periodic functions

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li
li 2011-4-20
I would like to know how to construct a 1 periodic function that is the same as the below link, http://www.flickr.com/photos/58977081@N02/5637186489/ the x1 is any values that could be adjusted by me. could anyone guide me to the result? thanks

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回答（5 个）

Paulo Silva
Paulo Silva 2011-4-20
amp=1; %amplitude of the wave
xp1=0.5; %this is your x1 value
r=0.1;n=3; %distance between points (r) and number of periods (n)
x0=0;y0=amp;x1=xp1;y1=0; %points to calculate m
m=(y1-y0)/(x1-x0); %declive of the wave
x=0:r:xp1; %x points of the wave where y value is not zero
xt=xp1:r:1; %x points of the wave where y value is zero
xx=[x xt]; %all the x points together
yy=[m*x+amp 0*xt]; %the y value of the wave
yyy=repmat(yy,1,n); %n periods of the wave
xxx=[]; %make the time value for all the wave periods
for a=0:n-1 %the use of the for can be avoided somehow
xxx=[xxx a+xx]; %I used the for just to make it work for now
end
%xxx=xxx-floor(n/2); %just in case you want to shift the x values
plot(xxx,yyy,'r') %plot it just to see if its working
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Matt Fig
Matt Fig 2011-4-20
Okie-dokie! I had to look that one up, it has been so long...
Paulo Silva
Paulo Silva 2011-4-20
We still need to know if li wants a nice graph or values for something else, maybe li wants a signal generator

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Clemens
Clemens 2011-4-20
I would write it as function:
function y = myfun(x,x1) %definition of function
xx = mod(x,1); % remainder to start of period 1
k=1/x1; % slope of first part
if xx<x1, y = 1-k*xx; % calculate value for part x<x1
else y=0; end % calculate for x>=x1
of course there would be many ways to describe the curve.
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Clemens
Clemens 2011-4-21
That is true. I would apply it as follows:
x = linspace(-100,100,10000);
y = arrayfun(@(a) myfun(a,0.3),x);
I suppose logical indexing will speed things up.
Clemens
Clemens 2011-4-21
Since I just read about producing infinite values. I'd do it:
x=0; step = 0.1;
while true
y = myfun(x,x1);
% so something with x,y
x = x+step;
end

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Matt Fig
Matt Fig 2011-4-20
Yet another approach:
x1 = .5; % x1 in image
amp = 3; % amplitude
n = 3; % number of periods.
x = [reshape(bsxfun(@plus,(0:n-1),[ 0 0 x1].'),1,3*n) n];
y = [0 repmat([amp 0 0],1,floor(length(x)/3))];
plot(x,y)
EDIT
In response to Paulo's comment, here is a version which preserves the exact point values at the integers and x1, yet supplies many more points.
x1 = .5; % x1 in image
amp = 3; % amplitude
n = 3; % number of periods.
r = 0.1; % max distance between points.
x = [0 (0:r:x1-eps) x1 x1+r-mod(x1,r):r:1-2*eps];
y = [repmat(max([0 -(amp/x1)*(x(2:end))+amp],0),1,n) 0];
x = [reshape(bsxfun(@plus,0:n-1,x.'),1,n*length(x)) n];
plot(x,y)
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Paulo Silva
Paulo Silva 2011-4-20
That does produce a nice graph but it's just a few points, not enough to be used for input on a function, for example the simulink models.
Matt Fig
Matt Fig 2011-4-20
Good point, Paulo!

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li
li 2011-4-21
o, thanks every one first. I think Paulo Silva's answer fit the most but there is another problem ha ha, I need the function to extend to infinity, which means it will repeat itself from -infinity to infinity, so how can I modify it? thanks
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li
li 2011-4-21
ok and how should I do with the minus part? thanks
li
li 2011-4-21
haha your modified answer is cool. It looks like you know what I am doing!

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li
li 2011-4-21
one last question, Paulo Silva how can I modify your code to get something like this?http://www.flickr.com/photos/58977081@N02/5639015965/ thanks
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li
li 2011-4-21
Or Matt can help to asnwer this too thanks
Paulo Silva
Paulo Silva 2011-4-21
amp=1; %amplitude of the wave
xp1=0.5; %this is your x1 value
r=0.1;n=4; %distance between points (r) and number of periods (n)
x0=0;y0=0;x1=xp1;y1=amp; %points to calculate m
m=(y1-y0)/(x1-x0); %declive of the wave
x=0:r:xp1; %x points of the wave where y value is not zero
xt=xp1:r:1; %x points of the wave where y value is zero
xx=[x xt]; %all the x points together
yy=[m*x 0*xt]; %the y value of the wave
yyy=repmat(yy,1,n); %n periods of the wave
xxx=[]; %make the time value for all the wave periods
for a=0:n-1 %the use of the for can be avoided somehow
xxx=[xxx a+xx]; %I used the for just to make it work for now
end
xxx=xxx-floor(n/2);
plot(xxx,yyy,'r') %plot it just to see if its working

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