Control Systems, Matlab, Electrical Engeering essay

ECE 415 Management Programs, Fall 2021

Pc Mission -1

Due: Saturday, Oct 16, 2021

Goal: To study among the helpful management design instruments supplied by MATLAB and

to place into follow the ideas seen at school.

Deliverable: A written report containing:

a) For every step, the corresponding block diagrams and another analytic

design instruments used, together with derivation of switch capabilities for open and

closed loop designs (in different phrases, present an evidence for the MATLAB

code given in Half I, and on your personal management design in Half II). b) Plots obtained at every step.

c) In Half II solely, the MATLAB code used at every step to resolve the given

downside.

The report ought to be typewritten, properly introduced (presentation counts

in the direction of the grade), and in good English. No hand-written report will probably be

accepted (besides, maybe, for diagrams). Submit it in PDF format by way of

Isidore.

It’s crucial that you simply strictly adhere to the respect code. This Laboratory will account for 35% of the Initiatives class.

Half I

On this first half, a motor velocity downside is introduced, and its resolution is given step

by step, along with the corresponding MATLAB directions. Please observe the

derivation and implement the given code, and, the place requested to, present a

justification for what the code does utilizing block diagrams or mathematical evaluation.

zero) Contemplate the DC motor described by the simplified first order differential equation

dy + 60 y = 600u − 1500w

dt

the place y is the motor velocity, u is the enter armature voltage, and w is a load (a

disturbance, from the management design standpoint).

Assume the preliminary circumstances are zero, and take the Laplace remodel:

Y (s) (s + 60) = 600U (s) − 1500W (s)

Y (s) = 600

s + 60 U (s) −

1500 W (s)

s + 60

We are going to design open and closed loop proportional (P) controllers for this plant. The

management goal is to attenuate steady-state error and supply good disturbance

rejection capabilities.

Open-loop proportional management

The controller is designed by assuming that w = zero (that’s, no exterior load is utilized on the motor). From the ultimate worth theorem, we set the controller acquire to

Okay = 60

600

(justify this selection!).

1) Discover the impulse response of the open-loop management system.

% Outline the plant: DC motor

num_motor = 600;

den_motor = [1 60];

motor = tf(num_motor, den_motor)

% Open-loop controller

Okay = 60/600;

ol_cont = tf(Okay,1)

% Compute the switch perform from reference to output, with out

disturbance

undist_plant = sequence(ol_cont, motor)

% Plot impulse response

impulse(undist_plant);

title(‘Open-loop management’)

2) Plot the undisturbed step response, for a step of magnitude 100.

% Plot response to a step of dimension 100

step(100*undist_plant);

title(‘Open-loop response to a step of dimension 100’)

three) Plot the response to a unit step disturbance, setting the reference to zero. Give the block diagram of what you might be doing right here and within the earlier merchandise.

% Compute switch perform from disturbance to output, with out

reference

dist_plant = (-1500/600)*motor

% Plot response to a unit step disturbance

step(dist_plant);

title(‘Open-loop response to a unit step disturbance’)

four) Plot the open-loop response when each reference and disturbance are current.

% Compute response to each reference and disturbance

t = [0:0.001:0.1]; % time vector

y_ref = step(100*undist_plant,t); % response attributable to reference

y_dist = step(dist_plant,t); % response attributable to disturbance

y = y_ref+y_dist; % whole response (making use of

%superposition!)

plot(t,y);

xlabel(‘Time’)

ylabel(‘Motor velocity’)

title(‘Open-loop response to a step reference of magnitude 100 and a

unit step disturbance’);

% Put undisturbed and disturbed responses in a single plot

maintain on;

plot(t,y_ref,’–‘);

legend(‘With disturbance’,’With out disturbance’)

% attempt transferring the legend with the mouse

grid; % add a grid

maintain off;

Closed-loop proportional management

We are going to first discover the switch perform from reference to output:

T RY

(s) = 600Okay

s + (60 + 600Okay )

(derive it, and supply a block diagram).

Subsequently, when W (s) = zero the output is given by Y (s) = TRY (s)R(s).

Additionally, the switch perform from disturbance to output is

T WY

(s) = − 1500

(derive it, and supply a block diagram).

s + (60 + 600Okay )

Compute the sensitivity of the closed-loop switch perform 𝑇𝑅𝑌 (𝑠) with respect to modifications in modifications in controller acquire 𝐾. How does it evaluate with 𝑇𝑊𝑌 (𝑠)?

When R(s) = zero, the output because of the disturbance is given by Y (s) = TWY (s)W (s).

Combining, if each reference and disturbance are current, the whole output is

Y (s) = TRY (s)R(s) + TWY (s)W (s).

Contemplating these outcomes, how would you need to choose the management fixed 𝐾? (Clarify your rationale!)

We are going to contemplate two design decisions, and evaluate their efficiency:

K1 = 10

K2 = 50

1) Plot the step responses for each controllers in a single plot.

% confirm motor switch perform

motor

% the 2 controller positive factors

K1 = 10;

K2 = 50;

% discover the switch perform from R(s) to Y(s) for every controller acquire

T1_ry = suggestions(K1*motor,1) % the 1 signifies unity suggestions

T2_ry = suggestions(K2*motor,1)

% discover the switch perform from disturbance W(s) to output Y(s)

T1_wy = suggestions(motor,K1)*(-1500/600)

T2_wy = suggestions(motor,K2)*(-1500/600)

% Plot the response to a step of magnitude 100 with out disturbance

clf; % clear determine

step(100*T1_ry); % for controller K1

title(‘Closed-loop response to a step of magnitude 100’)

maintain on;

step(100*T2_ry); % for controller K2

legend(‘Utilizing K_1′,’Utilizing K_2’);

maintain off;

grid;

2) Discover an approximation to the steady-state error for each designs from the plot.

zoom on;

% click on on the plot with the mouse to seek out the approximate error worth

zoom off;

three) Plot the response to a unit step disturbance for each designs.

% Contemplate a disturbance step enter, with R(s) = zero

step(T1_wy);

maintain on;

step(T2_wy);

title(‘Closed-loop response to a step disturbance’);

legend(‘Utilizing K_1′,’Utilizing K_2′);

maintain off;

grid;

four) Plot the closed-loop response when each reference and disturbance are current.

% Compute the general response, when R(s)=100/s and W(s)=1/s

t = [0:1e-6:1e-3];

y1_ref = step(100*T1_ry,t);

y2_ref = step(100*T2_ry,t);

y1_dist = step(T1_wy,t);

y2_dist = step(T2_wy,t);

y1 = y1_ref + y1_dist;

y2 = y2_ref + y2_dist;

plot(t,y1,’r’) % plot response utilizing K1 in crimson

maintain on;

plot(t,y2,’b’) % plot response utilizing K2 in blue

legend(‘Utilizing K_1′,’Utilizing K_2’)

grid;

xlabel(‘Time’)

ylabel(‘Motor velocity’)

title(‘Closed-loop response to a step reference of magnitude 100 and a

unit step disturbance’);

maintain off;

Half II

Contemplate the plant

𝑃(𝑠) = 1

(𝑠 + 1)(𝑠 + 5)

1) What’s the plant’s sort?

2) Let C(s) = Okay (a proportional controller). Discover the closed-loop switch perform

from reference to output utilizing unity suggestions.

three) Select totally different positive factors for Okay inside the vary 1 to 100. Plot the unit step response for the totally different positive factors. What occurs with the transient response of the

closed-loop as Okay will increase?

four) For 𝐾 = 20 discover the utmost worth attained by the output y(t) and the settling

time Ts for a unit step enter (the time it takes the output to settle inside a band

of  2% round its ultimate worth). Additionally discover what’s the steady-state worth of y(t)?

What’s the steady-state error equal to?

5) Design a controller that can enhance the system’s sort by 1 and that can yield

the smallest settling time you may acquire for a step enter. What’s the settling

time? What’s the steady-state error?

6) Plot the response of the closed-loop system to a unit ramp utilizing the controller

you designed partially (5).

There are a number of methods to do that. One is to make use of the sawtooth command collectively

with the lsim command to acquire the time response of the system. One other approach

is to implement the entire thing in Simulink, utilizing a sign generator block to

produce the ramp enter. And a 3rd approach nonetheless is to make use of the step command but

once more, noting that the Laplace remodel of a unit step is 1

𝑠 , and the Laplace

remodel of a unit ramp is 1

𝑠2 = (

1

𝑠 ) (

1

𝑠 ).

Report your observations. Is there a steady-state error? If that’s the case, what’s its

magnitude?

What would occur with the steady-state error if the enter have been, as a substitute, a prepare

of steps? You do not want to offer a plot for this query, simply reply primarily based on

the kind variety of the plant collectively together with your controller design.

7) Touch upon the efficiency limitations you discovered partially (5). Do that by

observing what occurs if you happen to make 𝐾 very small, or very massive.

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