I have written a simple code to calculate the phase and magnitude of a signal, based on the sinusoidal input given in my problem. I have already determined the magnitude of the signal corresponding to different values of w. More specifically, the phase I want is a vector calculated for different values of w. Notice that the signal I'm talking about is the output signal of a linear process. As matter of fact, I want the phase difference between the input u and the output y, defined for all values of w for all time steps. I have created the time and w vector in my code. Here is the main code I have written in MATAB2021a:
clc;clear;close all;
%{
Problem 2 Simulation, By M.Sajjadi
%}
%% Predifined Parameters
tMin = 0;
tMax = 50;
Ts = 0.01; % Sample Time
n = tMax/Ts; % #Number of iterations
t = linspace(tMin,tMax,n);
% Hint: Frequency Domain
wMin = 10^-pi;
wMax = 10^pi;
Tw = 10;
w = wMin:Tw:wMax; % Vector of Frequency
Nw = length(w);
a1 = 1.8;
a2 = -0.95;
a3 = 0.13;
b1 = 1.3;
b2 = -0.5;
%% Input Generation
M = numel(w);
N = length(t);
U = zeros(M,N);
Y = U; % Response to the sinusoidal Input, Which means the initial conditions are set to ZERO.
U(1,:) = sin(w(1)*t);
U(2,:) = sin(w(2)*t);
U(3,:) = sin(w(3)*t);
Order = 3; % The Order of the Differential Equation, Delay.
%% Main Loop for Amplitude and Phase
Amplitude = zeros(Nw,1); % Amplitude Values
for i=1:numel(w)
U(i,:) = sin(w(i)*t);
for j=Order+1:numel(t)
Y(i,j) = a1*Y(i,j-1) + a2*Y(i,j-2) + a3*Y(i,j-3) + ...
b1*U(i,j-1) + b2*U(i,j-2);
end
Amplitude(i) = max(abs(Y(i,:)));
end
I know I should use fft or findpeaks function in MATLAB, but I do not know how I should do it.
Related
The formula for the discrete double Fourier series that I'm attempting to code in MATLAB is:
The coefficient in front of the trigonometric sum (Fourier amplitude) is what I'm trying to extract from the fitting of the data through the double Fourier series seen above. Using my current code, the original function is not reconstructed, therefore my coefficients cannot be correct. I'm not certain if this is of any significance or insight, but the second term for the A coefficients (Akn(1))) is 13 orders of magnitude larger than any other coefficient.
Any suggestions, modifications, or comments about my program would be greatly appreciated.
%data = csvread('digitized_plot_data.csv',1);
%xdata = data(:,1);
%ydata = data(:,2);
%x0 = xdata(1);
lambda = 20; %km
tau = 20; %s
vs = 7.6; %k/s (velocity of CHAMP satellite)
L = 4; %S
% Number of terms to use:
N = 100;
% set up matrices:
M = zeros(length(xdata),1+2*N);
M(:,1) = 1;
for k=1:N
for n=1:N %error using *, inner matrix dimensions must agree...
M(:,2*n) = cos(2*pi/lambda*k*vs*xdata).*cos(2*pi/tau*n*xdata);
M(:,2*n+1) = sin(2*pi/lambda*k*vs*xdata).*sin(2*pi/tau*n*xdata);
end
end
C = M\ydata;
%least squares coefficients:
A0 = C(1);
Akn = C(2:2:end);
Bkn = C(3:2:end);
% reconstruct original function values (verification check):
y = A0;
for k=1:length(Akn)
y = y + Akn(k)*cos(2*pi/lambda*k*vs*xdata).*cos(2*pi/tau*n*xdata) + Bkn(k)*sin(2*pi/lambda*k*vs*xdata).*sin(2*pi/tau*n*xdata);
end
% plotting
hold on
plot(xdata,ydata,'ko')
plot(xdata,yk,'b--')
legend('Data','Least Squares','location','northeast')
xlabel('Centered Time Event [s]'); ylabel('J[\muA/m^2]'); title('Single FAC Event (50 Hz)')
I've spent quite some time trying to simulate a simple SISO system using two approaches:
1) Using lsim() in MATLAB
2) By writing down the difference equations myself and iterate over them in a loop.
I was never able to get the same simulation results from both approaches, and I have no idea what I am doing wrong.
I stacked my code in a single m-file so it's easier to follow. Here is the code:
function main()
clear all
clc
simulateUsing_lsim()
simulateUsing_loop()
end
%%%%%% Simulating using lsim %%%%%%%
function simulateUsing_lsim()
% Define the continuous-time closed-loop system
P = getContPlant();
[Kp,Ki,Kd] = get_PIDgains();
C = pid(Kp,Ki,Kd);
clSys_cont = feedback(C*P,1);
% Define the discrete-time closed-loop system
hk = get_sampling_time();
clSys_disc = c2d(clSys_cont,hk);
% Generate the reference signal and the time vector
[r,t] = getReference(hk);
%% Simulate and plot using lsim
figure
lsim(clSys_disc,r,t)
%% Finding and plotting the error
y = lsim(clSys_disc,r);
e = r - y;
figure
p = plot(t,e,'b--');
set(p,'linewidth',2)
legend('error')
xlabel('Time (seconds)')
ylabel('error')
% xlim([-.1 10.1])
end
%%%%%% Simulating using loop iteration (difference equations) %%%%%%%
function simulateUsing_loop()
% Get the cont-time ol-sys
P = getContPlant();
% Get the sampling time
hk = get_sampling_time();
% Get the disc-time ol-sys in SS representation
P_disc = ss(c2d(P,hk));
Ad = P_disc.A;
Bd = P_disc.B;
Cd = P_disc.C;
% Get the PID gains
[Kp,Ki,Kd] = get_PIDgains();
% Generate the reference signal and the time vector
[r,t] = getReference(hk);
%% Perform the system simulation
x = [0 0]'; % Set initial states
e = 0; % Set initial errors
integral_sum = 0; % Set initial integral part value
for i=2:1:length(t)
% Calculate the output signal "y"
y(:,i) = Cd*x;
% Calculate the error "e"
e(:,i) = y(:,i) - r(i);
% Calculate the control signal vector "u"
integral_sum = integral_sum + Ki*hk*e(i);
u(:,i) = Kp*e(i) + integral_sum + (1/hk)*Kd*(e(:,i)-e(:,i-1));
% Saturation. Limit the value of u withing the range [-tol tol]
% tol = 100;
% if abs(u(:,i)) > tol
% u(:,i) = tol * abs(u(:,i))/u(:,i);
% else
% end
% Calculate the state vector "x"
x = Ad*x + Bd*u(:,i); % State transitions to time n
end
%% Subplots
figure
plot(t,y,'b',t,r,'g--')
%% Plotting the error
figure
p = plot(t,e,'r');
set(p,'linewidth',2)
legend('error')
xlabel('Time (seconds)')
ylabel('error')
end
function P = getContPlant()
s = tf('s');
P = 1/(s^2 + 10*s + 20);
end
function [Kp,Ki,Kd] = get_PIDgains()
Kp = 350;
Ki = 300;
Kd = 50;
end
function hk = get_sampling_time()
hk = 0.01;
end
function [r,t] = getReference(hk)
[r,t] = gensig('square',4,10,hk);
end
I got the plant model P and its PID controller from this page (see equation 10), where the system is simulated against a step reference and the result looks pretty much exactly like the lsim() result (just for a single step peak).
However, the result of simulating the system using lsim() is this:
whereas, using the for loop, I got this performance:
I would highly appreciate any help or clarification why I am getting different results.
I have the following code:
Fs = 1000;
T = 1/Fs;
L = 1000;
t = (0:L-1)*T;
k = 25:1:50;
m = 1:1:25;
where k and m are corresponding. I want to plot the 25 sine waves resulting from:
x = m*sin(2*pi*k*t);
I thought about doing it using a for loop that takes one value from m and k each time, but I'm unsure how to proceed.
Below is a very basic plotting solution. You will notice that it's very difficult to see what's going on in the plot, so you might want to consider other ways to present this data.
function q45532082
Fs = 1000;
T = 1/Fs;
L = 1000;
t = (0:L-1)*T;
k = 26:1:50;
m = 1:1:25;
%% Plotting
assert(numel(m) == numel(k)); % We make sure that the number of elements is the same.
figure(); hold on; % "hold" is needed if you want to see all curves at the same time.
for ind1 = 1:numel(m)
plot(t,m(ind1)*sin(2*pi*k(ind1)*t));
end
This is the result:
Note that the number of elements in k and m in your code is different, so I had to change it.
Using the functionality of plot you can also plot all sine waves without a loop:
Fs = 1000;
T = 1/Fs;
L = 1000;
t = (0:L-1)*T;
k = 26:1:50;
m = 1:1:25;
x = m.*sin(2.*pi.*bsxfun(#times,t.',k)); %this results in an L*25 matrix, each column is data of one wave
% or, if you have version 2016b or newer:
% x = m.*sin(2.*pi.*t.'*k);
plot(t,x) % plot all sines at ones
and as #Dev-iL noted, I also had to change k.
The result with L = 1000 is too crowded, so I plot it here with L = 50:
I want to plot multiple realizations of a stochastic process in matlab. For a single realization I have the following code:
N = 80;
T = dt*N;
dWt = zeros(1,N);
S= repmat(S0,1,N);
S(1) = S0;
dWt = sqrt(dt) * randn;
for t=2:N
dWt(t) = sqrt(dt)*randn;
dSt = k*(mu-S(t-1))*dt + sigma*S(t-1)*dWt(t);
S(t) = S(t-1)+dSt;
end
plot(handles.pMeasure, [0:dt:T],[S0,S]);
I want to replicate this loop n times and plot the results in one plot.
You could add an additional for loop, but it would be best to vectorize everything and calculate all n instances at once:
k = ...
mu = ...
sigma = ...
S0 = ... % Initial condition
dt = ... % Time step
n = ... % Number of instances
N = 80; % Number of time steps, not counting initial condition
T = dt*N; % Final time
rng(1); % Always seed random number generator
dWt = sigma*sqrt(dt)*randn(n,N); % Calculate Wiener increments
S = zeros(n,N+1); % Allocate
S(:,1) = S0; % Set initial conditions
for t = 2:N+1
S(:,t) = S(:,t-1) + k*(mu-S(:,t-1))*dt + S(:,t-1).*dWt(:,t-1);
end
plot(handles.pMeasure,0:dt:T,S)
There are further ways to optimize this if want or you can also try sde_euler in my SDETools Matlab toolbox:
k = ...
mu = ...
sigma = ...
dt = ... % Time step
n = ... % Number of instances
N = 80; % Number of time steps, not counting initial condition
T = dt*N; % Final time
f = #(t,y)k*(mu-y); % Diffusion function
g = #(t,y)sigma*y; % Drift function
t = 0:dt:T; % Time vector
S0 = zeros(n,1); % Initial conditions
opts = sdeset('RandSeed',1,...
'SDEType','Ito'); % Set random seed, specify Ito SDE
S = sde_euler(f,g,t,S0,opts); % Simulate
plot(t,S)
How to make a certain pulse function with Matlab, like this function on the graph:
I have the function which generates a pulse, but I want to generate as the graph above
clear all; clc; close all
% Step 1: Calculate Fourier Transform Coefficients
N = 101; % Number of coefficients
D = [];
% Negative Coefficients
for n = -(N-1)/2 : -1
D = [D (1/(n*pi))*sin(n*pi/2)];
end
% Append D0
D0 = 0.5; D = [D D0];
% Positive Coefficients
for n = 1 : (N-1)/2
D = [D (1/(n*pi))*sin(n*pi/2)];
end
% Step 2: Calculate the series terms
series = [];
f0 = 1; % Frequency in Hz
T0 = 1/f0; % Period in seconds
dt = T0/100; % Sample interval
t = -T0 : dt : T0;
for n = -(N-1)/2 : (N-1)/2
r = n + (N+1)/2;
temp = D(r)*exp(j*n*2*pi*f0*t);
series = [series; temp];
end
% Step 3: Summation of the series terms
w = 8*sum(series);
% Step 4: Plot the approximated signal
plot(t,w); grid
xlabel('Time, seconds'); ylabel('signal, s(t)')
title('An approximation of a rectangular singal using Fourier series')'
You need to calculate the continuous Fourier series for your waveform, then modify the coefficients in D to match.
You can find information on general Fourier series elsewhere on the internet.