GNU octave graph - octave

I have plotted a simple velocity against time graph about vibration on GNU Octave and I would like to know the exact values when the time is 1 and 10s. I'm not very sure on this, please help me on this.

As long as your time array t contains those exact values, and assuming your velocity array is v, you could use
v(find(sum(t==[1 10].')))
and put as many values into that bracketed vector as you need.
If t doesn't have those exact values, like t = [0 0.9 10.1], check out the intelligent use of the unique function at this page
https://www.mathworks.com/matlabcentral/answers/375710-find-nearest-value-to-specific-number

Related

Making sense of soundMixer.computeSpectrum

All examples that I can find on the Internet just visualize the result array of the function computeSpectrum, but I am tasked with something else.
I generate a music note and I need by analyzing the result array to be able to say what note is playing. I figured out that I need to set the second parameter of the function call 'FFTMode' to true and then it returns sound frequencies. I thought that really it should return only one non-zero value which I could use to determine what note I generated using Math.sin function, but it is not the case.
Can somebody suggest a way how I can accomplish the task? Using the soundMixer.computeSpectrum is a requirement because I am going to analyze more complex sounds later.
FFT will transform your signal window into set of Nyquist sine waves so unless 440Hz is one of them you will obtain more than just one nonzero value! For a single sine wave you would obtain 2 frequencies due to aliasing. Here an example:
As you can see for exact Nyquist frequency the FFT response is single peak but for nearby frequencies there are more peaks.
Due to shape of the signal you can obtain continuous spectrum with peaks instead of discrete values.
Frequency of i-th sample is f(i)=i*samplerate/N where i={0,1,2,3,4,...(N/2)-1} is sample index (first one is DC offset so not frequency for 0) and N is the count of samples passed to FFT.
So in case you want to detect some harmonics (multiples of single fundamental frequency) then set the samplerate and N so samplerate/N is that fundamental frequency or divider of it. That way you would obtain just one peak for harmonics sinwaves. Easing up the computations.

Determination of formula for a 3 independent variable issue

I have 3 arrays of X, Y and Z. Each have 8 elements. Now for each possible combination of (X,Y,Z) I have a V value.
I am looking to find a formula e.g. V=f(X,Y,Z). Any idea about how that can be done?
Thank you in advance,
Astry
You have a function sampled on a (possibly nonuniform) 3D grid, and want to evaluate the function at any arbitrary point within the volume. One way to approach this (some say the best) is as a multivariate spline evaluation. https://en.wikipedia.org/wiki/Multivariate_interpolation
First, you need to find which rectangular parallelepiped contains the (x,y,z) query point, then you need to interpolate the value from the nearest points. The easiest thing is to use trilinear interpolation from the nearest 8 points. If you want a smoother surface, you can use quadratic interpolation from 27 points or cubic interpolation from 64 points.
For repeated queries of a tricubic spline, your life would be a bit easier by preprocessing the spline to generate Hermite patches/volumes, where your sample points not only have the function value, but also its derivatives (∂/∂x, ∂/∂y, ∂/∂z). That way you don't need messy code for the boundaries at evaluation time.

Octave specgram generates two diffent answers

I have a project where I have to recognize the frequency from an audio file. For this I use a single tone of 10 kHz to see if I can get it working.
Since I am pretty new to Octave, I tried this example with my own audio file.
I tried to understand what happens by doing some research to all functions.
My question here is; if I let specgram plot the figure when I do not specify it's output:
specgram(y,fftn,Fs,hanning(window),step);
it gives a line at 10kHz which is what I want.
But if I specify the output for the specgram function
[S,f,t]= specgram(y,fftn,Fs,hanning(window),step);
and let it plot, it plots the line at 18 kHz.
I figured it have to be in the inputs for the figure and I tried modifying these a bit, but every time I do that Octave gives an error.
I need the frequency as an given output, since I have to do some calculations with it, I figured I need to specify the frequency output.
This is the part of the code that specify the plot for the spectrogram:
step= fix(5*Fs/1000); % stepsize of the window
window= fix(90*Fs/1000); % window size
fftn =2^nextpow2(window); % Size of the FFT block
[S,f,t]= specgram(y,fftn,Fs,hanning(window),step);
S= abs(S(2:fftn*12000/Fs,:)); % Normalize the phase
S= S/max(S(:)); % Normalize the Energy
S= max(S, 10^(-40/10)); % Throw out values below -40 dB and above -3dB
S= min(S, 10^(-3/10));
figure
imagesc(t,f,(log(S)));
Can anyone help me here how to gain the frequency data from the audio file so I can use it in some calculations?
I have searched for answers already in the Octave manual for help and I tried it with various matlab sites. Also checked already many posts here such as:
How does Octave spectrogram 'specgram' from signal work?
Methodology of FFT for Matlab spectrogram / short time Fourier transform functions
P.S. Sorry for my bad English, it's not my native language
I found the answer myself, it turns out it is in this line of code:
S= abs(S(2:fftn*12000/Fs,:));
if I delete this line, the lines are placed on the right frequency in the figure. To me it looks like this line just takes a small space of the fft and replaces it with other frequencies but I'm not shure about that.

matlab function which is a function of an intergral

I need to write my own function which has the form f(x,y)=Integrate(g(x,y,z),z from 0 to inf). so the code I used was:
function y=f(x,y)
g=#(z)exp(-z.^2)./(z.^x).*(z.^2+y.^2).^(x/2);% as a function of x,y and z
y=quadgk(g,0,inf)
and if I call it for a single value like f(x0,y0), it works but if I try to calculate something like f([1:10],y0), then the error message says that there is something wrong with the times and dimension. In principle I can use for loops but then my code slows down and takes forever. Is there any help I can get from you guys? or references?
I'm trying to avoid the for loop since in matlab it's much faster to use matrix computation than to use for loop. I wonder if there is any trick that I can take advantage of this feature.
Thanks for any help in advance,
Lynn
Perhaps you can try to transpose the interval, creating row based values instead of column based f([1:10]',y0). Otherwise something in your function might be wrong, for example to get x^y to work with lists as input, you have to prefix with a dot x.^y. The same for mulitply and division I think..
If loop is no problem for you, you should do something like:
function y2=f(x,y)
y2=zeros(size(x));
for n=1:numel(x)
g=#(z)exp(-z.^2)./(z.^x(n)).*(z.^2+y.^2).^(x(n)/2);% as a function of x,y and z
y2(n)=quadgk(g,0,inf)
end
The problem here is that quadk itself uses vectors as argument for g. Then you have in g somethink like z.^x, which is the power of two vectors that is only defined if z and x have the same dimension. But this is not what you want.
I assume that you want to evaluate the function for all arguments in x and that the output vector has the same dimension as x. But this does not seem to be possible since even this simple example
g=#(x)[x;x.^2]
quad(g,0,1)
does not work:
Error using quad (line 79)
The integrand function must return an output vector of the same length as the
input vector.
A similar error shows when using quadgk. The documentation also says that this routine works only for scalar functions and this is not surprising since an adaptive quadrature rule would in general use different points for each function to evaluate the integral.
You have to use quadvinstead, which can integrate vector valued functions. But this gives wrong results since your function is integrated in the interval [0,\infty).

Finding the Maximum

How to find the following Maximum or supremum by computer software such as Mathematica and Matlab: $\sup\frac{(1+s)^{4}+(s+t)^{4}+t^{4}}{1+s^{4}+t^{4}}$?
Instead of numerical approximation, what is the accurate maximum?
Thanks.
Since the question seems a bit like homework, here's an answer that starts a bit like a lecture:
ask yourself what happens to the function as s and t go to small and to large positive and negative values; this will help you to identify the range of values you should be examining; both Mathematica and Matlab can help your figure this out;
draw the graph of your function over the range of values of interest, develop a feel for its shape and try to figure out where it has maxima; for this the Mathematic Plot3D[] function and the Matlab plot() function will both be useful;
since this is a function of 2 variables, you should think about plotting some of its sections, ie hold s (or t) constant, and make a 2D plot of the section function; again, develop some understanding of how the function behaves;
now you should be able to do some kind of search of the s,t values around the maxima of the function and get an acceptably accurate result.
If this is too difficult then you could use the Mathematica function NMaximize[]. I don't think that Matlab has the same functionality for symbolic functions built-in and you'll have to do the computations numerically but the function findmax will help.
In Matlab, one would create a vector/matrix with s and t values, and a corresponding vector with the function values. Then you can pinpoint the maximum using the function max
In Mathematica, use FindMaximum like this:
f[s_,t_]:= ((1+s)^4 + (s+t)^4 + t^4)/(1+s^4+t^4)
FindMaximum[ f[s,t],{s,0},{t,0} ]
This searches for a maximum starting from (s,t)=(0,0).
For more info, see http://reference.wolfram.com/mathematica/ref/FindMaximum.html