Challenge05_Quant_Arith0

Challenge05_Quant_Arith0 - The worst case accumulation is:...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Lesson Title: Quantization Case Study Lesson Number: 05 Challenge: A dedicated multiply-accumulate unit is to be used to perform the 16-sample DSP SAXPY (S=AX+Y) operation: = = 16 1 i i i Y X Z The input data sets X i and Y i are 8-bit signed numbers with magnitudes bounded from above by 4. The implementation architecture is shown below for a full-precision multiplier, an extended precision accumulator (20-btis), and a 16-bit rounder. Q: What is the approximate output error variance, in bits, due to the multiply-accumulate unit? Rround your answer to the nearest bit value. Consider what would happen if the accumulator wordwidth were set to 16-bits. Consider what would happen if the multiplier only outputted the 8 MSBs of the full precision product? 8 8 X i Y i 16 Z i 20 Round 16 MAC Z
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
The data format for the words X i and Y i is shown below. Since X i and Y i are assumed to be bounded between ± 4, the quantization step-size can be expressed as =2 -5 . This is also the weight of the LSB. The data format for the full precision product is shown below.
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The worst case accumulation is: 8 16 1 16 1 16 2 256 16 * 16 4 * 4 = = = = = = i i i i Y X Z The extended precision accumulator wordwidth has 4-additional bits of precision (headroom) to permit 16 full precision products to the accumulated without overflowThe data format for the accumulator output word is shown below. Retaining the 16-most significant bits results in a output data word having the format shown below. The weight of the LSB of the SAXPY result is now =2-7 . The error statistics, in bits, is therefore (assuming rounding) E ( e )=0; 2 = /12 log 2 ( = /sqrt(12)) = log 2 (2-7 /sqrt(12)) = -7 1.79 ~ -9 or, approximately 9 bits of fractional precision is left in the accumulators output. Why did Motorola call its 24-DSP fixed-point product the DSP56xxx? S F=8-3=5 I=2 S F=16-5=11 I=2+2=4 S I=4+4=8 F=20-9=11 S I=8 F=16-9=7...
View Full Document

This note was uploaded on 08/21/2010 for the course EEL 5525 taught by Professor Yang during the Summer '09 term at University of Florida.

Page1 / 2

Challenge05_Quant_Arith0 - The worst case accumulation is:...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online