0
\$\begingroup\$

'=Compliment

D=P'VST + PV'S'T + PV'ST'+ PV'ST + PVS'T' + PVS'T + PVST' + PVST I'm having trouble simplifying this.

\$\endgroup\$
  • 3
    \$\begingroup\$ Have you tried K-maps? \$\endgroup\$ – Chris Laplante Nov 19 '13 at 4:18
  • \$\begingroup\$ The Maxterms can be easily found out from your expression here, and when you have the maxterms, you know the minterms too (The rest of them). Now, as @ChrisLaplante said, use K-maps for SOP simplification. \$\endgroup\$ – Anshul Nov 19 '13 at 5:31
5
\$\begingroup\$

For a beginner this is best done with a 4-variable K-map.

See here: http://en.wikipedia.org/wiki/Karnaugh_map http://www.ece.rice.edu/~kmram/elec326/Notes/notes-326-set5.pdf

\$\endgroup\$
1
\$\begingroup\$

I'm a bit rusty on these things but this should help as a start: -

P'VST + PV'S'T + PV'ST'+ PV'ST + PVS'T' + PVS'T + PVST' + PVST becomes

(P'VST + PVST) + P(V'S'T + V'ST' + V'ST + VS'T' + VS'T + VST') which becomes

VST + P(S'T + ST' + V'ST + VS'T') etc..

\$\endgroup\$
0
\$\begingroup\$

If we break the terms apart, we have 8 (P'VST -> 1000), 6 (PV'S'T -> 0110), 5, 4, 3, 2, 1, and 0. From those we can combine (via Q-M, which combines values with single-bit differences):

3, 2, 1, and 0 (00XX -> PV)
5, 4, 1, and 0 (0X0X -> PS)
6, 4, 2, and 0 (0XX0 -> PT)
8 and 0 (X000 -> VST)

Putting those together we get PV + PS + PT + VST, which can be simplified even further if desired.

\$\endgroup\$
  • 2
    \$\begingroup\$ I don't think this answer is very useful to a beginner. \$\endgroup\$ – Chris Laplante Nov 19 '13 at 4:25

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.