03 Minimization of Logic Function

03 Minimization of Logic Function - 1 EE2000 Logic Circuit...

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Unformatted text preview: 1 EE2000 Logic Circuit Design 3 Minimization of Logic Function ¡ ¡ ¡ 2 Outline ¡ 3.1 The Meaning of Minimization ¡ 3.2 Minimization using Boolean Algebra ¡ 3.3 Introduction to Karnaugh Map ¡ 3.4 Minimization using Karnaugh Map ¡ 3.5 Boolean Function with Don’t-care Cases ¡ 3.6 Minimzation using Quine-McCluskey Method 3 3.1 The Meaning of Minimization 4 Minimizing Logic Circuits ¡ Boolean algebra is a useful tool for simplifying / minimizing logic circuits ¡ e.g. we can simplify f = x’yz + x’yz’ + xz using the postulates taught in last chapter ¡ f = x’yz + x’yz’ + xz ¡ = x’y + xz (adjacency) ¡ Reduce f from sum of 3 products to sum of 2 product terms ! 5 The Meaning of Minimization f = x’yz + x’yz’ + xz x y z f = x’y + xz x y z x’yz’ xz 4 gates, 11 inputs 3 gates, 6 inputs x’yz x’y xz equivalent = 6 Cost of a Logic Circuit ¡ How to measure the simplicity of a logic circuit? ¡ Gate-input Cost ¡ The number of inputs to the gates in the implementation ¡ Gate Cost ¡ The number of gates in the implementation ¡ Total cost = Gate cost + Gate-input cost 7 Why Gate-input Cost? ¡ The no. of gate-input proportional to the no. of transistors and wires used in a logic circuit ¡ Can be determined directly by looking at the logic diagram or the Boolean function ¡ A good measure for cost of the logic circuits 8 Example ¡ e.g. given that g 1 is equivalent to g 2 , which function has a lower cost to implement? ¡ g 1 ( a, b, c, d ) = abcd + a’b’c’d’ ¡ g 2 ( a, b, c, d ) = ( a’ + b )( b’ + c )( c’ + d )( d’ + a ) ¡ g 1 has 2 AND gates, 1 OR gate and 10 gate inputs ¡ Total cost = 13 ¡ g 2 has 4 OR gates, 1 AND gate and 12 gate inputs ¡ Total cost = 17 ¡ ∴ g 1 (SOP) has lower cost than g 2 (POS) in this example 9 SOP vs. POS ¡ SOP form better or POS form better? ¡ e.g. given that f 1 is equivalent to f 2 , which function has a lower cost to implement? ¡ f 1 ( a, b, c, d ) = ab’ + ad’ + ac’ + a’bd ¡ f 2 ( a, b, c, d ) = ( a + b )( a +¡d )( a’ + b’ + c’ + d’ ) ¡ f 1 = 18, f 2 = 11 ¡ ∴ f 1 (POS) has lower cost than f 2 (SOP) in this example ¡ Conclusion: SOP is not always superior than POS 10 Summary of 3.1 ¡ The complexity of a logic circuit is directly related to the Boolean function ¡ Total cost = Gate cost + Gate-input cost ¡ The key of simplifying logic function is to reduce the no. of terms and no. of literals ¡ ↓ no. of literals = ↓ no. of gate inputs ¡ ↓ no. of terms = ↓ no. of gates and next-level gate inputs ¡ Also ↓ space, power and the production cost 11 3.2 Minimization using Boolean Algebra 12 Remember This Example? ¡ e.g. Given 4 equivalent Boolean functions f 1 to f 4 expressed in SOP form already (to be proved in next chapter)....
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This note was uploaded on 02/06/2011 for the course EE 2000 taught by Professor Vancwting during the Spring '07 term at City University of Hong Kong.

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03 Minimization of Logic Function - 1 EE2000 Logic Circuit...

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