How to count a number of edges counting from canny?

count a number of edges counting from canny?

Answer 1

[Thorsten]

You can use bwlabel.

[Selva Karna]

Thiru Thorsten sir . i have attached which i use image ..please guide me which one is suitable for counting number of edges based [glass counting]

[Image Analyst]

Selva's "Answer" (which she accepted for some reason) is transferred here:

Thank you for useful time ...but bwlabel cannot be used....

My concept is counting bundle of glass. from wood box. the glass[images] from different like [bright,dark]. so the counting cannot use. The bundle of glass[images] have some time gab. So the bwlabel also count the gab, so only count glass edges only, so please guide me sir.

[Image Analyst]

It does count the number of contiguous edges. If that's not what you want, you haven't described what you want in sufficient detail for anyone to know what it is. We still don't. Why don't you attach an image to illustrate what you want?

[Selva Karna]

sorry sir my delay ...........

my need is counting number of glass[images].

[Selva Karna]

Sir I have attached image . how to count number of glass from above images...please help me ...sir...

[Image Analyst]

First you should absolutely work on getting a better photo. That should not be hard. I mean, why do you have all that clutter behind the glass? That's probably not necessary and just makes it harder. Maybe you can project a last beam across it and just look at deviations in the beam. In general it's so much better to start with a good image than to try to fix up a horrible image after capture. You need to do this to make your job easier.

[Selva Karna]

sir any algorithm or code available please refer me .. Guide me sir..................I am wait for your reply sir

[Selva Karna]

any body dont known about my question ?

[Image Analyst]

We're waiting for you to give a better image. That's a lot of work for us to do for you - to fix up a terrible image - so why should we spend our time to do that instead of just waiting until you improve your image?

[Selva Karna]

How to improve pixel intensity and also better edge detection...........

[Selva Karna]

Image Analyst sir still now u didn give solution for my question .sir i am wait for you sir,,,,,,,,

Answer 2

[John BG]

Hi Selva

1.- acquire image

      A=imread('count_ribbons.jpg')
      [im_1 im_2 im_3]=size(A)    % [y x 3] or [vertical horizontal 3] or [rows columns 3]  despite cursor on image shows [X Y]

to Image Analyst: let's work on the basis that there is no better image. We all see it's just a bunch of foil ribbons attached on a green cardboard, but think of it as a Radar or Sonar sample. It's what it is, right now you don't have anything better.

2.- check variance of RGB

      surf(var(double(A),0,3))
      surf(std(double(A),0,3))

3.- From var and std, there are heavily attenuated areas, especially on the upper half of the image.

Reading the image bottom up is as if target departing because the lines with ribbons get broader. To process a bunch of lines without taking into account Doppler, let's take a strip made of the lowest 40 lines, process them, and answer the amount that clearly shows up as the mode among these 40 lines.

Before processing the strip, let's find out what processing is required for each line

4.- acquire a line within the last 40 lines

      figure(1);imshow(A)
      p3=ginput(2); p3=uint64(floor(p3))                    % input 2 points excluding line left-right not containing ribbons
      % for instance:
      % p3 =             33                394
      %                   403                  38

5.- show the sample line

    py_ref=min(p3(2,2),p3(1,2))
    im_line3_1=A(py_ref,[min(p3(1,1),p3(2,1)):max(p3(1,1),p3(2,1))],1)  % Red
    im_line3_2=A(py_ref,[min(p3(1,1),p3(2,1)):max(p3(1,1),p3(2,1))],2)  % Green
    im_line3_3=A(py_ref,[min(p3(1,1),p3(2,1)):max(p3(1,1),p3(2,1))],3)  % Blue
    v3_ref=[1:1:(abs(double(p3(1,1))-double(p3(2,1))))+1]
    figure(2);plot(v3_ref,im_line3_1,'r',v3_ref,im_line3_2,'g',v3_ref,im_line3_3,'b')

 observing the sample line, the ribbons are the valleys. The peaks are the green/white thin lines separating ribbons.

6.- To count peaks coinciding with ribbons let's invert sample line:

im_line3_1=255-im_line3_1
im_line3_2=255-im_line3_2
im_line3_3=255-im_line3_3
    % figure(3);plot(v3_ref,im_line3_1,'r',v3_ref,im_line3_2,'g',v3_ref,im_line3_3,'b')

7.- Red component does not help that much, so carry on with Green and Blue only, (G+B) and amplify a bit

L1=3*(double(im_line3_2)+double(im_line3_3))  % amplify a bit
nL1=[1:1:length(L1)]
% figure(4);plot(L1)

8.- remove DC

L1=L1-mean(L1)
hold on;figure(4);plot(L1)

9.- 1st assault counting peaks, try different values and see how the amount of ribbons detected varies:

numel(findpeaks(L1,'minpeakheight',80))
numel(findpeaks(L1,'minpeakheight',100))
numel(findpeaks(L1,'minpeakheight',140))
numel(findpeaks(L1,'minpeakheight',170))
numel(findpeaks(L1,'minpeakheight',190))

sweep findpeaks parameter 'minpeakheight'

peaks_prob_1=0
for k=80:1:250 
    peaks_prob_1=[peaks_prob_1 numel(findpeaks(L1,'minpeakheight',k))]; 
end
figure(5);plot(peaks_prob_1);grid on
v_prob_1=[1:length(peaks_prob_1)]
 figure(6);hist(peaks_prob_1,v_prob_1)  % misleading, the true amount of ribbons does not show as the value to pick as answer

insert03

insert04

10.- It's a really short sample, spectrum analysis needs as many samples as possible, and the data of interest to repeat, yet let's see DTFT:

nL1=[1:length(L1)]           % just copied lines from another script with DTFT
x=L1;n=[1:1:length(L1)]
x=double(x)        

k=0:500;w=(pi/500)*k
X=x*(exp(-j*pi/500)).^(n'*k)
X=x*(exp(-j*pi/500)).^(n'*k)
magX=abs(X);angX=angle(X);realX=real(X);imagX=imag(X)

figure(15);subplot(2,2,1);plot(k/500,magX);grid
xlabel('freq[rad]');title('Magnitude(X)')
figure(15);subplot(2,2,3);plot(k/500,angX/pi);grid
xlabel('freq[rad]');title('Phase(X)')
figure(15);subplot(2,2,2);plot(k/500,realX);grid
xlabel('freq[rad]');title('Real(X)')
figure(15);subplot(2,2,4);plot(k/500,imagX);grid
xlabel('freq[pi]');title('Imag(X)')

 % count cycles:

   371/2*.164=30.42    % do we round up? or down? error by 1 or 2 ribbons too many?

11.- and how well does the famous FFT?

NFFT=2^nextpow2(length(L1))  % 512 freq points [0 2pi]
absL1=abs(fft(L1,NFFT))
figure(5);plot(absL1)

insert06

again count cycles:

371/2*43/256   % almost there but can't tell exactly

12.- try command FINDPEAKS and parameter 'minpeakprominence'

[pks,locs]=findpeaks(L1,'minpeakprominence',10)
[pks,locs]=findpeaks(L1,'minpeakprominence',15)
[pks,locs]=findpeaks(L1,'minpeakprominence',20)
[pks,locs]=findpeaks(L1,'minpeakprominence',25)
[pks,locs]=findpeaks(L1,'minpeakprominence',50)
[pks,locs]=findpeaks(L1,'minpeakprominence',60)

% minpeakprominence=60 still one ribbon too many
[pks,locs]=findpeaks(L1,'minpeakprominence',60);figure(7);plot(nL1,L1,locs,pks+4,'kv');grid on
 numel(pks)  %=31  and depending upon what line picked up

It is clear that more than one line has to be taken into account to make a decision.

13.- frequency contents with command PLOMB:

          Fs=1;Fmax=1/2;[Pxx,f]=plomb(double(L1),Fs,Fmax,10,'power');figure(7);plot(Pxx);grid on
          [Pxx_pks,f0]=findpeaks(Pxx,f,'MinPeakHeight',400)

depending upon the line chosen, when placing marker on spike tip, just above 300, there are 2 marker positions on top of the spike, there is need for higher frequency resolution.

You choose a different line and get a better result, just one sample on tip of spike achieved without telling plomb() what Fs and Fmax to use:

    [Pxx,f]=plomb(double(L1),nL1);figure(7);plot(Pxx);grid on

marker on f=123, started all over with different line, now length(L1)=379

tried with different windows, Welch for instance broadens peak of interest, and the frequency spike should be as sharp as possible

insert07-1

insert07-2

again count cycles:

123/758*379/2  % =30.75

Nyquist wrote: 9/10 you try something, there is a better way:

14.- correlate signal with expected pulse, have a look at a single received pulse

insert08

try different pulses, gradually make the pulse narrower and taller

     pulse1=[0 200 200 200 200 200 200 200 200 200 200 200 200 0]  % poor correlation
      pulse1=[0 50 100 100 100 100 100 100 100 100 100 50 0]  % a bit better
      pulse1=[0 0 50 100 100 100 100 100 100 100 50 0 0] % a bit better, just missing 1 pulse
       pulse1=[0 0 0 50 100 100 100 100 100 50 0 0 0]
      pulse1=[0 0 0 100 200 200 200 200 200 100 0 0 0]
      pulse1=[0 0 0 0 100 200 200 200 100 0 0 0 0]  %   the count of peaks is now correct, do not include start and end markers
      pulse1=[0 0 0 0 200 400 400 400 200 0 0 0 0]

if you manually check with findpeaks, at every pulse there is improvement. Yet, depending on chosen sample line, there are missed pulses. If you start the sample line too early and let it run too late you may have to remove the first and last peak counts

insert09

prob2=[]
for k=1:20
  [pks_xcorr1puls,locs_xcorr1puls]=findpeaks(r_L1_pulse1,'minpeakdistance',k)
  prob2=[prob2 numel(pks_xcorr1puls)]
end
% figure(9);plot(prob2);grid on

[pks_xcorr1puls,locs_xcorr1puls]=findpeaks(r_L1_pulse1,'minpeakdistance',6)
% remove start and end counts
 numel(pks_xcorr1puls)-2   % -2, as mentioned, for first and last of previous picture

% 1st half of r_L1_pulse1 does not have relevant information
r_L1_pulse1=r_L1_pulse1([375:750])
n_r_L1_pulse1=[1:1:length(r_L1_pulse1)]

spectrum of the autocorrelation

    [Pxcorr,f]=plomb(r_L1_pulse1,n_r_L1_pulse1);figure(10);plot(Pxcorr);grid on
    length(Pxcorr)=752
    122/752*376/2 % =30.5  pretty close, but still not as accurate as counting peaks

15.- Improve by using command FILTER input signal before XCORR.

Use the wonderful FDATOOL to build a direct FIR band pass filter equiripple that allows through the ribbons frequency only.

Try different normalized BPF stop1 pass stop2 configurations and use 'a' 'b' to filter the sample line

b: numerator

a: denominator

Mind the gap, once came across a book using b:den a:num

I found normalized band pass filter shoulders [.25 .3] and [.4 .45] reject 30dB enough

Tried to attach ribbon_filter.fda for the readers to import it directly into FDATOOL, but the browser editor refused arguing it was none of the file formats accepted, and .fda files are not accepted as attachments (!?).

_Could some one in Mathworks please make it possible for .fda files to be attached to answers/comments in the MATLAB CENTRAL forum?

thanks in advance._

So back on track, manually doesn't take that long once you know where are the band pass shoulders.

Filter mask

insert11-1 % filter pole/zero diagram insert11-2

b=[0.028  0.053 0.071  0.053 0.028]
a=[1.000 -2.026 2.148 -1.159 0.279]
L1_BPfilt=filter(b,a,L1)
nL1_BPfilt=[1:length(L1_BPfilt)]
hold all;figure(1);plot(nL1,L1,nL1_BPfilt,L1_BPfilt);grid on

Comparing sample line without (blue) and with (orange) filtering:

16.- Improve by using command DETREND: removes VLF. Detrend the sample line before filtering

L1_detrended=detrend(L1,'linear',[95 289]);hold all;figure(1);plot(L1_detrended)

detrend after filtering

L1_BPfilt_detr=detrend(L1_BPfilt,'linear',[95 289])

with

plot(abs(fft(L1_BPfilt_detr)))

now the highest spikes are around the area of interest, the frequency of the ribbons. Yet depending on the sample line chosen, the top spike may be the right one, or you may erroneously choose a nearby taller spike by mistake.

insert10

length(abs(fft(L1_BPfilt_detr)))  % =379

since we don't know the amount of ribbons in advance we cannot make the filter too narrow otherwise we might reject the right frequency.

17.- Improve by autocorrelating all received signal on itself, not on just a single pulse.

r_L1=xcorr(L1_BPfilt_detr,L1_BPfilt_detr)
r_L1=r_L1([374:end])  % self correlations are symmetric, we only one one side, taking right hand side
figure(2);plot(r_L1);grid on

I haven't done it here, I amplify a bit 3*(G+B), but the recommended place to amplify is AFTER filtering, not before.

After checking different FINDPEAKS 'minpeakheight', all and each ribbon has 1 and only 1 peak above 0 when 'minpeakheight'=0

[pks,locs]=findpeaks(r_L1,'minpeakheight',0)
numel(pks)   % =  29

note that this graph still includes start stop as peaks, 31-2

18.- So far it is only 1 line, let's remove luck out of the loop, with a for loop repeat for all lines in the sample strip. Recap:

        ribbon_count=zeros(1,40)
        for k=400:440                                                                                        % remove Doppler by taking strip where ribbons freq is almost constant
          im_line_2=A(k,:,2);im_line_3=A(k,:,3)                                        % remove RGB component showing poorest cyclic behaviour: Red                                          im_line_2=im_line_2([35:402]);im_line_3=im_line_3([35:402])     % take only line pixels containing ribbons burst
          im_line_2=255-im_line_2;im_line_3=255-im_line_3                      % invert R G to make ribbons show up as peaks
          L1=3*(double(im_line_2)+double(im_line_3))                               % 3*(R+G)
          L1=L1-mean(L1)                                                                        % MEAN: remove DC
          L1_BPfilt=filter(b,a,L1)                                                                % FILTER
          L1_BPfilt_detr=detrend(L1_BPfilt,'linear',[95 289])                         % DETREND: remove VLF, also called 'wander'
          r_L1=xcorr(L1_BPfilt_detr,L1_BPfilt_detr);r_L1=r_L1([374:end])      % XCORR: take rigth hand side half      
          [pks,locs]=findpeaks(r_L1,'minpeakheight',0)                                 % FINDPEAKS: with 'minpeakheight'
          ribbon_count(k-400+1)=numel(pks)                                              % NUMEL: count peaks
        end

have a look

      stem(ribbon_count)
      hist(ribbon_count)

the answer is

      mode(ribbon_count)   % overwhelming majority to: 29

If you find this answer of any help solving the question above, please click on the thumbs-up vote link,

thanks in advance

John

[e-mail]

UPC-ETSETB 99

comment 1: there are other tools and ways, for instance medfilt1, sgolay, sgolayfilt: all seem to be able to remove HF. I was about to try sgolay, and medfilt1, that both seem to remove HF, but fdatool went well enough calculating 'a' and 'b'.

comment 2: you see that some pulses have 'horns' and many have slope on top, and between pulses there are 'little pulses'? that can be caused by multipath, aka echo. Within certain limitations equalizers remove echo.

But equalizers need training sequences that precisely should have the right frequency, which is what the question is after

There is a built-in class to equalize signals.

I wanted to use the BER calculator with the QAM carrier sync recovery, by sweeping a frequency range with:

• guess a frequency
• build training sequence
• clean all lines in strip with equalizer
• filter, amplify, ..
• check BER

It would be reasonable to expect a BER notch on the right amount of ribbons but it would have taken far longer that the above lines.