Skip to main content

K-map Σm(0,1,2,8,10,14,15)+d, POS with NOR

Short answer

Simplified POS: F = (A+B')(B'+C)(A'+B+D'). Standard POS: ΠM(4,5,6,7,9,11,12,13). Realized with NOR gates using NOR-NOR form and inverters.

Simplify Σm(0,1,2,8,10,14,15) d = (3,7,11,13) using k-map, write its standard product of sum expression and realize it using NOR gates only.

Given F(A,B,C,D)=Σm(0,1,2,8,10,14,15)+d(3,7,11,13)F(A,B,C,D) = \Sigma m(0,1,2,8,10,14,15) + d(3,7,11,13). The 0-cells are 4, 5, 6, 9, 12.

K-map (grouping the 0s for POS)

        CD
AB      00  01  11  10
      +---+---+---+---+
 00   | 1 | 1 | X | 1 |
      +---+---+---+---+
 01   | 0 | 0 | X | 0 |
      +---+---+---+---+
 11   | 0 | X | 1 | 1 |
      +---+---+---+---+
 10   | 1 | 0 | X | 1 |
      +---+---+---+---+
  • Quad of 0s M(4,5,6,7) → (A+B‾)(A+\overline{B})
  • Quad of 0s M(4,5,12,13) → (B‾+C)(\overline{B}+C)
  • Pair of 0s M(9,11) → (A‾+B+D‾)(\overline{A}+B+\overline{D})
F=(A+B‾)(B‾+C)(A‾+B+D‾)F = (A+\overline{B})(\overline{B}+C)(\overline{A}+B+\overline{D})

Check: 4, 5, 6 by (A+B‾)(A+\overline{B}); 12 by (B‾+C)(\overline{B}+C); 9 by (A‾+B+D‾)(\overline{A}+B+\overline{D}). ((A‾+C+D‾)(\overline{A}+C+\overline{D}), cells 9 and 13, is an equally good third term.)

Standard product of sums

The given 0s are ΠM(4,5,6,9,12)\Pi M(4,5,6,9,12) with don't-cares d(3,7,11,13)d(3,7,11,13). The simplified POS also makes don't-cares 7, 11, 13 equal to 0, so its standard (canonical) POS is:

F=ΠM(4,5,6,7,9,11,12,13)F = \Pi M(4,5,6,7,9,11,12,13) F=(A+B‾+C+D)(A+B‾+C+D‾)(A+B‾+C‾+D)(A+B‾+C‾+D‾)(A‾+B+C+D‾)(A‾+B+C‾+D‾)(A‾+B‾+C+D)(A‾+B‾+C+D‾)\begin{aligned} F = {} & (A+\overline{B}+C+D)(A+\overline{B}+C+\overline{D})(A+\overline{B}+\overline{C}+D)(A+\overline{B}+\overline{C}+\overline{D}) \\ & (\overline{A}+B+C+\overline{D})(\overline{A}+B+\overline{C}+\overline{D})(\overline{A}+\overline{B}+C+D)(\overline{A}+\overline{B}+C+\overline{D}) \end{aligned}

Realization using NOR gates only

A POS expression maps directly to a two-level NOR-NOR circuit:

F=(A+B‾)‾+(B‾+C)‾+(A‾+B+D‾)‾‾F = \overline{\overline{(A+\overline{B})} + \overline{(\overline{B}+C)} + \overline{(\overline{A}+B+\overline{D})}}
A ──[NOR]── A'   (inputs tied)
B ──[NOR]── B'   (inputs tied)
D ──[NOR]── D'   (inputs tied)

A, B'     ──[NOR]── P1  ─┐
B', C     ──[NOR]── P2  ─┤──[NOR]── F
A', B, D' ──[NOR]── P3  ─┘

Gates: 3 NOR inverters (A‾,B‾,D‾\overline{A}, \overline{B}, \overline{D}), two 2-input NORs, one 3-input NOR and one 3-input output NOR: 7 NOR gates.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Digital Logic chapter 3 questions →