# Data Transmission Lesson 3 NETS2150/2850.

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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 attenuation, delay distortion and noise on signal propagation Appreciation of unit of decibel

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. McGraw-Hill The McGraw-Hill Companies, Inc., 2004

Signals Analogue signal Digital signal Periodic 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

Periodic Signals

In data communication, we commonly use periodic analogue signals and aperiodic digital signals.
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s(t) = A sin(2ft + ) A Sine Wave McGraw-Hill
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Sine Wave Peak Amplitude - A Frequency - f
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

Amplitude McGraw-Hill The McGraw-Hill Companies, Inc., 2004

Frequency is the rate of change with respect to time
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.

Period and frequency McGraw-Hill The McGraw-Hill Companies, Inc., 2004

Frequency and period are inverses of each other

Table 3.1 Units of periods and frequencies
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 McGraw-Hill The McGraw-Hill Companies, Inc., 2004

Example Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz Solution We make the following substitutions: 100 ms = 100  10-3 s = 100  10-3  106 ms = 105 ms 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

If a signal does not change at all, its frequency is zero
If a signal changes instantaneously, its frequency is infinite McGraw-Hill The McGraw-Hill Companies, Inc., 2004

Relationships between different phases
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Example A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians? 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 = rad McGraw-Hill The McGraw-Hill Companies, Inc., 2004

Sine wave examples McGraw-Hill The McGraw-Hill Companies, Inc., 2004

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)

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. When we change one or more characteristics of a single-frequency signal, it becomes a composite signal made of many frequencies. McGraw-Hill The McGraw-Hill Companies, Inc., 2004

According to Fourier analysis, any composite signal can be represented as a combination of simple sine waves with different frequencies, phases, and amplitudes. McGraw-Hill The McGraw-Hill Companies, Inc., 2004

Three odd harmonics

Frequency spectrum comparison

Analogue and Digital Data Transmission
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
Continuous values within some interval e.g. sound Digital Discrete values e.g. text, integers

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

Pro: Cheaper Less susceptible to noise Con: Greater attenuation Pulses become rounded and smaller Leads to loss of information

Attenuation of Digital Signals

Data vs Signal Analogue Analogue Analogue Analogue

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 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

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
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 and Dispersion (Delay Distortion)

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

Signal corruption McGraw-Hill The McGraw-Hill Companies, Inc., 2004

Delay Distortion Propagation velocity varies with frequency
Different signal component travel at different rate resulting in distortion

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

Spectrum & Bandwidth Spectrum Bandwidth
range of frequencies contained in signal Bandwidth width of spectrum band of frequencies containing most of the energy

Bandwidth McGraw-Hill The McGraw-Hill Companies, Inc., 2004

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. Solution B = fh - fl = = 800 Hz The spectrum has only five spikes, at 100, 300, 500, 700, and 900 McGraw-Hill The McGraw-Hill Companies, Inc., 2004

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Example A signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Solution B = fh - fl 20 = 60 - fl fl = = 40 Hz McGraw-Hill The McGraw-Hill Companies, Inc., 2004

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? 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. McGraw-Hill The McGraw-Hill Companies, Inc., 2004

A digital signal

Figure 3.17 Bit rate and bit interval

Example A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval) Solution The bit interval is the inverse of the bit rate. Bit interval = 1/ 2000 s = s = x 106 ms = 500 ms McGraw-Hill The McGraw-Hill Companies, Inc., 2004

A digital signal is a composite signal with 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

Baud rate and bit-rate (2)
2-level signal One signal element conveys 2 bit Multilevel signal VS One signal element conveys 1 bit

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 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) 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
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!

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

C = B log2 (1 + SNR) = B log2 (1 + 0) = B log2 (1) = B  0 = 0
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 McGraw-Hill The McGraw-Hill Companies, Inc., 2004

C = B log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163)
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 For this channel the capacity is calculated as C = B log2 (1 + SNR) = 3000 log2 ( ) = 3000 log2 (3163) C = 3000  = 34,860 bps McGraw-Hill The McGraw-Hill Companies, Inc., 2004

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? Solution C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps First, we use the Shannon formula to find our upper limit. Then we use the Nyquist formula to find the number of signal levels. 6 Mbps = 2  1 MHz  log2 L  L = 8 McGraw-Hill The McGraw-Hill Companies, Inc., 2004

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