Ahmed Saleh - 3 months ago 21

C Question

I have found that piece of code on blackfin533, which has fract32 which is from -1, 1, its in the format 1.31.

I can't get why the pre-shifting is required for calculating the amplitude of a complex number (re, img). I know if you want to multiply 1.31 by 1.31 fractional format then you need to shift right 31 bits.

**GO_coil_D[0].re**, **and GO_coil_D[0].im** are two **fract32**.

I can't get what the following code is doing :

`norm[0] = norm_fr1x32(GO_coil_D[0].re);`

norm[1] = norm_fr1x32(GO_coil_D[0].im);

shift = (norm[0] < norm[1]) ? (norm[0] - 1) : (norm[1] - 1);

vectorFundamentalStored.im = shl_fr1x32(GO_coil_D[0].im,shift);

vectorFundamentalStored.re = shl_fr1x32(GO_coil_D[0].re,shift);

vectorFundamentalStored.im = mult_fr1x32x32(vectorFundamentalStored.im, vectorFundamentalStored.im);

vectorFundamentalStored.re = mult_fr1x32x32(vectorFundamentalStored.re, vectorFundamentalStored.re);

amplitudeFundamentalStored = sqrt_fr16(round_fr1x32(add_fr1x32(vectorFundamentalStored.re,vectorFundamentalStored.im))) << 16;

amplitudeFundamentalStored = shr_fr1x32(amplitudeFundamentalStored,shift);

`round_`

`norm_fr1x32`

Answer

1) If the most significant *n* bits of the fractional part are all '0' bits, and they are followed by a '1' bit, then *n* behaves like a floating point binary exponent of value *n*, and the remaining 31-*n* bits behave like the mantissa. Squaring the number doubles the number of leading '0' bits to 2**n* and reduces the size of the mantissa to 31-2**n* bits. This can lead to a loss of precision in the result of the squaring operation.

2) `round_fr1x32`

converts the 1.31 fraction to a 1.15 fraction, losing up to 16 more bits of precision.

Hopefully you can see that steps 1 and 2 can remove a lot of precision in the number. Pre-scaling the number reduces the number of leading '0' bits *n* as much as possible, resulting in less precision being lost at step 1. In fact, for one of the two numbers being squared and added, the number of leading '0' bits *n* will be zero, so squaring that number will still leave up to 31 bits of precision before it is added to the other number. (Step 2 will reduce that precision to 15 bits.)

Lastly, you are wrong about the result of multiplying two 1.31 fraction format numbers - the result needs to be shifted right by 31 bits, not 62 bits.