periodic function with n cycles

11 vues (au cours des 30 derniers jours)
Rashmil Dahanayake
Rashmil Dahanayake le 8 Déc 2013
Commenté : Behrang Hoseini le 22 Mai 2022
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)
------
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
Image Analyst le 8 Déc 2013
Sounds like your homework. Is it?

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Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 8 Déc 2013
Modifié(e) : Andrei Bobrov le 10 Déc 2013
t = 0:pi/100:6*pi;
t1 = rem(t,2*pi/3);
l = t1 < pi/3 ;
F = @(t,l)sqrt(3)*l + sin((2*pi*l + ~l).*t -2*pi/3*l);
out = F(t1,l);
plot(t,out)
ADD
t = 2*pi*(0:.0005:1).';
t1 = rem(t,2*pi/3);
l1 = t1 < pi/3;
l0 = ~l1;
y = zeros(numel(t),2);
y(l1,1) = sqrt(3) + sin(t1(l1) - 2*pi/3);
y(l0,1) = sin(t1(l0));
y(l1,2) = sin(t1(l1) - 2*pi/3);
y(l0,2) = sin(t1(l0)) - sqrt(3);
yy = sin([t,bsxfun(@plus,t,[1, -1]*2*pi/3)]);
plot(t,[y,yy]);
  2 commentaires
Rashmil Dahanayake
Rashmil Dahanayake le 10 Déc 2013
Modifié(e) : Rashmil Dahanayake le 10 Déc 2013
Thanks. I modified further so that I can vary the frequency of the generated wave. Fyi. updated Code
f= 2; %frequency in Hz
x=linspace(0,1,1001);
t=x.';
w=2*pi*f;
T=1/f;
t1 = rem(t,T/3);
l1 = t1 < T/6;
l0 = ~l1;
y = zeros(numel(t),2);
y(l1,1) = sqrt(3) + sin(w*t1(l1) - 2*pi/3);
y(l0,1) = sin(w*t1(l0));
y(l1,2) = sin(w*t1(l1) - 2*pi/3);
y(l0,2) = sin(w*t1(l0)) - sqrt(3);
yy = sin([w*t,bsxfun(@plus,w*t,[1, -1]*2*pi/3)]);
plot(t,[y,yy]); grid on;
Behrang Hoseini
Behrang Hoseini le 22 Mai 2022
Hi,
I want to use this method to develop a periodic window to apply to a time function. The thing I could't understand is the second added part:
t = 2*pi*(0:.0005:1).';
t1 = rem(t,2*pi/3);
l1 = t1 < pi/3;
l0 = ~l1;
y = zeros(numel(t),2);
y(l1,1) = sqrt(3) + sin(t1(l1) - 2*pi/3);
y(l0,1) = sin(t1(l0));
y(l1,2) = sin(t1(l1) - 2*pi/3);
y(l0,2) = sin(t1(l0)) - sqrt(3);
yy = sin([t,bsxfun(@plus,t,[1, -1]*2*pi/3)]);
plot(t,[y,yy]);
do we need to add it?

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Plus de réponses (2)

Azzi Abdelmalek
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
Andrei Bobrov
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
zhenning li le 1 Nov 2020
truely thanks,it helps a lot!

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sixwwwwww
sixwwwwww le 8 Déc 2013
Modifié(e) : sixwwwwww le 8 Déc 2013
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
Rashmil Dahanayake le 9 Déc 2013
It seems like the variable count does not have any effect on the output.
ie If I want to have 5 cycles of the, count=5? but the output remains unchanged.
sixwwwwww
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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