Digital Signal on plot
Dave Smith
dpb
just make a data stream that matches the desired waveform...
x=[0:10];
y=rand(size(x));
y(y>.6)=1; y(y<.3)=-1,y(y>.3 && y<.6)=0;
plot(x,y)
--
Dave Smith
Thank you. This is very close to what i wanted but can you explain what is rand(size(x)) do? I ran this on matlab and I'm getting a diagonal line going from says 1-2. but I only want horizontal line so like 0-1 is a horizontal line and a vertical line going down from 2 and then another horizontal line from 2-3. I hope you understand what I'm trying to said.
dpb
> dpb <no...@non.net> wrote in message
> <hn0k4h$9c1$1...@news.eternal-september.org>...
> Thank you. This is very close to what i wanted but can you explain what
> is rand(size(x)) do?
Just generated a set of random numbers to start w/ of the same size as x...
...
> a horizontal line and a vertical line going down from 2 and then another
> horizontal line from 2-3. I hope you understand what I'm trying to said.
Yeah, you'll have to duplicate the x-coordinates at the break points
which the above doesn't do (didn't think thru the issue thoroughly enuf
before, sorry).
The fundamental idea is the same, just that you'll have to have the x
and y vectors contain the entries for both the (replicating) abcissa and
ordinates to draw a true square wave. What I showed would be a
triangle, indeed.
Of course, a true square wave is an idealization anyway, no matter how
good the rise/fall time of a wave generator... :)
If you perchance have the Signal Processing toolbox, it has a square()
function but it is "realizable" in the above sense as it doesn't
generate values at duplicated time points, either.
--
Dave Smith
duplicate x-coordinates? can you help me out, I been struggling with this all day, so frustrating
ImageAnalyst
This is very easy. Just use the line command. You give it the x1,y1
and the x2,y2 (end points of the line) and it does it. You just need
to figure out what they are (and it sounds like you already know that,
you just need to know how to plot it.)
Like this
X=rand(1,20);
plot(X);
set(gcf, 'Position', get(0,'Screensize')); % Maximize figure.
hold on;
line([2 7], [0.5 0.5], 'LineWidth', 3, 'Color', [0 .4 0]);
line([7 7], [0.5 0.3], 'LineWidth', 3, 'Color', [.8 0 0.1]);
line([7 17], [0.3 0.3], 'LineWidth', 3, 'Color', [0 .4 0]);
Dave Smith
I don't really understand this. Look like you're just drawing line on top on the plot where as I'm trying to draw digital signal with 0 and 1 from any binary input. Could it be I don't know how to apply this to my problem.
ImageAnalyst
the plot graph draw a line..." and then you gave a series of line
segments. If you want to do it all in one shot with the plot command,
then you're going to have to double up on some x values - insert an
coordinate at the same x value but a different y value. Plot can plot
a polyline but you have to give it all the coordinates (vertices) of
the polyline you want to plot.
dpb
Well, you have to decide the on/off positions but you need something like
x = [ 0 1 2 2 3 4 5 5 6 7 8 8 9 10 ];
y = [ -1 -1 -1 1 1 1 1 -1 -1 -1 -1 1 1 1 ];
plot(x,y)
axis([0 10 -2 2]) % fix y-axis so can see clearly now...
The trick is to determine programmatically what are the breakpoints
where you need the duplicated x values to create the vertical
transition. (Assuming that is what you really want irrespective of the
realizability issues raised earlier)
--
Walter Roberson
Presuming that you want the signal to start at 0 and to return to 0, then:
newx = reshape(repmat(1:length(y)+1,2,1),1,[]);
newy = [0 reshape(repmat(y(:),1,2).',1,[]) 0];
Dave Smith
Ok, I have test the duplicating x values and yes this is what I wanted. But now the trick like you said is to determine the breakpoints. Here is what I have so far:
XMIN=1;
XMAX=12;
YMIN=-2;
YMAX=2;
B=[0 1 0 0 1 1 0 0 0 1 1];
[row,column]=size(B);
for c=1:column;
if B(c)==1;
B(c)=-1;
end
if B(c)==0;
B(c)=1;
end
end
B(12)=B(11);
X=(1:11);
subplot(3,1,1);
plot(X,B);
axis([XMIN XMAX YMIN YMAX]);
I wanted the program to read the B matrix and if it see a 1, it will goes -1 and if it see a 0, it will goes 1. I had determine that everytime it switches between 0 and 1, I need to deplicate that x values and add that many to the matrix. For example, right now it's 12 and it changes from 1 to 0 or 0 to 1 5 times so I need to add 5 + 12 = 17 so in the end. it's a 17 columns matrix. I thought about it but I need some suggestion on what to do.
dpb
...
> Ok, I have test the duplicating x values and yes this is what I wanted.
> But now the trick like you said is to determine the breakpoints. Here is
> what I have so far:
>
...
> B=[0 1 0 0 1 1 0 0 0 1 1];
...
> I wanted the program to read the B matrix and if it see a 1, it will
> goes -1 and if it see a 0, it will goes 1.
Well, that part's simple enough...
B(B==0)=-1;
B(B==1)=1;
...
> everytime it switches between 0 and 1, I need to deplicate that x values
> and add that many to the matrix. For example, right now it's 12 and it
> changes from 1 to 0 or 0 to 1 5 times so I need to add 5 + 12 = 17 so in
> the end. it's a 17 columns matrix. I thought about it but I need some
> suggestion on what to do.
look at
doc diff
and see if that doesn't give you any ideas on how to determine where to
insert the added points...
--
Walter Roberson
> Ok, I have test the duplicating x values and yes this is what I wanted.
> But now the trick like you said is to determine the breakpoints. Here is
> what I have so far:
Was there a problem with the two-line solution I suggested?
Dave Smith
No, there were no problem actually. I just never got a chance to try your suggestion as I was busy spending hours on trying to duplicated the x-axis going by what dpb suggested. But your suggestion solved the problem the moment I tried it so thank you very much. Can you explain to me in as much details as possible on how those two lines worked? I looked up reshape and repmat in help file but I still don't fully understand it. This is just one type of scheme graphs I have to do, I still got 6 other more to do and I don't know if these two lines will work for them all so that is why I would like to understand the code.
Walter Roberson
> Walter Roberson <robe...@hushmail.com> wrote in message
> <hn3549$pig$1...@canopus.cc.umanitoba.ca>...
>> Was there a problem with the two-line solution I suggested?
> No, there were no problem actually. I just never got a chance to try
> your suggestion as I was busy spending hours on trying to duplicated the
> x-axis going by what dpb suggested. But your suggestion solved the
> problem the moment I tried it so thank you very much. Can you explain to
> me in as much details as possible on how those two lines worked? I
> looked up reshape and repmat in help file but I still don't fully
> understand it. This is just one type of scheme graphs I have to do, I
> still got 6 other more to do and I don't know if these two lines will
> work for them all so that is why I would like to understand the code.
Consider the x coordinates. If you start from a baseline (0 in the case
of the code) then you need the initial x value (1) for that, and then
you need the same x value to draw the up (or down) stroke to reach the
first y value. Then you will need the 2nd x value to draw the line
across to the second value, and the 2nd x again to draw the up or down
stroke to the second y value. Keep going in this manner, with 2 copies
of each x coordinate. When you reach the last y value, you have an
implicit draw across _past_ that y value to where the next y value would
start, and you have an implicit return to baseline, so you will need two
copies of the x coordinate which is length(y)+1 . Thus, what we do is
construct (1:length(y)+1), then we use repmat(,2,1) two take an
identical copy of that row. Matlab stores values down columns, so _in
memory_ the adjacent values at this point will be two copies of each of
the x coordinates. We could use a temporary variable to hold that and
then use (:) indexing to "unravel" that down the columns to x1 x1 x2 x2
etc., or we can do like I did and call reshape explicitly to do the
unravelling. reshape(,1,[]) means to take the data that is in memory and
change its dimensions to 1 by "however much you need", so after the
reshape, we wil be left with x1 x1 x2 x2 .... xn xn xn+1 xn+1 .
Now consider the y coordinates. Because we are starting from the
baseline, the adjusted y coordinates will need to start with a 0;
likewise because we return to the baseline at the end, they will need to
end with 0. That accounts for the [0 <something> 0] portion of the code.
Now, we need to draw the upstroke or downstroke to reach the first y
value, so the first y value comes next. We need to hold that y value
over to the next x value, so a duplicate of the first y value is needed.
After that, we need to draw up or down to the second y value, and then
need to hold that value to the next x coordinate, and so on. Thus we are
once more in the situation of needing to duplicate a vector y1 y2 y3 to
become y1 y1 y2 y2 y3 and so on, which uses the same duplication code as
we used for the x. But in the x case, we know that 1:length(y)+1 will
generate a row vector, but for the purposes of this two-liner, we cannot
assume that y is already a row vector. To force y to be a row vector, we
use y(:) which forces y to be a _column_ vector; once we have the column
vector, we repmat(,1,2) it to make a duplicate column, and then we use
.' to transpose it into two identical row vectors; the reshape(,1,[])
then converts it into a single long row vector. Then the [0 ... 0]
wrapper for the returns to baseline.
The length of the newy vector thus becomes 1 + 2*length(y) + 1, which is
2*length(y) + 2, which is 2*(length(y)+1) which is the same as for the
newx vector, so we have all the coordinates matched up.
This isn't the _optimal_ drawing for the line, but that probably doesn't
matter. If it does matter, then because the structure of it is so clear,
it should be fairly straight-forward to filter out redundant points; I
would suggest first removing the redundant up/down strokes (which will
show up as identical pairs of x and y coordinates adjacent to each
other), and then to remove the redundant x coordinates (which would show
up as (x1 y1) (x2 y1) (x3 y1), which is compressible to (x1 y1) (x3 y1))
Dave Smith
Thank you. I'm slowly understanding it the more I read it.
Kyung Ho Park
%Generate random binary data
N = 100; %number of binary values
a = 2*(randi(2,N,1)-1)-1; %polar sequence of -1/+1
%Plot preparation
c = (a*[1 1]).'; %repeat each data bit
%Construct time vector
t = zeros(2,N);
epsilon = 0.0001; %2*epsilon = "duration" of transition
t(1,:) = (1:N)+epsilon;
t(2,:) = t(1,:)-(2*epsilon-1);
%plot
plot(t(:), c(:), '-');
ylim([-2 2]);
Hope this helps.
Park Kyung Ho
Alan B
I'm not sure why this thread was revived, but since it was... None of this stuff is necessary at all. Use stairs().
"Stairstep plots are useful for drawing time history plots of zero-order-hold digital sampled-data systems."