ECE_15A_HW8_Soln_mms

# ECE_15A_HW8_Soln_mms - Homework 8 Solutions E CE 15A Winter...

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Homework 8 Solutions ECE 15A, Winter 2012, Jon Suen 1. 2. First, we note that every function has m(2). Drawing the K-maps: F1 AB\CD 00 01 11 10 00 1 01 11 10 F1=m(2)+AB’D’ F2 AB\CD 00 01 11 10 00 1 4-to-2 4-to-2 I3 I4 I5 I6 I7 V V O0 O1 O2 V V0 V1 I0 I1 I2 O0 O1 O0 O1

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01 1 11 10 F2=m(2)+A’CD+B’CD F3 AB\CD 00 01 11 10 00 1 01 11 10 F3= m(2)+A’BD’+A’BC The programming table is A A’ B B’ C C’ D D’ F1 F2 F3 0 1 0 1 1 0 0 1 1 1 1 1 0 1 0 0 0 0 1 1 0 0 0 1 0 0 1 0 1 0 0 1 0 0 0 0 1 1 0 1 0 0 1 0 0 1 1 0 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 0 1 A’BD’ A’BC F1 F2 F3 m(2) A’B’D’ A’CD B’CD
3. 4. Note that the multiplexer can be controlled by any other input with appropriate LUTs. (A or D would be the most logical) Where LUT1: A B C F 0 0 0 0 4 to 16 D C B A 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 F1 F2 F3 D LUT 1 A B C LUT 2

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0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 Which realizes A’B’C+A’BC and is active when D=0 LUT2: A B C F 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 1 Which realizes A’B’C’+A’B+AB’C’+AB and is active when D=1.
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ECE_15A_HW8_Soln_mms - Homework 8 Solutions E CE 15A Winter...

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