I'm trying to calculate the principal axis via principal component analysis. I have a pointcloud and use for this the Point Cloud Library (pcl). Furthermore, I try to visualize the principal axis I calculated in rviz with markers. Here is the code snipped I use:
void computePrincipalAxis(const PointCloud& cloud, Eigen::Vector4f& centroid, Eigen::Matrix3f& evecs, Eigen::Vector3f& evals) {
Eigen::Matrix3f covariance_matrix;
pcl::computeCovarianceMatrix(cloud, centroid, covariance_matrix);
pcl::eigen33(covariance_matrix, evecs, evals);
}
void createArrowMarker(Eigen::Vector3f& vec, int id, double length) {
visualization_msgs::Marker marker;
marker.header.frame_id = frameId;
marker.header.stamp = ros::Time();
marker.id = id;
marker.type = visualization_msgs::Marker::ARROW;
marker.action = visualization_msgs::Marker::ADD;
marker.pose.position.x = centroid[0];
marker.pose.position.y = centroid[1];
marker.pose.position.z = centroid[2];
marker.pose.orientation.x = vec[0];
marker.pose.orientation.y = vec[1];
marker.pose.orientation.z = vec[2];
marker.pose.orientation.w = 1.0;
marker.scale.x = length;
marker.scale.y = 0.02;
marker.scale.z = 0.02;
marker.color.a = 1.0;
marker.color.r = 1.0;
marker.color.g = 1.0;
marker.color.b = 0.0;
featureVis.markers.push_back(marker);
}
Eigen::Vector4f centroid;
Eigen::Matrix3f evecs;
Eigen::Vector3f evals;
// Table is the pointcloud of the table only.
pcl::compute3DCentroid(*table, centroid);
computePrincipalAxis(*table, centroid, evecs, evals);
Eigen::Vector3f vec;
vec << evecs.col(0);
createArrowMarker(vec, 1, evals[0]);
vec << evecs.col(1);
createArrowMarker(vec, 2, evals[1]);
vec << evecs.col(2);
createArrowMarker(vec, 3, evals[2]);
publish();
This results in the following visualization:
I'm aware that the scale is not very perfect. The two longer arrows are much too long. But I'm confused about a few things:
I think the small arrow should go either up, or downwards.
What does the value orientation.w of the arrow's orientation mean?
Do you have some hints what I did wrong?
Orientations are represented by Quaternions in ROS, not by directional vectors. Quaternions can be a bit unintuitive, but fortunately there are some helper functions in the tf package, to generate quaternions, for example, from roll/pitch/yaw-angles.
One way to fix the marker would therefore be, to convert the direction vector into a quaternion.
In your special case, there is a much simpler solution, though: Instead of setting origin and orientation of the arrow, it is also possible to define start and end point (see ROS wiki about marker types). So instead of setting the pose attribute, just add start and end point to the points attribute:
float k = 1.0; // optional to scale the length of the arrows
geometry_msgs::Point p;
p.x = centroid[0];
p.y = centroid[1];
p.z = centroid[2];
marker.points.push_back(p);
p.x += k * vec[0];
p.y += k * vec[1];
p.z += k * vec[2];
marker.points.push_back(p);
You can set k to some value < 1 to reduce the length of the arrows.
BOUNTY STATUS UPDATE:
I discovered how to map a linear lens, from destination coordinates to source coordinates.
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
1). I actually struggle to reverse it, and to map source coordinates to destination coordinates. What is the inverse, in code in the style of the converting functions I posted?
2). I also see that my undistortion is imperfect on some lenses - presumably those that are not strictly linear. What is the equivalent to-and-from source-and-destination coordinates for those lenses? Again, more code than just mathematical formulae please...
Question as originally stated:
I have some points that describe positions in a picture taken with a fisheye lens.
I want to convert these points to rectilinear coordinates. I want to undistort the image.
I've found this description of how to generate a fisheye effect, but not how to reverse it.
There's also a blog post that describes how to use tools to do it; these pictures are from that:
(1) : SOURCE Original photo link
Input : Original image with fish-eye distortion to fix.
(2) : DESTINATION Original photo link
Output : Corrected image (technically also with perspective correction, but that's a separate step).
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
My function stub looks like this:
Point correct_fisheye(const Point& p,const Size& img) {
// to polar
const Point centre = {img.width/2,img.height/2};
const Point rel = {p.x-centre.x,p.y-centre.y};
const double theta = atan2(rel.y,rel.x);
double R = sqrt((rel.x*rel.x)+(rel.y*rel.y));
// fisheye undistortion in here please
//... change R ...
// back to rectangular
const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta));
fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y);
return ret;
}
Alternatively, I could somehow convert the image from fisheye to rectilinear before finding the points, but I'm completely befuddled by the OpenCV documentation. Is there a straightforward way to do it in OpenCV, and does it perform well enough to do it to a live video feed?
The description you mention states that the projection by a pin-hole camera (one that does not introduce lens distortion) is modeled by
R_u = f*tan(theta)
and the projection by common fisheye lens cameras (that is, distorted) is modeled by
R_d = 2*f*sin(theta/2)
You already know R_d and theta and if you knew the camera's focal length (represented by f) then correcting the image would amount to computing R_u in terms of R_d and theta. In other words,
R_u = f*tan(2*asin(R_d/(2*f)))
is the formula you're looking for. Estimating the focal length f can be solved by calibrating the camera or other means such as letting the user provide feedback on how well the image is corrected or using knowledge from the original scene.
In order to solve the same problem using OpenCV, you would have to obtain the camera's intrinsic parameters and lens distortion coefficients. See, for example, Chapter 11 of Learning OpenCV (don't forget to check the correction). Then you can use a program such as this one (written with the Python bindings for OpenCV) in order to reverse lens distortion:
#!/usr/bin/python
# ./undistort 0_0000.jpg 1367.451167 1367.451167 0 0 -0.246065 0.193617 -0.002004 -0.002056
import sys
import cv
def main(argv):
if len(argv) < 10:
print 'Usage: %s input-file fx fy cx cy k1 k2 p1 p2 output-file' % argv[0]
sys.exit(-1)
src = argv[1]
fx, fy, cx, cy, k1, k2, p1, p2, output = argv[2:]
intrinsics = cv.CreateMat(3, 3, cv.CV_64FC1)
cv.Zero(intrinsics)
intrinsics[0, 0] = float(fx)
intrinsics[1, 1] = float(fy)
intrinsics[2, 2] = 1.0
intrinsics[0, 2] = float(cx)
intrinsics[1, 2] = float(cy)
dist_coeffs = cv.CreateMat(1, 4, cv.CV_64FC1)
cv.Zero(dist_coeffs)
dist_coeffs[0, 0] = float(k1)
dist_coeffs[0, 1] = float(k2)
dist_coeffs[0, 2] = float(p1)
dist_coeffs[0, 3] = float(p2)
src = cv.LoadImage(src)
dst = cv.CreateImage(cv.GetSize(src), src.depth, src.nChannels)
mapx = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
mapy = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
cv.InitUndistortMap(intrinsics, dist_coeffs, mapx, mapy)
cv.Remap(src, dst, mapx, mapy, cv.CV_INTER_LINEAR + cv.CV_WARP_FILL_OUTLIERS, cv.ScalarAll(0))
# cv.Undistort2(src, dst, intrinsics, dist_coeffs)
cv.SaveImage(output, dst)
if __name__ == '__main__':
main(sys.argv)
Also note that OpenCV uses a very different lens distortion model to the one in the web page you linked to.
(Original poster, providing an alternative)
The following function maps destination (rectilinear) coordinates to source (fisheye-distorted) coordinates. (I'd appreciate help in reversing it)
I got to this point through trial-and-error: I don't fundamentally grasp why this code is working, explanations and improved accuracy appreciated!
def dist(x,y):
return sqrt(x*x+y*y)
def correct_fisheye(src_size,dest_size,dx,dy,factor):
""" returns a tuple of source coordinates (sx,sy)
(note: values can be out of range)"""
# convert dx,dy to relative coordinates
rx, ry = dx-(dest_size[0]/2), dy-(dest_size[1]/2)
# calc theta
r = dist(rx,ry)/(dist(src_size[0],src_size[1])/factor)
if 0==r:
theta = 1.0
else:
theta = atan(r)/r
# back to absolute coordinates
sx, sy = (src_size[0]/2)+theta*rx, (src_size[1]/2)+theta*ry
# done
return (int(round(sx)),int(round(sy)))
When used with a factor of 3.0, it successfully undistorts the images used as examples (I made no attempt at quality interpolation):
Dead link
(And this is from the blog post, for comparison:)
If you think your formulas are exact, you can comput an exact formula with trig, like so:
Rin = 2 f sin(w/2) -> sin(w/2)= Rin/2f
Rout= f tan(w) -> tan(w)= Rout/f
(Rin/2f)^2 = [sin(w/2)]^2 = (1 - cos(w))/2 -> cos(w) = 1 - 2(Rin/2f)^2
(Rout/f)^2 = [tan(w)]^2 = 1/[cos(w)]^2 - 1
-> (Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
However, as #jmbr says, the actual camera distortion will depend on the lens and the zoom. Rather than rely on a fixed formula, you might want to try a polynomial expansion:
Rout = Rin*(1 + A*Rin^2 + B*Rin^4 + ...)
By tweaking first A, then higher-order coefficients, you can compute any reasonable local function (the form of the expansion takes advantage of the symmetry of the problem). In particular, it should be possible to compute initial coefficients to approximate the theoretical function above.
Also, for good results, you will need to use an interpolation filter to generate your corrected image. As long as the distortion is not too great, you can use the kind of filter you would use to rescale the image linearly without much problem.
Edit: as per your request, the equivalent scaling factor for the above formula:
(Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
-> Rout/f = [Rin/f] * sqrt(1-[Rin/f]^2/4)/(1-[Rin/f]^2/2)
If you plot the above formula alongside tan(Rin/f), you can see that they are very similar in shape. Basically, distortion from the tangent becomes severe before sin(w) becomes much different from w.
The inverse formula should be something like:
Rin/f = [Rout/f] / sqrt( sqrt(([Rout/f]^2+1) * (sqrt([Rout/f]^2+1) + 1) / 2 )
I blindly implemented the formulas from here, so I cannot guarantee it would do what you need.
Use auto_zoom to get the value for the zoom parameter.
def dist(x,y):
return sqrt(x*x+y*y)
def fisheye_to_rectilinear(src_size,dest_size,sx,sy,crop_factor,zoom):
""" returns a tuple of dest coordinates (dx,dy)
(note: values can be out of range)
crop_factor is ratio of sphere diameter to diagonal of the source image"""
# convert sx,sy to relative coordinates
rx, ry = sx-(src_size[0]/2), sy-(src_size[1]/2)
r = dist(rx,ry)
# focal distance = radius of the sphere
pi = 3.1415926535
f = dist(src_size[0],src_size[1])*factor/pi
# calc theta 1) linear mapping (older Nikon)
theta = r / f
# calc theta 2) nonlinear mapping
# theta = asin ( r / ( 2 * f ) ) * 2
# calc new radius
nr = tan(theta) * zoom
# back to absolute coordinates
dx, dy = (dest_size[0]/2)+rx/r*nr, (dest_size[1]/2)+ry/r*nr
# done
return (int(round(dx)),int(round(dy)))
def fisheye_auto_zoom(src_size,dest_size,crop_factor):
""" calculate zoom such that left edge of source image matches left edge of dest image """
# Try to see what happens with zoom=1
dx, dy = fisheye_to_rectilinear(src_size, dest_size, 0, src_size[1]/2, crop_factor, 1)
# Calculate zoom so the result is what we wanted
obtained_r = dest_size[0]/2 - dx
required_r = dest_size[0]/2
zoom = required_r / obtained_r
return zoom
I took what JMBR did and basically reversed it. He took the radius of the distorted image (Rd, that is, the distance in pixels from the center of the image) and found a formula for Ru, the radius of the undistorted image.
You want to go the other way. For each pixel in the undistorted (processed image), you want to know what the corresponding pixel is in the distorted image.
In other words, given (xu, yu) --> (xd, yd). You then replace each pixel in the undistorted image with its corresponding pixel from the distorted image.
Starting where JMBR did, I do the reverse, finding Rd as a function of Ru. I get:
Rd = f * sqrt(2) * sqrt( 1 - 1/sqrt(r^2 +1))
where f is the focal length in pixels (I'll explain later), and r = Ru/f.
The focal length for my camera was 2.5 mm. The size of each pixel on my CCD was 6 um square. f was therefore 2500/6 = 417 pixels. This can be found by trial and error.
Finding Rd allows you to find the corresponding pixel in the distorted image using polar coordinates.
The angle of each pixel from the center point is the same:
theta = arctan( (yu-yc)/(xu-xc) ) where xc, yc are the center points.
Then,
xd = Rd * cos(theta) + xc
yd = Rd * sin(theta) + yc
Make sure you know which quadrant you are in.
Here is the C# code I used
public class Analyzer
{
private ArrayList mFisheyeCorrect;
private int mFELimit = 1500;
private double mScaleFESize = 0.9;
public Analyzer()
{
//A lookup table so we don't have to calculate Rdistorted over and over
//The values will be multiplied by focal length in pixels to
//get the Rdistorted
mFisheyeCorrect = new ArrayList(mFELimit);
//i corresponds to Rundist/focalLengthInPixels * 1000 (to get integers)
for (int i = 0; i < mFELimit; i++)
{
double result = Math.Sqrt(1 - 1 / Math.Sqrt(1.0 + (double)i * i / 1000000.0)) * 1.4142136;
mFisheyeCorrect.Add(result);
}
}
public Bitmap RemoveFisheye(ref Bitmap aImage, double aFocalLinPixels)
{
Bitmap correctedImage = new Bitmap(aImage.Width, aImage.Height);
//The center points of the image
double xc = aImage.Width / 2.0;
double yc = aImage.Height / 2.0;
Boolean xpos, ypos;
//Move through the pixels in the corrected image;
//set to corresponding pixels in distorted image
for (int i = 0; i < correctedImage.Width; i++)
{
for (int j = 0; j < correctedImage.Height; j++)
{
//which quadrant are we in?
xpos = i > xc;
ypos = j > yc;
//Find the distance from the center
double xdif = i-xc;
double ydif = j-yc;
//The distance squared
double Rusquare = xdif * xdif + ydif * ydif;
//the angle from the center
double theta = Math.Atan2(ydif, xdif);
//find index for lookup table
int index = (int)(Math.Sqrt(Rusquare) / aFocalLinPixels * 1000);
if (index >= mFELimit) index = mFELimit - 1;
//calculated Rdistorted
double Rd = aFocalLinPixels * (double)mFisheyeCorrect[index]
/mScaleFESize;
//calculate x and y distances
double xdelta = Math.Abs(Rd*Math.Cos(theta));
double ydelta = Math.Abs(Rd * Math.Sin(theta));
//convert to pixel coordinates
int xd = (int)(xc + (xpos ? xdelta : -xdelta));
int yd = (int)(yc + (ypos ? ydelta : -ydelta));
xd = Math.Max(0, Math.Min(xd, aImage.Width-1));
yd = Math.Max(0, Math.Min(yd, aImage.Height-1));
//set the corrected pixel value from the distorted image
correctedImage.SetPixel(i, j, aImage.GetPixel(xd, yd));
}
}
return correctedImage;
}
}
I found this pdf file and I have proved that the maths are correct (except for the line vd = *xd**fv+v0 which should say vd = **yd**+fv+v0).
http://perception.inrialpes.fr/CAVA_Dataset/Site/files/Calibration_OpenCV.pdf
It does not use all of the latest co-efficients that OpenCV has available but I am sure that it could be adapted fairly easily.
double k1 = cameraIntrinsic.distortion[0];
double k2 = cameraIntrinsic.distortion[1];
double p1 = cameraIntrinsic.distortion[2];
double p2 = cameraIntrinsic.distortion[3];
double k3 = cameraIntrinsic.distortion[4];
double fu = cameraIntrinsic.focalLength[0];
double fv = cameraIntrinsic.focalLength[1];
double u0 = cameraIntrinsic.principalPoint[0];
double v0 = cameraIntrinsic.principalPoint[1];
double u, v;
u = thisPoint->x; // the undistorted point
v = thisPoint->y;
double x = ( u - u0 )/fu;
double y = ( v - v0 )/fv;
double r2 = (x*x) + (y*y);
double r4 = r2*r2;
double cDist = 1 + (k1*r2) + (k2*r4);
double xr = x*cDist;
double yr = y*cDist;
double a1 = 2*x*y;
double a2 = r2 + (2*(x*x));
double a3 = r2 + (2*(y*y));
double dx = (a1*p1) + (a2*p2);
double dy = (a3*p1) + (a1*p2);
double xd = xr + dx;
double yd = yr + dy;
double ud = (xd*fu) + u0;
double vd = (yd*fv) + v0;
thisPoint->x = ud; // the distorted point
thisPoint->y = vd;
This can be solved as an optimization problem. Simply draw on curves in images that are supposed to be straight lines. Store the contour points for each of those curves. Now we can solve the fish eye matrix as a minimization problem. Minimize the curve in points and that will give us a fisheye matrix. It works.
It can be done manually by adjusting the fish eye matrix using trackbars! Here is a fish eye GUI code using OpenCV for manual calibration.
I am trying to rotate the image manually using the following code.
clc;
m1 = imread('owl','pgm'); % a simple gray scale image of order 260 X 200
newImg = zeros(500,500);
newImg = int16(newImg);
rotationMatrix45 = [cos((pi/4)) -sin((pi/4)); sin((pi/4)) cos((pi/4))];
for x = 1:size(m1,1)
for y = 1:size(m1,2)
point =[x;y] ;
product = rotationMatrix45 * point;
product = int16(product);
newx =product(1,1);
newy=product(2,1);
newImg(newx,newy) = m1(x,y);
end
end
imshow(newImg);
Simply I am iterating through every pixel of image m1, multiplying m1(x,y) with rotation matrix, I get x',y', and storing the value of m1(x,y) in to `newImg(x',y')' BUT it is giving the following error
??? Attempted to access newImg(0,1); index must be a positive integer or logical.
Error in ==> at 18
newImg(newx,newy) = m1(x,y);
I don't know what I am doing wrong.
Part of the rotated image will get negative (or zero) newx and newy values since the corners will rotate out of the original image coordinates. You can't assign a value to newImg if newx or newy is nonpositive; those aren't valid matrix indices. One solution would be to check for this situation and skip such pixels (with continue)
Another solution would be to enlarge the newImg sufficiently, but that will require a slightly more complicated transformation.
This is assuming that you can't just use imrotate because this is homework?
The problem is simple, the answer maybe not : Matlab arrays are indexed from one to N (whereas in many programming langages it's from 0 to (N-1) ).
Try newImg( max( min(1,newX), m1.size() ) , max( min(1,newY), m1.size() ) ) maybe (I don't have Matlab at work so I can tell if it's gonna work), but the resulting image will be croped.
this is an old post so I guess it wont help the OP but as I was helped by his attempt I post here my corrected code.
basically some freedom in the implementation regarding to how you deal with unassigned pixels as well as wether you wish to keep the original size of the pic - which will force you to crop areas falling "outside" of it.
the following function rotates the image around its center, leaves unassigned pixels as "burned" and crops the edges.
function [h] = rot(A,ang)
rotMat = [cos((pi.*ang/180)) sin((pi.*ang/180)); -sin((pi.*ang/180)) cos((pi.*ang/180))];
centerW = round(size(A,1)/2);
centerH = round(size(A,2)/2);
h=255.* uint8(ones(size(A)));
for x = 1:size(A,1)
for y = 1:size(A,2)
point =[x-centerW;y-centerH] ;
product = rotMat * point;
product = int16(product);
newx =product(1,1);
newy=product(2,1);
if newx+centerW<=size(A,1)&& newx+centerW > 0 && newy+centerH<=size(A,2)&& newy+centerH > 0
h(newx+centerW,newy+centerH) = A(x,y);
end
end
end
I'm trying to convert the geomagnetic and accelerometer to rotate the camera in opengl ES1, I found some code from android and changed this code for iPhone, actually it is working more or less, but there are some mistakes, I´m not able to find this mistake, I put the code, also the call to Opengl Es1: glLoadMatrixf((GLfloat*)matrix);
- (void) GetAccelerometerMatrix:(GLfloat *) matrix headingX: (float)hx headingY:(float)hy headingZ:(float)hz;
{
_geomagnetic[0] = hx * (FILTERINGFACTOR-0.05) + _geomagnetic[0] * (1.0 - FILTERINGFACTOR-0.5)+ _geomagnetic[3] * (0.55);
_geomagnetic[1] = hy * (FILTERINGFACTOR-0.05) + _geomagnetic[1] * (1.0 - FILTERINGFACTOR-0.5)+ _geomagnetic[4] * (0.55);
_geomagnetic[2] = hz * (FILTERINGFACTOR-0.05) + _geomagnetic[2] * (1.0 - FILTERINGFACTOR-0.5)+ _geomagnetic[5] * (0.55);
_geomagnetic[3]=_geomagnetic[0] ;
_geomagnetic[4]=_geomagnetic[1];
_geomagnetic[5]=_geomagnetic[2];
//Clear matrix to be used to rotate from the current referential to one based on the gravity vector
bzero(matrix, sizeof(matrix));
//MAGNETIC
float Ex = -_geomagnetic[1];
float Ey =_geomagnetic[0];
float Ez =_geomagnetic[2];
//ACCELEROMETER
float Ax= -_accelerometer[0];
float Ay= _accelerometer[1] ;
float Az= _accelerometer[2] ;
float Hx = Ey*Az - Ez*Ay;
float Hy= Ez*Ax - Ex*Az;
float Hz = Ex*Ay - Ey*Ax;
float normH = (float)sqrt(Hx*Hx + Hy*Hy + Hz*Hz);
float invH = 1.0f / normH;
Hx *= invH;
Hy *= invH;
Hz *= invH;
float invA = 1.0f / (float)sqrt(Ax*Ax + Ay*Ay + Az*Az);
Ax *= invA;
Ay *= invA;
Az *= invA;
float Mx = Ay*Hz - Az*Hy;
float My = Az*Hx - Ax*Hz;
float Mz = Ax*Hy - Ay*Hx;
// if (mOut.f != null) {
matrix[0] = Hx; matrix[1] = Hy; matrix[2] = Hz; matrix[3] = 0;
matrix[4] = Mx; matrix[5] = My; matrix[6] = Mz; matrix[7] = 0;
matrix[8] = Ax; matrix[9] = Ay; matrix[10] = Az; matrix[11] = 0;
matrix[12] = 0; matrix[13] = 0; matrix[14] = 0; matrix[15] = 1;
}
Thank you very much for the help.
Edit: The iPhone it is permantly in landscape orientation and I know that something is wrong because the object painted in Opengl Es appears two times.
Have you looked at Apple's GLGravity sample code? It does something very similar to what you want here, by manipulating the model view matrix in response to changes in the accelerometer input.
I'm unable to find any problems with the code posted, and would suggest the problem is elsewhere. If it helps, my analysis of the code posted is that:
The first six lines, dealing with _geomagnetic 0–5, effect a very simple low frequency filter, which assumes you call the method at regular intervals. So you end up with a version of the magnetometer vector, hopefully with high frequency jitter removed.
The bzero zeroes the result, ready for accumulation.
The lines down to the declaration and assignment to Hz take the magnetometer and accelerometer vectors and perform the cross product. So H(x, y, z) is now a vector at right angles to both the accelerometer (which is presumed to be 'down') and the magnetometer (which will be forward + some up). Call that the side vector.
The invH and invA stuff, down to the multiplication of Az by invA ensure that the side and accelerometer/down vectors are of unit length.
M(x, y, z) is then created, as the cross product of the side and down vectors (ie, a vector at right angles to both of those). So it gives the front vector.
Finally, the three vectors are used to populate the matrix, taking advantage of the fact that the inverse of an orthonormal 3x3 matrix is its transpose (though that's sort of hidden by the way things are laid out — pay attention to the array indices). You actually set everything in the matrix directly, so the bzero wasn't necessary in pure outcome terms.
glLoadMatrixf is then the correct thing to use because that's how you multiply by an arbitrary column-major matrix in OpenGL ES 1.x.
I'm looking at the aurioTouch sample application for the iPhone SDK. It has a basic spectrum analyzer implemented when you choose the "FFT" option. One of the things the app is lacking is X axis labels (i.e. the frequency labels).
In the aurioTouchAppDelegate.mm file, in the function - (void)drawOscilloscope at line 652, it has the following code:
if (displayMode == aurioTouchDisplayModeOscilloscopeFFT)
{
if (fftBufferManager->HasNewAudioData())
{
if (fftBufferManager->ComputeFFT(l_fftData))
[self setFFTData:l_fftData length:fftBufferManager->GetNumberFrames() / 2];
else
hasNewFFTData = NO;
}
if (hasNewFFTData)
{
int y, maxY;
maxY = drawBufferLen;
for (y=0; y<maxY; y++)
{
CGFloat yFract = (CGFloat)y / (CGFloat)(maxY - 1);
CGFloat fftIdx = yFract * ((CGFloat)fftLength);
double fftIdx_i, fftIdx_f;
fftIdx_f = modf(fftIdx, &fftIdx_i);
SInt8 fft_l, fft_r;
CGFloat fft_l_fl, fft_r_fl;
CGFloat interpVal;
fft_l = (fftData[(int)fftIdx_i] & 0xFF000000) >> 24;
fft_r = (fftData[(int)fftIdx_i + 1] & 0xFF000000) >> 24;
fft_l_fl = (CGFloat)(fft_l + 80) / 64.;
fft_r_fl = (CGFloat)(fft_r + 80) / 64.;
interpVal = fft_l_fl * (1. - fftIdx_f) + fft_r_fl * fftIdx_f;
interpVal = CLAMP(0., interpVal, 1.);
drawBuffers[0][y] = (interpVal * 120);
}
cycleOscilloscopeLines();
}
}
From my understanding, this part of the code is what is used to decide which magnitude to draw for each frequency in the UI. My question is how can I determine what frequency each iteration (or y value) represents inside the for loop.
For example, if I want to know what the magnitude is for 6kHz, I'm thinking of adding a line similar to the following:
if (yValueRepresentskHz(y, 6))
NSLog(#"The magnitude for 6kHz is %f", (interpVal * 120));
Please note that although they chose to use the variable name y, from what I understand, it actually represents the x-axis in the visual graph of the spectrum analyzer, and the value of the drawBuffers[0][y] represents the y-axis.
I believe that the frequency of each bin it is using is given by
yFract * hwSampleRate * .5
I'm fairly certain that you need the .5 because yFract is a fraction of the total fftLength and the last bin of the FFT corresponds with half of the sampling rate. Thus, you could do something like
NSLog(#"The magnitude for %f Hz is %f.", (yFract * hwSampleRate * .5), (interpVal * 120));
Hopefully that helps to point you in the right direction at least.