An analogue signal is sampled every 160 µs to convert it into a digital representation. What is the corresponding sampling rate expressed in kHz?

According to the Sampling Theorem, for this sampling rate, what will approximately be the highest frequency present in the signal, in kHz, assuming the lowest frequency is very close to zero?

If each sample is quantised into 128 levels, what will be the resulting bitrate in bps?

Give your answer in scientific notation to 2 decimal places.

Hint: you will first need to determine the number of bits per sample that produces 128 quantisation levels.

Guest Aug 19, 2020

#1**0 **

1/160 x 10^{-6 }= 6250 hz = 6.25 kHz

Not sure about Sampling Theorem but I THINK the highest would be 6.25 kHz/2 = 3.125 kHz Highest freq in signal

2^{7 }= 128 so you will need 7 bits to represent

7 bits x 6.25 x 10^{3 }samples = 43750 bits per second = 4.38 x 10^{4 }bits per second

ElectricPavlov Aug 19, 2020