What’s the Use of It?
Channel capacity
Table of Contents
1.
2.
3.
4.
5.
6.
7.
8.
9.
Objectives
Competences
Channel Capacity
Nyquist Bandwidth
Shannon Capacity
Decibels
Example on channel capacity
Home Exercises
Bibliography/Sources
Objectives
Students should be able to:
1. Understand what is channel capacity
2. Understand what is noise and noise free
channels
3. Calculate the channel capacity of a given
channel
Competences
To know what is:
Channel Capacity,
bandwidth,
data rate,
Noise Channels
Noiseless Channels and
decibels
Channel Capacity
• There is a variety of impairments that can distort or corrupt a
signal. To what extent do these impairments limit the
maximum achievable data rate?
• Channel Capacity is the maximum rate at which data can be
transmitted over a communication channel.
• Data rate
– In bits per second (bps)
– Rate at which data can be communicated
• Bandwidth
– In cycles per second, or Hertz
– Constrained by transmitter and medium
Nyquist Bandwidth
• Assume a noise-free channel
• If rate of signal transmission is 2B, then a signal with
frequencies no greater than B is sufficient to carry
signal rate
• or, given bandwidth B, highest signal rate is 2B
• Given a binary signal, the maximum data rate
supported by a channel of bandwidth B Hz is 2B bps
• Maximum data rate, C, can be increased by using M
signal levels
• Nyquist formula: C= 2∙B∙log2M in bps (bits per second)
• However, receiver must be able to distinguish one of M
possible signal elements. Noise and other transmission
impairments limit the practical value of M.
Shannon Capacity Formula
• Nyquist’s formula indicates that doubling BW, doubles the data rate
in a noise-free channel.
• In practice, noise is always present. So, let us consider the
relationship between data rate, noise and error rate.
• Faster data rate shortens each bit duration so a burst of noise
affects more bits
– So, at a given noise level, the higher the data rate, the higher the error
rate
•
•
•
•
Signal-to-Noise ratio (SNR or S/N) expressed in decibels
SNRdB=10 log10 (Signal power/Noise power)
Max channel Capacity is C=B∙log2(1+SNR) in bps (bits per second)
This formula is for error-free capacity and assumes white noise. In
practice, data rate is lower than C.
A few things about Decibels (1)
• It is customary to express gains, losses and relative levels in
decibels because
– Signal strength often falls off exponentially, so loss is easily
expressed in terms of the decibel, which is a logarithmic
unit
– The net gain or loss in a cascaded transmission path can be
calculated with simple addition and subtraction
• The decibel (dB) is a measure of the ratio between two signal
levels. The decibel gain is given by
GdB=10∙log10 (Output power / Input power)
GdB=10∙log10 (Pout/Pin)
A few things about Decibels (2)
• Gain is expressed in positive dB values (GdB)
• Loss is expressed in negative dB values (LdB)
• E.g. A gain of –3dB means that the power has halved and this
is a loss of power.
Power Ratio
dB
Power Ratio
dB
101
10
10-1
-10
102
20
10-2
-20
103
30
10-3
-30
104
40
10-4
-40
105
50
10-5
-50
106
60
10-6
-60
A few things about Decibels (3)
• Note that dB is a measure of relative, not absolute difference.
• The dB is also used to measure the difference in Voltage
• Since
P = V2/R
Where,
P=Power dissipated across resistance R
v = Voltage across resistance R
Then GdB = 10 log10 (Pout/Pin)
= 10 log10 [(V2out/R) /(V2in/R)]
= 20 log10 (Vout/Vin)
Similarly LdB = 20 log10 (Vin/Vout)
Example on channel capacity
• Suppose that the spectrum of a noise-free
channel is between 3 MHz and 4 MHz and
SNRdB=24 dB.
– What is the maximum achievable data rate?
– How many signal levels are required to achieve
this rate?
Solution
• Bandwidth, B=4 MHz – 3 MHz = 1 MHz = 106 Hz.
• SNRdB=24 dB = 10log10(SNR)
• Therefore, SNR=10(24/10) = 102.4 = 251.2
• Using Shannon’s formula, C=B log2(1+SNR),
C=106 log2 (1+251.2) = 7.98 x 106 ~ 8 Mbps
• Based on Nyquist’s formula, C=2B log2M in order to achieve a
data rate of 8MBps in a channel bandwidth of 1MHz, then we
need M signal levels, where M is equal to:
8x106 = 2x106 log2M => 4 = log2M => M=24=16
Home Exercises
Q1. A modem to be used with a PSTN network uses a
modulation scheme with eight levels per signalling element.
Assuming the same channel bandwidth as in Q1, but a
noiseless channel, find the maximum possible data rate.
(Answer: 18.6kbps)
Q2. Given a channel with an intended capacity of 20 Mbps, the
bandwidth of the channel is 3 MHz. Assuming white
thermal noise, what signal to noise ratio in decibels is it
required to achieve this capacity? (Answer: 20 dB)
Useful log identities
• logaB=X => aX=B
• logaB = (log10 B)/(log10 a)
Bibliography/Sources
• Books:
• Σ. Ματακίας, Α. Τσιγκόπουλος. Επικοινωνίες και Δίκτυα.
Ελλάδα: ΑΘΗΝΑ, 2003.
• Web sites:
• Wikipedia, The Free Encyclopedia
http://en.wikipedia.org/wiki/Channel_capacity