Digital Electronics: Principles, Devices and Applications

60 Digital ElectronicsTable 3.3Multiplication using the repeated add and right-shift algorithm.10111 Multiplicand1 1 0 Multiplier00000 Start+ 0000000000 Result of first addition00000 0(Result of addition shifted one bit to right)+ 1011110111 Result of second addition01011 10(Result of addition shifted one bit to right)+ 10111100010 Result of third addition010001 010(Result of addition shifted one bit to right)Example 3.8Multiply (a) 10001 2 × 101 2 by using the ‘repeated add and left-shift’ algorithm and (b) (2B) 16 ×3 16 by using the ‘add and right-shift’ algorithm. Verify the results by showing equivalent decimalmultiplication.Solution(a) As a first step, we will multiply (10001) 2 by (101) 2 . The process is shown as follows:1 0 001× 1011 0 00100000100 0 1101 0 101The multiplication result is then given by placing the binary point three bits after the LSB, whichgives (1010.101) 2 as the final result. Also, (100.01) 2 = (4.25) 10 and (10.1) 2 = (2.5) 10 . Moreover,(4.25) 10 × (2.5) 10 = (10.625) 10 and (1010.101) 2 equals (10.625) 10 , which verifies the result.(b) (2B) 16 = 00101011 = 101011 and (3) 16 = 0011 = 11.Different steps involved in the multiplication process are shown in Table 3.4.The result of multiplication is therefore (10000001) 2 . Also, (2B) 16 = (43) 10 and (3) 16 = (3) 10 .Therefore, (2B) 16 × (3) 16 = (129) 10 . Moreover, (10000001) 2 = (129) 10 , which verifies the result.3.6 Binary DivisionWhile binary multiplication is the process of repeated addition, binary division is the process ofrepeated subtraction. Binary division can be performed by using either the ‘repeated right-shift and