Question

The output signal from an analogue sensor is sampled every 195 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 value, what approximately could be the highest frequency present in the signal, in kHz, assuming the lowest frequency is very close to zero?

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

Give your answer in scientific notation to 2 decimal places.

Hint: you need to determine the number of bits per sample that produces 512 quantisation levels.

Answer 1

Answer:

• A sample rate is just reciprocal of the sample period.
• According to sampling theorem the highest frequency in single would be $5.13/2=2.565 KHz.$
• $2^8=512$ So you will need 8 bits to represent.
• $8 bits \times5.128\times(10)^3$samples $=41.026$ bits per second.

Explanation:

Sample period of 1 millisecond $=1 kHz.$

Thus a sample period of $0.195$ milliseconds is

$1/0.195=5.12820513 Hz$

To determine the sample after $512$ levels quantization, we need to find bits per sample that produces $512$ quantization levels.

Thus,

$2^8=512$

So you will need 8 bits to represent.

8 bits $\times5.128\times(10)^3$ samples $=41.026$ bits per second.

Learn more about scientific notation, refer :

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