Digital Communications Fundamentals and Applications in CTE Department 2018-2019

رهاط 
ةزمحلا 
.م : 
ةداملا
 
سردم
1
-
2
01
8-
20
1
9
Dr. 
Hussam 
Dheaa
 
Kamel
Al-Mustafa University 
Collage 
CTE Department
2018-2019
Digital
 
Communications
CTE Department 
-3
rd
 
stage
Reference: 
Digital
Communications
Fundamentals
 
and
Applications,
2
nd 
Addition, 
by
 
FernardSklar
2
Digital
 
Communications
 
CTE 
Department -3
rd
 
stage
2- Binary 
Amplitude 
Shift 
Keying (BASK) 
or 
ON- OFF keying
 
(OOK):
ASK 
or 
OOK 
is the 
simplest digital 
modulation 
technique. The ASK 
waveform can 
be represented
 
as:
𝑠
(
𝑡
) 
= 
2𝑃
𝑠 
cos
(
2𝜋𝑓
0
𝑡
) 
to 
transmit symbol 
"1", 
and 
pulse is transmitted. 
To 
transmit symbol 
"0" 
𝑠
(
𝑡
) 
= 0
, 
that is  
no
signal transmitted 
for 
such symbol. Thus ASK waveform looks like 
an 
ON-OFF 
of 
the signal 
as 
shown in figure  
10.
Figure
 
10
The ASK 
wave form for 
symbol "1" can 
be 
represented
 
as:
𝑠
(
𝑡
) 
= 
𝑃
𝑠
𝑇
𝑏
. 
√2/𝑇
𝑏
 
cos
(
2𝜋𝑓
0
𝑡
)
=
 
𝑃
𝑠
𝑇
𝑏
1
(𝑡)
Digital
 
Communications
CTE Department -3
rd
 
stage
3
Thus there is only one 
carrier 
function 
1
(𝑡)
. 
If 
we take 
1
(
𝑡
) 
= 
2/𝑇
𝑏 
cos
(
2𝜋𝑓
0
𝑡
) 
as 
the orthonormal
 
basis
function. The signal space diagram will have two points 
on 
1
(𝑡)
. 
One will 
be 
at 
zero and other 
will 
be 
at 
𝑃
𝑠
𝑇
𝑏
. 
As
shown 
in
 
Fig.11:
Figure
 
11
Therefore 
the 
distance between the 
two 
signal points will
 
be,
𝑑  = 
𝑃
𝑠
𝑇
𝑏   
=
 
𝐸
𝑏
3-1 
Generation 
of
 
BASK:
Fig. 
12 shows 
BASK 
generator. The input 
binary 
sequence is applied to the product modulator. The modulator 
passes
the carrier 
when the input bits is "1", and block 
the 
carrier (zero output) when input bit is
 
"0".
Product
Modul
a
tor
𝐶𝑎𝑟𝑟𝑖𝑒𝑟
2𝑃
𝑠
 
cos
(
2𝜋𝑓
0
𝑡
)
Binary
 
ASK
signal
 
s(t)
Binary
Input
 
s(t)
Digital
 
Communications
CTE Department -3
rd
 
stage
Figure
 
12
The bandwidth of ASK
 
is
𝐵𝑊  = 
𝑅(1 
+
 
𝑟)
Where 
R 
is 
the 
bit 
rate and 
0 < 𝑟 < 1 
is 
related to how signal is
 
filtered.
3-2 
ASK
 
Detector:
The ASK signal is applied to the correlator consisting 
of 
multiplier 
and 
integrator. The locally generated 
coherent 
carrier  
and
applied to multiplier 
as 
shown 
in 
Fig.
 
13.
Figure
 
13
4
Digital
 
Communications
 
CTE 
Department -3
rd
 
stage
The output 
of 
multiplier 
is 
integrated over the bit period. The decision device takes the decision 
at 
the end 
of 
every 
bit
period. 
It 
compare the output 
of 
integrator with the
 
threshold.
3- 
Differential Phase Shift Keying
 
(DPSK):
DPSK
 
is
 
differentially
 
coherent
 
modulation
 
method.
 
It
 
does
 
not
 
need
 
synchronous
 
(coherent)
 
carrier
 
at
 
the
demodulator. The input sequence 
of 
binary bits is modified such that the next bit depends upon 
the 
previous bit.
Therefore 
in 
the receiver the previous received bits 
are 
used 
to 
detect the present
 
bit.
4-1 
Generator of
 
DPSK:
The input sequence d(t), the output sequence is b(t) 
and 
𝑏(𝑡 
𝑇
𝑏
) 
is the previous output delayed
 
by one 
bit period.
Depending upon values of d(t) 
and 
𝑏(𝑡 
𝑇
𝑏
)
, exclusive OR gate generates the output sequence b(t) 
as 
shown in Fig.
14.
Figure 
14: DPSK
 
generator
5
Digital
 
Communications
 
CTE 
Department -3
rd
 
stage
The truth table 
of 
exclusive OR gate is used 
to 
derive the level of waveforms shown in Fig. 15, which satisfied in 
any
interval b(t) is given
 
as,
𝑏
(
𝑡
) 
= 
𝑑(𝑡) 
𝑏(𝑡 
 
𝑇
𝑏
)
Figure 
15: DPSK
 
waveform
Note that the output sequence b(t) changes level 
at 
the beginning 
of 
each interval in which 
d(t)=1 
and does not change  level
when 
d(t)=0, 
so that symbol duration (T)= duration 
of 
two bits 
2𝑇
𝑏
. The sequence b(t) 
is 
applied 
to 
the 
balanced  
modulator
as 
shown in Fig. 14 with carrier 
2𝑃cos(2𝜋𝑓
0
𝑡)
, 
the 
modulator output
 
is:
𝑠
(
𝑡
) 
= 
𝑏
(
𝑡
)
√2𝑃
 
cos
(
2𝜋𝑓
0
𝑡
)
= ±√2𝑃
 
cos
(
2𝜋𝑓
0
𝑡
)
As 
shown 
in in Fig.15 the phase changes only when
 
d(t)=1.
6
Digital
 
Communications
CTE Department -3
rd
 
stage
4-2 
DPSK
 
Receiver:
During the transmission, the DPSK signal undergoes 
some 
phase 
shift 
𝜃
. 
Therefore 
the 
signal received 
at 
the input of
receiver
 
is
Received signal 
𝑦
(
𝑡
) 
= 
𝑏
(
𝑡
)
2𝑃 
cos
(
2𝜋𝑓
0
𝑡 
+
 
𝜃
)
The signal is multiplied with delayed version 
by one 
bit:
Multiplier output
= 
𝑏
(
𝑡
)
𝑏
(
𝑡 
− 𝑇
𝑏
)
(2𝑝) 
cos
(
2𝜋𝑓
0
𝑡 
+ 
𝜃
) 
cos
(
2𝜋𝑓
0
(
𝑡 
𝑇
𝑏
) 
+
 𝜃
)
Figure 
16a: QPSK
 
receiver
This signal is fed to integrator and
to 
the decision device. The
integrator and decision is controlled
by bit 
synchronizer 
to get 
the
output binary 
sequence. 
The block
diagram 
of 
such system 
is shown in
Fig.16a, and the equivalent of
 
this
system 
is shown 
in
 
Fig.16b.
7
Digital
 
Communications
CTE Department -3
rd
 
stage
4-3 
Bandwidth 
of
 
DPSK:
2
 
1
We 
know 
that one previous 
bit 
is used to decide the phase shift 
of 
next bit. Change in 
b(t) 
occurs only 
if 
input bitt is 
at 
level
'1'. 
No change occurs 
if 
input bit is 
at 
level '0'. Since one previous 
bit 
is 
always 
used to 
define the phase in next bit, 
the
symbol 
can be 
said to 
have 
two bits. Therefore 
one 
symbol duration (T) 
is 
equivalent to 
two 
bits duration 
(2𝑇
𝑏
)
, 
so
 
that
Symbol duration 
𝑇 =
 
2𝑇
𝑏
Bandwidth is 
given 
as, 
𝐵𝑊 = 
𝑇 
=
 
𝑇
𝑏
Or
 
𝐵𝑊 =
 
𝑓
𝑏
4-4 
Advantages and disadvantages 
of
 
DPSK:
Advantages:
1- 
DPSK does not need 
carrier at 
its receiver. 
Hence 
it is simple than
 
BPSK.
8
Digital
 
Communications
 
CTE 
Department -3
rd
 
stage
2- 
The bandwidth required of DPSK 
is 
reduced compared to that 
of
 
BPSK.
Disadvantages:
1
The probability 
of 
bit 
error 
rate 
of 
DPSK 
is 
higher than
 
BPSK.
2
Error 
in the first bit creates error 
in the 
second bit. 
Hence 
error propagation in DPSK 
is 
more, while in PSK 
each 
bit
is
 
independent.
3
Noise interference in DPSK 
is
 more.
4
Quadrature Phase Shift Keying
 
(QPSK):
There
 
are
 
two
 
important
 
points
 
in
 
communication
 
system,
 
transmission
 
power
 
and
 
channel
 
bandwidth.
 
The
 
channel
bandwidth depends upon the bit rate or signaling rate 
𝑓
𝑏
. 
The carrier 
is 
used 
as 
bandpass transmission over 
a 
channel.
If 
two 
or 
more bits 
are 
combined 
in 
some 
symbol instead of one bit, then the signaling rate is reduced and the
transmission channel bandwidth 
is 
reduced. 
In 
quadrature shift keying, two successive bits 
in the 
data sequence are
grouped
 
together.
In BPSK 
the phase shift 
occurs in 
two 
level only, 
the carrier is 
change 
by 
180
0
. 
In 
QPSK the combination 
of 
two
 
bits
form four 
distinct symbols, so that the change 
of 
carrier phase is 
45
0 
(
𝜋 
𝑟𝑎𝑑𝑖𝑎𝑛𝑠) 
from one 
symbol to 
the 
next
 
one.
9
4
Table 1: 
Symbol 
and 
corresponding phase shift in
 
QPSK
Digital
 
Communications
CTE Department -3
rd
 
stage
10
1
QPSK
 generator:
There are 
two type 
of
 
QPSK:
1
Offset 
QPSK 
(OQPSK):
Fig. 17 
shows the block diagram 
of 
OQPSK generator. The input signal is converted 
to 
NRZ and 
called 
b(t), the
demultiplexer divides 
b(t) 
into separated bit streams 
of 
odd numbered and 
even
 
numbered.
Digital
 
Communications
CTE Department -3
rd
 
stage
Figure 
17: OQPSK
 
generator
The symbol duration 
of 
both 
𝑏
𝑜
(𝑡)
and 
𝑏
𝑒
(𝑡) 
are 
2𝑇
𝑏
 
as 
shown 
in 
Fig. 
18. 
Note 
that the first 
symbol 
of 
𝑏
𝑜
(𝑡) 
is 
delayed 
by
one 
bit 
period with respect to symbol of 
𝑏
𝑒
(𝑡)
. 
This delay 
of 
𝑇
𝑏
is called offset 
so 
that the change 
in 
level cannot 
occur 
in
 
the
same time. Each 
𝑏
𝑒
(𝑡) 
and 
𝑏
0
(𝑡) 
are 
modulate 
a 
carrier 
2𝑝
𝑠
sin(2𝜋𝑓
0
𝑡) 
and 
2𝑝
𝑠
cos(2𝜋𝑓
0
𝑡)
 
respectivlly.
11
Digital
 
Communications
CTE Department -3
rd
 
stage
12
Digital
 
Communications
CTE Department -3
rd
 
stage
𝑠
𝑒
(
𝑡
) 
= 
𝑏
𝑒
(𝑡)
 
2𝑝
𝑠
sin(2𝜋𝑓
0
𝑡)
𝑠
0
(
𝑡
) 
= 
𝑏
0
(𝑡)
 
2𝑝
𝑠
sin(2𝜋𝑓
0
𝑡)
Thus 
𝑠
𝑒
(
𝑡
)
and 
𝑠
0
(
𝑡
) 
are 
basically 
BPSK 
signals but 
𝑇 = 
2𝑇
𝑏
. The output 
of 
adder is OQPSK, given
 
as:
𝑠
(
𝑡
) 
= 
𝑠
𝑒
(
𝑡
) 
+
 
𝑠
0
(
𝑡
)
13
Digital
 
Communications
CTE Department -3
rd
 
stage
=
𝑏
𝑒
(
𝑡
)
2𝑝
𝑠 
sin
(
2𝜋𝑓
0
𝑡
) 
+ 
𝑏
0
(𝑡)
 
2𝑝
𝑠
cos(2𝜋𝑓
0
𝑡)
Fig.19 shows the phasor diagram 
of 
QPSK 
signal of 
above equation.
Figure 
19: Phasor diagram 
of 
QPSK
 
signal
14
Digital
 
Communications
CTE Department -3
rd
 
stage
Offset 
QPSK is essentially 
the 
same 
as QPSK 
except that 
the 
I- 
and Q-channel pulse trains are staggered. The modulator
and the demodulator 
of 
OQPSK 
are 
shown 
in 
Figure 
20, 
which differs from 
the 
QPSK 
only by an 
extra 
delay 
of T/2
seconds in the Q-channel. Its 
power 
spectral density 
is 
the 
same 
as 
that 
of 
QPSK, and 
its 
error performance 
is 
also the same
as 
that 
of
 
QPSK.
Fig. 21: OQPSK
 
modulator
15
Digital
 
Communications
CTE Department -3
rd
 
stage
Fig. 21: OQPSK
 
demodulator
16
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This content delves into the Digital Communications course in the CTE Department during the academic year 2018-2019. It covers topics such as Binary Amplitude Shift Keying (BASK), Generation and Detection of BASK, Differential Phase Shift Keying (DPSK), DPSK Generator, and the application of Digital Modulation Techniques in communication systems.

  • Digital Communications
  • CTE Department
  • Modulation Techniques
  • Binary Amplitude Shift Keying
  • Differential Phase Shift Keying

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  1. . : 1- 2018-2019 Digital Communications CTE Department -3rdstage Reference: Digital Communications Fundamentalsand Applications, 2nd Addition, byFernardSklar Dr. Hussam Dheaa Kamel Al-Mustafa University Collage CTE Department 2018-2019

  2. CTE Department -3rdstage Digital Communications 2- BinaryAmplitude Shift Keying (BASK) or ON- OFFkeying(OOK): ASK or OOK is the simplest digital modulation technique.TheASK waveform can be represented as: ?(?) = 2??cos(2??0?) to transmit symbol "1", and pulse is transmitted. To transmit symbol "0" ?(?) = 0, that is no signal transmitted for such symbol. ThusASK waveform looks like an ON-OFF of the signal as shown in figure 10. Figure10 The ASK wave form for symbol "1" can be represented as: ?(?) = ????. 2/??cos(2??0?) = ???? 1(?) 2

  3. CTE Department -3rdstage Digital Communications Thus there is only one carrier function 1(?). If we take 1(?) = 2/??cos(2??0?) as the orthonormal basis function. The signal space diagram will have two points on 1(?). One will be at zero and other will be at ????. As shown in Fig.11: Figure11 Therefore the distance between the two signal points will be, ? = ????= ?? 3-1 Generation of BASK: Fig. 12 shows BASK generator. The input binary sequence is applied to the product modulator. The modulator passes the carrier when the input bits is "1", and block the carrier (zero output) when input bit is "0". Product Modulator Binary Inputs(t) BinaryASK signal s(t) ??????? 3 2??cos(2??0?)

  4. CTE Department -3rdstage Digital Communications Figure12 The bandwidth of ASK is ?? = ?(1 +?) Where R is the bit rate and 0 < ? < 1 is related to how signal is filtered. 3-2 ASK Detector: The ASK signal is applied to the correlator consisting of multiplier and integrator. The locally generated coherent carrier and applied to multiplier as shown in Fig. 13. Figure13 4

  5. CTE Department -3rdstage Digital Communications The output of multiplier is integrated over the bit period. The decision device takes the decision at the end of every bit period. It compare the output of integrator with the threshold. 3- Differential Phase Shift Keying (DPSK): DPSK is differentially coherent modulation method. It does not need synchronous (coherent) carrier at the demodulator. The input sequence of binary bits is modified such that the next bit depends upon the previous bit. Therefore in the receiver the previous received bits are used to detect the present bit. 4-1 Generator of DPSK: The input sequence d(t), the output sequence is b(t) and ?(? ??) is the previous output delayedby one bit period. Depending upon values of d(t) and ?(? ??), exclusive OR gate generates the output sequence b(t) as shown in Fig. 14. Figure 14: DPSKgenerator 5

  6. CTE Department -3rdstage Digital Communications The truth table of exclusive OR gate is used to derive the level of waveforms shown in Fig. 15, which satisfied in any interval b(t) is given as, ?(?) = ?(?) ?(? ??) Figure 15: DPSK waveform Note that the output sequence b(t) changes level at the beginning of each interval in which d(t)=1 and does not change level when d(t)=0, so that symbol duration (T)= duration of two bits 2??. The sequence b(t) is applied to the balanced modulator as shown in Fig. 14 with carrier 2?cos(2??0?), the modulator output is: ?(?) = ?(?) 2? cos(2??0?) = 2? cos(2??0?) As shown in in Fig.15 the phase changes only when d(t)=1. 6

  7. CTE Department -3rdstage Digital Communications 4-2 DPSK Receiver: During the transmission, the DPSK signal undergoes some phase shift ?. Therefore the signal received at the input of receiver is Received signal ?(?) = ?(?) 2? cos(2??0?+ ?) The signal is multiplied with delayed version by one bit: Multiplier output= ?(?)?(? ??)(2?) cos(2??0?+ ?) cos(2??0(? ??) + ?) Figure 16a: QPSK receiver This signal is fed to integrator and to the decision device. The integrator and decision is controlled by bit synchronizer to get the output binary sequence. The block diagram of such system is shown in Fig.16a, and the equivalent of this system is shown in Fig.16b. 7

  8. CTE Department -3rdstage Digital Communications 4-3 Bandwidth of DPSK: We know that one previous bit is used to decide the phase shift of next bit. Change in b(t) occurs only if input bitt is at level '1'. No change occurs if input bit is at level '0'. Since one previous bit is always used to define the phase in next bit, the symbol can be said to have two bits.Therefore one symbol duration (T) is equivalent to two bits duration (2??), so that Symbol duration ? =2?? 2 1 Bandwidth is given as, ?? =?=?? Or ?? =?? 4-4 Advantages and disadvantages of DPSK: Advantages: 1- DPSK does not need carrier at its receiver. Hence it is simple than BPSK. 8

  9. CTE Department -3rdstage Digital Communications 2- The bandwidth required of DPSK is reduced compared to that of BPSK. Disadvantages: 1 The probability of bit error rate of DPSK is higher than BPSK. 2 Error in the first bit creates error in the second bit. Hence error propagation in DPSK is more, while in PSK each bit is independent. 3 Noise interference in DPSK is more. 4 Quadrature Phase Shift Keying (QPSK): There are two important points in communication system, transmission power and channel bandwidth. The channel bandwidth depends upon the bit rate or signaling rate ??. The carrier is used as bandpass transmission over a channel. If two or more bits are combined in some symbol instead of one bit, then the signaling rate is reduced and the transmission channel bandwidth is reduced. In quadrature shift keying, two successive bits in the data sequence are grouped together. In BPSK the phase shift occurs in two level only, the carrier is change by 1800. In QPSK the combination of two bits form four distinct symbols, so that the change of carrier phase is 450 (????????) from one symbol to the next one. Table 1: Symbol and corresponding phase shift in QPSK Sr.No. Input successive bits Symbol 4 Phase shift in carrier ? = 1 1(1?) 0( 1?) ?1 ?/4 ? = 2 0( 1?) 0( 1?) ?2 3?/4 9

  10. CTE Department -3rdstage Digital Communications ? =3 0( 1?) 1(1?) ?3 5?/4 ? =4 1(1?) 1(1?) ?4 7?/4 1 QPSK generator: There are two type of QPSK: 1 Offset QPSK (OQPSK): Fig. 17 shows the block diagram of OQPSK generator. The input signal is converted to NRZ and called b(t), the demultiplexer divides b(t) into separated bit streams of odd numbered and even numbered. 10

  11. CTE Department -3rdstage Digital Communications Figure 17: OQPSK generator The symbol duration of both ??(?)and ??(?) are 2??as shown in Fig. 18. Note that the first symbol of ??(?) is delayed by one bit period with respect to symbol of ??(?). This delay of ??is called offset so that the change in level cannot occur in the same time. Each ??(?) and ?0(?) are modulate a carrier 2??sin(2??0?) and 2??cos(2??0?) respectivlly. 11

  12. CTE Department -3rdstage Digital Communications 12

  13. CTE Department -3rdstage Digital Communications ??(?) = ??(?) 2??sin(2??0?) ?0(?) = ?0(?) 2??sin(2??0?) Thus ??(?)and ?0(?) are basically BPSK signals but ? = 2??. The output of adder is OQPSK, givenas: ?(?) = ??(?) + ?0(?) 13

  14. CTE Department -3rdstage Digital Communications =??(?) 2??sin(2??0?) + ?0(?) 2??cos(2??0?) Fig.19 shows the phasor diagram of QPSK signal of above equation. Figure 19: Phasor diagram of QPSKsignal 14

  15. CTE Department -3rdstage Digital Communications Offset QPSK is essentially the same as QPSK except that the I- and Q-channel pulse trains are staggered. The modulator and the demodulator of OQPSK are shown in Figure 20, which differs from the QPSK only by an extra delay of T/2 seconds in the Q-channel. Its power spectral density is the same as that of QPSK, and its error performance is also the same as that of QPSK. Fig. 21: OQPSKmodulator 15

  16. CTE Department -3rdstage Digital Communications Fig. 21: OQPSKdemodulator 16

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