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**Physical Layer: Signals, Capacity, and Coding**

CS 4251: Computer Networking II Nick Feamster Fall 2008

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**This Lecture What’s on the wire? How much will fit?**

Frequency, Spectrum, and Bandwidth How much will fit? Shannon capacity, Nyquist How is it represented? Encoding

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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)

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**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?

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**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

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Skipping a Few Steps Any square wave with amplitude 1 can be represented as:

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**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…

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**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.]

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**Analog vs. Digital Signaling**

Analog signal: Continuously varying EM wave Digital signal: Sequence of voltage pulses Signal Analog Digital Signal occupies same spectrum as analog data Codec produces bitstream Digital data encoded using a modem Signal consists of two voltage levels Analog Data Digital

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**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

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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)

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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

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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

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**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

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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

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**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

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**Example Bandwidth: 3-4MHz S/N: 250 What is the capacity?**

How many signal levels required to achieve the capacity?

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**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

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**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)

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**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…

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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

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**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

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**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

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**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

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**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

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