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When I do a spectrogram in matlab / octave I can create a swept signal that looks like the RED plot line below. But how can I create a swept signal like the BLUE line in the 1st plot using the equation below.
thanks to Daniel and David for getting me this far with the code is below
startfreq=200;
fs=44100;
endfreq=20;
dursec= 10;%duration of signal in seconds
t=(0:dursec*fs)/fs; %Time vector
alpha=log(startfreq/endfreq)/dursec;
sig = exp(-j*2*pi*startfreq/alpha*exp(-alpha*t));
sig=(sig/max(abs(sig))*.8); %normalize signal
wavwrite([sig'] ,fs,32,strcat('/tmp/del.wav')); %export file
specgram(sig,150,400);
1st plot
2nd plot
How can I fix the the equation in the variable sig to get it to look like the BLUE line in the 1st plot?
3rd plot
This question is almost an month old, so you might have figured this out by now. Here's an answer in case you are still interested.
It appears that your current model for the frequency is
freq(t) = b*exp(-alpha*t)
with
freq(0) = b = startfreq
freq(dursec) = b*exp(-alpha*dursec) = endfreq
There are two free parameters (b and alpha), and two equations. The first equation, b = startfreq, gives us b (trivially).
Solving the last equation for alpha gives
alpha = -log(endfreq/startfreq)/dursec
= log(startfreq/endfreq)/dursec
So
freq(t) = startfreq * exp(-alpha*t)
To use this as the instantaneous frequency of a frequency-swept signal,
we need the integral, which I'll call phase(t):
phase(t) = -(startfreq/alpha) * exp(-alpha*t)
The (complex) frequency-swept signal is then
sig(t) = exp(2*pi*j * phase(t))
The real part of this signal is
sig(t) = cos(2*pi*phase(t))
That explains your current code. To generate a chirp whose frequency varies like the blue curve, you need a different model for the frequency. A more general model than the one used above is
freq(t) = a + b*exp(-alpha*t)
The requirements at t=0 and t=dursec are
freq(0) = a + b = startfreq
freq(dursec) = a + b*exp(-alpha*dursec) = endfreq
That's two equation, but we now have three parameters: a, b, and alpha. I'll use the two equations to determine a and b, and leave alpha as a free parameter. Solving gives
b = (startfreq - endfreq)/(1 - exp(-alpha*dursec))
a = startfreq - b
Integrating the model gives
phase(t) = a*t - (b/alpha)*exp(-alpha*t)
alpha is an arbitrary parameter. Following the formula from the first model, I'll use:
alpha = abs(log(startfreq/endfreq))/dursec
The following is a complete script. Note that I also changed the use of exp(-j*2*pi*...) to cos(2*pi*...). The factor 0.8 is there to match your code.
startfreq = 20;
endfreq = 200;
fs = 44100;
dursec = 10; % duration of signal in seconds
t = (0:dursec*fs)/fs; % Time vector
if (startfreq == endfreq)
phase = startfreq * t;
else
alpha = abs(log(endfreq/startfreq))/dursec;
b = (startfreq - endfreq)/(1 - exp(-alpha*dursec));
a = startfreq - b;
phase = a*t - (b/alpha)*exp(-alpha*t);
endif
sig = 0.8 * cos(2*pi*phase);
wavwrite([sig'] ,fs,32,strcat('del.wav')); % export file
specgram(sig,150,400);
there.
I am going to generate 10^6 random points in matlab with this particular characters.
the points should be inside a sphere with radious 25, the are 3-D so we have x, y, z or r, theta, phi.
there is a minimum distance between each points.
first, i decided to generate points and then check the distances, then omit points with do not have these condition. but, it may omit many of points.
another way is to use RSA (Random Sequential Addition), it means generate points one by one with this minimum distance between them. for example generate first point, then generate second randomly out of the minimum distance from point 1. and go on till achieving 10^6 points.
but it takes lots of time and i can not reach 10^6 points, since the speed of searching appropriate position for new points will take long time.
Right now I am using this program:
Nmax=10000;
R=25;
P=rand(1,3);
k=1;
while k<Nmax
theta=2*pi*rand(1);
phi=pi*rand(1);
r = R*sqrt(rand(1));
% convert to cartesian
x=r.*sin(theta).*cos(phi);
y=r.*sin(theta).*sin(phi);
z=r.*cos(theta);
P1=[x y z];
r=sqrt((x-0)^2+(y-0)^2+(z-0)^2);
D = pdist2(P1,P,'euclidean');
% euclidean distance
if D>0.146*r^(2/3)
P=[P;P1];
k=k+1;
end
i=i+1;
end
x=P(:,1);y=P(:,2);z=P(:,3); plot3(x,y,z,'.');
How can I efficiently generate points by these condition?
thank you.
I took a closer look at your algorithm, and concluded there is NO WAY it will ever work - at least not if you really want to get a million points in that sphere. There is a simple picture that explains why not - this is a plot of the number of points that you need to test (using your technique of RSA) to get one additional "good" point. As you can see, this goes asymptotic at just a few thousand points (I ran a slightly faster algorithm against 200k points to produce this):
I don't know if you ever tried to compute the theoretical number of points you could fit in your sphere when you have them perfectly arranged, but I'm beginning to suspect the number is a good deal smaller than 1E6.
The complete code I used to investigate this, plus the output it generated, can be found here. I never got as far as the technique I described in my earlier answer... there was just too much else going on in the setup you described.
EDIT:
I started to think it might not be possible, even with "perfect" arrangement, to get to 1M points. I made a simple model for myself as follows:
Imagine you start on the "outer shell" (r=25), and try to fit points at equal distances. If you divide the area of the "shell" by the area of one "exclusion disk" (of radius r_sub_crit), you get a (high) estimate of the number of points at that distance:
numpoints = 4*pi*r^2 / (pi*(0.146 * r^(2/3))^2) ~ 188 * r^(2/3)
The next "shell" in should be at a radius that is 0.146*r^(2/3) less - but if you think of the points as being very carefully arranged, you might be able to get a tiny bit closer. Again, let's be generous and say the shells can be just 1/sqrt(3) closer than the criteria. You can then start at the outer shell and work your way in, using a simple python script:
import scipy as sc
r = 25
npts = 0
def rc(r):
return 0.146*sc.power(r, 2./3.)
while (r > rc(r)):
morePts = sc.floor(4/(0.146*0.146)*sc.power(r, 2./3.))
npts = npts + morePts
print morePts, ' more points at r = ', r
r = r - rc(r)/sc.sqrt(3)
print 'total number of points fitted in sphere: ', npts
The output of this is:
1604.0 more points at r = 25
1573.0 more points at r = 24.2793037966
1542.0 more points at r = 23.5725257555
1512.0 more points at r = 22.8795314897
1482.0 more points at r = 22.2001865995
1452.0 more points at r = 21.5343566722
1422.0 more points at r = 20.8819072818
1393.0 more points at r = 20.2427039885
1364.0 more points at r = 19.6166123391
1336.0 more points at r = 19.0034978659
1308.0 more points at r = 18.4032260869
1280.0 more points at r = 17.8156625053
1252.0 more points at r = 17.2406726094
1224.0 more points at r = 16.6781218719
1197.0 more points at r = 16.1278757499
1171.0 more points at r = 15.5897996844
1144.0 more points at r = 15.0637590998
1118.0 more points at r = 14.549619404
1092.0 more points at r = 14.0472459873
1066.0 more points at r = 13.5565042228
1041.0 more points at r = 13.0772594652
1016.0 more points at r = 12.6093770509
991.0 more points at r = 12.1527222975
967.0 more points at r = 11.707160503
943.0 more points at r = 11.2725569457
919.0 more points at r = 10.8487768835
896.0 more points at r = 10.4356855535
872.0 more points at r = 10.0331481711
850.0 more points at r = 9.64102993012
827.0 more points at r = 9.25919600154
805.0 more points at r = 8.88751153329
783.0 more points at r = 8.52584164948
761.0 more points at r = 8.17405144976
740.0 more points at r = 7.83200600865
718.0 more points at r = 7.49957037478
698.0 more points at r = 7.17660957023
677.0 more points at r = 6.86298858965
657.0 more points at r = 6.55857239952
637.0 more points at r = 6.26322593726
618.0 more points at r = 5.97681411037
598.0 more points at r = 5.69920179546
579.0 more points at r = 5.43025383729
561.0 more points at r = 5.16983504778
542.0 more points at r = 4.91781020487
524.0 more points at r = 4.67404405146
506.0 more points at r = 4.43840129415
489.0 more points at r = 4.21074660206
472.0 more points at r = 3.9909446055
455.0 more points at r = 3.77885989456
438.0 more points at r = 3.57435701766
422.0 more points at r = 3.37730048004
406.0 more points at r = 3.1875547421
390.0 more points at r = 3.00498421767
375.0 more points at r = 2.82945327223
360.0 more points at r = 2.66082622092
345.0 more points at r = 2.49896732654
331.0 more points at r = 2.34374079733
316.0 more points at r = 2.19501078464
303.0 more points at r = 2.05264138052
289.0 more points at r = 1.91649661498
276.0 more points at r = 1.78644045325
263.0 more points at r = 1.66233679273
250.0 more points at r = 1.54404945973
238.0 more points at r = 1.43144220603
226.0 more points at r = 1.32437870508
214.0 more points at r = 1.22272254805
203.0 more points at r = 1.1263372394
192.0 more points at r = 1.03508619218
181.0 more points at r = 0.94883272297
170.0 more points at r = 0.867440046252
160.0 more points at r = 0.790771268402
150.0 more points at r = 0.718689381062
140.0 more points at r = 0.65105725389
131.0 more points at r = 0.587737626612
122.0 more points at r = 0.528593100237
113.0 more points at r = 0.473486127367
105.0 more points at r = 0.422279001431
97.0 more points at r = 0.374833844693
89.0 more points at r = 0.331012594847
82.0 more points at r = 0.290676989951
75.0 more points at r = 0.253688551418
68.0 more points at r = 0.219908564725
61.0 more points at r = 0.189198057381
55.0 more points at r = 0.161417773651
49.0 more points at r = 0.136428145311
44.0 more points at r = 0.114089257597
38.0 more points at r = 0.0942608092113
33.0 more points at r = 0.0768020649149
29.0 more points at r = 0.0615717987589
24.0 more points at r = 0.0484282253244
20.0 more points at r = 0.0372289153633
17.0 more points at r = 0.0278306908104
13.0 more points at r = 0.0200894920319
10.0 more points at r = 0.013860207063
8.0 more points at r = 0.00899644813842
5.0 more points at r = 0.00535025545232
total number of points fitted in sphere: 55600.0
This seems to confirm that you really can't get to a million, no matter how you try...
There are many things you could do to improve your program - both algorithm, and code.
On the code side, one of the things that is REALLY slowing you down is the fact that not only you use a for loop (which is slow), but in the line
P = [P;P1];
you append elements to an array. Every time that happens, Matlab needs to find a new place to put the data, copying all the points in the process. This quickly becomes very slow. Preallocating the array with
P = zeros(1000000, 3);
keeping track of the number N of points you have found so far, and changing your calculation of distance to
D = pdist2(P1, P(1:N, :), 'euclidean');
would at least address that...
The other issue is that you check new points against all previously found points - so when you have 100 points, you check about 100x100, for 1000 it is 1000x1000. You can see then that this algorithm is O(N^3) at least... not counting the fact that you will get more "misses" as the density goes up. A O(N^3) algorithm with N=10E6 takes at least 10E18 cycles; if you had a 4 GHz machine with one clock cycle per comparison, you would need 2.5E9 seconds = approximately 80 years. You can try parallel processing, but that's just brute force - who wants that?
I recommend that you think about breaking the problem into smaller pieces (quite literally): for example, if you divide your sphere into little boxes that are about the size of your maximum distance, and for each box you keep track of what points are in it, then you only need to check against points in "this" box and its immediate neighbors - 27 boxes in all. If your boxes are 2.5 mm across, you would have 100x100x100 = 1M boxes. That seems like a lot, but now your computation time will be reduced drastically, as you will have (by the end of the algorithm) only 1 point on average per box... Of course with the distance criterion you are using, you will have more points near the center, but that's a detail.
The data structure you would need would be a cell array of 100x100x100, and each cell contains the index of the good points found so far "in that cell". The problem with a cell array is that it doesn't lend itself to vectorization. If instead you have the memory, you could assign it as a 4D array of 10x100x100x100, assuming you will have no more than 10 points per cell (if you do, you will have to handle that separately; work with me here...). Use an index of -1 for points not yet found
Then your check would be something like this:
% initializing:
bigList = zeros(10,102,102,102)-1; % avoid hitting the edge...
NPlist = zeros(102, 102, 102); % track # valid points in each box
bottomcorner = [-25.5, -25.5, -25.5]; % boxes span from -25.5 to +25.5
cellSize = 0.5;
.
% in your loop:
P1= [x, y, z];
cellCoords = ceil(P1/cellSize);
goodFlag = true;
pointsSoFar = bigList(:, cellCoords(1)+(-1:1), cellCoords(2)+(-1:1), cellCoords(3)+(-1:1));
pointsToCheck = find(pointsSoFar>0); % this is where the big gains come...
r=sum(P1.^2);
D = pdist2(P1,P(pointsToCheck, :),'euclidean'); % euclidean distance
if D>0.146*r^(2/3)
P(k,:) = P1;
% check the syntax of this line...
cci = ind2sub([102 102 102], cellCoords(1), cellCoords(2), cellCoords(3));
NP(cci)=NP(cci)+1; % increasing number of points in this box
% you want to handle the case where this > 10!!!
bigList(NP(cci), cci) = k;
k=k+1;
end
....
I don't know if you can take it from here; if you can't, say so in the notes and I may have some time this weekend to code this up in more detail. There are ways to speed it up more with some vectorization, but it quickly becomes hard to manage.
I think that putting a larger number of points randomly in space, and then using the above for a giant vectorized culling, may be the way to go. But I recommend to take little steps first... if you can get the above to work at all well, you can then optimize further (array size, etc).
I found the reference - "Simulated Brain Tumor Growth Dynamics Using a Three-Dimensional Cellular Automaton", Ansal et al (2000).
I agree it is puzzling - until you realize one important thing. They are reporting their results in mm, but your code was written in cm. While that may seem insignificant, the formula for "critical radius", rc = 0.146r^(2/3) includes a constant, 0.146, that is dimensional - the dimensions are mm^(1/3), not cm^(1/3).
When I make that change in my python code to evaluate the number of possible lattice sites, it jumps by a factor 10. Now they claimed that they were using a "jamming limit" of 0.38 - the number where you really cannot find any more sites. If you include that limit, I predict no more than 200k points could be found - still short of their 1.5M, but not quite so crazy.
You might consider contacting the authors to discuss this with them? If you want to include me in the conversation, you can email me at: SO (just two letters) at my handle name dot united states. Same domain as where I posted links above...
This is easy to do in OpenCV however I would like a native Matlab implementation that is fairly efficient and can be easily changed. The method should be able to take the camera parameters as specified in the above link.
The simplest and most common way of doing undistort (also called unwarp or compensating for lens distortion) is to do a forward distortion on a chosen output photo size and then a reverse mapping using bilinear interpolation.
Here is code I wrote for performing this:
function I = undistort(Idistorted, params)
fx = params.fx;
fy = params.fy;
cx = params.cx;
cy = params.cy;
k1 = params.k1;
k2 = params.k2;
k3 = params.k3;
p1 = params.p1;
p2 = params.p2;
K = [fx 0 cx; 0 fy cy; 0 0 1];
I = zeros(size(Idistorted));
[i j] = find(~isnan(I));
% Xp = the xyz vals of points on the z plane
Xp = inv(K)*[j i ones(length(i),1)]';
% Now we calculate how those points distort i.e forward map them through the distortion
r2 = Xp(1,:).^2+Xp(2,:).^2;
x = Xp(1,:);
y = Xp(2,:);
x = x.*(1+k1*r2 + k2*r2.^2) + 2*p1.*x.*y + p2*(r2 + 2*x.^2);
y = y.*(1+k1*r2 + k2*r2.^2) + 2*p2.*x.*y + p1*(r2 + 2*y.^2);
% u and v are now the distorted cooridnates
u = reshape(fx*x + cx,size(I));
v = reshape(fy*y + cy,size(I));
% Now we perform a backward mapping in order to undistort the warped image coordinates
I = interp2(Idistorted, u, v);
To use it one needs to know the camera parameters of the camera being used.
I am currently using the PMD CamboardNano which according to the Cayim.com forums has the parameters used here:
params = struct('fx',104.119, 'fy', 103.588, 'cx', 81.9494, 'cy', 59.4392, 'k1', -0.222609, 'k2', 0.063022, 'k3', 0, 'p1', 0.002865, 'p2', -0.001446);
I = undistort(Idistorted, params);
subplot(121); imagesc(Idistorted);
subplot(122); imagesc(I);
Here is an example of the output from the Camboard Nano. Note: I artificially added border lines to see what the effect was of the distortion close to the edges (its much more pronounced):
You can now do that as of release R2013B, using the Computer Vision System Toolbox. There is a GUI app called Camera Calibrator and a function undistortImage.
I am using Gonzalez frdescp function to get Fourier descriptors of a boundary. I use this code, and I get two totally different sets of numbers describing two identical but different in scale shapes.
So what is wrong?
im = imread('c:\classes\a1.png');
im = im2bw(im);
b = bwboundaries(im);
f = frdescp(b{1}); // fourier descriptors for the boundary of the first object ( my pic only contains one object anyway )
// Normalization
f = f(2:20); // getting the first 20 & deleting the dc component
f = abs(f) ;
f = f/f(1);
Why do I get different descriptors for identical - but different in scale - two circles?
The problem is that the frdescp code (I used this code, that should be the same as referred by you) is written also in order to center the Fourier descriptors.
If you want to describe your shape in a correct way, it is mandatory to mantain some descriptors that are symmetric with respect to the one representing the DC component.
The following image summarize the concept:
In order to solve your problem (and others like yours), I wrote the following two functions:
function descriptors = fourierdescriptor( boundary )
%I assume that the boundary is a N x 2 matrix
%Also, N must be an even number
np = size(boundary, 1);
s = boundary(:, 1) + i*boundary(:, 2);
descriptors = fft(s);
descriptors = [descriptors((1+(np/2)):end); descriptors(1:np/2)];
end
function significativedescriptors = getsignificativedescriptors( alldescriptors, num )
%num is the number of significative descriptors (in your example, is was 20)
%In the following, I assume that num and size(alldescriptors,1) are even numbers
dim = size(alldescriptors, 1);
if num >= dim
significativedescriptors = alldescriptors;
else
a = (dim/2 - num/2) + 1;
b = dim/2 + num/2;
significativedescriptors = alldescriptors(a : b);
end
end
Know, you can use the above functions as follows:
im = imread('test.jpg');
im = im2bw(im);
b = bwboundaries(im);
b = b{1};
%force the number of boundary points to be even
if mod(size(b,1), 2) ~= 0
b = [b; b(end, :)];
end
%define the number of significative descriptors I want to extract (it must be even)
numdescr = 20;
%Now, you can extract all fourier descriptors...
f = fourierdescriptor(b);
%...and get only the most significative:
f_sign = getsignificativedescriptors(f, numdescr);
I just went through the same problem with you.
According to this link, if you want invariant to scaling, make the comparison ratio-like, for example by dividing every Fourier coefficient by the DC-coefficient. f*1 = f1/f[0], f*[2]/f[0], and so on. Thus, you need to use the DC-coefficient where the f(1) in your code is not the actual DC-coefficient after your step "f = f(2:20); % getting the first 20 & deleting the dc component". I think the problem can be solved by keeping the value of the DC-coefficient first, the code after adjusted should be like follows:
% Normalization
DC = f(1);
f = f(2:20); % getting the first 20 & deleting the dc component
f = abs(f) ; % use magnitudes to be invariant to translation & rotation
f = f/DC; % divide the Fourier coefficients by the DC-coefficient to be invariant to scale
I'm taking images using a tunneling microscope. However, the scope is drifting between successive images. I'm trying to use MatLab to calculate the offset between images. The code below calculates in seconds for small images (e.g. 64x64 pixels), but takes >2 hrs to handle the 512x512 pixel images I'm dealing with. Do you have any suggestions for speeding up this code? Or do you know of better ways to track images in MatLab? Thanks for your help!
%Test templates
template = .5*ones(32);
template(25:32,:) = 0;
template(:,25:64) = 0;
data_A = template;
close all
imshow(data_A);
template(9:32,41:64) = .5;
template(:,1:24) = 0;
data_B = template;
figure, imshow(data_B);
tic
[m n] = size(data_B);
z = [];
% Loop over all possible displacements
for x = -n:n
for y = -m:m
paddata_B = data_B;
ax = abs(x);
zerocols = zeros(m,ax);
if x > 0
paddata_B(:,1:ax) = [];
paddata_B = [paddata_B zerocols];
else
paddata_B(:,(n-ax+1):n) = [];
paddata_B = [zerocols paddata_B];
end
ay = abs(y);
zerorows = zeros(ay,n);
if y < 0
paddata_B(1:ay,:) = [];
paddata_B = vertcat(paddata_B, zerorows);
else
paddata_B((m-ay+1):m,:) = [];
paddata_B = vertcat(zerorows, paddata_B);
end
% Full matrix sum after array multiplication
C = paddata_B.*data_A;
matsum = sum(sum(C));
% Populate array of matrix sums for each displacement
z(x+n+1, y+m+1) = matsum;
end
end
toc
% Plot matrix sums
figure, surf(z), shading flat
% Find maximum value of z matrix
[max_z, imax] = max(abs(z(:)));
[xpeak, ypeak] = ind2sub(size(z),imax(1))
% Calculate displacement in pixels
corr_offset = [(xpeak-n-1) (ypeak-m-1)];
xoffset = corr_offset(1)
yoffset = corr_offset(2)
What you're calculating is known as the cross-correlation of the two images. You can calculate the cross-correlation of all offsets at once using Discrete Fourier Transforms (DFT or FFT). So try something like
z = ifft2( fft2(dataA) .* fft2(dataB).' );
If you pad with zeros in the Fourier domain, you can even use this sort of math to get offsets in fractions of a pixel, and apply offsets of fractions of a pixel to an image.
A typical approach to this kind of problem is to use the fact that it works quickly for small images to your advantage. When you have large images, decimate them to make small images. Register the small images quickly and use the computed offset as your initial value for the next iteration. In the next iteration, you don't decimate the images as much, but you're starting with a good initial estimate of the offset so you can constrain your search for solutions to a small neighborhood near your initial estimate.
Although not written with tunneling microscopes in mind, a review paper that may be of some assistance is: "Mutual Information-Based Registration of Medical Images: A Survey" by Pluim, Maintz, and Viergever published in IEEE Transactions on Medical Imaging, Vol. 22, No. 8, p. 986.
below link will help you find transformation between 2 images and correct/recover the distorted (in your case, image with offset)
http://in.mathworks.com/help/vision/ref/estimategeometrictransform.html
index_pairs = matchFeatures(featuresOriginal,featuresDistorted, 'unique', true);
matchedPtsOriginal = validPtsOriginal(index_pairs(:,1));
matchedPtsDistorted = validPtsDistorted(index_pairs(:,2));
[tform,inlierPtsDistorted,inlierPtsOriginal] = estimateGeometricTransform(matchedPtsDistorted,matchedPtsOriginal,'similarity');
figure; showMatchedFeatures(original,distorted,inlierPtsOriginal,inlierPtsDistorted);
The inlierPtsDistored, inlierPtsOriginal have attributes called locations.
These are nothing but matching locations of one image on another. I think from that point it is very easy to calculate offset.
The function below was my attempt to compute the cross-correlation of the two images manually. Something's not quite right though. Will look at it again this weekend if I have time. You can call the function with something like:
>> oldImage = rand(64);
>> newImage = circshift(oldImage, floor(64/2)*[1 1]);
>> offset = detectOffset(oldImage, newImage, 10)
offset =
32 -1
function offset = detectOffset(oldImage, newImage, margin)
if size(oldImage) ~= size(newImage)
offset = [];
error('Test images must be the same size.');
end
[imageHeight, imageWidth] = size(oldImage);
corr = zeros(2 * imageHeight - 1, 2 * imageWidth - 1);
for yIndex = [1:2*imageHeight-1; ...
imageHeight:-1:1 ones(1, imageHeight-1); ...
imageHeight*ones(1, imageHeight) imageHeight-1:-1:1];
oldImage = circshift(oldImage, [1 0]);
for xIndex = [1:2*imageWidth-1; ...
imageWidth:-1:1 ones(1, imageWidth-1); ...
imageWidth*ones(1, imageWidth) imageWidth-1:-1:1];
oldImage = circshift(oldImage, [0 1]);
numPoint = abs(yIndex(3) - yIndex(2) + 1) * abs(xIndex(3) - xIndex(2) + 1);
corr(yIndex(1),xIndex(1)) = sum(sum(oldImage(yIndex(2):yIndex(3),xIndex(2):xIndex(3)) .* newImage(yIndex(2):yIndex(3),xIndex(2):xIndex(3)))) * imageHeight * imageWidth / numPoint;
end
end
[value, yOffset] = max(corr(margin+1:end-margin,margin+1:end-margin));
[dummy, xOffset] = max(value);
offset = [yOffset(xOffset)+margin-imageHeight xOffset+margin-imageWidth];