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F(W,X,Y,Z) using 8:1 MUX

Short answer

Use an 8:1 MUX with W,X,Y as selects. Data inputs: I0=Z, I1=Z', I2=Z, I3=Z, I4=Z', I5=Z', I6=1, I7=Z. One NOT gate provides Z'. The output implements the given minterms.

Realize the logic circuit of the following using 8:1 MUX. F(W,X,Y,Z) = Σm(1,2,5,7,8,10,12,13,15)

Use W,X,YW, X, Y as the select lines S2S1S0S_2S_1S_0 of the 8:1 MUX and connect ZZ, Z′Z', 0 or 1 to the data inputs. Input IjI_j covers minterms 2j2j (Z=0Z = 0) and 2j+12j+1 (Z=1Z = 1).

Required minterms: 1, 2, 5, 7, 8, 10, 12, 13, 15.

I0I_0I1I_1I2I_2I3I_3I4I_4I5I_5I6I_6I7I_7
Z′Z'02468101214
ZZ13579111315
InputZZZ′Z'ZZZZZ′Z'Z′Z'1ZZ
             +--------+
  Z -------->| I0     |
  Z' ------->| I1     |
  Z -------->| I2     |
  Z -------->| I3 8:1 |----> F
  Z' ------->| I4 MUX |
  Z' ------->| I5     |
  1 -------->| I6     |
  Z -------->| I7     |
             +--------+
              W  X  Y
  (Z' from one NOT gate)

Check: WXYZ=1100WXYZ = 1100 (12): select 110 → I6=1I_6 = 1, so F=1F = 1. WXYZ=1001WXYZ = 1001 (9): select 100 → I4=Z′=0I_4 = Z' = 0, so F=0F = 0 (9 is not a minterm). Correct.

✓ Answer checked by Claude Opus 5.5 (high effort)

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