1 data transmission lesson 3 nets2150/2850. 2 lesson outline understand the properties a signal...

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Data Transmission

Lesson 3NETS2150/2850

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Lesson Outline

Understand the properties a signal Explain the difference of Data vs

Signal Understand the influence of

attenuation, delay distortion and noise on signal propagation

Appreciation of unit of decibel

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Position of the physical layer

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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

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

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Analogue & Digital Signals

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PeriodicSignals

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In data communication, we commonly use periodic analogue signals and

aperiodic digital signals.

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A Sine Wave

s(t) = A sin(2ft + )

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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

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Amplitude

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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.

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Period and frequency

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Frequency and period are inverses of each other

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Table 3.1 Units of periods and frequenciesTable 3.1 Units of periods and frequencies

Unit Equivalent Unit Equivalent

Seconds (s) 1 s hertz (Hz) 1 Hz

Milliseconds (ms) 10–3 s kilohertz (KHz) 103 Hz

Microseconds (s) 10–6 s megahertz (MHz) 106 Hz

Nanoseconds (ns) 10–9 s gigahertz (GHz) 109 Hz

Picoseconds (ps) 10–12 s terahertz (THz) 1012 Hz

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ExampleExample

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

SolutionSolution

We make the following substitutions:100 ms = 100 10-3 s = 100 10-3 10 s = 105 s

Now we use the inverse relationship to find the frequency, changing hertz to kilohertz100 ms = 10-1 s f = 1/10-1 Hz = 10 Hz = 10 10-3 KHz = 10-2 KHz

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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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Relationships between different phases

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ExampleExample

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

SolutionSolution

We know that one complete cycle is 360 degrees.

Therefore, 1/6 cycle is

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

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Sine wave examples

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Sine wave examples (continued)

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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)

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An analogue signal is best represented in the frequency domain

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Time and frequency domains

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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.

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

made of many frequencies.made of many frequencies.

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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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Three odd harmonics

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Adding first three harmonics

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Frequency spectrum comparison

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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

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Analogue and Digital Data

Analogue Continuous values within some

interval e.g. sound

Digital Discrete values e.g. text, integers

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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

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Advantages & Disadvantages

of Digital Pro:

Cheaper Less susceptible to noise

Con: Greater attenuation

Pulses become rounded and smaller Leads to loss of information

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Attenuation of Digital Signals

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Data vs Signal

Analogue Analogue

Analogue

Analogue

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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

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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

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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

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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

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Attenuation and Dispersion (Delay Distortion)

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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

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Signal corruption

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Delay Distortion

Propagation velocity varies with frequency Different signal component travel at

different rate resulting in distortion

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Noise

Additional unwanted signals inserted between transmitter and receiver e.g. thermal noise, crosstalk etc.

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Spectrum & Bandwidth

Spectrum range of frequencies contained in

signal Bandwidth

width of spectrum band of frequencies containing most

of the energy

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Bandwidth

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ExampleExample

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.

SolutionSolution

B = fh  fl = 900 100 = 800 HzThe spectrum has only five spikes, at 100, 300, 500, 700, and 900

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ExampleExample

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

SolutionSolution

B = fB = fhh f fll

20 = 60 20 = 60 ffll

ffll = 60 = 60 20 = 40 Hz20 = 40 Hz

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ExampleExample

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?

SolutionSolution

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

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A digital signal

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Figure 3.17 Bit rate and bit interval

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ExampleExample

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

SolutionSolution

The bit interval is the inverse of the bit rate.

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

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A digital signal is a composite signal A digital signal is a composite signal with an infinite bandwidthwith an infinite bandwidth

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Baud rate and bit-rate

bit rate is the number of bits transmitted per second

baud rate is 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

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Baud rate and bit-rate (2)

VS

One signal element conveys 1 bit

2-level signal

One signal element conveys 2 bit

Multilevel signal

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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

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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)

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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

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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!

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Example

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

B = 4 – 3 = 1 MHzSNRdB = 24 dB = 10log10(SNR)

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

8 106 = 8 Mbps

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ExampleExample

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 logC = B log22 (1 + SNR) = B log (1 + SNR) = B log22 (1 + 0) (1 + 0)

= B log= B log22 (1) = B (1) = B 0 = 0 0 = 0

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ExampleExample

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 logC = B log22 (1 + SNR) = 3000 log (1 + SNR) = 3000 log22 (1 + 3162) (1 + 3162)

= 3000 log= 3000 log22 (3163) (3163)

C = 3000 C = 3000 11.62 = 34,860 bps 11.62 = 34,860 bps

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ExampleExample

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?

Then we use the Nyquist formula to find the number of signal levels.

66 Mbps = 2 Mbps = 2 1 MHz 1 MHz log log22 LL L = 8 L = 8

SolutionSolution

C = B logC = B log22 (1 + SNR) = 10 (1 + SNR) = 1066 log log22 (1 + 63) = 10 (1 + 63) = 1066 log log22 (64) = 6 Mbps (64) = 6 Mbps

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

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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