chap 05 dsn
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TRANSCRIPT
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Computer Communications
Ch # 5 – Analog Signals
31 October 2012
By: Faiza Tariq
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Modulation of Digital Data
Digital-to-Analog ConversionAmplitude Shift Keying (ASK)Frequency Shift Keying (FSK)Phase Shift Keying (PSK)Quadrature Amplitude ModulationBit/Baud Comparison
Modulation Types
Amplitude Modulation If bit = 0 x(t) = 0 If bit = 1 x(t) = sin(2πft)
Frequency Modulation (FSK) If bit = 0 x(t) = sin(2πf1t) If bit = 1 x(t) = sin(2πf2t)
Phase Modulation If bit = 0 x(t) = sin(2πft) If bit = 1 x(t) = sin(2πft-θ)
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Figure 5.1 Digital-to-analog modulation
Figure 5.2 Types of digital-to-analog modulation
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Bit rate is the number of bits per second. Baud rate is the number of signal units per second. Baud rate is less than or
equal to the bit rate.
Bit Rate vs. Baud Rate Baud is like car and bit rate like passenger
If 1000 cars go from one place to another place, carrying only passenger i.e. driver then 1000 passengers could be transported
However if each car carries four passenger, then 4000 passengers are transported
No. of cars not no. of passengers determines traffic, i.e. may require wider high way
No. of bauds determines the required bandwidth, not number of bits
Baud Rate Baud rate is a measure of the number of discrete
signals that can be transmitted (or received) per unit of time
A modem’s baud rate measures the number of signals that it is capable of transmitting (or receiving) per second Baud rate represents the number of times per second that a
modem can modulate (or demodulate) the carrier signal to represent bits
Although baud rate and bit rate are sometimes used interchangeably to refer to modem data transfer speeds, these are only identical when each signal transmitted (or received) represents a signal bit A modem’s bit rate is typically higher than its baud rate
because each signal transmitted or received may represent a combination of two or more bits
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Bit Rates and Bandwidth The bandwidth of an analog channel is the difference
between the minimum and maximum frequencies it can carry A voice-grade dial-up circuit can transmit frequencies
between 300 and 3400 Hz and thus has a bandwidth of 3100 Hz
For digital circuits, bandwidth is a measure of the amount of data that can be transmitted per unit. Bits per second (bps) is the most widely used measure for digital circuits
Over time, bit rates (bps) have also become on of the key measures of modem performance (e.g. a 56 Kbps modem) However, modem bit rates are not necessarily an accurate
reflection of their data throughput rates
Carrier Signal
In analog transmission, the sending device produces a high frequency signal that acts as a basis for the information signal, this base signal is called the carrier signal or carrier frequency
The receiver device is tuned to the frequency of the carrier signal
Digital information then modulates the signal by changing one or more properties called modulation or shift keying
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Amplitude Shift Keying (ASK) Two binary values are represented by two different
amplitudes of the carrier frequency
Both frequency and phase remains constant while amplitude changes
Which voltage represents 1 and 0 is left on designers
ASK transmission is highly susceptible to noise interference because of unintentional voltages introduced onto a line (heat up)
ASK A popular ASK technique is called on/off keying
(OOK), one bit represented by no voltage and one by a voltage
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Frequency Shift Keying (FSK) Binary values are represented by two different
frequencies Resulting signal is: S(t) = A sin (2Лf1t) binary 1 S(t) = A sin (2Лf2t) binary 0 FSK avoids most of the noise problem as ASK, as no
voltage spikes, just frequency change Limiting factors of FSK are physical capabilities of the
carrier.
Phase Shift Keying (PSK) Phase of the signal is shifted to represent binary data
It is’nt susceptible to noise degradation that affects ASK
The main advantage of phase modulation is that only single frequency to send both zeros and ones in both direction (two phases in each total four phases required)
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Phase Shift Keying (PSK)Bit Phase0 01 180
Dibit Phase00 001 9010 18011 270
Tribit Phase000 0001 45010 90011 135100 180101 270111 315
000
001010
011
100
101111
0 0 1 0 1 0 0 0 1 0 1 0 1 1 0 0
Bit Baud Rate = n Bit Rate = n
00 10 10 00 10 10 11 10
Dibit Baud Rate = n Bit Rate = 2n
001 000 111 101 110
Tribit Baud Rate = n Bit Rate = 3n
Quadbit Baud Rate = n Bit Rate = 4n
0011 0001 1110 101 1101
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Figure 5.1 Digital-to-analog conversion
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Figure 5.2 Types of digital-to-analog conversion
Bit rate is the number of bits per second. Baud rate is the number of signal
elements per second.
In the analog transmission of digital data, the baud rate is less than
or equal to the bit rate.
Note
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An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate.
SolutionIn this case, r = 4, S = 1000, and N is unknown. We can find the value of N from
Example 5.1
Example 5.2
An analog signal has a bit rate of 8000 bps and a baudrate of 1000 baud. How many data elements arecarried by each signal element? How many signalelements do we need?
SolutionIn this example, S = 1000, N = 8000, and r and L areunknown. We find first the value of r and then the valueof L.
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Figure 5.3 Binary amplitude shift keying
Figure 5.4 Implementation of binary ASK
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Example 5.3
We have an available bandwidth of 100 kHz whichspans from 200 to 300 kHz. What are the carrierfrequency and the bit rate if we modulated our data byusing ASK with d = 1?
SolutionThe middle of the bandwidth is located at 250 kHz. Thismeans that our carrier frequency can be at fc = 250 kHz.We can use the formula for bandwidth to find the bit rate(with d = 1 and r = 1).
Example 5.4
In data communications, we normally use full-duplexlinks with communication in both directions. We needto divide the bandwidth into two with two carrierfrequencies, as shown in Figure 5.5. The figure showsthe positions of two carrier frequencies and thebandwidths. The available bandwidth for eachdirection is now 50 kHz, which leaves us with a datarate of 25 kbps in each direction.
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Figure 5.5 Bandwidth of full-duplex ASK used in Example 5.4
Figure 5.6 Binary frequency shift keying
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Example 5.5
We have an available bandwidth of 100 kHz whichspans from 200 to 300 kHz. What should be the carrierfrequency and the bit rate if we modulated our data byusing FSK with d = 1?
SolutionThis problem is similar to Example 5.3, but we aremodulating by using FSK. The midpoint of the band is at250 kHz. We choose 2Δf to be 50 kHz; this means
Figure 5.7 Bandwidth of MFSK used in Example 5.6
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Example 5.6
We need to send data 3 bits at a time at a bit rate of 3Mbps. The carrier frequency is 10 MHz. Calculate thenumber of levels (different frequencies), the baud rate,and the bandwidth.
SolutionWe can have L = 23 = 8. The baud rate is S = 3 MHz/3 =1000 Mbaud. This means that the carrier frequenciesmust be 1 MHz apart (2Δf = 1 MHz). The bandwidth is B= 8 × 1000 = 8000. Figure 5.8 shows the allocation offrequencies and bandwidth.
Figure 5.8 Bandwidth of MFSK used in Example 5.6
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Figure 5.9 Binary phase shift keying
Figure 5.10 Implementation of BASK
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Figure 5.11 QPSK and its implementation
Example 5.7
Find the bandwidth for a signal transmitting at 12Mbps for QPSK. The value of d = 0.
SolutionFor QPSK, 2 bits is carried by one signal element. Thismeans that r = 2. So the signal rate (baud rate) is S = N ×(1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6MHz.
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Figure 5.12 Concept of a constellation diagram
Example 5.8
Show the constellation diagrams for an ASK (OOK),BPSK, and QPSK signals.
SolutionFigure 5.13 shows the three constellation diagrams.
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Figure 5.13 Three constellation diagrams
Quadrature amplitude modulation is a combination of ASK and PSK.
Note
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Figure 5.14 Constellation diagrams for some QAMs
5-2 ANALOG AND DIGITAL
Analog-to-analog conversion is the representation ofanalog information by an analog signal. One may askwhy we need to modulate an analog signal; it isalready analog. Modulation is needed if the medium isbandpass in nature or if only a bandpass channel isavailable to us.
Amplitude ModulationFrequency ModulationPhase Modulation
Topics discussed in this section:
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Analog to Analog Conversion
A higher frequency may be needed for effective transmission of analog signal
Analog data (Voice) is combined (added) with carrier wave signal (carrier signal is used to convey the audio data)
Principal techniques used Amplitude Modulation (AM) Frequency Modulation (FM)
Figure 5.15 Types of analog-to-analog modulation
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Figure 5.16 Amplitude modulation
The total bandwidth required for AM can be determined
from the bandwidth of the audio signal: BAM = 2B.
Note
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Figure 5.17 AM band allocation
The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BFM = 2(1 + β)B.
Note
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Figure 5.18 Frequency modulation
Figure 5.19 FM band allocation
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Figure 5.20 Phase modulation
The total bandwidth required for PM can be determined from the bandwidth
and maximum amplitude of the modulating signal:
BPM = 2(1 + β)B.
Note
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Amplitude Modulation (AM)
(as shown in (a)
Amplitude Modulation (AM)
Analog Modulation
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Amplitude Modulation (AM) By changing amplitude of the carrier signal
The frequency and the phase of the carrier remains same
AM stations are allowed carrier frequencies anywhere between 530 and 1700 KHz
However each station’s carrier frequency must be separated from those on either side of it by atleast 10 KHz to avoid interference.
The total bandwidth required can be determined from the bandwidth of the audio (modulated) signal
Bw = 2 X Bw
Frequency Modulation (FM)
In FM transmission, the frequency of the carrier is modulated.
The bandwidth of the FM signal is 10 times the bandwidth of the modulated (audio) signal.
Bw = 10 X Bw
The bandwidth of an audio signal (speech and music) broadcast in stereo is 15 KHz.
Each FM station therefore needs a minimum bandwidth of 150 KHz.
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DTE – DCE Interface Data Terminal Equipment (DTE):
Data Terminal Equipment include (DTE) includes any unit that function either as a source of binary data or destination of digital data. E.g. PC, printer or fax.
Data Circuit Transmitting Equipment or Data Communication Equipment (DCE):
any functional unit that transmits or receives data in the form of an analog or digital signal
DCEs are simply instruments used to allow DTEs to communicate.
At physical layer level DCE takes data generated by a DTE, converts them to appropriate signal and transmits them across the network through a communication channel e.g. modems or multiplexers
Pixelization and Binary Representation Used in digital fax, bitmapped graphics
1-bit code: 0000000000111100011101100111111001111000011111100011110000000000
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Modems
Modulator/Demodulator
ITU-T Bell Baud Rate Bit Rate Modulation Technique
v.21 103 300 300 FSK
v.22 212 600 1200 PSK
v.23 202 1200 1200 FSK
v.26 201 1200 2400 PSK
v.27 208 1600 4800 PSK
Data Compression
In some cases data is compressed prior to transmission
For a call if transmission time is reduced, will automatically reduce the call charges.
Advantage minimum use of telephone line or less internet charges
Some modems, called intelligent modems offer compression feature, which select the algorithm to suit the type of the data being transmitted.
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Intelligent modems attempt to send fewer bits by searching for repeated pattern & sending some type of shorthand notation.
Receiving modem translates the shorthand back to the original bits creating the illusion that two modems are operating at higher speed.
ITU-T V.42 bis standard can theoretically compress to data to one – fourth its original size under ideal conditions
Since it is hard to predict if the data transmitted will have repeated pattern, the compression at this ideal level is not achieved
Character Suppression
When a frame consists of characters and a character is repeated twice or more times, may be replaced by another
The control device at the transmitter scans the frame contents prior to transmission, if contiguous string of three or more characters is located, replaces these sequence
¶¶¶¶¶¶¶ TO
7 ¶ TO
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Huffman Coding It exploits that not all the characters in a
transmitted frame occur with the same frequency. E.g.
In a frame comprising strings of character, certain characters occur more often than others
Instead of using fixed number of bits per character, we use a different encoding scheme in which the most common characters are encoded using fewer bits than less frequent characters
AAAABBCD
4 X 1 + 2 X 2 + 1 X 3 = 14 bits
v.34
Standard speed 28,800 bps
With data compression data rate can be twice the standard speed.
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Another Look at Compression
_ With 4:1 Compression, a V.34 Modem Can Receive Data at 115.2 kbps from the PC
_ However the ~30 kbps limit of the phone system is not exceeded. Still transmit at 33.6 kbps.
115.2 kbps 33.6 kbps
~35 kbpsMaximum
Compressionin Modem
56K Modem Bit rate = 56000 bps
Downloading max. speed = 56000 bps, as ISP uses digital line at other end.
Uploading max speed = 33,600 bps
Digital-Analog Conversion Upstream
Analog-Digital ConversionUpstream
DownstreamDigital-AnalogDownstream
No analog – digital conversion on downstream link makes download speed fast
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56 K Modem
56 K Standards (ITU-T) V.90
56 K dial-up modem
V.92 Less handshaking time
Supports CLI and Call Waiting, by sending IRQ
The quick connect method reduces the negotiation time from over 20 seconds to about 10 seconds.
The second additional feature is a Modem-on-Hold™ (MOH) feature. This codifies a method for the central site modem to request the client modem to go on hold, or vice versa, and is a mechanism whereby call-waiting tones can be better survived by voice-band dial-up
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28Modem Speed Standards
_ Most data modems are also fax modems_ V.14 14.4 kbps
_ V.29 9,600 bps
30Modem Training_ When two modems connect
_ They exchange information about themselves
_ This allows them to settle upon the highest supported standard in each category
_ They also do line testing during this period
_ This is called handshaking or training
_ Training takes 10 to 30 seconds
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31Modem Intelligence_ Computer Can Send Commands to Modem
_ Dial a number, including how long to wait, compression, etc.
_ Called intelligent modems
_ Hayes Developed the first Command Set_ Most modems follow the same command set
_ We call them “Hayes compatible”
_ Commands start with “AT”
_ Other Standards for Fax Modems_ Class 1 and Class 2: extensions to Hayes
36Telephone Bandwidth is Limited
_ Speed is Limited
_ Maximum speed is related to bandwidth (Shannon’s Law)
_ Maximum speed for phone lines for transmission is a little over 30 kbps
_ So modems can’t get much faster
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NULL Modem
To connect 2 PCs (DTE) with each other, modems are not needed
An interface is needed to connect these two DTEs to handle exchange
The solution by EIA standard – NULL Modem provides DTE – DTE without DCE
2 2
3 3
TX
RX
TX
RX
Example 1
An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate
Solution
Baud rate = 1000 bauds per second (baud/s)Bit rate = 1000 x 4 = 4000 bps
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Example 2
The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate?
Solution
Baud rate = 3000 / 6 = 500 baud/s
Figure 5.3 ASK
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Figure 5.4 Relationship between baud rate and bandwidth in ASK
Example 3
Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half-duplex.
Solution
In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz.
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Example 4
Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate?
Solution
In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps.
Example 5
Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the bandwidths in each direction. Assume there is no gap between the bands in the two directions.
Solution
For full-duplex ASK, the bandwidth for each direction isBW = 10000 / 2 = 5000 Hz
The carrier frequencies can be chosen at the middle of each band (see Fig. 5.5).
fc (forward) = 1000 + 5000/2 = 3500 Hzfc (backward) = 11000 – 5000/2 = 8500 Hz
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Figure 5.5 Solution to Example 5
Figure 5.6 FSK
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Figure 5.7 Relationship between baud rate and bandwidth in FSK
Example 6
Find the minimum bandwidth for an FSK signal transmitting at 2000 bps. Transmission is in half-duplex mode, and the carriers are separated by 3000 Hz.
Solution
For FSKBW = baud rate + fc1 fc0
BW = bit rate + fc1 fc0 = 2000 + 3000 = 5000 Hz
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Example 7
Find the maximum bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode.
Solution
Because the transmission is full duplex, only 6000 Hz is allocated for each direction. BW = baud rate + fc1 fc0 Baud rate = BW (fc1 fc0 ) = 6000 2000 = 4000But because the baud rate is the same as the bit rate, the bit rate is 4000 bps.
Figure 5.8 PSK
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Figure 5.9 PSK constellation
Figure 5.10 The 4-PSK method
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Figure 5.11 The 4-PSK characteristics
Figure 5.12 The 8-PSK characteristics
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Figure 5.13 Relationship between baud rate and bandwidth in PSK
Example 8
Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode.
Solution
For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.
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Example 9
Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate?
Solution
For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.
Quadrature amplitude modulation is a combination of ASK and PSK so that a
maximum contrast between each signal unit (bit, dibit, tribit, and so on)
is achieved.
Note:
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Figure 5.14 The 4-QAM and 8-QAM constellations
QAM
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Figure 5.15 Time domain for an 8-QAM signal
Figure 5.16 16-QAM constellations
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Figure 5.17 Bit and baud
Table 5.1 Bit and baud rate comparison
Modulation Units Bits/Baud Baud rate Bit Rate
ASK, FSK, 2-PSK Bit 1 N N
4-PSK, 4-QAM Dibit 2 N 2N
8-PSK, 8-QAM Tribit 3 N 3N
16-QAM Quadbit 4 N 4N
32-QAM Pentabit 5 N 5N
64-QAM Hexabit 6 N 6N
128-QAM Septabit 7 N 7N
256-QAM Octabit 8 N 8N
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Example 10
A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate?
Solution
The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is
4800 / 3 = 1600 baud
Example 11
Compute the bit rate for a 1000-baud 16-QAM signal.
Solution
A 16-QAM signal has 4 bits per signal unit since log216 = 4.
Thus, (1000)(4) = 4000 bps
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Example 12
Compute the baud rate for a 72,000-bps 64-QAM signal.
Solution
A 64-QAM signal has 6 bits per signal unit since log2 64 = 6.
Thus, 72000 / 6 = 12,000 baud
5.2 Telephone Modems
Modem Standards
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A telephone line has a bandwidth of almost 2400 Hz for data transmission.
Note:
Figure 5.18 Telephone line bandwidth
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Modem stands for modulator/demodulator.
Note:
Figure 5.19 Modulation/demodulation
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Figure 5.20 The V.32 constellation and bandwidth
Figure 5.21 The V.32bis constellation and bandwidth
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Figure 5.22 Traditional modems
Figure 5.23 56K modems
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5.3 Modulation of Analog Signals
Amplitude Modulation (AM)
Frequency Modulation (FM)
Phase Modulation (PM)
Figure 5.24 Analog-to-analog modulation
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Figure 5.25 Types of analog-to-analog modulation
The total bandwidth required for AM can be determined from the bandwidth
of the audio signal: BWt = 2 x BWm.
Note:
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Figure 5.26 Amplitude modulation
Figure 5.27 AM bandwidth
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Figure 5.28 AM band allocation
Example 13
We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM? Ignore FCC regulations.
Solution
An AM signal requires twice the bandwidth of the original signal:
BW = 2 x 4 KHz = 8 KHz
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The total bandwidth required for FM can be determined from the bandwidth
of the audio signal: BWt = 10 x BWm.
Note:
Figure 5.29 Frequency modulation
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Figure 5.30 FM bandwidth
The bandwidth of a stereo audio signal is usually 15 KHz. Therefore, an FM station needs at least a bandwidth of
150 KHz. The FCC requires the minimum bandwidth to be at least 200
KHz (0.2 MHz).
Note:
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Figure 5.31 FM band allocation
Example 14
We have an audio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM? Ignore FCC regulations.
Solution
An FM signal requires 10 times the bandwidth of the original signal:
BW = 10 x 4 MHz = 40 MHz