I have done an FFT on an audio signal and want to plot the results on a logarithmic scale like this
http://abletonuniverse.altervista.org/wp-content/uploads/2013/04/Spectrum.jpg
but for some reason I can't figure out how to scale the data to fit that logarithmic curve (starting from 0, ending at the nyquist frequency i.e. Samplingrate / block size).
Currently this is what I have
CGContextMoveToPoint(context, 0, rect.size.height);
// Skip the DC Offset (index 0)
for (int i = 1; i < _bins; i++) {
// Get frequency from bin
float binFreq = i * 44100.0 / (_bins * 2);
// Map to rect coordinate space
float x = log10f(binFreq) * rect.size.width / log10f(44100.0 / 2);
float y = _freqArray[i] * rect.size.height / -130.0;
// Draw line
CGContextAddLineToPoint(context, x, y);
}
My data comes in as 512 points of dB between 0 and ~-130 in _freqArray. Ignore the y component of the plotted point.
My data is not scaled correctly as my first bin (~43 hz) is log10(43)= 1.633 and the end of the graph is log10(22050) = 4.3, so the space between my first bin is taking up over a third of the window. It is important that I don't just put my first bin at the far left of the window the far left must be 0.
Does anybody know a correct way to scale the data into neat logarithmic bins such as in the picture? They have three neat bins representing orders of magnitude of 10, and it conveniently ends right at 22050.
For reference, this is what I have now, you can see the long straight line at the left of the spectrum. The first kink in the line is 43 hz i.e. My first bin. I am yet to put a grid up, but that will come when I figure out the scaling.
A 0Hz frequency cannot be shown on log-scale frequency axis. The left side shown on the graph you referenced corresponds to 10Hz. You can map the frequency coordinate such that this selected minimum frequency (e.g. minFrequencyToDisplay=10Hz) is drawn at a value of x=0, and the last bin drawn at a value of x=rect.size.width-1, like so:
// Map to rect coordinate space
float logMinFreq = log10f(minFrequencyToDisplay);
float x = (log10f(binFreq)-logMinFreq ) * (rect.size.width-1) /
(log10f(44100.0 / 2)-logMinFreq );
Gridlines at 100Hz, 1kHz and 10kHz can be obtained with the same formula.
Related
The first IMU (called S1) is placed on the shoulder and works as a reference; the other (S2) is placed on the arm. They provide the quaternion of their rotation relative to the absolute reference (magnetic north and gravity vector). The simple idea is that I need to show the Yaw, Pitch and Roll differences between these two (e.g. an ideal abduction/adduction movement should have pitch contributions only). I started using quaternions, by calculating the rotation between the two (conj(q1) * q2) and then converting it to YPR angles by using:
# Rotation quaternion q1 = (w,x,y,z)
# unit = sum of squared elements (sqx = x^2, etc.)
yaw = math.atan2(2 * q1[2] * q1[0] - 2 * q1[1] * q1[3], sqx - sqy - sqz + sqw)
pitch = math.asin(2 * (q1[1] * q1[2] + q1[3] * q1[0]) / sqx + sqy + sqz + sqw)
roll = math.atan2(2 * q1[1] * q1[0] - 2 * q1[2] * q1[3], -sqx - sqy + sqz + sqw)
but this doesn't work in my case, since pitch and roll are not consistent in different arm positions. E.g. if relative Yaw is 90 deg, pitch and roll angles are interchanged. E.g. if I apply a pitch rotation to S2, it appears to be a roll rotation (since the rotation is on the y-axis for the reference sensor S1).
How can I avoid this?
Should I simply convert both quaternion to YPR angles and then calculate the difference of each pair (without using the difference between quaternions)? Maybe the "rotation" approach is not correct, since I don't need the inverse transformation but only the actual rotation for each axis?
I had attached also the schematic to depict my question.
I need to rotate the vector V with the base point P by an angle and find the new vector V'.
The rotation axis is say for is about a local y axis at point P (which is parallel to global Y axis)
Subsequently, I need to rotate the initial vector V about x axis which is parallel to global Y axis.
The main reason for the rotation is to find the new vector V' at point P. Both the rotations are independent and each of the rotation provides a new V'. I'm programming this in VB.net and output is a double() of new vector V'.
Just apply the two rotations independently (see Wikipedia). The base point does not play any role in this because it is just a constant offset that never changes. If I got your description right, you want the following:
//rotation about y-axis
iAfterRot1 = cos(phi1) * i + sin(phi1) * k
jAfterRot1 = j
kAfterRot1 = -sin(phi1) * i + cos(phi) * k
//rotation about x-axis
iAfterRot2 = iAfterRot1
jAfterRot2 = cos(phi2) * jAfterRot1 - sin(phi2) * kAfterRot1
kAfterRot2 = sin(phi2) * jAfterRot1 + cos(phi2) * kAfterRot1
Does anybody knows how many pixels correspond for each millimeter of depth value in images taken from kinect for xbox360?
I'm using the standard resolution and settings...
Thanks!
1 pixel corresponds to a number of millimiters that depends on the depth value of that pixels (i.e. its level of gray).
The simplest way you can get the distance between two pixels in a depth image is to convert those pixels (which are expressed in Depth Space) in real world coordinates (i.e. in Skeleton Space)1. Then, you can calculate the distance between those points using a common euclidean distance formula.
So if you have two pixels P1 and P2, with depth values
respectively equal to D1 and D2, you can proceed as follows:
DepthImagePoint dip1 = new DepthImagePoint();
dip1.X = P1.x;
dip1.Y = P1.y;
dip1.Depth = D1;
DepthImagePoint dip2 = new DepthImagePoint();
dip2.X = P2.x;
dip2.Y = P2.y;
dip2.Depth = D2;
SkeletonPoint sp1 = CoordinateMapper.MapDepthPointToSkeletonPoint(DepthImageFormat.Resolution640x480Fps30, dip1);
SkeletonPoint sp2 = CoordinateMapper.MapDepthPointToSkeletonPoint(DepthImageFormat.Resolution640x480Fps30, dip2);
double dist = euclideanDistance(sp1, sp2);
1 See Coordinate Spaces for more information.
I have a straight line in space with an start and end point (x,y,z) and I am attempting to get the angle between this vector and the plane defined by z=0. I am using VB.NET
Here is a picture of the line in my 3d environment (the line I'm intersted in is circled in red) :
It is set to an angle of 70 degrees right now.
You need 2 rays to define an angle.
If you want the angle between a vector and a plane, it is defined for any vector in that plane. However, there is only one minimal value for that, which is the angle between a vector and its projection onto said plane.
Therefore, that minimal value is the one we take when we speak of the angle between a vector and a plane.
This value is also π/2 - the angle between your vector and the the vector that is normal to the plane.You can read more about it all on this site.
With v your vector (thus v.x = end.x - start.x and idem for y and z), n the normal to the plane and a the angle you are looking for, we know from the definition of a scalar product that:
<v,n> = ||v|| * ||n|| * cos(π/2 - a)
We know cos(π/2 - a) = sin(a), and the normal to the z=0 plane is simply the vector n = (0, 0, 1). Thus both the scalar product, v.x * n.x + v.y * n.y + v.z * n.z, and the norm of n, ||n|| = 1, can be simplified a lot. We get the following expression:
sin(a) = v.z / ||v||
Thus finally, the formula by taking the reciprocical of the sine and expliciting the norm of v:
a = Asin(v.z / sqrt( v.x*v.x + v.y*v.y + v.z*v.z ))
According to this documentation the Asin function exists in your System.Math class. It does, however, return the value in radians:
Return Value
Type: System.Double
An angle, θ, measured in radians, such that -π/2 ≤ θ ≤ π/2
-or-
NaN if d < -1 or d > 1 or d equals NaN.
Luckily the same System.Math class contains the value of π so that you can do the conversion:
a *= 180 / Math.PI
I need to find the average Edit: total 2D velocity given multiple 2D velocities (speed and direction). A few examples:
Example 1
Velocity 1 is 90° at a speed of 10 pixels or units per second.
Velocity 2 is 270° at a speed of 5 pixels or units per second.
The average velocity is 90° at 5 pixels or units per second.
Example 2
Velocity 1 is 0° at a speed of 10 pixels or units per second
Velocity 2 is 180° at a speed of 10 pixels or units per second
Velocity 3 is 90° at a speed of 8 pixels or units per second
The average velocity is 90° at 8 pixels or units per second
Example 3
Velocity 1 is 0° at 10 pixels or units per second
Velocity 2 is 90° at 10 pixels or units per second
The average velocity is 45° at 14.142 pixels or units per second
I am using JavaScript but it's mostly a language-independent question and I can convert it to JavaScript if necessary.
If you're going to be using a bunch of angles, I would just calculate each speed,
vx = v * cos(theta),
vy = v * sin(theta)
then sum the x velocities and the y velocities separately as vector components and divide by the total number of velocities,
sum(vx) / total v, sum(vy) / total v
and then finally calculate the final speed and direction with your final vx and vy. The magnitude of the speed can be found by a simple application of pythagorean theorem, and then final angle should just be tan-1(y/x).
Per example #3
vx = 10 * cos(90) + 10 * cos(0) = 10,
vy = 10 * sin(90) + 10 * sin(0) = 10
so, tan-1(10/10) = tan-1(1) = 45
then a final magnitude of sqrt(10^2 + 10^2) = 14.142
These are vectors, and you should use vector addition to add them. So right and up are positive, while left and down are negative.
Add your left-to-right vectors (x axis).
Example 1 = -10+5 = -5
Example 2 = -8 = -8
Example 3 = 10 = 10. (90 degrees is generally 90 degrees to the right)
Add you ups and downs similarly and you get these velocities, your left-to-right on the left in the brackets, and your up-to-down on the right.
(-5, 0)
(-8,0)
(10, 10)
These vectors contain all the information you need to plot the motion of an object, you do not need to calculate angles to plot the motion of the object. If for some reason you would rather use speeds (similar to velocity, but different) and angles, then you must first calculate the vectors as above and then use the Pythagorean theorem to find the speed and simple trigonometry to get the angle. Something like this:
var speed = Math.sqrt(x * x + y * y);
var tangeant = y / x;
var angleRadians = Math.atan(tangeant);
var angleDegrees = angleRadians * (180 / Math.PI);
I'll warn you that you should probably talk to someone who know trigonometry and test this well. There is potential for misleading bugs in work like this.
From your examples it sounds like you want addition of 2-dimensional vectors, not averages.
E.g. example 2 can be represented as
(0,10) + (0,-10) + (-8, 0) = (-8,0)
The speed is then equal to the length of the vector:
sqrt(x^2+y^2)
To get average:
add each speed, and then divide by the number of speeds.
10mph + 20mph / 2 = 15
12mph + 14mph + 13mph + 16mph / 4 = 14 (13,75)
This is not so much average as it is just basic vector addition. You're finding multiple "pixel vectors" and adding them together. If you have a velocity vector of 2 pixels to the right, and 1 up, and you add it to a velocity vector of 3 pixels to the left and 2 down, you will get a velocity vector of 1 pixel left, and 1 down.
So the speed is defined by
pij = pixels going up or (-)down
pii = pixels going right or (-)left
speedi = pii1 + pii2 = 2-3 = -1 (1 pixel left)
speedj = pij1 + pij2 = 1-2 = -1 (1 pixel down)
From there, you need to decide which directions are positive, and which are negative. I recommend that left is negative, and down is negative (like a mathematical graph).
The angle of the vector, would be the arctan(speedj/speedi)
arctan(-1/-1) = 45 degrees