# 10/4-6/05ELEC 5970-001/6970-001 Lecture 111 ELEC 5970-001/6970-001(Fall 2005) Special Topics in Electrical Engineering Low-Power Design of Electronic Circuits.

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10/4-6/05ELEC 5970-001/6970-001 Lecture 111 ELEC 5970-001/6970-001(Fall 2005) Special Topics in Electrical Engineering 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 http://www.eng.auburn.edu/~vagrawal vagrawal@eng.auburn.edu

10/4-6/05ELEC 5970-001/6970-001 Lecture 112 Basic Idea 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

10/4-6/05ELEC 5970-001/6970-001 Lecture 113 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.

10/4-6/05ELEC 5970-001/6970-001 Lecture 114 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 the transition density.

10/4-6/05ELEC 5970-001/6970-001 Lecture 115 Static Signal Probabilities Observe signal for interval t0 + t1 –Signal is 1 for duration t1 –Signal is 0 for duration t0 –Signal probabilities: p1 = t1/(t0 + t1) p0 = t0/(t0 + t1) = 1 – p1

10/4-6/05ELEC 5970-001/6970-001 Lecture 116 Static Transition Probabilities Transition probabilities: T01 = p0 Prob{signal is 1 | signal was 0} = p0 p1 T10 = p1 Prob{signal is 0 | signal was 1} = p1 p0 T = T01 + T10 = 2 p0 p1 = 2 p1 (1 – p1) Transition density: T = 2 p1 (1 – p1) Transition frequency: f = T/ 2 Power = CV 2 T/ 2 (correct formula)

10/4-6/05ELEC 5970-001/6970-001 Lecture 117 Static Transition Frequency 00.250.50.75 1.0 0.25 0.2 0.1 0.0 p1p1 f = p1(1 – p1)

10/4-6/05ELEC 5970-001/6970-001 Lecture 118 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.

10/4-6/05ELEC 5970-001/6970-001 Lecture 119 Cause for Error and Correction Probability of transition is not independent of the present state of the signal. Consider probability p01of a 0→1 transition, Then p01 ≠ p0 p1 We can write p1 = (1 – p1)p01 + p1 p11 p01 p1 = ───────── 1 – p11 + p01

10/4-6/05ELEC 5970-001/6970-001 Lecture 1110 Correction (Cont.) Since p11 + p10 = 1, i.e., given that the signal was previously 1, its present value can be either 1 or 0. Therefore, p01 p1 = ────── p10 + p01 This uniquely gives signal probability as a function of transition probabilities.

10/4-6/05ELEC 5970-001/6970-001 Lecture 1111 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

10/4-6/05ELEC 5970-001/6970-001 Lecture 1112 Probabilities: p0, p1, p00, p01, p10, p11 p01 + p00 =1 p11 + p10 = 1 p0 = 1 – p1 p01 p1 = ────── p10 + p01

10/4-6/05ELEC 5970-001/6970-001 Lecture 1113 Transition Density T = 2 p1(1 – p1) = p0 p01 + p1 p10 = 2 p10 p01/(p10 + p01) = 2 p1 p10 = 2 p0 p01

10/4-6/05ELEC 5970-001/6970-001 Lecture 1114 Power Calculation Power can be estimated if transition density is known for all signals. Calculation of transition density requires –Signal probabilities –Transition densities for primary inputs; computed from vector statistics

10/4-6/05ELEC 5970-001/6970-001 Lecture 1115 Signal Probabilities x1 x2 x1 x2 x1 x2 x1 + x2 – x1x2 x1 1 - x1

10/4-6/05ELEC 5970-001/6970-001 Lecture 1116 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.

10/4-6/05ELEC 5970-001/6970-001 Lecture 1117 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?

10/4-6/05ELEC 5970-001/6970-001 Lecture 1118 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?

10/4-6/05ELEC 5970-001/6970-001 Lecture 1119 Observation Numerical computation of signal probabilities is accurate for fanout-free circuits.

10/4-6/05ELEC 5970-001/6970-001 Lecture 1120 Remedies Use Shannon’s expansion theorem to compute signal probabilities. Use Boolean difference formula to compute transition densities.

10/4-6/05ELEC 5970-001/6970-001 Lecture 1121 Shannon’s Expansion Theorem C. E. Shannon, “A Symbolic Analysis of Relay and Switching Circuits,” Trans. AIEE, vol. 57, pp. 713-723, 1938. Consider: Boolean variables, X1, X2,..., Xn Boolean function, F(X1, X2,..., Xn) Then F = Xi F(Xi=1) + Xi’ F(Xi=0) Where Xi’ is complement of X1 Cofactors, F(Xi=j) = F(X1, X2,.., Xi=j,.., Xn), j = 0 or 1

10/4-6/05ELEC 5970-001/6970-001 Lecture 1122 Expansion About Two Inputs 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. Expansion about k variables will have 2 k terms.

10/4-6/05ELEC 5970-001/6970-001 Lecture 1123 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)

10/4-6/05ELEC 5970-001/6970-001 Lecture 1124 Correlated Signals When the output function is expanded about all reconverging input variables, All cofactors correspond to fanout-free circuits. Signal probabilities for cofactor outputs can be calculated without error. A weighted sum of cofactor probabilities gives the correct probability of the output. 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)

10/4-6/05ELEC 5970-001/6970-001 Lecture 1125 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

10/4-6/05ELEC 5970-001/6970-001 Lecture 1126 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.

10/4-6/05ELEC 5970-001/6970-001 Lecture 1127 Probability Calculation Algorithm 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. Identify reconverging and non-reconverging inputs of each supergate. Compute signal probabilities from PI to PO: –For a supergate whose input probabilities are known –Enumerate reconverging input states –For each input state do gate by gate probability computation –Sum up corresponding signal probabilities, weighted by state probabilities

10/4-6/05ELEC 5970-001/6970-001 Lecture 1128 Calculating Transition Density Boolean function 1 n x1, T1. xn, Tn y, T(Y) = ?

10/4-6/05ELEC 5970-001/6970-001 Lecture 1129 Boolean Difference 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. 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.

10/4-6/05ELEC 5970-001/6970-001 Lecture 1130 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.

10/4-6/05ELEC 5970-001/6970-001 Lecture 1131 Power Computation 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. Compute power, P =Σ0.5C Y V 2 T(Y) all Y where C Y is the capacitance of node Y and V is supply voltage.

10/4-6/05ELEC 5970-001/6970-001 Lecture 1132 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 )

10/4-6/05ELEC 5970-001/6970-001 Lecture 1133 Prob. Method vs. Logic Sim. Circuit No. of gates Probability methodLogic Simulation Error % Av. densityCPU s*Av. densityCPU 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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