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AnalogTransmission

Figure 5.1Digital-to-analog modulation

- Modulation of binary data or digital-to-analog modulation is the process of changing one of the characteristics of an analog signal based on the information in a digital signal (0s and 1s)
- Telephone wires carry analog signals.
- When we vary any one of these characteristics [amplitude or frequency or phase], we create a different version of that wave.
- Three mechanisms for modulating digital data into an analog signal: amplitude shift keying (ASK), frequency shift keying (FSK) , and phase shift keying (PSK).

A signal unit is composed of 1 or more bits.

Bit Rate and Baud Rate

Bit rate is the number of bits per second.

Baud rate is the number of signal units per second.

Bit rate equals the baud rate times the number of bits represented by each signal unit.

Baud rate equals the bit rate divided by the number of bits represented by each signal unit.

Baud rate is less than or equal to the bit rate.

Baud rate determines the bandwidth required to send the signal.

Aspects of Digital-to-Analog ConversionAn 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

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

In analog signal, 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.

Receiving device is tuned to the frequency of the carrier signal that it expects from the sender.

Digital information then modulates the carrier signal by modifying one or more of its characteristics (amplitude, frequency, or phase). This kind of modification is called modulation (or shift keying, and the information signal is called the modulating signal.

Carrier SignalFigure 5.3Amplitude Shift Keying (ASK)

- The strength of the carrier signal is varied to represent binary 1 or 0.
- Both frequency and phase remain constant while the amplitude changes.
- A bit duration is the period of time that defines 1 bit.
- Peak amplitude of the signal during each bit duration is constant, and its value depends on the bit (0 or 1).
- ASK is susceptible to noise interference. Noise refers to unintentional voltages introduced onto a line by various phenoma such as heat or electromagnetic induction created by other sources.

Figure 5.4Relationship between baud rate and bandwidth in ASK

- Bandwidth of a signal is the total range of frequencies occupied by that signal.
- When we decompose an ASK-modulated signal, we get a spectrum of many simple frequencies.
- Most significant ones are those between fc - Nbaud/2 and fc + Nbaud/2 with the carrier frequency fc at the middle.
- Bandwidth BW = (1 + d)* Nbaud
- BW is bandwidth, Nbaud is baud rate, and d is a factor related to the modulation process (with a minimum value of 0).

- Although there is only one carrier frequency, the process of modulation produces a complex signal that is a combination of many simple signals, each with a different frequency.

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.

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.

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 is

BW = 10000 / 2 = 5000 Hz

The carrier frequencies can be chosen at the middle of each band.

fc (forward) = 1000 + 5000/2 = 3500 Hz

fc (backward) = 11000 – 5000/2 = 8500 Hz

- Frequency of the carrier signal is varied to represent binary 1 or 0.
- Frequency of the signal during each bit duration is constant, and its value depends on the bit (0 or 1); both peak amplitude and phase remain constant.
- FSK avoids most of the problems from noise.
- Because the receiving device is looking for specific frequency changes over a given number of periods, it can ignore voltage spikes.

Figure 5.7Relationship between baud rate and bandwidth in FSK

- Although FSK shifts between two carrier frequencies, it is easier to analyse as two co-existing frequencies.
- FSK spectrum is a combination of two ASK spectra centered on fc0 and fc1. The bandwidth required for FSK transmission is equal to the baud rate of the signal plus the frequency shift (difference between the two carrier frequencies): BW = fc1 – fc0 + Nbaud
- Although there are only two carrier frequencies, the process of modulation produces a composite signal that is a combination of many simple signals, each with a different frequency.

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 FSK

BW = baud rate + fc1- fc0

BW = bit rate + fc1 - fc0 = 2000 + 3000 = 5000 Hz

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 = 4000

But because the baud rate is the same as the bit rate, the bit rate is 4000 bps.

- The phase of the carrier is varied to represent binary 1 or 0.
- Both peak amplitude and frequency remain constant as the phase changes.
- If we start with a phase of 00 to represent binary 0, then we can change the phase to 1800 to send binary 1.
- The phase of the signal during each bit duration is constant, and its value depends on the bit (0 or 1).
- 2-PSK or binary PSK: Two different phases (00 and 1800) are used. Constellation or phase-state diagram shows the relationship by illustrating only the phases.

- PSK is not susceptible to the noise degradation that affects ASK or to the bandwidth limitations of FSK.
- Even smaller variations in the signal can be detected reliably by the receiver.
- Instead of utilizing only two variations of a signal, each representing 1 bit, we can use four variations and let each phase shift represent 2 bits.
- A phase of 00 now represents 00; 900 represents 01; 1800 represents 10; and 2700 represents 11. This technique is called 4-PSK or Q-PSK. The pair of bits represented by each phase is called a dibit. We can transmit data twice as efficiently using 4-PSK as we can using 2-PSK.

Figure 5.12The 8-PSK characteristics

- 8-PSK: Instead of 900, we now vary the signal by shifts of 450.
- With eight different phases, each shift can represent 3 bits (1 tribit) at a time.
- 8-PSK is 3 times as efficient as 2-PSK.
- Minimum bandwidth required for PSK transmission is the same as that required for ASK transmission.
- Maximum bit rate in PSK transmission is much greater than that of ASK.
- While the maximum baud rates of ASK and PSK are the same for a given bandwidth, PSK bit rates using the bandwidth can be 2 or more times greater.

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.

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.

Figure 5.14The 4-QAM and 8-QAM constellations

- PSK is limited by the ability of the equipment to distinguish small differences in phase.
- Why not combine PSK, FSK and ASK?
- Bandwidth limitations make combinations of FSK with other changes practically useless.
- Combining ASK and PSK, we could have x variations in phase and y variations in amplitude, giving us x times y possible variations and the corresponding number of bits per variation. Quadrature amplitude modulation does that.
- 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.
- In 4-QAM and 8-QAM, number of amplitude shifts is fewer than the number of phase shifts. Because amplitude changes are susceptible to noise and require greater shift differences than do phase changes, the number of phase shifts used by a QAM system is always larger than the number of amplitude shifts.

Figure 5.1616-QAM constellations

- The first example, 3 amplitudes and 12 phases, handles noise best because of a greater ratio of phase shift to amplitude. It is ITU-T recommendation.
- The second example, four amplitudes and eight phases, is the OSI recommendation.
- It is to be noted that every intersection of phase and amplitude is utilized.
- In fact, 4 times 8 should allow for 32 possible variations. But by using only one-half of those possibilities, the measurable differences between shifts are increased and greater signal readability is ensured. In addition, several QAM designs link specific amplitudes with specific phases. This means that even with the noise problems associated with amplitude shifting, the meaning of a shift can be recovered from phase information.
- QAM has an advantage over ASK as is its less susceptibility to noise.
- Minimum bandwidth required for QAM transmission is the same as that required for ASK and PSK transmission. QAM has the same advantages as PSK over ASK.

- Each frequency shift represents a single bit; so it requires 1200 signal units to send 1200 bits. Its baud rate, therefore, is also 1200 bps.
- Each signal variation in an 8-QAM system, however, represents 3 bits. So a bit rate of 1200 bps, using 8-QAM, has a baud rate of only 400.
- A dibit system has a baud rate of one-half the bit rate, a tribit system has a baud rate of one-third the bit rate, and a quadbit system has a baud rate of one-fourth the bit rate.

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

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

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

Figure 5.18Telephone line bandwidth

- Traditional telephone lines can carry frequencies between 300 and 3300 Hz, giving them a bandwidth of 3000 Hz.
- All this range is used for transmitting voice, where a great deal of interference and distortion can be accepted without loss of intelligibility.
- Data signals require a higher degree of accuracy to ensure integrity. For safety’s sake, therefore, the edges of the bandwidth range are not used for data communications.
- We can say that the signal bandwidth must be smaller than the cable bandwidth. The effective bandwidth of a bandwidth line being used for data transmission is 2400 Hz, covering the range from 600 to 3000 Hz.
- Today some telephone lines are capable of handling more bandwidth than traditional lines.
- A telephone line has a bandwidth of almost 2400 Hz for data transmission.

Figure 5.19Modulation/demodulation

- Bandwidth defines a baseband nature, which means we need to modulate if we want to use this bandwidth for data transmission. Devices that were traditionally used to do so are called modems.
- Modem stands for modulator/demodulator
- Modulator creates a band-pass analog signal from binary data.
- Demodulator recovers the binary data from the modulated signal.
- The computer on left sends binary data to the modulator portion of the modem; the data is sent as an analog signal on the telephones lines. The modem on the right receives the analog signal, demodulates it through its demodulator, and delivers data to the computer on the right. The communication can be bidirectional, which means the computer on the right can also send data to the computer on the left using the same modulation/demodulation processes.

Figure 5.20The V.32 constellation and bandwidth

- Most popular modems available are base don the V-series standards published by ITU-T.
- V.32
- Uses a combined modulation and encoding technique called trelliscoded modulation.
- Trellis is essentially QAM plus a redundant bit.
- Data stream is divided into 4-bit sections. Instead of quadbit, however, a pentabit (5-bit pattern) is transmitted. The value of the extra bit is calculated from the values of the data bits.
- In any QAM system, the receiver compares each received signal point to all valid points in the constellation and selects the closest point as the intended value. A signal distorted by transmission noise can arrive closer in value to an adjacent point than to the intended point, resulting in a misidentification of the point and an error in the received data. The closer the points are in the constellation, the more likely it is that transmission noise can result in a signal’s being misidentified. By adding a redundant bit to each quadbit, trellis-coded modulation increases the amount of information used to identify each bit pattern and thereby reduces the number of possible matches. For this reason, a trellis-encoded signal is much less likely than a plain QAM signal to be misread when distorted by noise.

Figure 5.21The V.32bis constellation and bandwidth

- The V.32 calls for 32-QAM with a baud rate of 2400. Because only 4 bits of each pentabit represents data, the resulting speed is 4 * 2400 = 9600bps.
- V.32bits
- First introduced by ITU-T to support 14,400-bps transmission.
- V.32bis uses 128-QAM transmission (7 bits/baud with 1 bit for error control) at a rate of 2400 baud (2400*6 = 14,400 bps)
- V.32bis includes an automatic fall-back and fall-forward feature that enables the modem to adjust its speed upward or downward depending on the quality of the line or signal.
- V.34bis: Provides a bit rate of 28,800 with a 960-point constellation to a bit rate of 33,600 with a 1664-point constellation.
- V.90: Bit of 56,000bps, called 56K modems.
- These modems may be used only if one party is using digital signature.
- Supports asymmetric [downloading rate is 56 Kbps; uploading rate is 33.6Kbps].

- After modulation by the modem, an analog signal reaches the telephone company switching station, where it is sampled and digitized to be passed through the digital network. The quantization noise introduced into the signal at the sampling point limits the data rate according to the Shannon Capacity. This limit is 33.6 Kbps.
- 56K Modems:
- Uploading: Analog signal must still be sampled at the switching station, which means the data in uploading is limited to 33.6 Kbps.
- Downloading: No sampling involved. Signal is not affected by quantization noise and not subject to the Shannon capacity limitation. So, downloading is limited to 56 Kbps.

- 56 Kbps because telephone companies sample 8000 times per second with 8 bits per sample. One of the bits in each sample is used for control purposes, which means each sample is 7 bits. The rate is therefore 8000 * 7 or 56000 bps or 56 Kbps
- V.92
- Modems can adjust their speed, and if the noise allows, they can upload data at the rate of 48 Kbps. The downloading rate is still 56 Kbps.
- Modem can interrupt the Internet connection when there is an incoming call if the line has call-waiting service.

Figure 5.24Analog-to-analog modulation

- Modulation of an analog signal or analog-to-analog conversion is the representation of analog information by an analog signal.
- Modulation is needed if the medium has a band-pass nature or if only band-pass bandwidth is available to us.
- An example is radio. The government assigns a baseband bandwidth to each radio station. The analog signal produced by each station is a low-pass signal, all in the same range. To be able to listen to different stations, the low-pass signals need to be shifted, each to a different range.

Figure 5.25Types of analog-to-analog modulation

- Analog-to-analog modulation can be accomplished in three ways:
- Amplitude Modulation (AM)
- Frequency Modulation (FM)
- Phase Modulation (PM)
- Amplitude Modulation
- Carrier signal is modulated so that its amplitude varies with the changing amplitudes of the modulating signal.
- Frequency and phase of the carrier remains the same.

Figure 5.26Amplitude modulation

- Bandwidth of an AM signal is equal to twice the bandwidth of the modulating signal and covers a range centered on the carrier frequency BWt = 2 x BWm.
- Bandwidth of audio signal (speech and music) is usually 5 KHz. Therefore, an AM radio station needs a minimum bandwidth of 10 KHz. Federal Communications Commission (FCC) allows 10 KHz for each AM station.
- AM stations are allowed carrier frequencies anywhere between 530 and 1700 KHz. Each station’s carrier frequency must be separated from those on either side of it by at least 10 KHZ to avoid interference.

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

Figure 5.29Frequency modulation

- Frequency of the carrier signal is modulated to follow the changing voltage level (amplitude) of the modulating signal.
- Peak amplitude and phase of the carrier signal remains constant.
- Bandwidth of an FM signal is equal to 10 times the bandwidth of the modulating signal and, like AM bandwidths, covers a range centered on the carrier frequency. BWt = 10 x BWm.
- Bandwidth of an analog signal (speech and music) broadcast in stereo is almost 15 KHz. Each FM radio station, needs a minimum bandwidth of 150 KHz. FCC allows 200 KHz for each station to provide some room for guard bands.

- FM stations are allowed carrier frequencies anywhere between 88 and 108 MHz.
- Stations must be separated by at least 200 KHz to keep their bandwidths from overlapping.
- To create even more privacy, FCC requires that in a given area, only alternate bandwidth allocations may be used. Others remain unused to prevent any possibility of two stations interfering with each other.
- Given 88 to 108 MHz as a range, there are 100 potential FM bandwidths in an area, of which 50 can operate at any one time.

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

- Due to simpler hardware requirements, PM is used in some systems as an alternative to frequency modulation.
- Phase of the carrier signal is modulated to follow the changing voltage level (amplitude) of the modulating signal.
- Peak amplitude and frequency of the carrier signal remain constant, but as the amplitude of the information signal changes, the phase of the carrier changes correspondingly.
- Modulated signal of PM is similar to those of FM.

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