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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

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
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
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
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Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
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Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Source:  OpenStax, Virtual drum kit. OpenStax CNX. Dec 19, 2013 Download for free at http://cnx.org/content/col11607/1.1
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