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We round the angle to 0, 45, 90 and 135. With a specific angle value, we examine the maxima within the range of three pixel values in that direction (vertically, horizontally or diagonally). By picking up the maxima and discarding non maxima points, the edge will be one pixel wide. In order to enhance the detection, we use hysteresis thresholding.

Non-maximum suppression

s = padarray(s,[1 1]);theta1 = padarray(theta1,[1 1]);for k = 2:(size(m,1)+1) for l = 2:(size(m,1)+1)if ((theta1(k,l) == 90)&&(s(k,l)== max(max(s(k+1,l),s(k,l)),... s(k-1,l))))mge(k,l) = s(k,l); elseif ((theta1(k,l) == 0)&&(s(k,l)== max(max(s(k,l+1),... s(k,l)),s(k,l-1))))mge(k,l) = s(k,l); elseif ((theta1(k,l) == 135)&&(s(k,l) == max(max(s(k+1,l+1),... s(k,l)),s(k-1,l-1))))mge(k,l) = s(k,l); elseif ((theta1(k,l) == 45)&&(s(k,l) == max(max(s(k+1,l-1),... s(k,l)),s(k-1,l+1))))mge(k,l) = s(k,l); endend end
** When writing the MATLAB program for this part, finding the maxima at the boundary rows and columns require padding to the image matrix. So we padded zeros to the matrix to ensure that the boundary edges could also be detected.

Hysteresis thresholding

Here we set two threshold values, a low threshold and a high threshold. The threshold values are important for the resulting image since if the low value is too low there could be invalid edges not being discarded, and if the high value is too high the edge will be discontinuous and some valid edges could be mistakenly discarded. Moderate threshold values are required.

Hysteresis thresholding

% Hysteresis Thresholding % Hysteresis is one way of solving this problem. Instead of choosing a single% threshold, two thresholds thigh and tlow are used. high = 105;low = 35; tmg = mge;% find the edge to be discarted index1 = find(mge<=low); tmg(index1) = 0;for i = 3:(size(m,1)-2) for j = 3:(size(m,1)-2)% neighbors in a 3*3 region around it have gradient magnitudes % greater than threshold high ,keep the edgeif (mge(i,j)<high&&mge(i,j)>low) if ((mge(i-1:i+1,j-1:j+1)>high) == zeros(3)) tmg(i,j)=0;% If non of pixel (x; y)'s neighbors have high gradient % magnitude but at least one falls between low and high,% search the 5*5 region to see if any of these pixels have % a magnitude greater than high. If so, keep the edge.elseif ((mge(i-1:i+1,j-1:j+1)>low) ~= [0 0 0;0 1 0;0 0 0])if ((mge(i-2:i+2,j-2:j+2)>high) == zeros(5)) tmg(i,j)=0;end endend endend
It will be better if the webcam is placed in parallel with the paper; in this way the resulting edge detected will be a desired regular rectangular.

Image processing--template matching

After calibration, we are set up in the new coordinate system, which is the white paper region. We will detect drum stick position in this region.

To detect the position of drum stick, we used template matching. Since the stick head is black that has a high contrast from its surroundings (paper is white), we build a simple square template filled with black pixels. This template serves as a filter that helps us to find the position of stick head. We did 2D convolution between this template filter and the snapshot from the video. Euclidean distance is used to find the matching part. Below is a how we did template matching step by step:

Matching

As the template moves along x/y axis of the image, assume that it has size of N by N; the template will compare all its pixel values with N by N matrix of the image it moves to.

    Basic algorithm steps

  1. Set the minimum distance to infinity
  2. Iterate the template through the image
  3. Each iteration calculate for the euclidean distance between the template and the part of matrix the it iterates to
  4. If the distance is smaller than the current minimum distance, replace the minimum distance with that value
  5. Repeat the iteration until the 2D convolution is finished

Template matching code

i=0; while(i<100) % trigger fun: start logging nowtrigger(vid) frame = getsnapshot(vid); % record the current frameflushdata(vid); delete(vid);Im = double(frame); Im = Im/max(max(Im)); % normalize the image%%%%%%%%%%%%%%%%%%%%Processing Im%%%%%%%%%%%%%%%%%% for m = 1:size(Im,1)-N+1for n = 1:size(Im,2)-N+1 diff = sum(sum(Im(m:m+N-1,n:n+N-1)-temp));if diff<best_error loc_x = m+N/2;loc_y = n+N/2; endend endloc_y = loc_y-calib(1); loc_x = loc_x-calib(2);position(i,1) = loc_x;position(i,2) = loc_y; %%%%%%%%%%%%%%%%%%%%EndProcessing%%%%%%%%%%%%%%%%%%imshow(Im); hold onrectangle('Position',[size(Im,1)-loc_x,loc_y,N,N])i=i+1; end

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
Jude
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David Reply
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David
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emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
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Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
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hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Who can show me the full solution in this problem?
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Source:  OpenStax, Elec 301 projects fall 2013. OpenStax CNX. Sep 14, 2014 Download for free at http://legacy.cnx.org/content/col11709/1.1
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