DOC PREVIEW
Berkeley COMPSCI 150 - Lecture Notes

This preview shows page 1-2-3-4-5 out of 15 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Spring 2000 EECS150 Page 1Review - Canonical Forms• Standard form for a Boolean expression -unique algebraic expression from a TT.• Two Types:ϕ Sum of Productsκ Product of Sums• Sum of Products (disjunctive normalform, minterm expansion). Example:minterms a b c f f’a’b’c’ 0 0 0 0 1a’b’c 0 0 1 0 1a’bc’ 0 1 0 0 1a’bc 0 1 1 1 0ab’c’ 1 0 0 1 0ab’c 1 0 1 1 0abc’ 1 1 0 1 0abc 1 1 1 1 0One product (and) term for each 1 in f:f = a’bc + ab’c’ + ab’c +abc’ +abcf’ = a’b’c’ + a’b’c + a’bc’Spring 2000 EECS150 Page 2Canonical Forms• Product of Sums (conjunctive normal form,maxterm expansion). Example:maxterms a b c f f’a+b+c 0 0 0 0 1a+b+c’ 0 0 1 0 1a+b’+c 0 1 0 0 1a+b’+c’ 0 1 1 1 0a’+b+c 1 0 0 1 0a’+b+c’ 1 0 1 1 0a’+b’+c 1 1 0 1 0a’+b’+c’ 1 1 1 1 0One sum (or) term for each 0 in f:f = (a+b+c)(a+b+c’)(a+b’+c)f’ =(a+b’+c’)(a’+b+c)(a’+b+c’)(a’+b’+c)(a+b+c’)Mapping from one form to the other:Derive TT then proceed.Spring 2000 EECS150 Page 3Two-level Logic SimplicationKey tool: The Uniting Theorem x (y’ + y) = x (1) = xa b f f = ab’ + ab = a(b’+b) = a0 0 0 b values change within0 1 0 the on-set rows 1 0 1 a values don’t change1 1 1 b is eliminated, a remainsa b g g = a’b’+ab’ = (a’+a)b’ =b’0 0 1 b values stay the same0 1 0 a values changes1 0 11 1 0 b’ remains, a eliminatedSpring 2000 EECS150 Page 4Boolean CubesVisual technique for identifying when theUniting Theorem can be applied1-cube0 1x2-cube01 1100 10xy011 111110001000 100010101yxz3-cube• Sub-cubes of on nodes can be used forsimplification.– On-set - filled in nodes, off-set - empty nodesa b f g0 0 0 10 1 0 01 0 1 11 1 1 001 1100 10fbaa asserted & unchangedb varies ab + ab' = aSpring 2000 EECS150 Page 53-variable cube exampleFA carry out:a b c cout0 0 0 00 0 1 00 1 0 00 1 1 11 0 0 01 0 1 11 1 0 11 1 1 1011 111110001000 100010101bac(a' + a)bcab(c'+c)a(b+b')ccout = bc + ab + ac011 111110001000 100010101bacab’c’ + ab’c + abc’ + abc ac’ + ac + ab = a + ab = a• Both b & c change, a isasserted & remainsconstant.What about larger sub-cubes?Spring 2000 EECS150 Page 6Karnaugh Map Method• K-map is an alternative method ofrepresenting the TT and to help visual theadjacencies.0 101abcab00 01 11 1001abcd 00 01 11 10000111105 & 6 variable k-maps possibleg = b'0 101abcab00 01 11 10010 101abcab00 01 11 10010 10 1f = a0 0 1 00 1 1 1cout = ab + bc + ac1 10 00 0 1 10 0 1 1f = aExamples:Spring 2000 EECS150 Page 7K-maps (cont.)cab00 01 11 10011 0 0 10 0 1 1f = b'c' + acabcd 00 01 11 10000111101 0 0 10 1 0 01 1 1 11 1 1 1f = c + a'bd + b'd'(bigger groups are better)Circling Zerosabcd 00 01 11 10000111101 0 000 0111 1 1 11111f = (b' + c + d)(a' + c + d')(b + c + d')Spring 2000 EECS150 Page 8BCD incrementer examplea b c d w x y z0 0 0 0 0 0 0 10 0 0 1 0 0 1 00 0 1 0 0 0 1 10 0 1 1 0 1 0 00 1 0 0 0 1 0 10 1 0 1 0 1 1 00 1 1 0 0 1 1 10 1 1 1 1 0 0 01 0 0 0 1 0 0 11 0 0 1 0 0 0 01 0 1 0 - - - -1 0 1 1 - - - -1 1 0 0 - - - -1 1 0 1 - - - -1 1 1 0 - - - -1 1 1 1 - - - -00 01 110001111011abcd00 01 110001111011abcd00 01 110001111011abcd00 01 110001111011abcdw xy z0000 00000000000 00 000 0 0 00 011 11111 11 11 1 11 1--- ------ ---------------w = x =y = z =Spring 2000 EECS150 Page 9Higher Dimensional K-maps00 01 11 1000101110bcdea = 100 01 11 1000101110a = 000 01 11 1000101110cdefab = 1000 01 11 1000101110ab = 1100 01 11 1000101110ab = 0100 01 11 1000101110ab = 00Spring 2000 EECS150 Page 10Katz Chapter 3 - Multi-levelCombinational LogicŸ Example: reduced sum-of-products formx = adf + aef + bdf + bef + cdf + cef + gŸ implementation in 2-levels with gates:cost: 7-input OR, 6 3-input AND 50 transistors + 25 wires(19 linteral plus 6 internal)delay: 3-input AND gate delay + 7-input OR gate delayŸ Factored form:x = (a + b +c)(d + e)f + gcost: 1 3-input OR, 2 2-input OR, 1 3-input ANDdelay: 3-input OR + 3-input AND + 2-input ORWhich is faster?In general: Using multiple levels (more than 2) willreduce the cost. Sometimes also delay.Sometime a tradeoff between cost and delay.adfabcdefgxSpring 2000 EECS150 Page 11Multi-level Combinational LogicExample:F = abc + abd +a’c’d’ + b’c’d’let x = ab y = c+df = xy + x’y’Example: “result” output functions from 4-bitadder• No convenient hand methods for multi-levellogic simplication:1 CAD Tools, example misII (UCB)2 exploit some special structure, example adderabcabdacdbcdabcdfxySpring 2000 EECS150 Page 12NAND-NAND & NOR-NORNetworksDeMorgan’s Law:(a + b)’ = a’ b’ (a b)’ = a’ + b’ b + b = (a’ b’)’ (a b) = (a’ + b’)’push bubbles or introduce in pairs or remove pairs.Mapping from AND/OR to NAND/NAND= ===abcda)b)c) d)Spring 2000 EECS150 Page 13NAND-NAND & NOR-NORNetworks• Mapping AND/OR to NOR/NOR• Mapping OR/AND to NOR/NOR• OR/AND to NAND/NANDa) b)c)abcdabcdabcda) b)c)paira'b'c'd'Spring 2000 EECS150 Page 14Multi-level NetworksF = a(b + cd) + bc’Convert to NANDs (note fanout)cdbabc'fabcdSpring 2000 EECS150 Page 15EXOR FunctionParity, addition mod 2x y xor xnor0 0 0 1 x xor y = x’y + xy’0 1 1 01 0 1 01 1 0 1Another approach:x'yxy'xy01yxif x=0 then y else


View Full Document

Berkeley COMPSCI 150 - Lecture Notes

Documents in this Course
Lab 2

Lab 2

9 pages

Debugging

Debugging

28 pages

Lab 1

Lab 1

15 pages

Memory

Memory

13 pages

Lecture 7

Lecture 7

11 pages

SPDIF

SPDIF

18 pages

Memory

Memory

27 pages

Exam III

Exam III

15 pages

Quiz

Quiz

6 pages

Problem

Problem

3 pages

Memory

Memory

26 pages

Lab 1

Lab 1

9 pages

Memory

Memory

5 pages

Load more
Download Lecture Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?