I'm using WX GUI FFT to display a specific frequency range (38Hz for IR). I can't seem to get the plot to show the frequency range on the x axis. I have it set up like follows:
And here's what it looks like when it runs:
As can be seen there's no frequency range on the x axis.
Any idea what I'm doing wrong?
The display is absolutely correct.
You set the sampling rate, and hence the bandwidth to be displayed, to be nominally 10 MHz, so each of the ten divisions of the x-Axis should be 1 MHz wide.
Now, your center frequency is in fact of course not 0 MHz, but 38 kHz, but WX GUI rounds the numbers for display – it's really not that useful to see "0.038 MHz" as an axis label.
More importantly, you seem to be confused about what the sampling rate and what the center frequency are supposed to be; my gut feeling is that you want to observe an 38 kHz wide channel around 10 MHz. What you do is observe 10 MHz around 38 kHz (which, physically, doesn't make much sense, and you should see very clear warnings about that in the console).
Furthermore, WX is going away with the next release of GNU Radio. Use the Qt GUI instead.
Related
I am trying to extract a STL file from a volume (3D-matrix). Since I am working with a DICOM volume, I want to adapt the matrix elements to the voxel size. Everything seems to work fine, but when I check the stl output size (for example in a CAD software), it seems that a slice, a "voxel", is missing in each direction. This can be (maybe) reasonable for a DICOM, where the slices are counted considering the respective center (and therefore, in a certain sense, one slice may be missing in the finale measure), but not in the following example I tried.
I leave here a simple code: my voxel size is 0.723x0.723x0.8; I created a 10x10x10 matrix, so I would expect a 7.23 x 7.23 x 8 mm stl. Instead, I get a 6.507 x 6.507 x 7.2 mm stl, so I think something goes wrong.
R=randi(3,10,10,10);
R(R>1)=0;
[F1,V1]=isosurface(R,0);
[F2,V2]=isocaps(R,0);
faces=[F1;F2+length(V1(:,1))];
vertices=[V1;V2];
S=makehgtform('scale',[0.723,0.723,0.8]);
new_vertices=vertices*S(1:3,1:3);
p=patch('Vertices',new_vertices,'Faces',faces,'FaceColor','blue');
view(3)
camlight headlight; lighting gouraud
str=strcat(['R','.stl']);
stlwrite(str,faces,new_vertices);
How can isosurface, isocaps or stlwrite be responsible for this?
I am trying to implement an OCR / OCV algorithm for inspecting printed text in black ink on a white background. The text size is ranging from 3 pt. to 6 pt. I tried first to capture images with a 5 MP monochrome camera using an 8 mm, 12 mm and 16 mm lens but I could not get the characters with good clarity. I repeated the same test with 10 MP camera also considering that higher pixel depth will give more information but I got same results.
I'm not sure, how I can get a clearer image. Whether a 5 MP / 10 MP is enough and if there is any way to determine the lens to be used in such application.
The FOV for inspection is about 300 x 250 mm and the working distance I considered from approx. 400 mm to 650 mm.
Due to copyright concerns I cannot post the image of the object under inspection.
Any help or direction is greatly appreciated. Thanks.
It's simple geometry. It is:
3pt =~ 1mm.
Assuming you want to have 10 pixels to cover each character, your IFOV needs to be:
IFOV =~ (font_width / 10) / distance = 0.1 / 650 =~ 0.15 milliradians / pixel.
For the work area width you mention, the horizontal field of view is:
FOV = 2 * atan((300 / 2) / 650) =~ 453 milliradians =~ 26 deg
So the minimal (horizontal) sensor resolution you'd need is:
Width = 453 / 0.15 = 3020 pixels.
Thus a 10MP sensor should be quite sufficient, and 5MP one may be adequate.
To choose the lens, from the above spec for the FOV, and the format (width, height) of your choice of sensor, you can work out by the same simple trigonometry the needed focal length. Finally, among all lenses matching that focal length that are available for your camera mount, you need to choose one that (a) can be focused at the distance of interest and (b) has an adequate Optical Transfer Function such that one line can be resolved at the above IFOV.
In practice, after running the math and looking at catalogs, you'll end up with several candidate lenses. My advice would then be to get samples and try them out on your on setup, and specifically with the particular lighting rig you'll be using, before making a final decision. Depending on your particular project, factors influencing the choice, in addition (obviously) to the cost of the lens + sensor combination, may be size/weight, sensitivity to environment conditions (temperature, humidity, vibrations), availability and lead time for sourcing, etc.
I am trying to analyse the energy behaviour of n coupled harmonic oscillators where the initial displacements are distributed/defined by sine wave equation at various values of time. My total energy plot seems to appx. double from n= 4 to 8 to 16 and so on. Can someone please explain me the behaviour.
thanks.
To my understanding, the noise floor for each USRP may be different. I want to know how I can calculate the noise floor without physically going into the fft and spotting it out manually. I want to know if there is a block in GNU Radio that will calculate this, or if there is a stream of blocks I can use to calculate it. Please provide a block diagram in your answer ( block 1 ---> block 2 ---> ...etc.).
For my application, let's say I have a QT GUI frequency sink that is showing all noise at the moment. I want to calculate the noise floor so that I have a value that represents "no signal present" ie. noise. Once I have this value, I plan to set a threshold 5dB higher than the noise floor to indicate that a signal has been detected. I've been able to kind of eye ball the average noise value from the QT GUI Frequency Sink but that's not good enough for me. I want to be able to calculate it and not have to look into the plot every time to update the noise value every time I change USRPs.
For instance:
You can see the average noise value for this is around -55dB. I want to calculate this without having to eye ball it. This way, when a signal gets transmitted at (in this example) 0Hz, then the power of the signal will increase and I can see if a signal was detected.
I have a set of first 25 Zernike polynomials. Below are shown few in Cartesin co-ordinate system.
z2 = 2*x
z3 = 2*y
z4 = sqrt(3)*(2*x^2+2*y^2-1)
:
:
z24 = sqrt(14)*(15*(x^2+y^2)^2-20*(x^2+y^2)+6)*(x^2-y^2)
I am not using 1st since it is piston; so I have these 24 two-dim ANALYTICAL functions expressed in X-Y Cartesian co-ordinate system. All are defined over unit circle, as they are orthogonal over unit circle. The problem which I am describing here is relevant to other 2D surfaces also apart from Zernike Polynomials.
Suppose that origin (0,0) of the XY co-ordinate system and the centre of the unit circle are same.
Next, I take linear combination of these 24 polynomials to build a 2D wavefront shape. I use 24 random input coefficients in this combination.
w(x,y) = sum_over_i a_i*z_i (i=2,3,4,....24)
a_i = random coefficients
z_i = zernike polynomials
Upto this point, everything is analytical part which can be done on paper.
Now comes the discretization!
I know that when you want to re-construct a signal (1Dim/2Dim), your sampling frequency should be at least twice the maximum frequency present in the signal (Nyquist-Shanon principle).
Here signal is w(x,y) as mentioned above which is nothing but a simple 2Dim
function of x & y. I want to represent it on computer now. Obviously I can not take all infinite points from -1 to +1 along x axis and same for y axis.
I have to take finite no. of data points (which are called sample points or just samples) on this analytical 2Dim surface w(x,y)
I am measuring x & y in metres, and -1 <= x <= +1; -1 <= y <= +1.
e.g. If I divide my x-axis from -1 to 1, in 50 sample points then dx = 2/50= 0.04 metre. Same for y axis. Now my sampling frequency is 1/dx i.e. 25 samples per metre. Same for y axis.
But I took 50 samples arbitrarily; I could have taken 10 samples or 1000 samples. That is the crux of the matter here: how many samples points?How will I determine this number?
There is one theorem (Nyquist-Shanon theorem) mentioned above which says that if I want to re-construct w(x,y) faithfully, I must sample it on both axes so that my sampling frequency (i.e. no. of samples per metre) is at least twice the maximum frequency present in the w(x,y). This is nothing but finding power spectrum of w(x,y). Idea is that any function in space domain can be represented in spatial-frequency domain also, which is nothing but taking Fourier transform of the function! This tells us how many (spatial) frequencies are present in your function w(x,y) and what is the maximum frequency out of these many frequencies.
Now my question is first how to find out this maximum sampling frequency in my case. I can not use MATLAB fft2() or any other tool since it means already I have samples taken across the wavefront!! Obviously remaining option is find it analytically ! But that is time consuming and difficult since I have 24 polynomials & I will have to use then continuous Fourier transform i.e. I will have to go for pen and paper.
Any help will be appreciated.
Thanks
Key Assumptions
You want to use the "Nyquist-Shanon" theorem to determine sampling frequency
Obviously remaining option is find it analytically ! But that is time
consuming and difficult since I have 21 polynomials & I have to use
continuous Fourier transform i.e. done by analytically.
Given the assumption I have made (and noting that consideration of other mathematical techniques is out of scope for StackOverflow), you have no option but to calculate the continuous Fourier Transform.
However, I believe you haven't considered all the options for calculating the transform other than a laborious paper exercise e.g.
Numerical approximation of the continuous F.T. using code
Symbolic Integration e.g. Wolfram Alpha
Surely a numerical approximation of the Fourier Transform will be adequate for your solution?
I am assuming this is for coursework or research rather, so all you really care about as a physicist is a solution that is the quickest solution that is accurate within the scope of your problem.
So to conclude, IMHO, don't waste time searching for a more mathematically elegant solution or trick and just solve the problem with one of the above methods