ch11 - Chapter 11 Exercise Solutions E11.1 An analog signal...

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Chapter 11 Exercise Solutions E11.1 An analog signal has a continuous set of values over a given range. However, a digital signal has only discrete values. In reality, the digital signal still has a continuous set of values, people are only interested to know if a digital signal is higher than a specific value or lower than a specific value when dealing with digital signals. E11.2 (Skip) E11.3 (Skip) E11.4 It takes 129 E clock cycles to complete the conversion of 4 samples. Therefore it will take 129 ÷ 3000000 = 43 μ s. E11.5 The op amp circuit must have a gain of 500 (= 5V ÷ 10 mV). By choosing appropriate values of resistors, the following circuit can do the job: V in V out R R 1 2 + - V out V in = 1 + R 1 R 2 = 500 Figure S10.1 Voltage scaler circuit We need a resistance ratio of R 1 /R 2 = 499 which can be achieved by choosing R 1 = 750 K and R 2 = 1.5 K . E11.6 We can use the circuit in Figure 11.5c to perform the scaling and shifting. 0V = (R f ÷ R 1 ) × (- 40 mV) – R f ÷ R 2 × V 1 --- (1) 5V = (R f ÷ R 1 ) × (240 mV) – R f ÷ R 2 × V 1 --- (2) Set V 1 to and solve, V 1 = -5V, R 1 = 1.6K, R 2 = 200K, R f = 30K (28.7K exact), R 0 = 1.6K E11.7 Use the circuit shown in Figure 11.5c and choose appropriate resistors and V1. 0V = (R f ÷ R 1 ) × 2V) – R f ÷ R 2 × V 1 --- (1) 5V = (R f ÷ R 1 ) × 4V) – R f ÷ R 2 × V 1 --- (2) Set V1 = 5V, we can solve R 1 = 12K, R 2 = 30K, R f = 30K, R 0 = 12K E11.8 The value to be written into the ADCTL register is 00100011 2 . E11.9 The corresponding voltages are: 10- 1
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25: 5 × 25/255 = 0.49 V 64: 5 × 64/255 = 1.25 V 100: 5 × 100/255 = 1.96 V 150: 5 × 150/255 = 2.94 V 200: 5 × 200/255 = 3.92 V E11.10 The MAX1241 is a 12-bit A/D converter. Therefore, the corresponding voltages are Vx(128) = 5 × 128 ÷ 4095 = 0.16 V Vx(512) = 5 × 512 ÷ 4095 = 0.63 V Vx(768) = 5 × 768 ÷ 4095 = 0.94 V Vx(1024) = 5 × 1024 ÷ 4095 = 1.25 V Vx(2400) = 5 × 2400 ÷ 4095 = 2.93 V Vx(3072) = 5 × 3072 ÷ 4095 = 3.75 V E11.11 First, the ADCTL register must be properly configured: bit 7: read-only. Set to 0. bit 6: un-implemented. Set to 0. bit 5: set to 0 to select non-scan mode. bit 4: set to 0 to select single-channel mode. bit 3-0: set to 0000 to select channel AN0. We need to allocate two bytes to hold the sum of 100 samples. To sample the signal once every 10 ms, we will use the OC2 function to create a delay of 10 ms. The average of the sum is an 8- bit value. The voltage is to be displayed in two digits: one integral and one fractional. To convert to volt, the sum must be divided by 51. The quotient of this division will be an integer between 0 and 5. In order to obtain better accuracy, we will multiply the average by 10 before being divided by 51. The quotient of this division will contain two digits: one is the integral digit and the other is the fractional digit. They can be separated by dividing by 10. The procedure for solving this problem is as follows:
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ch11 - Chapter 11 Exercise Solutions E11.1 An analog signal...

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