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Digital Electronics Tutorial: Combinational Logic Design Examples Solutions ELEC1041 - Combinational Logic Examples - 1 Problem #1 Develop a minimized Boolean implementation of a “ones count” circuit that works as follows. The subsystem has four binary inputs A, B, C, D, and generates a 3-bit output, XYZ. XYZ is 000 if none of the inputs are 1, 001 if one input is 1, 010 if two are one, 011 if three inputs are 1, and 100 if all four inputs are 1. (a) Draw the truth tables for XYZ (A, B, C, D). (b) Minimize the functions X, Y, Z using 4-variable K-maps. Write down the Boolean expressions for the minimized Sum of Products form of each function. (c) Repeat the minimization process, this time deriving Product of Sums form. ELEC1041 - Combinational Logic Examples - 2 Problem #1 Solution (1/6) (a) Draw the truth tables for XYZ (A, B, C, D). ELEC1041 - Combinational Logic Examples - 3 Problem #1 Solution (2/6) (b) Minimize the functions X, Y, Z using 4-variable K-maps. Write down the Boolean expressions for the minimized Sum of Products form of each function. ELEC1041 - Combinational Logic Examples - 4 Problem #1 Solution (3/6) ELEC1041 - Combinational Logic Examples - 5 Problem #1 Solution (4/6) (c) Repeat the minimization process, this time deriving Product of Sums form. ELEC1041 - Combinational Logic Examples - 6 Problem #1 Solution (5/6) ELEC1041 - Combinational Logic Examples - 7 Problem #1 Solution (6/6) ELEC1041 - Combinational Logic Examples - 8 Problem #2 Consider a combinational logic subsystem that performs a two-bit addition function. It has two 2-bit inputs A B and C D, and forms the 3-bit sum X Y Z. (a) Draw the truth tables for XYZ(A,B,C,D). (b) Minimize the functions using 4-variable K-maps to derive minimized Sum of Products forms. (c) In your textbook and in class/lab we have introduced the Full Adder circuit. What is the relative performance to compute the resulting sum bits of the 2-bit adder compared to two full adders connected together? (Hint: which has the worst delay in terms of gates to pass through between the inputs and the final outputs, and how many gates is this?). ELEC1041 - Combinational Logic Examples - 9 Problem #2 Solution (1/3) (a) Draw the truth tables for XYZ(A,B,C,D). 0 1 2 3 ELEC1041 - Combinational Logic Examples - 10 00 01 10 11 Problem #2 Solution (2/3) (b) Minimize the functions using 4-variable K-maps to derive minimized Sum of Products forms. ELEC1041 - Combinational Logic Examples - 11 Problem #2 Solution (3/3) (c) In your textbook and in class/lab we have introduced the Full Adder circuit. What is the relative performance to compute the resulting sum bits of the 2-bit adder compared to two full adders connected together? (Hint: which has the worst delay in terms of gates to pass through between the inputs and the final outputs, and how many gates is this?). The above circuit is better because there are only two gate levels. But the adder circuit can be used to build adder circuit of more bits. I can’t imagine doing a K-map to construct a adder circuit of 64 bits using the method above. ELEC1041 - Combinational Logic Examples - 12 Problem #3 Show how to implement the full adder Sum(A, B, Cin) and Carry(A, B, Cin) in terms of: (a) Two 8:1 multiplexers; (b) Two 4:1 multiplexers; (c) If you are limited to 2:1 multiplexers (and inverters) only, how would you use them to implement the full adder and how many 2:1 multiplexers would you need? ELEC1041 - Combinational Logic Examples - 13 Problem #3 Solution (1/8) Ai 0 0 0 0 1 1 1 1 Bi 0 0 1 1 0 0 1 1 Cin 0 1 0 1 0 1 0 1 Sum 0 1 1 0 1 0 0 1 Cout 0 0 0 1 0 1 1 1 ELEC1041 - Combinational Logic Examples - 14 Problem #3 Solution (2/8) (a) Two 8:1 multiplexers (Sum); Ai 0 0 0 0 1 1 1 1 Bi 0 0 1 1 0 0 1 1 Cin 0 1 0 1 0 1 0 1 ELEC1041 - Combinational Logic Examples - 15 Sum 0 1 1 0 1 0 0 1 Problem #3 Solution (3/8) (a) Two 8:1 multiplexers (Carry); Ai 0 0 0 0 1 1 1 1 Bi 0 0 1 1 0 0 1 1 Cin 0 1 0 1 0 1 0 1 ELEC1041 - Combinational Logic Examples - 16 Cout 0 0 0 1 0 1 1 1 Problem #3 Solution (4/8) (b) Two 4:1 multiplexers (Sum); Ai 0 0 0 0 1 1 1 1 Bi 0 0 1 1 0 0 1 1 ELEC1041 - Combinational Logic Examples - 17 Cin 0 1 0 1 0 1 0 1 Sum 0 1 1 0 1 0 0 1 Problem #3 Solution (5/8) (b) Two 4:1 multiplexers (Carry); Ai 0 0 0 0 1 1 1 1 ELEC1041 - Combinational Logic Examples - 18 Bi 0 0 1 1 0 0 1 1 Cin 0 1 0 1 0 1 0 1 Cout 0 0 0 1 0 1 1 1 Problem #3 Solution (6/8) (c) 2:1 multiplexers (sum) Ai 0 0 0 0 1 1 1 1 ELEC1041 - Combinational Logic Examples - 19 Bi 0 0 1 1 0 0 1 1 Cin 0 1 0 1 0 1 0 1 Sum 0 1 1 0 1 0 0 1 Problem #3 Solution (7/8) (c) 2:1 multiplexers (Carry) Ai 0 0 0 0 1 1 1 1 ELEC1041 - Combinational Logic Examples - 20 Bi 0 0 1 1 0 0 1 1 Cin 0 1 0 1 0 1 0 1 Cout 0 0 0 1 0 1 1 1 Problem #3 Solution (8/8) (c) 2:1 multiplexers (Sum) Ver 2 - USE the same carry Ai 0 0 0 0 1 1 1 1 Bi 0 0 1 1 0 0 1 1 Cin 0 1 0 1 0 1 0 1 Sum 0 1 1 0 1 0 0 1 ELEC1041 - Combinational Logic Examples - 21 Problem #4 Design a combinational logic subsystem with three inputs, I3, I2, I1, and two outputs, O1, O0, that behaves as follows. The outputs indicate the highest index of the inputs that is driven high. For example, if I3 is 0, I2 is 1, I1 is 1, then O1, O0 would be 10 (i.e., I2 is the highest input set to 1). (a) Specify the function by filling out a complete truth table. (b) Develop the minimized gate-level implementation using the K-map method. (c) Develop an implementation using two 4:1 multiplexers. (d) Compare your implementation for (b) and (c). Which is better and under what criterion? ELEC1041 - Combinational Logic Examples - 22 Problem #4 Solution (1/5) (a) Specify the function by filling out a complete truth table. I3 I2 I1 O1 O0 0 0 0 0 0 0 0 1 0 1 0 1 x 1 0 1 x x 1 1 ELEC1041 - Combinational Logic Examples - 23 Problem #4 Solution (2/5) (b) Develop the minimized gate-level implementation using the K-map method. I3 I2 I1 O1 O0 0 0 0 0 0 0 0 1 0 1 0 1 x 1 0 1 x x 1 1 ELEC1041 - Combinational Logic Examples - 24 Problem #4 Solution (3/5) (c) Develop an implementation using two 4:1 multiplexers. I3 I2 I1 O1 O0 0 0 0 0 0 0 0 1 0 1 0 1 x 1 0 1 x x 1 1 ELEC1041 - Combinational Logic Examples - 25 Problem #4 Solution (4/5) (c) Develop an implementation using two 4:1 multiplexers. I3 I2 I1 O1 O0 0 0 0 0 0 0 0 1 0 1 0 1 x 1 0 1 x x 1 1 ELEC1041 - Combinational Logic Examples - 26 Problem #4 Solution (5/5) (d) Compare your implementation for (b) and (c). Which is better and under what criterion? ELEC1041 - Combinational Logic Examples - 27 Problem #5 You are to design a simple combinational subsystem to the following specification. The system has the ability to pass its inputs directly to its outputs when a control input, S, is not asserted. It interchanges its inputs when the control input S is asserted. For example, given four inputs A, B, C, D and four outputs W, X, Y, Z, when S=0, WXYZ=ABCD and when S=1, WXYZ=BCDA. Show how to implement this functionality using building blocks that are restricted to be 2:1 multiplexers and 2:1 demultiplexers. Draw your solution below, using black boxes for the mux/demux blocks. A B C D W X Y Z S ELEC1041 - Combinational Logic Examples - 28 Problem #5 Solution Each output is tied to the output of a 2:1 Multiplexer. S is the select signal. A W B X C Y D Z S ELEC1041 - Combinational Logic Examples - 29 Problem #6 Your task is to design a combinational logic subsystem to decode a hexadecimal digit in the range of 0 through 9, A through F to drive a seven-segment display. The hexadecimal numerals are as follows: Design a minimized implementation in PLA form. That is, look for common terms among the seven output functions. ELEC1041 - Combinational Logic Examples - 30 Problem #6 Solution (1/5) c1 c4 c0 c3 c2 c5 c6 c0 c1 c2 c3 c4 c5 c6 BCD to 7–segment control signal decoder A B C D ELEC1041 - Combinational Logic Examples - 31 Problem #6 Solution (2/5) c1 c4 c0 c3 c2 c5 c6 c0 c1 c2 c3 c4 c5 c6 BCD to 7–segment control signal decoder A B C D ELEC1041 - Combinational Logic Examples - 32 A B C D Problem #6 Solution (3/5) BCD BCD ACD BCD ABD BCD ACD ABC ABC ABD ABC ABC ABC ABD ABD ACD ACD ABC F0 F1 F2 F3 F4 F5 F6 ELEC1041 - Combinational Logic Examples - 33 A B C D Problem #6 Solution (4/5) BCD BCD ACD BCD ABD BCD ACD ABC ABC ABD ABC ABC ABC ABD ABD ACD ACD ABC 18 terms F0 F1 F2 F3 F4 F5 F6 ELEC1041 - Combinational Logic Examples - 34 A B C D Problem #6 Solution (5/5) ABCD ABCD ABCD ABCD ABCD ABCD ABCD ABCD ABCD ABCD ABCD ABCD ABCD ABCD ABCD ABCD F0 F1 F2 F3 F4 F5 F6 ELEC1041 - Combinational Logic Examples - 35 16 terms Problem #7 Calendar subsystem Determine number of days in a month (to control watch display) used in controlling the display of a wrist-watch LCD screen inputs: month, leap year flag outputs: number of days ELEC1041 - Combinational Logic Examples - 36 Problem #7 Solution (1/3) Calendar subsystem Use software implementation to help understand the problem integer number_of_days ( month, leap_year_flag) { switch (month) { case 1: return (31); case 2: if (leap_year_flag == 1) then return (29) else return (28); case 3: return (31); case 4: return (30); case 5: return (31); case 6: return (30); case 7: return (31); case 8: return (31); case 9: return (30); case 10: return (31); case 11: return (30); case 12: return (31); default: return (0); } } ELEC1041 - Combinational Logic Examples - 37 Problem #7 Solution (2/3) Formalize the problem Encoding: binary number for month: 4 bits 4 wires for 28, 29, 30, and 31 one-hot – only one true at any time Block diagram: month leap 28 29 30 31 month 0000 0001 0010 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 111– ELEC1041 - Combinational Logic Examples - 38 leap – – 0 1 – – – – – – – – – – – – 28 – 0 1 0 0 0 0 0 0 0 0 0 0 0 – – 29 – 0 0 1 0 0 0 0 0 0 0 0 0 0 – – 30 – 0 0 0 0 1 0 1 0 0 1 0 1 0 – – 31 – 1 0 0 1 0 1 0 1 1 0 1 0 1 – – Problem #7 Solution (3/3) Choose implementation target and perform mapping Discrete gates 28 = m8’ m4’ m2 m1’ leap’ 29 = m8’ m4’ m2 m1’ leap 30 = m8’ m4 m1’ + m8 m1 31 = m8’ m1 + m8 m1’ Can translate to S-o-P or P-o-S month 0000 0001 0010 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 111– leap – – 0 1 – – – – – – – – – – – – ELEC1041 - Combinational Logic Examples - 39 28 – 0 1 0 0 0 0 0 0 0 0 0 0 0 – – 29 – 0 0 1 0 0 0 0 0 0 0 0 0 0 – – 30 – 0 0 0 0 1 0 1 0 0 1 0 1 0 – – 31 – 1 0 0 1 0 1 0 1 1 0 1 0 1 – – Problem #8 Leap year flag Determine value of leap year flag given the year For years after 1582 (Gregorian calendar reformation), leap years are all the years divisible by 4, except that years divisible by 100 are not leap years, but years divisible by 400 are leap years. ELEC1041 - Combinational Logic Examples - 40 Problem #8 Solution (1/5) Leap year flag Encoding the year: binary – easy for divisible by 4, but difficult for 100 and 400 (not powers of 2) BCD – easy for 100, but more difficult for 4, what about 400? Parts: construct a circuit that determines if the year is divisible by construct a circuit that determines if the year is divisible by construct a circuit that determines if the year is divisible by combine the results of the previous three steps to yield the ELEC1041 - Combinational Logic Examples - 41 4 100 400 leap year flag Problem #8 Solution (2/5) Divisible-by-4 circuit BCD coded year YM8 YM4 YM2 YM1 – YH8 YH4 YH2 YH1 – YT8 YT4 YT2 YT1 – YO8 YO4 YO2 YO1 Only need to look at low-order two digits of the year all years ending in 00, 04, 08, 12, 16, 20, etc. are divisible by 4 if tens digit is even, then divisible by 4 if ones digit is 0, 4, or 8 if tens digit is odd, then divisible by 4 if the ones digit is 2 or 6. Translates into the following Boolean expression (where YT1 is the year's tens digit low-order bit, YO8 is the high-order bit of year's ones digit, etc.): YT1’ (YO8’ YO4’ YO2’ YO1’ + YO8’ YO4 YO2’ YO1’ + YO8 YO4’ YO2’ YO1’ ) + YT1 (YO8’ YO4’ YO2 YO1’ + YO8’ YO4 YO2 YO1’ ) Digits with values of 10 to 15 will never occur, simplify further to yield: YT1’ YO2’ YO1’ + YT1 YO2 YO1’ ELEC1041 - Combinational Logic Examples - 42 Problem #8 Solution (3/5) Divisible-by-100 and divisible-by-400 circuits Divisible-by-100 just requires checking that all bits of two low-order digits are all 0: YT8’ YT4’ YT2’ YT1’ • YO8’ YO4’ YO2’ YO1’ Divisible-by-400 combines the divisible-by-4 (applied to the thousands and hundreds digits) and divisible-by-100 circuits (YM1’ YH2’ YH1’ + YM1 YH2 YH1’) • (YT8’ YT4’ YT2’ YT1’ • YO8’ YO4’ YO2’ YO1’ ) ELEC1041 - Combinational Logic Examples - 43 Problem #8 Solution (4/5) Combining to determine leap year flag Label results of previous three circuits: D4, D100, and D400 leap_year_flag = D4 (D100 • D400’ ) ’ = D4 • D100’ + D4 • D400 = D4 • D100’ + D400 ELEC1041 - Combinational Logic Examples - 44 Problem #8 Solution (5/5) Implementation of leap year flag ELEC1041 - Combinational Logic Examples - 45 Problem #9 Consider the following functions, which are five different functions over the inputs A,B,C,D. (1) F(A,B,C) = m(1,2,6,7) (2) (3) (4) (5) F(A,B,C,D) = m(0,1,3,9,11,12,14,15) F’(A,B,C,D) = m(2,4,5,6,7,8,10,13) F(A,B,C,D) = (A B C + A’ B’) (C + D) F(A,B,C,D) = (A + B + C) (A + B + C’ + D) (A + B’ + C + D’) (A’ + B’) a) you were to implement these in a single PLA structure with four inputs, five outputs, and an unlimited number of product terms, how many unique product terms are there in this PLA implementation b) If you are trying to maximize the number of shared product terms across the five functions, rather than minimizing the literal count for each function independently, how many unique terms do you obtain? Draw the new Kmaps with your selection of implicants that minimizes the number of unique terms across all five functions ELEC1041 - Combinational Logic Examples - 46 Problem #9 Solution (1/2) Consider the following functions, which are five different functions over the inputs A,B,C,D. (1) F(A,B,C) = m(1,2,6,7) (2) F(A,B,C,D) = m(0,1,3,9,11,12,14,15) (3) F’(A,B,C,D) = m(2,4,5,6,7,8,10,13) (4) F(A,B,C,D) = (A B C + A’ B’) (C + D) (5) F(A,B,C,D) = (A’ + B + C) (A + B + C’ + D) (A + B’ + C + D’) (A’ + B’) a) you were to implement these in a single PLA structure with four inputs, five outputs, and an unlimited number of product terms, how many unique product terms are there in this PLA implementation F(A,B,C)= BC’+AB+A’B’C F(A,B,C,D)=ABD’+ACD+B’D+A’B’C’ F’(A,B,C,D)=A’B+BC’D+AB’D’+B’CD’ F(A,B,C,D)=A’B’D+A’B’C+ABC F(A,B,C,D)=A’C’D’+A’B’D+A’BC+AB’C From Revision Tutorial 17 unique terms ELEC1041 - Combinational Logic Examples - 47 Problem #9 Solution (2/2) b) If you are trying to maximize the number of shared product terms across the five functions, rather than minimizing the literal count for each function independently, how many unique terms do you obtain? Draw the new K-maps with your selection of implicants that minimizes the number of unique terms across all five functions 1. BC’ + AB + A’B’C => ABD' + ABD + A'BC' + ABC' + A'B'C 2. ABD’ + ACD + B’D + A’B’C’ => ABD' + ACD + AB'D + A'B'D + B'CD + B'C'D + A'B'C' 3. A’B + BC’D + AB’D’ + B’CD’ => A'BC + A'BC' + A'BD + A'BD' + BC'D + AB'D' + B'CD' 4. A’B’D + A’B’C + ABC => A’B’D + A’B’C + ABC 5. A’C’D’+A’B’D+A’BC+AB’C => A’C’D’+A’B’D+A’BC+AB’C shared terms: A'B'C is shared between 1 and 4. A’BC’ is shared between 1 and 3. ABD' is shared between 1 and 2. A’B’D is shared between 2, 4 and 5. A’BC is shared between 3 and 5. 5 shared terms +[22 – (4 2) – (1 3)] => 11 unique terms across all five functions NOTE: PLA Minimization is always carried out by CAD Tools ELEC1041 - Combinational Logic Examples - 48 Problem #10 Consider the following Boolean function in Product of Sums form: F(A,B,C,D) = (A + B’ + D) (A’ + B’ + D) (B’ + C’ + D’) (A’ + C + D) (A’ + C’ + D) Show how to implement this function with an 8:1 multiplexer, with A,B,C on the control inputs and D, its complement, and the constants 0 and 1 available as data inputs. Show your work: 0 1 2 3 F(A,B,C,D) 4 5 6 7 S2 S1 S0 A B C ELEC1041 - Combinational Logic Examples - 49 Problem #10 Solution (1/2) F(A,B,C,D) = (A + B’ + D) (A’ + B’ + D) (B’ + C’ + D’) (A’ + C + D) (A’ + C’ + D) 0 1 2 3 F(A,B,C,D) 4 5 6 7 S2 S1 S0 A B C ELEC1041 - Combinational Logic Examples - 50 Problem #10 Solution (2/2) F(A,B,C,D) = (A + B’ + D) (A’ + B’ + D) (B’ + C’ + D’) (A’ + C + D) (A’ + C’ + D) 1 1 D 0 D D D ) 0 1 2 3 4 5 6 7 F(A,B,C,D) S2 S1 S0 A B C ELEC1041 - Combinational Logic Examples - 51 Problem #11 Consider a variation on the calendar combinational subsystem that works as follows. Given the inputs MONTH (1-12), DAY (1-31), and LEAP_YEAR flag, the subsystem generates the output DAY_OF_YEAR (1-365 or 366). In this problem, you will design the subsystem to the block diagram level only. (a) One block maps the month into a day offset into the year. Identify the inputs and outputs and their bit widths. Use any formal specification method you wish (e.g., truth tables, ROM contents, equations, hardware description language, etc.) to describe the function of this block. (b) You may assume any width binary adder you may require. Indicate how the adder is composed with the block of part (a) and any other blocks or inputs to compute the correct output. Be sure to describe how you deal with the LEAP_YEAR input ELEC1041 - Combinational Logic Examples - 52 Problem #11 Solution (1/3) l (a) One block maps the month into a day offset into the year. Identify the inputs and outputs and their bit widths. Use any formal specification method you wish (e.g., truth tables, ROM contents, equations, hardware description language, etc.) to describe the function of this block Day m4 m3 m2 m1 m0 d8 d7 d6 d5 d4 d3 d2 d1 d0 Offset x 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 x 0 0 0 1 0 0 0 0 0 1 1 1 1 1 31 0 0 0 0 1 1 0 0 0 1 1 1 0 1 1 59 1 0 0 0 1 1 0 0 0 1 1 1 1 0 0 60 0 0 0 1 0 0 0 0 1 0 1 1 0 1 0 90 1 0 0 1 0 0 0 0 1 0 1 1 0 1 1 91 ----- ELEC1041 - Combinational Logic Examples - 53 Problem #11 Solution (2/3) (a) One block maps the month into a day offset into the year. Identify the inputs and outputs and their bit widths. Use any formal specification method you wish (e.g., truth tables, ROM contents, equations, hardware description language, etc.) to describe the function of this block m4 m3 m2 m1 m0 l ROM 26 x 10 d8 d7 d6 d5 d4 d3 d2 d1 d0 ELEC1041 - Combinational Logic Examples - 54 Problem #11 Solution (3/3) (b) You may assume any width binary adder you may require. Indicate how the adder is composed with the block of part (a) and any other blocks or inputs to compute the correct output. Be sure to describe how you deal with the LEAP_YEAR input Since leap has been taken care of in part a, you can just add whatever output from that block with day, which will give you the correct day of the year. m4 m3 m2 m1 m0 l ROM 26 x 10 day8 – day5 are always 0 day8 day7 day6 day5 day4 day3 day2 day1 day0 d8 d7 d6 d5 d4 d3 d2 d1 d0 d8-d0 day8 – day0 ELEC1041 - Combinational Logic Examples - 55 9-bit Adder d8 d7 d6 d5 d4 d3 d2 d1 d0 Problem #12 Design a two-bit comparator with the following inputs and outputs: Inputs: Numbers N1 and N2 to be compared N1 = A B N2 = C D Outputs: LT, GT, EQ LT = 1 when AB < CD example (A B = 01) < (C D = 10) GT = 1 when AB > CD example (A B = 11) > (C D = 10) EQ = 1 when AB = CD example (A B = 01) = (C D = 01) ELEC1041 - Combinational Logic Examples - 56 Problem #12 Solution (1/3) A B 0 0 N1 N2 A B C D LT EQ GT AB<CD AB=CD AB>CD block diagram and truth table 0 1 1 0 1 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 LT 0 1 1 1 0 0 1 1 0 0 0 1 0 0 0 0 EQ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 GT 0 0 0 0 1 0 0 0 1 1 0 0 1 1 1 0 we'll need a 4-variable Karnaugh map for each of the 3 output functions ELEC1041 - Combinational Logic Examples - 57 Problem #12 Solution (2/3) A C A 0 0 0 0 1 0 0 0 1 1 0 1 1 1 0 0 D C A 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 D C 0 1 1 1 0 0 1 1 0 0 0 0 0 0 1 0 B B B K-map for LT K-map for EQ K-map for GT D LT = A' B' D + A' C + B' C D EQ = A'B'C'D' + A'BC'D + ABCD + AB'CD’ = (A xnor C) • (B xnor D) GT = B C' D' + A C' + A B D' LT and GT are similar (flip A/C and B/D) ELEC1041 - Combinational Logic Examples - 58 Problem #12 Solution (3/3) A B C D two alternative implementations of EQ with and without XOR EQ EQ XNOR is implemented with at least 3 simple gates ELEC1041 - Combinational Logic Examples - 59 Problem #13 Design a 2X2 bit multiplier: Inputs: Numbers N1 and N2 to be multiplied N1 = A1 A0 N2 = B1 B0 Outputs: products: P8, P4, P2, P0 P0 = Product with weighting 20 = 1 P2 = Product with weighting 21 = 2 P4 = Product with weighting 22 = 4 P8 = Product with weighting 23 = 8 Example: 11 11 11 11 1001 P8 P4 P2 P0 ELEC1041 - Combinational Logic Examples - 60 A1A0 B1B0 B0(AA0) B1(AA0) Problem #13 Solution (1/2) A1 A2 B1 B2 P1 P2 P4 P8 block diagram and truth table A2 A1 B2 0 0 0 0 1 1 0 1 0 0 1 1 1 0 0 0 1 1 1 1 0 0 1 1 B1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 P8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 P4 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 P2 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 4-variable K-map for each of the 4 output functions ELEC1041 - Combinational Logic Examples - 61 P1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 Problem #13 Solution (2/2) A2 B2 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 K-map for P8 B1 K-map for P4 P4 = A2B2B1' + A2A1'B2 P8 = A2A1B2B1 B2 A2 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 A1 A1 A2 B2 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 A1 B1 K-map for P2 K-map for P1 P1 = A1B1 B1 P2 = A2'A1B2 + A1B2B1' + A2B2'B1 + A2A1'B1 B2 ELEC1041 - Combinational Logic Examples - 62 A2 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 A1 B1 Problem #14 Design a BCD increment by 1: Inputs: BCD Number N N = I8 I4 I2 I0 Outputs: products: O8, O4, O2, O0 O8 O4 O2 O0 = I8 I4 I2 I0 + 1 Example: 0011 0100 O8 O4 O2 O0 ELEC1041 - Combinational Logic Examples - 63 I8 I4 I2 I0 O8 O4 O2 O0 Problem #14 Solution (1/2) I1 I2 I4 I8 O1 O2 O4 O8 block diagram and truth table I8 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 I4 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 I2 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 I1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 O8 0 0 0 0 0 0 0 1 1 0 X X X X X X O4 0 0 0 1 1 1 1 0 0 0 X X X X X X O2 0 1 1 0 0 1 1 0 0 0 X X X X X X 4-variable K-map for each of the 4 output functions ELEC1041 - Combinational Logic Examples - 64 O1 1 0 1 0 1 0 1 0 1 0 X X X X X X Problem #14 Solution (2/2) I8 I2 0 0 X 1 0 0 X 0 0 1 X X 0 0 X X O8 I1 I2 0 0 X 0 1 1 X 0 0 0 X X 1 1 X X O8 = I4 I2 I1 + I8 I1' O4 = I4 I2' + I4 I1' + I4’ I2 I1I2 O2 = I8’ I2’ I1 + I2 I1' I4 I8 O4 O2 O1 = I1' 0 1 X 0 0 1 X 0 1 0 X X 0 1 X X O1 I2 I8 1 1 X 1 0 0 X 0 0 0 X X 1 1 X X I4 ELEC1041 - Combinational Logic Examples - 65 I1 I4 I1 I4 I8 I1 Problem #15 Analyse the behavior of the Circuit below when Input A changes from one logic state to another. A B C D F ELEC1041 - Combinational Logic Examples - 66 Problem #15 Solutions (1/1) pulse shaping circuit (Momentary Changes in outputs) A' xnor A = 0 delays matter in function 1 0A d 0 d 1B d 1 0 0 1D C 0 0 d 1 F 1 D remains high for three gate delays after A changes from low to high F is not always 0 pulse 3 gate-delays wide D remains low for three gate delays after A changes from high to low ELEC1041 - Combinational Logic Examples - 67 3d 0 Problem #16 Analyse the circuit below for static hazard A S B S' ELEC1041 - Combinational Logic Examples - 68 Problem #16 Solution (1/1) Due to a literal and its complement momentarily taking on the same value through different paths with different delays and reconverging we get Static 1 Hazard A S 1 1 1 0 B F 0 0 B d A 1 d 1 2d 0 1 1 2d S 0 3d 1 S' F S' Static-1 hazard static-0 hazard static-1 hazard ELEC1041 - Combinational Logic Examples - 69 Problem #17 Analyse the pulse shaping circuit below + open switch A B C D ELEC1041 - Combinational Logic Examples - 70 Problem #17 Solution (1/1) An Oscillator + open switch U = Undefined A 0 1 C U 1 2d close switch initially undefined open switch ELEC1041 - Combinational Logic Examples - 71 BU d 0 D U 3d 0 Problem #18 Which of the components below cab be used to build an inverter? ELEC1041 - Combinational Logic Examples - 72 Problem #18 Solution (1/1) 1 A’ A A nand 1 = A’ 0 A’ A A nor 0 = A’ 1 A’ A A xor 1 = A’ 0 A’ A A xnor 0 = A’ ELEC1041 - Combinational Logic Examples - 73 Problem #19 Consider the Equation: Z= 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 Use Shanon’s decomposition to implement this using 2-1 multiplexers. ELEC1041 - Combinational Logic Examples - 74 Problem #19 Solution (1/3) Z= 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 Z = A’ [B (C’ D’ + C D) + B’ (C’ D + C D’)] + A [B (C’ D + C D’) + B’ (C’ D’ + C D)] = A’ [B (C xor D)’ + B’ (C xor D)] + A [B (C xor D) + B’ (C xor D)’] = A’ [( B xor C xor D)] + A [( B xor C xor D)’] = A xor B xor C xor D ELEC1041 - Combinational Logic Examples - 75 Problem #19 Solution (2/3) Z= A xor B xor C xor D Decompose around A: Z = A’ (B xor C xor D) + A (B xor C xor D)’ ( B xor C xor D) Z Now Decompose (B xor C xor D) around B: B xor C xor D = B’ (C xor D) + B (C xor D)’ Z C xor D ELEC1041 - Combinational Logic Examples - 76 Problem #19 Solution (3/3) Now Decompose (C xor D) around C: C xor D = C’ D + B D’ D Z ELEC1041 - Combinational Logic Examples - 77