Question

Topic: ALUs

I am trying to create an ALU with an operation for multiplying by 15. The inputs are A (8 bit), 1 OP 3 bit, and an 8 bit output.

I need to create a circuit multiplying the 8 bit A value by 15 giving us an 8 bit output value, but I don't know what to do. If someone could

explain this to me, I would highly appreciate it.

Thank you so much! PLEASE INCLUDE EXPLANATIONS

Answer 1

To see how this works, let's examine the multiplication of the 8-bit binary numbers 111100102 and 100011002. These correspond to the decimal numbers 242 and 140

As a check, we see that 242 * 140 = 33880, which is equal to 10000100010110002. (See Appendix A to review base conversions.)

The hardware implementation follows directly from this observation. It requires four 4-by-4 multipliers, implemented as in, plus logic to sum the four-bit wide slices of the partial products.

Let's call the four 8-bit partial products PP0, PP1, PP2, and PP3. Then the final product bits are computed as follows:

Implementation:

The basic blocks of the implementation are (1) the calculation of partial products, (2) the summing of the 4-bit product slices, and (3) the carry look-ahead unit. We examine each of these in turn.

Calculation of Partial Products Each of the 8-bit partial products is implemented by a 74284/74285 pair. The subsystem has 16 inputs, the multiplicand and multiplier, and 32 outputs, constituting the four 8-bit partial products. The partial product subsystem.

Calculation of Sums The low-order 4 bits of the final product, P3-0, are the same as PP03-0 and do not participate in the sums. P7-4 and P11-8 are sums of three 4-bit quantities.