Given an n-bit natural X and an integer E represented under the form (-1)^SIGN.|E|, the shifting algorithm shown in figure 1 computes an n-bit natural M such that

When E 0, E trailing zeroes are added and E leading bits are removed.

When E

Example (n = 14):

If X = 01110111010010 and E = 5 then M=11101001000000;

If X = 01110111010010 and E = -5 then M=00000011101110;

Figure 2 shows a sequential machine that implements this algorithm for n=4 and |E|=3, where:

1. M and next_M are 4-bits numbers; count and next_count are 2-bits numbers and done and next_done are 1-bit numbers.

2. The bits of M, next_M, count and next_count are named as follows: M = {M3, M2,M1,M0}>; next_M = {next_M3,next_M2,next_M1,next_M0}; count={count1,count0}; next_count={next_count1,next_count0}

When reset=1, count, M and done are respectively set to |E|, X and 0. What are the Boolean functions for next_count and next_done?

M 0) and (done = 0) then if SIGN=0 then M

---------- -- Note1: This sentence computes M

Figure 1

Figure 2

A.

nextCount1=count1.count0+count1.done

nextCount0=count1.!{count0}.!{done}+count0.done

nextDone=!{count1}.!{count0}+done

B.

nextCount1=count1.done+!{count0}.!{done}

nextCount0=count1.!{count0}+count0.done

nextDone=!{count1}.!{count0}+done

C.

nextCount1=count1.(count0+done)

nextCount0=count1.!{count0}.!{done}+count0.done

nextDdone=!{count1}+!{count0}+done

D.

nextCcount1=count1.count0.!{done}

nextCount0=count1.!{count0}.!{done}

nextDone=!{count1}.!{count0}

SIGN reset X IE clk Done nextDone nextCount Count CIRCUITO COMBINACIONAL 2. -Registro 7-bit SIGN reset X IE clk Done nextDone nextCount Count CIRCUITO COMBINACIONAL 2. -Registro 7-bit