[PPT].Lesson 06-z_Transform

[PPT].Lesson 06-z_Transform - Challenge 05 05 z-Transform...

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z-Transform Challenge 5 Lesson 6: z-Transform – What is a z-transform? – How does one compute a z-transform? – What good is the z-transform? – What are the strange things? Challenge 6 Monday: Exam #1 review. Lesson 06 hallenge 05 Challenge 05 Q: What is the approximate multiply- accumulate (MAC) unit’s statistical utput error variance in bits? output error variance in bits? iven: - 6 16 Given: 16<X i <16 -16<Y i <16 16 Lesson 06 hallenge 05 Challenge 05 o prevent register overflow for To prevent register overflow for |X i |<16, and |Y i |< 16, the input data formats are shown below. it 8-bits S I=4 F=3 Since X i and Y i are assumed to be ounded between 6, the bounded between 16, the quantization step-size can be expressed as  2 8   /2 8 = 2 -3 . t h ih tf t hL S B is the weight of the LSB. Can be calibrated in volts/bit once DC specified 16 Lesson 06 ADC specified. hallenge 05 Challenge 05 What about the 16-bit full precision multiplier’s format?. he orst case m ltiplier o tp t is The worst case multiplier output is bounded from above by 16*16=2 8 . 16-bits S I=8 F=7 16 Lesson 06 hallenge 05 Challenge 05 What then is the worst case accumulator outcome? mming 16 consec ti e orst case Summing 16 consecutive worst case products result in: 16 16 1 16 1 i i i i i i Y X Y X Z 2 1 2 16 2 i i i Y X max r 16 times larger than 16 or 16 times larger than the maximum product Lesson 06 hallenge 05 Challenge 05 Therefore the extended precision accumulator needs 4 additional bits of “headroom” to avoid worst case gister overflow during run me register overflow during run-time. The accumulator wordwidth must erefore be 4+16 its = 20 its The therefore be 4+16-bits = 20-bits. The accumulator’s output format is: 0 its 20-bits S I=8+4=12 F=7 16 Lesson 06 hallenge 05 Challenge 05 Retaining the 16-most significant bits results in a output data word, results in the 16-bit format shown below: 16-bits S I=12 F=3 where the value (weight) of the output’s LSB is: out =2 -3 16 Lesson 06 hallenge 05 Challenge 05 Statistically (assuming rounding) out =2 -3 ; E(e)=0; e 2 = out 2 /12 or log 2 ( out / 12) = = log 2 (2 -3 / 12) = = -3 – 1.79 ~ -5 r (statistically) the output or, (statistically) the output contains about 5 bits of fractional fractional precision. 16 Lesson 06 hallenge 05 Challenge 05 The MAC studied in Challenge 5 was protected from run-time 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 4-bits of precision to guard against an unexpected event. Is it worth taking this chance?
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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.

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[PPT].Lesson 06-z_Transform - Challenge 05 05 z-Transform...

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