HW3_Solutions

HW3_Solutions - ECE 2030 E,F Homework #3 - Solutions (1)...

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Unformatted text preview: ECE 2030 E,F Homework #3 - Solutions (1) Textbook #2-20 – Optimize using 4-variable K-Maps, and finding PIs and EPIs. (a) F(W,X,Y,Z) = Σ m(0,2,3,5,7,8,10,11,12,13) Y YZ WX 00 00 01 1 m0 1 m5 m2 1 m7 m6 m15 m13 10 1 m8 1 m3 1 11 1 m12 10 1 m1 m4 01 W 11 m14 1 m9 F = X’Y + X’Z’ + WXY’ + W’XZ (first two terms are essential) m11 X 1m10 Z (b) F(A,B,C,D) = Σ m(3,4,5,7,9,13,14,15) C CD AB 00 00 m0 01 1 11 A 01 m4 m12 10 m8 11 m1 1 m5 1 m13 1 m9 10 1 m3 1 m7 1 m15 m11 F = A’BC’ + A’CD + AC’D + ABC (All four terms are essential) m2 m6 B 1 m14 m10 D (c) F(W,X,Y,Z) = Σ m(0,2,4,6,7,8,9,12,13,15) Y YZ WX 00 00 01 W 1 1 m0 m4 01 11 m1 m5 11 1 1 10 1 m3 1 m7 1 1 m2 m6 1 m12 m8 m13 1 10 Z m14 m11 m9 m15 m10 X F = W’Z’ + WY’ + XYZ (first two terms are essential) (2) Textbook #2-23 – Optimize as (1) SOP and (2) POS forms. (a) F(A,B,C,D) = Σ m(0,1,5,7,8,10,14,15) C CD AB 00 00 1 01 A 01 m0 m4 11 m12 10 1 m8 1 11 m1 m3 1 m5 1 m7 m6 00 01 A 11 1 m13 m9 m0 0 m4 0 m12 10 m8 C 10 0 0 m3 F’ = A’B’C+A’BD’+ABC’+AB’D F = (A+B+C’)(A+B’+D)(A’+B’+C)(A’+B+D’) m2 m7 0 m6 m15 0 m13 0 m9 OR m14 m5 0 F = B’C’D’+A’C’D+BCD+ACD’ m10 11 m1 OR 1 m11 01 B 1m14 m15 D 00 F = A’B’C’+A’BD+ABC+AB’D’ m2 CD AB 10 m11 B F’ = BC’D’+AC’D+B’CD+A’CD’ F = (B’+C+D)(A’+C+D’)(B+C’+D’)(A+C’+D) m10 D (b) F(W,X,Y,Z) = Π M(3,11,13,15) Y YZ WX 00 01 W 11 00 01 11 m0 m1 m4 m5 m12 0 m13 0 10 m3 m7 m6 0 m15 m14 0 10 m8 m9 m11 Z F = Z’+W’X+X’Y’ m2 m10 X F’ = WXZ+X’YZ F = (W’+X’+Z’)(X+Y’+Z’) (3) Textbook #2-24 – Optimize the functions together with don’t care conditions. (a) F(A,B,C,D) = Σ m(0,1,7,13,15), d(A,B,C,D) = = Σ m(2,6,8,9,10) C CD AB 00 00 1 m0 01 1 m4 11 A 01 m12 10 11 m1 m3 1 1 m13 m15 X m8 X m2 1m7 X m6 m5 X 10 m9 F = B’C’+ABD+BCD B m14 X m11 m10 D (b) F(W,X,Y,Z) = Σ m(2,4,9,12,15), d(W,X,Y,Z) = = Σ m(3,5,6,13) Y YZ WX 00 00 01 11 1 m4 1 1 X m5 11 X m1 01 W m0 10 X m13 m12 m3 m7 m2 Xm6 1 m15 m14 m11 m8 m9 X OR m10 1 10 F = XY’+WXZ+WY’Z+W’X’Y F = XY’+WXZ+WY’Z+W’YZ’ Z (c) F(A,B,C) = Σ m(1,2,4), d(A,B,C) = = Σ m(0,3,6,7) B BC A 00 0 A 1 01 11 10 X 1 X 1 1 m0 m1 m3 X m4 m5 C m2 X m7 m6 F = A’+C’ (4) Textbook #2-26 – Optimize the functions together with the don’t care conditions, in (1) SOP and (2) POS form: (a) F(A,B,C,D) = Π M(1,3,4,6,9,11), d(A,B,C,D) = = Σ m(0,2,5,8,10,12,14) C CD AB 00 01 11 10 0 0 X 00 01 0 Xm5 1 m7 0 m6 m4 11 X 1 m13 m15 10 X A X 0 0 m9 m11 m0 m12 m8 m1 m3 1 F = BD m2 X B Also: F’=B’+D’ m14 X m10 D (b) F(W,X,Y,Z) = Σ m(3,4,9,15), d(W,X,Y,Z) = = Σ m(0,1,2,5,10,12,14) Y YZ WX 00 00 X m0 01 11 10 X 1 X m1 01 W 1 X m5 m4 11 X m12 m13 m3 F = W’X’+W’Y’+WXY+X’Y’Z m7 1 m15 1 10 m8 m2 m6 X m14 X m9 m11 Z m10 X OR F’ = (WZ’ or X’Z’)+(XY’Z or WXY’)+W’XY+WX’Y F = (W’+Z)(X’+Y+Z’)(W+X’+Y’)(W’+X+Y’) (not unique) (5) Textbook #2-35a – Use tristate buffers and AND gates to realize the function: F = A’BC + ABD + AB’D’ C Solution: A B C D A’B D AB F D’ AB’ 0 A’B’ _____________________________________________________________________ (6) Design a CMOS transistor-level circuit that implements the following Boolean Expression: F(A,B,C,D) = A’B+ACD’+BC’D Solution: F(A’,B’,C’,D’) = AB’+A’C’D+B’CD’ (for the PMOS section) F’(A,B,C,D) = (A+B’)(A’+C’+D)(B’+C+D’) (for the NMOS section) VCC=“1” A A’ B’ B’ C’ C D D’ F A B’ A’ C’ D B’ C D’ GND=“0” ...
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