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CSE20 Lecture 16 Boolean Algebra: Applications

CSE20 Lecture 16 Boolean Algebra: Applications. Professor CK Cheng CSE Dept. UC San Diego. 1. Outlines. Introduction Light Switches Bit Counter Multiplication Multiplexer Priority Encoder Summary. Introduction. Language Description Truth Table Karnaugh Map Minimal Expression

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CSE20 Lecture 16 Boolean Algebra: Applications

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  1. CSE20 Lecture 16Boolean Algebra: Applications Professor CK Cheng CSE Dept. UC San Diego 1

  2. Outlines • Introduction • Light Switches • Bit Counter • Multiplication • Multiplexer • Priority Encoder • Summary

  3. Introduction • Language Description • Truth Table • Karnaugh Map • Minimal Expression • Digital Logic Networks

  4. Light Switches Given two switches A, B with two states (Up, Down), the switches control one light emitter Y. Initially A=B=Down, and Y=Off. The light Y is turned between Off and On when one of the switch changes its state. Describe Y as a function of A and B. Assignment: A, B are in {0, 1} (Down, Up) Y is in {0,1} (Off, On) 4

  5. Light Switches Truth Table Karnaugh Map 5

  6. Light Switches Minimal Expression in sum of products format. Y(A,B)=A’B+AB’ 6

  7. Bit Counter Input: Three binary bits (A, B, C) Output: (S1,S0) that counts the number of 1’s in (A, B, C). Derive minimal expressions of (S1,S0). For example: (A,B,C)=(0,0,0) => (S1,S0)=(0,0) 0: Binary code (A,B,C)=(1,1,0) => (S1,S0)=(1,0) 2: Binary code (A,B,C)=(1,1,1) => (S1,S0)=(1,1) 3: Binary code

  8. Bit Counter Truth Table Karnaugh Map 8

  9. Bit Counter Karnaugh Map S1=AB+BC+AC S0=AB’C’+ABC+ A’B’C+A’BC’ 9

  10. Multiplication Input: Two binary numbers (a1,a0), (b1,b0) Output: (s3,s2,s1,s0) product of the two numbers. Derive minimal expressions of (s3,s2,s1,s0). For example: (a1,a0)×(b1,b0) = (s3,s2,s1,s0) (0,0)×(0,0) = (0,0,0,0) (1,0)×(0,1) = (0,0,1,0) (1,1)×(1,0) = (0,1,1,0) (1,1)×(1,1) = (1,0,0,1)

  11. Multiplication: Truth TableKarnaugh Maps are left as exercises. 11

  12. Multiplication Input: Two binary numbers (a1,a0), (b1,b0) Output: (s3,s2,s1,s0) product of the two numbers. A minimal expression: s3=a1a0b1b0 s2=a1b1b0’+a1a0’b1 s1=a1’a0b1+a0b1b0’+a1a0’b0+a1b1’b0 s0=a0b0 Verification: (a1,a0), (b1,b0) are symmetric in the expressions.

  13. Multiplexer Input: Three binary bits S (select), A, B (data) Output: Y If S=0, then Y=B; else Y=A. For example: (S,A,B)=(0,1,0) => Y= 0. (S,A,B)=(1,1,0) => Y= 1. (S,A,B)=(1,0,1) => Y= 0.

  14. Multiplexer Truth Table Karaugh Map Y=S’B+SA

  15. Multiplexer Input: Three binary bits S (select), A, B (data) Output: Y If S=0, then Y=B; else Y=A. Minimal Expression: Y=S’B+SA

  16. Priority Encoder Input: Three binary bits E (Enable), D1, D0 (Devices IDs) Output: A, Y If E=0, then A=Y=0; Else if D0=1, A=1, Y=0; (Let Device 0 take higher priority) Else if D1=1, A=1, Y=1; Else A=0, Y=0. For example: (E,D1,D0)=(0,1,0) => Y= 0, A=0. (E,D1,D0)=(1,0,1) => Y= 0, A=1. (E,D1,D0)=(1,1,0) => Y= 1, A=1. (E,D1,D0)=(1,0,0) => Y= 0, A=0.

  17. Priority Encoder Truth Table Karaugh Maps

  18. Priority Encoder Input: Three binary bits E (Enable), D1, D0 Output: A, Y If E=0, then A=Y=0; Else if D0=1, A=1, Y=0; Else if D1=1, A=1, Y=1; Else A=0, Y=0. Minimal Expression: Y=ED1D0’ A=ED1+ED0

  19. Quiz • For the above Priority Encoder, let D1 take higher priority. Derive the minimal expression of the outputs A and Y.

  20. Summary • Language Description • Combinatorial Systems • Truth Table • #variables <7 • Karnaugh Maps • Two level optimization • Minimal Expressions • Logic Networks 20

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