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expanded by Jozef Goetz, 2009
The McGraw-Hill Companies, Inc., 2006
expanded by Jozef Goetz, 2009
PART IIPART II
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• Define Define transmissiontransmission..• Describe the concept of Describe the concept of frequencyfrequency, , spectrumspectrum, and , and bandwidthbandwidth..• Define Define analoganalog and and digital data transmissiondigital data transmission..• Describe the concept of Describe the concept of datadata, , signalssignals, and , and transmissiontransmission..• List the various impairmentsList the various impairments to effective transmission.to effective transmission.• Describe Describe attatteenuationnuation..• Describe Describe delay distortiondelay distortion..• Describe the concept of Describe the concept of noisenoise & & channel capacity.channel capacity.
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3-3 DIGITAL SIGNALS3-3 DIGITAL SIGNALS
In addition to being represented, In addition to being represented, informationinformation can also be can also be representedrepresented
•by an by an aanalognalog signal signal
•by a by a digital signaldigital signal •For example, For example,
•a 1 can be encoded as a a 1 can be encoded as a positivepositive voltage and voltage and •a 0 as a 0 as zerozero voltage. voltage.
•A digital signal can A digital signal can have more have more thanthan 2 levels 2 levels. . •In this case, we can send In this case, we can send more more thanthan 1 bit 1 bit forfor each level.each level.
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3-3 DIGITAL SIGNALS3-3 DIGITAL SIGNALS
•Bit Rate vs Bit Length•Digital Signal as a Composite Analog Signal•Application Layer
Topics discussed in this section:Topics discussed in this section:
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5 Figure 3.16 A digital signal
•The bit interval (bit duration) is the time required to send 1 bit.•The bit rate is the # of bit intervals per sec
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Example 6Example 6
A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)
SolutionSolution
The bit duration (interval) is the inverse of the bit rate.
Bit interval = 1/ 2000 [bps] = 0.000500 [s/bit] = 0.000500 x [106 s/bit] = 500 [s/bit]
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7Figure 3.16 Bit rate and bit interval
1 s
•The bit duration BD (interval) is the time required to send 1 bit.•The bit rate BR is the # of bit intervals per sec
R = 1 / D
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8 Bit length and bit duration p.73 and 64
•The bit length BL is the distance 1 bit occupies on the transmission medium.
•The bit duration BD (interval) is the time required to send 1 bit.
bit length = propagation speed * bit duration
L = S * D
Remember from your elementary school formula: D = S * T
where: D –distance, S – speed, t - Time
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Figure 3.16 Two digital signals: one with two signal levels and the other with four signal levels
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Appendix C reviews information about exponential and logarithmic
functions.
Note
Appendix C reviews information about exponential and logarithmic functions.
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A digital signal has 8 levels. How many bits are needed per level?
Example 3.16
Each signal level is represented by 3 bits.
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A digital signal has 9 levels. How many bits are needed per level? We calculate the number of bits by using the formula. Each signal level is represented by 3.17 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 2.
•For this example, 4 bits can represent one level.
Example 3.17
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•Assume we need to download text documents at the rate of 100 pages per seconds.
•What is the required bit rate of the channel?
SolutionA page is an average of 24 lines with 80 characters in each line. If we assume that one character requires 8 bits, the bit rate is
100 x 24 [line/sec] = 100 x 24 x 80[chr/sec] =
= 100 x 24 x 80 x 8 [bit/sec] = 1, 636, 000 [bit/sec] = 1.636 Mbps
Example 3.18
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A digitized voice channel, as we will see in Chapter 4, is made by digitizing a 4-kHz bandwidth analog voice signal.
We need to sample the signal at twice the highest frequency (2 samples per hertz).
We assume that each sample requires 8 bits. What is the required bit rate?
Solution bit rate = 2 x 4000 [Hz] x 8 [bits] = 2 x 4000 [1/sec] x 8 [bits] = 64,000 [bits/sec] = 64 [kbps]
Example 3.19
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What is the bit rate for high-definition TV (HDTV)?Solution•HDTV uses digital signals to broadcast high quality video signals. •The HDTV screen is normally a ratio of 16 : 9. •There are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second = 30 [1/sec]•24 [bits] represents one color pixel: 1920 x 1080 x 30 [1/sec] x 24 [bit] = = 1,492,992,000 [bit/sec] = 1.5 Gbps
Example 3.20
•The TV stations reduce this rate to 20 to 40 Mbps through compression.
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Figure 3.17 The time and frequency domains of periodic and nonperiodic digital signals
•Based on Fourier analysis, a digital signal is a composite analog signal with an infinite bandwidth B
•The periodic signal has discrete frequencies
•The nonperiodic signal has continues frequencies
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Figure 3.18 Baseband transmission
A digital signal can be transmitted by using 2 approaches:
• Baseband
• Sending a digital signal without changing to
an analog signal
• Broadband transmission => slide 33
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Figure 3.19 Bandwidths of two low-pass channels
• Baseband transmission requires a low-pass channel
with bandwidth that starts from 0
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Figure 3.20 Baseband transmission using a dedicated medium with
a. wide-bandwidth
Baseband transmission of a digital signal that preserves the shape of the digital signal is possible only if we have a low-pass channel with an infinite or very wide bandwidth.
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•An example of a dedicated channel where the entire bandwidth of the medium is used as one single channel is a LAN. •Almost every wired LAN today uses a dedicated channel for 2 stations communicating with each other. •In a bus topology LAN with multipoint connections, only two stations can communicate with each other at each moment in time (timesharing);
•the other stations need to refrain from sending data.
•In a star topology LAN, the entire channel between each station and the hub is used for communication between these two entities.
Example 3.21
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Assume digital bit rate = N, the worst case, max changes happens when the signal carries.
The sequence 01010101010101….or 1010101010101010. To simulate this we need an analog signal of f = N / 2.
Low-pass channels with Limited Bandwidth, we approximate the digital signal with an analog signal.
The level of approximation depends on the bandwidth available.
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22Low-pass channels with Limited Bandwidth
Figure 3.21 Rough approximation of a digital signal using the first harmonic N/2 and N/4 for worst case
Where: bit rate = N•1 is represented by positive amplitude,•0 is represented by negative amplitude
Here is a sequence ofall 3-bit combinations:
<= after analyzing, we need A channel with f = 0 , N/4, N/2.
p = phase
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23Low-pass channels with Limited Bandwidth
To make the shape of the analog signal look more like that of a digital signal, we need to add more harmonics to the first harmonics N/4, N/2.
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Figure 3.22 Simulating a digital signal with first three harmonicsfor all scenarios
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25 Digital versus analog
The best case when we use a single f signal to send it
Worst case: the most # of changes,Others are between 0 Hz and 3 Hz. So, to send 6 bits per sec we need a bandwidth of 3 Hz
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26Relationship between the bit rates and bandwidth
To send n bps (bit rate) through analog channel using the above approximation, we need bandwidth B such that
B = bit rate / 2 B = n / 2 Note: B represents f = 3 Hz of a single sine ( with only one
frequency which is the first harmonic)
The required bandwidth is proportional to the bit rate !
Using a single sine analog signal per each pattern as a “kind of digital signal” cannot be recognized on the receiver site correctly.
So we need to improve the shape of the signal by using a composite signals (i.e. adding more harmonics)
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A A digitaldigital signalsignal is a is a compositecomposite signal signal with an with an iinfinitenfinite bandwidth bandwidth..
Note:Note:
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Table 3.2 Bandwidth RequirementTable 3.2 Bandwidth Requirement
Bit
Rate
Harmonic
1
Harmonics
1, 3
Harmonics
1, 3, 5
Harmonics
1, 3, 5, 7
1 Kbps 500 Hz 1.5 KHz 2.5 KHz 3.5 KHz
10 Kbps 5 KHz 15 KHz 25 KHz 35 KHz
100 Kbps 50 KHz 150 KHz 250 KHz 350 KHz
How is it calculated:B = n / 2 is related to the first harmonicB = 3 n / 2 is related to the 2nd column etc.B = 5 n / 2 is related to the 3rd column etc.
So B >= n / 2 => n <= 2B
AddingAdding more more oddodd harmonicsharmonics gives us gives us better shapebetter shape of the of the signalsignal
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29Note
In baseband transmission, the required bandwidth is proportional to the bit rate;
if we need to send bits faster, we need more bandwidth.
The The bit ratebit rate and the and the bandwidthbandwidth are are proportionalproportional to each other to each other
andand
B >= n / 2 from the previous slide
A typical voice signal has a bandwidth of approximately 4 kHz, so we should be able send about 8000 bps if we use a single sine (with the 1st harmonic)
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What is the required bandwidth of a low-pass channel if we need to send 1 Mbps by using baseband transmission?
SolutionThe answer depends on the accuracy desired.a. The minimum bandwidth, is B = bit rate /2, or 500 kHz. b. A better solution is to use the first and the third harmonics with B = 3 × 500 kHz = 1.5 MHz.
c. Still a better solution is to use the first, third, and fifth harmonics with B = 5 × 500 kHz = 2.5 MHz.
Example 3.22
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We have a low-pass channel with bandwidth 100 kHz. What is the maximum bit rate of this channel?
SolutionSee note under table 3.2 From B >= bit rate / 2 => bit rate <= 2 x BThe maximum bit rate can be achieved if we use the first harmonic. The bit rate is 2 times the available bandwidth, or 200 kbps.
Example 3.23
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Figure 3.23 Bandwidth of a bandpass channel
If the available channel is a bandpass channel, we cannot send the digital signal directly to the channel;
•we need to convert the digital signal to an analog signal before transmission.
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Figure 3.24 Modulation of a digital signal for transmission on a bandpass channel
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34 Low-pass and band-pass
Used for digital bandwidth, Used for digital bandwidth, though it can be used by though it can be used by analog one tooanalog one too
Used for analog bandwidthUsed for analog bandwidth
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The The analog bandwidthanalog bandwidth of a of a mediummedium is is the the range of frequenciesrange of frequencies that medium that medium can pass and it is can pass and it is expressedexpressed in hertzin hertz; ;
the the digital bandwidthdigital bandwidth of a of a mediummedium is is the the maximum bit ratemaximum bit rate that medium can that medium can pass and itpass and it is is expressedexpressed in bitsin bits perper second.second.
Note:Note:
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DigitalDigital transmission transmission needsneeds a a low-passlow-pass channel channel (theoretically (theoretically between 0 and infinity).between 0 and infinity).•the the upper limitupper limit can be can be relaxedrelaxed if we if we lowerlower our standardsour standards by accepting by accepting a a limited # of harmonicslimited # of harmonics
Note:Note:
AnalogAnalog transmission transmission cancan use a use a band-band-passpass channel. channel.
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analog bandwidthanalog bandwidth
The The analog bandwidth analog bandwidth can always can always be be shiftedshifted The The bandwidthbandwidth of a of a mediummedium can be can be
divided into divided into several several band-passband-pass channelchannel to to carrycarry severalseveral analoganalog transmissiontransmission
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38ANALOG AND DIGITAL
DATA TRANSMISSION
Speech bandwidth ~ 300 Hz - 3.4 Speech bandwidth ~ 300 Hz - 3.4 kHzkHz
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39ANALOG AND DIGITAL
DATA TRANSMISSION
Data (information) can be analog or digital.The data can be transmitted using analog transmission or digital transmission.
This gives us 4 possibilities:
Analog Data, Analog Transmission
Analog Data, Digital Transmission
Digital Data, Analog Transmission
Digital Data, Digital Transmission
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40ANALOG AND DIGITAL
DATA TRANSMISSION
DATADATA TRANSMISSIONTRANSMISSION
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41ANALOG AND DIGITAL
DATA TRANSMISSION
DATADATA TRANSMISSIONTRANSMISSION
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42EFFECT OF BANDWIDTH ON A
DIGITAL SIGNAL
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•An example of broadband transmission using modulation is the sending of computer data through a telephone subscriber line, the line connecting a resident to the central telephone office.
•These lines are designed to carry voice with a limited bandwidth (0 – 4 kHz), so the max bit rate is only 8 kbps
•The channel is considered a bandpass channel. •We convert the digital signal from the computer to an analog signal, and send the analog signal.
•We can install two converters to change the digital signal to analog and vice versa at the receiving end. •The converter, in this case, is called a modem (modulator/demodulator)
Example 3.24
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•A second example is the digital cellular telephone. •For better reception, digital cellular phones convert the analog voice signal to a digital signal (see Chapter 16).
• Although the bandwidth allocated to a company providing digital cellular phone service is very wide, we still cannot send the digital signal without conversion.
•The reason is that we only have a bandpass channel available between caller and callee. •We need to convert the digitized voice to a composite analog signal before sending.
Example 3.25