ECE 231 - Fall 2023 - HW3 - Solutions

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George Mason University *

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231

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Electrical Engineering

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Dec 6, 2023

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Homework Assignment #3 – Solutions Fall 2023 Homework #3 Solutions R EADING : 1. ZyBook: 2.1 – 2.5, (2.6 – 2.7) 2. ZyBook: 3.1 – 3.5 M ULTIPLE C HOICE AND S HORT - ANSWER Q UESTIONS : [½ point each] 1. A combinational logic circuit is specified by the truth table given below. For three of the input combinations, the output of the circuit does not matter (ie. “don't care” outputs). The circuit that is realized implements the logic function F(A,B,C) = A'.C' + B . What will the output of the circuit for the input combination < A, B, C > = < 1, 1, 0 > ? (a) 0 (b) 1 (c) X (don't care) (d) unknown A B C output 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 X (don't care) 1 1 0 X (don't care) 1 1 1 X (don't care) ECE 231 – Digital System Design 1 of 13 Dr. Craig Lorie
Homework Assignment #3 – Solutions Fall 2023 2. Circle (M) corresponds to what simplified term? (a) a’b’c’ (b) a’bc’ (c) a’c’ (d) a’b’ 3. Which of the following results in the best simplification? Note: Each circle is denoted by parenthesis. (A, B) indicates that a circle includes cells A and B. (a) (m0, m2, m4, m6), (m7, m6), (m0, m1) (b) (m0, m4), (m2, m6), (m7, m6), (m0, m1) (c) (m0, m2, m4, m6), (m1, m7) (d) (m0, m2, m4, m6), (m7), (m1) 4. What is the value of 210 base 6 in decimal? (a) 78 (b) 38 (c) 8 (d) 18 ECE 231 – Digital System Design 2 of 13 Dr. Craig Lorie
Homework Assignment #3 – Solutions Fall 2023 L ONG -A NSWER P ROBLEMS : [Point value specified with each problem] 1. [2] Four-variable Karnaugh Maps (a) Derive the minimum-cost SOP expression for the logic function given below. F(A,B,C,D) = ° m( 2, 3, 5, 7, 11, 12, 13, 14, 15 ) F(A,B,C,D) = A'B'C + AB + CD + BD (b) Derive the min-cost SOP expression for the incompletely specified function given below. F(A,B,C,D) = ° m(0, 2, 8, 9, 10, 15) + ° d(1, 3, 6, 7) F(A,B,C,D) = B'D' + B'C' + BCD ECE 231 – Digital System Design 3 of 13 Dr. Craig Lorie 0 0 1 0 0 1 1 0 1 1 1 1 1 0 1 0 AB 00 01 00 01 11 10 11 10 CD F(A, B, C, D) 0 4 1 5 12 8 13 9 3 7 2 6 15 11 14 10 1 1 X 1 X X 1 1 X 1 AB 00 01 00 01 11 10 11 10 CD F(A, B, C, D) 0 4 1 5 12 8 13 9 3 7 2 6 15 11 14 10
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Homework Assignment #3 – Solutions Fall 2023 2. [2] Four-variable Majority Function. Design a minimum-cost AND-OR circuit to meet the specifications given below: Inputs: A, B, C, D Output: F Description: A four-variable majority function outputs a logic-1 when three or four of its input variables are equal to logic-1; otherwise, it outputs a logic-0. (a) Construct the truth table. Row A B C D F 0 0 0 0 0 0 1 0 0 0 1 0 2 0 0 1 0 0 3 0 0 1 1 0 4 0 1 0 0 0 5 0 1 0 1 0 6 0 1 1 0 0 7 0 1 1 1 1 8 1 0 0 0 0 9 1 0 0 1 0 10 1 0 1 0 0 11 1 0 1 1 1 12 1 1 0 0 0 13 1 1 0 1 1 14 1 1 1 0 1 15 1 1 1 1 1 ECE 231 – Digital System Design 4 of 13 Dr. Craig Lorie
Homework Assignment #3 – Solutions Fall 2023 (b) Use a K-map to derive a simplified SOP expression. F(A,B,C,D) = ABC + ABD + ACD + BCD (c) Draw the corresponding AND-OR circuit. ECE 231 – Digital System Design 5 of 13 Dr. Craig Lorie 0 0 0 0 0 0 1 0 0 1 1 1 0 0 1 0 AB 00 01 00 01 11 10 11 10 CD F(A, B, C, D) 0 4 1 5 12 8 13 9 3 7 2 6 15 11 14 10
Homework Assignment #3 – Solutions Fall 2023 3. [2] Number Conversion: Binary to/from Decimal. Show your work! Assume all numbers are unsigned . (a) Convert the binary number 10101101.01011 2 to decimal. 10101101.01011 = 2 0 + 2 2 + 2 3 + 2 5 + 2 7 + 2 -2 + 2 -4 + 2 -5 = 1 + 4 + 8 + 32 + 128 + 1/4 + 1/16 + 1/32 = 173.34375 10 (b) Convert the binary number 10011010.1101 2 to decimal. 10011010 2 = 2 1 + 2 3 + 2 4 + 2 7 = 2 + 8 + 16 + 128 = 154 10 0.1101 2 = 2 -1 + 2 -2 + 2 -4 = 1/2 + 1/4 + 1/16 = 0.8125 10 (c) Convert the decimal number 308.65 10 to binary. 308 / 2 = 154 0 154 / 2 = 77 0 77 / 2 = 38 1 38 / 2 = 19 0 19 / 2 = 9 1 9 / 2 = 4 1 4 / 2 = 2 0 2 / 2 = 1 0 1 / 2 = 0 1 308 10 = 100110100 2 0.65 * 2 = 1.30 1 0.30 * 2 = 0.60 0 0.60 * 2 = 1.20 1 0.20 * 2 = 0.40 0 0.40 * 2 = 0.80 0 0.80 * 2 = 1.60 1 0.60 * 2 = 1.20 1 0.20 * 2 = 0.40 0 0.40 * 2 = 0.80 0 0.65 10 = 0.1010011 2 308.65 10 = 100110100.1010011 2 ECE 231 – Digital System Design 6 of 13 Dr. Craig Lorie
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Homework Assignment #3 – Solutions Fall 2023 (d) Convert the decimal number 87.1 10 to binary. 87 / 2 = 43 1 43 / 2 = 21 1 21 / 2 = 10 1 10 / 2 = 5 0 5 / 2 = 2 1 2 / 2 = 1 0 1 / 2 = 0 1 87 10 = 1010111 2 87 = 64 + 16 + 4 + 2 + 1 = 2 6 + 2 4 + 2 2 + 2 1 + 2 0 87 10 = 1010111 2 0.1 * 2 = 0.2 0 0.2 * 2 = 0.4 0 0.4 * 2 = 0.8 0 0.8 * 2 = 1.6 1 0.6 * 2 = 1.2 1 0.2 * 2 = 0.4 0 0.4 * 2 = 0.8 0 0.1 10 = 00011 2 87.1 10 = 101011.00011 2 ECE 231 – Digital System Design 7 of 13 Dr. Craig Lorie
Homework Assignment #3 – Solutions Fall 2023 4. [1] Number Conversion: Binary to/from Hexadecimal. Show your work! Assume all numbers are unsigned . (a) Convert the binary number 1010111101.0101111 2 to hexadecimal. 00 10 1011 1101 . 0101 111 0 2 B D 5 E 1010111101.01011110 2 = 2BD.5E 16 (b) Convert the binary number 1101001011001.01011 2 to hexadecimal. 000 1 1010 0101 1001 . 0101 1 000 1 A 5 9 5 8 1101001011001.01011 2 = 1A59.58 16 (c) Convert the hexadecimal number 87E.3D 16 to binary. 8 7 E . 3 D 1000 0111 1110 . 0011 1101 87E.3D 16 = 100001111110.00111101 2 (d) Convert the hexadecimal number F4A.C1 16 to binary. F 4 A . C 1 1111 0100 1010 . 1100 0001 F4A.C1 16 = 111101001010.11000001 2 ECE 231 – Digital System Design 8 of 13 Dr. Craig Lorie
Homework Assignment #3 – Solutions Fall 2023 5. [1] Binary Codes and Parity. Show your work where necessary. Assume all numbers are unsigned . (a) Convert the BCD number 1011000.0111001 BCD to decimal. 0 101 1000 . 0111 001 0 5 8 . 7 2 1011000.0111001 BCD = 58.72 10 (b) Convert the decimal number 592.46 10 to BCD representation. 5 9 2 . 4 6 0101 1001 0010 . 0100 0110 592.46 10 = 010110010010.01000110 BCD (c) Convert the string RObot1 to it’s ASCII representation. Specify in hexadecimal. Since ASCII is a 7-bit code, assume that the MSb (ie. the left-most bit) is zero. R O b o t 1 0x52 0x4F 0x62 0x6F 0x74 0x31 0b01010010 0b01001111 0b01100010 0b01101111 0b01110100 0b00110001 (d) Specify the parity bit for each of the given 7-bit binary numbers. Complete the table given below. 7-bit binary number Parity Bit For even parity For odd parity 1101011 1 0 1010101 0 1 7-bit binary number Parity bit Total Number of Ones Parity 1101011 1 6 Even 1101011 0 5 Odd 1010101 0 4 Even 1010101 1 5 Odd ECE 231 – Digital System Design 9 of 13 Dr. Craig Lorie
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Homework Assignment #3 – Solutions Fall 2023 P RACTICE P ROBLEMS : (Do not submit solutions to these problems) 1. What is the number of the cell highlighted in the K-map below? (a) 7 (b) 6 (c) 10 (d) 13 (e) 9 2. Which binary code is used to identify the rows and columns in a 4-variable Karnaugh map? (a) Binary code (b) Gray code (c) One-hot code (d) ASCII (e) Unicode 3. Which property of Boolean algebra is used to combine cells in a Karnaugh map? (a) Distributive property (b) DeMorgan’s Law (c) Consensus property (d) Combining property (e) None of the above. ECE 231 – Digital System Design 10 of 13 Dr. Craig Lorie
Homework Assignment #3 – Solutions Fall 2023 4. (K) corresponds to which minterm? (a) a’b’c’ (b) abc (c) a’bc (d) a’bc’ 5. Circle (P) corresponds to what simplified term? (a) a’b’ (b) 1 (c) a (d) a’ 6. Which of the following results in the best simplification? Note: Each circle is denoted by parenthesis. (A, B) indicates that a circle includes cells A and B. (a) (m0, m1), (m3, m2), (m5, m7) (b) (m0, m2), (m1, m3, m5, m7) (c) (m0, m1, m3, m2), (m5, m7) (d) (m0, m1, m3, m2), (m1, m3, m5, m7) ECE 231 – Digital System Design 11 of 13 Dr. Craig Lorie
Homework Assignment #3 – Solutions Fall 2023 7. Number Systems (a) Which value is larger, 3.8 10 or 3.8 16 ? Justify your answer. 3.8 10 = 3 x 10 0 + 8 x 10 -1 = 3.8 10 3.8 16 = 3 x 16 0 + 8 x 16 -1 = 3 + 0.5 = 3.5 10 3.8 10 is larger than 3.8 16 . (b) Order the numbers given below from smallest to largest. Justify your answer. 1.101 2 , 1.5 10 , 1.6 16 smallest: 1.6 16 = 1.375 10 1.5 10 largest: 1.101 2 = 1.625 10 8. Number Conversion. Show your work! (a) Convert the hexadecimal number B2E 16 to decimal . B2E 16 = 1011 0010 1110 2 101100101101 2 = 2 1 + 2 2 + 2 3 + 2 5 + 2 8 + 2 9 + 2 11 = 2 + 4 + 8 + 32 + 256 + 512 + 2048 = 2862 10 B2E 16 = 11 x 16 2 + 2 x 16 1 + 14 x 16 0 = 2816 + 32 + 14 = 2862 10 (b) Convert 516 10 to Excess-3 representation. Not applicable. ECE 231 – Digital System Design 12 of 13 Dr. Craig Lorie
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Homework Assignment #3 – Solutions Fall 2023 9. Gray code is a bit encoding for a value sequence where successive values differ by only one bit. What are the values of X, Y, and Z in the given table? Gray Code 0001 0011 0010 X Y Z 0100 0000 (a) X = 0110, Y = 0111, and Z = 0101 (b) X = 1000, Y = 1001, and Z = 1010 (c) X = 1110, Y = 1111, and Z = 1101 (d) X = 1100, Y = 1101, and Z = 1111 10. How is 2048 bits written using the metric prefix? (a) 2 Gb (b) 2 Kb (c) 20 Kb (d) 20 Gb ECE 231 – Digital System Design 13 of 13 Dr. Craig Lorie