A-level Physics (Advancing Physics)/Digitisation/Worked Solutions

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1. Take samples for the signal below every 0.1ms, and then produce a reconstructed signal. How does it differ from the original?

Digitisation q1 solution.svg

The high frequency elements of the signal have been lost.

2. A signal is sampled for 5 seconds at a sampling rate of 20 kHz. How many samples were taken?

20 \times 10^3 = \frac{\mbox{No. of samples}}{5}

\mbox{No. of samples} = 20 \times 10^3 \times 5 = 100000

3. Most sounds created by human speech except for 'ss' and 'ff' have a maximum frequency of 4 kHz. What is a suitable sampling rate for a low-quality telephone?

4 \times 2 = 8 \mbox{ kHz}

4. Using a sampling rate of 20 kHz and 3 bits, sample the following signal, and then produce a reconstructed signal. What is the maximum frequency that can be perfectly reconstructed using this sampling rate?

First, calculate the length of each sample, by letting the number of samples equal 1:

20 \times 10^3 = \frac{1}{\mbox{Length of one sample}}

\mbox{Length of one sample} = \frac{1}{20 \times 10^3} = 0.00005\mbox{ s} = 0.05\mbox{ ms}

Then, we can sample the waveform and create a reconstructed signal:

Digitisation q2 solution.svg

To calculate the maximum frequency that can be perfectly reconstructed using this sampling rate (20 kHz):

\frac{20}{2} = 10\mbox{ kHz}