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Data Transmission. Lesson 3 NETS2150/2850. Lesson Outline. Understand the properties a signal Explain the difference of Data vs Signal Understand the influence of a ttenuation, d elay d istortion and n oise on signal propagation Appreciation of unit of d ecibel.

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Data transmission

Data Transmission

Lesson 3

NETS2150/2850


Lesson outline

Lesson Outline

  • Understand the properties a signal

  • Explain the difference of Data vs Signal

  • Understand the influence of attenuation, delay distortion andnoise on signal propagation

  • Appreciation of unit of decibel


Data transmission

Position of the physical layer

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Data transmission

To be transmitted, data must be transformed to electromagnetic signals

Signals can be analogue or digital. Analogue signals can have an infinite number of values in a range; Digital signals can have only a limited number of values.

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Signals

Signals

  • Analogue signal

    • Varies in a smooth way over time

  • Digital signal

    • Maintains a constant level then changes to another constant level

  • Periodic signal

    • Pattern repeated over time

  • Aperiodic signal

    • Pattern not repeated over time


Analogue digital signals

Analogue & Digital Signals


Periodic signals

PeriodicSignals


Data transmission

In data communication, we commonly use periodic analogue signals and aperiodic digital signals.

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Data transmission

A Sine Wave

s(t) = A sin(2ft + )

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Sine wave

Sine Wave

  • Peak Amplitude - A

    • maximum strength of signal

    • In volts (V)

  • Frequency - f

    • Rate of change of signal

    • Hertz (Hz) or cycles per second

    • Period = time for one repetition (T)

    • T = 1/f

  • Phase -  (in degree or radian)

    • the position of the waveform relative to time zero

    • How far from origin when voltage change from -ve to +ve


Data transmission

Amplitude

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Data transmission

Frequency is the rate of change with respect to time. Change in a short span of time means high frequency. Change over a long span of time means low frequency.


Data transmission

Period and frequency

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Data transmission

Frequency and period are inverses of each other


Data transmission

Table 3.1 Units of periods and frequencies

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Data transmission

Solution

We make the following substitutions:

100 ms = 100  10-3 s = 100  10-3 106ms = 105ms

Now we use the inverse relationship to find the frequency, changing hertz to kilohertz

100 ms = 10-1 s f = 1/10-1 Hz = 10 Hz = 10  10-3 KHz = 10-2 KHz

Example

Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz


Data transmission

If a signal does not change at all, its frequency is zero

If a signal changes instantaneously, its frequency is infinite

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Data transmission

Relationships between different phases

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Data transmission

Solution

We know that one complete cycle is 360 degrees.

Therefore, 1/6 cycle is

(1/6) 360 = 60 degrees = 60 x 2p /360 rad = 1.046 rad

Example

A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians?

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Data transmission

Sine wave examples

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Data transmission

Sine wave examples (continued)

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Wavelength

Wavelength

  • Distance occupied by one cycle (in meters)

  • Assuming signal velocity v

    •  = vT

    • f = v

    • c = 3*108 ms-1 (speed of light in free space)


Data transmission

An analogue signal is best represented in the frequency domain

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Data transmission

Time and frequency domains

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Data transmission

A single-frequency sine wave is not useful in data communications; we need to change one or more of its characteristics to make it useful.

When we change one or more characteristics of a single-frequency signal, it becomes a composite signal made of many frequencies.

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Data transmission

According to Fourier analysis, any composite signal can be represented as a combination of simple sine waves with different frequencies, phases, and amplitudes.

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Data transmission

Three odd harmonics


Data transmission

Adding first three harmonics


Data transmission

Frequency spectrum comparison


Analogue and digital data transmission

Analogue and Digital Data Transmission

  • Data

    • Entities that convey meaning

  • Signals

    • Electric or electromagnetic (EM) representations of data

  • Transmission

    • Communication of data by propagation and processing of signals


Analogue and digital data

Analogue and Digital Data

  • Analogue

    • Continuous values within some interval

    • e.g. sound

  • Digital

    • Discrete values

    • e.g. text, integers


Analogue and digital signals

Analogue and Digital Signals

  • Means by which data are propagated

  • Analogue

    • Continuously variable

    • Speech range 100Hz to 7kHz

    • Telephone range 300Hz to 3400Hz

    • Video bandwidth 4MHz

  • Digital

    • Use two DC components


Advantages disadvantages of digital

Advantages & Disadvantages of Digital

  • Pro:

    • Cheaper

    • Less susceptible to noise

  • Con:

    • Greater attenuation

      • Pulses become rounded and smaller

      • Leads to loss of information


Attenuation of digital signals

Attenuation of Digital Signals


Data vs signal

Data vs Signal

Analogue

Analogue

Analogue

Analogue


Analogue transmission

Analogue Transmission

  • Analogue signal transmitted without regard to content

  • May be analogue or digital data

  • Attenuated over distance

  • Use amplifiers to boost signal

    • But this also amplifies noise


Digital transmission

Digital Transmission

  • Concerned with content

  • Integrity endangered by noise, attenuation etc.

  • Repeaters used

  • Repeater extracts bit pattern from received signal and retransmits

    • Attenuation is overcome

    • Noise is not amplified


Advantages of digital transmission

Advantages of Digital Transmission

  • Digital technology

    • Low cost LSI/VLSI technology (smaller)

  • Data integrity

    • Longer distances over lower quality lines

  • Capacity utilization

    • High bandwidth links economical

    • High degree of multiplexing easier with digital techniques

  • Security & Privacy

    • Encryption


Transmission impairments

Transmission Impairments

  • Signal received may differ from signal transmitted

  • Analogue - degradation of signal quality

  • Digital - bit errors

  • Caused by

    • Attenuation and attenuation distortion

    • Delay distortion

    • Noise


Attenuation a nd dispersion delay distortion

Attenuation and Dispersion (Delay Distortion)


Attenuation

Attenuation

  • Signal strength falls off with distance

  • Depends on type of medium

  • Received signal strength:

    • must be enough to be detected

    • must be sufficiently higher than noise to be received without error

  • Attenuation is an increasing function of frequency


Data transmission

Signal corruption

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Delay distortion

Delay Distortion

  • Propagation velocity varies with frequency

    • Different signal component travel at different rate resulting in distortion


Noise

Noise

  • Additional unwanted signals inserted between transmitter and receiver

    • e.g. thermal noise, crosstalk etc.


Spectrum bandwidth

Spectrum & Bandwidth

  • Spectrum

    • range of frequencies contained in signal

  • Bandwidth

    • width of spectrum

    • band of frequencies containing most of the energy


Data transmission

Bandwidth

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Data transmission

Solution

B = fh-fl = 900 - 100 = 800 Hz

The spectrum has only five spikes, at 100, 300, 500, 700, and 900

Example

If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.

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Data transmission

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Data transmission

Solution

B = fh- fl

20 = 60 - fl

fl = 60 - 20 = 40 Hz

Example

A signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency?

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Data transmission

Solution

The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies between 3000 and 4000 Hz; the signal is totally lost.

Example

A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A medium can pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal faithfully pass through this medium?

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Data transmission

A digital signal


Data transmission

Figure 3.17Bit rate and bit interval


Data transmission

Solution

The bit interval is the inverse of the bit rate.

Bit interval = 1/ 2000 s = 0.000500 s = 0.000500 x 106ms = 500 ms

Example

A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)

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Data transmission

A digital signal is a composite signal with an infinite bandwidth

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Baud rate and bit rate

Baud rate and bit-rate

  • bit rate is the number of bits transmitted per second

  • baud rateis the number of signal units per second required to represent bits

    • An important measure in data transmission

    • Represents how efficiently we move data from place to place

    • Equals bit rate divided by the number of bits represented by each signal shift


Baud rate and bit rate 2

Multilevel signal

One signal element conveys 2 bit

Baud rate and bit-rate (2)

2-level signal

VS

One signal element conveys 1 bit


Channel capacity and nyquist bandwidth

Channel capacity and Nyquist Bandwidth

  • Given bandwidth B Hz, highest signal rate is 2B

  • For binary signal, data rate supported by B Hz is 2B bps in a noiseless channel

  • Can be increased by using M signal levels

  • C = 2B log2M


Example

Example

  • Assume voice channel (range 300-3400 Hz)

  • Thus, bandwidth is 3100 Hz (i.e. B)

  • This translates to capacity of 2B = 6200 bps

  • If M = 8 signal levels (3-bit word), capacity becomes 18,600 bps (2Blog2M)


Decibels db

Decibels (dB)

  • A measure of ratio between two signal levels

  • Gain is given by:

    GdB = 10 log10 Pout dB

    Pin

  • When gain is –ve, this means loss or attenuation

    Example 1: Pin = 100mW, Pout=1 mW

    Gain = 10 log10 (1/100) = -20 dB

    • implies attenuation is 20 dB


Shannon capacity formula

Shannon Capacity Formula

  • This considers data rate, noise and error rate in the channel

  • Faster data rate shortens each bit so burst of noise affects more bits

    • At given noise level, high data rate means higher error rate

  • Signal to noise ratio (SNR)

  • Thus, Shannon’s formula is:

    • C = B log2(1+SNR)

    • Represents theoretical max capacity!


Example1

Example

  • Assume spectrum of a channel is between 3 MHz and 4 MHz and the SNR is 24 dB

    B = 4 – 3 = 1 MHz

    SNRdB = 24 dB = 10log10(SNR)

     SNR = 251

    Thus, C = B log2(1+SNR) = 106 log2(1+251)

     8  106 = 8 Mbps


Data transmission

Example

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 = B log2 (1 + SNR) = B log2 (1 + 0)= B log2 (1) = B  0 = 0

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Data transmission

Example

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 log2 (1 + 3162) = 3000 log2 (3163)

C = 3000  11.62 = 34,860 bps

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Data transmission

Solution

First, we use the Shannon formula to find our upper limit.

C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps

Then we use the Nyquist formula to find the

number of signal levels.

6 Mbps = 2  1 MHz  log2L L = 8

Example

We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level?

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Summary

Summary

  • Analogue vs Digital Transmission

  • Transmission Impairments of a signal

  • Nyquist Formula to estimate channel capacity in a noiseless environment

  • Shannon Capacity Formula estimates the upper limit of capacity with noise effect

  • Next: Transmission Media


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