How to generate a square wave with integer values and fixed timespan?

I want to generate a square wave with integer values between 0 and 5 in a timespan of 30ms.

An example:

[1 1 1 1 2 2 2 4 4 4 4 4 5 5 5 5 5 5 5 5 5 0 0 0 0 0 0 0 2 2 2 2 3 3 3 ]

I've tried different things but I can't seem to find an effective way.

[David Goodmanson]

Hi Samuele, see the 'square' function.

[Adam Danz]

If those are the y-values, what are the corresponding x values?

Answer 1

[DGM]

Something like this. I'm assuming you want your timestep to be a uniform 30ms.

clf
% build the coarse signal
dt=0.03;
y=[1 1 1 1 2 2 2 4 4 4 4 4 5 5 5 5 5 5 5 5 5 0 0 0 0 0 0 0 2 2 2 2 3 3 3 ];
t=linspace(0,dt*numel(y),numel(y));
plot(t,y,'--b'); hold on

% but if you need better transition times, increase the resolution
tfine=linspace(0,dt*numel(y),numel(y)*100);
yfine=interp1(t,y,tfine,'nearest');
plot(tfine,yfine,'k')

While that's simple to get better transitions by interpolation, most of the signal is constant-valued. It's kind of a waste of space to have all those intermediate samples when all you really need are the points at the transitions. If you're starting from scratch, it'd be easy enough to make the vectors as needed, but let's say you're trying to work with an existing low-resolution step signal and you want to improve it without interpolating:

clf; clc
% build the coarse signal
dt=0.03;
y=[1 1 1 1 2 2 2 4 4 4 4 4 5 5 5 5 5 5 5 5 5 0 0 0 0 0 0 0 2 2 2 2 3 3 3 ];
t=linspace(0,dt*numel(y),numel(y));
plot(t,y,'--b'); hold on

% or you could reduce it to a simple series of points
% this allows a much higher effective resolution with a minimal number of samples
tt=0.0000001; % transition time
b=[t(find(diff(y)~=0)); t(find(diff(y)~=0)+1)]; % find times at steps
b=bsxfun(@plus,mean(b,1),[-tt/2;tt/2]); % reduce rise/fall time
tp=[t(1) b(:)' t(end)];
yp=interp1(t,y,tp,'nearest');
plot(tp,yp,'k')

• Original: 35 samples, 30ms transition time
• Full interpolation: 3500 samples, 300us transition time
• Edges only: 14 samples, 100ns transition time

The big point is that the number of required samples in the last case is independent of transition time.

[Adam Danz]

To get a true step with identical x values at each step,

y=[1 1 1 1 2 2 2 4 4 4 4 4 5 5 5 5 5 5 5 5 5 0 0 0 0 0 0 0 2 2 2 2 3 3 3 ];
dy = cumsum([false, diff(y(:).')==0]);
dt=0.03;
x = dy * dt;
plot(x,y)

[DGM]

I didn't even stop to think that you could do that. I mean, it makes sense in retrospect. I'm just way too used to trying to build step signals in programs where you can't actually have vertical lines.