EE1201_HW10_Soln

EE1201_HW10_Soln - P7.23 P7.24 1 O 1 1 1 1 1 (e) D=(A+BC) A...

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Unformatted text preview: P7.23 P7.24 1 O 1 1 1 1 1 (e) D=(A+BC) A B 6' D 0 0 0 1 O O 1 1 O 1 O 1 0 1 1 O 1 0 O O 1 O 1 O 1 1 0 O 1 1 1 0 (a) F=(A+B)? (b) F =A+B+Z§E§ (c) F=AB+(B—C)+D A A+B 7+AB (A+B)(7+AB) 0 1 O 1 HHOC\x c EA. + c A§C + A3 F either The tr L... O 0 :0 1 P729 (a) F=AB+€+AE=lA+§iCZ+Dl ‘ :1 1 '1. (b) F=A(§+c)+o= A+B? (c) F=A§C+A(B+C)= A+B+Z_' A+§Z‘ (d)* F=(A+B+C)(A+§+CXX+B+E)=X§E+ZBZ+A§1 (e)* F: ABC+ABC+A BC =(A +3 +C)(A +3+ch+3 +61 III-9. liiivies P7.30 NAND gates are said to be sufficient for combinatorial logic beca 5.. Boolean expression can be implemented solely with NAND gates. hue: Similarly, NOR gates are sufficient. closed and if either of the other two P731 In this circuit, if switch C is we can write: switches is closed, the output is high. Thus, D = (A + 3—)6' u The truth table is: A 3 . c o ' o o 0 o i I_ _ o o 1 1 0 1 0 0 0 1 1 0 | 1 o o o I 1 o 1 1 . 1 1 0 O 1 1 1 1 218 P7.36 6=A36+736 6=(A+B+C—)(A+B—+C)(A +B+C)(A + P7.37 H=Z§E+“ = Zm(0,1,2,4,5,6‘7) H=(A+§+?)=M(3) P738 1' = Zacmefmec = Zm(3,6,7) I=(A+B+C)(A+B+E)(A+§+C) = ZM(0,1,2,4,5) (Z+B+C)(X+B+C_) l 97.39 J=ZEC+ZBC+A§C+ABE+ABC ; = Zm(1,3,5,6,7) 'l J=(A+B+C)(A+§+C)(Z+3+C) = ZM(0,2,4) P7.4O K=Z§Z+Z§C+A3€+Aac = Zm(0,1,6,7) I K=(A+§+C)(A+§+ZXZ+B+€)(Z+B+E) I = ZM(2,3,4,5) I P7.41 (a) The. Truth Table is: I B E F 6 I 0 O 0 0 0 0 0 0 1 0 0 0 1 0 0 O O 1 1 0 O 1 0 O 1 0 1 0 1 1 O 1 1 0 0 0 1 1 1 0 1 0 0 0 1 220 P7.43 The fruit: table is: l A B A Q 3 I 0 0 0 0 1 1 '7 1 o 1 ll! 1 1 o Thus, we can write The product of sums expression and apply D '- | Laws to obtain: A+B) +(2—+B) A$3=(A+B)(Z+§)=( | The circuit is: P7.44 The T ...
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EE1201_HW10_Soln - P7.23 P7.24 1 O 1 1 1 1 1 (e) D=(A+BC) A...

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