1. Data Transmission
Lesson 3
NETS2150/2850

2. 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

3. Position of the physical layer
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4. 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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5. 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

6. Analogue & Digital Signals

7. Periodic Signals

8. In data communication, we commonly use periodic analogue signals and aperiodic digital signals.
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9. s(t) = A sin(2ft + ) A Sine Wave McGraw-Hill
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10. 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

11. Amplitude
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12. 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.

13. Period and frequency
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14. Frequency and period are inverses of each other

15. 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
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16. 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

17. 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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18. Relationships between different phases
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19. 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
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20. Sine wave examples
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21. Sine wave examples (continued)
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22. 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)

23. An analogue signal is best represented in the frequency domain
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24. Time and frequency domains
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25. 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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26. 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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27. Three odd harmonics

28. Adding first three harmonics

29. Frequency spectrum comparison

30. 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

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

32. 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

33. Advantages & Disadvantages of Digital
Pro:
Cheaper
Less susceptible to noise
Con:
Greater attenuation
Pulses become rounded and smaller
Leads to loss of information

34. Attenuation of Digital Signals

35. Data vs Signal
Analogue
Analogue
Analogue
Analogue

36. 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

37. 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

38. 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

39. 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

40. Attenuation and Dispersion (Delay Distortion)

41. 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

42. Signal corruption
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43. Delay Distortion Propagation velocity varies with frequency
Different signal component travel at different rate resulting in distortion

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

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

46. Bandwidth
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47. 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
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49. 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
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50. 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.
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51. A digital signal

52. Figure 3.17 Bit rate and bit interval

53. 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
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54. A digital signal is a composite signal with an infinite bandwidth
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55. 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

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

57. 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

58. 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)

59. 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

60. 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!

61. 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

62. 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
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63. 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
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64. 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
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65. 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