CSE140L Exercises Solutions1.aaa'VCCGND(a)VCCGNDabab(a+b)’(b)GNDabccabVCC(c)((a+b)c)’2. The 3-bit Johnson counter is shown in the following figure.D D DQ1QQ QQ0Q2clkclra) The possible states are: (0,0,0), (0,0,1), (0,1,1), (1,1,1), (1,1,0), (1,0,0)b) The possible states, with initial state (0,1,0) are: (0,1,0), (1,0,1)3. A and B both have incoming edges with different output values, so A is split into A0 and A1,and B is split into B0 and B1.PS x=a x = b OutputA0 A1 B0 0A1 A1 B0 1B0 C D 0B1 C D 1C A0 E 1D B0 E 1E B1 D 04.a) The truth table is as follows:Inst(7) Inst(6) R1 en R2 en R1 sel0 0 1 0 00 1 0 1 x1 0 1 0 11 1 1 0 1b)move1 a5a4a3a2a1a0 – move data into R1move2 001111 – move masking data into R2shift 000100 – left rotate R1 by 4mask 000000 – mask R1 and
View Full Document
Unlocking...