M lti l iMultiplexing

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OutlinesOutlines

1. Introduction

2. Binary Modulation Bandpass Signaling1. Amplitude Shift Keying (ASK)

2. Frequency Shift Keying (FSK) 

3. Phase Shift Keying (PSK)

3. Multilevel Modulated Bandpass Signaling1. Quadrature Phase Shift Keying (QPSK).

2. Offset QPSK (OQPSK).

3. π/4 QPSK.

4 M ary Phase Shift Keying (MPSK)4. M‐ary Phase Shift Keying (MPSK)

5. Quadrature Amplitude Modulation (QAM) 2

1. Introduction1. IntroductionMultiplexing is a set of techniques that allows the simultaneous transmission of multiple signals across a p gsingle data link.Multiplexer combines the incoming lines into one [single stream].stream].Demultiplexer (DEMUX) separates the stream back into its component transmissions (one to many) and directs them to their corresponding linesthem to their corresponding lines.

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1. Introduction1. IntroductionCategories of multiplexing.

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Figure 6.3 FDM

Frequency-division multiplexing is an analog techniqueFrequency division multiplexing is an analog technique that can be applied when the bandwidth of a link (in hertz) is greater than the combined bandwidths of the signals to be t itt dtransmitted.Signals generated by each sending device modulate different carrier frequencies.qFDM is an analog multiplexing technique that combines signals

Figure 6.4 FDM process

Example 1Example 1

Assume that a voice channel occupies a bandwidth of 4 KHz. We need to combine three voice channels into a link with a bandwidth of 12 KHz, from 20 to 32 KHz. Show the configuration using the frequency d i ith t th f d b ddomain without the use of guard bands.

S l tiS l tiSolutionSolution

Shift (modulate) each of the three voice channels to ( )a different bandwidth, as shown in Figure 6.6.

Figure 6.6 Example 1

Example 2Example 2

Five channels, each with a 100-KHz bandwidth, are to be multiplexed together. What is the minimum p gbandwidth of the link if there is a need for a guard band of 10 KHz between the channels to prevent i t f ?interference?SolutionSolution

For five channels, we need at least four guard bands. This means that the required bandwidth is at l tleast

5 x 100 + 4 x 10 = 540 KHz, as shown in Figure 6 7as shown in Figure 6.7.

Figure 6.7 Example 2

Example 3Example 3

Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an p gappropriate configuration using FDM

Sol tionSol tionSolutionSolutionThe satellite channel is analog. We divide it into four channels, each channel having a 250-KHz bandwidth. Each digital channel of 1 Mbps is

d l t d h th t h 4 bit d l t d t 1modulated such that each 4 bits are modulated to 1 Hz. One solution is 16-QAM modulation. Figure 6.8 shows one possible configurationshows one possible configuration.

Figure 6.8 Example 3

Figure 6.9 Analog hierarchy

In analog hierarchy, 12 voice channels are multiplexed g y, ponto a higher‐bandwidth line to create a group.The join them as group, supergroup, master group and j b  jumbo group.

Example 4Example 4

The Advanced Mobile Phone System (AMPS) uses two bands. The first band, 824 to 849 MHz, is used for sending; and 869 to 894 MHz is used for receiving. Each user has a bandwidth of 30 KHz in each direction The 3 KHz voice is modulated usingeach direction. The 3-KHz voice is modulated using FM, creating 30 KHz of modulated signal. How many people can use their cellular phonesmany people can use their cellular phones simultaneously?SolutionSolution

Each band is 25 MHz. If we divide 25 MHz into 30 KHz, we get 833.33. In reality, the band is divided into 832 channels.

Figure 6.10 WDM

Wave-Division Multiplexing [WDM] is an analog p g [ ] gmultiplexing technique to combine optical signalsOptical fiber data rate is higher than the data rate of

t lli t i i bl U i fib ti bl fmetallic transmission cable. Using a fiber-optic cable for one single line wastes the available bandwidth.

Figure 6.12 TDM

TDM is a digital multiplexing technique to combine dataTDM is a digital multiplexing technique to combine dataInstead of sharing a portion of the bandwidth as in FDM, time is shared.Each connection occupies a portion of time in the link.In a TDM, the data rate of the link is n times faster, and the nit d ration is n times shorterunit duration is n times shorter

Example 5Example 5

Four 1-Kbps connections are multiplexed together. A unit is 1 bit. Find (1) the duration of 1 bit before ( )multiplexing, (2) the transmission rate of the link, (3) the duration of a time slot, and (4) the duration of a f ?frame? SolutionSolutionWe can answer the questions as follows:We can answer the questions as follows:

1. The duration of 1 bit is 1/1 Kbps, or 0.001 s (1 p , (ms).2. The rate of the link is 4 Kbps.3. The duration of each time slot 1/4 ms or 250 μs. 4. The duration of a frame 1 ms.

Figure 6.14 Interleaving

TDM can be visualized as two fast rotating switches  TDM can be visualized as two fast rotating switches, one on the multiplexing side and the other on the demultiplexing side. The switches are synchronized and t t   t th     d  b t i   it  di ti  O  rotate at the same speed, but in opposite directions. On 

the multiplexing side, as the switch opens in front of a connection, that connection has the opportunity to send a unit onto the path. This process is called interleaving.

Example 6Example 6

Four channels are multiplexed using TDM. If each channel sends 100 bytes/s and we multiplex 1 byte y p yper channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the f t d th bit t f th li kframe rate, and the bit rate for the link. SolutionSolutionThe multiplexer is shown in Figure 6.15.

Example 7Example 7

A multiplexer combines four 100-Kbps channels using a time slot of 2 bits. Show the output with four g parbitrary inputs. What is the frame rate? What is the frame duration? What is the bit rate? What is the bit d ti ?duration?SolutionSolutionFigure 6.16 shows the output for four arbitrary inputs.

Figure 6.17 Framing bits

For synchronization between multiplexer and For synchronization between multiplexer and demultiplexer, one or more synchronization bits are usually added to the beginning of each frame. These bits, called framing bits, follow a pattern, frame to frame, that called framing bits, follow a pattern, frame to frame, that allows the demultiplexer to synchronize with the incoming stream so that it can separate the time slots accurately.accurately.In bit padding, the multiplexer adds extra bits to a device’s source stream to force the speed relationships among the various devices into integer multiples of each among the various devices into integer multiples of each other.

Example 8Example 8

We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (1) the data rate of each source (2) the duration ofthe data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of(4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link.

S l tiS l tiSolutionSolution

See next slideSee next slide.

Solution (continued)Solution (continued)

We can answer the questions as follows:

1. The data rate of each source is 2000 bps = 2 Kbps.p

2. The duration of a character is 1/250 s, or 4 ms.3. The link needs to send 250 frames per second.4 Th d i f h f i 1/2 0 44. The duration of each frame is 1/250 s, or 4 ms. 5. Each frame is 4 x 8 + 1 = 33 bits.6 The data rate of the link is 250 x 33 or 8250 bps6. The data rate of the link is 250 x 33, or 8250 bps.

Example 9Example 9

Two channels, one with a bit rate of 100 Kbps and another with a bit rate of 200 Kbps, are to be pmultiplexed. How this can be achieved? What is the frame rate? What is the frame duration? What is the bit t f th li k?bit rate of the link?SolutionSolution

W ll t l t t th fi t h l dWe can allocate one slot to the first channel and two slots to the second channel. Each frame carries 3 bits The frame rate is 100 000 frames per second3 bits. The frame rate is 100,000 frames per second because it carries 1 bit from the first channel. The frame duration is 1/100,000 s, or 10 ms. The bit rate , ,is 100,000 frames/s x 3 bits/frame, or 300 Kbps.

Figure 6.18 Digital Signal (DS) hierarchy

DS‐0 service is a single digital channel of 64 Kbps.g g 4 pDS‐1 is a 1.544‐Mbps service [24 times 64 Kbps plus 8Kbps of overhead]And so on.DS‐0, DS‐1, and so on are the names of services. To implement those services  the telephone companies use implement those services, the telephone companies use T lines (T‐1 to T‐4).

Table 6.1 DS and T lines ratesTable 6.1 DS and T lines rates

Service Line Rate (Mbps)

Voice Channels

DSDS--11 TT--11 1.5441.544 2424

DSDS--22 TT--22 6.3126.312 9696

DSDS--33 TT--33 44.73644.736 672672

DSDS--44 TT--44 274.176274.176 40324032

Figure 6.19 T-1 line for multiplexing telephone lines

T lines are digital lines designed for the transmission of g gdigital data, audio, or video. T lines also can be used for analog transmission ( l  t l h   ti )   id d th   l  (regular telephone connections), provided the analog signals are sampled first, then time‐division multiplexed.

Europeans use a version of T lines called E linesThe two systems are conceptually identical, but their capacities differ.

Table 6.2 E line ratesTable 6.2 E line rates

E Li Rate VoiceE Line (Mbps) Channels

EE--11 2.0482.048 3030

EE--22 8.4488.448 120120

EE--33 34.36834.368 480480

EE--44 139.264139.264 19201920

Figure 6.21 Multiplexing and inverse multiplexing

Inverse multiplexing is the opposite of multiplexingInverse multiplexing is the opposite of multiplexing.Inverse multiplexing takes the data stream from one high‐speed line and breaks it into portions that can be sent across several lower‐speed lines simultaneously, with no loss in the collective data rate.

1. Introduction1. Introduction

Digital modulation bandpass signals are used to g p gtransmit the digital binary or multilevel line code signal through analog communication channel.

Figure: Digital-to-analog modulationFigure: Digital to analog modulation

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OutlinesOutlines

1. Introduction

2. Binary Modulation Bandpass Signaling1. Amplitude Shift Keying (ASK)

2. Frequency Shift Keying (FSK) 

3. Phase Shift Keying (PSK)

3. Multilevel Modulated Bandpass Signaling1. Quadrature Phase Shift Keying (QPSK).

2. Offset QPSK (OQPSK).

3. π/4 QPSK.

4 M ary Phase Shift Keying (MPSK)4. M‐ary Phase Shift Keying (MPSK)

5. Quadrature Amplitude Modulation (QAM) 31