Computer Communication & Networks Lecture # 16. Today’s Menu ϞModulation/Demodulation ϞAmplitude...
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Transcript of Computer Communication & Networks Lecture # 16. Today’s Menu ϞModulation/Demodulation ϞAmplitude...
Computer Communication & Networks
Lecture # 16
Today’s Menu
Ϟ Modulation/Demodulation
Ϟ Amplitude Modulation
Ϟ Frequency Modulation
Ϟ Phase Modulation
Ϟ Quadrature Amplitude Modulation
Modulation
Ϟ Process of changing one of the characteristic of analog signal based on some digital input
Ϟ By changing one aspect of a simple electrical signal back and forth, we can use it to represent digital data
Ϟ Example: using internet over the telephone line
Modulation
There are three characteristics of a sine wave, so we can change
each of them to modulate a signal
Ϟ Amplitude Modulation
Ϟ Frequency Modulation
Ϟ Phase Modulation
We can also combine them to get another method knows as QAM
Quadrature Amplitude Modulation
Bit Rate VS Baud Rate
Bit rate
Number of bits transmitted during one second
Baud rate
Number of signal units/elements per second that are required to
represent that bit
Ϟ To measure the efficiency of computer we use bit rate
Ϟ To measure the efficiency of a data communication
system we use baud rate
Bit Rate VS Baud Rate
Lets use the transportation example we used to understand the
bandwidth concepts here to better understand bit/baud rate
Think of a baud as a car and bit as a passenger
Ϟ If 100 cars travel from one place to another carrying only one passenger (driver only), than 100 passenger are transported
Ϟ However if each car carry 4 passengers than 400 passengers are transported using 100 cars
Ϟ Note the number of cars (bauds) not the number of passengers (bits) determine the traffic and therefore need for wider highways
Bit Rate VS Baud Rate
An analog signal carries 4 bits in each signal element. If 1000 signal
elements are sent per second, find the baud rate and bit rate?
Baud rate = number of signal elements per second
Baud rate = 1000 bauds
Bit rate = baud rate * number of bits per signal element
Bit rate = 1000 * 4 = 4000 b/s
Amplitude Modulation
Ϟ Amplitude of the signal is varied to represent binary 1 or 0Ϟ Both frequency and phase remains constant, while the
amplitude changes
Ϟ Highly vulnerable to the noise interferenceϞ Noise usually affects the amplitude
Amplitude Modulation
1 bit (1) 1 bit (1) 1 bit (0) 1 bit (1) 1 bit (0)
1 baud 1 baud 1 baud 1 baud 1 baud
1 Second
Frequency Modulation
Ϟ Frequency of the signal is varied to represent binary 1 or 0Ϟ Both amplitude and phase remains constant, while the
frequency changes
Ϟ Avoids most of the noise problemsϞ We are looking for specific frequency changes over a given
number of periods, it can ignore voltage spikes
Frequency Modulation
1 bit (1) 1 bit (1) 1 bit (0) 1 bit (1) 1 bit (0)
1 baud 1 baud 1 baud 1 baud 1 baud
1 Second
Phase Modulation
Ϟ Phase of the signal is varied to represent binary 1 or 0Ϟ Both amplitude and frequency remains constant, while the
phase changes
Ϟ It is not susceptible to the noise degradationϞ Higher degree variations are not easily detected by the receiver
Phase Modulation
1 bit(1) 1 bit(1) 1 bit(0) 1 bit(1) 1 bit(0)
1 baud 1 baud 1 baud 1 baud 1 baud
1 Second
Phase Modulation
2 Phase Modulation
Ϟ There are multiple variations of phase modulation
Ϟ The version we just studied was 2 phase modulation
Ϟ The reason is that it as two phase shifts 0o and 180o
Ϟ We can see it in the constellation diagram
Constellation diagram
Bit Phase
0 0
1 180
Phase Modulation
4 Phase Modulation
Ϟ As noise does not affect this type of modulation, why not utilize it
more by adding more phases and representing more bits over
one phase
Constellation diagram
Baud rate = 4 (remember it is the number of signal elements)
Bit rate = 8 (remember it is the number of bit elements)
Bit Phase
00 0
01 90
10 180
11 270
Phase Modulation
8 Phase Modulation
Constellation diagram
Baud rate = 8
Bit rate = 24
Bit Phase
000 0
001 45
010 90
011 135
100 180
101 225
110 270
111 315