Transcript Chapter 12 - XSL: Extensible Stylesheet Language

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PART III
expanded by Jozef Goetz, 2009
The McGraw-Hill Companies, Inc., 2006
Jozef Goetz, 2009
expanded by Jozef Goetz, 2009
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3.4 Transmission Impairment
Attenuation
Distortion
Noise
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Figure 3.25
Impairment types
Attenuation = means loss of energy
Distortion means the signal changes its form or shape
Noise is another cause of impairment
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Data Rate of a Channel.
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• Theory: A perfect (noiseless) channel will still have a finite
transmission capacity.
• Introducing noise into a channel will further reduce the capacity of that
channel.
• Max Bit Rate for noiseless channel(is rarely achieved):
If bandwidth = B and the signal-power-to-noise-power ratio is
SNR = S/N
The ratio itself is not quoted, instead the quantity is given
SNRdB = 10 log SNR [decibels]
= 10 log10 (S/N) [decibels]
A ratio of 100 is 20 dB, ratio of 1000 is 30 dB
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log  log10
Figure 3.21
Attenuation
Attenuation = means loss of energy
to overcome the resistance of the medium (converted to heat)
We can use the decibel to measure the changes in the strength
of a signal
The decibel (dB) measures the relative strength of 2 signals
A = 10 log10 (P2/P1) [dB]
where P1 and P2 are the powers of a signal at points 1 and 2, respectively.
If A < 0 a signal is attenuated
If A > 0 a signal is amplified
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Example 3.26
Imagine a signal travels through a transmission medium and
its power is reduced to half.
This means that P2 = 1/2 P1.
In this case, the attenuation (loss of power) can be calculated
as
Solution
A = 10 log10 (P2/P1) = 10 log10 (0.5P1/P1) = 10 log10 (0.5)
= 10(–0.3) = –3 dB
A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.
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Example 3.27
Imagine a signal travels through an amplifier and its
power is increased ten times.
This means that P2 = 10 x P1.
In this case, the amplification (gain of power) can be
calculated as
A = 10 log10 (P2/P1) = 10 log10 (10P1/P1)
= 10 log10 (10) = 10 (1) = 10 dB
What is A when its power is increased 2 times?
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Example 3.28
One reason that engineers use the decibel to measure the changes
in the strength of a signal is that decibel numbers can be added
(or subtracted) when we are talking about several points instead
of just two (cascading).
In Figure 3.27 a signal travels a long distance from point 1 to point 4.
•The signal is attenuated by the time it reaches point 2.
•Between points 2 and 3, the signal is amplified.
•Again, between points 3 and 4, the signal is attenuated.
•We can find the resultant decibel for the signal just by adding the decibel measurements
between each set of points.
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Figure 3.27
Example 3.28
strength of signal = –3dB + 7dB – 3dB
= +1 dB
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Example 3.29
Sometimes the decibel is used to measure signal power in
milliwatts.
•In this case, it is referred to as dBm and is calculated as
dBm = 10 log10 Pm ,
•where Pm is the power in milliwatts.
Find/calculate the power of a signal with dBm = −30.
Solution
We can calculate the power in the signal as
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Example 3.30
The loss in a cable is usually defined in decibels per
kilometer (dB/km). If the signal at the beginning of a
cable with −0.3 dB/km has a power of 2 mW,
what is the power of the signal at 5 km?
Solution
The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB.
We can calculate the power as
XXXXXX
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dB
Transmission of Light through Fiber
•
Attenuation of light through fiber in the infrared region of the
spectrum (visible light is 0.4 – 0.7 microns)
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•
3 wavelength bands (25,000 – 30,000 GHz wide) are used for optical
communication.
They are centered at .85, 1.30, 1.55 microns
•
Last 2 have attenuation property less than 5% loss per km
(kilometer)
•
.85 microns band has higher attenuation but at that wavelength
the laser and electronics can be made the same material (gallium
arsenide)
Attenuation = 10 log (R / T) [decibels]
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Where: T = transmitted power, R = received power
e.g. a factor of 2 loss gives an attenuation of
A = 10 log10 0.5 = -3 dB
•
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Figure 3.28
Distortion
Distortion means the signal changes its form or shape.
Each signal has its own
propagation speed through a medium,
so its own delay
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Figure 3.29 Noise is another cause of impairment
Noise types:
•thermal
•the random motion of electrons
•induced noise
•comes from e.g. motor and appliances
•crosstalk
•one wire on the other
•impulse noise
• comes from a signal with high energy in very
short period of time: e.g. power lines, lightning
can corrupt the signal.
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Example 3.31
The power of a signal is 10 mW (miliwatt) and the power
of the noise is 1 μW;
what are the values of SNR (signal-power-to-noisepower ratio) and SNRdB ?
Solution
The values of SNR and SNRdB can be calculated as
follows:
μW
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Example 3.32
The values of SNR and SNRdB for a noiseless channel
are infinite
We can never achieve this ratio in real life; it is an ideal.
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Figure 3.30 Two cases of SNR: a high SNR and a low SNR
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3.5 Data Rate Limits
Noiseless Channel: Nyquist Bit Rate
Noisy
Channel: Shannon Capacity
Using Both Limits
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3.5 Data Rate Limits
A very important consideration in data
communications is how fast we can send data,
in bits per second, over a channel.
Data Rate depends on 3 factors:
1. The bandwidth available
2. The levels of signals we can use
3. The quality (level of the noise) of
the channel
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Noiseless Channel: Nyquist Bit Rate
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Henry Nyquist proved that if an arbitrary signal has been run through a low-pass filter
of bandwidth B, the filtered signal can be completely reconstructed by making only (at least)
2B [samples/sec] (data encoded rate)
Nyquist theoretical maximum bit rate:
Nyquist Bit Rate [bps] = 2 B log2 L
= sampling rate [sample/sec] x log2 L[#bits/sample] =
2B [1/sec] x log2 L [bit] = 2B log2 L [bit/sec]
B – bandwidth
L - # of signal levels
Each signal may convey several bits –
for example if 8 voltages are possible per signal, then 3 bits are sent on every signal
(per sample).
If the signal is BINARY (only two voltage levels), then the bit rate is equal to
2B (or the baud rate).
•e.g. a noiseless B = 3-kHz channel cannot transmit binary (i.e. two-level) signals at a
bit rate exceeding 2B = 6000 bps.
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Note
Increasing the levels of a signal L may
reduce the reliability of the system
•b/c it is very unreliable to distinguish
signals for high value of L e.g. 64 by the
receiver
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Example 3.33
Does the Nyquist theorem bit rate agree with the intuitive
bit rate described in baseband transmission?
Solution
They match when we have only two levels. We said, in
baseband transmission, the
bit rate = 2 x B
if we use only the first harmonic in the worst case.
However, the Nyquist formula is more general than what
we derived intuitively;
it can be applied
1. to baseband transmission and modulation
2. it can be applied when we have two or more levels
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Example 3.34
Consider a noiseless channel with a bandwidth of 3000 Hz
transmitting a signal with two signal levels.
The maximum bit rate can be calculated as
Bit Rate [bps] = 2 B log2 L
Bit Rate = 2  3000[1/sec]  log2 2 [bit] = 6000 [bps]
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Example 3.35
Consider the same noiseless channel, transmitting a
signal with 4 signal levels (for each level, we send 2
bits). The maximum bit rate can be calculated as:
Bit Rate [bps] = 2 B log2 L
Bit Rate = 2 x 3000 [1/sec] x log2 4 [bit]= 12,000 [bps]
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Example 3.36
We need to send 265 kbps over a noiseless channel with
a bandwidth of 20 kHz. How many signal levels do we
need?
Solution
We can use the Nyquist formula as shown:
Since this result is not a power of 2, we need to
•either increase the number of levels or
•reduce the bit rate.
If we have 128 levels, the bit rate is 280 kbps.
If we have 64 levels, the bit rate is 240 kbps.
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Noisy Channel: Shannon Capacity
Theory: A perfect (noiseless) channel will still have a finite
transmission capacity.
• Introducing noise (thermal or white noise) into a channel will further
reduce the capacity of that channel.


Theoretical maximum data rate for a noisy
channel = Capacity C
C [bps] = B log2 (1 + SNR)
SNR - signal-to-noise ratio

the formula defines characteristic of the channel, not the
method of transmission b/c there is no indication of the signal
level
•
E.g. a channel of B = 3000 Hz bandwidth with the a signal to thermal noise
ratio of 3162 dB (typical for analog tel. system) can never transmit much more
than 30,000 bps
•
•
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No matter how many or how few signal levels are used and
No matter how often or how infrequently samples are taken
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Example 3.37
Consider an extremely noisy channel in which the value
of the signal-to-noise ratio is almost zero.
In other words, the noise is so strong that the signal is
faint. For this channel the capacity is calculated as
C [bps] = B log2 (1 + SNR) = B log2 (1 + 0)
= B log2 (1) = B  0 = 0
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Example 3.38
We can calculate the theoretical highest bit rate of a
regular telephone line.
A telephone line normally has a bandwidth of 3000 Hz
(300 Hz to 3300 Hz).
The signal-to-noise ratio is usually 3162. For this
channel the capacity is calculated as
C = B log2 (1 + SNR) = 3000 [1/s] log2 (1 + 3162) [bit]
= 3000 log2 (3163) [bit/s]
C = 3000  11.62 = 34,860 bps
This is the highest bit rate. If we want to send faster than
this, we can either increase B or SNR
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Example 3.39
The signal-to-noise ratio is often given in decibels.
Assume that SNRdB = 36 and the channel bandwidth is 2
MHz.
The theoretical channel capacity can be calculated as
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Example 3.40
C = B log2 (1 + SNR)
For practical purposes, when the SNR is very high, we
can assume that SNR + 1 is almost the same as SNR. In
these cases, the theoretical channel capacity can be
simplified to
For example, we can calculate the theoretical capacity of
the previous example as
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Example 3.41 Using both formulas
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We have a channel with a 1 MHz bandwidth. The SNR
for this channel is 63;
•what is the appropriate max bit rate and signal level?
Solution
First, we use the Shannon formula to find our upper
limit.
C = B log2 (1 + SNR) = 106 [1/s] log2 (1 + 63) [bit] = 106 log2 (64) [bit/s] = 6 Mbps
Then we use the Nyquist formula to find the number of
signal levels.
Bit Rate [bps] = 2  B  log2 L
For better performance we choose 4 Mbps, the closest power of 2 but less than 6 Mbps (see C)
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4 [Mbps] = 2  1 [MHz]  log2 L [bit]  L = 4
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Note
•The Shannon capacity gives us the
upper limit of data rate;
C [bps] = B log2 (1 + SNR)
•the Nyquist formula tells us how many
signal levels we need.
Bit Rate [bps] = 2  B  log2 L
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3-6 PERFORMANCE
•One important issue in networking is the performance of
the network—how good is it?
•We discuss quality of service, an overall measurement
of network performance, in greater detail in Chapter 24.
•In this section, we introduce terms that we need for future
chapters.
Topics discussed in this section:
Bandwidth
Throughput
Latency (Delay)
Bandwidth-Delay Product
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Note
In networking, we use the term
bandwidth in two contexts.
❏ The first, bandwidth in hertz, refers to
•the range of frequencies in a
composite signal or
•the range of frequencies that a channel
can pass.
❏ The second, bandwidth in bits per
second, refers to the speed of bit
transmission in a channel or link.
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Note
An increase in bandwidth in hertz means
an increase in bandwidth in bits/second
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Example 3.42
The bandwidth of a subscriber line is 4 kHz for voice or
data.
The bandwidth of this line for data transmission
can be up to 56,000 bps (b/c of 2 B log2 L) using a
sophisticated modem to change the digital signal to
analog.
Example 3.43
If the telephone company improves the quality of the line
and increases the bandwidth to 8 kHz, we can send
112,000 bps by using the same technology as mentioned in
Example 3.42.
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Throughput
1. Throughput the measurement of how fast data can pass through an
entity
The bandwidth is a potential measurement of a link,
but throughput is an actual measurement.
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Example 3.44
A network with bandwidth of 10 Mbps can pass only an
average of 12,000 frames per minute with each frame
carrying an average of 10,000 bits.
What is the throughput of this network?
Solution
We can calculate the throughput as
The throughput is almost one-fifth of the bandwidth in
this case.
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Propagation
2. Propagation time measures the distance
a signal can travel through a medium in 1 sec
In a vacuum, light is propagated with a speed 3 x 10 ^ 8 m/s = 300,000 km/s
Distance = Propagation speed x Propagation time
D=SxT
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Example 3.45
What is the propagation time if the distance between the
two points is 12,000 km? Assume the propagation speed
to be 2.4 × 10^8 m/s in cable.
Distance = Propagation speed x Propagation time
Solution
We can calculate the propagation time as
The example shows that a bit can go over the Atlantic
Ocean in only 50 ms if there is a direct cable between the
source and the destination.
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Transmission time
3. Transmission time measures the time required
for transmission of a message
•A time between the first bit leaving the sender and
the last bit arriving at the receiver.
Transmission time = Message size / Bandwidth
or
Message size = Bandwidth x Transmission time
In a vacuum, light is propagated with a speed 3 x 10 ^ 8 m/s = 300,000 km/s
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Example 3.46
•What are the propagation time and the transmission
time for a 2.5-kbyte message (an e-mail) if the bandwidth
of the network is 1 Gbps?
•Assume that the distance between the sender and the
receiver is 12,000 km and that light travels at 2.4 × 10^8
m/s.
Solution
We can calculate the propagation and transmission time
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Example 3.46 (continued)
•Note that in this case, because the message is short and
the bandwidth is high, the dominant factor is the
propagation time, not the transmission time.
•The transmission time can be ignored.
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Example 3.47
What are the propagation time and the transmission time
for a 5-Mbyte message (an image) if the bandwidth of the
network is 1 Mbps?
•Assume that the distance between the sender and the
receiver is 12,000 km and that light travels at 2.4 × 108
m/s.
Solution
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Example 3.47 (continued)
Note that in this case, because the message is very long
and the bandwidth is not very high, the dominant factor
is the transmission time, not the propagation time.
The propagation time can be ignored.
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4. Delay (Latency)
Delay defines how long it takes for an entire msg to completely arrive
at the destination from the time the 1st bit is sent out from the source.
4. Delay = Propagation time +
Transmission time +
Queuing
time +
Processing delay
Queuing time – the time needed to hold
the message before processing
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Figure 3.31 Filling the link with bits for case 1
Delay defines how long it takes for an entire msg to completely arrive
at the destination from the time the 1st bit is sent out from the source.
Product 1 x 5 is the maximum bits can fill the link.
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Figure 3.32 Filling the link with bits in case 2
5
5
p.93 correct it
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Note
5. The bandwidth-delay product defines
the number of bits that can fill the link.
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Concept of bandwidth-delay product
We can think about the link between two points as a pipe.
The cross section of the pipe represents the bandwidth,
and the length of the pipe represents the delay.
volume of the
bandwidth-delay product, as
We can say the
3.33.
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pipe defines the
shown in Figure
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Wavelength
Wavelength is the distance a simple signal can travel in one period
V =λ/T =λ f
λ - wavelength
T – period
f - frequency
T = λ/V
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or
in air V = c speed of light,
so
c = λ f or
λ=c/f =cT
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Wavelength
Wavelength λ binds the period T (or the frequency f ) of a simple sine wave
to the propagation speed V of the medium.
The frequency of a signal is independent of the medium
the wavelength depends on both the frequency and the medium:
λ=c/f =cT
V =λ/T =λ f
λ - wavelength
T – period
f - frequency
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in air V = c speed of light,
so
c = λ f or
λ=c/f =cT
Summary
The physical layer is responsible for
transmitting a bit stream over a physical
medium. It is concerned with
a. physical characteristics of the media
b. representation of bits
c. type of encoding
d. synchronization of bits
e. transmission rate and mode
f. the way devices are connected with each
other and to the links
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ANALOG AND DIGITAL
DATA TRANSMISSION SUMMARY
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Analog = continuous
Digital = discrete
Data - entities that convey meaning, or information
Signals - are electromagnetic representations of
data.
Signaling - is the physical propagation of signal
along a suitable medium.
Transmission - communication of data by the
propagation and processing of signals.
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Metric Units
The metric prefixes
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Analog and Digital Signals:
Summary
•Signals are means by which data is propagated
•Analog
• Continuously variable
• Various media
• wire, fiber optic,
• Speech bandwidth 100Hz to 7kHz
• Telephone bandwidth 300Hz to 3400Hz
• Video bandwidth 4MHz
Digital
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Summary
* Data must be transformed into electromagnetic
signals prior to transmission across a network.
* Data and signals can be either analog or digital.
* A signal is periodic if it consists of a continuously
repeating pattern.
* Each sine wave can be characterized by its amplitude,
frequency, and phase.
* Frequency and period are inverses of each other.
* A time-domain graph plots amplitude as a function of
time.
* A frequency-domain graph plots each sine wave’s
peak amplitude against its frequency.
* By using Fourier analysis, any composite signal can be
represented as a combination of simple sine waves.
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Summary
* The spectrum of a signal consists of the sine
waves that make up the signal.
* The bandwidth of a signal is the range of
frequencies the signal occupies.
* Bandwidth is determined by finding the
difference between the highest and lowest
frequency components.
• Bit rate (number of bits per second) and bit
interval (duration of 1 bit) are terms used to
describe digital signals.
* A digital signal is a composite signal with an
infinite bandwidth.
* Bit rate and bandwidth are proportional to
each other.
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Summary
* The Nyquist formula determines the theoretical data
rate for a noiseless channel.
* The Shannon capacity determines the theoretical
maximum data rate for a noisy channel.
* Attenuation, distortion, and noise can impair a signal.
* Attenuation is the loss of a signal’s energy due to the
resistance of the medium.
* The decibel measures the relative strength of two
signals or a signal at two different points.
* Distortion is the alteration of a signal due to the
differing propagation speeds of each of the
frequencies that make up a signal.
* Noise is the external energy that corrupts a signal.
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Wavelength – ignore it
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• Data is transmitted through a wave
•
The higher frequency, the more data that
can be encoded and transmitted
•
•
•
•
A few bits per Hertz at low frequencies to 8 bits at high frequencies (*)
•
more harmonics can be transmitted
A coaxial cable with a 750 MHz bandwidth can carry several Gb/sec
λ f = c , solve it for f and differentiate with
respect to λ, we get
df / dλ = - c / λ²
Go to finite differences, instead of differentials and
only look at absolute values we get
Δf = - c Δλ / λ²
(1)
Where: Δλ - the width of a wavelength band,
Jozef Goetz, 2009 Δf - the frequency band
Wavelength
Wavelength
- ignore it
Δf = - c Δλ / λ²
(1)
Where: Δλ - the width of a wavelength band,
Δf - the frequency band
See fig on slide Transmission of Light through Fiber :
λ = 1.3 micrometer, Δλ = 0.17 micrometer,
So Δf = 30 THz and data encoded rate per Hertz at
8bits/sec, we get 240 Tbps
Conclusion from (1) and (*):
•
The wider the wavelength band, the higher
the data rate.
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