l6_ece15a_6

l6_ece15a_6 - Last time: Implementing Boolean functions ECE...

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1 ECE 15A Fundamentals of Logic Design Lecture 6 Malgorzata Marek-Sadowska Electrical and Computer Engineering Department UCSB 2 Last time: Implementing Boolean functions ± Representations of Boolean functions are not unique: the same functions can be expressed in many different ways ± Different representations yield different realizations ± Optimization in terms of gate count, path length, wiring, power, etc. 3 Last time: Expressions and circuits ± Any Boolean expression can be converted into a circuit by combining basic gates in a relatively straightforward way. ± The diagram below shows the inputs and outputs of each gate. ± The precedences are explicit in a circuit. Clearly, we have to make sure that the hardware does operations in the right order! (x + y’)z + x’ 4 Last time example: half adder x y Sum Carry 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 HA x y Sum Carry Sum= x’y + xy’ Carry = xy x y sum x y carry (dnf: 22 transistors) (6 transistors) Dnf realization: 22+6 transistors 5 Different realizations of a half adder x y sum (14 transistors) Sum=x’y+y’x = (x+y)(xy)’ x y carry (6 transistors) Total_2 = 20 transistors Total_1= 20 transistors Sum=x’y+y’x = (x+y)(xy)’= [(x+y)(xy)’]’’=[(x+y)’+(xy)’’]’ x y Sum Carry Total_3=14 transistors 6 Example ± abc c ab c b a bc a c b a f + + + + = a b c f Direct representation. 9 gates, 11 nets, 42 transistors.
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2 7 Simplification ac b ac b a a ac ab b a b b ac c c ab c c b a c b a abc abc c ab bc a c b a abc c ab c b a bc a c b a f + = + + = + + = + + + + + = + + + + + = + + + + = ) ( ) ( ) ( ) ( f b a c 2 gates, 4 nets, 12 transistors 8 Equivalence check abc c ab c b a bc a c b a f + + + + = 1 ac b f + = 2 ? 2 1 f f Is = a b c f1 f2 0 0 0 0 0 0 0 1 0 0 0 1 0 1 1 0 1 1 1 1 1 0 0 0 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 Both expressions produce the same truth tables -> they are equivalent. 9 Uniting theorem ± Powerful tool in Boolean simplification ± Uniting theorem: AB + AB’ = A (B+B’) = A ± Example: ab’c’+ab’c+abc’+abc = ab’(c’+c) + ab(c + c’)= ab’+ab = a(b+b’) = a 10 Boolean Cubes Visual technique for identifying when the Uniting Theorem can be applied 1-cube 0 1 x 2-cube 01 11 00 10 xy 011 111 110 001 000 100 010 101 y x z 3-cube ± Sub-cubes of on nodes can be used for simplification. ² On-set - filled in nodes, off-set - empty nodes
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l6_ece15a_6 - Last time: Implementing Boolean functions ECE...

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