Question

An analogue sensor has a bandwidth which extends from very low frequencies up to 8.75 kHz. Using the Sampling Theorem (Section 3.3.1), what is the minimum sampling rate (number of samples per second) required to convert the sensor output signal into a digital representation without incurring any aliasing?

If each sample is now quantised into 512 levels, what will be the resulting sensor output bitrate in kbps?

Give your answer in scientific notation to one decimal place.

Hint: you need to determine the number of bits per sample that allows for 512 quantisation levels (see Sections 2.4 (Block 1) and 3.3.2 (Block 3)).

Answer 1

Explanation:

According to the Sampling Theorem, the minimum sampling rate required is at least twice the highest frequency component in the signal. In this case, the highest frequency component is 8.75 kHz, so the minimum sampling rate required is:

2 x 8.75 kHz = 17.5 kHz

Therefore, the minimum sampling rate required to avoid aliasing is 17.5 kHz.

To determine the resulting sensor output bitrate in kbps, we need to calculate the number of bits per sample. Since the signal is quantised into 512 levels, we need at least 9 bits per sample to represent all possible levels (2^9 = 512).

The sensor output bitrate is the product of the sampling rate and the number of bits per sample. Using the minimum sampling rate of 17.5 kHz and 9 bits per sample, we get:

bitrate = 17.5 kHz x 9 bits/sample = 157.5 kbps

Expressing the result in scientific notation to one decimal place, we get:

bitrate = 1.6 x 10^5 kbps