filling in between two curves with different x values

Hello.

I have a plot

plot(x,y,'r--','LineWidth', 2);

that I want to display with boundaries such that:

plot(x+5,y,'k--','LineWidth', 0.5); 
plot(x-5,y,'k--','LineWidth', 0.5);

I want to shade the area in between and have tried:

patch([k-5 k+5], [y y], 'g')

but I can't get it to work.

Thanks for any help.

Answer 1

[Guillaume]

I would think that

patch([x-5, fliplr(x+5)], [y fliplr(y)], 'g')

should work.

[Jason]

Thanks, but I get:

Error using patch
Vectors must be the same length.

[Star Strider]

You didn’t tell us the dimensions of the vectors.

Try this:

patch([(x(:)'+5) fliplr(x(:)'-5)], [y(:)' fliplr(y(:)')], 'g', 'FaceAlpha',0.5)

It forces the vectors to be column vectors (using (:)), then transposes them to row vectors (note the transpose operator, (')), then plots them. Either plot the patch object before the others, or add 'FaceAlpha'. If you want to eliminate the patch object border lines, add 'EdgeColor','none'.

[Guillaume]

What is

size(x)
size(y)

It is very puzzling that you're getting an error on the patch call but not on the plot calls.

[Jason]

Here it is:

size k

ans =

            1.00        151.00

size y

ans =

            1.00        151.00

[Jason]

Hmm. I've just seen the first element in y is a NaN. Its appearing as I have:

y=abs(sin(X)./X);

as I don't have the sinc function. x=0, y=Nan.

It seems to work if I force:

y(:,1)=1

[Star Strider]

Not necessary to ‘force’ anything. You can imply L’Hospital’s Rule and avoid NaN values this way:

y = abs(sin(X+1E-8)./(X+1E-8));

[Jason]

Thankyou for reminding me about Mr Hosptial!

Jason

[Star Strider]

We’re all in this together!