A hidden passage is controlled by a redstone circuit for an adder which performs binary addition two bits at a time. He notes that it adds the inputs (X2, ya) and (X2-1, Ya+1), the (2i)th and (2i+1)th bits of two binary numbers, and returms the sum as outputs s2, Sas1. The addition is performed each clock cycle, starting from i = 0 until i == - 1, wheren is the number of bits and presumed to be even. For example, when adding x = 0110102 and y = 0110012, we take the first two bits of each number as x1,9 and y;,9 respectively. In this case, x1,9 is 10 and y:0 = 01, and the carry bit c, is 1. Our sum, s0 is 00 and the next carry bit c, is 1. On the next clock cycle, we look at the next two bits, X32 = 10 and y32 = 10, so our sum s32 is 01, but we have a carry c, = 1. Finally, our last addition to perform is on xs4 = 01 and ys4 = 01, with a carry bit c, = 1. Our sum therefore is ss.4 = 11 with carry cg = 0. Our final result of the addition is 110100. You are only responsible for the addition of each two-bit partition of the number and carry bits, and are not responsible for the final sum or selecting two-bit partitions of the numbers. a) Fill in the state table for the new adder, the first row has been provided.

A hidden passage is controlled by a redstone circuit for an adder which performs binary addition two bits at a time. He notes that it adds the inputs (X2, ya) and (X2-1, Ya+1), the (2i)th and (2i+1)th bits of two binary numbers, and returms the sum as outputs s2, Sas1. The addition is performed each clock cycle, starting from i = 0 until i == - 1, wheren is the number of bits and presumed to be even. For example, when adding x = 0110102 and y = 0110012, we take the first two bits of each number as x1,9 and y;,9 respectively. In this case, x1,9 is 10 and y:0 = 01, and the carry bit c, is 1. Our sum, s0 is 00 and the next carry bit c, is 1. On the next clock cycle, we look at the next two bits, X32 = 10 and y32 = 10, so our sum s32 is 01, but we have a carry c, = 1. Finally, our last addition to perform is on xs4 = 01 and ys4 = 01, with a carry bit c, = 1. Our sum therefore is ss.4 = 11 with carry cg = 0. Our final result of the addition is 110100. You are only responsible for the addition of each two-bit partition of the number and carry bits, and are not responsible for the final sum or selecting two-bit partitions of the numbers. a) Fill in the state table for the new adder, the first row has been provided.

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter4: Processor Technology And Architecture

Section: Chapter Questions

Problem 9VE

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