Lesson 06-z_Transform - z-Transform Challenge 5 Lesson 6...

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Lesson 06 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.
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Lesson 06 Challenge 05 Q: What is the approximate multiply- accumulate (MAC) unit’s statistical output error variance in bits? Given: 16 -16<X i <16 -16<Y i <16
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Lesson 06 Challenge 05 To prevent register overflow for |X i | <16, and |Y i |< 16, the input data formats are shown below. 8-bits Since X i and Y i are assumed to be bounded between ± 16, the quantization step-size can be expressed as ∆= 2(1629/ 2 8 = 32 /2 8 = 2 -3 . is the weight of the LSB. Can be calibrated in volts/bit once ADC specified. 16 S   I=4    F=3    
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Lesson 06 Challenge 05 What about the 16-bit full precision multiplier’s format?. The worst case multiplier output is bounded from above by 16*16=2 8 . 16-bits 16 S  I=8       F=7    
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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
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Lesson 06 Challenge 05 Therefore the extended precision accumulator needs 4 additional bits of “headroom” to avoid worst case register overflow during run-time. The accumulator wordwidth must therefore be 4+16-bits = 20-bits. The accumulator’s output format is: 20-bits 16 S  I=8+4=12    F=7   
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Lesson 06 Challenge 05 Retaining the 16-most significant bits results in a output data word, results in the 16-bit format shown below: 16-bits where the value (weight) of the output’s LSB is: out =2 -3 16 S        I=12      F=3
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Lesson 06 Challenge 05 Statistically (assuming rounding)  out =2 -3 ; E(e)=0;  σ e 2 = out 2 /12   or log 2 ( out / 12) =            = log 2 (2 -3  12) = = -3 – 1.79 ~ -5  or, (statistically) the output contains  about 5 bits of fractional  fractional   precision.  16
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Lesson 06 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? If we decide to take the chance we normally would want to protect the system, somehow, from the ravages of overflow. But how? Saturating arithmetic
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Lesson 06 Challenge 05 National Instruments What happens if you don’t use saturating arithmetic?.  What happens if you do use saturating arithmetic?  
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Lesson 06 What is a z-transform?
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