hw6_soln_2009

hw6_soln_2009 - ’ b d ’(a c ’ d(a b ’ d 6 A = B = C...

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Homework #6 Solutions ECE 15A, Winter 2009 1) Z = BF + CEF + ACDF B F E C A D = 2) NAND Only a b c d e g h z 3) NOR Only a b c d e g h z
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4) Realize Z=A[BC +D+E(F +GH)] using NOR gates. A B C G H D E F Z 5) F(a,b,c,d) = Σ m(0,1,5,7,10,11,14,15) a) F = a b c + a c d + bcd + acd hazard transitioning from a c d to bcd (m5  m7) and from bcd to acd (m15  m14) and F = a b c + a bd + abc + acd hazard transitioning from a b c to a bd (m1  m5) and from a bd to abc (m7  m15) b) F = a b c + a bd + abc + acd + a c d + bcd c) F = ac + bc d + ab d + a b c + a cd + a bd F = (a + c)(b + c + d)(a
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Unformatted text preview: ’ + b + d ’ )(a + c ’ + d)(a + b ’ + d) 6) A = B = C = 1, so F = (A + B ’ + C ’ )(A ’ + B + C ’ )(A ’ + B ’ + C) = 1 But, in the figure, gate 4 outputs F = 0, indicating that something is wrong. For the last NOR gate, F = 1 only when all its inputs are 0. But the output of the gate is 1. Therefore, gate 4 is working properly, but gate 1 is connected incorrectly or is malfunctioning. 7) x y z w a b c d e F T = 0 1 2 3 4 5 6 7 1.0 0.0 2.4 3.4 4.8 5.8 6.2 7.2 e is the output of the inverter...
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This note was uploaded on 03/02/2009 for the course ECE 15A taught by Professor M during the Winter '08 term at UCSB.

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hw6_soln_2009 - ’ b d ’(a c ’ d(a b ’ d 6 A = B = C...

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