Digital and Analog Transmission
-
Upload
bradpit106 -
Category
Documents
-
view
224 -
download
0
Transcript of Digital and Analog Transmission
-
7/30/2019 Digital and Analog Transmission
1/36
1Dr. L. Christofi Spring 2009
Lecture 3
Digital TransmissionAnalog Transmission
ACOE412
Data Communications
Spring 2009
2Dr. L. Christofi Spring 2009
0. Overview0. OverviewIn this lecture we will cover the following topics:
4. Digital Transmission4.1 Digital to digital conversion4.2 Analog to digital conversion4.3 Transmission modes4.4 Summary (part 4)
5. Analog Transmission5.1 Digital to analog conversion5.2 Analog and digital5.3 Summary (part 5)
-
7/30/2019 Digital and Analog Transmission
2/36
3Dr. L. Christofi Spring 2009
4.1 DIGITAL4.1 DIGITAL --TOTO --DIGITAL CONVERSIONDIGITAL CONVERSION
In this section, we see how we can represent digital data byIn this section, we see how we can represent digital data byusing digital signals. The conversion involves threeusing digital signals. The conversion involves threetechniques:techniques: line codingline coding andand scramblingscrambling . Line coding is. Line coding isalways needed; scrambling may or may not be needed.always needed; scrambling may or may not be needed.
Line CodingLine Coding SchemesScrambling
Topics discussed in this section: Topics discussed in this section:
4Dr. L. Christofi Spring 2009
Line coding and decodingLine coding is the process of converting binary data i.e. asequence of bits, into a digital signal
-
7/30/2019 Digital and Analog Transmission
3/36
5Dr. L. Christofi Spring 2009
Signal element versus data element
6Dr. L. Christofi Spring 2009
Pulse Rate vs Bit Rate The pulse rate defines the number of pulses per second
A pulse is the minumum amount of time required totransmit a symbol
The bit rate define sthe maximum number of bits persecond
The relationship between pulse rate and bit rate is
BitRate = PulseRate x log 2L
where, L is the number of data levels of the signal
-
7/30/2019 Digital and Analog Transmission
4/36
7Dr. L. Christofi Spring 2009
A signal has four data levels with a pulse duration of 1ms. Findthe Pulse Rate and Bit Rate.
Solution
Example
Pulse Rate = 1/1ms=1000 pulses per second
Bit Rate = Pulse Rate x log 2L =1000 x log 24 = 2000bps
8Dr. L. Christofi Spring 2009
Effect of lack of synchronization
-
7/30/2019 Digital and Analog Transmission
5/36
9Dr. L. Christofi Spring 2009
In a digital transmission, the receiver clock is 0.1% faster thanthe sender clock. How many extra bits per second does thereceiver receive if the data rate is 1 kbps? How many if thedata rate is 1 Mbps?
SolutionAt 1 kbps, the receiver receives 1000(1+0.001) = 1001 bpsinstead of 1000 bps.
Example
At 1 Mbps, the receiver receives 1,000,000(1+0.001)=1,001,000 bps instead of 1,000,000 bps.
10Dr. L. Christofi Spring 2009
A self-synchronising digital signal includes timinginformation in the data being transmitted. This can beachieved if there are transitions in the signal that alertthe receiver to the beginning, middle or end of pulse.If the receivers clock is out of synchronisation, these
alerting points can reset the clock.
Note
-
7/30/2019 Digital and Analog Transmission
6/36
11Dr. L. Christofi Spring 2009
Line coding schemes
Line coding
12Dr. L. Christofi Spring 2009
Unipolar NRZ scheme
Unipolar encoding uses only one voltage level
bit 0 = zero bit 1=high
-
7/30/2019 Digital and Analog Transmission
7/36
13Dr. L. Christofi Spring 2009
Polar NRZ-L and NRZ-I schemes
Polar encoding uses two voltage levels (+ve and ve)
In NRZ-L encoding: bit 0 = high bit 1=low
In NRZ-I encoding the signal is inverted if 1 is encountered
14Dr. L. Christofi Spring 2009
In NRZ-L the level of the voltage determines the valueof the bit.
In NRZ-I the inversionor the lack of inversion
determines the value of the bit.
Note
-
7/30/2019 Digital and Analog Transmission
8/36
15Dr. L. Christofi Spring 2009
Polar RZ scheme
Polar encoding uses three voltage levels (+ve, zero & ve)
In RZ encoding the signal changes during each bit, notbetween bits: bit 1 = high-to-zero bit 0 = low-to-zero
Main disadvantage of RZ encoding is that it requires twosignal changes to encode 1 bit => occupies more bandwidth
16Dr. L. Christofi Spring 2009
Polar biphase: Manchester anddifferential Manchester schemes
In Manchester encoding: bit 1 = low-to-high bit 0 = high-to-low
In Differential Manchester encoding: bit 1 = no transition bit 0 = transition
-
7/30/2019 Digital and Analog Transmission
9/36
17Dr. L. Christofi Spring 2009
In Manchester and differential Manchester encoding,the transition
at the middle of the bit is used for synchronization.
Note
The minimum bandwidth of Manchester and
differential Manchester is 2 times that of NRZ.
Note
18Dr. L. Christofi Spring 2009
Bipolar schemes: AMI and pseudoternary
Bipolar encoding uses three voltage levels (+ve, zero & ve) Alternate Mask Inversion (AMI):
bit 0 = zero bit 1 = alternating +ve and ve pulses Pseudoternary :
bit 0 = alternating +ve and ve pulses bit 1 = zero
-
7/30/2019 Digital and Analog Transmission
10/36
19Dr. L. Christofi Spring 2009
Summary of line coding schemes
20Dr. L. Christofi Spring 2009
4.2 ANALOG4.2 ANALOG --TOTO --DIGITAL CONVERSIONDIGITAL CONVERSION
As we have seen, line coding can be used to convert binaryAs we have seen, line coding can be used to convert binarydata to a digital signal. Sometimes however, our data isdata to a digital signal. Sometimes however, our data isanalog, such as audio. If we want to store the audio byanalog, such as audio. If we want to store the audio byrecording it in a computer so that we can send it digitally,recording it in a computer so that we can send it digitally,we need to change it through a process called sampling.we need to change it through a process called sampling.
Pulse Amplitude ModulationPulse Code Modulation
Topics discussed in this section: Topics discussed in this section:
-
7/30/2019 Digital and Analog Transmission
11/36
21Dr. L. Christofi Spring 2009
Pulse Amplitude Modulation (PAM)
PAM is an analog-to-ditigal conversion method. It takes an analog signal, samples it and generates a series of pulsesbased on the results of sampling. Sampling means measuring the amplitude of the signal at equalintervals However PAM is not useful to data communications because eventhough it translates the original waveform to a series of pulses, thesepulses are still of any amplitude (still an analog, not a digital signal). Tomake them digital we modify them using Pulse Code Modulation (PCM ).
22Dr. L. Christofi Spring 2009
Components of PCM encoder
-
7/30/2019 Digital and Analog Transmission
12/36
23Dr. L. Christofi Spring 2009
Three different sampling methods for PCM
24Dr. L. Christofi Spring 2009
According to the Nyquist theorem, the sampling ratemust be
at least 2 times the highest frequency contained in thesignal.
Note
-
7/30/2019 Digital and Analog Transmission
13/36
25Dr. L. Christofi Spring 2009
Nyquist sampling rate for low-pass andbandpass signals
26Dr. L. Christofi Spring 2009
For an intuitive example of the Nyquist theorem, let ussample a simple sine wave at three sampling rates: f s = 4f(2 times the Nyquist rate), f s = 2f (Nyquist rate), andfs = f (one-half the Nyquist rate). Figure in the next slideshows the sampling and the subsequent recovery of thesignal.
It can be seen that sampling at the Nyquist rate can create
a good approximation of the original sine wave (part a).Oversampling in part b can also create the sameapproximation, but it is redundant and unnecessary.Sampling below the Nyquist rate (part c) does not producea signal that looks like the original sine wave.
Example
-
7/30/2019 Digital and Analog Transmission
14/36
27Dr. L. Christofi Spring 2009
Recovery of a sampled sine wave for different sampling rates
28Dr. L. Christofi Spring 2009
Telephone companies use PCM to digitize voice byassuming a maximum frequency of 4000 Hz. The samplingrate therefore is 8000 samples per second.
Example
-
7/30/2019 Digital and Analog Transmission
15/36
29Dr. L. Christofi Spring 2009
A complex low-pass signal has a bandwidth of 200 kHz. Whatis the minimum sampling rate for this signal?
SolutionThe bandwidth of a low-pass signal is between 0 and f, where fis the maximum frequency in the signal. Therefore, we cansample this signal at 2 times the highest frequency (200 kHz).The sampling rate is therefore 400,000 samples per second.
Example
30Dr. L. Christofi Spring 2009
A complex bandpass signal has a bandwidth of 200 kHz. Whatis the minimum sampling rate for this signal?
SolutionWe cannot find the minimum sampling rate in this casebecause we do not know where the bandwidth starts or ends.
We do not know the maximum frequency in the signal.
Example
-
7/30/2019 Digital and Analog Transmission
16/36
31Dr. L. Christofi Spring 2009
Quantization and encoding of a sampledsignal
32Dr. L. Christofi Spring 2009
We want to digitize the human voice. What is the bit rate,assuming 8 bits per sample?
SolutionThe human voice normally contains frequencies from 0 to4000 Hz. So the sampling rate and bit rate are calculated asfollows:
Example
-
7/30/2019 Digital and Analog Transmission
17/36
33Dr. L. Christofi Spring 2009
Components of a PCM decoder
34Dr. L. Christofi Spring 2009
We have a low-pass analog signal of 4 kHz. If we send theanalog signal, we need a channel with a minimum bandwidthof 4 kHz. If we digitize the signal and send 8 bits per sample,we need a channel with a minimum bandwidth of 8 4 kHz =32 kHz.
Example
-
7/30/2019 Digital and Analog Transmission
18/36
35Dr. L. Christofi Spring 2009
4.3 TRANSMISSION MODES4.3 TRANSMISSION MODES
The transmission of binary data across a link can beThe transmission of binary data across a link can beaccomplished in either parallel or serial mode. In parallelaccomplished in either parallel or serial mode. In parallelmode, multiple bits are sent with each clock tick. In serialmode, multiple bits are sent with each clock tick. In serialmode, 1 bit is sent with each clock tick. While there is onlymode, 1 bit is sent with each clock tick. While there is onlyone way to send parallel data, there are two subclasses ofone way to send parallel data, there are two subclasses ofserial transmission: asynchronous and synchronous.serial transmission: asynchronous and synchronous.
Parallel TransmissionSerial Transmission
Topics discussed in this section: Topics discussed in this section:
36Dr. L. Christofi Spring 2009
Data transmission modes
Asynchronous Synchronous
-
7/30/2019 Digital and Analog Transmission
19/36
37Dr. L. Christofi Spring 2009
Parallel transmission
38Dr. L. Christofi Spring 2009
Serial transmission
-
7/30/2019 Digital and Analog Transmission
20/36
39Dr. L. Christofi Spring 2009
In asynchronous transmission, we send 1 start bit (0)at the beginning and 1 or more stop bits (1s) at the
end of each byte. There may be a gap betweeneach byte.
Note
Asynchronous here means asynchronous at the byte
level,but the bits are still synchronized;their durations are the same.
Note
Asynchronous transmission
40Dr. L. Christofi Spring 2009
Asynchronous transmission
-
7/30/2019 Digital and Analog Transmission
21/36
41Dr. L. Christofi Spring 2009
In synchronous transmission, we send bits one afteranother without start or stop bits or gaps. It is the
responsibility of the receiver to group the bits.
Note
Synchronous transmission
42Dr. L. Christofi Spring 2009
Synchronous transmission
-
7/30/2019 Digital and Analog Transmission
22/36
43Dr. L. Christofi Spring 2009
4.4 SUMMARY (part 4) Line coding is the process of converting binary data to a digital signal. The number of different values allowed in a signal is the signal level. The number of symbols
that represent data is the data level. Bit rate is a function of the pulse rate and data level. Line coding methods must eliminate the dc component and provide a means of synchronization
between the sender and the receiver. Line coding methods can be classified as unipolar, polar, or bipolar. NRZ, RZ, Manchester, and differential Manchester encoding are the most popular polar encoding
methods. AMI is a popular bipolar encoding method. Analog-to-digital conversion relies on PCM (pulse code modulation). PCM involves sampling, quantizing, and line coding. The Nyquist theorem says that the sampling rate must be at least twice the highest-frequency
component in the original signal. Digital transmission can be either parallel or serial in mode. In parallel transmission, a group of bits is sent simultaneously, with each bit on a separate line. In serial transmission, there is only one line and the bits are sent sequenti ally. Serial transmission can be either synchronous or asynchronous. In asynchronous serial transmission, each byte (group of 8 bits) is framed with a start bit and a
stop bit. There may be a variable-length gap between each byte. In synchronous serial transmission, bits are sent in a continuous stream without start and stop
bits and without gaps between bytes. Regrouping the bits into meaningful bytes is theresponsibility of the receiver.
44Dr. L. Christofi Spring 2009
5.1 DIGITAL5.1 DIGITAL --TOTO --ANALOG CONVERSIONANALOG CONVERSION
DigitalDigital --toto --analoganalog conversion is the process of changingconversion is the process of changingone of the characteristics of an analog signal based onone of the characteristics of an analog signal based onthe information in digital data.the information in digital data.
Aspects of Digital-to-Analog Conversion
Amplitude Shift KeyingFrequency Shift KeyingPhase Shift KeyingQuadrature Amplitude Modulation
Topics discussed in this section: Topics discussed in this section:
-
7/30/2019 Digital and Analog Transmission
23/36
45Dr. L. Christofi Spring 2009
Digital-to-analog conversion
46Dr. L. Christofi Spring 2009
Types of digital-to-analog conversion
-
7/30/2019 Digital and Analog Transmission
24/36
47Dr. L. Christofi Spring 2009
Bit rate is the number of bits per second.
Baud rate is the number of signal elements persecond.
In the analog transmission of digital data, the
Baud Rate = Bit Rate / Number of bits per signal unitBaud Rate = Bit Rate / Number of bits per signal unit
Note
spects of digital- o-analog conversion
48Dr. L. Christofi Spring 2009
An analog signal carries 4 bits per signal element. If 1000signal elements are sent per second, find the bit rate.
SolutionIn this case, r = 4, S = 1000, and N is unknown. We can findthe value of N from
Example
-
7/30/2019 Digital and Analog Transmission
25/36
49Dr. L. Christofi Spring 2009
Example
An analog signal has a bit rate of 8000 bps and a baud rateof 1000 baud. How many data elements are carried byeach signal element? How many signal elements do weneed?
SolutionIn this example, S = 1000, N = 8000, and r and L are unknown.We find first the value of r and then the value of L.
50Dr. L. Christofi Spring 2009
Amplitude Shift Keying (ASK) In ASK the amplitude of the carrier is varied to represent binary 1 or 0.Frequency and phase remain constant.
ASK is highly susceptible to noise interference
Minimum bandwidth required for transmission is equal to the baudrate. BW = (1+d) S baud , where d=modulation factor
-
7/30/2019 Digital and Analog Transmission
26/36
51Dr. L. Christofi Spring 2009
Implementation of ASK
52Dr. L. Christofi Spring 2009
Example
We have an available bandwidth of 100 kHz which spansfrom 200 to 300 kHz. What are the carrier frequency andthe bit rate if we modulated our data by using ASK withd = 1?
SolutionThe middle of the bandwidth is located at 250 kHz. This meansthat our carrier frequency can be at f c = 250 kHz. We can use
the formula for bandwidth to find the bit rate (with d = 1 and r =1).
-
7/30/2019 Digital and Analog Transmission
27/36
53Dr. L. Christofi Spring 2009
Example
In data communications, we normally use full-duplex linkswith communication in both directions. We need to dividethe bandwidth into two with two carrier frequencies, asshown in figure below. The figure shows the positions of twocarrier frequencies and the bandwidths. The availablebandwidth for each direction is now 50 kHz, which leaves uswith a data rate of 25 kbps in each direction.
54Dr. L. Christofi Spring 2009
Frequency Shift Keying (FSK) In FSK the frequency of the carrier is varied to represent binary 1 or 0.Peak amplitude and phase remain constant.
FSK avoids most of problems with noise.
The bandwidth required for FSK transmission is equal to the baud rate plusthe frequency difference between the two carriers: BW = (f c2 -fc1) + S baud
-
7/30/2019 Digital and Analog Transmission
28/36
55Dr. L. Christofi Spring 2009
Example
We have an available bandwidth of 100 kHz which spans from 200 to300 kHz. What should be the carrier frequency and the bit rate if wemodulated our data by using FSK with d = 1?
SolutionThis problem is similar to Example 5.3, but we are modulating by usingFSK. The midpoint of the band is at 250 kHz. We choose 2 f to be 50 kHz;this means
56Dr. L. Christofi Spring 2009
Phase Shift Keying (PSK) In PSK the phase of the carrier is varied to represent binary 1 or 0. Peakamplitude and frequency remain constant.
This method is often called 2-PSK or binary PSK
The minimum bandwidth required for PSK transmission is the same as thatfor ASK: BW = (1+d) S baud , where d=modulation factor
-
7/30/2019 Digital and Analog Transmission
29/36
57Dr. L. Christofi Spring 2009
Implementation of binary PSK
58Dr. L. Christofi Spring 2009
QPSK (4-PSK) and its implementation
-
7/30/2019 Digital and Analog Transmission
30/36
59Dr. L. Christofi Spring 2009
Example
Find the bandwidth for a signal transmitting at 12 Mbps forQPSK. The value of d = 0.
SolutionFor QPSK, 2 bits is carried by one signal element. This meansthat r = 2. So the signal rate (baud rate) is S = N (1/r) = 6Mbaud. With a value of d = 0, we have B = S = 6 MHz.
60Dr. L. Christofi Spring 2009
Concept of a constellation diagram
-
7/30/2019 Digital and Analog Transmission
31/36
61Dr. L. Christofi Spring 2009
Show the constellation diagrams for an ASK, BPSK, andQPSK signals.
Solution
Example
62Dr. L. Christofi Spring 2009
Quadrature Amplitude Modulation is a combination ofASK and PSK.
Note
QAM
-
7/30/2019 Digital and Analog Transmission
32/36
63Dr. L. Christofi Spring 2009
Constellation diagrams for some QAMs
64Dr. L. Christofi Spring 2009
5.2 ANALOG AND DIGITAL5.2 ANALOG AND DIGITAL
AnalogAnalog --toto --analog conversion is the representation ofanalog conversion is the representation ofanalog information by an analog signal. One may ask whyanalog information by an analog signal. One may ask whywe need to modulate an analog signal; it is already analog.we need to modulate an analog signal; it is already analog.Modulation is needed if the medium isModulation is needed if the medium is bandpassbandpass in naturein natureor if only aor if only a bandpassbandpass channel is available to us.channel is available to us.
Amplitude ModulationFrequency ModulationPhase Modulation
Topics discussed in this section: Topics discussed in this section:
-
7/30/2019 Digital and Analog Transmission
33/36
65Dr. L. Christofi Spring 2009
Types of analog-to-analog modulation
66Dr. L. Christofi Spring 2009
Amplitude Modulation (AM)
In AM radio, the bandwidth of the modulated signal must betwice the bandwidth of the modulating signal.
-
7/30/2019 Digital and Analog Transmission
34/36
67Dr. L. Christofi Spring 2009
AM band allocation
68Dr. L. Christofi Spring 2009
Frequency Modulation (FM)
In FM radio, the bandwidth of the modulated signal must be10 times the bandwidth of the modulating signal.
-
7/30/2019 Digital and Analog Transmission
35/36
69Dr. L. Christofi Spring 2009
FM band allocation
70Dr. L. Christofi Spring 2009
Phase Modulation (PM)
-
7/30/2019 Digital and Analog Transmission
36/36
71Dr. L. Christofi Spring 2009
5.3 SUMMARY (part 5) Digital-to-analog modulation can be accomplished using the following:
*Amplitude shift keying (ASK)the amplitude of the carrier signal varies.*Frequency shift keying (FSK)the frequency of the carrier signal varies.*Phase shift keying (PSK)the phase of t he carrier signal varies.*Quadrature amplitude modulation (QAM)both the phase and amplitude of the carrier signal
vary. QAM enables a higher data transmission rate than other digital-to-analog methods. Baud rate and bit rate are not synonymous. Bit rate is the number of bits transmit-ted per second.
Baud rate is the number of signal units transmitted per second. One signal unit can represent oneor more bits.
The minimum required bandwidth for ASK and PSK is the baud rate. The minimum required bandwidth (BW) for FSK modulation is BW =f c1-f c0 + N baud , where f c1 is
the frequency representing a 1 bit, f c0 is the frequency representing a 0 bit, and N baud is the baudrate.
ASK modulation is especially susceptible to noise. Because it uses two carrier frequencies, FSK modulation requires more bandwidth than ASK and
PSK. PSK and QAM modulation have two advantages over ASK:
*They are not as susceptible to noise.*Each signal change can represent more than one bit.
Analog-to-analog modulation can be implemented by using the following:* Amplitude modulation (AM)
* Frequency modulation (FM)* Phase modulation (PM) In AM radio, the bandwidth of the modulated signal must be twice the bandwidth of the modulating
signal. In FM radio, the bandwidth of the modulated signal must be 10 times the bandwidth of the
modulating signal.
72Dr. L. Christofi Spring 2009
References
W. Stalling, Local and Metropolitan Area Networks ,6 th edition, Prentice Hall, 2000
F. Halsall, Data Communications, Computer Networks andOpen Systems , 4 th edition, Addison Wesley, 1995
B.A. Forouzan, Data Communications and Networking ,4th edition, McGraw-Hill, 2007
W. Stallings, Data and Computer Communications ,7 th edition, Prentice Hall, 2004