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Microcontrollers, Second Edition by J. W. Bruce, Robert Reese, Bryan A. Jones

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APPENDIX CPROBLEM SOLUTIONS

Answers to all the odd-numbered problems are provided in this appendix.

Chapter 1

1. 25 = 32, 26 = 64, so 6 bits

3. 120 = 0x78 = 0b0111 1000

5. 0xF4 = 0b1111 0100

7. 0b1011 0111 = 0xB7 = 11*16 + 7 = 183

9. 0xB2 - 0x9F = 0x13. ~0x9F = ~(0b1001 1111) = 0b0110 0000 = 0x60. So 0xB2 - 0x9F = 0xB2 + ~0x9F + 0x01 = 0xB2 + 0x60 + 0x01 = 0x13

11. See Figure C.1.

Figure C.1Problem 1.11

image

13. 0x2A << 1 = 0x54

15. 0.3 * period = 20 μs; so period = 20 μs/0.3 = 66.7 μs. Frequency = 1/(66.7 μs) = 15 kHz

17. See Figure C.2.

Figure C.2Problem 1.17

19. See Figure C.3.

Figure C.3Problem 1.19

Chapter 2

1. See Table C.1.

Table C.1: Problem ...

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