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Fall 2006, Oct. 31, Nov. 2 ELEC 5270-001/6270-001 Lecture 10 1 ELEC 5270-001/6270-001(Fall 2006) Low-Power Design of Electronic Circuits Power Analysis: Probabilistic Methods Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering Auburn University, Auburn, AL 36849 http://www.eng.auburn.edu/~vagrawal vagrawal@eng.auburn.edu
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 102 Basic Idea View signals as a random processes View signals as a random processes Prob{s(t) = 1} = p1 p0 = 1 – p1 C 0→1 transition probability = (1 – p1) p1 Power, P = (1 – p1) p1 CV 2 f ck
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 103 Source of Inaccuracy 1/f ck p1 = 0.5 P = 0.5CV 2 f ck p1 = 0.5 P = 0.33CV 2 f ck p1 = 0.5 P = 0.167CV 2 f ck Observe that the formula, Power, P = (1 – p1) p1 CV 2 f ck, is not Correct.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 104 Switching 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.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 105 Static Signal Probabilities Observe signal for interval t 0 + t 1 Observe signal for interval t 0 + t 1 Signal is 1 for duration t 1 Signal is 1 for duration t 1 Signal is 0 for duration t 0 Signal is 0 for duration t 0 Signal probabilities: Signal probabilities: p 1 = t 1/(t 0 + t 1) p 1 = t 1/(t 0 + t 1) p 0 = t 0/(t 0 + t 1) = 1 – p 1 p 0 = t 0/(t 0 + t 1) = 1 – p 1
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 106 Static Transition Probabilities Transition probabilities: Transition probabilities: T 01 = p 0 Prob{signal is 1 | signal was 0} = p 0 p1 T 01 = p 0 Prob{signal is 1 | signal was 0} = p 0 p1 T 10 = p 1 Prob{signal is 0 | signal was 1} = p 1 p 0 T 10 = p 1 Prob{signal is 0 | signal was 1} = p 1 p 0 T = T 01 + T 10 = 2 p 0 p 1 = 2 p 1 (1 – p 1) T = T 01 + T 10 = 2 p 0 p 1 = 2 p 1 (1 – p 1) Transition density: T = 2 p 1 (1 – p 1) Transition density: T = 2 p 1 (1 – p 1) Transition frequency: f = T/ 2 Transition frequency: f = T/ 2 Power = CV 2 T/ 2 (correct formula) Power = CV 2 T/ 2 (correct formula)
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 107 Static Transition Frequency 00.250.50.75 1.0 0.25 0.2 0.1 0.0 p1p1 f = p1(1 – p1)
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 108 Inaccuracy in Transition Density 1/f ck 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.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 109 Cause for Error and Correction Probability of transition is not independent of the present state of the signal. Probability of transition is not independent of the present state of the signal. Consider probability p 01 of a 0 → 1 transition, Consider probability p 01 of a 0 → 1 transition, Then p 01 ≠ p 0 × p 1 Then p 01 ≠ p 0 × p 1 We can write p 1 = (1 – p 1)p 01 + p 1 p 11 We can write p 1 = (1 – p 1)p 01 + p 1 p 11 p 01 p 01 p 1 = ───────── 1 – p 11 + p 01 1 – p 11 + p 01
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1010 Correction (Cont.) Since p 11 + p 10 = 1, i.e., given that the signal was previously 1, its present value can be either 1 or 0. Since p 11 + p 10 = 1, i.e., given that the signal was previously 1, its present value can be either 1 or 0. Therefore, Therefore, p 01 p 01 p 1 = ────── p 10 + p 01 p 10 + p 01 This uniquely gives signal probability as a function of transition probabilities.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1011 Transition and Signal Probabilities 1/f ck p01 = p10 = 0.5 p1 = 0.5 p01 = p10 = 1/3 p1 = 0.5 p01 = p10 = 1/6
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1012 Probabilities: p0, p1, p00, p01, p10, p11 p 01 + p 00 =1 p 01 + p 00 =1 p 11 + p 10 = 1 p 11 + p 10 = 1 p 0 = 1 – p 1 p 0 = 1 – p 1 p 01 p 01 p 1 = ─────── p 10 + p 01 p 10 + p 01
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1013 Transition Density T = 2 p 1 (1 – p 1) = p 0 p 01 + p 1 p 10 T = 2 p 1 (1 – p 1) = p 0 p 01 + p 1 p 10 = 2 p 10 p 01 / (p 10 + p 01) = 2 p 1 p 10 = 2 p 0 p 01
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1014 Power Calculation Power can be estimated if transition density is known for all signals. Power can be estimated if transition density is known for all signals. Calculation of transition density requires Calculation of transition density requires Signal probabilities Signal probabilities Transition densities for primary inputs; computed from vector statistics Transition densities for primary inputs; computed from vector statistics
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1015 Signal Probabilities x1 x2 x1 x2 x1 x2 x1 + x2 – x1x2 x1 1 - x1
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1016 Signal Probabilities x1 x2 x3 x1 x2 y = 1 - (1 - x1x2) x3 = 1 - x3 + x1x2x3 = 0.625 X1X2X3Y 0001 0010 0101 0110 1001 1010 1101 1111 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.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1017 Correlated Signal Probabilities x1 x2 x1 x2 y = 1 - (1 - x1x2) x2 = 1 – x2 + x1x2x2 = 1 – x2 + x1x2 = 0.75 X1X2Y 001 010 101 111 0.5 0.250.625?
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1018 Correlated Signal Probabilities x1 x2 x1 + x2 – x1x2 y = (x1 + x2 – x1x2) x2 = x1x2 + x2x2 – x1x2x2 = x1x2 + x2 – x1x2 = x2 = 0.5 X1X2Y 000 011 100 111 0.5 0.75 0.375?
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1019 Observation Numerical computation of signal probabilities is accurate for fanout-free circuits. Numerical computation of signal probabilities is accurate for fanout-free circuits.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1020 Remedies Use Shannon’s expansion theorem to compute signal probabilities. Use Shannon’s expansion theorem to compute signal probabilities. Use Boolean difference formula to compute transition densities. Use Boolean difference formula to compute transition densities.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1021 Shannon’s Expansion Theorem C. E. Shannon, “A Symbolic Analysis of Relay and Switching Circuits,” Trans. AIEE, vol. 57, pp. 713-723, 1938. C. E. Shannon, “A Symbolic Analysis of Relay and Switching Circuits,” Trans. AIEE, vol. 57, pp. 713-723, 1938. Consider: Consider: Boolean variables, X1, X2,..., Xn Boolean 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) Where Where Xi’ is complement of X1 Xi’ is complement of X1 Cofactors, F(Xi=j) = F(X1, X2,.., Xi=j,.., Xn), j = 0 or 1 Cofactors, F(Xi=j) = F(X1, X2,.., Xi=j,.., Xn), j = 0 or 1
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1022 Expansion About Two Inputs F = XiXj F(Xi=1, Xj=1) + XiXj’ F(Xi=1, Xj=0) F = XiXj F(Xi=1, Xj=1) + XiXj’ F(Xi=1, Xj=0) + Xi’Xj F(Xi=0, Xj=1) + Xi’Xj’ F(Xi=0, Xj=0) In general, a Boolean function can be expanded about any number of input variables. In general, a Boolean function can be expanded about any number of input variables. Expansion about k variables will have 2 k terms. Expansion about k variables will have 2 k terms.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1023 Correlated Signal Probabilities X1 X2 X1 X2 X1X2Y 001 010 101 111 Y = X1 X2 + X2’ Shannon expansion about the reconverging input: Y = X2 Y(X2=1) + X2’ Y(X2=0) = X2 (X1) + X2’ (1)
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1024 Correlated Signals When the output function is expanded about all reconverging input variables, When the output function is expanded 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 calculated without error. Signal probabilities for cofactor outputs can be calculated without error. A weighted sum of cofactor probabilities gives the correct probability of the output. A weighted sum of cofactor probabilities gives the 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) + (1-xi)xj f(Xi=0, Xj=1) + (1-xi)(1-xj) f(Xi=0, Xj=0)
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1025 Correlated Signal Probabilities X1 X2 X1 X2 X1X2Y 001 010 101 111 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
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1026 Example Point of reconv. Supergate 0.5 0.25 1010 0.5 0.0 1.0 0.5 1.0 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 of Probabilistic Testability in Combinational Circuits,” Integration, the VLSI Journal, vol. 7, no. 1, pp. 49-75, April 1989.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1027 Probability Calculation Algorithm Partition circuit into supergates. Partition circuit into supergates. Definition: A supergate is a circuit partition with a single output such that all fanouts that reconverge at the output are contained within the supergate. Definition: A supergate is a circuit partition with a single output such that all fanouts that reconverge at the output are contained within the supergate. Identify reconverging and non-reconverging inputs of each supergate. Identify reconverging and non-reconverging 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 known For a supergate whose input probabilities are known Enumerate reconverging input states Enumerate reconverging input states For each input state do gate by gate probability computation For each input state do gate by gate probability computation Sum up corresponding signal probabilities, weighted by state probabilities Sum up corresponding signal probabilities, weighted by state probabilities
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1028 Calculating Transition Density Boolean function 1 n x1, T1. xn, Tn y, T(Y) = ?
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1029 Boolean Difference Boolean diff(Y, Xi) = 1 means that a path is sensitized from input Xi to output Y. Boolean diff(Y, Xi) = 1 means that a path is sensitized from input Xi to output Y. Prob(Boolean diff(Y, Xi) = 1) is the probability of transmitting a toggle from Xi to Y. Prob(Boolean diff(Y, Xi) = 1) is the probability of transmitting a toggle from Xi to Y. Probability of Boolean difference is determined from the probabilities of cofactors of Y with respect to Xi. Probability of Boolean difference is determined from the probabilities of cofactors of Y with respect to Xi. ∂ Y Boolean 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.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1030 Transition Density n T(y) =Σ T(Xi) Prob(Boolean diff(Y, Xi) = 1) i=1 F. Najm, “Transition Density: A New Measure of Activity in Digital Circuits,” IEEE Trans. CAD, vol. 12, pp. 310-323, Feb. 1993.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1031 Power Computation For each primary input, determine signal probability and transition density for given vectors. For each primary input, determine signal probability and transition density for given vectors. For each internal node and primary output Y, find the transition density T(Y), using supergate partitioning and the Boolean difference formula. For each internal node and primary output Y, find the transition density T(Y), using supergate partitioning and the Boolean difference formula. Compute power, Compute power, P =Σ0.5C Y V 2 T(Y) all Y all Y where C Y is the capacitance of node Y and V is supply voltage.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1032 Transition Density and Power X1 X2 X3 0.2, 1 0.3, 2 0.4, 3 0.06, 0.7 0.436, 3.24 Transition density Signal probability Y CiCi CYCY Power = 0.5 V 2 (0.7C i + 3.24C Y )
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1033 Prob. Method vs. Logic Sim. Circuit No. of gates Probability method Logic Simulation Error% Av. density CPU s* Av. density CPU s* C4321603.460.523.3963+2.1 C49920211.360.588.57241+29.8 C8803832.781.063.25132-14.5 C13553464.191.396.18408-32.2 C19088802.972.005.01464-40.7 C267011933.503.454.00619-12.5 C354016694.473.774.491082-0.4 C531523073.526.414.791616-26.5 C6288240625.105.6734.1731057-26.5 C755235123.839.855.082713-24.2 * CONVEX c240
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1034 Probability Waveform Methods F. Najm, R. Burch, P. Yang and I. Hajj, “CREST – A Current Estimator for CMOS Circuits,” Proc. IEEE Int. Conf. on CAD, Nov. 1988, pp. 204-207. F. Najm, R. Burch, P. Yang and I. Hajj, “CREST – A Current Estimator for CMOS Circuits,” Proc. IEEE Int. Conf. on CAD, Nov. 1988, pp. 204-207. C.-S. Ding, et al., “ Gate-Level Power Estimation using Tagged Probabilistic Simulation, ” IEEE Trans. on CAD, vol. 17, no. 11, pp. 1099-1107, Nov. 1998. C.-S. Ding, et al., “ Gate-Level Power Estimation using Tagged Probabilistic Simulation, ” IEEE Trans. on CAD, vol. 17, no. 11, pp. 1099-1107, Nov. 1998. F. Hu and V. D. Agrawal, “ Dual-Transition Glitch Filtering in Probabilistic Waveform Power Estimation, ” Proc. IEEE Great Lakes Symp. VLSI, Apr. 2005, pp. 357-360. F. Hu and V. D. Agrawal, “ Dual-Transition Glitch Filtering in Probabilistic Waveform Power Estimation, ” Proc. IEEE Great Lakes Symp. VLSI, Apr. 2005, pp. 357-360. F. Hu and V. D. Agrawal, “ Enhanced Dual-Transition Probabilistic Power Estimation with Selective Supergate Analysis, ” Proc. IEEE Int. Conf. Computer Design, Oct. 2005. pp. 366-369. F. Hu and V. D. Agrawal, “ Enhanced Dual-Transition Probabilistic Power Estimation with Selective Supergate Analysis, ” Proc. IEEE Int. Conf. Computer Design, Oct. 2005. pp. 366-369.
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Fall 2006, Oct. 31, Nov. 2ELEC 5270-001/6270-001 Lecture 1035 Power Estimation by Prob. Waveform Circuit TPSDualTransSupergate method E avg %σE tot %E avg %σE tot %E avg %σE tot % c172.32.60.12.32.60.12.32.60.1 c43229.938.835.89.511.86.511.516.611.5 c4996.814.07.03.68.20.62.33.0 c8808.315.31.68.015.75.24.89.00.0 c135524.231.632.95.811.25.45.09.50.5 c190815.023.14.117.727.911.27.016.32.0 c267016.629.87.216.728.39.913.223.66.2 c354013.826.39.810.325.62.410.526.43.7 c531511.824.42.313.431.510.111.327.03.4 c628827.427.532.115.718.84.112.715.40.2 c755214.527.53.214.831.47.814.127.61.3 Avg.15.523.712.410.719.45.78.616.12.9 E avg = average node error, σ = av. node standard deviation, E tot = total error
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