Answer:
The MATLAB Code for this PI Controller will be:
Kp = 350;
Ki = 300;
Kd = 50;
C = pid(Kp,Ki,Kd)
T = feedback(C*P,1);
t = 0:0.01:2;
step(T,t)
Explanation:
When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.
Obtain an open-loop response and determine what needs to be improved
Add a proportional control to improve the rise time
Add a derivative control to reduce the overshoot
Add an integral control to reduce the steady-state error
Adjust each of the gains $K_p$, $K_i$, and $K_d$ until you obtain a desired overall response.
The further explanation is attached in the Word File.
Answer:
it is not possible to place the wires in the condui
Explanation:
given data
total area = 2.04 square inches
wires total area = 0.93 square inches
maximum fill conduit = 40%
to find out
Can it is possible place wire in conduit conduit
solution
we know maximum fill is 40%
so here first we get total area of conduit that will be
total area of conduit = 40% × 2.04
total area of conduit = 0.816 square inches
but this area is less than required area of wire that is 0.93 square inches
so we can say it is not possible to place the wires in the conduit
Answer:
Explanation gives the answer
Explanation:
% Using MATLAB,
% Matlab file : fieldtovar.m
function varargout = fieldtovar(S)
% function that accepts single structure as input, assigning each
% of the field values to user-defined variables
fields = fieldnames(S); % get the field names of the input structure
% check if number of user-defined variables and number of fields in
% structure are equal
if nargout == length(fields)
% if equal assign each value of structure to user-defined varable
for i=1:nargout
varargout{i} = getfield(S,fields{i});
end
else
% if not equal display an error message
error('The number of output variables does not equal the number of fields');
end
end
%This brings an end to the program
So what happens is the host will not kill the y no se que hacer para no one can see it in
Answer:
D. Both pull-in and hold-in windings are energized.
Explanation:
The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.
The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.
