Question

(1) Convert the decimal numbers +61 and +27 to their 8-bit 2's complement representations. Then perform the binary equivalent of (b) (27)+ (+61); and Convert the answers back to decimal and verify that they are correct.

Answer 1

If the decimal number is positive, the two’s-complement number is the binary equivalent of the decimal number. If the decimal number is negative, the two's complement number is as follows: • Find l’s complement i.e., complementing each bit of the true binary equivalent of the decimal number. Add 1 to the result of l's complement number to get the magnitude bits. Convert 2710 to binary number. 2 127 2 13 rem. =1=a, 2 16 rem.=1=a, 2 |3 rem. = 0 = a, | 2 |1_ rem. =1=az 0 rem. =1=24 The binary equivalent of decimal number (2710 )is, 2710 = (araza,a,a, ), =110112 The 8-bit representation of decimal number(271.)is, 00011011,

Determine the 2's complement of (270). -2710 0111001002 + 1's complement number + 1 Add 1 to the l's complement number -2710 11100101, → 2's complement number Convert 61,0 to binary number. 2 161 2 30 rem.=1= a. 2 [15 rem.= 0 = a 2 17 rem. =1= a 23 rem. =1= a; 211_ rem. =1= a rem. =1= az 0 The binary equivalent of decimal number(6110 )is, 6110 = (a-a_aza 4a), = 1111012 The 8-bit representation of decimal number(6110 )is, 00111101,

Determine the 2's complement of (610). -6110 0110000102 → l's complement number + 1 Add 1 to the l's complement number -61,0 110000112 → 2's complement number (a) Perform the addition (+27)+(-61) in the 2's complement form. +27 H 00011011 +(-61) >11000011 -34 H 11011110 Here, the sign bit (MSB bit) is 1, indicating a negative number. Verify the above value is correct or not. The 2's complement of (11011110) is, 00100001, → l's complement number + 1 Add 1 to the 1's complement number 001000102 → 2's complement number

Therefore, [(ox 2?)+(0x2°)+(1x 2°)+(0x24) ) 001000102 = | +(0x2°)+(0x22)+(1x2-)+0x2°) = 0+0+32 +0+0+0+2+0 = 34 The equivalent decimal value of (11011110) is, -34. Hence, verified. (b) Perform the addition (-2 1 the 2's complement form. -27 + 11100101 +(+61) + 00111101 34 H 100100010 Discard the carry that is 9h bit of resultant value. 34 H 00100010 Here, the sign bit (MSB (gth-bit)) is 0, indicating a positive number.

001000 Verify the above value is correct or not. on f(0x2?)+(0x 20)+(1x 2°) +(0x 2^) | +(0x2?)+(0x22)+(1x2-)+(0x 2°) = 0+0+32 +0+0+0+2+0 = 34 The equivalent decimal value of (00100010) is, 34. Hence, verified. (c) Perform the addition (-27)+(-61) in the 2’s complement form. -27 H 11100101 +(-61) +11000011 -88 H110101000 Discard the carry that is 9" bit of resultant value. -88 H10101000 Here, the sign bit (MSB (8th-bit)) is 1, indicating a negative number.

Verify the above value is correct or not. The 2’s complement of (10101000) is, 01010111, → l's complement number + 1 Add 1 to the l's complement number 010110002 → 2's complement number Therefore, f(0x2°)+(1x 2°)+(0x2°)+ (1x24) 01011000, 1 + (1x2°)+(0x2?)+ (0x2°) + (0x2°)ſ = 0 +64 +0+16+8+0+0+0 = 88 The equivalent decimal value of (10101000) is, -88. Hence, verified.