ME_Digital1.1

Question

Comments

i think ur right but what are the inputs of OR gate if o/p of OR leads to 1 then its just A xor B since 1 inpput is free it may be taken as 1 i think

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can u explain on this?

Answer 1

 S2 select line is output OR gate .So  C+1=1   , C+0 = C  whatever value of C will be output will be 1. So  S2 will always be 1.

In the above MUX     : X0,X3,X5,X6 are active high .

By using characteristic equation of MUX - 

 Y = E' ( S2'S1'S0' X0 + S2'S1S0 X3 + S2S1'S0 X5 + S2S1S0' X6 )  I have not included rest all becoz they are active low .

In the above equation  E =0 and S2 will always be 1. so X0 and X3 terms become 0 , X5=1,X6=1

So the final output will be ->

Y = S1'S0 + S1S0' => A'B + AB'    i.e   A XOR B.