Physical Layer:Signals, Capacity, and Coding

CS 4251: Computer Networking IINick 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:

• 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?

xAsin(

Fourier Transform• Continuous Fourier transform:

• Discrete Fourier transform:

• F is a function of frequency – describes how much of each frequency f contains

• Fourier transform is invertible

dxexfk xikxf

2)()(F )(F

1

0

2kF

n

x

xix

nk

ef

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 occupies same spectrum as analog data

Codec produces bitstream

Digital data encoded using a modem

Signal consists of two voltage levels

Analog Digital

Analog

Digital

Data

Signal

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