Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: THEend8_
Advanced Control
PROJECT 1
DC Motor Drive
Introduction
This control problem concerns the control of current and speed of a DC motor with rigid shaft. These systems
are usually called direct-drive system and are used in high performance industrial application.
Fig. 1 shows the schematic of a DC machine driving a load:
Fig. l
where:
V Actuation (control) Voltage Vmax = 400V
i Armature current irated = 42A
ω motor speed (rad/s) ωrated = 1000rpm
ka motor constant ka = 3.4
E back emf = kaω
T motor torque = kaI Trated = 143Nm
Ra motor armature resistance Ra = 1.020
La motor armature inductance La = 10.2mH
J moment of inertia of motor + load J = 0.4kg/m2
B friction coefficient of motor + load B = 0.682Nm/rads-1
TL Additional unknown load
Note that at rated speed (1000rpm), the motor operates at half rated load (half rated torque) in order to
overcome the friction.
Exercise 1
Defining x = [I ω], v = [I ω], and u = [V TL] show that:
and draw a Transfer function Block diagram of the plant in the form of:
Fig. 2
Note that the above relates the speed of the load to the actuating voltage. If TL = 0, then it can be reduced to one
transfer block relating ω to V. What is this single transfer function?
Using the state-space block, enter the system into Simulink. Create the system matrices in Matlab and check
eigenvalues. Why are they not the poles of the transfer function block of Fig. 2?
Control Philosophy
In Fig.2, V is variable and provided by a power electronic converter. It is possible to control speed by
comparing the speed with a reference speed, taking the error through a controller whose output is V. If you do
this then the motor current is uncontrolled — it could reach high values and blow up the power converter. In
power electronics and drives, one should always control current first.
Remember, current is proportional to torque, therefore a current demand is also a torque demand. When we
come to control speed, the speed error will be passed through a controller to yield a current demand i.e. if speed
error then a change torque (or current) is demanded. We are going to have an outer speed loop and an inner
current loop. The inner current loop will be much faster than the outer speed loop. This will become evident as
we develop the control.
So first you will design a state space current controller. Then this closed loop system will be the plant for a state
space speed controller.
Current loop design
For the current control we wish the current to respond to a step change in reference within about 20ms (e.g. 2%
settling in about 20ms). We also want the current to follow the reference with zero state error. We need
integral control (ref "integral control" in lecture notes). From the notes we see that we need to augment the
system so that the integral of current is an extra state.