Sequential Circuits

Q3: Binary Multiplier

Q-Implement binary multiplication using shifter:

Eg. If we are multiply 11 * 4

Then 11 = 1011                  4 = 0100

Algorithm: For multiplication we first multiply the LSB of 4 (multiplier) with multiplicand and then shift it towards right. Then we multiply the next bit and then add it to the shifted result. Again we MULTIPLY, ADD & Shift or if bit of multiplier is 1 then ADD multiplicand and SHIFT and if bit of multiplier is 0 then ADD zero (or don’t perform ADD but just) SHIFT. We store multiplier in register Q & multiplicand in A and use adder as:

Now I’ll show the contents of shifter at every clock tick if we have to find A* B =  1011 * 0100

Clock tick                             contents of register                                        Function

1st tick                                   0              0000       0100                       Initial data is stored in register from inputs

2nd tick                                  0              0000       0100                       Result of adder is stored

2nd tick                                  0              0000       0010                       it is shifted towards right

3rd tick                                   0              0000       0010                       Result of adder is stored

3rd tick                                   0              0000       0001                       shifted right again

4th tick                                   0              1011       0001                       firstly result of adder is stored

4th tick                                   0              0101       1000                       now right shifted

5th tick                                   0              0101       1000                       Result of adder is stored

5th tick                                   0              0010       1100                       Again right shifted

And we get the answer as 001011002

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