So I plot sine(w*time) vs cosine(w*time)
w being angular frequency.
Hope I'm not wasting anyone's time if I ask:
Would this look like a circle?
I've researched a whole bunch but most websites only graph sine and cosine side-by-side and show comparisons.
I got it to look like a circle and I was just wondering if this is correct.
Also, What can I call this plot? I just gave it a title "plot of a circle". But I am wondering if that is professional enough since I am doing it for class.
Thanks for your time and answers. Greatly appreciated.
My MATLAB code for anyone interested:
clear all; clc; % clear the Workspace and the Command Window
f = 2; w = 2*pi*f; % specify a frequency in Hz and convert to rad/sec
T = 0.01; % specify a time increment
time = 0 : T : 0.5; % specify a vector of time points
x = sin(w*time); % evaluate the sine function for each element of the vector time
y = cos(w*time);
plot(x,y)
axis equal
grid on
xlabel('sin(w*time)');ylabel('cos(w*time)');title('Plot of a Circle');
axis([-1.1 1.1 -1.1 1.1]);
print
Here is a link to a Wolfram Alpha query I just did:
http://www.wolframalpha.com/input/?i=x%3Dsin%28t%29%2C+y%3Dcos%28t%29
I am not sure if it what you want to see, but that site (WolframAlpha.com) is a great place to explore and challenge mathematical concepts that are new to you.
Also, I would call it a plot of a circle since that is what the output looks like.
You are making a Lissajous curve. Keep in mind that a cosine is just a sine offset by pi/2 radians, and so plotting a sine against a cosine will indeed result in a circle. Changing the frequency and/or relative phase between x(t) and y(t) will result in many different interesting patterns.
Related
I have a signal 's' of voice of which you can see an extract here:
I would like to plot the zero crossing points in the same graph. I have tried with the following code:
zci = #(v) find(v(:).*circshift(v(:), [-1 0]) <= 0); % Returns Zero-Crossing Indices Of Argument Vector
zx = zci(s);
figure
set(gcf,'color','w')
plot(t,s)
hold on
plot(t(zx),s(zx),'o')
But it does not interpole the points in which the sign change, so the result is:
However, I'd like that the highlighted points were as near as possible to zero.
I hope someone can help me. Thanks you for your responses in advanced.
Try this?
w = 1;
crossPts=[];
for k=1:(length(s)-1)
if (s(k)*s(k+1)<0)
crossPts(w) = (t(k)+t(k+1))/2;
w = w + 1;
end
end
figure
set(gcf,'color','w')
plot(t,s)
hold on
plot(t, s)
plot(crossPts, zeros(length(crossPts)), 'o')
Important questions: what is the highest frequency conponent of the signal you are measuring? Can you remeasure this signal? What is your sampling rate? What is this analysis for? (Schoolwork or scholarly research). You may have quite a bit of trouble measuring the zeros of this function with any significance or accurracy because it looks like your waveform has a frequency greater than half of your sampling rate (greater than your Nyquist frequency). Upsampling/interpolating your entire waveform will allow you to find the zeros much more precisely (but with no greater degree of accurracy) but this is a huge no-no in the scientific community. While my method may not look super pretty, it's the most accurate method that doesn't make unsafe assumptions. If you just want it to look pretty, I would recommend interp1 and using the 'Spline' method. You can interpolate the whole waveform and then use the above answer to find more accurate zeros.
Also, you could calculate the zeros on the interpolated waveform and then display it on the raw data.
A remotely possible solution to improve your data;
If you're measuring a human voice, why not try filtering at the range of human speech? This should be fine mathematically and could possibly improve your waveform.
I'm just learning Matlab and the fast fourier transform algorithm.
As a first step I tried to duplicate this example: https://en.wikipedia.org/wiki/Fourier_transform#Example
I use the following code:
t = -6:0.01:6;
s = cos(2 * pi * 3 * t) .* exp(-pi * t.^2);
figure(1);
plot(t, s);
xlim([-2 2]);
r = fft(s);
figure(2);
plot(t, abs(r));
And I obtained the following picture:
Figure 2:
Figure 1 is OK, but Figure 2 is not. I see one of the problem is that in Figure 2 I should plot vector r against frequency, not against time. Another problem in Figure 2 is the scale in the Y-axis.
Thus, I have 2 questions in order to duplicate the example:
How can I obtain the frequency domain (X-axis in Figure 2)?
How should I scale vector r (Y-axis in Figure 2)?
Your issue is that you aren't actually creating a frequency vector to plot the fft against. The reason that the fft is plotted against time is because that is what you specified in your plot command.
Here is a working fft outline:
N=length(t);
index=0:N-1;
FrequencyResolution=SamplingRate/N;
Frequency=index.*FrequencyResolution;
data_fft=fft(detrend(data));
%the detrend isn't necessary but it does look nicer because it focuses the plot on changes around the mean of the data
data_FFTmagnitude=abs(data_fft);
plot(Frequency, data_FFTmagnitude)
I remember once for the first time that I wanted to use DFT and FFT for one of my study projects I used this webpage, it explains in detail with examples on how to do so. I suggest you go through it and try to replicate for your case, doing so will give you insight and better understanding of the way one can use FFt as you said you are new to Matlab. Do not hesitate to ask again if you need more detailed help.
And also keep in mind that for FFT it is better to have signal length of a power of 2, that way you will get the most exact results, and if you cannot control your signal length you can take the largest power of 2 close to that length, as everyone usually does.
I am new to MATLAB and I wrote some code to generate a sine wave. However the graph is not correct. Here is the screenshot of my code and the plot
What is the problem? Please help!
MATLAB plots discrete points and simply draws a straight line to connect neighbouring points together. Your time points are one second (1s) in between, and you are specifying a frequency of 100 Hz. In addition, because your sampling time is a multiple of the period of your sine wave, substituting all of those values of t would thus make the sin result equal to 0, though there is some numerical imprecision. Specifically, if you look at the y-axis, you'll see that the magnitude of your numbers is around 10^{-13}. However even if you escape this, the sampling time is TOO LARGE for the specified frequency of your wave and so this huge gap in between points is visualized as that jagged wave that you see in your graph.
The solution is to simply make your sampling time smaller. Try something small, like 1e-4 for example:
t = 0:1e-4:0.05;
f = 100;
A = 2;
x = A*sin(2*pi*f*t);
plot(t,x);
We get this now:
I have been working on DFT in Matlab recently, here is my code in Matlab. which part of my code has problem, my sampling is wrong??? I'll be grateful if you answer my question:
dt = 0.01; %sampling time interval
Fs = 1/dt; %sampling rate
t = 0:dt:45; %Time vector
t0 = 5; %duration of applied stress
N = length(t); %number of sample points
y_timedomain = heaviside(t)-heaviside(t-t0); %the step function
figure (1)
plot(y_timedomain)
axis([-100,1000,-0.2,1.2]);
y_freqDomain=abs(fft(y_timedomain)); % fft of step funcion, y(t)
z = fftshift(y_freqDomain); % DFT and shift center to zero
figure (2)
plot(linspace(-10,10,length(y_freqDomain)),z)
xlabel('Sample Number')
ylabel('Amplitude')
title('Using the Matlab fft command')
grid
axis([-.3,.3,0,1000]);
meanwhile, I have 2 question about this code:
1- my step function at 0 time, has magnitude of 1/2, but i want my step function at 0 time be 0 instead of 1/2,( such as rectangle shape), but i don't know how to correct it???
2- when we do DFT, should we use "shift FFT" always????
if you give me your advice about this code i will be really thankful.
Heaviside Step Function
MATLAB does define the step function with 1/2 for x=0 (see http://de.mathworks.com/help/symbolic/heaviside.html?refresh=true), if you wish otherwise you could define your own function, maybe like this here?
mystep = #(x) sign(max(x, 0));
fftshift
No you don't have to use fftshift always, that really depends on what exactly you want to do. In principle, the fft is not very suited for signals like the step function (if you want to do some frequency analysis, in that particular case, I recommend to do analytical ft!). There are many side effects (I don't want to go into that field right now) and fftshift is one of the tools to help deal with those. Read the doc (doc fftshift) and learn more about the ft in general.
I'm trying to display a spectrum of a sound sample with the correct frequency-axis, in Hertz, and a
log-amplitude y-axis. I thought I had the frequency display right, but the graphs that it produces look rather weird and I've found absolutely nothing regarding displaying the log amplitude on an axis.
The code that I'm using (heavily borrowed from mathsworks fft example) is:
y=x(100:200);
Fs = 10000; % Sampling frequency
numsamples = 20000; % Number of samples in the signal
NFFT = 2^nextpow2(numsamples); % Next power of 2 from length of y
Y = fft(y,NFFT)/numsamples;
f = Fs/2*linspace(0,1,NFFT/2+1);
plot(f,2*abs(Y(1:NFFT/2+1)))
And here's one of the graphs that it produced:
I don't know if that's right or not, but it doesn't look anything like the examples I've seen....
I'm aware this is probably a really stupid question and I'm missing something obvious, or maybe I've actually got it right and don't understand enough to realise it, but this stuff is really doing my head in and I'm not finding the documentation particularly illuminating.
I think the x-axis is correct.
To obtain a logarithmic y-axis replace the plot bysemilogy:
semilogy(f,2*abs(Y(1:NFFT/2+1)))
or equivalently use plot as it stands followed by
set(gca,'YScale','log')