Sending
The Best Place to Get MATLAB Project Help In the UKSuccessful Completion of MATLAB projects has proven to be a daunting task for many students in the UK but as a student, you don’t have to worry anymore about complex MATLAB projects causing you trouble because we are here to ensure all your projects...
Where to Find Matlab Assignment Experts in Singapore?The Matlab computing environment is one of the most significant requirements when it comes to the fields of Science, Engineering, Economics, or statistics. This makes MatLab one of the fundamental courses one has to take in the process of their ed...
Matlab assignment solutions tips: How to create a simple GUI Learn how to create a GUI from our Matlab assignment help experts. We also provide outstanding solutions for projects on this topic. Get professional Matlab assignment help on how to create a simple GUI Graphical User Interface commonl...
Texas, USA
Doctor of Philosophy, Electrical Engineering, University of Houston, USA
Profession
Full time online electrical engineering tutor
Skills
I have been teaching electrical engineering for close to ten years now to college and university students. Before 2016 I worked full-time as a lecturer at a local university but during my free time, I would offer online tuition as an independent freelancer. But the more I did freelancing, the more I loved working remotely and so I decided to apply for a tutoring position at Matlab Assignment Experts. Since then, I have been administering online lessons on various electrical engineering topics such as circuit theory, electrical installations, microprocessors, electrical control systems, and power engineering, to name a few. If you are looking for convenient and pocket-friendly tutoring services on any of these topics, then I am the person for the job.
%% Question 2
% Input values
vxi = [0 -0.2478 -0.4943 -0.7384 -0.9786];
vyi = [4 3.9915 3.9659 3.9234 3.8644];
ti = [0 0.04 0.08 0.12 0.17];
% Initial values
x1 = 4;
y1 = 0;
% Calculate x2, x3, x4, y2, y3, y4
x2 = x1 + vxi(1)*(ti(2)-ti(1));
x3 = x1 + vxi(2)*(ti(3)-ti(2));
x4 = x1 + vxi(3)*(ti(4)-ti(3));
y2 = y1 + vyi(1)*(ti(2)-ti(1));
y3 = y1 + vyi(2)*(ti(3)-ti(2));
y4 = y1 + vyi(3)*(ti(4)-ti(3));
%% Question 3
% Read file into a table
data = readtable('A1_input.txt');
% Do not take into account the first element of the data because that one
% contains the units
for k = 2:size(data,1)
t(k-1) = str2double(data.time{k});
vx(k-1) = str2double(data.vx{k});
vy(k-1) = str2double(data.vy{k});
end
% Display the first four values
[t(1:4)', vx(1:4)', vy(1:4)']
%% Question 4
% Plot vx and vy in the same graph
figure
plot(t, vx, t, vy);
grid on
legend('vx', 'vy');
xlabel('Time (s)');
ylabel('Velocity (m/s)');
grid on
%% Question 5
% Calculate x and y positions using for loop
x(1) = x1;
y(1) = y1;
for k = 2:length(t)
x(k) = x1;
y(k) = y1;
for j = 1:k-1
x(k) = x(k) + vx(j)*(t(j+1)-t(j));
y(k) = y(k) + vy(j)*(t(j+1)-t(j));
end
end
%% Question 6
% Calculate x and y positions with vector operations
% x and y using vectors
% vectors of tj+1 tj
tvec = t(2:end) - t(1:end-1);
xv = cumsum(vx(1:end-1).*tvec);
xv = xv + x1;
xv = [x1, xv];
yv = cumsum(vy(1:end-1).*tvec);
yv = yv + y1;
yv = [y1, yv];
%% Question 7
% Plot the original datapoints and the calculated ones
figure
hold on
plot(t, x);
plot(t, y);
plot(t, xv, 'k--', 'linewidth', 3);
plot(t, yv, 'g--', 'linewidth', 3);
grid on
xlabel('Time (s)');
ylabel('Position');
legend('X-position from Q5', 'Y-position from Q5', 'X-position from Q6', 'Y-position from Q6');
%% Question 8
% Plot the trajectory
figure
plot(xv, yv), grid on
xlabel('X-Position (m)');
ylabel('Y-Position (m)');
%% Question 9
% Write a new file
file = fopen('output.txt', 'wt');
fprintf(file, 'Time\tx\ty\n');
fprintf(file, '(s)\t(m)\t(m)\n');
for k = 1:length(t)
fprintf(file, '%.3f\t%.3f\t%.3f\n', t(k),xv(k), yv(k));
end
fclose(file);
%% Question 10
% Write a new file
file = fopen('output2.txt', 'wt');
fprintf(file, 'Time\tx\ty\n');
fprintf(file, '(s)\t(m)\t(m)\n');
fprintf(file, '%.3f\t%.3f\t%.3f\n', t',xv', yv');
fclose(file);
P = 0; % initial money in the account
deposit_per_year = [288, 345, 355, 382, 392]; % amounts deposited each month, per year (year 1, year2, ... year5)
withdraw = 2331; % amount to be withdrawn each year
r = 0.04; % interes of the saving account
rcd = 0.08; % interest of the CD
n_CD = 0; % number of CDs bought
year = 1;
deposited = 0;
for i = 1:5*12 % 5 years, 12 months per year: total 60 months
% if i < 60
% deposited = deposit_per_year(year);
% else
% deposited = deposit_per_year(5);
% end
deposited = deposit_per_year(year);
P = P*(1+r);
P = P + deposited;
fprintf("Deposited %.2f at month %.0f. Current amount: %.4f\n", deposited, i, P);
if mod(i,12) == 0 % the end of a year
if P > 2811
fprintf('Amount in account: %.4f. Withdrawed %.2f at the end of year %0.f\n', P, withdraw, year);
P = P - withdraw;
n_CD = n_CD + 1;
% Formula of compound anually. We use the interest rate of the
% CD multiplied by the number of CDs
P = P*(1+(n_CD-1)*rcd);
end
year = year + 1;
end
end
fprintf("\nThe final amount in the account after %.0f years is: %.4f\n", 5, P)
| File | Actions |
|---|