BASEBAND DATA
TRANSMISSION
IMPORTANT DEFINITIONS
Baud rate – symbol transmission rate in bauds/sec.
Bit rate – binary information transmission rate in bits/sec.
Bandwidth-the different in frequency between the maximum and
minimum frequency present in the signal.
Bit-error rate (BER) – the number of error present within the period
of data transmission.
Packet error rate (PER) – the number of packets with at least an error
present.
Signal-to-noise ratio (SNR) – the ratio of signal power to noise power.
SIGNAL MODEL
If x(t) is the signal, then y(t) is the signal corrupted by noise n(t) is
y (t ) = x (t ) + n (t )
The probability density function (pdf) for noise can be represented
as a Gaussian distribution
 n02 
1
PN (n0 ) =
exp−

2πN0
 2N0 
where n0 is noise at time n(t0) and E[n(t0)]2 =N0 is noise power at
time t0.
PROBABILITY OF ERRORSINGLE OBSERVATION
Consider x(t) to be two possibilities, x0(t) and x1(t), at time t=t0,
x0=x0(t0) and x1=x1(t0). Conditional probability for symbol x0 and x1
can be represented as
Py | x0 ( y ) =
 ( y − x0 ) 2 
1

exp −
2 N0 
2πN 0

Py| x1 ( y ) =
 ( y − x1 ) 2 

exp −
2N 0 
2πN 0

1
The priori probability for both symbols is P(x0)=P(x1)=1/2.
PROBABILITY OF ERRORSINGLE OBSERVATION
Probability of
error
reference
point
x0
x1
T=
x0 + x1
2
Probability of error (area under the graph) can be calculated as
∞
T
Pe = ∫ Py| x0 ( y ) P ( x0 ) dy + ∫ Py| x1 ( y ) P ( x1 ) dy
T
 (y−x0)
1 1
= ∫
exp−
2T 2πN0  2N0
∞
2
−∞
T

 (y−x1)2 
1 1
dy + ∫
dy
exp−
2−∞ 2πN0  2N0 

PROBABILITY OF ERRORSINGLE OBSERVATION
Since priori probability are equal, then the probability of error is
∞
 ( y − x0 )2 
1
1
dy
Pe = 2. ∫
exp −
2 T 2πN0
2 N0 

If x0 is shifted to 0, probability of error can be written as
∞
Pe =
∫
x −x
T= 1 0
2
 y2 
1
dy
exp −
2πN0
 2 N0 
Substituting z2=y2/N0, the probability of error is
γ 
Pe = Q   =
2
∞
∫
γ
x1− x 0
=
2 2 N0
 z2 
1
 dz
exp  −
2π
 2 
Q( ) is referred as the Q function and is given in table form.
PROBABILITY OF ERRORMULTIPLE OBSERVATIONS
To maximize the probability of correct detection, value of γ2
has to be increased
γ
2
2
[
x1 − x0 ]
=
N0
For multiple detection within a symbol Tb, then the time factor
is included such that the value of γ2
1
2
γ =
N0
Tb
2
[
x
(
t
)
−
x
(
t
)
]
dt
0
∫ 1
0
The probability of correct detection is obtained from Q(γ/2).
To maximize probability of correct detection, the match filter
structure is adopted.
OPTIMUM RECEIVER
STRUCTURE
The output of the receiver is
Tb
z (t ) =
∫ [x (t ) − x (t )]y(t )dt
1
0
0
The optimum receiver structure based on the matched filter is
x
y (t ) = x (t ) + n (t )
∫
Tb
0
dt
x1 (t )
x
x 0 (t )
z (t )
z (n )
-
∫
Tb
0
Tb
dt
ISI PROBLEM
Intersymbol interference (ISI) is due to significant path delay in
the transmission medium and band limited bandwidth of the
channel.
Channel
Transmitted data
- 1 bit
Received data
Channel
Transmitted data
- 2 bits
Received data
At receiver, the magnitude of the second sample is lower compared
to the first sample.
CONDITION FOR NO ISI
Intersymbol interference (ISI) can be eliminated by a proper choice
of bit rate or transmission rate and the proper choice of the received
pulse
If the signal is sampled at intervals Tb , then to obtain zero ISI x(t)
should satisfy
1 for n = 0
x(n) = 
0 for n ≠ 0
Nyquist’s pulse shaping criterion or theorem for distortionless
transmission, gives a condition on the Fourier transform,which
results in a pulse shaping having the zero ISI property.
NYQUIST’S PULSE SHAPING
CRITERIA
To ensure no ISI, Nyquist’s pulse shaping criterion or theorem
for distortionless transmission states that the transmission rate is
rb ≤ 2 f BW
where fBW is the bandwidth of the channel.
FREQUENCY RESPONSE MINIMUM ROLL-OFF
The frequency domain representation of a pulse is
X ( f ) =1
=0
− f b < f < fb
elsewhere
Its time domain representation by taking the inverse Fourier
transform is
sin 2π f b t
x (t ) = 2 f b
−∞ ≤t ≤ ∞
2π f b t
FREQUENCY RESPONSE MINIMUM ROLL-OFF
FREQUENCY RESPONSE MAXIMUM ROLL-OFF
The frequency domain representation of a pulse is
 πf 

X ( f ) = 1+ cos 
 2 fb 
=0
− 2 fb ≤ f ≤ 2 fb
elsewhere
Similarly, its time domain representation by taking the inverse
Fourier transform is
 1 
 1 
sin4π  −t 
sin4π  +t 
2 fb 
2 fb 
sin4π fbt


x(t) = 4 fb
+ 2 fb
+ 2 fb
−∞ ≤ t ≤ ∞
4π fbt
 1 
 1 
4π  −t 
4π  +t 
 2 fb 
 2 fb 
FREQUENCY RESPONSE MAXIMUM ROLL-OFF
PASSBAND DATA
TRANSMISSION
INTRODUCTION
Passband data transmission refers transmission through a passband
channel such as the PSTN and radio channels.
The basic modulation techniques for data transmission are :
PSK (Phase Shift-Keying), FSK (Frequency Shift-Keying) and ASK
(Amplitude Shift-Keying)
Other modulation techniques such as DPSK (Differential PSK),
CPFSK (Continuous Phase FSK),QPSK (Quadrature PSK),
QAM (Quadrature Amplitude Modulation), Multiphase PSK etc are
derived from these 3 basic modulation techniques
PSK (PHASE SHIFT-KEYING)
The PSK signal is defined as
 x1 ( t ) = A cos 2π f 1 t 
x (t ) = 

 x 0 ( t ) = − A cos 2π f 1 t 
x
y (t ) = x (t ) + n (t )
∫
Tb
0
x
x 0 (t ) = − A cos 2πf 1 t
0 ≤ t ≤ Tb
dt
x1 (t ) = A cos 2πf 1 t
s =1
s=0
z (t )
z (n )
-
∫
Tb
0
Tb
dt
Receiver structure
PSK (PHASE SHIFT-KEYING)
The BER is derived by first calculating the γ2 that is then substituted
in the Q function.
γ max 2 =
1
N0
γ max
1
=
2
2
2 A 2Tb
=
N0

BER = Q 


A 2Tb 
2N0 

∫
Tb
0
[ x1 ( t ) − x 0 ( t )] 2 dt
A 2Tb
2N0
2 A2Tb
=
N0
COHERENT ASK (AMPLITUDE
SHIFT-KEYING)
The ASK signal is defined as
 x1 (t ) = A cos 2πf 1t 
x (t ) = 

 x 0 (t ) = 0

y (t ) = x (t ) + n (t )
x
x1 (t ) = A cos 2πf 1 t
∫
Tb
0
0 ≤ t ≤ Tb
s =1
s=0
dt
z (t )
Tb
z (n )
Receiver structure
COHERENT ASK (AMPLITUDE
SHIFT-KEYING)
The BER is derived by first calculating the γ2 that is then substituted
in the Q function.
γ max
γ max
2
2
1
=
N0
=
∫
Tb
0
[ x1 ( t )] 2 dt
A 2 Tb
8N 0

BER = Q 


A 2 Tb
8N 0




1 A 2Tb
=
N0 2
COHERENT FSK (FREQUENCY
SHIFT-KEYING)
The FSK signal is defined as
 x 1 ( t ) = A cos 2 π f 1 t 
x (t ) = 

 x 0 ( t ) = A cos 2 π f 0 t 
x
y (t ) = x (t ) + n (t )
0 ≤ t ≤ Tb
∫
Tb
0
dt
x1 (t ) = cos 2πf 1t
x
x0 (t ) = cos 2πf 0 t
s =1
s =0
z (t )
z (n )
-
∫
Tb
0
Tb
dt
Receiver structure
COHERENT FSK (FREQUENCY
SHIFT-KEYING)
The BER is derived by first calculating the γ2 that is then substituted
in the Q function.
γ max
2
1
=
N0
∫
Tb
0
[ x1 (t ) − x 0 (t )] 2 dt =
1 2  sin 2π ( f1 − f 0 )Tb 
A Tb 1 −

N0
2
π
(
f
−
f
)
T
1
0
b 

1
1
< Tb , ( f1 − f 0 ) → large γ 2 = A2Tb
( f1 − f 0 )
N0
γ
max
2
1
=
2

BER = Q 


A 2Tb
=
N0
A 2Tb
4N 0




A 2Tb
4N 0
PROBLEMS WITH
COHERENT DETECTION
y (t ) = x (t ) + n (t )
x
∫
Tb
0
dt
z (t )
Tb
z (n )
x1 (t ) = A cos 2πf 1 t
Exact synchronization is required between x(t) and x1(t).
The BER increases by cosφ where φ is phase difference between
x(t) and x1(t).
Carrier recovery detection is required additional complexity
Non coherent detection is simpler but suboptimum in BER
performance.
NONCOHERENT ASK
y (t ) = x (t ) + n (t )
h1(t)
Bandpass filter
Center freq=f1
BW=2rb
∫
Tb
0
2
dt
z (t )
Tb
z (n )
Receiver structure
BER NON COHERENT ASK
The BER for non coherent ASK is
 A2
1
Pe = exp −
 8N p
2





The filtered noise power is
Np=2N0rb=2N0/Tb
where rb is the bit-rate, Tb is the bit-duration, and
N0 power of the additive white noise.
NON COHERENT FSK
Bandpass filter
Center freq=f1
BW=2rb
h1(t)
∫
Tb
2
0
dt
z (t )
z (n )
y (t ) = x (t ) + n (t )
h0(t)
Bandpass filter
Center freq=f0
BW=2rb
∫
Tb
0
Tb
2
dt
Receiver structure
BER NONCOHERENT FSK
The BER for non coherent FSK is
 A2 
1

Pe = exp −
 4N 
2
p 

The filtered noise power is
Np=2N0rb=2N0/Tb
where rb is the bit-rate, Tb is the bit-duration, and
N0 power of the additive white noise.
CPFSK (CONTINUOUSPHASE FSK)
To minimize sidelobes in the power spectrum, phase transition
from one binary value to another is made continuous resulting
in the CPFSK. The CPFSK signal is
x1 (t) = Acos2πf1t = Acos(2π ( fc + f dev)t) 
s =1
x(t) = 
 0 ≤ t ≤ Tb
s=0
x0 (t) = Acos2πf0t = Acos(2π ( fc − f dev)t)
where fc and fdev are
1
1
h
f c = ( f1 + f 0 ), f dev = ( f1 − f 0 ) =
, h = Tb ( f1 − f 0 )
2
2
2Tb
The CPFSK signal defined in terms of phase is
x(t) = cos(2πf ct ± φ(t)) = cos(2πf ct) cosφ(t) m sin(2πf ct) sinφ(t)
φ(t) = φ(0) ±
πh
Tb
t
PHASE TRELLIS DIAGRAM
FOR CPFSK
2πh
2πh
πh
πh
−πh
−πh
-2πh
-2πh
Sequence : 1011
Sequence : 1001
MSK (MINIMUM SHIFTKEYINGS)
MSK is a special case of CPFSK where the modulation index
h is ½.
Since a square wave that represents binary data has large
number harmonics, this is reduced by filtering the input binary
sequence with Gaussian filter that is realizable by having a raised
cosine function or Hanning function before MSK modulation is
Performed. The power spectrum broadened the mainlobe
but minimized the sidelobes. The resulting modulation technique
is known as GMSK and at present used in the GSM/PCN digital
cellular system.
DPSK-TRANSMIT
The DPSK developed to simplify the PSK by not requiring
carrier recovery circuit. The input sequence is encoded as
a0 = 1
a k = a k −1 s k ⊕ a k −1 s k
where ak is the encoded sequence and sk is the actual data
sequence.
Example of encoded sequence and transmitted signal
k
0
1
2
3
4
5
sk
1
1
0
1
0
0
ak
1
1
1
0
0
1
0
x(t)
c1
c1
c1
-c1
-c1
c1
-c1
c1 = cos 2πf1t
DPSK-RECEIVE
The process for detecting the transmitted sequence
is shown in the table.
k
0
1
2
3
4
5
x(t)
c1
c1
c1
-c1
-c1
c1
-c1
φk
0
0
0
π
π
0
π
Phase
diff
+
+
-
+
-
+
sk
1
1
0
1
0
0
DPSK RECEIVER
y (t ) = x (t ) + n (t )
x
h1(t)
h2(t)
z (t )
Bandpass filter
Center freq=f1
BW=2rb
Tb
Tb
Lowpass filter
BW=fc
Receiver structure
z (n )
BER DPSK
The BER for DPSK is
 A2 
1

Pe = exp −
 2N 
2
p 

The filtered noise power is
Np=2N0rb=2N0/Tb
where rb is the bit-rate, Tb is the bit-duration, and N0 power
of the additive white noise.
QPSK (QUADRATURE PSK)
For PSK, the symbol and bit rates because each symbol represents
one bit of information.To increase spectrum efficiency, two bits of
information represent a symbol for QPSK.
2 bits 1 symbol
Binary sequence s0
s1
|_____|
Symbol
b0
s2
s3
|_____|
b1
s4
s5
|_____|
b2
s6
s7
|_____|
b3
Within a symbol duration, even order bit transmit at cos(2πf1t)
and odd order bit as sin(2πf1t)
QUADRATURE PSK
All the possible combinations for transmit signal.
s0
s1
0
0
0
1
1
1
1
0
x(t)
− ACos2πf1t − A sin 2πf1t
− ACos2πf1t + A sin 2πf1t
+ ACos2πf1t + A sin 2πf1t
ACos2πf1t − A sin 2πf1t
inphase
quadrature
QUADRATURE PSK
Constellation diagram
BER QPSK
The BER for QPSK is calculated as follows
Pe = Pe,11P(s =11) + Pe,10P(s =10) + Pe,01P(s = 01) + Pe,00P(s = 00)
where Pe,11, Pe,10, Pe,01, and Pe,00 error derived from the
conditional probability, and P(s=11), P(s=10), P(s=01), and
P(s=00) are priori probabilities that is equal to 1/4.
For example, Pe,11 is probability of getting s=01 or s=10 if actual
data transmitted is s=11. This is calculated as follows
P
e,11
=
0
0
−∞
−∞
∫ ∫ ρ
( x, y)dxdy
X ,Y | s =11
BER QPSK
The conditional probability density function (pdf) is a 2-D
Gaussian that is defined as
 − (x 2 + y 2 ) 
ρ X ,Y |S =11 ( x, y) = 2πσ 2 exp 2σ 2 


1
This function is difficult to integrate. Alternative this can be
evaluated using the union bound.
UNION BOUND ON
PROBABILITY OF ERROR
When the geometry of the signal set is difficult to analyze,
upper bound for the error probability is used instead. For M=4,
the error Pe,11 is the union for the probability of error events
e10 and e01. Pe,11 can be written as
Pe,11 = P[e10 + e01 ] = P[e10 ] + P[e01 ] − P[e10 , e01 ] ≤ P[e10 ] + P[e01 ]
For arbitrary M, the union bound is
M
Pe, j ≤
∑
i =1
P[eij ]
i ≠ j,
Pe, j ≤
(M −1)
Q(d j / 2N0 )
M
i≠ j
where dj is the distance between symbol sj and its neighbour.
BER QPSK – UNION BOUND
Given that s=11. Error Pe,11 will occur if s=01 or s=10.
s=01
s=11
s=10
Pe ,11
γ 
γ 
γ 
= Q  + Q , Q  =
2
2
2
A 2 Tb
since signal per direction is PSK
2N 0
BER QPSK – UNION BOUND
Since Pe,11 = Pe,10 = Pe,01= Pe,00, then the BER is
Pe = Pe,11P(s = 11) + Pe,10 P(s = 10) + Pe,01P(s = 01) + Pe,00 P(s = 00)
Pe = 4 Pe ,11 • ρ (s = 11)

γ 
Pe = 2Q   = 2Q 

2

 γ 
 γ  1
= 4 Q  + Q  •
 2  4
 2
A 2 Tb
2N 0




POWER SPECTRUM FOR QPSK
S xx ( f ) =
2
A Tb  sin 2π ( f 1 − f )τ 
A Tb  sin 2π ( f 1 + f )τ 
+
2  2π ( f1 − f )τ 
2  2π ( f 1 + f )τ 
2
2
2
QPSK RECEIVER
∫
x
y (t ) = x (t ) + n (t )
Tb
0
dt
xI (t) = Acos2πf1t
Tb
∫
x
Tb
0
dt
xQ (t ) = A sin 2πf1t
Receiver structure
16 QAM (QUADRATURE
AMPLITUDE MODULATION)
Q
I
I
Q
0
1
3
2
0000
0001
0011
0010
6
7
5
4
0110
0111
0101
0100
12
13
15
14
1100
1101
1111
1110
10
11
9
8
1010
1011
1001
1000
BER 16 QAM
Derivation of the BER based on the union bound
Error in region
Total error in region
γ 
Pe , 0111 = 4Q 
2
1
P (s = 0111) =
16
γ  1
4 Pe,0111 • P(s = 0111) = 4 • 4Q  •
 2  16
γ 
Pe , 0101 = Pe , 0111 = 4Q 
2
1
P (s = 0111) =
16
is:
γ 
= Q 
2
BER 16 QAM
Error in region
:
γ 
Pe,1000 = 2Q 
2
1
(
)
P s = 1000 =
16
Pe,1010 = Pe,1000
Total error in region
4 Pe ,1000
:
γ  1
• ρ (s = 1000 ) = 4 • 2Q  •
 2  16
=
1 γ 
Q 
2 2
P(s = 1010) = P(s = 1001)
Overall result:
γ  3 γ  1 γ  6 γ 
γ 
Pe = Q  + Q  + Q  = Q = 3Q 
2 2 2 2 2 2 2
2
MULTICARRIER
MODULATION
Single frequency f0
Binary sequence
Multicarrier-4 frequencies
s0
s1
f3
f2
f1
f0
s2
s3
s3
s2
s1
s0
s4
s5
s6
s7
s7
s6
s5
s4
Major advantage is bit-rate>baud-rate, and overcome effect of frequency
selectivity in transmission channel.
Major disadvantage is high transmit power and PAPR
(Peak to Average Power Ratio)
MULTICARRIER
MODULATION
Modulation
Transmit
Binary
sequence
Serial to parallel
conversion
Frequency
encoding
IDFT
Serial to parallel
conversion
Frequency
decoding
DFT
Receive
Binary
sequence
Demodulation
For each frequency, any modulation technique can be used from the
basic such as ASK to advanced such as QPSK and QAM.
MULTICARRIER MODULATIONENCODING & IDFT
Signal per carrier is PSK. Number of carriers is 16 at 500, 1000,1500,
and 2000. IDFT points is 16 and sampling frequency.
Sequence is s0=1, s1=1, s2=0, s3=1.
x ( n ) = cos( 2π 500 nT s ) + cos( 2π 1000 nT s )
− cos( 2π 1500 nT s ) + cos( 2π 2000 nT s )
0 ≤ n ≤ 15
MULTICARRIER MODULATIONDECODING & DFT
DFT is performed and transmitted sequence estimated from the
peaks
X (1) = 8
→
s0 = 1
X ( 2) = 8
→
s1 = 1
X ( 3) = − 8 →
s2 = 0
X ( 4) = 8
s3 = 1
→
MULTICARRIER MODULATIONENCODING & IDFT
Signal per carrier is QPSK. Number of carriers is 16 at 500, 1000,1500,
and 2000. IDFT points is 16 and sampling frequency.
Sequence is s0=1, s1=0, s2=1, s3=1, s4=0 s5=1, s6=0, s7=1.
x(n) = cos(2π 500nTs ) − sin(2π 500nTs ) + cos(2π1000nTs ) + sin(2π1000nTs )
− cos(2π1500nTs ) + sin(2π1500nTs ) − cos(2π 2000nTs ) + sin(2π 2000nTs )
Tim e R e p re s e n t a t io n o f S ig n a l
R e a l a n d Im a g in a ry S p e c t ru m
Amplitude
0 ≤ n ≤15
10
4
5
3
0
2
-5
0
5
10
15
F re q s a m p le s k
Amplitude
10
Amplitude
1
-1 0
0
-1
5
-2
0
-3
-5
-1 0
0
5
10
F re q s a m p le s k
15
-4
0
5
10
Tim e s a m p le s n
15
MULTICARRIER MODULATIONDECODING & DFT
DFT is performed and transmitted sequence estimated from the
peaks
Tim e R e p re s e n t a t io n o f S ig n a l
R e a l a n d Im a g in a ry S p e c t ru m
10
3
5
Amplitude
4
2
0
-5
-1 0
0
5
0
15
10
15
10
-1
-2
-3
-4
10
F re q s a m p le s k
Amplitude
Amplitude
1
0
5
10
15
Tim e s a m p le s n
5
0
-5
-1 0
0
5
F re q s a m p le s k
real[X (1)] = 8
→
s 0 = 1, imag [X (1)] = 8
→ s1 = 0
real[X (2)] = 8
→
s 2 = 1, imag [X (2)] = −8
→ s3 = 1
real[X (3)] = −8 →
s 4 = 0, imag [X (3)] = −8
→
s5 = 1
real[X (4)] = −8 →
s 6 = 0, imag [X (4)] = −8
→
s7 = 1