Fall 2006, Oct. 31, Nov. 2 ELEC 5270-001/6270-001 Lecture 10 1 ELEC 5270-001/6270-001(Fall 2006)...

Fall 2006, Oct. 31, Nov. 2Fall 2006, Oct. 31, Nov. 2 ELEC 5270-001/6270-001 Lecture 10ELEC 5270-001/6270-001 Lecture 10 11

ELEC 5270-001/6270-001(Fall ELEC 5270-001/6270-001(Fall 2006)2006)

Low-Power Design of Electronic CircuitsLow-Power Design of Electronic Circuits

Power Analysis: Probabilistic MethodsPower Analysis: Probabilistic Methods

Vishwani D. AgrawalVishwani D. AgrawalJames J. Danaher ProfessorJames J. Danaher Professor

Department of Electrical and Computer Department of Electrical and Computer EngineeringEngineering

Auburn University, Auburn, AL 36849Auburn University, Auburn, AL 36849http://www.eng.auburn.edu/~vagrawalhttp://www.eng.auburn.edu/~vagrawal

[email protected]@eng.auburn.edu

http://www.eng.auburn.edu/~vagrawal

mailto:[email protected]


Basic IdeaBasic Idea

View signals as a random processesView signals as a random processesProb{s(t) = 1} = p1 p0 = 1 – p1

C

0→1 transition probability = (1 – p1) p1

Power, P = (1 – p1) p1 CV2fck


Source of InaccuracySource of Inaccuracy

1/fck

p1 = 0.5 P = 0.5CV 2 fck

p1 = 0.5 P = 0.33CV 2 fck

p1 = 0.5 P = 0.167CV 2 fck

Observe that the formula, Power, P = (1 – p1) p1 CV2fck, is notCorrect.


Switching FrequencySwitching Frequency

Number of transitions per unit time:

N(t)T = ───

t

For a continuous signal:

N(t)T = lim ───

t→∞ t

T is defined as transition density.


Static Signal ProbabilitiesStatic Signal Probabilities

Observe signal for interval Observe signal for interval t t 0 + 0 + t t 11 Signal is 1 for duration Signal is 1 for duration t t 11 Signal is 0 for duration Signal is 0 for duration t t 00 Signal probabilities:Signal probabilities:

p p 1 = 1 = t t 1/(1/(t t 0 + 0 + t t 1)1) p p 0 = 0 = t t 0/(0/(t t 0 + 0 + t t 1) = 1 – 1) = 1 – p p 11


Static Transition Static Transition ProbabilitiesProbabilities

Transition probabilities:Transition probabilities: T T 01 = 01 = p p 0 Prob{signal is 1 | signal was 0} = 0 Prob{signal is 1 | signal was 0} = p p 0 0

pp11 T T 10 = 10 = p p 1 Prob{signal is 0 | signal was 1} = 1 Prob{signal is 0 | signal was 1} = p p 1 1

p p 00 TT = = T T 01 + 01 + T T 10 = 2 10 = 2 p p 0 0 p p 1 = 2 1 = 2 p p 1 (1 – 1 (1 – p p 1)1)

Transition density: Transition density: TT = 2 = 2 p p 1 (1 – 1 (1 – p p 1)1) Transition frequency: Transition frequency: f = T/ f = T/ 22 Power = Power = CV CV 22T/ T/ 2 (correct formula)2 (correct formula)


Static Transition FrequencyStatic Transition Frequency

0 0.25 0.5 0.75 1.0

0.25

0.2

0.1

0.0

p1

f =

p1

(1 –

p1

)


Inaccuracy in Transition Inaccuracy in Transition DensityDensity

1/fck

p1 = 0.5 T = 1.0

p1 = 0.5 T = 4/6

p1 = 0.5 T = 1/6

Observe that the formula, T = 2 p1 (1 – p1), is not correct.


Cause for Error and Cause for Error and CorrectionCorrection

Probability of transition is not independent Probability of transition is not independent of the present state of the signal.of the present state of the signal.

Consider probability Consider probability p p 01 of a 0→1 01 of a 0→1 transition,transition,

Then Then p p 01 ≠ 01 ≠ p p 0 × 0 × p p 11 We can write We can write p p 1 = (1 – 1 = (1 – p p 1)1)p p 01 + 01 + p p 1 1 p p 1111

p p 0101p p 1 = ─────────1 = ─────────

1 – 1 – p p 11 + 11 + p p 0101


Correction (Cont.)Correction (Cont.) Since Since p p 11 + 11 + p p 10 = 1, i.e., given that the 10 = 1, i.e., given that the

signal was previously 1, its present value signal was previously 1, its present value can be either 1 or 0.can be either 1 or 0.

Therefore,Therefore, p p 0101

p p 1 = ──────1 = ────── p p 10 + 10 + p p 0101

This uniquely gives signal probability as a This uniquely gives signal probability as a function of transition probabilities.function of transition probabilities.


Transition and Signal Transition and Signal ProbabilitiesProbabilities

1/fck

p01 = p10 = 0.5 p1 = 0.5

p01 = p10 = 1/3 p1 = 0.5

p1 = 0.5p01 = p10 = 1/6


Probabilities: p0, p1, p00, p01, p10, Probabilities: p0, p1, p00, p01, p10, p11p11

p p 01 + 01 + p p 00 =100 =1 p p 11 + 11 + p p 10 = 110 = 1 p p 0 = 1 – 0 = 1 – p p 11 p p 0101

p p 1 = ───────1 = ───────

p p 10 + 10 + p p 0101


Transition DensityTransition Density

TT = 2 = 2 p p 1 (1 – 1 (1 – p p 1) = 1) = p p 0 0 p p 01 + 01 + p p 1 1 p p 1010

= 2 = 2 p p 10 10 p p 01 / (01 / (p p 10 + 10 + p p 01)01)

= 2 = 2 p p 1 1 p p 10 = 2 10 = 2 p p 0 0 p p 0101


Power CalculationPower Calculation

Power can be estimated if transition Power can be estimated if transition density is known for all signals.density is known for all signals.

Calculation of transition density Calculation of transition density requiresrequires Signal probabilitiesSignal probabilities Transition densities for primary inputs; Transition densities for primary inputs;

computed from vector statisticscomputed from vector statistics


Signal ProbabilitiesSignal Probabilities x1

x2

x1 x2

x1

x2

x1 + x2 – x1x2

x1 1 - x1


Signal ProbabilitiesSignal Probabilities x1

x2 x3

x1 x2

y = 1 - (1 - x1x2) x3 = 1 - x3 + x1x2x3 = 0.625

X1 X2 X3 Y0 0 0 10 0 1 00 1 0 10 1 1 01 0 0 11 0 1 01 1 0 11 1 1 1

0.5

0.5

0.5

0.25 0.625

Ref: K. P. Parker and E. J. McCluskey,“Probabilistic Treatment of General Combinational Networks,” IEEE Trans. on Computers, vol. C-24, no. 6, pp. 668-670, June 1975.


Correlated Signal Correlated Signal ProbabilitiesProbabilities

x1

x2

x1 x2

y = 1 - (1 - x1x2) x2 = 1 – x2 + x1x2x2 = 1 – x2 + x1x2 = 0.75

X1 X2 Y0 0 10 1 01 0 11 1 1

0.5

0.5 0.25 0.625?



x1

x2

x1 + x2 – x1x2

y = (x1 + x2 – x1x2) x2 = x1x2 + x2x2 – x1x2x2 = x1x2 + x2 – x1x2 = x2 = 0.5

X1 X2 Y0 0 00 1 11 0 01 1 1

0.5

0.5 0.75 0.375?


ObservationObservation

Numerical computation of signal Numerical computation of signal probabilities is accurate for fanout-probabilities is accurate for fanout-free circuits.free circuits.


RemediesRemedies

Use Shannon’s expansion theorem to Use Shannon’s expansion theorem to compute signal probabilities.compute signal probabilities.

Use Boolean difference formula to Use Boolean difference formula to compute transition densities.compute transition densities.


Shannon’s Expansion Shannon’s Expansion TheoremTheorem

C. E. Shannon, “A Symbolic Analysis of Relay C. E. Shannon, “A Symbolic Analysis of Relay and Switching Circuits,” and Switching Circuits,” Trans. AIEETrans. AIEE, vol. 57, , vol. 57, pp. 713-723, 1938.pp. 713-723, 1938.

Consider:Consider: Boolean variables, X1, X2, . . . , XnBoolean variables, X1, X2, . . . , Xn Boolean function, F(X1, X2, . . . , Xn)Boolean function, F(X1, X2, . . . , Xn)

Then F = Xi F(Xi=1) + Xi’ F(Xi=0)Then F = Xi F(Xi=1) + Xi’ F(Xi=0) WhereWhere

Xi’ is complement of X1Xi’ is complement of X1 Cofactors, F(Xi=j) = F(X1, X2, . . , Xi=j, . . , Xn), j = 0 or 1Cofactors, F(Xi=j) = F(X1, X2, . . , Xi=j, . . , Xn), j = 0 or 1


Expansion About Two InputsExpansion About Two Inputs F = XiXj F(Xi=1, Xj=1) + XiXj’ F(Xi=1, F = XiXj F(Xi=1, Xj=1) + XiXj’ F(Xi=1,

Xj=0)Xj=0)

+ Xi’Xj F(Xi=0, Xj=1)+ Xi’Xj F(Xi=0, Xj=1)

+ Xi’Xj’ F(Xi=0, Xj=0)+ Xi’Xj’ F(Xi=0, Xj=0) In general, a Boolean function can be In general, a Boolean function can be

expanded about any number of input expanded about any number of input variables.variables.

Expansion about k variables will have 2Expansion about k variables will have 2kk terms.terms.



X1

X2

X1 X2

X1 X2 Y0 0 10 1 01 0 11 1 1

Y = X1 X2 + X2’

Shannon expansion about the reconverging input:

Y = X2 Y(X2=1) + X2’ Y(X2=0) = X2 (X1) + X2’ (1)


Correlated SignalsCorrelated Signals When the output function is expanded When the output function is expanded

about all reconverging input variables,about all reconverging input variables, All cofactors correspond to fanout-free circuits.All cofactors correspond to fanout-free circuits. Signal probabilities for cofactor outputs can be Signal probabilities for cofactor outputs can be

calculated without error.calculated without error. A weighted sum of cofactor probabilities gives the A weighted sum of cofactor probabilities gives the

correct probability of the output.correct probability of the output.

For two reconverging inputs:For two reconverging inputs:f = xixj f(Xi=1, Xj=1) + xi(1-xj) f(Xi=1, Xj=0)f = xixj f(Xi=1, Xj=1) + xi(1-xj) f(Xi=1, Xj=0)

+ (1-xi)xj f(Xi=0, Xj=1) + (1-xi)(1-xj) f(Xi=0, + (1-xi)xj f(Xi=0, Xj=1) + (1-xi)(1-xj) f(Xi=0, Xj=0)Xj=0)


Correlated Signal Correlated Signal ProbabilitiesProbabilities X1

X2

X1 X2

X1 X2 Y0 0 10 1 01 0 11 1 1

Y = X1 X2 + X2’

Shannon expansion about the reconverging input:

Y = X2 Y(X2=1) + X2’ Y(X2=0) = X2 (X1) + X2’ (1)

y = x2 (0.5) + (1-x2) (1) = 0.5 (0.5) + (1-0.5) (1) = 0.75


ExampleExample

Point of reconv.

Supergate0.5

0.5

0.5

0.5

0.25

10

0.50.0

0.01.0

0.51.0

Signal probability for supergate output = 0.5 Prob{rec. signal = 1} + 1.0 Prob{rec. signal = 0} = 0.5 × 0.5 + 1.0 × 0.5 = 0.75

0.375

Reconv. signal

S. C. Seth and V. D. Agrawal, “A New Model for Computation ofProbabilistic Testability in Combinational Circuits,” Integration, the VLSI Journal, vol. 7, no. 1, pp. 49-75, April 1989.


Probability Calculation Probability Calculation AlgorithmAlgorithm

Partition circuit into supergates.Partition circuit into supergates. Definition: A supergate is a circuit partition with a single Definition: A supergate is a circuit partition with a single

output such that all fanouts that reconverge at the output output such that all fanouts that reconverge at the output are contained within the supergate. are contained within the supergate.

Identify reconverging and non-reconverging Identify reconverging and non-reconverging inputs of each supergate.inputs of each supergate.

Compute signal probabilities from PI to PO:Compute signal probabilities from PI to PO: For a supergate whose input probabilities are For a supergate whose input probabilities are

knownknown Enumerate reconverging input statesEnumerate reconverging input states For each input state do gate by gate probability computationFor each input state do gate by gate probability computation Sum up corresponding signal probabilities, weighted by state Sum up corresponding signal probabilities, weighted by state

probabilitiesprobabilities


Calculating Transition Calculating Transition DensityDensity

Boolean function

1

n

x1, T1..... xn, Tn

y, T(Y) = ?


Boolean DifferenceBoolean Difference

Boolean diff(Y, Xi) = 1 means that a path is sensitized Boolean diff(Y, Xi) = 1 means that a path is sensitized from input Xi to output Y.from input Xi to output Y.

Prob(Boolean diff(Y, Xi) = 1) is the probability of Prob(Boolean diff(Y, Xi) = 1) is the probability of transmitting a toggle from Xi to Y.transmitting a toggle from Xi to Y.

Probability of Boolean difference is determined from the Probability of Boolean difference is determined from the probabilities of cofactors of Y with respect to Xi. probabilities of cofactors of Y with respect to Xi.

∂YBoolean diff(Y, Xi) = ── = Y(Xi=1) ⊕ Y(Xi=0)

∂Xi

F. F. Sellers, M. Y. Hsiao and L. W. Bearnson, “Analyzing Errors with the Boolean Difference,” IEEE Trans. on Computers, vol. C-17, no. 7, pp. 676-683, July 1968.


Transition DensityTransition Density

nT(y) = Σ T(Xi) Prob(Boolean diff(Y, Xi) = 1)

i=1

F. Najm, “Transition Density: A New Measure of Activity in DigitalCircuits,” IEEE Trans. CAD, vol. 12, pp. 310-323, Feb. 1993.


Power ComputationPower Computation For each primary input, determine signal For each primary input, determine signal

probability and transition density for given vectors.probability and transition density for given vectors. For each internal node and primary output Y, find For each internal node and primary output Y, find

the transition density T(Y), using supergate the transition density T(Y), using supergate partitioning and the Boolean difference formula.partitioning and the Boolean difference formula.

Compute power,Compute power,

P =P = ΣΣ 0.5C0.5CYY V V22 T(Y) T(Y)

all Yall Y

where Cwhere CYY is the capacitance of node Y and V is is the capacitance of node Y and V is supply voltage.supply voltage.


Transition Density and Transition Density and PowerPower

X1

X2 X3

0.2, 1

0.3, 2

0.4, 3

0.06, 0.7

0.436, 3.24

Transition densitySignal probability

YCi

CY

Power = 0.5 V2 (0.7Ci + 3.24CY)


Prob. Method vs. Logic Sim. Prob. Method vs. Logic Sim.

CircuitCircuit No. of No. of gatesgates

Probability Probability methodmethod Logic SimulationLogic Simulation

ErrorError

%%Av. Av. densitydensity CPU s*CPU s* Av. Av.

densitydensity CPU s*CPU s*

C432C432 160160 3.463.46 0.520.52 3.393.39 6363 +2.1+2.1

C499C499 202202 11.3611.36 0.580.58 8.578.57 241241 +29.8+29.8

C880C880 383383 2.782.78 1.061.06 3.253.25 132132 -14.5-14.5

C1355C1355 346346 4.194.19 1.391.39 6.186.18 408408 -32.2-32.2

C1908C1908 880880 2.972.97 2.002.00 5.015.01 464464 -40.7-40.7

C2670C2670 11931193 3.503.50 3.453.45 4.004.00 619619 -12.5-12.5

C3540C3540 16691669 4.474.47 3.773.77 4.494.49 10821082 -0.4-0.4

C5315C5315 23072307 3.523.52 6.416.41 4.794.79 16161616 -26.5-26.5

C6288C6288 24062406 25.1025.10 5.675.67 34.1734.17 3105731057 -26.5-26.5

C7552C7552 35123512 3.833.83 9.859.85 5.085.08 27132713 -24.2-24.2* CONVEX c240


Probability Waveform Probability Waveform MethodsMethods

F. Najm, R. Burch, P. Yang and I. Hajj, “CREST – A F. Najm, R. Burch, P. Yang and I. Hajj, “CREST – A Current Estimator for CMOS Circuits,” Current Estimator for CMOS Circuits,” Proc. IEEE Int. Proc. IEEE Int. Conf. on CADConf. on CAD, Nov. 1988, pp. 204-207., Nov. 1988, pp. 204-207.

C.-S. Ding, C.-S. Ding, et alet al., “Gate-Level Power Estimation using ., “Gate-Level Power Estimation using Tagged Probabilistic Simulation,” Tagged Probabilistic Simulation,” IEEE Trans. on CADIEEE Trans. on CAD, , vol. 17, no. 11, pp. 1099-1107, Nov. 1998.vol. 17, no. 11, pp. 1099-1107, Nov. 1998.

F. Hu and V. D. Agrawal, “Dual-Transition Glitch F. Hu and V. D. Agrawal, “Dual-Transition Glitch Filtering in Probabilistic Waveform Power Estimation,” Filtering in Probabilistic Waveform Power Estimation,” Proc. IEEE Great Lakes Symp. VLSIProc. IEEE Great Lakes Symp. VLSI, Apr. 2005, pp. , Apr. 2005, pp. 357-360.357-360.

F. Hu and V. D. Agrawal,F. Hu and V. D. Agrawal, “ “Enhanced Dual-Transition Enhanced Dual-Transition Probabilistic Power Estimation with Selective Probabilistic Power Estimation with Selective Supergate Analysis,” Supergate Analysis,” Proc. IEEE Int. Conf. Computer Proc. IEEE Int. Conf. Computer DesignDesign, Oct. 2005. pp. 366-369., Oct. 2005. pp. 366-369.


Power Estimation by Prob. Power Estimation by Prob. WaveformWaveform

CircuitTPS DualTrans Supergate method

Eavg % σ Etot % Eavg % σ Etot % Eavg % σ Etot %

c17 2.3 2.6 0.1 2.3 2.6 0.1 2.3 2.6 0.1

c432 29.9 38.8 35.8 9.5 11.8 6.5 11.5 16.6 11.5

c499 6.8 14.0 7.0 3.6 8.2 0.6 2.3 3.0 3.0

c880 8.3 15.3 1.6 8.0 15.7 5.2 4.8 9.0 0.0

c1355 24.2 31.6 32.9 5.8 11.2 5.4 5.0 9.5 0.5

c1908 15.0 23.1 4.1 17.7 27.9 11.2 7.0 16.3 2.0

c2670 16.6 29.8 7.2 16.7 28.3 9.9 13.2 23.6 6.2

c3540 13.8 26.3 9.8 10.3 25.6 2.4 10.5 26.4 3.7

c5315 11.8 24.4 2.3 13.4 31.5 10.1 11.3 27.0 3.4

c6288 27.4 27.5 32.1 15.7 18.8 4.1 12.7 15.4 0.2

c7552 14.5 27.5 3.2 14.8 31.4 7.8 14.1 27.6 1.3

Avg. 15.5 23.7 12.4 10.7 19.4 5.7 8.6 16.1 2.9

Eavg = average node error, σ = av. node standard deviation, Etot = total error

Fall 2006, Oct. 31, Nov. 2 ELEC 5270-001/6270-001 Lecture 10 1 ELEC 5270-001/6270-001(Fall 2006)...

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Transcript of Fall 2006, Oct. 31, Nov. 2 ELEC 5270-001/6270-001 Lecture 10 1 ELEC 5270-001/6270-001(Fall 2006)...