Applied Physics Expert
1953 Order Completed 99 % Response Time 63 Reviews Since 2015

Applied Physics Expert

Malacca, Malaysia

Frank P

Bachelor’s Degree, Applied Physics, Sunway University, Malaysia

Profession

I am an expert signal processing assignment helper with deep knowledge of Matlab and several other digital processing programs. 

Skills

After obtaining my bachelor’s degree in applied physics, I was looking for an opportunity to utilize the skills I had acquired in college. I, therefore, decided to venture into academic writing because, with this career path, I knew I would be subjected to challenges that would help me cement my skills. Luckily, Matlab Assignment Experts was looking for signal processing experts and that’s how I landed a job as a signal processing assignment help expert. It’s almost six years since I started working with this company and so far, I have delivered over 1950 successful orders. My area of expertise is speech signal processing, music signal processing, compressed sensing, and cognitive radio, though I can handle any topic related to signal processing.

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Curve Plotting
clc, clear all, close all % Load dat file load crazy_func % Plot the curve figure subplot(1,2,1); plot(eks, why), grid on xlabel('eks'); ylabel('why'); title('why vs. eks'); % Determine d(why)/d(eks) N = length(why); % length of data % The length of d(why)/d(eks) will be N - 1 for i = 1:N-1 dwhy(i) = why(i+1)-why(i); deks(i) = eks(i+1) - eks(i); end % Now, compute d(why)/d(eks) dwhydeks = dwhy./deks; % Plot the resulting curve subplot(1,2,2) plot(eks(1:N-1), dwhydeks), grid on xlabel('deks'); ylabel('dwhy'); title('dwhy vs. deks'); %% NOTE FOR THE REPORT: % eks < 20 -> why is a sine function, so its derivative must be a cosine % function % eks >=20 and eks < 40 -> why is an exponential function so its % derivative must be also an exponential functionç % eks >=40 and eks < 60 -> why is a collection of random points so its % derivative must be also a collection of random points % eks >= 60 and eks < 80 -> why is a parabola, so its derivative must be % a straight line with positive slope % eks >= 80 and eks < 100 -> why is a line, so its derivative must be a % constant value
Vector Operations Using Matlab
clc, clear all, close all %% 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 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 % 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]; 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'); figure plot(xv, yv), grid on xlabel('X-Position (m)'); ylabel('Y-Position (m)'); 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); 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);