Signals and Communication Systems

Lessons

Senior 5 · Physics

Signals and Communication Systems

4 lessons

Lesson 2 of 4

Converting Signals: Sampling, Quantisation and Bit Rate

25 min

A voice note as a list of numbers

A voice note you record on a phone is stored and sent as nothing but a long list of binary numbers. Yet when it is played back, the listener hears a continuous, natural voice. Turning a continuous signal into numbers is done by an analogue-to-digital converter (ADC), and turning the numbers back into a voltage by a digital-to-analogue converter (DAC). Two questions decide the quality: how often the signal is measured, and how finely each measurement is recorded.

Sampling

An ADC measures the analogue voltage at regular instants. Each measurement is a sample, and the number of samples per second is the sampling rate (or sampling frequency) $f_{s}$ . If the samples are too far apart, rapid changes in the signal fall between them and are lost. The sampling theorem (Nyquist and Shannon) states that a signal containing frequencies up to $f_{\max}$ can be reconstructed exactly provided

f_s \ge 2f_{\max}

If the signal is sampled more slowly than this, high frequencies are not just lost: they reappear as false low frequencies, called aliases, which spoil the signal. To prevent this, a low-pass filter removes frequencies above $\frac{f_s}{2}$ before sampling.

Quantisation

Each sample is then rounded to the nearest of a fixed set of levels, a process called quantisation, and the level number is written in binary. An ADC with $n$ bits has $2^{n}$ levels. If the levels run from 0 to $V_{\max}$ , the step between them, the resolution, is

\Delta V = \frac{V_{\max}}{2^n - 1}

Rounding to the nearest level makes an error of up to half a step, the quantisation error. Each extra bit doubles the number of levels and halves the error.

Sample	Time / ms	Analogue voltage / V	Nearest level	3-bit code
1	0.000	4.75	5	101
2	0.125	6.05	6	110
3	0.250	6.68	7	111
4	0.375	6.45	6	110
5	0.500	5.43	5	101
6	0.625	3.89	4	100

The diagram shows a 3-bit ADC (8 levels, 0 V to 7 V, so $\Delta V = 1$ V) sampling a 667 Hz tone 8000 times per second. The first six samples are in the table. The staircase is a rough copy of the smooth curve; with more bits and faster sampling the steps become finer and the copy better.

Definition

Sampling rate $f_{s}$ : the number of samples taken per second. Quantisation: rounding each sample to the nearest of $2^{n}$ levels, where $n$ is the number of bits per sample (the bit depth). Bit rate: the number of bits produced per second, $\text{bit rate} = f_s \times n \times (\text{number of channels})$ . Example: telephone speech sampled at 8 kHz with 8 bits gives $8000 \times 8 = 64\,000$ bit s⁻¹.

Worked example 1: the telephone channel

Telephone speech is filtered to below 3.4 kHz, sampled at 8.0 kHz and coded with 8 bits per sample. (a) Check that the sampling rate satisfies the sampling theorem. (b) Find the number of quantisation levels. (c) Find the bit rate.

Solution

(a) $2f_{\max} = 2 \times 3.4 = 6.8$ kHz, and $8.0 \ge 6.8$ : satisfied, with a margin that makes the filter easier to build.

(b) $2^{8} = 256$ levels.

(c) Bit rate $= 8000 \times 8 = 64\,000$ bit s⁻¹ $= 64$ kbit s⁻¹. This 64 kbit s⁻¹ channel is the basic building block of digital telephone networks.

Worked example 2: music on a CD

CD audio is sampled at 44.1 kHz with 16 bits per sample, in two channels (stereo). (a) What is the highest frequency that can be recorded? (b) Find the bit rate. (c) How many megabytes does a 3.0-minute song need, uncompressed? (1 byte = 8 bits.)

Solution

(a) $\frac{f_s}{2} = 22.05$ kHz, just above the upper limit of human hearing (about 20 kHz).

(b) $44\,100 \times 16 \times 2 = 1.41 \times 10^{6}$ bit s⁻¹.

(c) $1.4112 \times 10^{6} \times 180 = 2.54 \times 10^{8}$ bits $= \frac{2.54 \times 10^{8}}{8} = 3.2 \times 10^{7}$ bytes $\approx 32$ MB. Compressed formats such as MP3 reduce this about ten-fold by discarding sounds the ear cannot notice.

Back to analogue: the DAC

At the receiver, a DAC turns each binary number back into a voltage, held until the next sample arrives. The result is the orange staircase. A low-pass reconstruction filter then smooths away the sharp steps, which contain only frequencies above $\frac{f_s}{2}$ , leaving a close copy of the original. The copy is never perfect, because of the quantisation error, but with 16 bits the error is about $\frac{1}{65\,536}$ of the full range, far below what the ear can hear.

How fast can a channel carry bits?

Claude Shannon showed in 1948 that a channel of bandwidth $B$ and signal-to-noise power ratio $\frac{S}{N}$ can carry information, without errors, at no more than

C = B\log_2\left(1 + \frac{S}{N}\right)

This channel capacity links the two ideas of this lesson and the last: to send more bits per second you need either more bandwidth or a better signal-to-noise ratio.

Worked example 3: the limit of an old telephone line

An analogue telephone line has a bandwidth of 3.1 kHz and an SNR of 30 dB. Find its Shannon capacity, and comment on the fact that the fastest dial-up modems of the 1990s reached 33.6 kbit s⁻¹ on such lines.

Solution

30 dB is a power ratio of $10^{3} = 1000$ .

$C = 3100 \times \log_2(1001) = 3100 \times 9.97 = 3.1 \times 10^{4}$ bit s⁻¹ $= 31$ kbit s⁻¹.

The modems were working at the Shannon limit of the line; slightly better lines (a little more bandwidth or SNR) allowed 33.6 kbit s⁻¹. No cleverer modem could do much better on that channel. Much faster internet needed channels with far more bandwidth: fibre, cable and modern radio.

Warning

Twice the highest frequency, not twice the bandwidth of something else. The sampling rate must be at least twice the highest frequency in the signal. And the number of levels is $2^{n}$ , not $2 n$ : 8 bits give 256 levels, not 16.

Tip

Bit rate = how often × how finely × how many channels. Write $f_s \times n \times \text{channels}$ first, then substitute; it avoids most bit-rate errors.

A/D Conversion, Sample Rate & Bit Depth for audio explained, Mixed Signals. Watch how the sampling rate limits the highest frequency and how the bit depth sets the number of levels.

✦

Practice

State the sampling theorem, and explain what happens if it is not obeyed.
An ADC has 4 bits and an input range of 0 V to 15 V. Find (a) the number of levels, (b) the resolution, (c) the binary code for an input of 9.3 V.
Find the minimum sampling rate for a signal containing frequencies up to 15 kHz.
A digital radio signal is sampled at 48 kHz with 24 bits, in stereo. Find the bit rate.
A 10-bit ADC has an input range of 0 V to 5.0 V. Find the maximum quantisation error.
A voice note lasts 60 s and is recorded at 8.0 kHz with 8 bits, mono. Find its size in kilobytes (1 kB = 1000 bytes).
Find the Shannon capacity of a radio channel of bandwidth 20 MHz with an SNR of 20 dB.
An engineer proposes to double the quality of a recording by doubling the sampling rate of a speech signal from 8 kHz to 16 kHz. Evaluate this, assuming the speech has been filtered to below 3.4 kHz.

Show answersHide answers

A signal whose highest frequency is $f_{\max}$ can be reconstructed exactly only if $f_s \ge 2f_{\max}$ . If sampled more slowly, high frequencies appear as false low frequencies (aliasing) and the signal is distorted.
(a) $2^{4} = 16$ ; (b) $\frac{15}{16 - 1} = 1.0$ V; (c) nearest level 9 = 1001.
$2 \times 15 = 30$ kHz.
$48\,000 \times 24 \times 2 = 2.30 \times 10^{6}$ bit s⁻¹.
Resolution $= \frac{5.0}{2^{10} - 1} = \frac{5.0}{1023} = 4.9$ mV; maximum error = half a step $= 2.4$ mV.
$8000 \times 8 \times 60 = 3.84 \times 10^{6}$ bits $= 4.8 \times 10^{5}$ bytes $= 480$ kB.
SNR = 100; $C = 20 \times 10^{6} \times \log_2 101 = 20 \times 10^{6} \times 6.66 = 1.3 \times 10^{8}$ bit s⁻¹.
Little benefit. 8 kHz already exceeds $2 \times 3.4 = 6.8$ kHz, so all the frequencies present are captured; doubling $f_{s}$ doubles the bit rate and storage without adding information. To improve quality, either keep higher frequencies (filter at, say, 7 kHz and sample at 16 kHz, which does help) or use more bits per sample to reduce quantisation error.

Key points

ADC: sample at $f_s \ge 2f_{\max}$ , then quantise to $2^{n}$ levels; resolution $\frac{V_{\max}}{2^n - 1}$ ; maximum error half a step.
Bit rate $= f_s \times n \times$ channels; telephone 64 kbit s⁻¹, CD 1.41 Mbit s⁻¹.
DAC plus a reconstruction filter turns the codes back into a smooth signal.
Shannon: $C = B\log_2\left(1 + \frac{S}{N}\right)$ ; more bits per second need more bandwidth or a better SNR.

Sources and credits

Video: A/D Conversion, Sample Rate & Bit Depth for audio explained, Mixed Signals, embedded from YouTube.

Written by Ganzaa. The conversion diagram (drawn to scale from the tabulated values), the worked examples and the practice exercises are by Ganzaa.

1A voice note as a list of numbers

2Sampling

3Quantisation

4Back to analogue: the DAC

5How fast can a channel carry bits?

A voice note as a list of numbers

Sampling

Quantisation

Back to analogue: the DAC

How fast can a channel carry bits?