Transcript Lektion 1-Introduktion

Datornätverk A – lektion 4
MKS B – lektion 4
•Kapitel 5: Modulation.
4.3 Sampling
Pulse Amplitude Modulation
Pulse Code Modulation
Sampling Rate: Nyquist Theorem
How Many Bits per Sample?
Bit Rate
PCM = Pulse Code Modulation = Digitalisering
av analoga signaler och seriell överföring
Sifferexempel från PSTN = publika telefonnätet:
011011010001...
1
0
Antivikningsfilter
Sampler
AD-omvandlare med
seriell utsignal
DAomvandlare
Interpolationsfilter
Högtalare
Mikrofon
34004000Hz
filter
8000
sampels
per sek
8 bit per sampel
dvs 64000 bps
per tfnsamtal
28 = 256
spänningsnivåer
Exempel
En 6 sekunder lång ljudinspelning digitaliseras. Hur stor är
inspelningens informationsmängd?
a) 22000 sampels/sekund, 256 kvantiseringsnivåer.
22000sampels * 6 s * 8 bit = 1056000bit.
b) 22000 sampels/sekund, 16 kvantiseringsnivåer.
22000sampels * 6 s * 4 bit = 528000bit.
c) 5500 sampels/sekund, 256 kvantiseringsnivåer.
5500sampels * 6 s * 8 bit = 264000bit.
Samplingsteoremet
•
•
•
•
•
•
f < fs/2
Den högsta frekvens som kan samplas är halva samplingsfrekvensen.
Om man samplar högre frekvens än fs/2 så byter signalen frekvens, dvs det
uppstår vikningsdistorsion (aliasing).
För att undvika vikningsdistorsion så har man ett anti-vikningsfilter innan
samplingen, som tar bort frekvenser över halva samplingsfrekvensen.
Interpolationsfiltret används vid rekonstruktion av den digitala signalen för att
”gissa” värden mellan samplen.
Ett ideal interpolationsfilter skulle kunna återskapa den samplade signalen
perfekt om den uppfyller samplingsteoremet. I verkligheten finns inga ideala
filter.
Följdregel: Nyqvist’s sats säger att max datahastighet = 2B2log M, där M är
antal nivåer, och B är signalens bandbredd, oftast lika med signalens övre
gränsfrekvens.
Figure 4.18
PAM
Note:
Pulse amplitude modulation has some
applications, but it is not used by itself
in data communication. However, it is
the first step in another very popular
conversion method called
pulse code modulation.
Figure 4.19
Quantized PAM signal
Figure 4.20
Quantizing by using sign and magnitude
Note:
According to the Nyquist theorem, the
sampling rate must be at least 2 times
the highest frequency.
Example 4
What sampling rate is needed for a signal with a
bandwidth of 10,000 Hz (1000 to 11,000 Hz)?
Solution
The sampling rate must be twice the highest frequency in
the signal:
Sampling rate = 2 x (11,000) = 22,000 samples/s
Example 5
A signal is sampled. Each sample requires at least 12
levels of precision (+0 to +5 and -0 to -5). How many bits
should be sent for each sample?
Solution
We need 4 bits; 1 bit for the sign and 3 bits for the value.
A 3-bit value can represent 23 = 8 levels (000 to 111),
which is more than what we need. A 2-bit value is not
enough since 22 = 4. A 4-bit value is too much because 24
= 16.
Example 6
We want to digitize the human voice. What is the bit rate,
assuming 8 bits per sample?
Solution
The human voice normally contains frequencies from 0
to 4000 Hz.
Sampling rate = 4000 x 2 = 8000 samples/s
Bit rate = sampling rate x number of bits per sample
= 8000 x 8 = 64,000 bps = 64 Kbps
Distorsion till följd av digitalisering
• Vikningsdistorsion
○ Inträffar om man inte filtrerar bort frekvenser som är högre än
halva samplingsfrekvensen.
• Kvantiseringsdistorsion (kvantiseringsbrus)
○ Avrundningsfelet låter ofta som ett brus.
○ Varje extra bit upplösning ger dubbelt så många spänningsnivåer,
vilket ger en minskning av kvantiseringsdistorsionen med 6 dB. 16
bit upplösning ger ett signal-brus-förhållande på ca 16*6 = 96 dB
(beroende på hur man mäter detta förhållande.)
○ Svaga ljud avrundas bort, eller dränks i kvantiseringsbruset.
Informationsmängd
• N bit kan representera 2N alternativa värden eller koder.
Ex: ASCII-kodens 7 bitar kan representera 27 = 2·2 ·2 ·2 ·2 ·2 ·2 = 128
tecken.
• En kod som kan anta M alternativa värden har
informationsmängden
log M
.
2 log M 
log 2
Ex: ISO-latinkodens 256 tecken kräver 2log 256 = 8 bit per tecken.
Figure 5.25
Types of analog-to-analog modulation
Figure 5.26
Amplitude modulation
Figure 5.29
Frequency modulation
Modulation och demodulation
• Baudrate = antal symboler per sekund. Enhet: baud eller
symboler/sekund.
• Bitrate = datahastighet. Enhet: bps eller bit/s.
• Vid många modulationsformer t.ex. s.k. ASK, PSK, och QAM
är signalens bandbredd = symbolhastigheten.
• Vid FSK är bandbredden vanligen större.
Digitala modulationsmetoder
Binär signal
ASK = Amplitude
Shift Keying (AM)
FSK = Frequency
Shift Keying (FM)
PSK = Phase Shift
Keying (PSK)
Exempel 1:
Till höger visas fyra symboler som används av ett s.k. 4PSKmodem (PSK=Phase Shift Keying). De fyra symbolerna
representerar bitföljderna 00, 01, 11 resp 10.
1
0
-1
1
a) Nedan visas utsignalen från det sändande modemet. Vilket
meddelande, dvs vilken bitsekvens, överförs?
0
0.5
1
time [milliseconds]
01
0
-1
Svar: 11 00 10 10.
b) Tidsaxeln är graderad i tusendels sekunder. Vad är
symbolhastigheten i baud eller symboler/sekund?
00
0
0.5
1
time [milliseconds]
11
1
0
Svar: 1/1ms = 1000 symber per sekund = 1kbaud.
c) Vad är bithastigheten i bit per sekund (bps)?
-1
Svar: 2000bps.
0
0.5
1
time [milliseconds]
1
1
0
10
0
-1
0
0.5
1
1.5
2
2.5
time [millisecond]
3
3.5
4
-1
0
0.5
1
time [milliseconds]
Exempel 2:
Nedan visas åtta symboler som används av ett s.k. 8QAM-modem
(QAM=Quadrature Amplitude Modulation). Symbolerna i övre raden
representerar bitföljderna 000, 001, 011 resp 010 (från vänster till höger).
Undre raden representerar 100, 101, 111 resp 110.
000
001
011
010
2
2
2
2
0
0
0
0
-2
0
0.005
100
-2
0.01 0
0.005
101
-2
0.01 0
0.005
111
-2
0.01 0
2
2
2
2
0
0
0
0
-2
0
0.005
-2
0.01 0
0.005
-2
0.01 0
0.005
-2
0.01 0
0.005
110
0.01
0.005
0.01
Forts exempel 2:
a) Nedan visas utsignalen från det sändande modemet. Vilket
meddelande, dvs vilken bitsekvens, överförs?
Modulatorns utsignal
Spänning [Volt]
2
0
-2
0
0.005
0.01
0.015 0.02 0.025
Tid [sekunder]
0.03
0.035
0.04
b) Tidsaxlarna är graderad i sekunder. Vad är symbolhastigheten i baud
eller symboler/sekund?
c) Vad är bithastigheten i bit per sekund (bps)?
Example 1
An analog signal carries 4 bits in each signal unit. If 1000
signal units are sent per second, find the baud rate and the
bit rate
Solution
Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bps
Figure 5.4
Relationship between baud rate and bandwidth in ASK
Figure 5.3
ASK
Example 3
Find the minimum bandwidth for an ASK signal
transmitting at 2000 bps. The transmission mode is halfduplex.
Solution
In ASK the baud rate and bit rate are the same. The baud
rate is therefore 2000. An ASK signal requires a
minimum bandwidth equal to its baud rate. Therefore,
the minimum bandwidth is 2000 Hz.
Example 4
Given a bandwidth of 5000 Hz for an ASK signal, what
are the baud rate and bit rate?
Solution
In ASK the baud rate is the same as the bandwidth,
which means the baud rate is 5000. But because the baud
rate and the bit rate are also the same for ASK, the bit
rate is 5000 bps.
Example 6
Find the minimum bandwidth for an FSK signal
transmitting at 2000 bps. Transmission is in half-duplex
mode, and the carriers are separated by 3000 Hz.
Solution
For FSK
BW = baud rate + fc1 - fc0
BW = bit rate + fc1 - fc0 = 2000 + 3000 = 5000 Hz
Example 7
Find the maximum bit rates for an FSK signal if the
bandwidth of the medium is 12,000 Hz and the difference
between the two carriers is 2000 Hz. Transmission is in
full-duplex mode.
Solution
Because the transmission is full duplex, only 6000 Hz is
allocated for each direction.
BW = baud rate + fc1 - fc0
Baud rate = BW - (fc1 - fc0 ) = 6000 - 2000 = 4000
But because the baud rate is the same as the bit rate, the
bit rate is 4000 bps.
Figure 5.8 PSK
Figure 5.5
Solution to Example 5
Figure 5.6
FSK
Figure 5.9
PSK constellation
Figure 5.10
The 4-PSK method
Figure 5.11 The 4-PSK characteristics
Figure 5.12
The 8-PSK characteristics
Figure 5.13
Relationship between baud rate and bandwidth in PSK
Example 9
Given a bandwidth of 5000 Hz for an 8-PSK signal, what
are the baud rate and bit rate?
Solution
For PSK the baud rate is the same as the bandwidth,
which means the baud rate is 5000. But in 8-PSK the bit
rate is 3 times the baud rate, so the bit rate is 15,000 bps.
Note:
Quadrature amplitude modulation is a
combination of ASK and PSK so that a
maximum contrast between each
signal unit (bit, dibit, tribit, and so on)
is achieved.
Figure 5.14
The 4-QAM and 8-QAM constellations
Figure 5.15
Time domain for an 8-QAM signal
Figure 5.16
16-QAM constellations
Figure 5.17
Bit and baud
Table 5.1 Bit and baud rate comparison
Modulation
Units
Bits/Baud
Baud rate
Bit Rate
Bit
1
N
N
4-PSK, 4-QAM
Dibit
2
N
2N
8-PSK, 8-QAM
Tribit
3
N
3N
16-QAM
Quadbit
4
N
4N
32-QAM
Pentabit
5
N
5N
64-QAM
Hexabit
6
N
6N
128-QAM
Septabit
7
N
7N
256-QAM
Octabit
8
N
8N
ASK, FSK, 2-PSK
Example 10
A constellation diagram consists of eight equally spaced
points on a circle. If the bit rate is 4800 bps, what is the
baud rate?
Solution
The constellation indicates 8-PSK with the points 45
degrees apart. Since 23 = 8, 3 bits are transmitted with
each signal unit. Therefore, the baud rate is
4800 / 3 = 1600 baud
Example 11
Compute the bit rate for a 1000-baud 16-QAM signal.
Solution
A 16-QAM signal has 4 bits per signal unit since
log216 = 4.
Thus,
1000·4 = 4000 bps
Example 12
Compute the baud rate for a 72,000-bps 64-QAM signal.
Solution
A 64-QAM signal has 6 bits per signal unit since
log2 64 = 6.
Thus,
72000 / 6 = 12,000 baud
5.2 Telephone Modems
Modem Standards
Note:
A telephone line has a bandwidth of
almost 2400 Hz for data transmission.
Figure 5.18
Telephone line bandwidth
Note:
Modem stands for
modulator/demodulator.
Figure 5.19
Modulation/demodulation
Figure 5.20
The V.32 constellation and bandwidth
Figure 5.21
The V.32bis constellation and bandwidth
Figure 5.22
Traditional modems
Figure 5.23
56K modems
5.3 Modulation of Analog Signals
Amplitude Modulation (AM)
Frequency Modulation (FM)
Phase Modulation (PM)
Figure 5.24
Analog-to-analog modulation
Note:
The total bandwidth required for AM
can be determined from the bandwidth
of the audio signal:
BWt = 2 x BWm.
Example 7
Consider a noiseless channel with a bandwidth of 3000
Hz transmitting a signal with two signal levels. The
maximum bit rate can be calculated as
Bit Rate = 2  3000  log2 2 = 6000 bps
Shannons regel
Kanalkapaciteten C är max antal bit per sekund vid bästa
möjliga modulationsteknik och felrättande kodning:
C = B log2 (1+S/N),
där B är ledningens bandbredd i Hertz (oftast ungefär lika med
övre gränsfrekvensen), S är nyttosignalens medeleffekt i Watt
och N (noice) är bruseffekten i Watt.
Example 8
Consider the same noiseless channel, transmitting a signal
with four signal levels (for each level, we send two bits).
The maximum bit rate can be calculated as:
Bit Rate = 2 x 3000 x log2 4 = 12,000 bps
Example 9
Consider an extremely noisy channel in which the value
of the signal-to-noise ratio is almost zero. In other words,
the noise is so strong that the signal is faint. For this
channel the capacity is calculated as
C = B log2 (1 + SNR) = B log2 (1 + 0)
= B log2 (1) = B  0 = 0
Example 10
We can calculate the theoretical highest bit rate of a
regular telephone line. A telephone line normally has a
bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signalto-noise ratio is usually 3162. For this channel the
capacity is calculated as
C = B log2 (1 + SNR) = 3000 log2 (1 + 3162)
= 3000 log2 (3163)
C = 3000  11.62 = 34,860 bps
Example 11
We have a channel with a 1 MHz bandwidth. The SNR
for this channel is 63; what is the appropriate bit rate and
signal level?
Solution
First, we use the Shannon formula to find our upper
limit.
C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps
Then we use the Nyquist formula to find the
number of signal levels.
4 Mbps = 2  1 MHz  log2 L  L = 4