value from graph

theta= 0:.1:360

theta=deg2rad(theta)

w=B*d.*cos(theta)

AF=1/N*(sin(N*(w./2))./sin(w./2))

plot((theta),AF)

here i need the value of theta at at a poin of AF

i wrote that x(y(0.7)) in command window but it gives

"Empty matrix: 1-by-0"

how can i get the value of theta at y=0.7

Answer 1

[Matt Fig]

I am surprised you didn't get an error, indexing into y like that. What is y anyway, you don't define it. All I see is theta, w, and AF. And what are B,d and N? Also, why no semi-colons?

.

.

EDIT

I don't know why you would do that when you can just do this:

x = theta;
y = AF;

but anyway...

[mn,I] = min(abs(.7-y));
x(I)

Note that you are using a cyclical function, so there might be more than one value of y 'equal' to .7. If you want to get them all, use:

tol = 1e-6;  % decide how close is close enough.
I = abs(.7-y)<tol
x(I)

[nitya ...]

h = plot((theta),AF)

x=(get(h,'Xdata'));

y=(get(h,'Ydata'));

i wrote like this in my code

[Sean de Wolski]

I was thinking y must be a function to not throw an error.

[nitya ...]

here 'theta' varies from 0 to 360 with an intervel of 1 degree

so 'AF' also having 361 values for the corresponding 'theta' values and it can plot on x-y axis

and i need theta value at AF=0.7

[Matt Fig]

@nitya

Well yes, everyone can see theta and AF defined. What we didn't see in your initial post are x, y, B, d, and N. Now we know what x and y are, but not the rest of the unknowns. Did you try the code?

[Matt Fig]

@ Sean de

I thought of that too, but guessed y must be a variable given the context. It is still a mystery how that line didn't error.