Download Digital Fundamentals Tenth Edition Instructor Solution by Floyd, Buchla PDF

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It is unlikely that both inputs are open. The most probable fault is that the output is stuck in the LOW state (shorted to ground, perhaps) although it could be open. (b) Pin 4 input or pin 6 output internally open. 31 Chapter 3 36. The timer input to the AND gate is open. Check for 30-second HIGH level on this input when ignition is turned on. 37. An open seat-belt input to the AND gate will act like a constant HIGH just as if the seat belt were unbuckled. 38. Two possibilities: An input stuck LOW or the output stuck HIGH.

See Figure 5-2 for the circuit corresponding to each expression. (a) X = (A + B)(C + D) = AC + AD + BC + BD (b) (c) X = ABC + CD = ( ABC )(CD ) = ( A + B)CCD = ACD + BCD X = (AB + C)D + E = ABD + CD + E (d) X = ( A + B)( BC ) + D = ( A + B)( BC ) + D = A + B + BC + D = A + B + D (e) X = ( AB + C ) D + E = ( AB + C ) D + E = ABD + C D + E (f) X = ( AB + CD )( EF + GH ) = ( AB + CD)( EF + GH ) = ( AB + CD) + ( EF + GH ) = ( AB)(CD ) + ( EF )(GH ) = ( A + B )(C + D) + ( E + F )(G + H ) = AC + BC + A D + B D + E G + F G + E H + F H 60 Chapter 5 5.

FIGURE 3-30 47. See Figure 3-31. FIGURE 3-31 Multisim Troubleshooting Practice 48. Input A shorted to output. 49. Inputs shorted together. 50. No fault. 51. Output open. 34 CHAPTER 4 BOOLEAN ALGEBRA AND LOGIC SIMPLIFICATION Section 4-1 Boolean Operations and Expressions 1. X=A+B+C+D This is an OR configuration. 2. Y = ABCDE 3. X = A +B +C 4. (a) (c) (e) 0+0+1=1 1⋅0⋅0=1 1⋅0⋅1=0 5. (a) (b) (c) (d) (e) (f) (g) AB = 1 when A = 1, B = 1 ABC = 1 when A = 1, B = 0, C = 1 A + B = 0 when A = 0, B = 0 A + B + C = 0 when A = 1, B = 0, C = 1 A + B + C = 0 when A = 1, B = 1, C = 0 A + B = 0 when A = 1, B = 0 ABC = 1 when A = 1, B = 0, C = 0 6.

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