of the DC motor is controlled by a proportional-integral-derivative controller (PID) and is designed according to specifications. The closed-loop PID control block diagram is shown in Figure 1. 


Figure 2.  DC motor PID position control block diagram

where the parameters are the same as in Figure 1 and  kd  is the derivative gain.

Dynamics of the DC motor, i.e. the relationship between its speed and voltage in time can be represented by the first-order transfer function. As a reminder, τ and K, denote, respectively, the process time constant and the steady-state gain.


However, this time the position dynamics of the DC motor are described by Equation (3), which features an additional integrator s/1,  i.e. zero pole in the denominator.  Where  does  this  integrator  come  from?  Please  refer  to  the lecture notes on speed and position control for clues.

c.  Performance metrics

The second-order transfer function in a standard (canonical) form is given in Equation (4).


where  ωn  is  called  the   natural  undamped  frequency   and  ζ  is  called  the damping  ratio.  These   two parameters  determine  the  shape  of  its  step response. A standard system response of a second order system to step input 

R(s) = s/R0  where R0  is the step amplitude and R0 = 1.5 is shown in Figure 3 .     The red trace shows the response (output), y(t), and the blue trace is the step input r(t). 


Figure 3. Standard second order step response

As you  may observe, the step  response of a second-order system  in Figure 3 looks quite different from the step response of a first order system which we analysed in the previous lab.

The first thing to notice is that the response y(t) is oscillatory and at some time goes above the reference input r(t). The maximum value of the response is denoted by the variable ymax  and it occurs at a time tmax.

The maximum `distance' that the response travels above the reference signal relative to the magnitude of the step is called percentage  overshoot (PO). It is usually measured in % and can be calculated with Equation (5)  given below. 


The time it takes for the response to reach its maximum value is called the peak time of the system. As can be deduced fromFigure 3, peak time tp  can be calculated as follows.


As the system may develop oscillations (visible inFigure 3), which we will later see happens when the damping ratio ζ < 1, it is also important to know how long it takes for the system to settle to a new reference setpoint.

This property of the system is measured by a parameter called the settling time (ts ) which is defined as the time required for the response curve to reach and stay within a range of a certain percentage (usually 5% or 2%) of the final value.

Another parameter we might measure is the steady-state error ess  which is defined as the difference between the input (command) and the output of a system as time goes to infinity (i.e. when the response has reached a steady state). Hence ess  can be calculated as follows.


Each  one  of  the  above  parameters  characterises  different  aspects  of  our system response and hence the above parameters are often used to define the design criteria for a controller to be designed.

Therefore, it is vital that we can estimate or, better, to calculate them before the controller is installed and then commissioned. For a second order system, such as the ones described by Equation (2) and Equation (4), we will see that Os, tp, and ts  depend on the values of the  natural undamped frequency ωn and the damping ratio ζ .

If  the  transfer  function  of  our  system  can  be  represented  in  a  standard (canonical) form, i.e. with Equation (4), then we can calculate Os, tp  and ts analytically.


As we can see the amount of overshoot depends solely on the damping ratio parameter ζ .

The peak time depends on both the damping ratio ζ and natural frequency of the system ωn.


The settling time ts for a 2nd order, underdamped system responding to a step response cannot be calculated exactly but can be approximated if the damping ratio ζ ≪ 1. If we assume a tolerance level of 5%, ts  can be approximated with the equation below.


The  damping  ratio  affects  the  shape   of  the   response  while  the   natural frequency affects the speed of the response.

Now, let's go back to our closed-loop transfer function of a DC-motor and PI controller described with Equation (2). If we try to match the denominator of our transfer function, i.e. τ s 2  + (kkp  + 1) s + kki  with the denominator of the second order canonical system, i.e. s 2  + 2 ζ ωns + ωn(2) we will find that both polynomials will be equal if:


and


d.  Response to disturbance signal

Based on the block diagram in Figure 1we can derive two open-loop and two closed-loop transfer functions for the system. The open-loop transfer function Gθ(o),Td(s) describes  the  open-loop  response  of  the  system  to  disturbance, whilst the transfer function Gθ(o),r (s) describes the open-loop response of the system to the reference signal.

In general, we can derive a closed-loop transfer function from the open-loop transfer function using the following equation:


where H(s) represents the system on the feedback path. Hence, with help of Equation  (13)  the  closed-loop  transfer  functions  Gθ,Td(s)  and  Gθ,r (s)  are derived below:


Using the Final-Value Theorem to evaluate Equation (14)


given a step disturbance with an amplitude of Td0


the steady state angle of the closed-loop system is


If we solve Equation (18) for a PD controller, i.e. when ki   = 0, we will see that the steady state angle θss  is non-zero.


When integral action is added, i.e. full PID controller is used then ki   ≠ 0 and


We can carry out the same reasoning for the step change in the reference signal. Given a step disturbance with an amplitude of R0


the steady state angle of the closed-loop system is


If we solve Equation (22) for a PD controller, i.e. when ki   = 0 we will see that


Therefore, we can see that in order to reject disturbances we need to have a PID controller in place but to track the reference signal with zero error the PID controller is not needed. The reason for that is that the integral action required to eliminate the steady state error is already featured in the DC motor transfer function - see Equation (3).

3.  Tasks

The DC motor parameters

The  parameters of the  DC  motor are  listed  in Table 1, you will need them in the subsequent exercises.

Table 1 DC motor parameters

Parameter

Symbol

Value

Unit

Model gain used

K

31.0

rad/s

Model time constant

τ

0.155

s

a.  Task 1 - Speed control modelling in Simulink/MATLAB

The first task is to create a Simulink model of the system depicted in Figure 4. During the laboratory session, the instructor will explain how to create such model  however  you  are  encouraged  to  create  this  model  yourself.  As the diagram in Figure 4 should represent the actual system, i.e.  DC  Motor Control Trainer, the modelling process needs to account for an offset in the reference   signal,   a   limitation   in   the   input   signal  <-10,  +10>   and   the measurement noise around  ±5  rad/s.  The  final  model  should  be  similar  to the  model shown in Table 1.


Figure 4 . Speed control model

Once the  model  is  ready  you  need  to  understand  how  to  simulate  it  and adjust  its  parameters,  e.g.  kp   or  ki   gains.  The  model  should  be  simulated with the fixed step of 0.01s.

Important skills to acquire during this activity are generating and annotating figures, measuring values from graphs, exporting graphs and data. These skills are crucial for the successful completion of the subsequent tasks.

b.  Task 2 - Speed control: Qualitative PI Control

1.   To complete this task, you need the model created in Task 1.

2.   Generate a reference signal according to the following specification:

•   Signal Type = ‘square wave’

•   Amplitude = 25 rad/s

•    Frequency = 0.25 Hz

•         Offset = 25 rad/s

3.   Set the gains and the model parameters as follows:

•   kp  = 0.02 (Vs)/rad

•   ki  = 1.00 V/rad

•   vd  = 0 V

•   K = 31 rad/s

•   τ = 0.155 s

4.   Examine  the   behaviour  of  the   measured   speed,  with   respect  to  the reference speed in the scope. Measure percentage overshoot, peak time and settling time and put the values in Table 2. Prepare a graph of the step response to be included in the report. Make sure you include both the Speed (rad/s) and the control signal Voltage (V) scopes.

Table 2. Step response results.


Parameter

Symbol

Value

Unit

Proportional gain

kp

 

Vs/rad

Integral gain

ki

 

V/rad

Measured peak time

tp

 

s

Measured percentage overshoot

PO

 

%

Measured settling time

ts

 

s

5.   Increment kp   by  small  steps, e.g. 0.01  (V·s)/rad. Record step responses (i.e. save figures and export data) for values of kp  (0.05, 0.1  and 0.3). Make  sure  you  include  both  the  Speed  (rad/s)  and  the  control  signal Voltage (V) scopes.

6.   Look at the changes in the measured signal with respect to the reference signal. Investigate the performance difference of changing kp  in terms of the   percent   overshoot,   settling   time,   peak   time.    Note   down   your observations.

7.   Set kp  to 0.00 (V·s)/rad and ki  to 0.00 V/rad.

8.   Increment the integral gain, ki  ,  by steps of 0.05 V/rad. Vary the integral gain  between 0.05 V/rad  and  2.00 V/rad.  Record  step  responses  (save figures) for values of ki  (0.2, 0.5 and 2.00).  Make sure you include both the Speed (rad/s) and the control signal Voltage (V) scopes.

9.   Examine the response of the measured speed in the Speed (rad/s) scope and  compare  the   results  when  ki    is  set   low  to  when   it   is  set   high. Investigate  the  performance  difference  of  changing  ki   in  terms  of  the percent overshoot, settling time, peak time. Note down your observations.

c.  Task 3 - Speed control: PI Control According to Specifications

Pre-experiment calculations

1.   Using the equations for the overshoot  (Equation (8)) and the peak time (Equation  (9)),  calculate  the  expected  peak  time, tp ,  and  percent overshoot, PO, given the following closed loop system specifications.

•    ζ = 0.75

•    ωn  = 16.0 rad/s

2.   Calculate  the  proportional, kp  ,  and  integral, ki   using  Equation  (11) and  Equation  (12),  respectively.  The  electric  DC  motor  parameters provided  by the  manufacturer  are as follows: steady-state gain K = 31.0 rad/s, time constant τ = 0.155 s. Alternatively, the parameters K and τ for the specific board can be found by carrying out the bump test described in the first laboratory session handouts.


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