Understanding the CORDIC algorithm with Python

You're calculating y[i+1] from x[i+1], when it should be x[i].

Rewriting the above as:

(x,y) = (x - (d*(2.0**(-i))*y), (d*(2.0**(-i))*x) + y)

to update x and y concurrently gives the correct result.

y)

Thank you so much for the help (whoever you are), those are little subtle bugs that can drive you crazy. Now I know why our Prof. in Functional programming was making a big deal out of assignment statements! :)

As for the domain between -pi/2 and pi/2 do I just keep adding/ subtracting until I get it within that domain? Thanks.

Usually with embedded trig algorithms you just implement a single quadrant (0 to pi/2), but you can use a half-circle if you want. And yes, it's mostly a matter of adding/subtracting. But for some quadrants you might need to negate the angle, or the result - I'll let you figure out the details yourself.

This is the first time I've heard someone describe Python syntax as being "similar to C" - it's about as different as you could get while still remaining an imperative programming language. But a well-written Python program should be fairly easily understood by C programmers.

When posting sample code in a newsgroup (or other public forum), it's a good idea to cut out unnecessarily or old code - keep it to a minimum. Including the nested function K_vals(), for example, serves only to confuse.

For any trigonometric function f, start by using the 2*pi periodicity, f(x+2*n*pi) = f(x) to coerce the value into the range 0..2*pi, e.g. x = x % 2*pi or x = x - 2*pi * floor(x / 2*pi) (but check how % handles negative values).

Then, for values outside of the range 0..pi/2, use the identities:

sin(x) = sin(pi - x) for pi/2

Also known as the ASTC rule

All, Sin, Tan, Cos

For the quadrants from 0 (rad/degree) that results are positive for.

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es

cos, tan =3D -tan

n, cos =3D -cos

an =3D -tan

At school we learned that the trig functions that are positive in each quadrant are:

Quadrant +ve Mnemonic

0..pi/2 sin, cos, tan All pi/2..pi sin Stations pi..3*pi/2 tan To 3*pi/2..2*pi cos Crewe

Might mean something if you are English ;->

If you represent angle such that 0 to pi maps to 0 to 2^n then getting within the domain only requires a bit-mask (no iteration).

Bob

... snip ...

Much simpler, and not language dependent, is just think of the signs applied to the x and y axis when you draw a rt triangle in it. x is positive with displacement to right, y with displacement up. Now all you need to remember is what sin, cos, tan mean, i.e. the ratio of which sides.

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Your basic "unit circle" definitions. Sometimes, I wonder if it is even taught, anymore. Few younger folks in today's college calculus classes seem to even know about it -- and my latest experience with them was this last Winter term. Yet it's what I always go back to if I want to gain some insight on a new question (or one I have long since forgotten about.) You can't go wrong with that and Pythagorean. You also get all your half-angle and double-angle equivalences, etc. It's all there to be had with merely a gander.

Also, there are symmetries to exploit beyond the 180 or 90 degree ones that I've seen mentioned in this thread. At a minimum, octants (or 45 degrees) are your basic brain-dead demarcation for calculation. More is still possible. For those whose approximation exposure has been sadly limited only to the very easily understood Taylor's theorem, then if for no other reason than the fact that staying closer in range to where the approximation was taken improves retained precision per cpu cycle consumed.

Jon