periodic function with n cycles
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Hi, I need to create a periodic function and plot it.
F(x)=sqrt(3) + *Sin(t -2*pi/3) --> 0<t<pi/3
F(x)=Sin(t) --> pi/3 <t<2*pi/3
repeat the signal 0<t<3*pi with the period 2*pi/3 Then plot(t,Fx)
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At the moment I use the following code
>> t1=0:.01:pi/3;
>> t2=pi/3:.01:2*pi/3;
A=sqrt(3) + sin(t1*2*pi- 2*pi/3);
B=sin(t2);
plot(t1,A,t2,B)
This method is produce the answer a one cycle. However it is quite difficult to repeat the pattern for multiple times.
Can any one n please suggest way of doing this
1 commentaire
Image Analyst
le 8 Déc 2013
Sounds like your homework. Is it?
Réponse acceptée
Plus de réponses (2)
Azzi Abdelmalek
le 8 Déc 2013
t1=0:.01:pi/3;
t2=pi/3:.01:2*pi/3;
A=sqrt(3) + sin(t1*2*pi- 2*pi/3);
B=sin(t2);
t=[t1 t2],
y=[A,B]
plot(t,y)
m=5 % Repetition
n=numel(t);
tt=0:0.01:n*m*0.01-0.01
yy=repmat(y,1,m)
plot(tt,yy)
4 commentaires
Rashmil Dahanayake
le 9 Déc 2013
Rashmil Dahanayake
le 9 Déc 2013
Modifié(e) : Rashmil Dahanayake
le 10 Déc 2013
Andrei Bobrov
le 10 Déc 2013
Modifié(e) : Andrei Bobrov
le 10 Déc 2013
Hi Rashmil! See my variant of your problem (after ADD in my answer)
zhenning li
le 1 Nov 2020
truely thanks,it helps a lot!
you can do it as follow:
count = 1;
for t = 0:pi/3:pi - pi/3
if mod(count, 2) == 1
x = linspace(t, t + pi/3);
y = sqrt(3) + sin(x * 2 * pi - 2 * pi/3);
plot(x, y), hold on
count = count + 1;
else
x = linspace(t, t + pi/3);
y = sin(x);
plot(x, y), hold on
count = count + 1;
end
end
Maybe following link is also helpful for you:
2 commentaires
Rashmil Dahanayake
le 9 Déc 2013
sixwwwwww
le 9 Déc 2013
It was selected to choose between the plots curve should be plotted. It doesn't have effect on output actually. The output is controlled by the range in the for loop:
for t = 0:pi/3:pi - pi/3
changing pi - pi/3 to pi - pi/3 will give more periods of the plot
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