EE319K_FINAL_SOL_2006

# EE319K_FINAL_SOL_200 - EE319K Fall 2006 Final A Page 1(4 Question 1 Precision(alternatives is range divided by resolution Therefore the number of

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EE319K Fall 2006 Final A Page 1 Jonathan W. Valvano December 14, 2006 9am-12n (4) Question 1. Precision (alternatives) is range divided by resolution. Therefore the number of bits is log 2 (200/0.1) = log 2 (2000) = 11 bits (4) Question 2. The digital result is (Vin-Vmin)*(Nmax-Nmin)/(Vmax-Vmin) + Nmin. In this case the ADC result = 0.625*256/2.5 = 64 (or = 0.625*255/2.5 = 64) (2) Question 3. C=0 (because it fits 160-140 = 20) (2) Question 4. V=1 (because it doesn’t fit -90 + -40 = -130 (4) Question 5. The value = I * = 1152/256 = 4.5 (4) Question 6. J) Hardware sets it when there is no data in the transmit data register, because TDRE means transmit data register empty. (4) Question 7. D) Hardware sets it when there is data in the receive data register, because RDRF means receive data register full (4) Question 8. The sequence length determines how many samples will be taken. The MULT bit is zero, so the same channel is sampled multiple times. ATDDR0 always received the first conversion. C)

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## This note was uploaded on 11/03/2010 for the course EE 319K taught by Professor Bard during the Spring '08 term at University of Texas at Austin.

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EE319K_FINAL_SOL_200 - EE319K Fall 2006 Final A Page 1(4 Question 1 Precision(alternatives is range divided by resolution Therefore the number of

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