Matlab Project Online Assignment Help

The following report made for Matlab project online assignment help illustrates the capabilities of our dedicated experts. It is based on the impact of nonlinear systems and realistic models on motor design equations and was done using Simulink model. System-block state-feedback diagram of this system was also developed as one of the deliverables. The conclusion was the chosen tuning method, backed by the reasons in the given constraints. The Matlab support solution also contains the screenshots of the Simulink model used in the modus operandi. It’s a classic case of how an electrical engineering problem can be solved using Matlab/Simulink. We hope the students find it useful and it gives them some insights on how to go about solving such assignments.

The impact of nonlinear systems and more realistic models on our motor design equations-

Signal Linearisation

The Whitworth Art Gallery at the University of Manchester is hosting a 6 month Pre-Raphaelite Artist exhibition of sculptures and paintings. Matlab project online assignment help The main hall contains £100 million of art on loan and the air in the central hall must be kept within a precisely defined set of temperature and humidity limits to avoid damage to it. An environmental control system monitors the hall, sampling two times per minute. These values are processed through a look-up table to drive a variable speed drive fan. We have to programme the control system for this unit.

L = 22e-3;
R = 0.8;
Ti = L / R;
w1 = 2*pi*500;
Kp = L*w1;
Ti = L / R;
fp = 6e3;
Tp = 1 / fp;
Kt = 2.42;
Ke = Kt;
Kea = Ke;
Bm = 0.001;
Jm = 0.11;
Bf = 0.0015;
Jf = 4.2;
J = Jf + Jm;
B = Bf + Bm;
Kf = 0.005168;

The motor drive/system comprises a 20kW (nominal steady-state rating) brushed DC motor, with a 4-quadrant PWM inverter using Matlab coder H-bridges per phase. The DC link may be Matlab project online assignment help assumed to be derived from rectified three-phase mains and is 560V DC. The inverter can only supply a limited current and its current set-point should be limited signal linearisation to an output current of 83A max. Other parameters for the motor are:

L=22mH                      Bmotor     =0.001 Nm s/rad                      KT=2.42Nm /A

R=0.8W                                    Jmotor      =0.11 Nm s2 /rad

We  may assume in this example that the inner loop can use back-emf decoupling. The inverter uses PWM at 6kHz and signals take two PWM periods to propagate from the input to the output. We may wish to model this as a function

although it may be easier to model it as a ‘transport delay’ block in SIMULINK.

A fan load is added to our motor drive system. This load torque can be modelled as extra damping, inertia and a fan load:

BFAN = 0.0015 Nm s/rad,         JFAN =4.2 Nm s2 /rad    KFAN=0.005168 Nm s2/rad2 

The load speed range is 0 to 1500rpm.

We must design an appropriate control system for the inner and outer loops, with full justification for our choices (or if suitable control is not possible, explain why, and what are the problems and how the system could be changed to overcome them).

T1 = 1 / w1;

sysCurrentLoop = tf(1, [T1 1]);

sysMechanical = tf(1, [J B]);

sysG = sysMechanical * sysCurrentLoop;

% bode(sysG)

Consider the system-block state-feedback diagram of this system, including disturbance torque input, back-EMF nulling and controllers. Linearise the system Matlab project online assignment help showing your full working. Draw the system-block state-feedback diagram of this system, again including disturbance torque.

Our electrical ventilation system consists of fan(plant), motor(actuator), inverter(power amplifier), some sensors(current sensor, and angular speed sensor such as tachometer), and Matlab project online assignment help controller.

Part 1.

We showed more detailed system configuration in Fig.


                    More Detailed System Configuration

As you can see in the figure, our control system consists of current loop and speed loop. In order to make the design of each loop independent, we use back EMF nulling technique. Then, the bandwidth of the current loop must be sufficiently Matlab project online assignment help wide enough to facilitate the design of the speed loop. The current loop consists of inverter and motor except controller. Because the PWM switching frequency of inverter is 6kHz, and the electrical time constant of the motor is
T_e=L/R=(22×〖10〗^(-3))/0.8=27.5×〖10〗^(-3) (sec)=27.5(msec),
we will take the bandwidth of the current loop as 500Hz. For speed loop, we must design speed controller so that its settling time is about 100ms. Finally both of loops(i.e., current loop and speed loop) must have zero steady-state-error.
The constraint that must be met is
|i^* (t) |≤83 A
|v(t) |≤560 V
Part 2.
After analyzing the system, we can find that the plant of the current loop is first order system. We must design a current controller which admits zero steady-state error, we will use PI tuning for current controller. For speed loop, time constant of the plant is so much larger than desired time constant(or equivalently, settling time), we will use integrator + lead compensator.
Part 3.
1. Current Controller Design
Because we use back EMF nulling technique, we can ignore the influence of back EMF in the design of current loop. Ignoring back EMF, the differential equation describing current loop becomes
di/dt=-R/L i+1/L u_1 (t)
where u_1 (t)=u(t)+K_e ω(t). Therefore
i(s)/(u_1 (s) )=1/(R+Ls)
On the other hand, our current controller will be expressed in the form of
C(s)=K_p (1+1/(T_i s)).


                                  Simplified Current Loop

We showed the simplified diagram of current loop. From this figure, we can write the closed loop transfer function of the current loop as

If we set , it can be simplified as

Thus bandwidth of the closed loop is found as

From the part 1 of this report, the bandwidth of the closed loop must be

Therefore the parameters of the PI controller for current loop are found as

We showed the unit step response of simplified current loop in Fig.


Figure. Unit Step Response of Simplified Current Loop

From this plot, we can see that it has very good transient response. Because physical constraint that output of the controller must be less than inverter input voltage(±560 V), We check whether the output of the controller Simulink Matlab help exceeds the limit when input to current loop is set at its maximum(83A). We showed the controller output in Fig. As we can see in this figure, the output of the controller exceeds the Matlab project online assignment helpphysical constraints(its output is about 6000 V at the beginning of the controller). So, we added limiter at the output of the current controller. We showed its step response when its input is set its maximum(83A) in Fig. 8. It shows still good response. Then, because inverter can be modeled as an element with delay, we added transport delay in the Simulink model of the current controller. The output of the current loop with regarding delay and limit was shown in Fig.


Figure. Output of Current Controller with Maximum Current as Input


Figure. Output of the Current Loop with Limit Considered

From this Fig., we can conclude that designed current controller meets the design requirement.


Figure. Output of the Current Loop with Physical Constraints Considered

2. Speed Controller Design


Figure. Open Loop Bode Diagram of the Speed Loop

The plant for speed controller consists of current loop, and load. Its transfer function is


We showed the open loop Bode diagram of speed loop in Figure 10. We will design our speed controller using sisotool in MATLAB. First of all, we need to add integrator, because the steady-state error of our system must be zero. The open loop Bode diagram after we added integrator was shown in Fig.17
Figure. Open Loop Bode Diagram after Integrator Added
As you can see in this figure, the open loop became almost unstable after adding an integrator. As is well known, the lead compensator can be expressed as
K (s+z)/(s+p),z<p
In fact, the phase margin of the speed loop is only 0.0602°. In order to design a stable system, p<ω_1 must be met. Therefore we take p=1.5⋅〖10〗^3 rad/s. zcan be found so that the compensated system has more phase margin than 60°. Finally, K can be found so that crossover frequency is about
The transfer function of speed controller found was


Figure. Open Loop Bode Diagram of Compensated Speed Loop

We showed the step response of the compensated system in Fig.


Figure. Step Response of the Compensated Speed Loop

sysK = tf(45000*[0.025 1], [0.00066 1 0]);

sysL = sysG * sysK;


sysC = feedback(sysL, 1);


grid on

We can conclude that it meets the design specification about its settling time(less than 100 msec), and steady-state error(zero steady-state error). Finally we will consider the closed loop response with regarding the maximum current and air drag(which was dealt as disturbance). We showed the unit step response of the speed loop in Fig.


Step Response of the Compensated Speed Loop(with Current Limit)

Then we showed the response when its possible maximum(1500rpm) was set as input in Fig. In this figure, it takes about 22 sec to reach 1500 rpm. It is due to large disturbance which


Step Response of the Compensated Speed Loop(with Current Limit, Possible Maximum Input)

is proportional to square of speed. But even in this case, the speed loop outperforms the severe conditions showing zero steady-state error.

Simulink Diagram of the System

We showed the Simulink diagram of the whole system in Figure 16. It has two subsystems for speed controller and current controller(PI controller). We showed Matlab support their diagram in Fig. 17 and Fig. 18, respectively.


Simulink Diagram for Speed Controller


              Simulink Diagram for PI Controller