1
$\begingroup$

I am designing a 4-bit comparator with a look ahead unit using a bit slice approach. I have to break the propagation of the Logical expressions for (A<B)i and (A>B)i. The main question is, could I use the following to simplify the boolean expressions.. The i is a subscript meaning the ith bit. The i-1 is a subscript meaning the bit before the ith bit.

(A<B)i = ~Ai*Bi + ~Ai*(A<B)i-1 + Bi*(A<B)i-1

Could I Take ~Ai common from the first two expressions leaving me with

~Ai*(Bi + (A<B)i-1) + Bi(A<B)i-1

Let's call Bi + (A<B)i-1) = Q

And let's call ~Ai = P

I'm left with P*Q + Q

Using the boolean law that Q + Q*P = Q

Can I simplify the expression to just

(A<B)i = Bi + (A<B)i-1

$\endgroup$
  • $\begingroup$ I would have preferred a simpler notation. $\endgroup$ – Shreesh Feb 17 '16 at 8:16
0
$\begingroup$

No, you cannot simplify to $(A<B)_i = B_i + (A<B)_{i-1}$, Take for example $A=11$ and $B = 10$.

Then $(A<B)_2$ = False, $B_2$ = True, and $(A<B)_1$ = False. If you substitute the values above then the equation is not satisfied. The original equation(s) is correct.

$(A<B)_i = \neg A_i B_i + (\neg A_i+B_i)(A<B)_{i-1}$
$(A<B)_1 = \neg A_1 B_1$

In your derivation of simpler formula, $B_i + (A<B)_{i-1} = Q$ but you are incorrectly substituting $B_i (A<B)_{i-1} = Q$ in rightmost term.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.