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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)
------
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

Accepted Answer

Andrei Bobrov
Andrei Bobrov on 8 Dec 2013
Edited: Andrei Bobrov on 10 Dec 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 Comments
Rashmil Dahanayake
Rashmil Dahanayake on 10 Dec 2013
Edited: Rashmil Dahanayake on 10 Dec 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 on 22 May 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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More Answers (2)

Azzi Abdelmalek
Azzi Abdelmalek on 8 Dec 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 Comments
Andrei Bobrov
Andrei Bobrov on 10 Dec 2013
Edited: Andrei Bobrov on 10 Dec 2013
Hi Rashmil! See my variant of your problem (after ADD in my answer)
zhenning li
zhenning li on 1 Nov 2020
truely thanks,it helps a lot!

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sixwwwwww
sixwwwwww on 8 Dec 2013
Edited: sixwwwwww on 8 Dec 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 Comments
Rashmil Dahanayake
Rashmil Dahanayake on 9 Dec 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 on 9 Dec 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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