# 0.3 Implementation: detect position of drum stick  (Page 2/2)

 Page 2 / 2

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

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
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Abhi
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can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
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oops. ignore that.
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hmm
Abhi
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Abhi
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salma
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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