periodic function with n cycles
    6 visualizzazioni (ultimi 30 giorni)
  
       Mostra commenti meno recenti
    
    Rashmil Dahanayake
      
 il 8 Dic 2013
  
    
    
    
    
    Commentato: Behrang Hoseini
 il 22 Mag 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
Risposta accettata
  Andrei Bobrov
      
      
 il 8 Dic 2013
        
      Modificato: Andrei Bobrov
      
      
 il 10 Dic 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 Commenti
  Rashmil Dahanayake
      
 il 10 Dic 2013
				
      Modificato: Rashmil Dahanayake
      
 il 10 Dic 2013
  
			
		
  Behrang Hoseini
 il 22 Mag 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?
Più risposte (2)
  Azzi Abdelmalek
      
      
 il 8 Dic 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 Commenti
  Andrei Bobrov
      
      
 il 10 Dic 2013
				
      Modificato: Andrei Bobrov
      
      
 il 10 Dic 2013
  
			Hi Rashmil! See my variant of your problem (after ADD in my answer)
  sixwwwwww
      
 il 8 Dic 2013
        
      Modificato: sixwwwwww
      
 il 8 Dic 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 Commenti
  sixwwwwww
      
 il 9 Dic 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
Vedere anche
Categorie
				Scopri di più su Logical in Help Center e File Exchange
			
	Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!






