How can I plot the periodic function? please help

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how can I plot(t,Tg) or interpret Tg in t=(0:0.001:0.1n),n is a given positive integer ?
but it didn't work. plz help.

Accepted Answer

Walter Roberson
Walter Roberson on 6 Jun 2019
Tg = 2900 * size(t);
tmod = mod(t, 0.1);
mask = tmod > 0.007;
Tg(mask) = A * exp(-B * tmod(mask));
  3 Comments
Torsten
Torsten on 7 Jun 2019
Replace
Tg = 2900 * size(t);
by
Tg = 2900 * ones(size(t));

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More Answers (1)

mousa alsaeed
mousa alsaeed on 25 Nov 2019
here is a code for a function that can give you the plot for any periodic function first thing you need to define a variable t as symbolic variable
syms t
second thing define your function
f = log(t)-t
the function needs a period and a specific number of cycles to draw your function in
T = 3; , N = 10;
here is the code for the function :
% done by mousa alsaeed
% email : mousa276alsaeed98@gmail.com
function periodic(fun,period,number_of_cycles)
fp = 0;
syms t
for i=0:number_of_cycles-1
fp = fp + subs(fun,t,(t-i*period))*(heaviside(t-i*period)-heaviside(t-(i+1)*period));
end
fplot(fp,[0 number_of_cycles*period])
grid on
title('the periodic function')
xlabel('t')
set(gca,'XMinorTick','On','YMinorTick','On');
end
here is what you gonna write on the commmand window :
periodic(f,T,N)
and here is the results :
3.png
  3 Comments
mousa alsaeed
mousa alsaeed on 4 Aug 2020
sorry for the delay
same inputs as before and here is the new code that plots periodic function to be symmetric around the y-axis (i.e. starts from -N/2*T to N/2*T)
% done by mousa alsaeed
% email : mousa276alsaeed98@gmail.com
function periodic(fun,period,number_of_cycles)
fp = 0;
syms t
for i= -number_of_cycles/2 : number_of_cycles/2
fp = fp + subs(fun,t,(t-(2*i+1)*0.5*period))*(heaviside(t-i*period)-heaviside(t-(i+1)*period));
end
fplot(fp,[(-number_of_cycles/2)*period (number_of_cycles/2)*period])
grid on
title('the periodic function')
xlabel('t')
set(gca,'XMinorTick','On','YMinorTick','On');
end
here is the results
Nabeel Bakr
Nabeel Bakr on 5 Aug 2020
Thank you very much for your help ... Regards

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