On Fri, 19 Oct 2001, Bernard Ebinu wrote:
> Hi Dr. Amaral,
>
> I was wondering whether you could clarify slide 51 of lecture 8 for me.
> In that slide the first half sum is supposed to be computed by the
> generate line and then pass through the xor at the end. My question to
> you is that if you were to start by letting A=1 and B=1 you should get a
> carry of 1( 0 in the case of the circuit) and a sum of 0, correct. I
> understand the carrry but why is the circuit producing a one as the sum
> of the first term if you use the above values.
> Thanks for your time,
>
Bernard:
With C0 = 0, A0 = 1, and B0 = 1, I get:
g0' = (A0.B0)' = 0
p0' = (A0+B0)' = 0
hS0 = (g0'.(p0')') = (0.(0)')=(0.1) = 0
c0 = c0' = 0
S0 = hs0 xor c0 = 0 xor 0 = 0
Your mistake is probably because you looked at the (X xor Y)' column
of slide 31, that is the truth table for an XNOR, and not an XOR.
I hope you don't mind me forwarding this to the class, as I just
felt in the same trap myself... (Will have to change slide 31
into two separate truth tables, one for XOR and another for XNOR
to avoid this in the future).
Hope this helps
Cheers,
Nelson
/
\ / / Jose Nelson Amaral - amaral@cs.ualberta.ca
) / ( Associate Professor
/ / \ Dept. of Computing Science - University of Alberta
( / ) Edmonton, Alberta, Canada, T6G 2E8
\ O / Phone: (780)492-5411 Fax: (780)492-1071
\ / http://www.cs.ualberta.ca/~amaral
`----'
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