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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
slide3

Position of the physical layer

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slide4

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
slide8

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

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slide9

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
slide11

Amplitude

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slide12

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.

slide13

Period and frequency

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slide15

Table 3.1 Units of periods and frequencies

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slide16

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

slide17

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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slide18

Relationships between different phases

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slide19

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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slide20

Sine wave examples

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slide21

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)
slide23

An analogue signal is best represented in the frequency domain

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slide24

Time and frequency domains

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slide25

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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slide26

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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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
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
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
slide42

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
slide46

Bandwidth

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slide47

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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slide48

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slide49

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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slide50

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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slide53

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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slide54

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

slide62

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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slide63

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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slide64

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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