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MSwann
Posts: **10**

I must be misunderstanding how to use the cordic divider. I can find now way to finagle it to divide a 32-bit number into another 32-bit number in the manner needed for my solution.

In my use case, both dividend and divisor are left justified so that bit 31 of both is always set to 1. What differs between them is expressed in the remainder of the bits after bit 31, and also in the two exponent registers that hold the scaling factor for each.

Lets take a simple example without reference to scaling:

A = 0x80000000

B = 0x80000000

X = A/B

The result I need for my solution is:

X=0x80000000

Remainder=0x00000000

What I get instead is:

X=0x00000001

Remainder=0x00000000

The code for this is somewhat like:

QDIV A,B

GETQX X

GETQY Remainder

I also tried QFRAC, with no luck.

My current bit-banging code works well but consumes a huge number of cycles. I will try to post that next, if I can figure out how to format it so it's not all squished up against the left column.

(Edited for corrections)

In my use case, both dividend and divisor are left justified so that bit 31 of both is always set to 1. What differs between them is expressed in the remainder of the bits after bit 31, and also in the two exponent registers that hold the scaling factor for each.

Lets take a simple example without reference to scaling:

A = 0x80000000

B = 0x80000000

X = A/B

The result I need for my solution is:

X=0x80000000

Remainder=0x00000000

What I get instead is:

X=0x00000001

Remainder=0x00000000

The code for this is somewhat like:

QDIV A,B

GETQX X

GETQY Remainder

I also tried QFRAC, with no luck.

My current bit-banging code works well but consumes a huge number of cycles. I will try to post that next, if I can figure out how to format it so it's not all squished up against the left column.

(Edited for corrections)

## Comments

1012,18415,681> I must be misunderstanding how to use the cordic divider. I can find now way to finagle it to divide a 32-bit number into another 32-bit number in the manner needed for my solution.

>

> In my use case, both dividend and divisor are left justified so that bit 31 of both is always set to 1. What differs between them is expressed in the remainder of the bits after bit 31, and also in the two exponent registers that hold the scaling factor for each.

>

> Lets take a simple example without reference to scaling:

>

> A = 0x80000000

> B = 0x80000000

>

> X = A/B

>

> The result I need for my solution is:

>

> X=0x80000000

> Remainder=0x00000000

>

> What I get instead is:

> X=0x00000001

> Remainder=0x00000000

>

> The code for this is somewhat like:

>

> QDIV A,B

> GETQX X

> GETQY Remainder

>

> I also tried QFRAC, with no luck.

>

> My current bit-banging code works well but consumes a huge number of cycles. I will try to post that next, if I can figure out how to format it so it's not all squished up against the left column.

>

> (Edited for corrections)

80000000 / 80000000 = 00000001 r=00000000

3,814A = 2^31 * a (where a is a real number 0 <= a < 2

B = 2^31 * b (where b is a real number 0 <= b < 2

and you want to calculate

X = 2^31 * (a/b)

The hardware will calculate A/B exactly, but

A/B = (2^31*a) / (2^31*b) = a/b

whereas you actually want the result represented as a 1.31 fixed point number, i.e. 2^31 * (a/b).

There are two ways to fix this:

(1) Scale the result after the division. This is very prone to error; a/b itself is probably just 0 or 1, so you'd have to figure out based on b and the remainder what the rest of the bits should be.

(2) Scale A before division; mathematically 2^31 * (a / b) = 2^31 * (A / = (2^31 * A) / B. In this case you'll have to represent 2^31*A as a 64 bit number, but that's OK, the QDIV instruction can do 64 bit by 32 bit division.

You can get fancy and check the remainder if you want to round X correctly.

If you happen to know that A/B is less than 1 (so A < B ) then you could simplify this quite a bit by just doing qfrac A,B and shifting the result right by 1, i.e. calculating 2^32*A / B and then dividing by 2 to get 2^31 * A / B. Or you could shift A right by 1 and calculate 2^31 * A/B == 2^32 * (A/2) / B. You'll lose a little bit of precision, but it's marginally faster. The big cost is the qfrac/qdiv anyway.

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