b5 Radio Digital Modulation -comm theorem
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Transcript of b5 Radio Digital Modulation -comm theorem
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SZE 3533SZE 3533
Topic V Radio Digital ModulationTopic V Radio Digital Modulation
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5.0 Digital Transmission5.0 Digital Transmission
Digital signal can also be transmitted through free space if theanalog carrier signal is used.
There are three techniques that can be used:
(ASK Amplitude-Shift Keying)
(FSK) Frequency-Shift Keying) (PSK Phase-Shift Keying)
Pemodulatan Digi
Pulse signals from the previous modulation or coding method mormallyare not transmitted in their original form (baseband signal). They will
modulate a carriers that are suitable with the channels used.
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5.1 Radio Digital Modulation
This modulation technique is similar to the analog modulation in which
the modulating signal will vary the ampltud, frequency or phase of the
carrier signal.
The only difference is that the modulating signal is digital signal :
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Analog ADCLine
coding
Digital
transmission
Sampling
QuantizationCoding
RZ, NRZ, AMI ASK, FSK,PSK
Block diagram for digital transmission system
Digital data is used to
modulate the carrier.The task of the carrier is to
shift the baseband signal
spectrum (digital data) to a
higher spectrum (around
the carrier signal).
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5.2 Amplitude Shift Keying (ASK)5.2 Amplitude Shift Keying (ASK)
Pemodulatan Digi
ASK
1 0 1 1 0
X
=
Unipolar
m m3 m5
Amplitud, V
w(rad/s)
c
Amplitud, V
w(rad/s)
mc 5
mc 5+
c
Amplitud, V
w(rad/s)
-Digital signal is
used to switch thecarrier ampiltude
(low and high).
-Also called on-off
keying (OOK) andinterrupted
continous wave
(ICW).
The string of pulses
from digital signal
will change the
amplitude of the
carrier signal. 5
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5.2.1 ASK signal generation5.2.1 ASK signal generation
Mathematical Analysis :
Vm(t)
Vc(t)
VASK (t)
ASK signal generation
+++= ...)5cos
5
13cos
3
1cos
2
2
1)( ttttv mmmm
mcccc wheretEtv >>= cos)(
+++= ....5cos5
13cos
3
1cos
2
2
1cos)( ttttEtv mmmccASK
Therefore :
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If only the 1st 5 harmonics are considered ;
BWASK
= 5 x fb
=> Bit rate,; fb
= 2 fm
Spectrum of ASK signal
Time domain
0 1 1 0 0 1 1 0 0 1 1 0
For unperiodic signal : BWASK
= fb
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vm(t)
vc(t)
vASK
(t)
ASK generation/trasmission :
Using multiplier
vm(t)
vc(t)
vASK
(t)
Using ON and OFF switch
ASK receiver :
vASK
(t)vm(t)
Using envelope detector Using coherent detector
vASK
(t)vm(t)
vc(t)
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Noise or error in digital communication system is measured using Bit Error
Rate (BER).
BER is measured based on the differences between the send and received
bits in period To .
Normally the BER will depends on other factors such the modulation
techniques and SNR (Eb/No) as shown in igure 5.19 (text book).
( )
( )sidedsingle:densityspectrumpowernoise
bitperenergy
oN
ESNR b=
o
be
N
EerfcP
42
1=
jalursatuhingarkuasaspektrumketumpatanadalah
bitsetiapbagitenagapurataadalah
oN
Eb
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5.3 Pemodulatan Anjakan Frekuensi (FSK)5.3 Pemodulatan Anjakan Frekuensi (FSK)
Pemodulatan Digi
FSK
1 0 1
X
=
0 1 0
X
=
+ASK
1
ASK2
Unipolar
m m3 m5
Amplitud, V
w(rad/s)
Amplitud, V
w(rad/s)
mc 5
2+2c
Amplitud, V
w(rad/s)
2c
1c
mc 5
1
1c
The generation of an
FSK waveform at the
Transmitter can be
achieved by generatingtwo ASK waveform and
adding them together
with a summing
amplifier.
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5.3.1 Penjanaan FSK5.3.1 Penjanaan FSK
)()()( 21 tvtvtvFSK +=
v1(t)
v2(t)
Mathematical Analysis :
( ) ( ) ( ) ( ) ( )1dan 12211 tvtvtvtvtv mmm ===Where :
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1jalurdasaruntuk....5cos5
13cos
3
1cos
2
2
1)(1
+++= ttttv mmmm
Taking Fourier series :
++= ....5cos5
13cos
3
1cos
2
2
1)(2 ttttv mmmm
( ) ( )1 12 tvtv mm =And ;
Therefore :
mcccc tEtv >>= ;cos)(And we have ;
Therefore :
+++= ...5cos
5
13cos
3
1cos
2
2
1cos)( 1 ttttEtv mmmccFSK
+++ ...5cos5
13cos
3
1cos
2
2
1cos 2 ttttE mmmcc
)()()(21
tvtvtvmmFSK
+=
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Spectrum for FSK signal
fff
fffBW
b
m
mmFSK
+==
++=
236
323
BW for FSK signal if up to 3rd harmonics are considered is
given by:
For unperiodic signal :
ffBW bFSK += 2
Where: fb
= 2 fm
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Generation of FSK signal :
Using switch
fc1
fc2
fm
vFSK
Using multiplier
The swichting actions will
produce 2 different
frequencies accordingly
LPF VCOfm vFSK
Using VCO
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Example 5.1 :
An unsynchronous FSK system is modulated by a digital signal and
operates at 10 MHz. The frequency deviation is 850 Hz and bit rate is 110
bit/s. The peak to peak carrier signal is 2 V and noise power spectrumdensity (2 bands) is 1 x 10-4 volts2/Hz. Calculate the BER for the system.
Solution :
voltsE
Epp
c 0.12
0.2
2===
sf
Tb
31009.9110
11 ===
( ) ( )
Joule
TEE cb
3
3
1055.4
2
1009.91
2
=
==
HzvoltsNo /1012
24=
HzvoltsNo /10224=
( ) 38.1110221055.4
2
1
2
1 43
===
eeBERPe
61074.5 =16
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Synchronous FSK receiver (coherent detector)
cos(c+ )t
cos( c- )t
vPSK
(t)LPF vm(t)-
+
The output fromthe amplifier is : ( ) ( )[ ]ttt
EEtv cc
ccD ++= 2cos2cos2cos
22
( ) ( )tEtv ccFSK += cosIf the FSK signalreceived is:
( ) ( )tEtv ccFSK = cos
( ) ( )[ ]tttEE
tv cccc
D ++
= 2cos2cos2cos22
LPF will allow eitherEc/2or
Ec/2which will represent the
output digital signal.
=o
be
NEerfcP 6.0
21
Probability of error (BER)
for synchronous detector :
Eb
, energy per bit
/ 2noise powerspectrum density :double
sided
erfc, error function
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Example
5.2 :FSK signal in Example 5.1 is fed to synchronous receiver. Calculate the new
BER and compare with that of BER which uses unsynchronous receiver.
Solution :
43 1021055.4 == ob NdanE
( ) ( )69.32
1
102
1055.46.0
2
14
3
erfcerfcBERPe =
==
From erfc table (3.69) ~ 1.9 x 10-7
87
105.92
109.1
=
=eP
(From Example 5.1)
6107.5 =eP
For unsyhcronous receiver :
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5.4 Phase Shift Keying (PSK)5.4 Phase Shift Keying (PSK)
Pemodulatan Digi
PSK
1 0 1 1 0
X
=
Bipolar
m
m3
m5
Amplitud, V
w(rad/s)c
Amplitud, V
w(rad/s)
mc 5
mc 5+
c
Amplitud, V
w(rad/s)
The phase of the carrier is
set to 0o
or 180o
depending on the digital
signal.
( ) ( )[ ]ttEtv ccPSK += cos
FSK signal can be
represented by :
(t) = 0 => 1
(t) = 180o => 0 , so ;
( ) 1cos
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BWPSK
= fb
= 2fm
PSK signal PSK spectrum
BWPSK
= BWBPSK
= fb
= 2fm The same as BWASK;
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vPSK
(t)
5.4.1 PSK generation5.4.1 PSK generation
Uses inverter to convert binary 1 to -1signal.
Both signals will be fed to one switch to
produce PSK signal.
vPSK
(t)
fc
vm(t)
-1
Using switch
LPF
vm(t)
fc
Using multiplier
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5.4.2 PSK receiver5.4.2 PSK receiverFKS demodulator must use coherent detector.
Advantages of PSK are :
Immune to noise
The same BW with ASK
Multilevel
Using multiplier
+1 atau -1
vPSK
(t) vm(t)
fc
The probability of error in the receiver
of PSK :
=
o
be
N
EerfcP
2
1
2
TEE cb =
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Example 5.3 :
Calculate the probability of error (BER) for Example 5.1 using PSK and
compare the BER with the synchronous and unsynchronous FSK.
Solution :
43 1021055.4 == ob NdanE From Example 5.1
( )77.421
1021055.4
21
4
3
erfcerfcPe =
=
From the table :
erfc (4.77) ~ 1.55 x 10-11
1211
1075.72
1055.1
=
=eP
8108 =eP6107.5 =eP
Comparison :
Synchronous
Unsynchronous
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5.4.3 Summary of Digital Modulation5.4.3 Summary of Digital Modulation
)(1
)(0
2
1
tkosA
tkosA
cc
cc
)(1
00
tkosA cc
Pemodulatan Digi
)(1
)(0
+
tkosA
tkosA
cc
cc
1 0 1 1 0
ASK
FSK
PSK
=
=
=
[
+
+=
...55
13
3
1
1
2
1)(
tkostkos
tkostm
mm
m
=
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Dibit Phase11 45
01 135
00 225
10 315
5.5 Quadrature Phase Shift Keying (QPSK)
One symbol represens 2 bits.
For example, for 4 different symbols they can be represented by
combination of 2 bits (00, 01, 10 & 11) and the phase will vary from 00 to
3600.
For binary PSK the phase changes from 00 dan 1800.
QPSK changes from ( /4) 450, (3 /4) 1350, (5 /4) 2250 to (7 /4) 3150. QPSK is better than PSK because of the efficeicy in the frequency
spectrum, = 2 bps/Hz. The code used in QPSK is Grey code .
Mapping of QPSK signal
using Grey code
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Mathematically they can be written as follow :
( ) ( )4/cos11 += tEtv cc
( ) ( )4/3cos01 += tEtv cc
( ) ( )4/5cos00 += tEtv cc
( ) ( )4/7cos10
+= tEtvcc
Mapping is done to separate the
input bits into 2 components, Iand Q.
I => Inphase
Q => Quadrature
QPSK signal is given
as :S
QPSK(t) = A cos (
ct + [2m 1]/4)
Where m = 1 , 2 , 3 , 4 dan 0 t Ts= 2T
b
The carrier
phase varies forevery 2T
b
11 01 00 10
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x(t)
t
ao a1 a2 a3 a4 a5
event
ao a2 a4 a1(t)
todd
a1 a3 a5a2(t)
I => Inphase
Q => Quadrature
2Tb
QPSK is actually 2 BPSK system
which has a 90o phase shift
between them.
i) Channel I is BPSKIwith
the phase of 0o and 180o
ii) Channel Q is BPSKQ withthe phase of 90o and
270o
180o 0o
I
R
90o
270o
R
I
+
Channel I Channel Q
(-1) (1)
(1)
(-1)
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135o 45o
315o225o
(-1,1) 10
(-1,-1) 00 01 (1,-1)
11 (1,1)CONSTELLATION
DIAGRAM
BPSKI
+ BPSKQ
= QPSK
QPSK spectrum is half of that of
BPSK with the same bit rate.
sin ct - /4
sin ct
sin ct + /4cos ct
28
-sin ct
-cos ct
sin ct +3 /4
sin ct -3 /4
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QPSK TransmitterQPSK Transmitter QPSK ReceiverQPSK Receiver
BPF is used to reduce the unwanted signals (noise, etc). The output from
BPF => I and Q signals. Both signals will be demodulated with oscillator
ofcos ct and sin
ct signals.
LPF will filter out the high frequency signals after demodulation process.
Output from the comparator is logic 1 if the sample value is positive and
logic 0 if negative.
Binary signal will be produced by the parallel to serial converter.
QPSK Receiver Operation :
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=
o
be
N
EerfcP
2
1
Probability of erroror BER for QPSK :
bo
b
f
BW
N
C
N
E=
BER is considered as the ratio of carrier to noise power (C/N) at the
receiver input
The relationship between / and C/N is given by the followingequation :
C/Nis the ration of carrier and
noise power
BWis the bandwidth for noise atthe receiver
fbis the bit rate
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Example 5.4 :
Compare the ratio of carrier to noise power (C/N) needed to send data at 120
Mb/s using BPSK and QPSK if BER required is 10-7 .
Solution :
BPSK :
=
o
be
N
EerfcP
2
1
=
o
b
N
Eerfc2
1107
5.13
o
b
N
EFrom the table:
spectrum BPSK : = 1bps/Hz
BW = fb
/ = 120 / 1 = 120 MHz
bo
b
f
BW
N
CE=
So :
=
o
bb E
BW
f
N
C
[ ] dBN
C3.115.135.13
120
120==
=
From :
For QPSK : 5.13o
b
N
E(= BPSK)
spectrum QPSK : = 2bps/HzBW = f
b/ = 120 / 2 = 60 MHz
[ ] dBN
C3.140.275.13
60
120==
=
QPSK system > 3 dB than BPSK to have
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5.6 Phase Shift Keying M-ARY
Phase shift keying M-ary refers to the symbol used for modulation system.
M-ary system used includes 8 PSK, 16 PSK, 32 PSK, 64 PSK and so on.
In general every symbol can be represented by several bits :
M= 2n M = symbol or leveln = number of bits
For 16 levels system, every level or symbol can be represented by 4 bits
as follows :
0000, 0001, 0010 ..............1111
The larger the no. of level => more complex circuit & higherC/N.
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Table 5.2 is the summary of BER with 10-7 for M-ary system. In the M-ary
system bit rate is normally written as symbols/s orbaud rate.
Modulation
techniques
Spectrum
efficiency,C/N (dB) (P
e= 10-7)
BPSK 1 b/s/Hz 11.5
QPSK 2 b/s/Hz 14.5
8PSK 3 b/s/Hz 19.5
16PSK 4 b/s/Hz 25.5
32PSK 5 b/s/Hz 32.5
The differences for different modulation techniques, spectrum efficiency, &C/N
The increases in energy per bit, Ebwill reduce the BER and
increases the performance of the system.
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