Demodulation

• Coherent (synchronous) demodulation/detection

– Use a BPF to reject out-of-band noise
– Multiply the incoming waveform with a cosine of the carrier
frequency
– Use a LPF
– Requires
q carrier regeneration
g ((both frequency
q y and p
phase
synchronization by using a phase-lock loop)
• Noncoherent demodulation (envelope detection etc.)
– Makes no explicit efforts to estimate the phase
159
 ASK
• Amplitude shift keying (ASK) = on-off keying (OOK)
s0((t)) = 0
s1(t) = A cos(2 fct)
or s(t) = A(t) cos(2fct), A(t) ‫{ א‬0, A}
• Coherent detection

• Assume an ideal band-pass filter with unit gain on [fc−W, fc

+W]. For a practical band-pass filter, 2W should be
interpreted as the equivalent bandwidth.
160
 Coherent Demodulation
• Pre-detection signal:
x(t )  s (t )  n(t )
 A cos(2 f c t )  nc (t ) cos(2 f c t )  ns (t ) sin(2 f c t )
 [ A(t )  nc (t )]cos(2 f c t )  ns (t ) sin(2 f c t )

• After multiplication with 2cos(2 fct):

y(t )  [ A(t )  nc (t )]2cos2 (2 fct )  ns (t )2sin(2 fct )cos(2 fct )

 [ A(t )  nc (t )](1  cos(4 fct ))  ns (t )sin(4 fct )
• After low
low-pass
pass filtering:
y (t )  A(t )  nc (t )

161
 Bit Error Rate
• Reminder: The in-phase noise component nc(t) has the
g
same variance as the original band-pass
p noise n(t)
()
– The received signal is identical to that for baseband digital
transmission
~ (t )
– The
Th sample l values
l off y (t will
ill h
have PDF
PDFs th
thatt are id
identical
ti l tto
those of the baseband case
• For ASK the statistics of the receiver signal are identical to
those of a baseband system
• Theep
probability
obab y o of e
error
o for
o ASK
S is
s the
e sa
same e as for
o the
e
baseband case
• Assume equiprobable
q p transmission of 0s and 1s.
• Then the decision threshold must be A/2 and the
probability of error is given by:  A
Pe, ASK  Q  
 2 
162
 PSK
• Phase shift keying (PSK)
s(t) = A(t) cos(2 fct), A(t) ‫{ א‬−A, A}
• Use
U coherent
h td
detection
t ti again, i tto eventually
t ll gett th
the
detection signal:
y (t )  A(t )  nc (t )
• Probability density functions for PSK for equiprobable 0s
and 1s in noise (use threshold 0 for detection):
– (a): symbol 0 transmitted
– ((b):
) symbol
y 1 transmitted

163
 Analysis
• Conditional error probabilities:
 1  (n  A)2 
Pe0   exp   dn
 2  2 
0 2

0 1  (n  A)2 
Pe1   exp   dn
 2  2 
 2

• In the first set n  n  A  dn  dn and when n  0,

0 n  A
1   n~ 2  ~
Pe 0 
 2 A exp  2 2 dn
• In the second set n  (n  A)  n  A  dn  dn
when n  0,
0 n  A, and when n  
, n   :
1 A  n 2  1   n 2 
Pe1 
 2  exp   2 2 (1)dn   2 A exp   2 2 dn
164
 Bit Error Rate
• So:
 1  n2 
Pe0  Pe1  Pe,PSK   exp  2 dn
A  2
 2 
• Change variable of integration to z ≡ n/σ ֜ dn = σdz
and when n = A, z = A/σ. Then:
1  z2 / 2  A
Pe,PSK  
2 
Ae d  Q 
dz
 
• Remember that
1 
Q x    exp( t 2 / 2)dt
2 x

165
 FSK
• Frequency Shift Keying (FSK)
(  f0t),
s0((t)) = A cos(2 ), if symbol
y 0 is transmitted
s1(t) = A cos(2 f1t), if symbol 1 is transmitted
• Symbol recovery:
– Use two sets of coherent detectors, one operating at a frequency f0
and the other at f1.

Coherent FSK demodulation

demodulation.
The two BPF’s are non- y
overlapping in frequency
spectrum

166
 Output
• Each branch = an ASK detector
 A + noise
i if symbol
b l presentt
LPF output on each branch = 
noise if symbol not present

• n0(t): the noise output of the top branch

• n1(t): the noise output of the bottom branch
• Each of these noise terms has identical statistics to
nc(t).
(t)
• Output if a symbol 1 were transmitted
y = y1((t)) = A + [[n1((t)) − n0((t)]
)]
• Output if a symbol 0 were transmitted
y = y0((t)) = − A + [n1((t)) − n0((t)])

167
 Bit Error Rate for FSK
• Set detection threshold to 0
• Difference from PSK: the noise term is now n1(t) − n0(t).

• The noises in the two channels are independent because

their spectra are non-overlapping.
– the p
proof is done in the p
problem sheet.
– the variances add.
– the noise variance has doubled!

• Replace σ2 in (172) by 2σ2 (or σ by 2 σ )

 A 
Pe ,FSK  Q 
 2 

168
 The Sum of Two R.V.
• Noise is the sum or difference of two independent zero
mean random variables:
– x1: a random variable with variance σ12
– x2: a random variables with variance σ22
• What is the variance of y ≡ x1 ± x2?
• By definition
 y2  E{ y 2 }  E{ y}2  E{( x1  x2 ) 2 }
 E{x12  2 x1 x2  x22 }  E{x12 }  E{x1 x2 }  E{x22 }
• For independent variables: E{x1x2} = E{x1}E{x2}
• For zero-mean random variables:
E{x1} = E{x2} = 0 ֜ E{x1x2} = 0
• So
 y2  E{x12 }  E{x22 }   12   22
169
Comparison of Three Schemes

170
 Comment
• To achieve the same error probability (fixed Pe):

• PSK can be reduced by 6 dB compared with a baseband

or ASK system (a factor of 2 reduction in amplitude)

• FSK can be reduced by 3 dB compared with a baseband

or ASK (a factor of 2 reduction in amplitude)

• Caution: The comparison is based on peak SNR. In terms

of average SNR, PSK only has a 3 dB improvement over
ASK, and FSK has the same performance as ASK

171
Q-function

172
 Examples
• Consider binary PSK modulation. Assume the carrier
amplitude
p A = 1 v,, and noise standard deviation  = 1/3.
Determine the bit error probability.
– Answer: Pe = 1.03  10-3.

• Now suppose the bit error probability is 10-5. Determine

the value of A/.
– Answer: A/ = 4.3.

173
 Application: GMSK
• Gaussian minimum shift keying (GMSK), a special form of
FSK p preceded byy Gaussian filtering,
g, is used in GSM
(Global Systems for Mobile Communications), a leading
cellular phone standard in the world.
– Also known as digital FM, built on some
of FM-related advantages of AMPS,
the first-generation
g analogg system
y
(30 KHz bandwidth).
– Binary data are passed through a
Gaussian filter to satisfy stringent
requirements of out-of-band radiation.
– Minimum Shift Keying: its spacing between the two frequencies of
FSK is minimum in a certain sense (see problem sheet).
– GMSK is allocated bandwidth of 200 kHz, shared among 32 users.
This provides a (30/200)x32
(30/200)x32=4 4.8
8 times improvement over AMPS
AMPS.

174
Lecture 11: Noncoherent Demodulation

175
 Outline

• Noncoherent demodulation of ASK

• Noncoherent demodulation of FSK

• Differential demodulation of DPSK

• References
– Haykin & Moher, Communication Systems, 5th ed., Chap. 9
– Lathi, Modern Digital and Analog Communication Systems,
3 d ed.,
3rd d ChChap. 13

176
 Noncoherent Demodulation
• Coherent demodulation assumes perfect synchronization.
– Needs a p
phase lock loop.
p
• However, accurate phase synchronization may be difficult
in a dynamic channel.
– Phase synchronization error is due to varying propagation delays,
frequency drift, instability of the local oscillator, effects of strong
noise ...
– Performance of coherent detection will degrade severely.
• When
e the
e ca
carrier
e pphase
ase is
suunknown,
o ,o one
e must
us rely
eyo on non-
o
coherent detection.
– No provision is made for carrier phase recovery.
• The phase  is assumed to be uniformly distributed on [0,
2].
• Circuitry is simpler, but analysis is more difficult!
177
 Noncoherent Demodulation of ASK
Signal
plus Bandpass Envelope Threshold Decision
noise filter detector device
in t = nTb

• Output of the BPF

y(t) = n(t) when 0 is sent
y(t) = n(t) + A cos(2 fct) when 1 is sent
• Recall
n(t )  nc (t ) cos(2 f c t )  ns (t ) sin(2 f c t )
• Envelope
p
R  nc2 (t )  ns2 (t ) when 0 is sent
R  ( A  nc (t )) 2  ns2 (t ) when
h 1 iis sentt
178
 Distribution of the Envelope
• When symbol 0 is sent, the envelope (that of the bandpass
noise alone)) has Rayleigh
y g distribution
r r2 /(2 2 )
f (r)  e , r 0
 2

• When symbol 1 is sent, the envelope (that of a signal +

bandpass
p noise)) has Rician distribution

r ( r 2  A2 ) /( 2 2 )  Ar 
f (r)  e I 0  2 , r  0
 2
 

• The first case dominates the error p

probability
y when
A/σ >> 1.

179
 Rayleigh Distribution
• Define a random variable R  X  Y where X and Y
2 2

are independent
p Gaussian with zero mean and variance  2
• R has Rayleigh distribution:
 r 2 / ( 2 2 )
f R (r )  r
2
e , r0

Normalized Rayleigh distribution

v = r/
fV(v) =  fR(r)

• Proving it requires change into polar coordinates:

Y
R  X 2 Y2 , t 1
  tan
X
180
 Derivation
• Consider a small area dxdy = rdrd. One has
f R , (r ,  )drd  f X ,Y ( x, y )dxdy  f X ,Y ( x, y ) rdrd
x2  y 2
1 
 e 2 2
rdrd
2 2

• Hence x2  y 2 r2
r  r 
f R , ( r ,  )  e 2 2
 e 2 2
2 2
2 2

• pdf of R:
r2
2 r 
f R (r )   f R , (r ,  )d  e 2 2
, r0
0  2

• pdf of   1
f  ( )   f R , (r ,  )dr  ,   [0, 2 ]
0 2

181
 Rician Distribution
• If X has nonzero mean A, R has Rician distribution:
(r2  A2 )/(2 2 )
fR (r)   2 e
r
I0 (Ar
A
2 ), r 0
where 2
I 0 ( x)  1
2 0
e x cos d
is the modified zero-order Bessel function of the first kind.

Normalized Rician distribution

v = r/
a = A/
fV(v) =  fR(r)

182
 Derivation
• Similarly,
f R , (r ,  )drd  f X ,Y ( x, y )rdrd
( x  A )2  y 2
1 
 e 2 2
rdrd
2 2

r 2  A2  2 Ar cos
1 
 e 2 2
rdrd
2 2

• Hence
r 2  A2  2 Ar cos
r 
f R , ( r ,  )  e 2 2
2 2

Bessel function
• pdf of R:
r 2  A2 Ar cos 
2 r  1 2
f R (r)   f R , ( r,  )d  e 2 2
 e 2
d , r0
0  2
2 0

183
 Error Probability
• Let the threshold be A/2 for simplicity.
• The error pprobability
y is dominated by
y symbol
y 0,, and
is given by
1  r  r 2 /(2 2 )
Pe   e dr
2 A /2  2

• The final expression

1  A2 /(8 2 )
Pe, ASK ,noncoherent  e
2
• Cf.
Cf coherent demodulation
 A  1  A2 /(8 2 )
Pe, ASK ,Coherent  Q  e
 2  2
• Noncoherent demodulation results in some performance
degradation.
g Yet,, for a large
g SNR,, the p
performances of
coherent and noncoherent demodulation are similar.
184
Noncoherent Demodulation of FSK

Bandpass
filter
centered at f1
R1

If R1 > R2,
choose 0
If R1 < R2,
choose 1

R2
Bandpass
filter
centered at f2

185
 Distribution of Envelope
• When a symbol 1 is sent, outputs of the BPFs
y1((t)) = n1((t))
y2(t) = n2(t) + A cos(2 f2t)
• Again, the first branch has Rayleigh distribution
2 2
r1 r1 /(2 )
f R1 (r1 )   2 e , r1  0

• while the second has Rice distribution

r2  ( r22  A2 )/(2 2 )
f R2 (r2 )   2 e I 0 ( Ar22 ),
) r2  0

• Note the envelopes

p R1 and R2 are statisticallyy independent.
p

186
 Error Probability
• Error occurs if Rice < Rayleigh
Pe  P( R2  R1 )
 
  2
r2
e  ( r22  A2 )/(2 2 ) Ar2
I0 (  2 )  2 e r1  r12 /(2 2 )
dr1dr2
0 r2 
 
  ( r2 2  A2 )/(
)/(2 2 )
I0 (  2 )
2 2
r2 Ar2 r1 /(  )
 r1 /(2
2
e  2 e dr
d 1dr
d2
0 r2

 r2
 2 e  (2 r22  A2 )/(2 2 )
I 0 ( Ar22 )dr2
0


 A2 /(4 2 )  ( x 2  2 )/(2 2 )
 e
1
2
x
 2 e I 0 (  x2 )dx x  2r2 ,   A / 2
0

• Observe the integrand is a Rician density

1  A2 /(4 2 )
Pe , FSK ,noncoherent  e
2
• Cf. coherent demodulation
 A  1  A 2 / ( 4 2 )
Pe , FSK ,Coherent  Q  2e
 2 
187
 DPSK: Differential PSK
• It is impossible to demodulate PSK with an envelop
detector,, since PSK signals
g have the same frequency
q y and
amplitude.
• We can demodulate PSK differentially, where phase
reference is provided by a delayed version of the signal in
the previous interval.
• Differential encoding is essential: bn = bn−1  an, where an,
bn ‫ א‬1.

bn
input DPSK
an signal
bn-1
Delay Tb cos(ct)

188
 Differential Demodulation
signal
plus Bandpass Lowpass Threshold
noise filter filter device
Decision
D i i
on an
Delay Tb

• Computing the error probability is cumbersome but

fortunately the final expression is simple:
1 A2 /(2 2 )
Pe,DPSK  e
2
– The derivation can be found in Haykin, Communication Systems,
4th ed., Chap. 6.
– Performance is degraded in comparison to coherent PSK.
• Cf. coherent demodulation
 A  1  A2 /( 2 2 )
Pe ,PSK  Q   e
  2
189
 Illustration of DPSK

Information symbols {an} 1 -1 -1 1 -1 -1 1 1

{bn-1} 1 1 -1 1 1 -1 1 1

Differentially encoded
1 1 -1 1 1 -1 1 1 1
sequence {bn}
Transmitted phase
0 0  0 0  0 0 0
(radians)
Output of lowpass filter
+ – – + – – + +
(polarity)
ec s o
Decision 1 -1 -1 1 -1 -1 1 1

Note: The symbol 1 is inserted at the beginning of the differentially encoded

sequence is the reference symbol.

190
 Summary and Comparison
Scheme Bit Error Rate
Noncoherent
ASK, FSK
Coherent ASK Q(A/2)
Coherent
Coherent FSK Q(A/2) ASK, FSK

Coherent PSK Q(A/)

Coherent PSK
Noncoherent ½ exp( A2/82)
exp(-A
ASK
Noncoherent p( 2/42)
½ exp(-A
FSK
DPSK ½ exp(-A2/22)

Caution: ASK and FSK have the same bit

error rate if measured by average SNR.
Average SNR, dB
191
 Conclusions
• Non-coherent demodulation retains the hierarchy of
p
performance.

• Non-coherent
Non coherent demodulation has error performance slightly
worse than coherent demodulation, but approaches
coherent performance at high SNR.

• Non-coherent demodulators are considerably easier to

build.

192
 Application: DPSK
• WLAN standard IEEE 802.11b
• Bluetooth2
• Digital audio broadcast (DAB): DPSK + OFDM (orthogonal
frequency division multiplexing)
• Inmarsat (International Maritime Satellite Organization):
now a London-based mobile satellite company

193
Revision Lecture

267
 Noise performance of bandpass digital
communications
• Modulation formats: ASK, FSK, PSK

• Coherent detection and its error probability

• Noncoherent detection and its error probability (including

differential detection for DPSK)

• Q-function
Q function Q(x): computation by using the graph or
approximation
1  x2 / 2
Q( x)  e ,x0
2 x

275
 Performance of digital modulation
Scheme Bit Error Rate
Noncoherent
ASK, FSK
Coherent ASK Q(A/2)
Coherent
Coherent FSK Q(A/2) ASK, FSK

Coherent PSK Q(A/)

Coherent PSK
Noncoherent ½ exp( A2/82)
exp(-A
ASK
Noncoherent p( 2/42)
½ exp(-A
FSK
DPSK ½ exp(-A2/22)

Caution: ASK and FSK have the same bit

error rate
t if measured
d by
b average SNR.
SNR
Average SNR, dB
276