chapter 6 대역폭 활용 : 다중화와 확장 (bandwidth utilization: multiplexing and...

HANNAM UNIVERSITYHttp://netwk.hannam.ac.kr

Chapter 6대역폭 활용 : 다중화와 확장(Bandwidth Utilization:

Multiplexing and Spreading)

HANNAM UNIVERSITYHttp://netwk.hannam.ac.kr 2

6 6 장 장 다중화 (Multiplexing)

6.1 다중화

6.2 확산 대역 방식

6.3 요 약


Bandwidth utilization is the wise use of available bandwidth to achieve specific goals.

Efficiency can be achieved by multiplexing; privacy and anti-jamming can be achieved by spreading.


6.1 Multiplexing6.1 Multiplexing

Whenever the bandwidth of a medium linking two devices Whenever the bandwidth of a medium linking two devices is greater than the bandwidth needs of the devices, the link is greater than the bandwidth needs of the devices, the link can be shared. Multiplexing is the set of techniques that can be shared. Multiplexing is the set of techniques that allows the simultaneous transmission of multiple signals allows the simultaneous transmission of multiple signals across a single data link. As data and telecommunications across a single data link. As data and telecommunications use increases, so does traffic.use increases, so does traffic.

Topics discussed in this section:Topics discussed in this section:주파수 분할주파수 분할 (Frequency-Division) Multiplexing(Frequency-Division) Multiplexing

파장 분할파장 분할 ((Wavelength-Division) MultiplexingWavelength-Division) Multiplexing

동기 시분할동기 시분할 ((Synchronous Time-Division) MultiplexingSynchronous Time-Division) Multiplexing

통계 시분할통계 시분할 (Statistical Time-Division) Multiplexing(Statistical Time-Division) Multiplexing


6.1 6.1 다중화다중화

다중화 시스템 기본 형식■ 단일 데이터 링크를 통해 여러 개의 신호를 동시에 전송하기

위한 기술


다중화 다중화 (( 계속계속 ))

다중화기 (MUX, Multiplexer)■ 전송 스트림을 단일 스트림으로 결합 (many to one)

다중복구기 (DEMUX, Demultiplexer) ■ 스트림을 각각의 요소로 분리 (one to many)■ 전송 스트림을 해당 수신장치에 전달

링크 (Link)■ 물리적인 경로

채널 (Channel)■ 한 쌍의 장치간에 전송을 위한 하나의 경로


다중화 다중화 (( 계속계속 ))

다중화의 범주


주파수 분할 다중화■ FDM : Frequency-division Multiplexing

■ 링크의 대역폭이 전송되는 조합 신호의 대역폭 보다 클 때 적용할 수 있는 아날로그 기술

■ 신호가 겹치지 않도록 보호대역 (guard band) 만큼 떨어져 있어야 한다 .

주파수 분할 다중화주파수 분할 다중화


주파수 분할 다중화주파수 분할 다중화 (( 계속계속 ))

FDM 은 신호들을 합성하는 아날로그 다중화 기술이다 .

FDM is an analog multiplexing technique that combines analog signals.



FDM 처리 과정■ 각 전화기는 비슷한 범위의 주파수 대역의 신호 발생■ 이 신호는 서로 다른 반송 주파수로 변조된다 (f1, f2, f3)



다중화 풀기 (Demultiplexing)■ 개개의 신호를 분리하여 수신기에 전달


Example 6.1Example 6.1

AAssume that a voice channel occupies a bandwidth of 4 kHz. We ssume that a voice channel occupies a bandwidth of 4 kHz. We need to combine three voice channels into a link with a bandwidth 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 of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain. Assume there are no guard bands.frequency domain. Assume there are no guard bands.

SolutionSolutionWe shift (modulate) each of the three voice channels to a different We shift (modulate) each of the three voice channels to a different bandwidth, as shown in Figure 6.6. We use the 20- to 24-kHz bandwidth, as shown in Figure 6.6. We use the 20- to 24-kHz bandwidth for the first channel, the 24- to 28-kHz bandwidth for bandwidth for the first channel, the 24- to 28-kHz bandwidth for the second channel, and the 28- to 32-kHz bandwidth for the third the second channel, and the 28- to 32-kHz bandwidth for the third one. Then we combine them as shown in Figure 6.6. one. Then we combine them as shown in Figure 6.6.



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

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

5 × 100 + 4 × 10 = 540 kHz,5 × 100 + 4 × 10 = 540 kHz, as shown in Figure 6.7.as shown in Figure 6.7.


Example 6.2


Example 6.3

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

SolutionThe satellite channel is analog. We divide it into four channels, each channel having a 250-kHz bandwidth. Each digital channel of 1 Mbps is modulated such that each 4 bits is modulated to 1 Hz. One solution is 16-QAM modulation. Figure 6.8 shows one possible configuration.


Example 6.3



아날로그 계층구조



FDM 의 다른 응용

■ 라디오

▶ AM : 방송국당 10 kHz

▶ FM : 방송국당 200 kHz

■ TV : 채널당 6MHz

■ 1 세대 이동전화 : 사용자마다 60 kHz ( 수신 : 30 kHz, 발신 : 30 kHz)



The Advanced Mobile Phone System (AMPS) uses two bands. The Advanced Mobile Phone System (AMPS) uses two bands. The first band of 824 to 849 MHz is used for sending, and 869 to The first band of 824 to 849 MHz is used for sending, and 869 to 894 MHz is used for receiving. Each user has a bandwidth of 30 894 MHz is used for receiving. Each user has a bandwidth of 30 kHz in each direction. How many people can use their cellular kHz in each direction. How many people can use their cellular phones simultaneously?phones simultaneously?

SolutionSolutionEach band is 25 MHz. If we divide 25 MHz by 30 kHz, we get Each band is 25 MHz. If we divide 25 MHz by 30 kHz, we get 833.33. In reality, the band is divided into 832 channels. Of these, 833.33. In reality, the band is divided into 832 channels. Of these, 42 channels are used for control, which means only 790 channels 42 channels are used for control, which means only 790 channels are available for cellular phone users. are available for cellular phone users.


WDM : Wavelength-division Multiplexing

기본 개념은 FDM 과 같으며 , 광섬유의 고속 전송률을 이용하기 위해 설계

파형 분할 다중화파형 분할 다중화


파형 분할 다중화파형 분할 다중화 (( 계속계속 ))

WDM

■ 다중 빛 소스를 단일 빛으로 결합

■ 단일 빛은 다중 빛 소스로 분리

■ 프리즘 이용 : 임계각과 주파수 기반


파형 분할 다중화파형 분할 다중화 (( 계속계속 ))

WDM is an analog multiplexing WDM is an analog multiplexing technique to combine optical signals.technique to combine optical signals.

WDM 은 광섬유 신호를 묶기 위한 아날로그 다중화 기법이다 .


■ 송신과 수신장치에 의해 요구되는 데이터 전송률 보다 전송

매체의 데이터 전송률이 클 때 적용되는 디지털 처리 기술

시분할 다중화시분할 다중화


시분할 다중화시분할 다중화 (( 계속계속 ))

TDM 은 여러 개의 저속 채널을 하나의 고속 채널로 조합하는 다중화 기술이다 .

TDM is a digital multiplexing technique TDM is a digital multiplexing technique for combining several low-rate for combining several low-rate

channels into one high-rate one.channels into one high-rate one.



시간틈새 (time-slot) 와 프레임 (frame)



동기 TDM 에서는 링크의 전송률은 n 배 빠르고 , 단위 기간은 n 배 짧다 .

In synchronous TDM, the data rate of the link is n In synchronous TDM, the data rate of the link is n times faster, and the unit duration is n times shorter.times faster, and the unit duration is n times shorter.



In Figure 6.13, the data rate for each input connection is 3 kbps. If In Figure 6.13, the data rate for each input connection is 3 kbps. If 1 bit at a time is multiplexed (a unit is 1 bit), what is the duration of 1 bit at a time is multiplexed (a unit is 1 bit), what is the duration of ((aa) each input slot, () each input slot, (bb) each output slot, and () each output slot, and (cc) each frame?) each frame?

SolutionSolutionWe can answer the questions as follows: We can answer the questions as follows: a.a. The data rate of each input connection is 1 kbps. This means The data rate of each input connection is 1 kbps. This means

that the bit duration is 1/1000 s or 1 ms. The duration of the that the bit duration is 1/1000 s or 1 ms. The duration of the input time slot is 1 ms (same as bit duration).input time slot is 1 ms (same as bit duration).


Example 6.5 (continued)Example 6.5 (continued)

b.b. The duration of each output time slot is one-third of the input The duration of each output time slot is one-third of the input time slot. This means that the duration of the output time slot is time slot. This means that the duration of the output time slot is 1/3 ms.1/3 ms.

c.c. Each frame carries three output time slots. So the duration of a Each frame carries three output time slots. So the duration of a frame is 3 × 1/3 ms, or 1 ms. The duration of a frame is the frame is 3 × 1/3 ms, or 1 ms. The duration of a frame is the same as the duration of an input unit.same as the duration of an input unit.



Figure 6.14 shows synchronous TDM with a data stream for each Figure 6.14 shows synchronous TDM with a data stream for each input and one data stream for the output. The unit of data is 1 bit. input and one data stream for the output. The unit of data is 1 bit. Find (Find (aa) the input bit duration, () the input bit duration, (bb) the output bit duration, () the output bit duration, (cc) the ) the output bit rate, and (output bit rate, and (dd) the output frame rate.) the output frame rate.

SolutionSolutionWe can answer the questions as follows:We can answer the questions as follows:a.a. The input bit duration is the inverse of the bit rate: The input bit duration is the inverse of the bit rate:

1/1 Mbps = 1 μs.1/1 Mbps = 1 μs.

b.b. The output bit duration is one-fourth of the input bit duration, or The output bit duration is one-fourth of the input bit duration, or ¼ μs.¼ μs.



c.c. The output bit rate is the inverse of the output bit duration or The output bit rate is the inverse of the output bit duration or 1/(4μs) or 4 Mbps. This can also be deduced from the fact that 1/(4μs) or 4 Mbps. This can also be deduced from the fact that the output rate is 4 times as fast as any input rate; so the output the output rate is 4 times as fast as any input rate; so the output rate = 4 × 1 Mbps = 4 Mbps. rate = 4 × 1 Mbps = 4 Mbps.

d.d. The frame rate is always the same as any input rate. So the The frame rate is always the same as any input rate. So the frame rate is 1,000,000 frames per second. Because we are frame rate is 1,000,000 frames per second. Because we are sending 4 bits in each frame, we can verify the result of the sending 4 bits in each frame, we can verify the result of the previous question by multiplying the frame rate by the number previous question by multiplying the frame rate by the number of bits per frame.of bits per frame.


Figure 6.14 Example 6.6



Four 1-kbps connections are multiplexed together. A unit is 1 bit. Four 1-kbps connections are multiplexed together. A unit is 1 bit. Find (Find (aa) the duration of 1 bit before multiplexing, () the duration of 1 bit before multiplexing, (bb) the ) the transmission rate of the link, (transmission rate of the link, (cc) the duration of a time slot, and () the duration of a time slot, and (dd) ) the duration of a frame.the duration of a frame.

SolutionSolutionWe can answer the questions as follows:We can answer the questions as follows:aa. The duration of 1 bit before multiplexing is 1 / 1 kbps, or 0.001 s . The duration of 1 bit before multiplexing is 1 / 1 kbps, or 0.001 s

(1 ms).(1 ms).

b.b. The rate of the link is 4 times the rate of a connection, or 4 kbps. The rate of the link is 4 times the rate of a connection, or 4 kbps.



c.c. The duration of each time slot is one-fourth of the duration of The duration of each time slot is one-fourth of the duration of each bit before multiplexing, or 1/4 ms or 250 μs. Note that we each bit before multiplexing, or 1/4 ms or 250 μs. Note that we can also calculate this from the data rate of the link, 4 kbps. can also calculate this from the data rate of the link, 4 kbps. The bit duration is the inverse of the data rate, or 1/4 kbps or The bit duration is the inverse of the data rate, or 1/4 kbps or 250 μs.250 μs.

d.d. The duration of a frame is always the same as the duration of a The duration of a frame is always the same as the duration of a unit before multiplexing, or 1 ms. We can also calculate this in unit before multiplexing, or 1 ms. We can also calculate this in another way. Each frame in this case has four time slots. So the another way. Each frame in this case has four time slots. So the duration of a frame is 4 times 250 μs, or 1 ms.duration of a frame is 4 times 250 μs, or 1 ms.



끼워넣기 (interleaving)

■ 스위치가 장치들을 일정한 비율로 정해진 순서대로 이동한다 .



Four channels are multiplexed using TDM. If each channel sends Four channels are multiplexed using TDM. If each channel sends 100 bytes /s and we multiplex 1 byte per channel, show the frame 100 bytes /s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate for the link.the frame rate, and the bit rate for the link.

SolutionSolutionThe multiplexer is shown in Figure 6.16. Each frame carries 1 byte The multiplexer is shown in Figure 6.16. Each frame carries 1 byte from each channel; the size of each frame, therefore, is 4 bytes, or from each channel; the size of each frame, therefore, is 4 bytes, or 32 bits. Because each channel is sending 100 bytes/s and a frame 32 bits. Because each channel is sending 100 bytes/s and a frame carries 1 byte from each channel, the frame rate must be 100 carries 1 byte from each channel, the frame rate must be 100 frames per second. The bit rate is 100 × 32, or 3200 bps. frames per second. The bit rate is 100 × 32, or 3200 bps.


Figure 6.16 Figure 6.16 Example 6.8Example 6.8



A multiplexer combines four 100-kbps channels using a time slot A multiplexer combines four 100-kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? What is the frame duration? What is the bit rate? frame rate? What is the frame duration? What is the bit rate? What is the bit duration?What is the bit duration?

SolutionSolutionFigure 6.17 shows the output for four arbitrary inputs. The link Figure 6.17 shows the output for four arbitrary inputs. The link carries 50,000 frames per second. The frame duration is therefore carries 50,000 frames per second. The frame duration is therefore 1/50,000 s or 20 μs. The frame rate is 50,000 frames per second, 1/50,000 s or 20 μs. The frame rate is 50,000 frames per second, and each frame carries 8 bits; the bit rate is 50,000 × 8 = 400,000 and each frame carries 8 bits; the bit rate is 50,000 × 8 = 400,000 bits or 400 kbps. The bit duration is 1/400,000 s, or 2.5 μs. bits or 400 kbps. The bit duration is 1/400,000 s, or 2.5 μs.


Figure 6.17 Figure 6.17 Example 6.9Example 6.9


빈 틈새 (Empty slots)

■ 발신자가 전송할 데이터가 없다면 해당 틈새가 비게 된다 .



① 다단계 다중화 (multilevel multiplexing)

② 복수 틈새 할당 (multiple-slot allocation)

③ 펄스 채워 넣기 (pulse stuffing)

데이터율 관리데이터율 관리 (data rate management)(data rate management)

입력 측 데이터율 이 서로 다른 경우 해결 방법


데이터율 관리데이터율 관리 (( 계속계속 ))

다단계 다중화

■ 어느 입력의 데이터 율이 다른 것들에 비해 정수 배 만큼

빠를 때 사용하는 기술



복수 틈새 할당 (multiple-slot allocation)

■ 입력회선에 한 개 보다 더 많은 틈새를 할당하는 것



펄스 채우기 (pulse stuffing)

■ 가장 높은 데이터 율에 맞추기 위해 공 비트를 끼워 넣는 것

■ 비트 패딩 , 비트 채우기 라고도 함



프레임 동기화 (Frame synchronization)

■ 프레임 구성비트 (framing bits)



We have four sources, each creating 250 characters per second. If We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (to each frame, find (aa) the data rate of each source, () the data rate of each source, (bb) the ) the duration of each character in each source, (duration of each character in each source, (cc) the frame rate, () the frame rate, (dd) ) the duration of each frame, (the duration of each frame, (ee) the number of bits in each frame, ) the number of bits in each frame, and (and (ff) the data rate of the link.) the data rate of the link.

SolutionSolutionWe can answer the questions as follows:We can answer the questions as follows:a.a. The data rate of each source is 250 × 8 = 2000 bps = 2 The data rate of each source is 250 × 8 = 2000 bps = 2

kbps.kbps.



b.b. Each source sends 250 characters per second; therefore, the Each source sends 250 characters per second; therefore, the duration of a character is 1/250 s, or duration of a character is 1/250 s, or 4 ms.4 ms.

c.c. Each frame has one character from each source, which means Each frame has one character from each source, which means the link needs to send 250 frames per second to keep the the link needs to send 250 frames per second to keep the transmission rate of each source.transmission rate of each source.

d.d. The duration of each frame is 1/250 s, or 4 ms. Note that the The duration of each frame is 1/250 s, or 4 ms. Note that the duration of each frame is the same as the duration of each duration of each frame is the same as the duration of each character coming from each source.character coming from each source.

e.e. Each frame carries 4 characters and 1 extra synchronizing bit. Each frame carries 4 characters and 1 extra synchronizing bit. This means that each frame is This means that each frame is 4 × 8 + 1 = 33 bits.4 × 8 + 1 = 33 bits.



Two channels, one with a bit rate of 100 kbps and another with a Two channels, one with a bit rate of 100 kbps and another with a bit rate of 200 kbps, are to be multiplexed. How this can be bit rate of 200 kbps, are to be multiplexed. How this can be achieved? What is the frame rate? What is the frame duration? achieved? What is the frame rate? What is the frame duration? What is the bit rate of the link?What is the bit rate of the link?

SolutionSolutionWe can allocate one slot to the first channel and two slots to the We can allocate one slot to the first channel and two slots to the second channel. Each frame carries 3 bits. The frame rate is second channel. Each frame carries 3 bits. The frame rate is 100,000 frames per second because it carries 1 bit from the first 100,000 frames per second because it carries 1 bit from the first channel. The bit rate is 100,000 frames/s × 3 bits per frame, or 300 channel. The bit rate is 100,000 frames/s × 3 bits per frame, or 300 kbps. kbps.

SolutionSolutionWe can allocate one slot to the first channel and two slots to the We can allocate one slot to the first channel and two slots to the second channel. Each frame carries 3 bits. The frame rate is second channel. Each frame carries 3 bits. The frame rate is 100,000 frames per second because it carries 1 bit from the first 100,000 frames per second because it carries 1 bit from the first channel. The bit rate is 100,000 frames/s × 3 bits per frame, or 300 channel. The bit rate is 100,000 frames/s × 3 bits per frame, or 300 kbps. kbps.



DS (Digital Signal) 서비스

■ 디지털 신호의 계층 구조



DS 와 T 회선 전송속도



아날로그 전송용 T 회선



T-1 프레임 구조



E 회선

■ 유럽은 E 라인으로 불리는 T- 라인 버전을 사용


통계적 TDM

■ TDM 틈새 비교



6.2 6.2 확산 대역 방식확산 대역 방식 (SPREAD SPECTRUM)(SPREAD SPECTRUM)

In spread spectrum (SS), we combine signals from In spread spectrum (SS), we combine signals from different sources to fit into a larger bandwidth, but our different sources to fit into a larger bandwidth, but our goals are to prevent eavesdropping and jamming. To goals are to prevent eavesdropping and jamming. To achieve these goals, spread spectrum techniques add achieve these goals, spread spectrum techniques add redundancy.redundancy.

Topics discussed in this section:Topics discussed in this section:Frequency Hopping Spread Spectrum (FHSS)Frequency Hopping Spread Spectrum (FHSS)Direct Sequence Spread Spectrum Synchronous (DSSS)Direct Sequence Spread Spectrum Synchronous (DSSS)


무선 통신 응용을 위해 설계 여분의 정보 추가

6.2 6.2 확산 대역 방식확산 대역 방식


주파수 뛰기 대역 확산 방식 (Frequency hopping spread spectrum, FHSS)■ 발신지 M 개의 신호를 서로 다른 반송파를 사용하여 변조한다 .

확산 대역 방식확산 대역 방식 (( 계속계속 ))


FHSS 에서 주파수 선택



FHSS 사이클



대역폭 공유



직접 순열 확산 방식 (DSSS)

■ 각 데이터 비트를 확산코드를 사용하여 n 개의 bit 로 대체



DSSS 예



6.3 6.3 요 약요 약

chapter 6 대역폭 활용 : 다중화와 확장 (bandwidth utilization: multiplexing and...

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