Simple automobile model

[نواف العصيمي]

Example: CRUISE
Simple automobile model
Purpose
To understand the dynamics of a simple automobile model.
System


Figure 1: Model diagram
The model assumes a single mass for the car with motion along a level road and opposed by viscious friction. The differential equation relating the speed v to the input force u, which may be assumed proportional to accelerator position, is m + bv = u. Thus the transfer function from u to v is given by V(s)/U(s) = 1/(ms + b) which can be written as (1/b)/(1 + sT), where the time constant T equals m/b. The response to a step input of magnitude U is thus an exponential with time constant T rising to the steady state speed of U/b.
System Parameters
m = 1000 [kg] mass of the car
b = 50 [Nsec/m] friction
u = 500 [N] constant excitation of the car
System excitation
The input force if u, which is constant, and applied at t = 0.
Task
Default system parameters are given but you can change these in the program before submitting to DYNAST for the solution. The run time is TR. Check from the simulation that the time constant and the final speed have the correct values for the parameters chosen.
Solution
Data









Figure 2: Simulation results




Example: CRUISE-P
Cruise control of the car
Purpose
To understand the implementation of a simple speed (cruise) control on the car.


Figure 1: Model diagram
System Parameters
m = 1000 [kg] mass of the car
b = 50 [Nsec/m] friction
System excitation
The input is a constant reference speed value applied at t = 0 and denoted, vdes, in the program, which is set at the default value of 10 in the line below `control' in the program. Since the system is linear changing this value only rescales the graph, however, in practice there will be a limit to the maximum value of u (see nonlinear control examples) which you can show increases with increase in the reference speed setting.
Task
The default automobile parameters are set at m = 1000 and b = 50 and the default controller parameters Kp, Kd, Ki and are set at 400, 0, 0, 0, respectively. The controller parameters are located in the program on the line beginning "control" and is the time constant associated with the derivative term. Since the model of the car has no integral term a steady state error to a speed demand will exist if there is no integral term included in the controller.
For Kd = 0 and Ki = 0, calculate the steady state speed error and the maximum value of the control signal u for m = 1000, b = 50, vdes = 10, for the two cases of Kp = 400 and 800, respectively. Repeat with Ki set to 40 rather than zero. Check your calculations from the simulation results.
Solution
Data







Links to used submodel(s):
• PID controller
• Difference block



Figure 2: Simulation results



DC Motor position control
Purpose
To study closed loop position constrol using a DC motor and PID controller.
System
The motor has the model considered in the previous example. It is constrolled by a PID controller which has the position error as input. Note that the motor transfer function constains an integration so that there will be no steady state error to a step reference input with the PID controller in the proportional mode. On the other hand a non zero value for the disturbance input W will cause a steady state error.


Figure 1: Model diagram
System Parameters
J = 3.2284 . 10-6 [kg.m2] moment of inertia of the rotor
B = 3.5077 . 10-6 [N.m.s/rad] damping ratio of the mechanical system
K = 0.0274 [Nm/Amp] electromotive force constant
R = 4 [ ]
electric resistance
L = 2.75 . 10-6 [H] electric inductance
P = 17
I = 600
D = 0.15
tau = 0.0015
S = 1 desired position of shaft
W = 0 disturbance
System excitation
There are two inputs, the reference input, S, and the disturbance input, W.
Task
Check that there is a position error with the controller in the P mode when W is not zero. You may wish to calculate this for specific parameters and compare with the simulation results. Include an integral term in the controller to show that with finite W there is now no position error. By examining the controller output signal, note that to avoid pure differentiation of the step input and therefore a very high value of controller output at t = 0+ a non zero value of is required.
Solution
Data







Links to used submodel(s):
• 2-pin integrator block
• Permanent magnet DC machine
• PID controller
• Summator block
• Difference block



Figure 2: Simulation results




DC Motor speed
For more information see http://virtual.cvut.cz/ctm/analog.html.


Figure 1: Model diagram
System Parameters
J = 0.01 [kg.m2] moment of inertia of the rotor
b = 0.1 [N.m.s/rad] damping ratio of the mechanical system
K = 0.01 [Nm/Amp] electromotive force constant
R = 1 [ ]
electric resistance
L = 0.5 [H] electric inductance
Solution
Data







Links to used submodel(s):
• Permanent magnet DC machine
• 2-pin integrator block



Figure 2: Simulation results
















Integrator block
Description
Outputs the integral of scaled input signal

where t0 is the starting time of transient analysis.
Symbol


Figure INT-1: Symbol, the first variant.


Figure INT-2: Symbol, the second variant.
Interface
in input
out output
External Parameters
c = 1 coefficient
Data
:: Integrator block
Int
in, ::input
out/ :utput
c = 1; ::coefficient
BI out = c*in;
EO@;




Summator block
Outputs the sum of both input values



Figure SUMMATOR-1: Symbol.
Interface
in1 first input
in2 second input
out output
Data
:: Summator block
Summator
in1, ::first input
in2, ::second input
out; :utput
BS out = in1 + in2;
EO@;


لتحميله بملف نصي مع الالوان والاكواد

http://www.yy44.net/download.php?fil...241e0b4754.zip

نموذج في تحليل الانظمه

التوقيع