I'm trying to solve the following problem using MATLAB but I faced multiple issues. The plot I obtained doesn't seem right even though I tried to obtain the steady-state solution, I got a plot that doesn't look steady.
The problem I'm trying to solve
The incorrect plot I got.
and here is the code
% system parameters
m=1; k=1; c=.1; wn=sqrt(k/m); z=c/2/sqrt(m*k); wd=wn*sqrt(1-z^2);
% initial conditions
x0=0; v0=0;
%% time
dt=.001; tMax=8*pi; t=0:(tMax-0)/999:tMax;
% input
A=1
omega=(2*pi)/10
F=A/2-(4*A/pi^2)*cos(omega*t); Fw=fft(F);
F=k*A*cos(omega*t); Fw=fft(F);
% normalize
y = F/m;
% compute coefficients proportional to the Fourier series coefficients
Yw = fft(y);
% setup the equations to solve the particular solution of the differential equation
% by the method of undetermined coefficients
N=1000
T=10
k = [0:N/2];
w = 2*pi*k/T;
A = wn*wn-w.*w;
B = 2*z*wn*w;
% solve the equation [A B;-B A][real(Xw); imag(Xw)] = [real(Yw); imag(Yw)] equation
% Note that solution can be obtained by writing [A B;-B A] as a scaling + rotation
% of a 2D vector, which we solve using complex number algebra
C = sqrt(A.*A+B.*B);
theta = acos(A./C);
Ywp = exp(j*theta)./C.*Yw([1:N/2+1]);
% build a hermitian-symmetric spectrum
Xw = [Ywp conj(fliplr(Ywp(2:end-1)))];
% bring back to time-domain (function synthesis from Fourier Series coefficients)
x = ifft(Xw);
figure()
plot(t,x)
Your forcing function doesn't look like the triangle wave in the problem. I edited the %% time section of your code into the following and appeared to give a steady state response.
%% time
TP = 10; % forcing time period (10 s)
dt=.001;
tMax= 3*TP; % needs to be multiple of the time period
t=0:(tMax-0)/999:tMax;
% input
A=1; % Forcing amplitude
omega=(2*pi)/TP;
% forcing is a triangle wave
% generate a triangle wave with min/max values of 0/1.
F = 0*t;
for i = 1:length(t)
if mod(t(i), TP) <= TP/2
F(i) = mod(t(i), TP)/(TP/2);
else
F(i) = 2 - mod(t(i), TP)/(TP/2);
end
end
F = F*A; % scale triangle wave by amplitude
% you can also use MATLAB's sawtooth() function if you have the signal
% processing toolbox
Related
I am using the following code to generate a fft and mathematical Fourier transform of a signal. I want to then mathematically recreate the original signal of the fft. This works on the mathematical signal but not on the fft since it is a Discrete Transform. Does anyone know what change I can make to my inverse transform equation that will make it work for fft?
clear all; clc;
N = 1024;
N2 = 1023;
SNR = -10;
fs = 1024;
Ts = 1/fs;
t = (0:(N-1))*Ts;
x = 0.5*sawtooth(2*2*pi*t);
x1 = fft(x);
Magnitude1 = abs(x1);
Phase1 = angle(x1)*360/(2*pi);
for m = 1:1024
f(m) = m; % Sinusoidal frequencies
a = (2/N)*sum(x.*cos(2*pi*f(m)*t)); % Cosine coeff.
b = (2/N)*sum(x.*sin(2*pi*f(m)*t)); % Sine coeff
Magnitude(m) = sqrt(a^2 + b^2); % Magnitude spectrum
Phase(m) = -atan2(b,a); % Phase spectrum
end
subplot(2,1,1);
plot(f,Magnitude1./512); % Plot magnitude spectrum
......Labels and title.......
subplot(2,1,2);
plot(f,Magnitude,'k'); % Plot phase spectrum
ylabel('Phase (deg)','FontSize',14);
pause();
x2 = zeros(1,1024); % Waveform vector
for m = 1:24
f(m) = m; % Sinusoidal frequencies
x2 = (1/m)*(x2 + Magnitude1(m)*cos(2*pi*f(m)*t + Phase1(m)));
end
x3 = zeros(1,1024); % Waveform vector
for m = 1:24
f(m) = m; % Sinusoidal frequencies
x3 = (x3 + Magnitude(m)*cos(2*pi*f(m)*t + Phase(m)));
end
plot(t,x,'--k'); hold on;
plot(t,x2,'k');
plot(t,x3,'b');```
There are a few comments about the Fourier Transform, and I hope I can explain everything for you. Also, I don't know what you mean by "Mathematical Fourier transform", as none of the expressions in your code is resembles the Fourier series of the sawtooth wave.
To understand exactly what the fft function does, we can do things step by step.
First, following your code, we create and plot one period of the sawtooth wave.
n = 1024;
fs = 1024;
dt = 1/fs;
t = (0:(n-1))*dt;
x = 0.5*sawtooth(2*pi*t);
figure; plot(t,x); xlabel('t [s]'); ylabel('x');
We can now calculate a few things.
First, the Nyquist frequency, the maximum detectable frequency from the samples.
f_max = 0.5*fs
f_max =
512
Also, the minimum detectable frequency,
f_min = 1/t(end)
f_min =
1.000977517106549
Calculate now the discrete Fourier transform with MATLAB function:
X = fft(x)/n;
This function obtains the complex coefficients of each term of the discrete Fourier transform. Notice it calculates the coefficients using the exp notation, not in terms of sines and cosines. The division by n is to guarantee that the first coefficient is equal to the arithmetic mean of the samples
If you want to plot the magnitude/phase of the transformed signal, you can type:
f = linspace(f_min,f_max,n/2); % frequency vector
a0 = X(1); % constant amplitude
X(1)=[]; % we don't have to plot the first component, as it is the constant amplitude term
XP = X(1:n/2); % we get only the first half of the array, as the second half is the reflection along the y-axis
figure
subplot(2,1,1)
plot(f,abs(XP)); ylabel('Amplitude');
subplot(2,1,2)
plot(f,angle(XP)); ylabel('Phase');
xlabel('Frequency [Hz]')
What does this plot means? It shows in a figure the amplitude and phase of the complex coefficients of the terms in the Fourier series that represent the original signal (the sawtooth wave). You can use this coefficients to obtain the signal approximation in terms of a (truncated) Fourier series. Of course, to do that, we need the whole transform (not only the first half, as it is usual to plot it).
X = fft(x)/n;
amplitude = abs(X);
phase = angle(X);
f = fs*[(0:(n/2)-1)/n (-n/2:-1)/n]; % frequency vector with all components
% we calculate the value of x for each time step
for j=1:n
x_approx(j) = 0;
for k=1:n % summation done using a for
x_approx(j) = x_approx(j)+X(k)*exp(2*pi*1i/n*(j-1)*(k-1));
end
x_approx(j) = x_approx(j);
end
Notice: The code above is for clarification and does not intend to be well coded. The summation can be done in MATLAB in a much better way than using a for loop, and some warnings will pop up in the code, warning the user to preallocate each variable for speed.
The above code calculates the x(ti) for each time ti, using the terms of the truncated Fourier series. If we plot both the original signal and the approximated one, we get:
figure
plot(t,x,t,x_approx)
legend('original signal','signal from fft','location','best')
The original signal and the approximated one are nearly equal. As a matter of fact,
norm(x-x_approx)
ans =
1.997566360514140e-12
Is almost zero, but not exactly zero.
Also, the plot above will issue a warning, due to the use of complex coefficients when calculating the approximated signal:
Warning: Imaginary parts of complex X and/or Y arguments ignored
But you can check that the imaginary term is very close to zero. It is not exactly zero due to roundoff errors in the computations.
norm(imag(x_approx))
ans =
1.402648396024229e-12
Notice in the codes above how to interpret and use the results from the fft function and how they are represented in the exp form, not on terms of sines and cosines, as you coded.
I am using following MATLAB code for implementing 1D diffusion equation along a rod with implicit finite difference method. `
xsize = 10; % Model size, m
xnum = 10; % Number of nodes
xstp = xsize/(xnum-1); % Grid step
tnum = 504; % number of timesteps
kappa = 833.33; % Thermal diffusivity, m^2/s
dt = 300; % Timestep
x = 0:xstp:xsize; %Creating vector for nodal point positions
tlbc = sin(linspace(0.1,2.9,tnum)); % left boundary condition
%Define initial temperature profile
tback = 0; % background temperature, K
for i=1:1:xnum
% Background profile
t0imp(i) = tback; % profile for implicit solving
end
% Time cycle
timesum=0; % Elapsed time
for t=1:1:tnum
% Matrix of coefficients initialization for implicit solving
L = sparse(xnum,xnum);
% Vector of right part initialization for implicit solving
R = zeros(xnum,1);
% Implicit solving of 1D temperature equation: dT/dt=kappa*d2T/dx2
% Composing matrix of coefficients L()
% and vector (column) of right parts R()
% First point: T=tback
L(1,1) = 1;
R(1,1) = tlbc(t);
% Intermediate points
for i=2:1:xnum-1
% dT/dt=kappa*d2T/dx2
% Tnew(i)/dt-kappa*(Tnew(i-1)-2*Tnew(i)+Tnew(i+1))/dx^2=Told(i)/dt
L(i,i-1) = -kappa/xstp^2;
L(i,i) = 1/dt+2*kappa/xstp^2;
L(i,i+1) = -kappa/xstp^2;
R(i,1) = t0imp(i)/dt;
end
% Last point:T=tback
L(xnum,xnum) = 1;
R(xnum,1) = tback;
% Obtaining solution for implicit temperature profile
t1imp = L\R;
end `
I want to change right boundary condition to infinite boundary condition so that presence of boundary at 10 m has no effect on heat flow and temperature flows in 10 m domain as if there was no boundary at 10 m.
If you have localized phenomenon that is far away from the boundary then I suggest you use a non-uniform grid for your x-space discretization. The mesh should have more points in the region of interest. You can then choose a large xsize but keep the number of mesh points reasonable.
Non-uniform grids are trickier to implement (not that bad for 1-d) but fortunately Matlab has pdepe that solves non-uniform equations of the type
\begin{equation}
c\left(x , t,u, \frac{\partial u}{\partial x} \right) \frac{\partial u}{\partial t} = x^{-m} \frac{\partial}{\partial t} \left( x^m f\left(x , t,u, \frac{\partial u}{\partial x} \right)\right) + S\left(x , t,u, \frac{\partial u}{\partial x} \right)
\end{equation}
with user defined meshes. You can read more about pdepe here
https://www.mathworks.com/help/matlab/ref/pdepe.html
They also solve the equation you are interested in.
I have my data in arrays which are exponential like e^(ax+c)+d. I'm trying to draw a fit to them.
a = data1 (:,1);
b = data1 (:,2);
log(b);
p = polyfit (a,log(b),1);
But I don't know what to do now. I found an equation by polyfit and I was hoping to take the exponential of the equation I got from polyfit with
exp (0.5632x+2.435)
But I figured out that it doesn't work like that. Does anyone have any suggestions?
try with nonlinear fitting:
%% PARAMETERS (you need this part)
clear all;
clc, clf;
N = 128; % number of datapoints
Nint = N*10; % number of datapoints for curve interpolation
fun = #(prms,x) prms(4).^(prms(1)*x+prms(2))+prms(3); % write your function
iniPrm = rand(4,1); % find some initial values for the parameters (choose meaningful values for better results)
%% SIMULATE DATA (this is only for testing purposes)
SNR = .01; % signal to noise ratio for simulated data
noise = (rand(1,N)-.5)*SNR; % create some random noise
x = linspace(0,10,N); % create the x axis
y = fun(iniPrm,x) + noise; % simulate a dataset that follows the given function
x = x(:); % reshape as a vector
y = y(:); % reshape as a vector
X = linspace(x(1),x(end),Nint); % interpolate the output to plot it smoothly
plot(x,y,'.r','markersize',10); hold on; % plot the dataset
%% FIT AND INTERPOLATE YOUR MODEL
[out.BETA,out.RESID,out.J,out.COVB,out.MSE] = nlinfit(x,y,fun,iniPrm,[]); % model your data
[out.YPRED,out.DELTA] = nlpredci(fun,X,out.BETA,out.RESID,'Covar',out.COVB); % interpolate your model
out.YPREDLOWCI = out.YPRED - out.DELTA; % find lower confidence intervals of your fitting
out.YPREDUPCI = out.YPRED + out.DELTA; % find upper confidence intervals of your fitting
out.X = X; % store the interpolated X
%% PLOT FITTING
plotCI = #(IO,spec) patch([IO.X(:);flipud(IO.X(:))],[IO.YPREDLOWCI(:);flipud(IO.YPREDUPCI(:))],spec{:}); % create patches: IE: patch(0:10,10:-1:0,ones(10,1)-1,1,{'r','facealpha',0.2})
plot(X,out.YPRED,'-b','linewidth',3);
plotCI(out,{'r','facealpha',.3,'edgealpha',0})
The error
File: parameter_estimation_1.m Line: 8 Column: 10
Unexpected MATLAB expression.
crops up when I run the following MATLAB code:
T = 0:0.25:5; % time vector (row)
T = T'; % time vector (column)
seed = [3;0.5]; % seed for noise generation
randn('state',seed); % using the same seed each time
uu = 0.5 1 0.25*randn(length(T),1); % mean of 0.5 with variance
% of 0.25
U = 2*round(uu)-1; % creates PRBS with -1 and 1 values %
sim('est_vdv'); % runs simulation of linear van de vusse % diagram
figure(1); % plot input-output data
subplot(2,1,1),plot(tp,yp,'k',t,y,'ko');
xlabel('time, min'), ylabel('y')
title('PRBS estimation example')
subplot(2,1,2),plot(tp,up,'k'); xlabel('time, min'),ylabel('u')
axis([0 5 -1.1 1.1])
% % generate phi matrix for estimation
for j = 4:22;
phi(j-3,:) = [y(j-2) y(j-3) u(j-2) u(j-3)];
end
%
theta = inv(phi'*phi)*phi'*y(3:21) % estimate parameters
num = [theta(3) theta(4)]; % numerator of discrete transfer function
den = [1 -theta(1) -theta(2)]; % denominator of discrete transfer function
sysd = tf(num,den,0.25) % create discrete tf object
tzero(sysd) % calculate zeros
pole(sysd) % calculate poles
syszpk = zpk(sysd) % zero-pole-k form
This code is supposed to run in tandem with a SIMULINK model titled "est_vdv" to estimate the parameters of a model.
How should I deal with this error?
Thanks Friends for your suggestions, I have been able to figure out what went wrong with the 5th line, it should have been
uu=0.5+0.25.*randn(length(T),1)
Sorry, it was wrongly indicated that the error was in the 8th line.
I have written a matlab function (Version 7.10.0.499 (R2010a)) to evaluate incoming FT signal and calculate the morlet wavelet for the signal. I have a similar program, but I needed to make it more readable and closer to mathematical lingo. The output plot is supposed to be a 2D plot with colour showing the intensity of the frequencies. My plot seems to have all frequencies the same per time. The program does make an fft per row of time for each frequency, so I suppose another way to look at it is that the same line repeats itself per step in my for loop. The issue is I have checked with the original program, which does return the correct plot, and I cannot locate any difference beyond what I named the values and how I organized the code.
function[msg] = mile01_wlt(FT_y, f_mn, f_mx, K, N, F_s)
%{
Fucntion to perform a full wlt of a morlet wavelett.
optimization of the number of frequencies to be included.
FT_y satisfies the FT(x) of 1 envelope and is our ft signal.
f min and max enter into the analysis and are decided from
the f-image for optimal values.
While performing the transformation there are different scalings
on the resulting "intensity".
Plot is made with a 2D array and a colour code for intensity.
version 05.05.2016
%}
%--------------------------------------------------------------%
%{
tableofcontents:
1: determining nr. of analysis f, prints and readies f's to be used.
2: ensuring correct orientation of FT_y
3:defining arrays
4: declaring waveletdiagram and storage of frequencies
5: for-loop over all frequencies:
6: reducing file to manageable size by truncating time.
7: marking plot to highlight ("randproblemer")
8: plotting waveletdiagram
%}
%--------------------------------------------------------------%
%1: determining nr. of analysis f, prints and readies f's to be used.
DF = floor( log(f_mx/f_mn) / log(1+( 1/(8*K) ) ) ) + 1;% f-spectre analysed
nr_f_analysed = DF %output to commandline
f_step = (f_mx/f_mn)^(1/(DF-1)); % multiplicative step for new f_a
f_a = f_mn; %[Hz] frequency of analysis
T = N/F_s; %[s] total time sampled
C = 2.0; % factor to scale Psi
%--------------------------------------------------------------%
%2: ensuring correct orientation of FT_y
siz = size(FT_y);
if (siz(2)>siz(1))
FT_y = transpose(FT_y);
end;
%--------------------------------------------------------------%
%3:defining arrays
t = linspace(0, T*(N-1)/N, N); %[s] timespan
f = linspace(0, F_s*(N-1)/N, N); %[Hz] f-specter
%--------------------------------------------------------------%
%4: declaring waveletdiagram and storage of frequencies
WLd = zeros(DF,N); % matrix of DF rows and N columns for storing our wlt
f_store = zeros(1,DF); % horizontal array for storing DF frequencies
%--------------------------------------------------------------%
%5: for-loop over all frequencies:
for jj = 1:DF
o = (K/f_a)*(K/f_a); %factor sigma
Psi = exp(- 0*(f-f_a).*(f-f_a)); % FT(\psi) for 1 envelope
Psi = Psi - exp(-K*K)*exp(- o*(f.*f)); % correctional element
Psi = C*Psi; %factor. not set in stone
%next step fits 1 row in the WLd (3 alternatives)
%WLd(jj,:) = abs(ifft(Psi.*transpose(FT_y)));
WLd(jj,:) = sqrt(abs(ifft(Psi.*transpose(FT_y))));
%WLd(jj,:) = sqrt(abs(ifft(Psi.*FT_y))); %for different array sizes
%and emphasizes weaker parts.
%prep for next round
f_store (jj) = f_a; % storing used frequencies
f_a = f_a*f_step; % determines the next step
end;
%--------------------------------------------------------------%
%6: reducing file to manageable size by truncating time.
P = floor( (K*F_s) / (24*f_mx) );%24 not set in stone
using_every_P_point = P %printout to cmdline for monitoring
N_P = floor(N/P);
points_in_time = N_P %printout to cmdline for monitoring
% truncating WLd and time
WLd2 = zeros(DF,N_P);
for jj = 1:DF
for ii = 1:N_P
WLd2(jj,ii) = WLd(jj,ii*P);
end
end
t_P = zeros(1,N_P);
for ii = 1:N_P % set outside the initial loop to reduce redundancy
t_P(ii) = t(ii*P);
end
%--------------------------------------------------------------%
%7: marking plot to highlight boundary value problems
maxval = max(WLd2);%setting an intensity
mxv = max(maxval);
% marks in wl matrix
for jj= 1:DF
m = floor( K*F_s / (P*pi*f_store(jj)) ); %finding edges of envelope
WLd2(jj,m) = mxv/2; % lower limit
WLd2(jj,N_P-m) = mxv/2;% upper limit
end
%--------------------------------------------------------------%
%8: plotting waveletdiagram
figure;
imagesc(t_P, log10(f_store), WLd2, 'Ydata', [1 size(WLd2,1)]);
set(gca, 'Ydir', 'normal');
xlabel('Time [s]');
ylabel('log10(frequency [Hz])');
%title('wavelet power spectrum'); % for non-sqrt inensities
title('sqrt(wavelet power spectrum)'); %when calculating using sqrt
colorbar('location', 'southoutside');
msg = 'done.';
There are no error message, so I am uncertain what exactly I am doing wrong.
Hope I followed all the guidelines. Otherwise, I apologize.
edit:
my calling program:
% establishing parameters
N = 2^(16); % | number of points to sample
F_s = 3.2e6; % Hz | samplings frequency
T_t = N/F_s; % s | length in seconds of sample time
f_c = 2.0e5; % Hz | carrying wave frequency
f_m = 8./T_t; % Hz | modulating wave frequency
w_c = 2pif_c; % Hz | angular frequency("omega") of carrying wave
w_m = 2pif_m; % Hz | angular frequency("omega") of modulating wave
% establishing parameter arrays
t = linspace(0, T_t, N);
% function variables
T_h = 2*f_m.*t; % dimless | 1/2 of the period for square signal
% combined carry and modulated wave
% y(t) eq. 1):
y_t = 0.5.*cos(w_c.*t).*(1+cos(w_m.*t));
% y(t) eq. 2):
% y_t = 0.5.*cos(w_c.*t)+0.25*cos((w_c+w_m).*t)+0.25*cos((w_c-w_m).*t);
%square wave
sq_t = cos(w_c.*t).*(1 - mod(floor(t./T_h), 2)); % sq(t)
% the following can be exchanged between sq(t) and y(t)
plot(t, y_t)
% plot(t, sq_t)
xlabel('time [s]');
ylabel('signal amplitude');
title('plot of harmonically modulated signal with carrying wave');
% title('plot of square modulated signal with carrying wave');
figure()
hold on
% Fourier transform and plot of freq-image
FT_y = mile01_fftplot(y_t, N, F_s);
% FT_sq = mile01_fftplot(sq_t, N, F_s);
% Morlet wavelet transform and plot of WLdiagram
%determining K, check t-image
K_h = 57*4; % approximation based on 1/4 of an envelope, harmonious
%determining f min and max, from f-image
f_m = 1.995e5; % minimum frequency. chosen to showcase all relevant f
f_M = 2.005e5; % maximum frequency. chosen to showcase all relevant f
%calling wlt function.
name = 'mile'
msg = mile01_wlt(FT_y, f_m, f_M, K_h, N, F_s)
siz = size(FT_y);
if (siz(2)>siz(1))
FT_y = transpose(FT_y);
end;
name = 'arnt'
msg = arnt_wltransf(FT_y, f_m, f_M, K_h, N, F_s)
The time image has a constant frequency, but the amplitude oscillates resempling a gaussian curve. My code returns a sharply segmented image over time, where each point in time holds only 1 frequency. It should reflect a change in intensity across the spectra over time.
hope that helps and thanks!
I found the error. There is a 0 rather than an o in the first instance of Psi. Thinking I'll maybe rename the value as sig or something. besides this the code works. sorry for the trouble there