[PDF] ECE-223 Solutions for Assignment #1





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ECE-223 Solutions for Assignment #1

(198)12 = 1*122 + 9*12 + 8*120 = 144 + 108 + 8 = 260 1.9) Convert the hexadecimal number 68BE to binary and then from binary convert it to octal.



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[PDF] A Binary – to – Hexadecimal conversion: 4-bit groups starting at the

24 mar 2020 · C Hexadecimal – to –Decimal Conversion:By two method * First method: Example (19): (A85)16====(2693)10 1- Convert to binary number

  • How to convert a hexadecimal to binary in C?

    To convert (213AFE)H to BCD, first it has to be converted to binary which gives (2177790)D . Now each digit is converted to its BCD code which gives (0010 0001 0111 0111 0111 1001 0000)BCD . The final result in the processes above are different.

ECE-223, Solutions for Assignment #1

Chapter 1, Digital Design, M. Mano, 3rd Edition

1.4) Convert the following numbers with the indicated bases to decimal: (4310)

5, and (198)12.

(4310)

5 = 4*53 + 3*52 +51+0 = 500 + 75 + 5 = 580

(198)

12 = 1*122 + 9*12 + 8*120 = 144 + 108 + 8 = 260

1.7) Express the following numbers in decimal: (10110.0101)

2, (16.5)16, and (26.24)8

(10110.0101)

2 = 16 + 4 + 2 + 1/4 + 1/16 = 22+ 5/16 = 22.3125

(16.5)

16 = 16 + 6+ 5/16 = 22 + 5/16 = 22.3125

(26.24)

8 = 16+ 6 + 2/8 + 4/64 = 20 + 20/64 = 22 + 5/16 = 22.3125

1.8) Convert the following numbers to hexadecimal and to decimal

a) 1.11010 b) 1110.10 Explain why the decimal answer in (b) is 8 times that of (a). (a) 1.11010 = 0001.1101 = 1.D = 1+ 13/16 = 1.8125 (b) 1110.10 = 1110.1000 = E.8 = 14+8/16 = 14.5 (14.5 = 8 * 1.8125, Reason is that 1110.10 is the same as 1.11010 shifted left 3 times; i.e. multiplies by 8.)

1.9) Convert the hexadecimal number 68BE to binary and then from binary convert it to octal.

68BE = 0110 1000 1011 1110 = 110 100 010 111 110 = 64276 (octal)

1.10) Convert the decimal number 345 to binary in two ways: (a) convert directly to binary; (b)

convert first to hexadecimal, then from hexadecimal to binary, Which method is faster? Results of repeated division by 2: quotients are followed by reminders: a) 345/2 = 172 (1); 86(0); 43(0); 21(1); 10(1); 5(0); 2(1); 1(0); 0(1)

Answer 101011001.

b) Results of repeated by division by 16(much shorter):

345/16 = 21(9); 1(5); 0(1) answer = 159 (hex) = 0001 0101 1001 = 101011001 (bin)

1.16) Obtain the 1's and 2's complements of the following binary numbers:

a) 11101010 b) 01111110 c) 00000001 d) 10000000 e) 00000000 a) 1' comp. = 00010101; 2's comp. = 00010110 b) 1' comp. = 10000001; 2's comp. = 10000010 c) 1' comp. = 11111110; 2's comp. = 11111111 d) 1' comp. = 01111111; 2's comp. = 10000000 e) 1' comp. = 11111111; 2's comp. = 00000000

1.18) Perform subtraction on the following unsigned binary number using 2's-complement of the

subtrahend. Where the result should be negative, 2's complement it and affix a minus sign. a) 11011 - 11001 b) 110100 - 10101 c) 1011 - 110000 d) 101010 - 101011 a) 11011 - 11001 = 11011 + 00111 = 00010 (27 -25 = 2) b) 110100 - 10101 = 110100 + 101011 = 011111 (52 -21 = 31) c) 1011 - 110000 = 011011 -> 100101 (11 -48= -37) (No Carry implies negative.) d) 101010 - 101011 = 111111 -> - 000001 (42 -43 = -1)

1.24) Represent decimal number 6027 in (a) BCD (b) excess-3 code, (c) 2421 code.

6 0 2 7

a) BCD 0110 0000 0010 0111 b) EXCESS-3 1001 0011 0101 1010 c) 2421 1100 0000 0010 1101quotesdbs_dbs14.pdfusesText_20
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