Digital Electronics: Principles, Devices and Applications

66 Digital Electronicsquotient is given by the division of the two mantissas (i.e. dividend mantissa divided by divisormantissa) and the exponent of the quotient is given by subtraction of the two exponents (i.e. dividendexponent minus divisor exponent).IfthenN 1 = m 1 × 2 e1 and N 2 = m 2 × 2 e2N 1 × N 2 = m 1 × m 2 × 2 e1+e2andN 1 /N 2 = m 1 /m 2 × 2 e1−e2Again, post-normalization may be required after multiplication or division, as in the case of additionand subtraction operations.Example 3.12Add (a) (39) 10 and (19) 10 and (b) (1E) 16 and (F3) 16 using floating-point numbers. Verify the answersby performing equivalent decimal addition.Solution(a) (39) 10 = 100111 = 0.100111 × 2 6 .(19) 10 = 10011 = 0.10011 × 2 5 = 0.010011 × 2 6 .Therefore, (39) 10 + (19) 10 = 0.100111 × 2 6 + 0.010011 × 2 6= (0.100111 + 0.010011) × 2 6 = 0.111010 × 2 6= 111010 = (58) 10and hence is verified.(b) (1E 16 = (00011110) 2 = 0.00011110 × 2 8 .(F3) 16 = (11110011) 2 = 0.11110011 × 2 8 .(1E 16 + F3) 16 = (0.00011110 + 0.11110011) × 2 8 = 100010001= 000100010001= (111) 16 .Also, (1E 16 + (F3) 16 = (111) 16 and hence is proved.Example 3.13Subtract (17) 8 from (21) 8 using floating-point numbers and verify the answer.Solution• (21) 8 = (010001) 2 = 0.010001 × 2 6 .• (17) 8 = (001111) 2 = 0.001111 × 2 6 .• Therefore, (21) 8 − (17) 8 = (0.010001 − 0.001111) × 2 6= 0.000010 × 2 6 = 000010 = (02) 8 .• Also, (21) 8 − (17) 8 = (02) 8 and hence is verified.