periodic functions

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

Answer 1

[Paulo Silva]

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

[Matt Fig]

This introduces some strange artifacts into the function. Try, for example,

x1 = .75;

n = 4;

amp = 6;

[Paulo Silva]

Inded it does but that can be interpreted as Slew Rate

[Matt Fig]

Okie-dokie! I had to look that one up, it has been so long...

[Paulo Silva]

We still need to know if li wants a nice graph or values for something else, maybe li wants a signal generator

Answer 2

[Clemens]

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.

[Matt Fig]

This does not produce the graph. You are using an IF statement to pick out elements of an array, which it does not do... Logical indexing could be made to work.

[Clemens]

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]

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

Answer 3

[Matt Fig]

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)

[Paulo Silva]

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]

Good point, Paulo!

Answer 4

[li]

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

[Matt Fig]

What do you mean? Are you saying you want an infinite number of function values stored on your computer? That is going to be tough to do! If you mean that you need to be able to specify an arbitrary ending point, look at my edited answer. You can plug in for n the ending point.

[li]

http://www.flickr.com/photos/58977081@N02/5639550024/

here it is thanks

[Paulo Silva]

Infinity is too much but in my code you can choose a bigger n and shift xxx values (xxx-floor(n/2))

[li]

ok and how should I do with the minus part? thanks

[li]

haha your modified answer is cool. It looks like you know what I am doing!

Answer 5

[li]

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

[li]

Or Matt can help to asnwer this too thanks

[Paulo Silva]

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