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10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 1
ELEC 5970-001/6970-001(Fall 2005)Special Topics in Electrical EngineeringLow-Power Design of Electronic Circuits
Power Analysis: Probabilistic Methods
Vishwani D. AgrawalJames J. Danaher Professor
Department of Electrical and Computer EngineeringAuburn University
http://www.eng.auburn.edu/[email protected]
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 2
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 CV2fck
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 3
Source of Inaccuracy
1/fck
p1 = 0.5 P = 0.5CV2fck
p1 = 0.5 P = 0.33CV2fck
p1 = 0.5 P = 0.167CV2fck
Observe that the formula, Power, P = (1 – p1) p1 CV2fck, is notCorrect.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 4
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.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 5
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
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 6
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 = CV2T/ 2 (correct formula)
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 7
Static Transition Frequency
0 0.25 0.5 0.75 1.0
0.25
0.2
0.1
0.0
p1
f =
p1
(1 –
p1
)
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 8
Inaccuracy in Transition Density
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.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 9
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
p01p1 = ─────────
1 – p11 + p01
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 10
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.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 11
Transition and Signal Probabilities
1/fck
p01 = p10 = 0.5 p1 = 0.5
p01 = p10 = 1/3 p1 = 0.5
p1 = 0.5p01 = p10 = 1/6
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 12
Probabilities: p0, p1, p00, p01, p10, p11
• p01 + p00 =1
• p11 + p10 = 1
• p0 = 1 – p1
• p01
p1 = ──────
p10 + p01
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 13
Transition Density
• T = 2 p1(1 – p1) = p0 p01 + p1 p10
= 2 p10 p01/(p10 + p01)
= 2 p1 p10 = 2 p0 p01
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 14
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
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 15
Signal Probabilities
x1
x2
x1 x2
x1
x2
x1 + x2 – x1x2
x1 1 - x1
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 16
Signal 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.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 17
Correlated Signal Probabilities
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?
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 18
Correlated Signal Probabilities
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?
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 19
Observation
• Numerical computation of signal probabilities is accurate for fanout-free circuits.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 20
Remedies
• Use Shannon’s expansion theorem to compute signal probabilities.
• Use Boolean difference formula to compute transition densities.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 21
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
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 22
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 2k terms.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 23
Correlated Signal Probabilities
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)
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 24
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)
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 25
Correlated Signal Probabilities 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
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 26
Example
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.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 27
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
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 28
Calculating Transition Density
Boolean function
1
n
x1, T1..... xn, Tn
y, T(Y) = ?
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 29
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.
∂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.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 30
Transition 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.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 31
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 = Σ 0.5CY V2 T(Y)
all Y
where CY is the capacitance of node Y and V is supply voltage.
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 32
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 densitySignal probability
YCi
CY
Power = 0.5 V2 (0.7Ci + 3.24CY)
10/4-6/05 ELEC 5970-001/6970-001 Lecture 11 33
Prob. Method vs. Logic Sim.
CircuitNo. of gates
Probability method Logic SimulationError
%Av. density CPU s* Av. density CPU s*
C432 160 3.46 0.52 3.39 63 +2.1
C499 202 11.36 0.58 8.57 241 +29.8
C880 383 2.78 1.06 3.25 132 -14.5
C1355 346 4.19 1.39 6.18 408 -32.2
C1908 880 2.97 2.00 5.01 464 -40.7
C2670 1193 3.50 3.45 4.00 619 -12.5
C3540 1669 4.47 3.77 4.49 1082 -0.4
C5315 2307 3.52 6.41 4.79 1616 -26.5
C6288 2406 25.10 5.67 34.17 31057 -26.5
C7552 3512 3.83 9.85 5.08 2713 -24.2* CONVEX c240