chapter 3 데이터와 신호 (data and signals)

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Chapter 3데이터와 신호

(Data and Signals)


3 3 장 신호장 신호

3.1 아날로그와 디지털

3.2 주기 아날로그 신호

3.3 디지털 신호

3.4 전송 장애

3.5 데이터 전송률의 한계

3.6 성 능

3.7 요 약


3.1 3.1 아날로그와 디지털 데이터아날로그와 디지털 데이터

정보가 전송되기 위해서는 전자기 신호로 변환되어야 한다 .

데이터는 아날로그나 디지털이 될 수 있다 .아날로그 데이터는 연속적이고 연속된 값을 갖는다 .

디지털 데이터는 이산적인 상대를 가지며 이산 값을 갖는다 .

To be transmitted, data must be transformed to electromagnetic To be transmitted, data must be transformed to electromagnetic signals.signals.Data can be analog or digital. Analog data are continuous and take Data can be analog or digital. Analog data are continuous and take

continuous values. Digital data have discrete states and take discrete continuous values. Digital data have discrete states and take discrete values.values.


아날로그와 디지털 신호아날로그와 디지털 신호

아날로그 신호 (analog signal) 는 연속적인 파형 디지털 신호 (digital signal) 는 이산적이며 1, 0 과 같이

제한된 수의 정의된 값만을 가질 수 있음 아날로그와 디지털 신호의 비교


주기 신호와 비주기 신호주기 신호와 비주기 신호

주기 신호 (Periodic signals)

■ 연속적으로 반복된 패턴으로 구성

■ 사이클 (cycle) – 하나의 완성된 패턴

■ 신호의 주기 (T) 는 초 단위로 표현

비주기 신호 (Aperiodic signals)■ 시간에 따라 반복된 패턴이나 사이클이 없이 항상 변한다

■ 신호는 반복된 패턴이 없다


주기 신호와 비주기 신호주기 신호와 비주기 신호

데이터 통신에서는 흔히 주기 아날로그 신호를 사용하거나 비주기 디지털 신호를 사용한다 .

In data communications, we commonly use periodic In data communications, we commonly use periodic analog signals and nonperiodic digital signals.analog signals and nonperiodic digital signals.


3.2 3.2 주기 아날로그 신호주기 아날로그 신호

싸인파 (sine wave, 정현파 ) 는 아날로그 주기 신호의 가장 기본적인 형태

단순 아날로그 신호 ( 정현파 ) s 는 순간 진폭 , A 는 최대 진폭 , f 는 주파수 , 는 위상이라 할때


We discuss a mathematical approach to sine We discuss a mathematical approach to sine waves in Appendix C.waves in Appendix C.


최대 진폭최대 진폭

전송하는 신호의 에너지에 비례하는 가장 큰 세기의 절대값 전기 신호의 경우 , 최대 진폭은 전압 (v) 으로 측정 위상과 주파수는 같지만 진폭이 서로 다른 두 신호 .


The power in your house can be represented by a sine wave with a peak The power in your house can be represented by a sine wave with a peak amplitude of 155 to 170 V. However, it is common knowledge that the amplitude of 155 to 170 V. However, it is common knowledge that the voltage of the power in U.S. homes is 110 to 120 V. This discrepancy is voltage of the power in U.S. homes is 110 to 120 V. This discrepancy is due to the fact that these are due to the fact that these are root mean squareroot mean square (rms) values. The signal (rms) values. The signal is squared and then the average amplitude is calculated. The peak is squared and then the average amplitude is calculated. The peak

value is equal to value is equal to 22½½ × rms × rms value. value.

미국의 경우 , 가정의 전기는 155 에서 170v 의 최대 진폭을 갖는 정현파로

표시할 수 있다 . 그러나 미국 가정의 전압은 110 에서 120v 로 알려져

있다 . 이 차이는 이 전압이 평균 제곱값 (rms, root mean sguare) 을

택하기 때문이다 . 신호는 제곱을 하여 평균 진폭을 계산 한다 . 최대값은

2½ × rms 값과 같다 .

Example 3.1


Example 3.2

The voltage of a battery is a constant; this constant value can The voltage of a battery is a constant; this constant value can be considered a sine wave, as we will see later. For example, be considered a sine wave, as we will see later. For example, the peak value of an the peak value of an AAAA battery is normally battery is normally1.5 V1.5 V..

건전지의 전압은 일정하다 . 이 일정한 값은 하나의 정현파로 볼 수 있다 .예를 들어 , AA 배터리의 최대값은 보통 1.5v 이다 .


주기와 주파수주기와 주파수

주기 (Period) 와 주파수 (Frequency)■ 주기 (T)

▶ 하나의 사이클을 완성하는데 필요한 시간 ( 초 단위 )

■ 주파수▶ 주기의 역수 (1 / t), 1 초 동안 생성되는 신호 주기의 수▶ 주파수 = 1 / 주기 , 주기 = 1 / 주파수▶ f = 1 / T , T = 1 / f

Frequency and period are the inverse of each otherFrequency and period are the inverse of each other..


진폭과 위상은 같지만 주파수가 서로 다른 신호 .



주기와 주파수 단위


Example 3.3Example 3.3

The power we use at home has a frequency of The power we use at home has a frequency of 60 Hz60 Hz. . The period of this sine wave can be determined as The period of this sine wave can be determined as follows:follows:



예제 3.4 , 예제 3.5

100ms 를 마이크로 초로 나타내고 상응하는 주파수를 KHz 로 나타내라 .

풀이 > 100ms 를 마이크로 초로 나타내자 . 표 3.1 로부터 1ms 와 1 초에 상응하는 값을 찾아서 값을 대입하면 다음과 같다 .

100 ms = 100 10-3 s = 100 10-3 106 ℳ s = 105 ℳ s 이제 주파수를 찾기 위해 역을 취하고 다시 kHz 로 변환하면 아래와 같

다 .

100 ms = 100 10-3 s = 10-1 s

f = 1/10-1 Hz = 10 10-3 KHz = 10-2 KHz



SolutionSolutionFrom Table 3.1 we find the equivalents of 1 ms (1 ms is 10From Table 3.1 we find the equivalents of 1 ms (1 ms is 10−3−3 s) s) and 1 s (1 s is 10and 1 s (1 s is 1066 μs). We make the following substitutions:. μs). We make the following substitutions:.

Express a period of 100 ms in microseconds.Express a period of 100 ms in microseconds.



The period of a signal is 100 ms. What is its frequency in The period of a signal is 100 ms. What is its frequency in kilohertz?kilohertz?

SolutionSolutionFirst we change 100 ms to seconds, and then First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10we calculate the frequency from the period (1 Hz = 10−3−3 kHz). kHz).



만약 신호가 전혀 변화하기 않으면 주파수는 0 이다 . 신호가 순간적으로 변화하면 주파수는 무한대이다 .

Frequency is the rate of change with respect to time. Frequency is the rate of change with respect to time. Change in a short span of time means high frequency. Change in a short span of time means high frequency. Change over a long span of time means low frequency.Change over a long span of time means low frequency.

If a signal does not change at all, its frequency is zero.If a signal does not change at all, its frequency is zero.If a signal changes instantaneously, its frequency is infinite.If a signal changes instantaneously, its frequency is infinite.

주파수는 시간에 대한 신호의 변화율이다 . 짧은 기간 내의 변화는 높은 주파수를 의미한다 .

긴 기간에 걸친 변화는 낮은 주파수를 의미한다 .


위상위상

시간 0 시에 대한 파형의 상대적인 위치

Phase describes the position of the waveform relative to time 0.

시간축을 따라 앞뒤로 이동될 수 있는 파형에서 그 이동된 양

첫 사이클의 상태를 표시

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진폭과 주파수는 같지만 위상이 서로 다른 정현파


위상위상

예제 2정현파는 시간 0 의 점에서 1/6 사이클 만큼 벗어나 있다 . 위상은 얼마인

가 ?풀이 > 하나의 완전한 원은 360 도이다 . 그러므로 원의 1/6 은 다음과

같다 . (1/6) 360 = 60 degrees = 60 x 2p /360 rad = 1.046 radA sine wave is offset 1/6 cycle with respect to time 0.

What is its phase in degrees and radians?

SolutionWe know that 1 complete cycle is 360°. Therefore, 1/6 cycle is


위상위상

정현파의 예

최대 진폭 : 5, 주파수 : 4, 위상 : 0

최대 진폭 : 10, 주파수 : 8, 위상 : 0

최대 진폭 : 5, 주파수 : 2, 위상 : ∏ /2


■ 단순 정현파의 주기나 주파수가 전송 매체를 통과하는 전파속도 (propagation speed) 와 연관

■ 단순신호가 한 주기 동안 진행 할 수 있는 거리 .

■ 파장과 주기

파장파장 (wavelength)(wavelength)


■ 파장은 전파속도와 주기가 주어지면 계산 가능

■ 파장을 x, 전파속도를 c( 빛의 속도 ), 주파수를 f 라 하면

■ 진공에서의 빛의 전파속도 3*108 m/s, 빨간색 빛 ( 주파수 = 4 * 1014) 의 파장은

cfx

frequencynpropagatio

period speedn propagatio wavelength

mmcfx 75.01075.0

104103 6

14

8

파장파장 (wavelength)(wavelength)


시간 영역과 주파수 영역

시간 영역 도면 (time-domain plot) - 시간에 대한 순간적인 진폭

주파수 영역 도면 (freguency-domain plot) - 주파수에 대한 최대 진폭


시간 영역과 주파수 영역시간 영역과 주파수 영역


정현파의 시간 영역과 주파수 영역


A complete sine wave in the time domain can A complete sine wave in the time domain can be represented by one single spike in the be represented by one single spike in the

frequency domain.frequency domain.

시간 영역에서 완전한 정현파는 주파수 영역에서 뾰족점 하나로 나타낸다 .


3 개의 정현파의 시간 영역과 주파수 영역


A single-frequency sine wave is not useful in data A single-frequency sine wave is not useful in data communications; we need to send a composite signal, communications; we need to send a composite signal,

a signal made of many simple sine wavesa signal made of many simple sine waves..

단일 주파수 정현파는 데이터 통신에 유용하지 않다 . 여러 개의 단일 정현파로 만들어진 복합 신호가 필요하다 .


According to Fourier analysis, any composite signal According to Fourier analysis, any composite signal is a combination of simple sine waves with different is a combination of simple sine waves with different

frequencies, amplitudes, and phases. Fourier analysis frequencies, amplitudes, and phases. Fourier analysis is discussed in Appendix C.is discussed in Appendix C.

퓨리에 분석에 따라서 , 임의의 복합 신호는 서로 다른 주파수 , 진폭 , 위상을 갖는 단순 정현파의 조합으로

나타낼 수 있다 . 퓨리에 분석은 부록 C 에 있다 .


If the composite signal is periodic, the decomposition If the composite signal is periodic, the decomposition gives a series of signals with discrete frequencies; gives a series of signals with discrete frequencies;

if the composite signal is nonperiodic, if the composite signal is nonperiodic, the decomposition gives a combination of sine waves the decomposition gives a combination of sine waves

with continuous frequencieswith continuous frequencies..



Figure 3.9 shows a periodic composite signal with frequency Figure 3.9 shows a periodic composite signal with frequency f. This type of signal is not typical of those found in data f. This type of signal is not typical of those found in data communications. We can consider it to be three alarm communications. We can consider it to be three alarm systems, each with a different frequency. The analysis of systems, each with a different frequency. The analysis of this signal can give us a good understanding of how to this signal can give us a good understanding of how to decompose signals.decompose signals.


복합 주기 신호


복합 주기 신호의 시간 영역과 주파수 영역 분해



Figure 3.11 shows a nonperiodic composite signal. It can Figure 3.11 shows a nonperiodic composite signal. It can be the signal created by a microphone or a telephone set be the signal created by a microphone or a telephone set when a word or two is pronounced. In this case, the when a word or two is pronounced. In this case, the composite signal cannot be periodic, because that implies composite signal cannot be periodic, because that implies that we are repeating the same word or words with exactly that we are repeating the same word or words with exactly the same tone.the same tone.


비주기 신호의 시간과 주파수 영역


대역폭대역폭

복합신호의 대역폭은 신호에 포함된 최고 주파수와 최저 주파수의 차이이다 .

The bandwidth of a composite signal is the difference The bandwidth of a composite signal is the difference between the highest and the lowest frequencies between the highest and the lowest frequencies

contained in that signal.contained in that signal.


주기와 비주기 복합 신호의 대역폭


대역폭대역폭

예제 3.10 만약 주기적 신호가 주파수 100, 300, 500, 700, 900Hz 를 갖는 5 개의 정현파로 분해된다면 , 그 대역폭은 얼마인가 ? 모든 구성요소가 10 볼트의 최대 진폭을 갖는다고 가정하고 스펙트럼을 그리시오 .

풀이 > fh 를 최고 주파수 , fl 을 최저 주파수라 하고 , B 는 대역폭이라 하자 . B = fh - fl = 900 - 100 = 800 Hz 스펙트럼은 주파수 100, 300, 500, 700, 900 의 5 개 막대만을 가진다 .


대역폭대역폭

예제 3.11 어떤 신호가 20Hz 의 대역폭을 가지며 , 최고 주파수는 60Hz 이다 .

가장 낮은 주파수는 얼마인가 ? 신호가 같은 진폭의 모든 정수 주파수를 포함할 때 , 스펙트럼을 그리시오 .

풀이 > fh 를 최고 주파수 , fl 을 최저 주파수라 하고 , B 는 대역폭이라 하자 . B = fh - fl 20 = 60 – fl fl = 60 - 20 = 40 Hz



A nonperiodic composite signal has a bandwidth of 200 kHz, A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have an amplitude of 0. Draw V. The two extreme frequencies have an amplitude of 0. Draw the frequency domain of the signal.the frequency domain of the signal.

SolutionSolutionThe lowest frequency must be at 40 kHz and the highest at The lowest frequency must be at 40 kHz and the highest at 240 kHz. Figure 3.15 shows the frequency domain and the 240 kHz. Figure 3.15 shows the frequency domain and the bandwidth.bandwidth.


Figure 3.15 The bandwidth for Example 3.12The bandwidth for Example 3.12



An example of a nonperiodic composite signal is the signal An example of a nonperiodic composite signal is the signal propagated by an AM radio station. In the United States, propagated by an AM radio station. In the United States, each AM radio station is assigned a 10-kHz bandwidth. The each AM radio station is assigned a 10-kHz bandwidth. The total bandwidth dedicated to AM radio ranges from 530 to total bandwidth dedicated to AM radio ranges from 530 to 1700 kHz. We will show the rationale behind this 10-kHz 1700 kHz. We will show the rationale behind this 10-kHz bandwidth in Chapter 5.bandwidth in Chapter 5.



Another example of a nonperiodic composite signal is the Another example of a nonperiodic composite signal is the signal propagated by an FM radio station. In the United signal propagated by an FM radio station. In the United States, each FM radio station is assigned a 200-kHz States, each FM radio station is assigned a 200-kHz bandwidth. The total bandwidth dedicated to FM radio bandwidth. The total bandwidth dedicated to FM radio ranges from 88 to 108 MHz. We will show the rationale ranges from 88 to 108 MHz. We will show the rationale behind this 200-kHz bandwidth in Chapter 5.behind this 200-kHz bandwidth in Chapter 5.



Another example of a nonperiodic composite signal is the Another example of a nonperiodic composite signal is the signal received by an old-fashioned analog black-and-white signal received by an old-fashioned analog black-and-white TV. A TV screen is made up of pixels. If we assume a TV. A TV screen is made up of pixels. If we assume a resolution of 525 × 700, we have 367,500 pixels per screen. If resolution of 525 × 700, we have 367,500 pixels per screen. If we scan the screen 30 times per second, this is 367,500 × 30 = we scan the screen 30 times per second, this is 367,500 × 30 = 11,025,000 pixels per second. The worst-case scenario is 11,025,000 pixels per second. The worst-case scenario is alternating black and white pixels. We can send 2 pixels per alternating black and white pixels. We can send 2 pixels per cycle. Therefore, we need 11,025,000 / 2 = 5,512,500 cycles cycle. Therefore, we need 11,025,000 / 2 = 5,512,500 cycles per second, or Hz. The bandwidth needed is 5.5125 MHz. per second, or Hz. The bandwidth needed is 5.5125 MHz.


3.3 3.3 디지털 신호디지털 신호 (DIGITAL SIGNALS)(DIGITAL SIGNALS)

In addition to being represented by an analog signal, information can In addition to being represented by an analog signal, information can also be represented by a also be represented by a digital signaldigital signal. For example, a 1 can be encoded . For example, a 1 can be encoded as a positive voltage and a 0 as zero voltage. A digital signal can have as a positive voltage and a 0 as zero voltage. A digital signal can have more than two levels. In this case, we can send more than 1 bit for each more than two levels. In this case, we can send more than 1 bit for each level.level.

Bit RateBit RateBit LengthBit LengthDigital Signal as a Composite Analog SignalDigital Signal as a Composite Analog SignalApplication LayerApplication Layer

Topics discussed in this section:Topics discussed in this section:


2 개의 준위와 4 개의 준위를 갖는 신호 .


신호가 L 개의 준위 (level) 을 가지면 각 준위는 log2L 비트가 필요하다 .

Appendix C reviews information about exponential and logarithmic

functions.Appendix C reviews information about Appendix C reviews information about exponential and logarithmic functions.exponential and logarithmic functions.



A digital signal has eight levels. How many bits are needed A digital signal has eight levels. How many bits are needed per level? We calculate the number of bits from the per level? We calculate the number of bits from the formulaformula

Each signal level is represented by 3 bits.Each signal level is represented by 3 bits.



A digital signal has nine levels. How many bits are needed A digital signal has nine levels. How many bits are needed per level? We calculate the number of bits by using the per level? We calculate the number of bits by using the formula. Each signal level is represented by 3.17 bits. formula. Each signal level is represented by 3.17 bits. However, this answer is not realistic. The number of bits However, this answer is not realistic. The number of bits sent per level needs to be an integer as well as a power of sent per level needs to be an integer as well as a power of 2. For this example, 4 bits can represent one level.2. For this example, 4 bits can represent one level.


비트율 (Bit rate)-1 초 동안 전송된 비트의 수-bps(bit per second)

비트길이 (bit length)- 한 비트가 전송매체를 통해 차지하는 길이- Bit length = propagation speed x bit duration



Assume we need to download text documents at the rate of Assume we need to download text documents at the rate of 100 pages per minute. What is the required bit rate of the 100 pages per minute. What is the required bit rate of the channel?channel?

SolutionSolutionA page is an average of 24 lines with 80 characters in each A page is an average of 24 lines with 80 characters in each line. If we assume that one character requires 8 bits, the bit line. If we assume that one character requires 8 bits, the bit rate israte is



A digitized voice channel, as we will see in Chapter 4, is A digitized voice channel, as we will see in Chapter 4, is made by digitizing a 4-kHz bandwidth analog voice made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest signal. We need to sample the signal at twice the highest frequency (two samples per hertz). We assume that each frequency (two samples per hertz). We assume that each sample requires 8 bits. What is the required bit rate?sample requires 8 bits. What is the required bit rate?

SolutionSolutionThe bit rate can be calculated asThe bit rate can be calculated as



What is the bit rate for high-definition TV (HDTV)?What is the bit rate for high-definition TV (HDTV)?

SolutionSolutionHDTV uses digital signals to broadcast high quality video HDTV uses digital signals to broadcast high quality video signals. The HDTV screen is normally a ratio of 16 : 9. There signals. The HDTV screen is normally a ratio of 16 : 9. There are 1920 by 1080 pixels per screen, and the screen is renewed are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second. Twenty-four bits represents one color 30 times per second. Twenty-four bits represents one color pixel. pixel.

The TV stations reduce this rate to 20 to 40 Mbps through compression. The TV stations reduce this rate to 20 to 40 Mbps through compression.


복합 아날로그 신호로서의 디지털 신호 .- 디지털 신호는 무한대의 주파수를 갖는 복합 신호- 대역폭은 무한대- 주기 및 비주기 디지털 신호의 시간 및 주파수 영역


디지털 신호의 전송

- 기저대역 (baseband, 베이스밴드 )- 광대역 (wideband)


기저대역 (baseband) 전송- 디지털 신호를 아날로그 신호로 바꾸지 않고 있는 그대로 채널을통해 전송


저대역 통과 채널 (low-pass channel)- 주파수 0 부터 시작하는 대역폭을 갖는 채널


경우 1: 광대역폭 (widebandwidth) Low-pass 채널- 전용 매체를 사용하는 banseband 전송

- 약간의 오차는 추론으로 가능- 동축이나 광섬유에서 사용-LAN 에 많이 사용


경우 2 : 제한된 대역폭 (Limited bandwidth), Low-pass 채널

대략적 근사값 (Rough approximation)- 비트율 N 의 디지털 신호

- 주파수 의 아날로그 신호 필요 .

- 요구 대역폭

2Nf

2N


최악의 경우에 대해 일차 조파를 이용한 디지털 신호의 대략적 근사값


보다 나은 근사 (better Approximation)- 보다 많은 수의 조파 (harmonics) 이용- 처음 3 개의 조파를 이용한 디지털 신호의 시뮬레이션


In baseband transmission, the required In baseband transmission, the required bandwidth is proportional to the bit rate;bandwidth is proportional to the bit rate;

if we need to send bits faster, we need more if we need to send bits faster, we need more bandwidth.bandwidth.


Table 3.2 Bandwidth requirementsBandwidth requirements



What is the required bandwidth of a low-pass channel if we need to send What is the required bandwidth of a low-pass channel if we need to send 1 Mbps by using baseband transmission?1 Mbps by using baseband transmission?

Solution The answer depends on the accuracy desired.The answer depends on the accuracy desired. a. The minimum bandwidth, is B = bit rate /2, or 500 kHz.The minimum bandwidth, is B = bit rate /2, or 500 kHz. b. A better solution is to use the first and the thirdA better solution is to use the first and the third harmonics with B = 3 × 500 kHz = 1.5 MHz. harmonics with B = 3 × 500 kHz = 1.5 MHz.

c. Still a better solution is to use the first, third, and fifthStill a better solution is to use the first, third, and fifth

harmonics with B = 5 × 500 kHz = 2.5 MHz harmonics with B = 5 × 500 kHz = 2.5 MHz..



We have a low-pass channel with bandwidth 100 kHz. What We have a low-pass channel with bandwidth 100 kHz. What is the maximum bit rate of this channel?is the maximum bit rate of this channel?

SolutionThe maximum bit rate can be achieved if we use the first The maximum bit rate can be achieved if we use the first harmonic. The bit rate is 2 times the available bandwidth, or harmonic. The bit rate is 2 times the available bandwidth, or 200 kbps.200 kbps.


광대역 전송 ( 변조 이용 )- 디지털 신호로 전송하기 위해 아날로그 신호로 전환 사용- 변조를 하면 띠대역 통과 채널 사용 전송- 띠대역 통과 (bandpass) 채널의 대역폭


If the available channel is a bandpass If the available channel is a bandpass channel, we cannot send the digital signal channel, we cannot send the digital signal

directly to the channel; directly to the channel; we need to convert the digital signal to an we need to convert the digital signal to an

analog signal before transmission.analog signal before transmission.


띠대역 채널에서 전송을 위한 디지털 신호의 변조



An example of broadband transmission using modulation An example of broadband transmission using modulation is the sending of computer data through a telephone is the sending of computer data through a telephone subscriber line, the line connecting a resident to the subscriber line, the line connecting a resident to the central telephone office. These lines are designed to carry central telephone office. These lines are designed to carry voice with a limited bandwidth. The channel is considered voice with a limited bandwidth. The channel is considered a bandpass channel. We convert the digital signal from a bandpass channel. We convert the digital signal from the computer to an analog signal, and send the analog the computer to an analog signal, and send the analog signal. We can install two converters to change the digital signal. We can install two converters to change the digital signal to analog and vice versa at the receiving end. The signal to analog and vice versa at the receiving end. The converter, in this case, is called a converter, in this case, is called a modemmodem which we discuss which we discuss in detail in Chapter 5.in detail in Chapter 5.



A second example is the digital cellular telephone. For A second example is the digital cellular telephone. For better reception, digital cellular phones convert the better reception, digital cellular phones convert the analog voice signal to a digital signal (see Chapter 16). analog voice signal to a digital signal (see Chapter 16). Although the bandwidth allocated to a company Although the bandwidth allocated to a company providing digital cellular phone service is very wide, we providing digital cellular phone service is very wide, we still cannot send the digital signal without conversion. still cannot send the digital signal without conversion. The reason is that we only have a bandpass channel The reason is that we only have a bandpass channel available between caller and callee. We need to convert available between caller and callee. We need to convert the digitized voice to a composite analog signal before the digitized voice to a composite analog signal before sending.sending.


3.4 3.4 전송 장애 전송 장애

- 신호가 매체를 통해 전송할 때 생기는 장애


감쇠감쇠 (attenuation) (attenuation) 에너지 손실을 의미

매체를 통해 이동할 때 매체의 저항을 이겨내기 위해 약간의 에너지가 손실

증폭기를 이용하여 신호를 다시 증폭


감쇠감쇠 (dB, decibel) (dB, decibel) 데시벨

■ 신호의 손실된 길이나 획득한 길이를 보이기 위해 사용■ 2 개의 다른 점에서 두 신호 또는 하나의 신호의 상대적 길이를 측정■ 신호가 감쇠하면 음수 , 증폭되면 양수

dB=10 log10(p2/p1) *p1과 p2 는 신호의 전력

예제 3.26 신호가 전송매체를 통해 이동하고 있고 전력이 반으로 줄었다고

상상해보자 이것은 P2 = 1/2 P1을 의미한다 . 이 경우 감쇠 ( 전력 손실 ) 는 다음과 같이 계산할 수 있다 .

10 log10 (P2/P1) = 10 log10 (0.5P1/P1)

= 10 log10 (0.5)

= 10(–0.3) = –3 dB



Suppose a signal travels through a transmission medium Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) is (1/2)P1. In this case, the attenuation (loss of power) can be calculated ascan be calculated as

A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.


신호가 증폭기를 통해 이동하고 전력이 10 배 늘었다고 상상해보자 . 이것은 P2=10 × P1 을 의미한다 . 이 경우 증폭 ( 전력 증가 ) 은 다음과 같이 계산할 수 있다 .

10 log10 (P2/P1) = 10 log10 (10P1/P1) = 10 log10 (10) = 10 (1) = 10 dB

A signal travels through an amplifier, and its power is increased A signal travels through an amplifier, and its power is increased 10 times. This means that P10 times. This means that P22 = 10P = 10P1 1 . In this case, the . In this case, the amplification (gain of power) can be calculated asamplification (gain of power) can be calculated as



공학자가 신호의 길이 변화를 측정하는 데 데시벨을 사용하는 이유 중 하나는 단지 공학자가 신호의 길이 변화를 측정하는 데 데시벨을 사용하는 이유 중 하나는 단지 22 개개 (( 직렬직렬 )) 의 점 대신 의 점 대신 여러 점에 관하여 이야기할 때 데시벨 숫자가 더해지거나 빼질 수 있기 때문이다여러 점에 관하여 이야기할 때 데시벨 숫자가 더해지거나 빼질 수 있기 때문이다 . . 다음 그림에서 신호는 다음 그림에서 신호는 점 점 1 1 에서 점 에서 점 44 까지의 긴 거리를 이동한다까지의 긴 거리를 이동한다 . . 신호는 점 신호는 점 22 에 도착한 시간에 의해 감쇠도고 점 에 도착한 시간에 의해 감쇠도고 점 22 와 점 와 점 3 3 사이에서 증폭되었다가 다시 점 사이에서 증폭되었다가 다시 점 33 과 점 과 점 4 4 사이에서 감쇠된다사이에서 감쇠된다 . . 신호의 데시벨 결과를 구하기 위해 각 신호의 데시벨 결과를 구하기 위해 각 지점 사이에서 측정된 지점 사이에서 측정된 dB dB 값은 더하기만 하면 된다값은 더하기만 하면 된다 . . 이 경우 데시벨은 다음처럼 계산될 수 있다이 경우 데시벨은 다음처럼 계산될 수 있다 ..

dB = –3 + 7 – 3 = +1




Sometimes the decibel is used to measure signal power in milliwatts. Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as In this case, it is referred to as dBdBmm and is calculated as dB and is calculated as dBmm = 10 = 10 log10 Plog10 Pm m , where P, where Pmm is the power in milliwatts. Calculate the power is the power in milliwatts. Calculate the power of a signal with dBof a signal with dBmm = −30. = −30.

SolutionSolutionWe can calculate the power in the signal asWe can calculate the power in the signal as



The loss in a cable is usually defined in decibels per kilometer The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?has a power of 2 mW, what is the power of the signal at 5 km?

SolutionSolutionThe loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power ascalculate the power as


일그러짐일그러짐 (distortion) (distortion) 신호의 모양이나 형태가 변하는 것

반대되는 신호나 다른 주파수 신호로 만듬


잡음잡음 (noise) (noise)

열잡음 , 유도된 잡음 , 혼선 , 충격잡음등의 여러 형태의 잡음


신호대 잡음비 (SNR)-SNR : signal-to-noise ration- 잡음과 신호 전력의 비율

-SNR 은 데시벨로 표시 SNRdB = 10 log10SNR

powernoiseaeragepowersignalaerageSNR



The power of a signal is 10 mW and the power of the noise The power of a signal is 10 mW and the power of the noise is 1 μW; what are the values of SNR and SNRis 1 μW; what are the values of SNR and SNRdB dB ??

SolutionSolutionThe values of SNR and SNRdB can be calculated as The values of SNR and SNRdB can be calculated as follows:follows:



The values of SNR and SNRdB for a noiseless channel areThe values of SNR and SNRdB for a noiseless channel are

We can never achieve this ratio in real life; it is an ideal.We can never achieve this ratio in real life; it is an ideal.


Figure 3.30 Two cases of SNR: a high SNR and a low SNR


3.5 3.5 데이터 전송률의 한계데이터 전송률의 한계

데이터 전송률의 세 요소■ 가역 대역폭■ 사용 가능한 신호 준위■ 채널의 품질 ( 잡음의 정도 )

데이터 전송률을 계산하는 두가지 이론적 수식■ 나이퀴스트 수식 (Nyquist bit rate) : 잡음이 없는 채널에서

사용■ 새논 수식 (Shannon capacity) : 잡음이 있는 채널에서 사용


무잡음 채널 무잡음 채널 : : 나이퀴스트 전송률나이퀴스트 전송률 (Nyquist bit (Nyquist bit rate)rate)

나이퀴스트 전송률■ 잡음이 없는 채널의 경우 사용■ 대역폭은 채널의 대역폭 , L 은 데이터를 나타내는 데 사용한 신호준위의

개수 , 전송률은 초당 비트수라고 할때

■ 예제 3.34두 개의 신호 준위를 갖는 신호를 전송하는 3,000Hz 의 대역폭을 갖는 무잡음 채널이 있다 . 최대 전송률은 다음과 같이 계산된다 . 전송률 = 2 × 3,000 = log22=6,000bps

■ 예제 3.35네 개의 신호 준위 ( 각 준위는 2 비트를 나타낸다 ) 를 사용하는 신호를 위의

예제와 동일한 채널을 사용하여 보낸다고 하자 . 최대 전송률은 다음과 같다 . 전송률 = 2 × 3,000 = log24 = 12,000bps

전송률 = 2 × 대역폭 × log2L



We need to send 265 kbps over a noiseless channel with a bandwidth of We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?20 kHz. How many signal levels do we need?

SolutionSolutionWe can use the Nyquist formula as shown:We can use the Nyquist formula as shown:

Since this result is not a power of 2, we need to either increase the Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps. If we have 64 levels, the bit rate is 240 kbps.is 280 kbps. If we have 64 levels, the bit rate is 240 kbps.


잡음 있는 채널 잡음 있는 채널 : : 섀논 용량섀논 용량 (Shannon capacity)(Shannon capacity) 섀논 용량

■ 잡음이 있는 채널에서의 최대 전송률을 결정하는 수식■ 대역폭은 채널의 대역폭 , SNR 은 신호에 대한 잡음 비율 , 용량은 bps

단위의 채널 용량이라고 하면

■ 예제 3.37신호 대 잡음의 비율값이 거의 0인 , 거의 잡음에 가까운 채널을 생각해보자 . 다시 말해 , 잡음이 너무 강해서 신호가 약해진다 . 이 채널에 대한 용량을 계산하면 다음과 같다 . C = B log2(1+SNR)=B log2(1+0) = B log2(1) = B × 0 = 0이것은 채널의 용량이 0이다 . 대역폭은 고려되지 않았다 . 다른 말로 하자면 이 채널로는 어떤 데이터도 보낼 수 없다 .

■ 예제 3.38일반 전화선의 이론적인 최고 데이터 전송률을 계산할 수 있다 . 전화선은 일반적으로

3,000Hz(300Hz 에서 3,300Hz) 의 대역폭을 갖는다 . 신호 대 잡음의 비율이 보통 3,162(35dB) 이다 . 이 채널에 대한 용량을 계산하면 다음과 같다 .

C = B log2(1+SNR) = 3,000 log2(1+3,162) = 3,000 log2(3,163) C = 3,000 × 11.62 = 34,860bps이는 전화선의 최대 비트율이 34,860Kbps 임을 의미한다 .

용량 = 대역폭 × log2(1+SNR)



The signal-to-noise ratio is often given in decibels. The signal-to-noise ratio is often given in decibels. Assume that SNRAssume that SNRdBdB = 36 and the channel bandwidth is = 36 and the channel bandwidth is 2 MHz. The theoretical channel capacity can be 2 MHz. The theoretical channel capacity can be calculated ascalculated as



For practical purposes, when the SNR is very high, we For practical purposes, when the SNR is very high, we can assume that SNR + 1 is almost the same as SNR. can assume that SNR + 1 is almost the same as SNR. In these cases, the theoretical channel capacity can be In these cases, the theoretical channel capacity can be simplified tosimplified to

For example, we can calculate the theoretical capacity For example, we can calculate the theoretical capacity of the previous example asof the previous example as


두 가지 한계를 사용하기두 가지 한계를 사용하기

실제로는 어떤 신호 준위의 어떤 대역폭이 필요하진 알기 위해 두가지 방법을 모두 사용

■ 예제 3.411 MHz 의 대역폭을 갖는 채널이 있다 . 이 채널의 SNR 은 63 이다 . 적절한

전송률과 신호 준위는 무엇인가 ?

풀이 > 우선 상한을 구하기 위해 새논 수식을 사용한다 . C = B log2(1+SNR) = 106log2(1+63)=106log2(64)=6Mbps

비록 섀논 수식으로부터 6Mbps 의 전송률을 구했으나 이는 상한일 뿐이다 . 더 나은 성능을 위해 조금 낮은 값 , 예를 들어 4Mbps 를 택한다 . 그 후에 신호의 준위를 구하기 위해 나이퀴스트 식을 사용한다 .

4Mbps = 2 × 1MHz × log2L L=4


One important issue in networking is the One important issue in networking is the performanceperformance of the of the network—how good is it? We discuss quality of service, an overall network—how good is it? We discuss quality of service, an overall measurement of network performance, in greater detail in Chapter measurement of network performance, in greater detail in Chapter 24. In this section, we introduce terms that we need for future 24. In this section, we introduce terms that we need for future chapters.chapters.

BandwidthBandwidthThroughputThroughputLatency (Delay)Latency (Delay)Bandwidth-Delay ProductBandwidth-Delay Product

Topics discussed in this section:Topics discussed in this section:

3.6 3.6 성능성능 (PERFORMANCE)(PERFORMANCE)


대역폭 (bandwidth)

In networking, we use the term In networking, we use the term bandwidth in two contexts.bandwidth in two contexts.

❏❏ The first, bandwidth in hertz, refers to the range of The first, bandwidth in hertz, refers to the range of frequencies in a composite signal or the range of frequencies in a composite signal or the range of frequencies that a channel can pass.frequencies that a channel can pass.

❏❏ The second, bandwidth in bits per second, refers to the The second, bandwidth in bits per second, refers to the speed of bit transmission in a channel or linkspeed of bit transmission in a channel or link..



The bandwidth of a subscriber line is 4 kHz for voice or The bandwidth of a subscriber line is 4 kHz for voice or data. The bandwidth of this line for data transmissiondata. The bandwidth of this line for data transmissioncan be up to 56,000 bps using a sophisticated modem to can be up to 56,000 bps using a sophisticated modem to change the digital signal to analog.change the digital signal to analog.



If the telephone company improves the quality of the If the telephone company improves the quality of the line and increases the bandwidth to 8 kHz, we can send line and increases the bandwidth to 8 kHz, we can send 112,000 bps by using the same technology as mentioned 112,000 bps by using the same technology as mentioned in Example 3.42.in Example 3.42.


처리율 (throughput)■ 어떤 지점을 데이터가 얼마나 빨리 지나가는가를 측정



A network with bandwidth of 10 Mbps can pass only an average of A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?10,000 bits. What is the throughput of this network?

SolutionSolutionWe can calculate the throughput asWe can calculate the throughput as

The throughput is almost one-fifth of the bandwidth in this case.The throughput is almost one-fifth of the bandwidth in this case.


전파속도 (propagation speed)■ 신호나 비트가 매체를 통해 1초 동안 이동할 수 있는 거리

전파시간 (propagation time)■ 신호 ( 또는 비트 ) 가 전송매체의 한 지점에서 다른 지점까지

이동하는데 요구되는 시간을 측정하여 거리를 전파속도로 나눈 값

파장 (wavelength) ■ 한 주기에서 간단한 신호가 이동할 수 있는 거리


지연 (Delay, Latency)- 발신지에서 첫번째 비트를 보낸 시간부터 전체 메시지가 목적지에 도착할 때까지 걸린 시간

-Lantency = propagation time + transmission time

+queuing Time + processing delay

전파시간 (propagation time)- 비트가 발신지에서 목적지까지 이동하는데 걸리는 시간

- 전파시간 = 거리 /전파속도

- 진공에서 빛의 전파 속도 = 3 x 108 m/s



What is the propagation time if the distance between the two points is What is the propagation time if the distance between the two points is 12,000 km? Assume the propagation speed to be 2.4 × 108 m/s in 12,000 km? Assume the propagation speed to be 2.4 × 108 m/s in cable.cable.

SolutionSolutionWe can calculate the propagation time asWe can calculate the propagation time as

The example shows that a bit can go over the Atlantic Ocean in only The example shows that a bit can go over the Atlantic Ocean in only 50 ms if there is a direct cable between the source and the destination.50 ms if there is a direct cable between the source and the destination.


전송시간 (transmission time)

- 메시지를 전송하는데 걸리는 시간은 메시지 크기와 채널의

대역폭에 좌우

대역폭크기메시지전송시간



What are the propagation time and the transmission time for a 2.5-What are the propagation time and the transmission time for a 2.5-kbyte message (an e-mail) if the bandwidth of the network is 1 kbyte message (an e-mail) if the bandwidth of the network is 1 Gbps? Assume that the distance between the sender and the receiver Gbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s.is 12,000 km and that light travels at 2.4 × 108 m/s.

SolutionSolutionWe can calculate the propagation and transmission time as shown We can calculate the propagation and transmission time as shown on the next slide:on the next slide:


Example 3.46(continued)Example 3.46(continued)

Note that in this case, because the message is short and the Note that in this case, because the message is short and the bandwidth is high, the dominant factor is the propagation time, bandwidth is high, the dominant factor is the propagation time, not the transmission time. The transmission time can be ignored.not the transmission time. The transmission time can be ignored.



What are the propagation time and the transmission time for a 5-What are the propagation time and the transmission time for a 5-Mbyte message (an image) if the bandwidth of the network is 1 Mbyte message (an image) if the bandwidth of the network is 1 Mbps? Assume that the distance between the sender and the receiver Mbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 10is 12,000 km and that light travels at 2.4 × 1088 m/s. m/s.

SolutionSolutionWe can calculate the propagation and transmission times as shown We can calculate the propagation and transmission times as shown on the next slide.on the next slide.


Example 3.47(continued)Example 3.47(continued)

Note that in this case, because the message is very long and the Note that in this case, because the message is very long and the bandwidth is not very high, the dominant factor is the transmission bandwidth is not very high, the dominant factor is the transmission time, not the propagation time. The propagation time can be time, not the propagation time. The propagation time can be ignored.ignored.


큐 시간 (queuing time)- 중간 또는 종단 장치들이 메시지를 전송하기 까지 붙들고 있는 시간


대역폭 – 지연 곱 (bandwidth-Delay product)-bandwidth x delay-Case 1( 대역폭 1bps)


Case 2( 대역폭 4bps)


대역폭 - 지연 곱의 개념- 링크를 두 지점을 연결하는 파이프로 생각

- 파이프 단면 : 대역폭 , 파이프 길이 : 지연

The bandwidth-delay product defines the number of bits that can fill the link.


파형 난조 (jitter)

- 서로 다른 데이터 패킷이 서로 다른 지연 시간을 갖게

되어 생기는 현상 .- 29 장에서 설명


3.7 3.7 요약요약

chapter 3 데이터와 신호 (data and signals)

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