Pulstran function to generate biphasic pulses
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How can I correct following code to get pulses like on the picture? I don't get correct shape of pulses.
Parameters: N = 400 (number of pulses), Tp = 1 (pulse width), d1 = 1 (interphase delay [us]), d2 = 1 (interpulse delay [us]). Let's say the U0=1V. It should look like this:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1708221/image.jpeg)
number_of_points_on_interval = 1000;
t = linspace(0, 1.6, 10000);
Tp = 0.001; %pulse width [ms]
d1 = 0.001; %interphase delay [ms]
d2 = 0.001; %interpulse delay [ms]
dPos = 0.001:(d1+Tp+d2):1.6; % times when positive pulses start
dNeg = dPos + Tp + 0.001; % times when negative pulses start
dT = [dPos dNeg]'; % times of pulses
% Amplitude of positive and negative pulses
U=1;
dAmp = [U*ones(numel(dPos),1); -1 * U*ones(numel(dNeg),1)];
% generating a signal with rectangular pulses
y = pulstran(t,[dT dAmp],'rectpuls',Tp);
plot(t,y)
ylim padded
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回答(1 个)
Mathieu NOE
2024-6-3
hello
have to admit I almost never used pulstran yet , so I simply prefer to come back to a very simple solution with repmat if that suffice for the job
NB that to test the code I reduced the final time for the plot - simply put back t_final = 1.6 on your side
also I changed interpulse from 1 to 3 ms just to check the code - here also change back to your own settings
t_final = 1.6/50; % <= put back t_final = 1.6 on your side
dt = 1e-4;
t = (0:dt:t_final);
Tp = 0.001; %pulse width [ms]
d1 = 0.001; %interphase delay [ms]
d2 = 0.003; %interpulse delay [ms]
% Amplitude of positive and negative pulses
U=1;
% d2 period (zero amplitude)
t1 = (0:dt:d2-dt);
y1 = zeros(size(t1));
% pos pulse (+1 amplitude)
t2 = t1(end) + (dt:dt:Tp);
y2 = U*ones(size(t2));
% interphase (zero amplitude)
t3 = t2(end) + (dt:dt:d1);
y3 = zeros(size(t3));
% neg pulse (-1 amplitude)
t4 = t3(end) + (dt:dt:Tp);
y4 = -U*ones(size(t4));
dT = [t1 t2 t3 t4]'; % times of pulses
dAmp = [y1 y2 y3 y4]';
% how many cycles do we need to repeat the pattern ?
repet = fix(t_final/t4(end));
y = repmat(dAmp,repet,1);
t = dt*(0:numel(y)-1);
% generating a signal with rectangular pulses
plot(t,y)
% ylim padded
5 个评论
Mathieu NOE
2024-6-3
ok - maybe it's because it's monday .... I was slow to understand your comment
now , you simply pass t as argument to my newly created function and there you go ...
t_final = 1.6/40;
dt = 1e-4;
t = (0:dt:t_final);
% generating a signal with rectangular pulses
y = mypulsetrain(t);
plot(t,y)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function y = mypulsetrain(t)
dt = mean(diff(t));
t_final = t(end);
Tp = 0.001; %pulse width [ms]
d1 = 0.001; %interphase delay [ms]
d2 = 0.003; %interpulse delay [ms]
% Amplitude of positive and negative pulses
U=1;
% d2 period (zero amplitude)
t1 = (0:dt:d2-dt);
y1 = zeros(size(t1));
% pos pulse (+1 amplitude)
t2 = t1(end) + (dt:dt:Tp);
y2 = U*ones(size(t2));
% interphase (zero amplitude)
t3 = t2(end) + (dt:dt:d1);
y3 = zeros(size(t3));
% neg pulse (-1 amplitude)
t4 = t3(end) + (dt:dt:Tp);
y4 = -U*ones(size(t4));
dT = [t1 t2 t3 t4]'; % times of pulses
dAmp = [y1 y2 y3 y4]';
% how many cycles do we need to repeat the pattern ?
repet = fix(t_final/t4(end));
y = repmat(dAmp,repet,1);
% padd y with zeros to match t length
p = numel(t) - numel(y);
y = [y;zeros(p,1)];
end
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