EE462: AI Enabled Digital

Communications
Week # 8

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 Contents

• Noncoherent Detection of MPSK and BFSK

• How to calculate the average probability of symbol error for different

modulation schemes?

• How to compare different modulation schemes based on their error

performances?

The contents of this material are based on

• Chapter 4, Sklar, B., 2001. Digital communications (Vol. 2, p. 1011). Upper Saddle
River, NJ, USA:: Prentice hall.

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 Noncoherent Detection
Introduction
• There are some applications in which it would very difficult and expensive to
use coherent detection.
• For example, the propagation delay on some radio channels changes too
rapidly to permit accurate tracking of the carrier phase at the receiver, and
the unsynchronized or noncoherent detection becomes the only viable
recourse.
• In the noncoherent detection, it is not required that the reference signal at
the receiver (could be i(t)) to be in phase with the received signal, r(t).

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 Noncoherent Detection
Introduction
If the transmitted signal is

The received signal takes the following form

where  is random variable. It can be assumed to be uniformly distributed over [0 2 ].

Furthermore, the noncoherent detector compares (correlates) the received signal with a
reference that might be noisy reference.
Notice that the phase of the received signal is not known due to the noise added to it.
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 In MPSK, if we assume that the random variabl 
Differential varies slowly relative to 2T (The period of two
successive pulses), then
PSK (DPSK)
The basic idea of DPSK is: The carrier phase of the previous signaling can be used
as a phase reference estimate to the current phase.
In general, DPSK is less efficient than coherent PSK, but for higher signal to noise
ratio, the efficiency of DPSK approaches the one for PSK.
ADVANTAGE: Less complexity in the receiver.

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 Binary Binary Source Diff. Encoder BPSK
Differential
Phase Shift
Keying DEMOD.
(detector)

(BDPSK)
Ik is the information message. In case of binary, the
message is {0, 1}.
Let Ck be defined as:

Ck can be chosen as:

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 Consider the information message {1100110110}.

Find the encoded and decoded messages, Ck and I^k.

Example: We will use the exclusive NOR to compute Ck

k 0 1 2 3 4 5 6 7 8 9 10

Ik 1 1 0 0 1 1 0 1 1 0

ck 1 1 1 0 1 1 1 0 0 0 1

k    0    0 0 0 
I^ k 1 1 0 0 1 1 0 1 1 0

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 In the above example, we chose the initial Ck to be 1.
This is an arbitrary choice. Moreover, instead of the
exclusive NOR, we could choose exclusive OR. If the
later is chosen, then above condition is reversed.
Example: Notice the difference between the PSK and DPSK. As we
recall, in PSK, the received signal is compared with a
clean reference. However, in DPSK, two noisy signals are
compared with each other
There are other ways to come up with differentional
pulse shift keying (DPSK). For example, let us assume
that another encoder is

Ik XOR Ck

Ck-1 8
 k 0 1 2 3 4 5 6 7 8 9 10

Ik 1 0 1 1 0 1 0 0 0 1

Example: ck 1 0 0 1 0 0 1 1 1 1 0

The decoder process is the same as the encoder.

Thus, the decoded sequences are:
k 0 1 2 3 4 5 6 7 8 9 10

ck 1 0 0 1 0 0 1 1 1 1 0

Ik 1 0 1 1 0 1 0 0 0 1

Again, the process is Exclusive OR.

When the initial bit is 0, the process will be the same. 9
Detection of Sent and received signal in FSK is

Noncoherent
FSK
where  is a r.v. uniformly distributed over [0, 2 ].
Notice here that the phase measurement is not exploited here.
All what we need is an energy measurement.
Since our receiver is an energy detector, and we don’t know the phase, it
requires twice as many channel branches as the coherent receiver.
Two in phase (I) and two quadrature(Q) channels to detect the signal associated
with the frequency 1 and 2 .

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 Detection of
Noncoherent
FSK
The bases signals are:

Now, suppose that

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Detection of
Noncoherent
FSK

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 The lower branch
Detection of
Noncoherent
FSK
Now,

The input to the decision

circuit is

For Mary FSK, we need 2M correlators and 2M squaring circuits.

In the noncoherent FSK, the separation between 1 and 2 is a multiple of 2/T.
Self Study (Example 4.3) 13
 2 (t)
16QAM 8PSK s3 “011”
“0 10” “ 001”
 2 (t) s4 s2
“0000” “0001” “0011” “0010”
s1 s2 s3 s4 Es
3
“110” s“000”
1
“1000” “100 1” “1011 ” “1010 ” s5  1 (t)
s5 s6 s7 s8
Example of -3 -1
1
1 3
 1 (t)
“111 ”
s 6
s“100”
8

two- s9
“1100”
s10
“1101”
-1
s
11 12
“1111” “1110”
s “101” s7
2 (t)

dimensional QPSK “00”

s 2“01” s1
s13 s14 -3
s
15 16s
modulation “0100” “0101” “0111” “0110” Es

 1 (t)

s3 “11” “10”
s4
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 Error  Before evaluating the error probability, it is

probability important to remember that:

 The type of modulation and detection ( coherent or non-

of bandpass coherent) determines the structure of the decision circuits

and hence the decision variable, denoted by z.

modulation
 The decision variable, z, is compared with M-1 thresholds,
corresponding to M decision regions for detection purposes.

 1 (t )
T r1

r (t )

0  r1  Decision
⁝ r
r Circuits mˆ
 N (t)   Compare z
T rN  with threshold.

0 rN

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 Error  The matched filters output (observation vector= r ) is
the detector input and the decision variable is a z  f (r)
probability function of r , i.e.
 For MPAM, MQAM and MFSK with coherent detection z  r

 For MPSK with coherent detection z  r

 For non-coherent detection (M-FSK and DPSK), z | r|

• We know that for calculating the average


probability of symbol error, we need to determine
• Pr(r liesinside Zi | si sent)  Pr(z satisfies condition Ci | si sent)
 Hence, we need to know the statistics of z, which
depends on the modulation scheme and the
detection type.

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 Error  AWGN channel model: r  s i  n
 The signal vector si  (ai1, ai2 ,..., aiN ) is deterministic.
probability  The elements of the noise vector n  (n1 , n2 ,..., nN ) are
i.i.d Gaussian random variables with zero-mean and
variance N0 / 2 . The noise vector's pdf is
1  n2
pn (n)  exp  
N 0 N /2
 N0 
 
 The elements of the observed vector r  (r1 , r2 ,..., rN )
are independent Gaussian random variables. Its pdf is
1  r  s
2

pr (r | s i )  exp  i 
N 0 N /2


N0 


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 Error  BPSK and BFSK with coherent detection:
probability  s s /2
P  Q  
N /2  B
1 2

2 (t)  0   1 (t)
“0” “1”

BPSK s1 s2 BFSK
“0”
s1 Eb
s1  s 2  2Eb

 Eb Eb  1 (t)
s 2“1”
 2 (t)
s1  s 2  2 Eb Eb

 2Eb   Eb 
PB Q   PB Q  
 N   N 
 0   0 

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 Error  Non-coherent detection of BFSK
probability
2 / T c o s (  1t )
Decision variable:
Difference of envelopes
T r11
 2 z  z1  z 2
0
z  r 2 r 2
1 11 12
2 / T s in (  1 t )
T r12
r(t)
0
 2 + z
Decision rule:
m̂
2 / T c o s ( 2 t ) if z(T )  0, m̂  1
r 21 if z(T )  0, m̂  0
T
 2 -
0

2 / T sin (  2 t ) z  r 2
r 2
2 21 22
T r 22
0
 2 19
 Error  Non-coherent detection of BFSK …
probability PB 
1
2
1
Pr(z1  z 2 | s 2 )  Pr(z 2  z1 | s1 )
2
Pr(z1  z 2 | s 2 )  E Pr(z1  z 2 | s 2 , z 2 )

 0
 
  p(z | s )dz  p(z | s )dz
Pr(z1  z 2 | s 2 , z 2 ) p(z 2 | s 2 )dz 2  
0  z2 1 2 1
 2 2 2

1  Eb 
PB  exp 
 Rayleigh pdf Rician pdf
2  2N 0 

 Similarly, non-coherent detection of DBPSK

1  Eb 
PB  exp 

2  N0 
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 Error  Coherent detection of M-PAM
probability Decision variable:


z  r1
“00” “01” “11” “10”
s1 s2 s3 s4
4-PAM   1 (t)
 3 Eg  Eg 0 Eg 3 Eg

 1 (t)
T
r1

ML detector
r(t) (Compare with M-1 thresholds) m̂
0

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 Error  Coherent detection of M-PAM ….
probability
 Error happens if the noise, n1  r1  sm , exceeds in amplitude
one-half of the distance between adjacent symbols. For symbols
on the border, error can happen only in one direction. Hence:

Pe (s m )  Pr | n1 || r1  s m | E g  for 2  m  M 1;

P (s )  Pr n  r  s
e 1 1 1 1  Eg  and 
Pe (s M )  Pr n 1  r1  s M   E g 
1 M 2
     
M
1 1
PE (M ) 
M
 P (s
m1
e m )
M
Pr |n1 | Eg 
M
Pr n1  E g  Pr n1   E g
M


2(M 1)

Pr n1  E g 
2(M 1) 

 Eg n1
p (n)dn 
2(M 1)  2Eg 
Q
 N0 
M M M  
(M 1)
2
Es  (log2 M )Eb  Eg
3
Gaussian pdf with
2(M 1)  6 log 2 M Eb  zero mean and variance N0 / 2
PE (M )  Q 
 
M  M 1 N 0 
2

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 Error  Coherent detection
probability of M-QAM
2 (t)
“0 000”
s1 “s0001”
2
s 3“0011s” 4 “0010”

“1s000” “s1001” s 7“1011”s 8 “1010”

5 6

16-QAM  1(t)
s9 s10 s11 s12
“1100” “1101” “1111” “1110”

 1 (t) s13 s14 s15 s16

T
r1 ML detector “0100” “0101” “0111” “0110”

0 (Compare with M 1 thresholds)

r(t) Parallel-to-serial
m̂
converter
 2 (t)
T
r2 ML detector

0 (Compare with M 1 thresholds)

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 Error probability …
 Coherent detection of M-QAM …
 M-QAM can be viewed as the combination of two M  PAM
modulations on I and Q branches, respectively.
 No error occurs if no error is detected on either the I or the Q
branch.
 Considering the symmetry of the signal space and the
orthogonality of the I and Q branches:

PE (M ) 1 PC (M ) 1 Pr(no error detected on I and Q branches)

Pr(no error detected on I and Q branches) Pr(no error on I)Pr(no error on Q)


Pr(no error on I) 2  1  PE  M 
2

1 Q 3log2 M Eb 
PE (M )  41    Average probability of
 M   M 1 N 0 
symbol error for M  PAM
 Error probability …
 Coherent detection
of MPSK 2 (t)
s 3 “011”
“s010” s“001”
4 2
Es

s“000”
8-PSK “110” 1
s5  1(t)

“11s1”
 1 (t) 6
s“100”
8

T r1 “101”s7

r(t)

0
r1 ˆ
arctan Compute Choose m̂
 2 (t) r2 | i ˆ | smallest
T

0
r2 Decision variable
z  ̂  r
 Error probability …
 Coherent detection of MPSK …
 The detector compares the phase of observation vector to M-1
thresholds.
 Due to the circular symmetry of the signal space, we have:
1 M / M
PE (M )  1 PC (M )  1
M
 Pc (s m )  1 Pc (s1 )  1 
m1
 / M
pˆ ( )d

where
2 Es cos( ) exp  Es sin 2  ; |  | 
pˆ ( ) 
 N0  N0  2
 It can be shown that

 2Es      2log 2 M Eb  

PE (M )  2Q sin    or PE (M )  2Q  sin   
 N0
  M    N0  M  

 Error probability …
 Coherent detection of M-FSK

 1 (t)


T r1
0
 r1  ML detector:
r(t) ⁝  r
r Choose
the largest element m̂
 M (t)   in the observed vector

rM 

T
0 rM
 Error probability …
 Coherent detection of M-FSK …
 The dimension of the signal space is M. An upper
bound for the average symbol error probability can be
obtained by using the union bound. Hence:
 Es 
PE (M )  M 1Q 
 N 
 0 

or, equivalently


PE (M )  M 1Q 
log 2 M Eb 

 N0 
 
Bit error probability versus symbol error probability

 Number of bits per symbol k  log2 M

 For orthogonal M-ary signaling (M-FSK)
PB 2 k 1 M /2
 k 
PE 2 1 M 1
PB  1
lim
k  P 2
E

 For M-PSK, M-PAM and M-QAM

PE
PB  for PE  1
k
Probability of symbol error for binary modulation

Note!
• “The same average symbol
PE energy for different sizes of
signal space”

Eb / N 0 dB
Probability of symbol error for M-PSK

Note!
• “The same average symbol
energy for different sizes of
PE signal space”

Eb / N 0 dB
Probability of symbol error for M-FSK

Note!
• “The same average symbol
energy for different sizes of
PE signal space”

Eb / N 0 dB
Lecture 8
 Probability of symbol error for M-PAM

Note!
• “The same average symbol
energy for different sizes of
PE signal space”

Eb / N0 dB
 Probability of symbol error for M- QAM

Note!
• “The same average symbol
energy for different sizes of
PE signal space”

Eb / N0 dB
Example of samples of matched filter output
for some bandpass modulation schemes