Zero Crossing of Signal
이전 댓글 표시
Hello everyone.
I want to find the zero crossing of my Signals. I used below code but it works incorrectly.
%%%%%%%%%%%Time Domain DA3 %%%%%%%%%%%
da3=da2*tf3;
da3s=ilaplace(da3);
da3t=subs(da3s,{t},{time});
//da3t is signal of which i need zero corssings
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);
zx = zci(da3t); //to find zero crossings
close all
figure;
plot(time(zx), da3t(zx), 'bp')
plot(time*1e9,da3t ,'-k','LineWidth',2);
xlabel('Time - [ns]');ylabel('Amplitude - [V]');
title('DA Transient Output ');
hold on
grid on
plot(time(zx), da3t(zx), 'bp') // to plot zero crossing
채택된 답변
Star Strider
2020년 4월 11일
The ‘zci’ funciton works correctly, although the latest version of it is:
zci = @(v) find(v(:).*circshift(v(:), 1, 1) <= 0); % Returns Approximate Zero-Crossing Indices Of Argument Vector (>= R2016b)
to work with more recent MATLAB releases. Note that it returns the indices of the approximate zero-crossings. If you want the exact zero-croissings, you need to interpolate to find them.
I cannot run your code, so I cannot correct the errors.
댓글 수: 22
Star Strider
2020년 4월 11일
편집: Star Strider
2020년 4월 11일
I do not understand what you are doing with these.
I will delete my Answer in a few minutes.
Syed Adeel
2020년 4월 11일
Syed Adeel
2020년 4월 11일
Star Strider
2020년 4월 11일
When I run it, I get:
Check for missing argument or incorrect argument data type in call to function 'solve'.
Error in cspreamp_model_function (line 51)
tf=solve(eq1,eq2,eq3,'vout,vgs1,vs6');
Error in Model_AnalogSection_ASD_ (line 56)
[tfcsp,tfvin,x1t,y1t,y2t,ys2]=cspreamp_model_function(QIN,QTIME,CD);
I am running R2020a. I believe quoted string arguments in the Symbollic Math Toolbox have been deprecated for the last several releases.
I will help you with the zero-crossings problem (since I wrote the ‘zci’ function), however I cannot correct everything that may be wrong with your code.
It would very much help of we could simply narrow the zero-crossing problem down to just that portion of the code, with everything else provided. I will revisit this when you have provided ‘da3’ ‘time’ and everything else, and the required syms statement necessary to run it.
.
Syed Adeel
2020년 4월 11일
편집: Syed Adeel
2020년 4월 11일
Star Strider
2020년 4월 11일
I ran that function, as the error mesages demonstrate.
The errors thrown are:
Check for missing argument or incorrect argument data type in call to function 'solve'.
Error in cspreamp_model_function (line 51)
tf=solve(eq1,eq2,eq3,'vout,vgs1,vs6');
Error in Model_AnalogSection_ASD_ (line 56)
[tfcsp,tfvin,x1t,y1t,y2t,ys2]=cspreamp_model_function(QIN,QTIME,CD);
The code will not work in R2020a. I can go no further.
Please provide whatever ‘da3’ and ‘time’ are, and the syms statement to use them, as in your original post.
.
Syed Adeel
2020년 4월 11일
Star Strider
2020년 4월 11일
My pleasure.
I can help with your problem if I have ‘da3’ and ‘time’. You should be able to provide ‘da3’ and ‘time’ by using disp(da3), and disp(time), just after you create them, then copying them and pasting them here. The problem is that I cannot run your code (it is incompatible with R2020a and I no longer have access to R2016a), so I cannot create them myself.
Syed Adeel
2020년 4월 11일
편집: Syed Adeel
2020년 4월 11일
Syed Adeel
2020년 4월 11일
Star Strider
2020년 4월 12일
We still have a problem.
The ‘da3’ output ends with: ‘... Output truncated. Text exceeds maximum line length of 25,000 characters for Command Window display.’
Try it again with:
disp(vpa(da3))
and copy that to a Comment here.
The ‘time’ vector is fine, so you only need to try it again with ‘da3’.
.
Star Strider
2020년 4월 12일
Unrecognized function or variable 's'.
Error in
x1s=(exp(-s*to1))*1/s-(exp(-s*to2))*(1/s);
That loooks like it is about 15 lines from the start.
Syed Adeel
2020년 4월 12일
편집: Syed Adeel
2020년 4월 12일
You are right My mind got dumb on this problem since morning. my apology boss. I just reconfirmed and its working, i correctd and have attached the output Graph attached.
Rf = 40e3; Cf = 680e-15;
R2 = 16e3;
R1 = 16e3;
wu = 1 / ((R1 * Rf / (R1 + Rf)) * Cf);
R1da1=1.35e3/2; %CHANGE FROM PAPER e3/2
R2da1=6.26e3/2;
fda1lp=150e6;%changed from 6 to 5
tau1da1=1/(2*pi*fda1lp);
%%% DA2 Parameters %%% %%%% %%% %%% %%% %%% %%% %%% %%% %%% %%% %%% %%% %%%
R1da2=2.43e3/2;
R2da2=6.26e3/2;
C1da2=84e-12;
%%% DA3 Parameters %%% %%%% %%% %%% %%% %%% %%% %%% %%% %%% %%% %%% %%% %%%
R1da3=2.45e3/2;
R2da3=9.77e3/2;
C1da3=92e-12;
T = 30 / wu;
N = 100;
time = 1: 1: N;
time=T*time/N;
Q = 5e-15;
QTIME = 4e-9; % Max 11ns
I= Q / QTIME;
to1=1e-9;
to2=to1+QTIME;
syms s;
X1s = (exp (-s * to1)) * 1 / s- (exp (-s * to2)) * (1 / s);
I = Q / QTIME;
ip = I * X1s;
%%%%%%%%%%%%%%%%%%%%%%%%% Time Domain DA1-DA3 %%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%
syms gda1 t1da1 ;
tf_da1 = gda1 / (1 + s * t1da1);
tf = subs (tf_da1, {gda1, t1da1}, {R2da1 / R1da1, tau1da1});
pretty (tf);
%%%%% DA2 Transfer Function %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%
syms r1da2 r2da2 c1da2;
tf_da2=(r2da2/r1da2)*(1+s*c1da2*2*r1da2)/(1+s*c1da2*r1da2);
tf_da2=tf_da2./(1+s/(2*pi*150e6));
tf2=subs(tf_da2,{r1da2,r2da2,c1da2},{R1da2,R2da2,C1da2});
pretty(tf2);
syms r1da3 r2da3 c1da3;
tf_da3=s*c1da3*r2da3/(1+s*c1da3*r1da3);
tf_da3=0.1+tf_da3./(1+s/(2*pi*150e6));
tf3=subs(tf_da3,{r1da3,r2da3,c1da3},{R1da3,R2da3,C1da3});
pretty(tf3);
%%%%%%%%%%% Time Domain DA1 %%%%%%%%%%%
syms t ;
da1 = ip * tf;
DA11 = ilaplace (da1);
da1t = subs (DA11, {t}, {time});
%%%%%%%%%%% Time Domain DA2 %%%%%%%%%%%
da2 = da1 * tf2;
da2s = ilaplace (da2);
da2t = subs (da2s, {t}, {time});
%%%%%%%%%%% Time Domain DA3 %%%%%%%%%%%
da3 = da2 * tf3;
da3s = ilaplace (da3);
da3t = subs (da3s, {t}, {time}); % zerocrossing for this signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%
close all
figure;
plot (time * 1e9, da3t, '-k' , 'LineWidth' , 2);
xlabel ( 'Time - [ns]' ); ylabel ( 'Amplitude - [V]' );
title ( 'DA Transient Output' );
hold on
grid on
legend ( 'DA1' , 'DA2' , 'DA3' );
Star Strider
2020년 4월 12일
The problem remains:
Unrecognized function or variable 's'.
Error in
X1s = (exp (-s * to1)) * 1 / s- (exp (-s * to2)) * (1 / s);
Syed Adeel
2020년 4월 12일
편집: Syed Adeel
2020년 4월 12일
Star Strider
2020년 4월 12일
The end of your code now becomes:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%
close all
figure;
zci = @(v) find(v(:).*circshift(v(:), [1 0]) <= 0); % Returns Approximate Zero-Crossing Indices Of Argument Vector
da3td = double(da3t);
idx = zci(da3td);
if da3td(1) * da3td(end) < 0
idx = idx(2:end);
end
for k = 1:numel(idx)
t_exact(k) = interp1(da3td(idx(k))+[-1 +1]*1E-5, time(idx(k))*1E9+[-1 +1], 0);
end
plot (time * 1e9, da3t, '-k' , 'LineWidth' , 2);
hold on
plot(time(idx)*1E9, da3td(idx), 'r+')
hold off
xlabel ( 'Time - [ns]' ); ylabel ( 'Amplitude - [V]' );
title ( 'DA Transient Output' );
hold on
grid on
legend ( 'DA1' , 'DA2' , 'DA3' );
text(t_exact, 0, sprintf('\\downarrow time = %.4f',t_exact), 'HorizontalAlignment','left', 'VerticalAlignment','bottom')
You can eliminate plotting the red ‘+’ if you want.
.
Syed Adeel
2020년 4월 12일
Star Strider
2020년 4월 12일
Thank you!
As always, my pleasure!
The code I posted will detect all the zero-crossings of a specific function and will interpolate all of them to find the exact values of the ‘time’ value of the zero-crossings. (That is the reason I included the for loop, even though it is not needed here.) The code may need to be modified to show the zero-crossings on the plot, however the essence of the code I posted should work for all of them.
If you have several different curves, plotted on the same or different axes, it would be necessary to run my code to find all the zero-crossings for each one.
.
Syed Adeel
2020년 4월 15일
Star Strider
2020년 4월 15일
The indices returned by ‘zci’ will remain as a vector. The more precise values will be in the ‘t_exact’ vector in the comment I previously posted.
Syed Adeel
2020년 4월 15일
Star Strider
2020년 4월 16일
It shows NaN because it is trying to extrapolate, although I am not certain what the reason is for that.
One solution for that is to let it extrapolate, and see what it returns:
t_exact(k) = interp1(da3td(idx(k))+[-1 +1]*1E-5, time(idx(k))*1E9+[-1 +1], 0, 'linear','extrap');
That should solve the extrapolation problem.
Another option (if the data permit it, so that it does not reference indices beyond the vector limits) is:
t_exact(k) = interp1(da3td(idx(k)+[-1 +1])*1E-5, time(idx(k)+[-1 +1])*1E9, 0);
If that needs to extrapolate, add those arguments.
추가 답변 (1개)
Image Analyst
2020년 4월 11일
Did you try to use find() to find out the first index where da3t goes negative?
firstNegIndex = find(da3t <= 0, 1, 'first')
firstNegTime = time(firstNegIndex)
By the way, time is a built-in function so you should not use that as the name of your variable.
댓글 수: 3
Syed Adeel
2020년 4월 11일
Image Analyst
2020년 4월 12일
It only crosses once. Once it's crossed, the values are now below the x axis. So you'll only have one crossing per signal, or possibly more if it later goes from below to above the axis. But you won't have 3 consecutive points where there is a crossing unless the signal has a value of 0 for 3 elements consecutively, which your signals don't do.
Syed Adeel
2020년 4월 15일
Thankyou so much for your reply. Your option gives me accrate first zero crossing. Now as I have 3 zero corssings can you please tell me how to extend your method?Thankyou
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