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Review for Exam 1

Review for Exam 1. Chapters 1 through 3. Chapter 1 Overview. Information Representation Number Systems [binary, octal and hexadecimal] Base Conversion Decimal Codes [BCD (binary coded decimal)] Alphanumeric Codes Parity Bit Gray Codes. 1-2 Number Systems.

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Review for Exam 1

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  1. Review for Exam 1 • Chapters 1 through 3

  2. Chapter 1 Overview • Information Representation • Number Systems [binary, octal and hexadecimal] • Base Conversion • Decimal Codes [BCD (binary coded decimal)] • Alphanumeric Codes • Parity Bit • Gray Codes

  3. 1-2 Number Systems • Positive radix, positional number systems • Examples: • Decimal (radix r =10) • Binary (radix r =2) • Octal (radix r = ) • Hexadecimal (r = ) Ex: 24.3 = 2x101 + 4x100+3x10-1 Digits (0-9) Bits (0-1) Ex: 1101.01 = ( . )10 Digits: 1,2,…9, A, B, C, D, E, F

  4. Range of numbers • Binary number: ex. a 3-bit number: n=3 • 000, 001 … ,111 or in decimal system: 0, 1 … 7 • Total of 8 numbers (=23) • Range: from 0 to 7 (0 to 23-1) • In general a n-bit number represents: • 2ndifferentnumbers • Min: 0 • Max number: 2n-1 • For fractions: m bits after the radix point: • Min: 0 • Max number: (2m -1)/2m

  5. Use of HEX system 9 E . 5 • Short hand notation of large binary numbers: • Each HEX digits can be represented by exactly 4 bits (16=24) • Thus (10011110.0101)2 Conversion from binary to HEX and HEX to binary is very easy: (10011101)2= ( )16 (1010110110.11)2 = ( )16 B39.716 = ( )2

  6. Octal system • Radix r = 8 • 8 digits: • 0, 1, 2,…7 • Ex: 2758 = 2x82 + 7x8 + 5x1 = 128 + 56 + 5 = 18910 • Each octal digit can be represented by 3 bits

  7. 1-3 Conversion Between Bases To convert from one base to another: 1) Convert the Integer Part 2) Convert the Fraction Part 3) Join the two results with a radix point

  8. Example: convert (325.65)10 to hex Least significant digit Most significant Most significant Least significant 325.6510 = 145.A6616 • Integer part: 32510 = ( . )16 • Fractional part: .65 325/16 = 20 and rem = 5 20/16 = 1 and rem = 4 1/16 = 0 and rem = 1 Thus 32510 = 14516 0.65x16 = 10.4 thus int = 10= A 0.4x16 = 6.4 thus int = 6 0.4x16 = 6.4 thus int = 6 Etc. Thus .6510 = A6616

  9. Conversion - Summary Divisions (or x) by 16 Hexadecimal Ai.16i Decimal Divisons by 2 Group in bits of 4 SAi.2i Binary Divisons by 8 Octal Hex: through the binary representation Group in bits of 3 Ai.8i Octal

  10. 1-4 Binary Codes • A n-bit binary code is a n-bit word which can represent up to 2ndifferent elements. • Example: 3-bit code can represent up to 8 different elements”

  11. Binary Coded Decimal (BCD) • The BCD code is the 8,4,2,1 code. • This code only encodes the first ten values from 0 to 9. • Each decimal digit is coded separately by 4 bits • Example: • (325)10= (0011 0010 0101)BCD • Exercise: (856)10 = ( )BCD 3 2 5

  12. Overview Chapter 2 • Binary Logic and Gates • Boolean Algebra • Standard Forms • Two-Level Optimization • Map Manipulation • Other Gate Types • Exclusive-OR Operator and Gates • High-Impedance Outputs

  13. Operator Definitions and Truth Tables AND NOT X Y Z = X·Y X = Z X 0 0 0 0 1 0 1 0 1 0 1 0 0 1 1 1 • Truth table- a tabular listing of the values of a function for all possible combinations of values on its arguments • Example: Truth tables for the basic logic operations:

  14. 2-2 Boolean Algebra + 0 X X = X + X X . X 1. 2. . 1 X X = 5. 6. X + X X X . X X = = Existence complements 7. 1 8. 0 = = Dual Boolean algebra deals with binary variables and a set of three basic logic operations: AND (.), OR (+) and NOT ( ) that satisfy basic identities Basic identities Existence 0 and 1 or operations with 0 and 1 3. 1 1 4. . 0 0 X X + = = Idempotence Involution 9. X = X Replace “+” by “.”, “.” by +, “0” by “1” and “1’’ by”0”

  15. Boolean Algebra 10. = X + Y Y + X 12. (X + Y) Z X + (Y Z) + = + X (Y+ Z) XY XZ = + 14. 16. X + Y X . Y = 11. XY YX = 13. (XY) Z X(Y Z ) = X + YZ = (X + Y) (X + Z) 15. 17. X . Y X + Y = Boolean Theorems of multiple variables Commutative Associative Distributive DeMorgan’s Dual

  16. Other useful Theorems X(X + Y) = XY X + XY = X + Y XY + XY = Y (X + Y)(X + Y) = Y XY + XZ + YZ = XY + XZ (X + Y)( X + Z)(Y + Z) = (X + Y)( X + Z) Dual Minimization X(X + Y) = X Absorption X + XY = X Simplification Consensus

  17. 2-3 Standard (Canonical) Forms • It is useful to specify Boolean functions in a form that: • Allows comparison for equality. • Has a correspondence to the truth tables • Canonical Forms in common usage: • Sum of Products (SOP), also called Sum or Minterms (SOM) • Product of Sum (POS), also called Product of Maxterms (POM)

  18. Maxterms and Minterms • Examples: Two variable minterms and maxterms. • The index above is important for describing which variables in the terms are true and which are complemented.

  19. Index Examples – Four Variables = m M = M m i i i i Index Binary MintermMaxterm iPattern miMi 0 0000 1 0001 3 0011 5 0101 7 0111 10 1010 13 1101 15 1111 a b c d ? a b c d Notice: the variables are in alphabetical order in a standard form ? + + + a b c d + + + a b c d a b c d + + + ? a b c d + + + a b c d a b c d ? a b c d + + + a b c d Relationship between min and MAX term?

  20. Minterm Function Example • F(A, B, C, D, E) = m2 + m9 + m17 + m23 • F(A, B, C, D, E) write in standard form: • Sum of Product (SOP) expression: • F = Σm(2, 9, 17, 23) A’B’C’DE’ + A’BC’D’E + AB’C’D’E + AB’CDE m2 m9 m17 m23

  21. Expressing a function with Maxterms = + + + + + + F (x y z) ·(x y z)·(x y z ) 1 + + + + ·( x y z )·( x y z) • Start with the SOP: F1(x,y,z) =m1 + m4 + m7 • Thus its complement F1can be written as • F1 = m0 +m2 +m3 + m5 + m6 (missing term of F1) • Apply deMorgan’s theorem on F1: • (F1 = (m0 +m2 +m3 + m5 + m6) = m0.m2.m3.m5.m6 = M0.M2.M3.M5.M6 = ΠM(0,2,3,5,6) Thus the Product of Sum terms (POS): also called, Big M notation

  22. 2-4 Circuit Optimization • Goal: To obtain the simplest implementation for a given function • Optimization requires a cost criterion to measure the simplicity of a circuit • Distinct cost criteria we will use: • Literal cost (L) • Gate input cost (G) • Gate input cost with NOTs (GN)

  23. Literal Cost • Literal – a variable or its complement • Literal cost – the number of literal appearances in a Boolean expression corresponding to the logic circuit diagram • Examples (all the same function): • F = BD + AB’C + AC’D’ L = 8 • F = BD + AB’C + AB’D’ + ABC’ L = • F = (A + B)(A + D)(B + C + D’)( B’ + C’ + D) L = • Which solution is best?

  24. Karnaugh Maps (K-map) • A K-map is a collection of squares • Each square represents a minterm • The collection of squares is a graphical representation of a Boolean function • Adjacent squares differ in the value of one variable • Alternative algebraic expressions for the same function are derived by recognizing patterns of squares • The K-map can be viewed as • A reorganized version of the truth table • A topologically-warped Venn diagram as used to visualize sets in algebra of sets

  25. 2-5 Map Manipulation: Systematic Simplification A Prime Implicant is a product term obtained by combining the maximum possible number of adjacent squares in the map into a rectangle with the number of squares a power of 2. A prime implicant is called an Essential Prime Implicant if it is the only prime implicant that covers (includes) one or more minterms. Prime Implicants and Essential Prime Implicants can be determined by inspection of a K-Map.

  26. Don't Cares in K-Maps • Sometimes a function table or map contains entries for which it is known: • the input values for the minterm will never occur, or • The output value for the minterm is not used • In these cases, the output value need not be defined • Instead, the output value is defined as a “don't care” • By placing “don't cares” ( an “x” entry) in the function table or map, the cost of the logic circuit may be lowered. • Example 1: A logic function having the binary codes for the BCD digits as its inputs. Only the codes for 0 through 9 are used. The six codes, 1010 through 1111 never occur, so the output values for these codes are “x” to represent “don’t cares.”

  27. Other Gate Types: overview A B A B A B A B A A B BUF NAND NOR XOR XNOR 0 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 1 1 0 0 01

  28. The Tri-State Buffer IN OUT EN EN IN OUT 0 X Hi-Z 1 0 0 1 1 1 Symbol • For the symbol and truth table, IN is the data input, and EN, the control input. • For EN = 0, regardless of the value on IN (denoted by X), the output value is Hi-Z. • For EN = 1, the output value follows the input value. • Variations: • Data input, IN, can be inverted • Control input, EN, can be inverted by addition of “bubbles” to signals. Truth Table OUT= IN.EN

  29. NAND Mapping Algorithm • Replace ANDs and ORs: • Repeat the following pair of actions until there is at most one inverter between : • A circuit input or driving NAND gate output, and • The attached NAND gate inputs.

  30. NOR Mapping Algorithm • Replace ANDs and ORs: • Repeat the following pair of actions until there is at most one inverter between : • A circuit input or driving NAND gate output, and • The attached NAND gate inputs.

  31. Enabling Function • Enabling permits an input signal to pass through to an output • Disabling blocks an input signal from passing through to an output, replacing it with a fixed value • The value on the output when it is disable can be Hi-Z (as for three-state buffers and transmission gates), 0 , or 1 • When disabled, 0 output • When disabled, 1 output • See Enabling App in text

  32. 3-7 Decoding D0 D1 : : Dm-1 A0 : : An-1 m-elements ≤ 2n n-2n decoder n bits • A n-bit binary code can represent up to m=2n elements: m elements n-bit binary code • Decoding - the conversion of an n-bit input code to an m-bit output code withn ≤ m ≤ 2n such that each valid code word produces a unique output code encoding (ex. 256 alpha-num. chars) (ex. 8-bit ASCII code) decoding

  33. 2-to-4 Line Decoder circuit D A A = 0 1 0 D A A = 1 1 0 D A A = 2 1 0 D A A = 3 1 0 A 0 A 1 Notice that the outputs of the decoder correspond to the minterms: Di=mi

  34. Decoder Expansion • Larger decoders can be realized by implementing each minterm using a single AND gate: • However for large decoders this requires multiple input AND gates which is not always feasible. • Better to use a hierarchical approach: build larger ones from smaller decoders. • Approach: • Output AND gates have only 2 inputs and implement the minterms. • The output AND gates are driven by two decoders with their numbers of inputs either equal or differing by 1.

  35. Rule for building large decoders • k-to-2k decoder: • One needs 2k output AND gates • If k can be divided by 2: • use two k/2-to-2k/2 decoders • If k cannot divided by 2: • use a (k+1)/2 and • use a (k-1)/2 decoder. • Previous example: 3-to-8 decoder (k=3): • Use a 2-to-4 and a 1-to-2 decoder

  36. Combinational Logic Implementation- Decoder and OR Gates • Implement m functions of n variables with: • Sum-of-minterms expressions • One n-to-2n-line decoder • mOR gates, one for each output

  37. Example A B C F 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 0 1 2 3 4 5 6 7 A B C F • Design and implement a majority function F(ABC) using a 3-to-8 decoder • Truth table: • Minterms: • F=m(3,5,6,7) • Implementation using decoder: Indicate MSB, LSB 2 1 0

  38. Encoding • Typically, an encoder converts a code containing exactly one bit that is 1 to a binary code corresponding to the position in which the 1 appears: ex. D1=1  output 0001 • Examples: Octal-to-Binary encoder • Other examples? 1 0 0 0 0 1 0 0 0 D0 D1 : : Dm-1 0 1 2 3 m-1 0 1 2 n-1 A0 : : An-1 encoder

  39. Priority Encoder • If more than one input value is 1, then the encoder just designed does not work. • An encoder that can accept all possible combinations of input values and produce a meaningful result is a priority encoder. • Among the 1s that appear, it selects the most significant input position (or the least significant input position) containing a 1 and responds with the corresponding binary code for that position. D0 D1 D2 D3 0 1 2 3 1 0 A1 A0 To processor ? V

  40. 3-9 Selecting (multiplexers) I0 I1 I2 I3 In-1 0 1 2 3 : : n-1 OUT • Selecting of data or information is a critical function in digital systems and computers • Circuits that perform selecting have: • A set of n information inputs from which the selection is made • A set of k control (select) lines for making the selection • A single output n ≤ 2k inputs k-1 .. 1 0 Sk-1..S1 S0 k select lines

  41. 4:1 MUX realization S1 S0 OUT 0 0 I0 0 1 I1 1 0 I2 1 1 I3 OUT = S1S0I0+ S1S0I1+ S1S0I2+ S1S0I3 m0 m1 m2 m3 2k-1 or OUT = Σmi Ii i=0 • Expression for OUT • Circuit implementation: SOP • 4 AND gates (4 product terms) • 2-to-4 line decoder (to generate the minterms)

  42. Exercise • Build a 8:1 MUX using two 4:1 and one 2:1 muxes 4:1 I0 I1 I2 I3 0 1 2 3 1 0 0 1 OUT S1 S0 4:1 I4 I5 I6 I7 0 1 2 3 S2 1 0 Ex: S2S1S0=110 : select I6 S1 S0

  43. Multiplexer-based combinational circuitsrealization- Approach 1 4:1 0 1 2 3 F 1 0 Example m0 m1 A B OUT =F 0 0 I0 0 0 1 I11 1 0 I21 1 1 I30 2k-1 m2 F = Σmi Ii m3 i=0 F= Σm(1,2) • A mux can be easily used to implement a function defined by a truth table (lookup table) • Indeed the output F of a mux is equal to: Give the input Ii the value of 0 or 1 as shown in the truth table 0 1 1 0 A B

  44. Combinational Logic Implementation- Multiplexer Approach 2 X • Implement any m functions of n + 1 variables by using: • An m-wide 2n-to-1-line multiplexer • Design: • Find the truth table for the functions. • Based on the values of the first n variables, separate the truth table rows into pairs • For each pair and output, define a rudimentary function of the final variable (0, 1, X, )

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