I am using matlab to plot complex exponential function.But i am not getting the required waveform as output.
My Signal is exp ( j2πmf ) where m takes various positive values.
My code is shown below.
close all;
clc;
f= -0.5:0.5;
Rez = cos(2*pi*1*f);
Imz = sin(2*pi*1*f)*j;
z = Rez + Imz;
z_n = exp(z);
plot(f,z_n);
xlabel('Frequency ->');
ylabel('Amplitude->');
grid on
axis tight
My Output Signal
But I want the signal shown below as my output
First of all, you try to make a plot of complex numbers z_n.
This does not make any sense.
You can plot the real part (real(z_n), the imaginary part (imag(z_n) or the absolute values (abs(z_n).
However, your exceptions in the second diagram are wrong too.
Your function exp(j2πmf) is a rotating vector with absolute amplitude 1.
This results in:
real(exp(j2πmf) = cos(2πmf)
imag (exp(j2πmf) = sin(2πmf); By the way, imag does not include j, it is the value which has j as a factor.
abs(exp(j2πmf) = 1
Related
I wrote code below in MATLAB to plot cosine function from derivative of sine function but output plot was not what I expect to be!
clear;
clc;
close all;
delta = 1e-15;
t = linspace(0, 20, 1000);
y_derived = (sin(t + delta) - sin(t)) / delta;
y_expected = cos(t);
hold on
plot(y_derived)
plot(y_expected)
legend('y_{derived}', 'y_{expected}')
grid on
output plot is like this :
Can any one help me whats happen?
MATLAB plots exactly what you tell it to plot. The issue is in the way you compute the derivative: Your finite difference quotient uses a delta = 1e-15 which is very close to the machine precision eps = 2.2e-16, which is why you get lots of rounding errors. Actually the staircase-ness nicely shows the discrete nature of the number type you're using. Set e.g. delta = 1e-6 and it will probably look a lot better.
Theoretical result for convolution of an exponential function with a sinusoidal function is shown below.
When I plot the function directly using matlab, I get this,
However Matlab conv command yields this,
The two plots look similar but they are not the same, see the scales. The matlab yield is ten times of the theoretical result. Why?
The matlab code is here.
clc;
clear all;
close all;
t = 0:0.1:50;
x1 = exp(-t);
x2 = sin(t);
x = conv(x1,x2);
x_theory = 0.5.*(exp(-t) + sin(t) - cos(t));
figure(1)
subplot(313), plot(t, x(1:length(t))); subplot(311), plot(t, x1(1:length(t))); subplot(312), plot(t, x2(1:length(t)))
figure(2)
subplot(313), plot(t, x_theory); subplot(311), plot(t, x1(1:length(t))); subplot(312), plot(t, x2(1:length(t)))
conv does discrete-time convolution, it does not do the mathematical integral function. Numerically, this basically means multiplying and adding the result of the two signals many times, once for each point, with a small shift of one of the signals.
If you think about this, you will realize that the sampling of the signals will have an effect. i.e. if you have a point every 0.1 values, or 0.001 values, the amount of points you multiply is different, and thus the result is different in value (not in shape).
Therefore, every time you do numerical convolution, you always need to multiply with the sampling rate, to "normalize" the operation.
Just change your code to do
sampling_rate= 0.1;
t = 0:sampling_rate:50;
x = conv(x1,x2)*sampling_rate;
I have adapted the code in Comparing FFT of Function to Analytical FT Solution in Matlab for this question. I am trying to do FFTs and comparing the result with analytical expressions in the Wikipedia tables.
My code is:
a = 1.223;
fs = 1e5; %sampling frequency
dt = 1/fs;
t = 0:dt:30-dt; %time vector
L = length(t); % no. sample points
t = t - 0.5*max(t); %center around t=0
y = ; % original function in time
Y = dt*fftshift(abs(fft(y))); %numerical soln
freq = (-L/2:L/2-1)*fs/L; %freq vector
w = 2*pi*freq; % angular freq
F = ; %analytical solution
figure; subplot(1,2,1); hold on
plot(w,real(Y),'.')
plot(w,real(F),'-')
xlabel('Frequency, w')
title('real')
legend('numerical','analytic')
xlim([-5,5])
subplot(1,2,2); hold on;
plot(w,imag(Y),'.')
plot(w,imag(F),'-')
xlabel('Frequency, w')
title('imag')
legend('numerical','analytic')
xlim([-5,5])
If I study the Gaussian function and let
y = exp(-a*t.^2); % original function in time
F = exp(-w.^2/(4*a))*sqrt(pi/a); %analytical solution
in the above code, looks like there is good agreement when the real and imaginary parts of the function are plotted:
But if I study a decaying exponential multiplied with a Heaviside function:
H = #(x)1*(x>0); % Heaviside function
y = exp(-a*t).*H(t);
F = 1./(a+1j*w); %analytical solution
then
Why is there a discrepancy? I suspect it's related to the line Y = but I'm not sure why or how.
Edit: I changed the ifftshift to fftshift in Y = dt*fftshift(abs(fft(y)));. Then I also removed the abs. The second graph now looks like:
What is the mathematical reason behind the 'mirrored' graph and how can I remove it?
The plots at the bottom of the question are not mirrored. If you plot those using lines instead of dots you'll see the numeric results have very high frequencies. The absolute component matches, but the phase doesn't. When this happens, it's almost certainly a case of a shift in the time domain.
And indeed, you define the time domain function with the origin in the middle. The FFT expects the origin to be at the first (leftmost) sample. This is what ifftshift is for:
Y = dt*fftshift(fft(ifftshift(y)));
ifftshift moves the origin to the first sample, in preparation for the fft call, and fftshift moves the origin of the result to the middle, for display.
Edit
Your t does not have a 0:
>> t(L/2+(-1:2))
ans =
-1.5000e-05 -5.0000e-06 5.0000e-06 1.5000e-05
The sample at t(floor(L/2)+1) needs to be 0. That is the sample that ifftshift moves to the leftmost sample. (I use floor there in case L is odd in size, not the case here.)
To generate a correct t do as follows:
fs = 1e5; % sampling frequency
L = 30 * fs;
t = -floor(L/2):floor((L-1)/2);
t = t / fs;
I first generate an integer t axis of the right length, with 0 at the correct location (t(floor(L/2)+1)==0). Then I convert that to seconds by dividing by the sampling frequency.
With this t, the Y as I suggest above, and the rest of your code as-is, I see this for the Gaussian example:
>> max(abs(F-Y))
ans = 4.5254e-16
For the other function I see larger differences, in the order of 6e-6. This is due to the inability to sample the Heaviside function. You need t=0 in your sampled function, but H doesn't have a value at 0. I noticed that the real component has an offset of similar magnitude, which is caused by the sample at t=0.
Typically, the sampled Heaviside function is set to 0.5 for t=0. If I do that, the offset is removed completely, and max difference for the real component is reduced by 3 orders of magnitude (largest errors happen for values very close to 0, where I see a zig-zag pattern). For the imaginary component, the max error is reduced to 3e-6, still quite large, and is maximal at high frequencies. I attribute these errors to the difference between the ideal and sampled Heaviside functions.
You should probably limit yourself to band-limited functions (or nearly-band-limited ones such as the Gaussian). You might want to try to replace the Heaviside function with an error function (integral of Gaussian) with a small sigma (sigma = 0.8 * fs is the smallest sigma I would consider for proper sampling). Its Fourier transform is known.
I'm trying to find two x values for each y value on a plot that is very similar to a Gaussian fn. The difficulty is that I need to be able to find the values of x for several values of y even when the gaussian fn is very close to zero.
I can't post an image due to being a new user, however think of a gaussian function and then the regions where it is close to zero on either side of the peak. This part where the fn is very close to reaching zero is where I need to find the x values for a given y.
What I've tried:
When the fn is discrete: I have tried interp1, however I get the error that it is not strictly monotonic increasing because of the many values that are close to zero.
When I fit a two-term gaussian:
I use fzero (fzero(function-yvalue)) however I get a lot of NaN's. These might be from me not having a close enough 'guess' value??
Does anyone have any other suggestions for me to try? Or how to improve what I've already attempted?
Thanks everyone
EDIT:
I've added a picture below. The data that I actually have is the blue line, while the fitted eqn is in red. The eqn should be accurate enough.
Again, I'm trying to pick out x values for a given y where y is very small (approaching 0).
I've tried splitting the function into left and right halves for the interpolation and fzero method.
Thanks for your responses anyway, I'll have a look at bisection.
Fitting a Gaussian seems to be uneffective, as its deviation (in the x-coordinate) from the real data is noticeable.
Since your data is already presented as a numeric vector y, the straightforward find(y<y0) seems adequate. Here is a sample code, in which the y-values are produced from a perturbed Gaussian.
x = 0:1:700;
y = 2000*exp(-((x-200)/50).^2 - sin(x/100).^2); % imitated data
plot(x,y)
y0 = 1e-2; % the y-value to look for
i = min(find(y>y0)); % first entry above y0
if i == 1
x1 = x(i);
else
x1 = x(i) - y(i)*(x(i)-x(i-1))/(y(i)-y(i-1)); % linear interpolation
end
i = max(find(y>y0)); % last entry above y0
if i == numel(y)
x2 = x(i);
else
x2 = x(i) - y(i)*(x(i)-x(i+1))/(y(i)-y(i+1)); % linear interpolation
end
fprintf('Roots: %g, %g \n', x1, x2)
Output: Roots: 18.0659, 379.306
The curve looks much like your plot.
I have a data-set which is loaded into matlab. I need to do exponential fitting for the plotted curve without using the curve fitting tool cftool.
I want to do this manually through executing a code/function that will output the values of a and b corresponding to the equation:
y = a*exp(b*x)
Then be using those values, I will do error optimization and create the best fit for the data I have.
Any help please?
Thanks in advance.
Try this...
f = fit(x,y,'exp1');
I think the typical objective in this type of assignment is to recognize that by taking the log of both sides, various methods of polynomial fit approaches can be used.
ln(y) = ln(a) + ln( exp(x).^b )
ln(y) = ln(a) + b * ln( exp(x) )
There can be difficulties with this approach when errors such as noise are involved due to the behavior of ln as it approaches zero.
In this exercise I have a set of data that present an exponential curve and I want to fit them exponentially and get the values of a and b. I used the following code and it worked with the data I have.
"trail.m" file:
%defining the data used
load trialforfitting.txt;
xdata= trialforfitting(:,1);
ydata= trialforfitting(:,2);
%calling the "fitcurvedemo" function
[estimates, model] = fitcurvedemo(xdata,ydata)
disp(sse);
plot(xdata, ydata, 'o'); %Data curve
hold on
[sse, FittedCurve] = model(estimates);
plot(xdata, FittedCurve, 'r')%Fitted curve
xlabel('Voltage (V)')
ylabel('Current (A)')
title('Exponential Fitting to IV curves');
legend('data', ['Fitting'])
hold off
"fitcurvedemo.m" file:
function [estimates, model] = fitcurvedemo(xdata, ydata)
%Call fminsearch with a random starting point.
start_point = rand(1, 2);
model = #expfun;
estimates = fminsearch(model, start_point);
%"expfun" accepts curve parameters as inputs, and outputs
%the sum of squares error [sse] expfun is a function handle;
%a value that contains a matlab object methods and the constructor
%"FMINSEARCH" only needs sse
%estimate returns the value of A and lambda
%model computes the exponential function
function [sse, FittedCurve] = expfun(params)
A = params(1);
lambda = params(2);
%exponential function model to fit
FittedCurve = A .* exp(lambda * xdata);
ErrorVector = FittedCurve - ydata;
%output of the expfun function [sum of squares of error]
sse = sum(ErrorVector .^ 2);
end
end
I have a new set of data that doesn't work with this code and give the appropriate exponential fit for the data curve plotted.