Written Hwk10 - Written Hwk #10 Solutions Problem 10.1. For...

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Written Hwk #10 Solutions Problem 10.1. For each instruction, we have three options: we can use FS = 0000 (F = A), 0111 (F = A), or 1100 (F = B). We will use the first for all of these; nothing besides FS changes if 0111 is used, but if 1100 is used MB must be 0 and BA is set to the source register instead of AA. R0 <-- R7: 000 111 xxx x 0000 0 1 R7 <-- R6: 111 110 xxx x 0000 0 1 R6 <-- R5: 110 101 xxx x 0000 0 1 R5 <-- R4: 101 100 xxx x 0000 0 1 R4 <-- R3: 100 011 xxx x 0000 0 1 R3 <-- R2: 011 010 xxx x 0000 0 1 R2 <-- R1: 010 001 xxx x 0000 0 1 R1 <-- R0: 001 000 xxx x 0000 0 1 Problem 10.2. We can use the adder for the first four operations (S2=0) and a simple logic unit for the last four (S2=1). Adder: We want the A input to be A when S1=0 and 0 when S1=1. The B input should be B when S1S0=00 and B' otherwise. Finally, the carry in to the lowest stage is S0. The logic/shift unit can be done with a 4:1 MUX, selected by S[1:0]. Finally, we select the output of the adder when S2=0 and the logic/shift unit output when S2=1.
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Problem 10.3.
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This note was uploaded on 02/26/2010 for the course ECE 290 taught by Professor Brown during the Fall '08 term at University of Illinois at Urbana–Champaign.

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Written Hwk10 - Written Hwk #10 Solutions Problem 10.1. For...

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