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Unformatted text preview: Lesson 06 zTransform Challenge 5 Lesson 6: zTransform – What is a ztransform? – How does one compute a ztransform? – What good is the ztransform? – What are the strange things? Challenge 6 Monday: Exam #1 review. Lesson 06 Challenge 05 Q: What is the approximate multiply accumulate (MAC) unit’s statistical output error variance in bits? Given: 1616<X i <1616<Y i <16 Lesson 06 Challenge 05 To prevent register overflow for X i  <16, and Y i < 16, the input data formats are shown below. 8bits Since X i and Y i are assumed to be bounded between ± 16, the quantization stepsize can be expressed as ∆= 2 (1 6 29/ 2 8 = 3 2 /2 8 = 23 . ∆ is the weight of the LSB. Can be calibrated in volts/bit once ADC specified. 16 S I=4 F=3 Lesson 06 Challenge 05 What about the 16bit full precision multiplier’s format?. The worst case multiplier output is bounded from above by 16*16=2 8 . 16bits 16 S I=8 F=7 Lesson 06 Challenge 05 What then is the worst case accumulator outcome? Summing 16 consecutive worst case products result in: 12 8 16 1 16 1 16 1 2 16 2 = = < ≤ = ∑ ∑ ∑ = = = i i i i i i i i i Y X Y X Y X Z max 16 or 16 times larger than the maximum product Lesson 06 Challenge 05 Therefore the extended precision accumulator needs 4 additional bits of “headroom” to avoid worst case register overflow during runtime. The accumulator wordwidth must therefore be 4+16bits = 20bits. The accumulator’s output format is: 20bits 16 S I=8+4=12 F=7 Lesson 06 Challenge 05 Retaining the 16most significant bits results in a output data word, results in the 16bit format shown below: 16bits where the value (weight) of the output’s LSB is: ∆ out =23 16 S I=12 F=3 Lesson 06 Challenge 05 Statistically (assuming rounding) ∆ out =23 ; E(e)=0; σ e 2 = ∆ out 2 /12 or log 2 ( ∆ out / √ 12) = = log 2 (23 / √ 12) = = 3 – 1.79 ~ 5 or, (statistically) the output contains about 5 bits of fractional fractional precision. 16 Lesson 06 Challenge 05 The MAC studied in Challenge 5 was protected from runtime overflow because the integer field was choosen to be sufficiently large to handle worst case outcomes. Suppose in investigating the proposition that X i <16, and Y i < 16 it was discovered that except on rare occasions, the input bounds are normally X i <1, and Y i < 1. We have essentially lost 4bits of precision to guard against an unexpected event. Is it worth taking this chance? If we decide to take the chance we normally would want to protect the system, somehow, from the ravages of overflow. But how?...
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This note was uploaded on 09/18/2011 for the course EEL 5718 taught by Professor Janisemcnair during the Fall '11 term at University of Florida.
 Fall '11
 janisemcnair
 Computer Networks

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