The following signal is sampled by 12-bit ADC triggered by timer TIM7. The timer clock frequency is 80 MHz. The value of TIM7_PSC register is 79. The value of TIM7_ARR register is 99. Assuming that sampling starts at t = 0, determine the value of the sample x[80]. Show your work.
x(t) = 1.5 + 0.6sin(70t )
Answer:
The sampling frequency is given by:
f_s = 80 MHz / (79 * 99) = 100.125 kHz
The sample x[80] is taken at t = 80/100.125 kHz = 0.799 ms.
At t = 0.799 ms, the value of the signal is:
x[80] = 1.5 + 0.6sin(70 * 0.799 ms) = 1.5 + 0.6sin(55.93) = 1.5 + 0.6(0.945) = 2.1425
The 12-bit ADC will convert the analog signal to a digital value with a resolution of 2^12 = 4096. Therefore, the digital value of x[80] is:
x[80] = 2.1425 * 4096 = 8753.6