ch03_complete - F.3 Chapter 3 Solutions 3.1 Gate=1 Gate=0...

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F.3 Chapter 3 Solutions 3.1 N-Type P-Type Gate=1 closed open Gate=0 open closed 3.2 3.3 There can be 16 different two input logic functions. 3.4 A B C 0 0 1 0 1 0 1 0 0 1 1 0 B P Type N Type P Type N Type A C = 1 IN = 1 OUT = 0 P Type N Type
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F.3. CHAPTER 3 SOLUTIONS 2 3.5 A B C OUT 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 0 3.6 C = A’; D = B’; Z = (C+D)’ = (A’+B’)’ = A . B A B C D Z 0 0 1 1 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 1 3.7 There is short circuit (path from Power to Ground) when either A = 1 and B = 0 or A = 0 and B = 1. B P Type N Type P Type N Type A C = 0, A=1, B= 0
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F.3. CHAPTER 3 SOLUTIONS 3 3.8 Correction: Please correct the logic equation to Y = NOT ( A AND (B OR C ) ) 3.9 A B NOT(NOT(A) OR NOT(B)) 0 0 0 0 1 0 1 0 0 1 1 1 AND gate has the same truth table. 3.10 A B A NOR B 0 0 1 0 1 0 1 0 0 1 1 0 A A B B C Y = NOT(A AND (B OR C)) C
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F.3. CHAPTER 3 SOLUTIONS 4 3.11 a. Three input And-Gate Three input OR-Gate
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F.3. CHAPTER 3 SOLUTIONS 5 b. (1) A = 1, B = 0, C = 0. AND Gate OR Gate
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F.3. CHAPTER 3 SOLUTIONS 6 b. (2) A = 0, B = 0, C = 0 AND Gate OR Gate
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F.3. CHAPTER 3 SOLUTIONS 7 b. (3) A = 1, B = 1, C = 1 AND Gate OR Gate
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F.3. CHAPTER 3 SOLUTIONS 8 3.12 3.13 A five input decoder will have 32 output lines. 3.14 A 16 input multiplexer will have one output line (ofcourse!). It will have 4 select lines. 3.15 C in 1 1 1 0 A 0 1 1 1 B 1 0 1 1 S 0 0 1 0 C out 1 1 1 1 A = 7, B = 11, A + B = 18. In the above calculation, the result (S) is 2 !! This is because 18 is too large a number to be
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This homework help was uploaded on 04/07/2008 for the course ECE 495K taught by Professor Vijaykumar during the Spring '08 term at Purdue University-West Lafayette.

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ch03_complete - F.3 Chapter 3 Solutions 3.1 Gate=1 Gate=0...

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