R2020b Matlab Editor is running very slow

Good day everyone,

Please, I wish if someone can kindly help me to solve the following problem. I wrote the code in the attached file on Matlab R2020a to generate a trapezoidal wave, then analyze it , it was running perfectly, then suddenly when I started to run the code, it sarted to take an extemely long time to finsh the running process. I tried to check if the problem is from my lap top, that's why, I ran the same code on another lap top with a different matlab version R2017b , however the result was the same and the running time can be up to two or three hours. I wish if someone can kindly offer some help as soon as possible .

Thank you so much .

[Mostafa Abotaleb]

Can someone kindly answer me or give me a hint as soon as possible ?

[dpb]

Hint: Attach the m-file as code text -- use the CODE buttons to format it to be legible.

Guess: You didn't preallocate an output vector/array and are building it by dynamic addressing. If the problem size is large, this can be the cause of such a symptom.

[Mostafa Abotaleb]

clc;
f=input('enter the frequency ...');
x = 0:0.000001:10;
y1 = 1*(asind(sin((2*pi*f)*x+1)))-0.5;
y2 = 1*(asind(sin((2*pi*f)*x+1))+acosd(cos((2*pi*f)*x+1)))-1/2;
y3 = 1*(acosd(cos((2*pi*f)*x+1)))-1/2;
massimo = max(y2);
minimo = min(y2);
y4 = y2/massimo;

subplot(4,1,1);
plot(x,y1);
subplot(4,1,2);
plot(x,y2);
subplot(4,1,3);
plot(x,y3);
y7 = y4;
for dd=1:length(y4)
  if y4(dd)<0
    y4(dd)= 0;
  end
end

subplot(4,1,4);
plot(x,y4);
figure;
plot(x,y4);
y5 = y4;
y6 = y4;
for uu = 1:length(y4)
  tt = uu + 1
  if tt < length(y4)
    subtra = y4(uu) + y4(tt);
    if subtra >= 1 && subtra < 2; 
      y4(uu) = sym(y4(uu));
    end
  end
end
counterones = 0;
counterzeros = 0;
fc1 = [];
fc0 = [];
insert1 = 0;
insert0 = 0;

for ww = 1:length(y4)
    qq = ww + 1;
    if qq < length(y4)
      if y4(ww)==1 && y4(qq) == 1 
        counterones = counterones + 1
      end
      if y4(ww)==0 && y4(qq)==0 
        counterzeros = counterzeros + 1
      end
      if y4(ww)==1 && y4(qq)<1
        insert1 = counterones + 1;
        counterones =0;
        fc1 = [fc1 insert1];
      end
      if y4(ww)==0 && y4(qq)>0
        insert0 = counterzeros + 1;
         counterzeros =0;
        fc0 = [fc0 insert0];
      end
    end
    insert0 = 0;
    insert1 = 0;
end

[Mostafa Abotaleb]

dp , Thank you so much for your reply , I attached the code as in the answers as you said , If you have time , I wsh if you kindly check it and tell me the problem , Really Thanks

[Mostafa Abotaleb]

dp, in the line x = 1:0.000001:10 , I used this extremely small increment because I would like to simulate a frequency of 31.25 kHz and if I used less increments, the output signal became distorted.

[dpb]

for dd=1:length(y4)
  if y4(dd)<0
    y4(dd)= 0;
  end
end

is just

y4=max(y4,0);

instead of the loop...will save a little.

But, the real problem is precisely what I guessed -- you're building a new array dynamically by concatnenation in the following code segment. This causes a reallocation and recopy of the data every time that is terribly time consuming.

 if y4(ww)==1 && y4(qq)<1
    insert1 = counterones + 1;
    counterones =0;
    fc1 = [fc1 insert1];
  end
  if y4(ww)==0 && y4(qq)>0
    insert0 = counterzeros + 1;
     counterzeros =0;
    fc0 = [fc0 insert0];
  end

I didn't try to read the code to decipher precisely what you're wanted result is, but your problem is in the above that doesn't preallocate the two new vectors and copy into them the data wanted at the locations instead appending a new value onto the end.

Describe the output in words that is wanted here...

[Mostafa Abotaleb]

Thank you so much for your reply another time , regarding the purpose of this part of the code that you mentioned in your answer, its purpose is to count the number of ones and zeros in each single trapezoidal wave form and accumulate each of them in two arrays fc1 and fc0, however I noticed that the code takes longer time in this attached part of the code, the purpose of this part is to convert all the values of 1.0000 into only '1' . I tried many functions like 'round' , 'fix' and others , but unfortunately , it didn't workout . it worked out only with the function 'sym', in the line

' y4(uu) = sym(y4(uu));'

for uu = 1:length(y4)
  tt = uu + 1
  if tt < length(y4)
    subtra = y4(uu) + y4(tt);
    if subtra >= 1 && subtra < 2; 
      y4(uu) = sym(y4(uu));
    end
  end
end

[dpb]

if y4(ww)==1 && y4(qq)<1

You just above turned all y4<1 to 0 so the above can never be true

for uu = 1:length(y4)
  tt = uu + 1
  if tt < length(y4)
...

Don't use the inner if, just use

for uu = 1:numel(y4)-1
  tt = uu + 1
  ...

Also, what's the point of using sym in sym(y4(uu))? It serves no purpose but will definitely add a lot of overhead as symbolic. What is the point of this in the end?

[Mostafa Abotaleb]

regarding this part :

if y4(ww)==1 && y4(qq)<1

qq = ww+1, so iam checking the condition when the falling edge of the signal will start , detecting the last element of 1

Regarding the second part :

for uu = 1:numel(y4)-1

tt = uu + 1

Thank you so much , I will modify it.

regarding the use of sym(y4(uu) , I mentionede in mylast answer that I would like to convert all the values of 1.0000 into only '1' , because otherwise I am not able to make any comparison operation on them , I tried to use 'round' and 'fix' , but it didn't work out

[dpb]

No. You're confusing the display format of the value with the value. And, if y4(uu) isn't identically the integer 1, then sym(y4(uu)) won't be, either.

If there is some rounding in the initial values that you want defuzzified, then round or fix is definitely the operation you're looking for.

None of this, however, fixes the big problem of building the arrays fc0 and fc1 dynamically which is where the time bottleneck of the first order is.

If you're simply looking for (+/-) crossings, then start with

y=sign(y); y(y==0)=1;   % convert to +/- inidicators, count 0 as +ive
dy=diff(y);             % gives sign change [0, +/-2]

which gives you a vector first of [-1,1] and then the sign changes of dy==2 or -2 for +ive, -ive crossing locations. Remember to use an offset of 1 because numel(dy) is one less than numel(y) after take difference.

No loops are needed at all...

[Mostafa Abotaleb]

regarding using round or fix , I used them as I mentioned and many times , but the problem was still the sme , and value of 1.0000 remained the same , right now I am using the same code with less frequency of 10Hz and horizontal axis increments of 0.01 and it is working fine , I tried fixing and rounding but the value remained 1.0000, and it changed to '1' only when I used 'sym'. I will try to minimize the loops as you said but fixing and rounding , it works for the whole array or vector , yes I agree , for example if wanted to make fixing for a = [0.7 , 0.8, 0.9, 1.0000] it will result into a= [1,1,1,1], but I am trying only to convert the values of 1.0000 and keep the other values as they are. maybe I didn't completely understand you, but really thank you for answering and for your help

[dpb]

SHOW US!!! DON'T JUST TELL US!!!

I don't believe you understand and that there's really a difference other than the formatting of the display at the command window of the value. If, indeed sym() turned it into a visible "1" then that proves that the value itself internally is, indeed, identically the integer 1.

And, again, you've not solved the problem that is the time crunch; the code I showed above will run in milliseconds over the whole vector.

[dpb]

>> x = 0:0.000001:10; y=rand(size(x))-0.5;
>> tic,y=sign(y);y(y==0)=1;dy=diff(y);fp=find(dy==2);fn=find(dy==-2);toc
Elapsed time is 0.230929 seconds.
>> 

fn, fp above are the +/-xings for the given y that is the size of your original vector. This is a 15-yr old lower-middle-class machine that wasn't at all fast for its time, what more now and it took only 1/4-sec.

[Mostafa Abotaleb]

I said thank you so much for your help and I needed it , and I will not shout as you did , the same code will work if you tried lower time increments as I said and lower frequency like 10 Hz , I don't need to show you anything since you already have the whole code. I thought I was discussing with you in a respectful manner because I am the one who asked for help !!!!!!!!!!!!!!!!, but since you will write in such a disrespectful manner , I will have to wish you good luck and stop writing and another time , I don't need to show anything since you already have the code , regards

[dpb]

No disrespect intended -- just to emphasize it helps to show us the data itself, not to just describe a symptom. "Help us help you!"

There's no rounding code anywhere in what you have posted so we can't see what you tried, specifically that you thought didn't work, and I don't have the symbolic TB so can't run your code as is, even if I were to download it.

Just attach a section of the y4() array that has an element that is one that you think you need sym() for as a .mat file so can see just what it is that you're geting...

As for the code running, sure it'll run -- but you already pointed out in the beginning you needed the resolution to model a very high sampling rate accurately -- so running it at a much reduced rate doesn't solve the real problem while the solution I showed works for the sizes of vectors you really have an eliminates the rounding issues entirely, whatever they might be.

[dpb]

for uu = 1:length(y4)
  tt = uu + 1
  if tt < length(y4)
    subtra = y4(uu) + y4(tt);
    if subtra >= 1 && subtra < 2; 
      y4(uu) = sym(y4(uu));
    end
  end
end

is

subtra=sum([y4(1:end-1);y4(2:end)]);
ix=(subtra>=1) && (subtra<2); 
y4(uu) = sym(y4(uu));

but I still don't see the point/purpose of any of it.

It just looks like an effort to code sign in a round about way to me. What's the point of the addition of the two adjacent elements instead of just using the signal itself?

[dpb]

Maybe what you're looking for instead of round or fix is ceil...but it's not clear just what the intent is/what the problem is that you're trying to cure.