Physical Layer - NOISE | Network Operations and Internet
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Transcript Physical Layer - NOISE | Network Operations and Internet
Physical Layer:
Signals, Capacity, and Coding
CS 4251: Computer Networking II
Nick Feamster
Fall 2008
This Lecture
• What’s on the wire?
– Frequency, Spectrum, and Bandwidth
• How much will fit?
– Shannon capacity, Nyquist
• How is it represented?
– Encoding
Digital Domain
• Digital signal: signal where intensity maintains
constant level for some period of time, and then
changes to some other level
– Amplitude: Maxumum value (measured in Volts)
– Frequency: Rate at which the signal repeats
– Phase: Relative position in time within a single period
of a signal
– Wavelength: The distance between two points of
corresponding phase ( = velocity * period)
Any Signal: Sum of Sines
• Our building block:
A sin( x
• Add enough of them to
get any signal f(x) you
want!
• How many degrees of
freedom?
• What does each
control?
• Which one encodes the
coarse vs. fine structure
of the signal?
Fourier Transform
• Continuous Fourier transform:
F(k ) F f ( x)
• Discrete Fourier transform:
n 1
Fk f x e
f ( x) e
2ik x
dx
2i kn x
x 0
• F is a function of frequency – describes how much of
each frequency f contains
• Fourier transform is invertible
Skipping a Few Steps
• Any square wave with amplitude 1 can be
represented as:
Spectrum and Bandwidth
• Any time domain signal can be represented in
terms of the sum of scaled, shifted sine waves
• The spectrum of a signal is the range of
frequencies that the signal contains
– Most signals can be effectively represented in finite
bandwidth
• Bandwidth also has a direct relationship to data
rate…
Relationship: Data Rate and Bandwidth
• Goal: Representation of square wave in a form
that receiver can distinguish 1s from 0s
• Signal can be represented as sum of sine waves
• Increasing the bandwidth means two things:
– Frequencies in the sine wave span a wider spectrum
– “Intervals” in the original signal occur more often
• [Include representation of square wave as sum
of sine waves here. Derive data rate from
bandwidth.]
Analog vs. Digital Signaling
• Analog signal: Continuously varying EM wave
• Digital signal: Sequence of voltage pulses
Signal
Analog
Analog
Data
Digital
Digital
Signal occupies same Codec produces
spectrum as analog
bitstream
data
Digital data encoded
using a modem
Signal consists of two
voltage levels
Transmission Impairments
• Attenuation
– The strength of a signal falls off with distance over
any transmission medium
• Delay distortion
– Velocity of a signal’s propagation varies w/ frequency
– Different components of the signal may arrive at
different times
• Noise
Attenuation
• Signal strength attentuation is typically
expressed as decibel levels per unit distance
• Signal must have sufficient strength to be:
– Detected by the receiver
– Stronger than the noise in the channel to be received
without error
• Note: Increasing frequency typically increases
attentuation (often corrected with equalization)
Sources of Noise
• Thermal noise: due to agitation of electrons,
function of temperature, present at all
frequencies
• Intermodulation noise: Signals at two different
frequencies can sometimes produce energy at
the sum of the two
• Crosstalk: Coupling between signals
Channel Capacity
• The maximum rate at which data can be transmitted over
a given communication path
• Relationship of
– Data rate: bits per second
– Bandwidth: constrained by the transmitter, nature of
transmission medium
– Noise: depends on properties of channel
– Error rate: the rate at which errors occur
• How do we make the most efficient use possible of a
given bandwidth?
– Highest data rate, with a limit on error rate for a given bandwidth
Nyquist Bandwidth
• Consider a channel that has no noise
• Nyquist theorem: Given a bandwidth B, the
highest signal rate that can be carried is 2B
• So, C = 2B
– But (stay tuned), each signal element can represent
more than one bit (e.g., suppose more than two signal
levels are used)
– So … C = 2B lg M
• Results follow from signal processing
– Shannon/Nyquist theorem states that signal must be
sampled at twice its highest rate to avoid aliasing
Shannon Capacity
• All other things being equal, doubling the
bandwidth doubles the data rate
• What about noise?
– Increasing the data rate means “shorter” bits
– …which means that a given amount of noise will
corrupt more bits
– Thus, the higher the data rate, the more damage that
unwanted noise will inflict
Shannon Capacity, Formally
• Define Signal-to-Noise Ratio (SNR):
– SNR = 10 log (S/N)
• Then, Shannon’s result says that, channel
capacity, C, can be expressed as:
– C = B lg (1 + S/N)
• In practice, the achievable rates are much lower,
because this formula does not consider impulse
noise or attenuation
Example
• Bandwidth: 3-4MHz
• S/N: 250
• What is the capacity?
• How many signal levels required to achieve the
capacity?
Modulation
• Baseband signal: the input
• Carrier frequency: chosen according to the
transmission medium
• Modulation is the process by which a data
source is encoded onto a carrier signal
• Digital or analog data can be modulated onto
digital and analog signals
Data Rate vs. Modulation Rate
• Data rate: rate, in bits per second, that a signal
is transmitted
• Modulation rate: the rate at which the signal
level is changed (baud)
Digital Data, Digital Signals
• Simplest possible scheme: one voltage level to
“1” and another voltage level to “0”
• Many possible other encodings are possible,
with various design considerations…
Aspects of a Signal
• Spectrum: a lack of high-frequency components
means that less bandwidth is required to
transmit the signal
– Lack of a DC component is also desirable, for various
reasons
• Clocking: Must determine the beginning and
end of each bit position.
– Not easy! Requires either a separate clock lead, or
time synchronization
• Error detection
• Interference/Noise immunity
• Cost and complexity
Nonreturn to Zero (NRZ)
• Level: A positive constant voltage represents
one binary value, and a negative contant voltage
represents the other
• Disadvantages:
– In the presence of noise, may be difficult to
distinguish binary values
– Synchronization may be an issue
Improvement: Differential Encoding
• Example: Nonreturn to Zero Inverted
– Zero: No transition at the beginning of an interval
– One: Transition at the beginning of an interval
• Advantage
– Since bits are represented by transitions, may be
more resistant to noise
• Disadvantage
– Clocking still requires time synchronization
Biphase Encoding
• Transition in the middle of the bit period
– Transition serves two purposes
• Clocking mechanism
• Data
• Example: Manchester encoding
– One represented as low to high transition
– Zero represented as high to low transition
Aspects of Biphase Encoding
• Advantages
– Synchronization: Receiver can synchronize on the
predictable transition in each bit-time
– No DC component
– Easier error detection
• Disadvantage
– As many as two transitions per bit-time
• Modulation rate is twice that of other schemes
• Requires additional bandwidth