EET105 Fall 2018 Digital Electronics Charles Rubenstein Ph
EET-105 – Fall 2018 Digital Electronics Charles Rubenstein, Ph. D. Adjunct Professor of Engineering EXAM 1 REVIEW Session 9: MON 10/01/18 /*/ FRI 09/28/18 Mon/Wed: 9: 25 am – 11: 15 am // Wed/Fri: 11: 40 am – 1: 30 pm Lupton Hall 247
EET 105 Class Schedule - 1 Mon/Wed: Wed/Fri: 01 02 5 Sep 31 Aug 10 Sep 5 Sep 12 Sep 7 Sep 12 Sep 19 Sep 14 Sep 24 Sep 19 Sep 21 September 26 September Class Session Topic 1. Syllabus Overview, Decimal and Binary Number Systems 2. Octal and Hex Number Systems; OR, AND, NOT Gates 3. Boolean Algebra; Logic Circuits; Truth Tables; NOR, NAND 4 L 1. Equipment Review; –– IC Pin Diagrams; LAB 1: Logic Gates 5 L 2. LAB 2: Simple Logic Circuits 6 L 3. LAB 3: Logic circuits using NAND/NOR Gates 7. OPEN LAB SESSION 8. EXAM 1 - Wednesday 26 September Closed Book – No Notes – No Calculators Copyright © 2018 C. P. Rubenstein 2
Review of Exam 1 NOTE: This review of Exam 1 will NOT be posted online Copyright © 2018 C. P. Rubenstein 3
Exam 1 : Problem 1 (5 Points) Perform each of the following conversions. Show your work (steps) for full credit. Each of the following 4 exercises is 5 points each. 1. 1110112 to decimal Six bit Binary Number, using 8 -bit Table: 27 = 128 26 = 64 25 = 32 24 = 16 23 = 8 22 = 4 21 = 2 20 = 1 - - 1 1 1 0 1 1 32 + 16 + 8 + 0 + 2 + 1 = 5910 Copyright © 2018 C. P. Rubenstein 4
Exam 1 : Problem 2 (5 Points) Perform each of the following conversions. Show your work (steps) for full credit. Each of the following 4 exercises is 5 points each. 2. 5810 to binary Realizing from Question 1 that 1110112 yields 5910 we can immediately note that 5810 would be one less or 1110102 Alternate method uses the Binary Table noting that 58 is less than 64, thus the first – MSB – is in the 6 th bit (25) place; 58 – 32 = 26; greater than 16 thus the next 1 is in the 5 th bit (24) place; 26 – 16 = 10; greater than 8 thus the next 1 is in the 4 th bit (23) place; 10 – 8 = 2; NOT greater than or equal to 4 so 0 is in the 3 rd bit (22) place; 10 – 8 = 2; greater than (or equal to) 2 thus the next 1 is in the 2 nd bit (21); And since we have an EVEN decimal number, 0 is in the 1 st bit (20) place: This yields the Binary Number 1110102 Copyright © 2018 C. P. Rubenstein 5
Exam 1 : Problem 3 (5 Points) Perform each of the following conversions. Show your work (steps) for full credit. Each of the following 4 exercises is 5 points each. 3. D 01 C 16 to decimal Four bit Hexadecimal Number, using 4 -bit Table: 163 = 162 = 161 = 160 = 4096 256 16 1 D 0 1 C C = 12; D = 13 therefore: D 01 C 16 = (13)(4096) + (0)(256) + (1)(16) + (12)(1) = 53, 248 + 0 + 16 + 12 = 53, 27610 Copyright © 2018 C. P. Rubenstein 6
Exam 1 : Problem 4 (5 Points) Perform each of the following conversions. Show your work (steps) for full credit. Each of the following 4 exercises is 5 points each. 4. AE 1 C 16 to binary A= 10; C = 12; E = 14 therefore: Breaking the Hex into four-bit groups of binary AE 1 C 16 Becomes A E 1 C 23 = 22 = 21 = 20 = 8 4 2 1 1 0 1 1 1 0 0 For the binary equivalent of 1010 1110 0001 11002 Copyright © 2018 C. P. Rubenstein 7
Exam 1 : Problem 5 a (5 Points) 5 a. What is the largest binary value that can be represented using 5 bits? Largest five bit Binary Number is = 111112 Copyright © 2018 C. P. Rubenstein 8
Exam 1 : Problem 5 a (5 Points) 5 b. What is the value in decimal? (of the largest binary value that can be represented using 5 bits) – – five bits – – / 27 = 128 26 = 64 25 = 32 24 = 16 23 = 8 22 = 4 21 = 2 20 = 1 - - - 1 1 1 Thus; 16 + 8 + 4 + 2 + 1 = 3110 Copyright © 2018 C. P. Rubenstein 9
Exam 1 : Problem 6 (10 Points) 6. Write the next 3 consecutive numbers in binary. (Hint: counting sequence. ) 1011 Starting Binary Number = 1011 Next Number (1) = 1011 + 1 = 1100 Next Number (2) = 1100 + 1 = 1101 Next Number (3) = 1101 + 1 = 1110 Copyright © 2018 C. P. Rubenstein 10
Exam 1 : Problem 7 a (5 Points) 7 a. ) Draw the logic circuit to implement the function: Z = Ax (/B+/C) + Bx. C + /Ax. D = A(/B+/C) + BC + /AD using AND, OR and Inverter gates. Find the value of Z for the following input combination: A=B=0, C=D=1 Copyright © 2018 C. P. Rubenstein 11
Exam 1 : Problem 7 b, c (5 Points @) 7 b. ) Construct the Truth Table A B C 0 0 0 (for the function: Z = Ax(/B+/C)+ Bx. C+/Ax. D ) 0 0 0 1 (/B+/C)=1 except when B=C=1 0 0 1 A(/B+/C)=1 0 only when A=1 unless B=C=1 D BC=1 only when B=C=1 /AD = 1 only when A=0, D=1 7 c. ) Find the value of Z for the following input combination: A=B=0 and C=D=1 For 0011 (= 3); Z=1 A*(/B+/C) B*C /A*D Z 1 1 1 0 1 0 1 0 1 1 1 0 0 0 1 1 1 1 0 1 1 1 0 0 1 1 1 0 1 1 1 1 Copyright © 2018 C. P. Rubenstein 1 12
Exam 1 : Problem 8 a (5 Points) 8 a. ) Draw the logic circuit to implement the function: Z = Ax/(B+C)+ Bx. C+/(Ax. D) = A/(B+C)+ BC+/(AD) using AND, OR and Inverter gates. NAND and NOR gates can be used also. Copyright © 2018 C. P. Rubenstein 13
Exam 1 : Problem 8 b, c (5 Points @) 8 b. ) Construct the Truth Table (for the function: Z = Ax/(B+C) + Bx. C + /(Ax. D) /(B+C)=1 when B=C=0 A/(B+C)=1 when A=1 & B=C=0 A B C D A*/(B+C) BC=1 only when B=C=1 /(AD) = 1 except when A=D=1 8 c. ) Find the value of Z for the following input combination: A=B=C=0 and D=1 For 0001 (= 1); Z=1 B*C /(A*D) Z 0 0 1 1 0 0 0 1 1 1 0 0 1 1 0 1 0 0 1 1 1 0 1 1 1 1 0 0 0 1 1 1 0 1 0 1 1 0 0 1 1 1 1 0 1 1 1 Copyright © 2018 C. P. Rubenstein 1 1 1 1 14
Exam 1 : Problem 9 (15 Points) Fill the Truth Table for the function Z= Ax. B+ /C+Bx. C A B C AB 0 0 0 1 1 1 0 0 1 1 1 /C BC Z 1 1 1 1 1 Copyright © 2018 C. P. Rubenstein 1 1 15
Exam 1 : Problem 10 a (5 Points) 10 a. Given the circuit below, find the expression for the output G. Evaluate the Node Outputs: AND the resulting Nodes: G = (A+/B+C) ● (B+C) Copyright © 2018 C. P. Rubenstein 16
Exam 1 : Problem 10 b (10 Points) 10 b. Given the circuit below, write the truth table Truth Table for G = (A+/B+C) ● (B+C) A B C /B 0 0 0 1 1 1 0 0 1 1 1 0 1 A+/B+C 1 1 1 1 (B+C) 1 1 1 Copyright © 2018 C. P. Rubenstein G 0 1 0 1 17
Questions? Copyright © 2018 C. P. Rubenstein 18
Any Questions? Send me an email … c. rubenstein@ieee. org Copyright © 2018 C. P. Rubenstein 19
End Copyright © 2018 C. P. Rubenstein 20
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