Chapter 5
Noncoherent Receivers
In the previous chapters we assumed the receiver had perfect knowledge of the desired received signal except thetransmitted data bits which we want to detect. In this chapter we consider the case where the receiver knowseverything about the received signal except the data bits transmitted and the carrier phase. The approach is totreat the unknown phase as a random variable and then to average the likelihood of the received signal given thedata bits and unknown phase with respect to the distribution of the phase. We begin by assuming a general carriermodulated waveform with M signals with arbitrary correlation. For this general signal set we derive the optimalreceiver in the presence of additive Gaussian noise. We then derive the important special case of additive whiteGaussian noise. It will be shown in the case of additive white Gaussian noise that the receiver can be implementedusing an envelope detector. Next we derive an expression for the error probability of two signals in additive whiteGaussian noise. In section 3 we derive the error probability for M-orthogonal signals in additive white Gaussiannoise. Finally we derive expressions for the error probability of repetition codes with noncoherent reception.
1. Optimal Receiver in AGN
Assume additive stationary Gaussian noise and that sk�t � for 0 � k � M � 1 has the form
sk�t ��� ak
�t � cos
�ωct � βk
�t ���� 0 � k � M � 1
with s2
k�t � dt � 1
2
a2
k�t � dt � E
where ak�t � and βk
�t � are lowpass waveforms with respect to ωc. When sk
�t � is transmitted the received waveform
has the formr�t ��� ak
�t � cos
�ωct � βk
�t ��� θk ��� n
�t �
where θk is a random phase. If θk � 0 with probability 1 then we have the usual coherent reception situationalready discussed. We will for this section assume that θk is uniformly distributed on the interval 0 � 2π � and thatthe receiver does not know the value of θk. We can use the representation of bandpass signals and noise in derivingthe optimal receiver.
r�t ��� Re uk
�t � e jθk � jωct ��� n
�t �
whereuk�t ��� ak
�t � cosβk
�t ��� jak
�t � sinβk
�t �
and j ��� � 1. Assuming the noise is also narrow band we can express the noise as
n�t ��� nc
�t � cosωct � ns
�t � sin ωct �
Then the lowpass representation of the received signal becomes
r̃�t ��� uk
�t � e jθk � z
�t �
5-1
5-2 CHAPTER 5. NONCOHERENT RECEIVERS
wherez�t ��� nc
�t ��� jns
�t ��
Now assume that z�t � is a Gaussian process with covariance function K
�s � t � which has eigenfunctions ϕl
�t � and
eigenvalues λl . Then we can express the received lowpass signal as
r̃�t � �
∞
∑l � 0
�uk � le jθk � zl � ϕl
�t �
�∞
∑l � 0
r̃lϕl�t �
whereuk � l �
uk�t � ϕ �l � t � dt
zl �
z�t � ϕ �l � t � dt
and zl is a complex Gaussian random variable with mean zero and with
E Re�zl � 2 � � λl
E Im �zl � 2 � � λl
E Re�zl � Im �
zl � � � 0 �We can now calculate the probability density of r̃ � �
r̃1 ��� � � � r̃N � conditioned on the value of θk. Let r̃l � uk � le jθk � zl
then
pk�r̃l�θk ��� 1
2πλlexp
�� 1
2λl
�r̃l � e jθkuk � l � 2 �
pk�r̃�θk � � 1
∏Nl � 1 2πλl
exp � � 12
N
∑l � 1
�r̃l � e jθkuk � l � 2
λl �� 1
∏Nl � 1 2πλl
exp � � 12
N
∑l � 1
�r̃l� 2 � �
uk � l � 2 � r̃le jθku�k � l � r̃
�l e jθk uk � l
λl ��
N
∏l � 1
�2πλl � 1 exp � � 1
2
N
∑l � 1
�r̃l� 2 � �
uk � l � 2 � 2Re�r̃le jθk u
�k � l �
λl ��
N
∏l � 1
�2πλl � 1 exp � � 1
2 N
∑l � 1
�r̃l� 2 � �
uk � l � 2λl � � N
∑l � 1
Re�r̃lu�k � l � cosθk � Im
�r̃lu�k � l � sin θk
λl ��
N
∏l � 1
�2πλl � 1 exp � � 1
2 N
∑l � 1
�r̃l� 2 � �
uk � l � 2λl � � Re
N
∑l � 1
r̃lu�k � l
λl� cos
�θk � � Im
N
∑l � 1
r̃lu�k � l
λl� sin
�θk � �
�N
∏l � 1
�2πλl � 1 exp � � 1
2 N
∑l � 1
�r̃l� 2 � �
uk � l � 2λl � � �������
N
∑l � 1
�r̃lu�k � l �
λl����� cos
�θk � ψ ��� �
where
ψ � tan 1 �� Im ∑Nl � 1
r̃l u �k � lλl
�Re ∑N
l � 1r̃l u �k � lλl
���� �The joint density given signal k transmitted is then obtained by averaging with respect to θk. The joint densitygiven signal k transmitted is then
pk�r̃1 ��� � � � r̃N ���
2π
θ � 0
12π
pk�r̃1 ��� � � � r̃N
�θk � dθk
1. 5-3
�
2π
θ � 0
12π
N
∏l � 1
�2πλl � 1 exp � � 1
2 N
∑l � 1
�r̃l� 2 � �
uk � l � 2λl � � �������
N
∑l � 1
�r̃lu�k � l �
λl����� cos
�θk � ψ � � � dθ
�N
∏l � 1
�2πλl � 1 exp � � 1
2
N
∑l � 1
�r̃l� 2 � �
uk � l � 2λl � 2π
θ � 0
12π
exp � � �����N
∑l � 1
�r̃lu�k � l �
λl����� � cosθ � dθ
�N
∏l � 1
�2πλl � 1 exp � � 1
2
N
∑l � 1
�r̃l� 2 � �
uk � l � 2λl � I0 � �����
N
∑l � 1
�r̃lu�k � l �
λl����� �where
I0�x � ∆� 1
2π
2π
θ � 0exp
�xcosθ � dθ
is the modified Bessel function of order 0. Now let us calculate the likelihood ratio between hypothesis Hk and thehypothesis that no signal was present.
Λk�N ��� pk
�r̃1 � � � � � r̃N �
p0�r̃1 � � � � � r̃N � �
∏Nl � 1
�2πλl � 1 exp
�� 1
2 ∑Nl � 1
�r̃l
� 2 � �uk � l � 2
λl
� I0
� �∑N
l � 1r̃ku �k � lλl
� �∏N
l � 1
�2πλl � 1 exp
� � 12 ∑N
l � 1
�r̃l
�2
λl�
� exp � � 12
N
∑l � 0
uk � lu �k � lλl � I0 � �����
N
∑l � 1
r̃lu�k � l
λl����� � �
Now
limN � ∞
N
∑l � 1
uk � lu �k � lλl
�
uk�t � q �k � t � dt
where
qk�t ��� lim
N � ∞
N
∑l � 1
uk � lλl
ϕl�t �
is the solution of the integral equation
uk�s ���
K�s � t � qk
�t � dt �
SimilarlyN
∑l � 1
r̃lu�k � l
λl�
r̃�t � q �k � t � dt �
Thus the optimal receiver computes the following likelihood ratios
Λk � 0 � r̃ � t � ��� limN � ∞
�r�t ����� exp
� � 12
uk�t � q �k � t � dt � I0 � � r̃
�t � q �k � t � dt
� �and chooses k for which Λk � 0 is maximum.
Special Case: White Gaussian Noise
In this case the integral equation is easily solved:
qk�t ��� 2
N0uk�t �
so that
Λk � 0 � r̃ � t � ��� exp� � 1
N0
�uk�t � � 2dt � I0 � 2
N0
� r̃�t � u �k � t � dt
� � �
5-4 CHAPTER 5. NONCOHERENT RECEIVERS
For equi-energy signals this reduces to choosing k that maximizes
I0 � 2N0
� r̃�t � u �k � t � dt
� �and since I0
�x � is an increasing function of x the optimal receiver chooses k to maximize�
r̃�t � u �k � t � dt
� 2 � �Re
r̃�t � u �k � t � dt � 2 � �
Im
r̃�t � u �k � t � dt � 2 �
Now note thatr�t ��� Re � r̃ � t � e jωct �
and consider the following integralr�t � ak
�t � cos
�ωct � βk
�t � � dt �
Re r̃ � t � e jωct � Re uk
�t � e jωct � dt �
Since (can you show this) for any two complex numbers a and b
Re a � Re b ��� 12
Re ab� ��� 1
2Re ab �
the above integral becomes
r�t � ak
�t � cos
�ωct � βk
�t ��� dt � 1
2
Re
�r̃�t � � u �k � t � � dt � 1
2
Re
�r̃�t � � uk
�t � e j2ωct � dt �
That the second term is zero is due to the fact that both r̃ and uk are lowpass processes. Thus
r�t � ak
�t � cos
�ωct � βk
�t � � dt � 1
2
Re
�r̃�t � � u �k � t � � dt �
Similarly r�t � ak
�t � sin
�ωct � βk
�t ��� dt � 1
2
Im
�r̃�t � � u �k � t � � dt �
Thus an equivalent form of the optimal receiver computes the following for each k
� r�t � ak
�t � cos
�ωct � βk
�t � � dt � 2 � �
r�t � ak
�t � sin
�ωct � βk
�t ��� dt � 2
and decides that signal k was transmitted if k maximizes the above expression.
2. Performance of Binary Signals in AWGN
Consider a binary communication system with noncoherent reception. Assume the two transmitted signals are
sk�t ��� ak
�t � cos
�ωct � βk
�t � � � k � 1 � 2
with1
2E
s0�t � s1
�t � dt � ρ0 � 1 � ρ
Let H1 denote the event that signal s1 is transmitted and H2 the event that s2 is transmitted. The received signaldiffers from the transmitted signal in that there is a random phase term included and because of the noise. If s1 istransmitted the received signal then is
r�t ��� a1
�t � cos
�ωct � β1
�t ��� θ1 ��� n
�t �
where n�t � is a white Gaussian noise process. If s2 is transmitted the received signal then is
r�t ��� a2
�t � cos
�ωct � β2
�t ��� θ2 ��� n
�t � �
2. 5-5
We would like to compute the error probability of the optimal receiver. The optimal receiver processes the receivedsignal by correlating with two signals and then sums the squares. That is the receiver first computes
Xk � c ∆�
r�t � ak
�t � cos
�ωct � βk
�t � � dt
andXk � s ∆�
r�t � ak
�t � sin
�ωct � βk
�t � � dt
ThenXk � X2
k � c � X2k � s �
The receiver decides signal 2 was transmitted if X2 � X1 and otherwise decides s1 transmitted. The probability oferror given signal s2 is transmitted is then
P�error
�H1 � � P
�X2 � X1
�H1 �
To calculate the error probability we need to know the density of Xk. It is easy to see that Xk � c and Xk � s are aGaussian random variables with mean
E Xk � c �Hm � θm � � 12
ρk �m cosθm
E Xk � s �Hm � θm � � 12
ρk �m sinθm
and variance
Var Xk � c �Hm � θm ��� 14
N0E
where E is the energy of the transmitted signal. The density of Xk given Hm can then be calculated in a straightfor-ward manner as
pm�xk ��� 1
2σ2 exp
���µ2 � xk �
2σ2� I0 � � xkµ
σ2
� � x � 0
where
µ∆� µ2
c � µ2s
µc∆� 1
2ρk �m cosθk
µs∆� 1
2ρk �m sinθk
σ2 ∆� 14
N0E
Involved calculation then yields
P�error
�H1 � � P
�X2 � X1
�H1 � � Q
�a � b � � 1
2exp
� � � a2 � b2 ��� 2 � I0�ab �
where
a ��
E2N0
�1 ��� 1 � � ρ � 2 �
b ��
E2N0
�1 � � 1 � � ρ � 2 �
and
Q�a � b ���
∞
b2 � 2 exp� � � a2
2� x � � I0
� � 2xa � dx
and is called Marcum’s Q function.
5-6 CHAPTER 5. NONCOHERENT RECEIVERS
4 6 8 10 12 14 16 1810
−6
10−5
10−4
10−3
10−2
10−1
100
EB/N
0 (dB)
Pe,b ρ=1
ρ=.2ρ=.4
ρ=.6
ρ=.8
ρ=0
Figure 5.1: Performance of Nonorthogonal Signals with Noncoherent Demodulation
3. Error Probability for M-orthogonal Signals in AWGN
In this section we derive the error probability for M-ary orthogonal signal in additive white Gaussian noise. Thetransmitted signal is one of M signals, s0
�t � � � � � � sM 1
�t � where
si�t ��� � 2Pa j
�t � cos
�2π fct �� 0 � t � T �
where T
0a j�t � ai
�t � dt �
�0 i �� jT i � j �
The optimal receiver computes
Xc � j �
T
0r�t ��
2T
a j�t � cos
�2π fct � dt
Xs � j �
T
0r�t � � 2Ta j
�t � sin
�2π fct � dt
for j � 0 � 1 ��� � � � M � 1. These are further processed by computing Y j � X2c � j � X2
s � j.The statistics for the receiver output (assuming signal si
�t � is transmitted are
X jc � � E δi j cosφi � nc � N � � Eδi j cosφi � N0
2
�X js � � E δi j sinφi � n � � N � � Eδi j sinφi � N0
2
�j � 0 � 2 � � ��� � M � 1 �
3. 5-7
Let Yj � X2jc � X2
js . Then we need to determine the probability that Y0 which corresponds to nonzero mean randomvariables is less than Y1 � Y2 ��� � � � YM 1 which correspond to zero mean random variables.
Pc � i � P�Yj � Yi � j �� i
�si transmitted �
� E P �Yj � Yi � j �� i
�Yi � si trans � �
� E ∏j �� i
P�Yj � Yi
�si trans � Yi � �
�
fs�yi � P �
Yj � yi�sitransmitted � � M 1 dyi
�
fs�yi � Fn
�yi � �
�M 1 � dyi �
where Fs�y � is the distribution function of Y j given signal s j
�t � was transmitted and Fn
�y � ( fn
�y � ) is the distribution
(density) function of Yi given signal s j�t � was transmitted with i �� j. Doing an integration by parts we find
Pc � i � �M � 1 �
∞
0Fs�y � Fn
�y � � � M 2 � fn
�y � dy
Thus from the appendix we find that the error probability is given as
Pc � i � �M � 1 �
∞
0 1 � Q
� � 2EN0
� � 2y � � 1 � e y � � M 2 � e ydy �
This expression is not too difficult to evaluate for most values of M. For small values of M and alternative (andbetter known) expression can be derived as follows.
Define Z j � � X2jc � X2
js . Then
Pc � i � P�Z j � Zi � j �� i
�si transmitted �
� E P �Z j � Zi � j �� i
�Zi � si trans � �
� E ∏j �� i
P�Z j � Zi
�si trans � Zi � �
�
p�zi � P �
Z j � zi � � M 1 dzi
Doing a change of variables (ν � u2
2σ2 � dν � uduσ2 ) we obtain
P�Z j � zi � �
u � zi
uσ2 e u2 � 2σ2
du �
ν � z2i� 2σ2
e νdν
� 1 � e z2i� 2σ2
Pc � i �
∞
0p�zi � 1 � e z2
i� 2σ2 � M 1 dzi
�
∞
0p�zi ��� M 1
∑l � 0
� � 1 � l � M � 1l
�e lz2
i� 2σ2 � dzi
�M 1
∑l � 0
� � 1 � l � M � 1l
� ∞
0p�zi � e lz2
i� 2σ2
dzi
∞
0p�zi � e lz2
i� 2σ2
dzi �
∞
0
zi
σ2 e ���z2i µ2 2σ2 I0
� µzi
σ2
�e lz2
i� 2σ2
dzi
� e µ2 � 2σ2
∞
0
zi
σ2 e � l � 1 � z2i� 2σ2
I0
� µzi
σ2
�dzi
5-8 CHAPTER 5. NONCOHERENT RECEIVERS
Do another change of variables, (wi � � l � 1 zi, dwi � � l � 1 dzi) we get
∞
0p�zi � e lz2
i� 2σ2
dzi � e µ2 � 2σ2
∞
0
wi
σ2 � l � 1e w2
i� � 2σ2 � I0 � µwi
� l � 1 σ2
� dwi
� l � 1
� e µ2 � 2σ2 1�l � 1 �
wi
σ2 e w2i
2σ2 I0 � µwi
� l � 1 σ2
�dwi
Let µ̂ � µ�l � 1
� Then
∞
0p�zi � e lz2
i� 2σ2
dzi � e µ2 � 2σ2eµ̂2 � 2σ2
l � 1
wi
σ2 e � w2i µ̂2
2σ2 � I0 � µ̂wi
σ2
�dwi� ��� �� 1
�exp � µ̂2 µ2
2σ2 �l � 1 � �
exp � µ2
l 1 µ2
2σ2 �l � 1
� exp
�� lµ2
2�l � 1 � σ2
� 1l � 1
�
Substituting this into the expression for the probability of correct, we obtain
Pc � i � M 1
∑l � 0
� � 1 � l � M � 1l
� exp � lµ2
2�l � 1 � σ2 �
l � 1 �where
µ2
2σ2 � E2
2 N0E2
� EN0
�
Thus
Pc � i � 1 �M 1
∑l � 1
� � 1 � l M 1l �
l � 1exp
�� l�
l � 1 �EN0
�Pe � i � 1 � Pc � i � M 1
∑l � 1
� � 1 � l � 1 M 1l �
l � 1exp
�� l
l � 1EN0
�� 1 � Pc � i � M 1
∑l � 1
� � 1 � l � 1 M 1l �
l � 1exp
�� l
l � 1Eb log2
�M �
N0
�Pe � 1
M
M
∑i � 2
Pe � i � Pe � iWe can also determine the bit error probability from the symbol error probability. If M � 2k then for each
symbol transmitted k bits of information are being transmitted. Because of the symmetry
Pe � b � 12
MM � 1
Pe � iFor M � 2 the error probability takes a particularly simple form, namely
Pe � b � Pe � i � 12
e Eb� 2N0 � M � 2 �
4. 5-9
Eb � N0 (dB)
Pb
1
10 1
10 2
10 3
10 4
10 5
10 6
10 7
10 8
10 9
10 10
� 5 � 1 � 59 0 5 10 15
M � 2
4
8
16
32
64M � ∞
Figure 5.2: Bit error probability for M-ary orthogonal signal with noncoherent detection
The limiting behavior of the error probability for M-ary orthogonal signals for M large with noncoherent demodu-lation is the same as the limiting performance of coherent demodulation.
limM � ∞
Pe�M � �
�1 � Eb � N0 � ln
�2 ��� � 1 � 59dB
0 � Eb � N0 � ln�2 ��� � 1 � 59dB
limM � ∞
Pe � b � M � ��
1 � 2 � Eb � N0 � ln�2 ��� � 1 � 59dB
0 � Eb � N0 � ln�2 ��� � 1 � 59dB
4. Error Estimates for Repetition Codes with Noncoherent Reception
Consider transmitting one of two codewords of length L using binary frequency shift keying (orthogonal) and non-coherent reception. The optimum receiver would compute the log-likelihood ratio and compare to zero (for equallylikely codewords) to determine which of the two codewords was transmitted. Assume that the first codeword isthe all zero vector (0,0,...,0) of length L and the other codeword is the all one vector (1,1,...,1) of length L. If thetwo codewords agreed in some positions then we would not need to process the received signal over the interval oftime where they agreed since no information can be gained about which signal was transmitted from the receivedsignal in that interval.
The transmitted signal would be
5-10 CHAPTER 5. NONCOHERENT RECEIVERS
s0�t ���
L 1
∑l � 0
� 2Pcos�ω0t � θ0 � l � pT
�t � lT �
or
s1�t ���
L 1
∑l � 0
� 2Pcos�ω1t � θ1 � l � pT
�t � lT �
where θi � l are independent identically distributed random variables with uniform distribution unknown to the re-ceiver. The received signal is
r�t ���
�s0�t ��� n
�t � H0 true
s1�t ��� n
�t � H1 true
The receiver processes the signals using noncoherent matched filters; that is the received signal is multiplied byexp jω0t then passed through a filter with impulse response pT
�t � , sampled and then the magnitude is formed.
Denote by Y0 � l the output of the noncoherent matched filter
Y0 � l � � ∞ ∞r�s � pT
�s � lT � exp
�jω0
�s � lT � � ds
� 2and
Y1 � l � � ∞ ∞r�s � pT
�s � lT � exp
�jω1
�s � lT � � ds
� 2The statistics of Y0 � l � Y1 � l were calculated in the previous section. Because of orthogonality of the received signalsand that the noise is white the joint statistics of Y � �
Y0 � 1 ��� � � � Y0 � L � Y1 � 1 ��� � � � Y1 � L � factor (conditioned on either of thetwo hypotheses) as
p�y0 � 1 ��� � � � y0 � L � y1 � 1 � � � � � y1 � L � �Hk ���
L
∏l � 1
p�y0 � l �Hk � p � y1 � l �Hk � k � 0 � 1 �
The log-likelihood ratio is then
Λ � logp�y�H1 �
p�y�H0 � � log
L
∏l � 1
p�y0 � l �H1 � p � y1 � l �H1 �
p�y0 � l �H0 � p � y1 � l �H0 �
�L
∑l � 1
logp�y0 � l �H1 � p � y1 � l �H1 �
p�y0 � l �H0 � p � y1 � l �H0 �
�L
∑l � 1
log�
p�y0 � l �H1 � p � y1 � l �H1 � � � log
�p�y0 � l �H0 � p � y1 � l �H0 � �
Substituting in the appropriate density functions yields
Λ �L
∑l � 1
log I0� µ � y1 � l
σ2 � � � log I0� µ � y0 � l
σ2 � �
The optimum receiver is thus
L
∑l � 1
log I0� µ � y1 � l
σ2 � �H1�
�
H0
L
∑l � 1
log I0� µ � y0 � l
σ2 � �
which is quite complex in implementation. It would be useful to have suboptimum receivers which are easierto implement but have nearly optimum performance. Before we suggest some suboptimal receivers is would be
4. 5-11
useful to get estimates of the performance of the optimum receiver. The error probability (given H0 say) is easy towrite down but hard to evaluate except for L small. It is
Pe �
yI R1 � p � y �H0 � dy
where I R1 � is the region where the log-likelihood ratio is positive. This is a 2L dimensional integral. The Chernoffbound to the error probability can be expressed as
Pe � DL
where
D �
y
�p�y�H0 � p � y �H1 � dy �
For the additive white Gaussian channel
D � ∞
0
12σ2 exp
� � 12
� yσ2 �
µ2
σ2 � � � I0� µ � y
σ2 � dy� 2
�� ∞
0exp
� � � w � Γ � 2 � � � I0� � 4Γw � dw � 2
and Γ � E � N0. This is much more computationally attractive than the exact expression. An asymptotic form forlow signal-to-noise ratios is not to difficult to compute.
D � 1 � � Γ2
� 2
Γ small �For hard decisions
D � 2�
p�1 � p � � p � 1
2e Γ
2
which for low signal-to-noise ratios becomes
D � 1 � � Γ2 � 2
� 2
Γ small �
Thus for low signal-to-noise ratios hard decisions cost � 2=1.5dB.Now let us consider some suboptimal receivers. First note that µ � σ2 � 2 � N0. So the argument to the Bessel
function will be small if the noise power is large (low signal-to-noise ratio) and will be large if the noise poweris small (high signal-to-noise ratio). Also note that (see Abramowitz and Stegun Handbook of MathematicalFunctions)
I0�x ��� 1 � x2
4� x small
log I0�x ��� x2
4� x small
I0�x ��� ex
� 2πx� x large
Using the approximation for log I0 in the optimum decision rule we get
L
∑l � 1
� µ2y1 � l4σ4 �
H1� �H0
L
∑l � 1
� µ2y0 � l4σ4 �
L
∑l � 1
y1 � l H1� �H0
L
∑l � 1
y0 � l
5-12 CHAPTER 5. NONCOHERENT RECEIVERS
So for small signal-to-noise ratios the optimum receiver is the square-law combining receiver.Now consider when the argument to the Bessel function is large. The optimum decision rule then becomes
L
∑l � 1
� µ � y1 � l2σ2 � � log � 2πµ � y1 � l � σ2 � � µ � y0 � l
2σ2 ��� log � 2πµ � y0 � l � σ2
H1�
�
H0
0
Let wi � l � yi � l � � 2σ2 � then the decision rule is
L
∑l � 1
� �4Γw1 � l � � log � 2π
�4Γw1 � l � � �
4Γw0 � l ��� log � 2π�
4Γw0 � l H1�
�
H0
0
Note that the average value of W given signal present is Γ � 1 while the average value of W given no signal is 1.For very large Γ the terms
� �4Γw � dominates the terms of the form log
�2π � 4Γw and thus the optimum decision
rule is
L
∑l � 1
� � w1 � l � H1�
�
H0
L
∑l � 1
� � w0 � l �Thus the decision rule for very high signal-to-noise ratio is to add the square-roots of the noncoherent matchedfilter outputs.
It is of interest to analyze the performance of these suboptimal receivers. The receiver for very low signal-to-noise ratios is (relatively) easy to analyze. First let us normalize the density for the sum of the squares of 2Lrandom variables. Let
W0 � 12σ2
L
∑l � 1
X2c � 0 � l � X2
s � 0 � land similarly for W1. Then the density of W is given by
fW0
�w0�H0 � �
� w0
Γ� �
L 1 � � 2exp
� � � w0 � Γ � � IL 1� �
4w0Γ � w � 0
fW0
�w0�H1 � � w
�L 1 �
0�L � 1 � ! exp
� � w0 � w � 0
and similarly for W1.
Pe � 1 � P�W0 � W1
�H0 �
� 1 � ∞
0fW0
�w0�H0 � P �
W1 � w0�H0 � dw0
� 1 �
∞
0fW0
�w0�H0 � 1 �
L 1
∑m � 0
wm0
m!e w0 � dw0
Pe � 12
exp� � Γ
2� L 1
∑i � 0
�Γ � 2 � i
i!�L � i � 1 � !
L 1
∑j � i
�j � L � 1 � !�
j � i � !2 j � L 1
For L � 1 the above becomes
Pe � 12
e Γ � 2where Γ � E � N0. The Chernoff bound can also be calculated for square-law combining.
5. Primer on sums of squares of Gaussian random variables
First we derive the density for the sum of the squares of two Gaussian random variables. Let
Xc � N�µc � σ2 �
Xs � N�µs � σ2 �
5. 5-13
with Xc � Xs independent. Let µ2 � µ2c � µ2
s and
Y � X2c � X2
s �Then
P�Y � y � �
1
2πσ2 exp
�� 1
2σ2 � � xc � µc � 2 � �xs � µs � 2 � � dxcdxs
x2c � x2
s � y
�
12πσ2 exp
�� 1
2σ2 x2c � x2
s � 2�xcµc � xsµs ��� µ2 � � dxcdxs
x2c � x2
s � y
�
12πσ2 exp � � 1
2σ2 x2c � x2
s � 2µ � x2c � x2
s
x2c � x2
s � y
cos�φ � γ ��� µ2 � � dxcdxs
where φ � tan 1 xsxc
and γ � tan 1� µs
µc
�.
�
r2 � � y
2π
φ � 0
r2πσ2 exp
�� r2
2σ2 �µrσ2 cos
�φ � γ ��� µ2
2σ2 � � drdφ
�
r � � y
rσ2 exp
� � r2
2σ2� e µ2 � 2σ2 1
2π
2π
φ � 0exp � µr
σ2 cos�φ � γ � dφdr� ��� �
I0 � µrσ2 �
xcµc � xsµs � βcos�φ � γ �
� β cosφcosγ � sinφsin γ �
φ � tan 1 xs
xc
tanφ � xsxc
cosφ � xc� x2c � x2
s
sinφ � xs� x20 � x2
s
cosγ � µc�µ2
c � µ2s
sinγ � � µs�µ2
c � µ2s
β � � µ2c � µ2
s � x2c � x2
s
xcµc � xsµs � � µ2c � µ2
s � x2c � x2
s cos�φ � γ � �
φ � tan 1 � xs
xc
�γ � tan 1 � � µs
µc
�P
�Y � y � �
r� �
y
rσ2 exp
�� r2 � µ2
2σ2� I0
� µrσ2
�dr
Let u � r2 then 0 � r � � y is equivalent to u � y. Also du � 2rdr.
P�Y � y � �
u � y
12σ2 exp
�� u � µ2
2σ2� I0 � µ � u
σ2
�du
fY�y � � 1
2σ2 exp
�� y � µ2
2σ2� I0 � µ � y
σ2
�
5-14 CHAPTER 5. NONCOHERENT RECEIVERS
A change of variables makes for a cleaner expression: Let W � Y � � 2σ2 � . Then fW�w ��� 2σ2 fY
�2σ2w �
fW�w ��� exp
� � � w � Γ � � I0
�� 4Γw
�where Γ � µ2 � � 2σ2 � . (If the receiver does this normalization then it must know the power density of the noise).Now let Z � � Y . Then
P�Z � z � � P � � Y � z � P � Y � z2 �
FZ�z � � FY
�z2 �
fZ�z � � fY
�z2 � � 2z �
� zσ2 exp
�� z2 � µ2
2σ2� I0
� µzσ2
�µ � 0 � fY
�y ��� 1
2σ2 e y � 2σ2
fZ�z ��� z
σ2 e z2 � 2σ2
Using the fact that a density must integrate to one we can derive an useful integral.
∞
0
rσ2 exp
� � r2 � 2σ2 � exp� � αr2 � I0
�rβ � dr � 1
1 � 2σ2αexp
� σ2β2
1 � 2σ2α�
Generalization:
Xc � i � N�µc � i � σ2 � i � 1 � 2 � � � � � L
Xs � i � N�µs � i � σ2 � i � 1 � 2 ��� � � � L
with Xc � i � Xs � i independent. Let Λ � ∑Li � 1 µ2
c � i � µ2s � i and
Y �L
∑i � 1
X2c � i � X2
s � i �Then
fY�y ��� 1
2σ2 exp� � � y � Λ
2σ2
� � � yΛ
� �L 1 � � 2
IL 1� � yΛ
σ2 �
FY�y ��� 1 � QL � � Λ
σ� � y
σ�
where
QL�a � b ��� Q
�a � b ��� exp
���a2 � b2 � � 2 � L 1
∑k � 1
� ba� kIk
�ab �
and
Q�a � b ��� exp
� � � a2 � b2 ��� 2 � ∞
∑k � 1
� ba� kIk
�ab �
For Λ � 0
fY�y � � 1
2σ2 exp� � � y
2σ2
� � � y2σ2
� �L 1 � 1�
L � 1 � !FY�y � � 1 � exp
� � � y2σ2
� � L 1
∑k � 0
1k!
� y2σ2
� k
5. 5-15
Let Z � � Y then
fZ�z � � zL
σ2Λ�L 1 � � 2 exp
� � � z2 � Λ2σ2
� � IL 1� z � Λ
σ2 �
FZ�z � � 1 � QL � � Λ
σ� zσ�
For Λ � 0 we obtain
fZ�z � � z2L 1
2L 1σ2L�L � 1 � ! exp
� � z2
2σ2 �FZ�z � � 1 � exp
� � z2
2σ2 � L 1
∑l � 0
�z2 � � 2σ2 ��� l
l!
Consider two random variables Z1 and Z2 where Z1 has distribution given above with Λ1 � 0 and with differentvariances σ1 and σ2. Assume that they are independent. We wish to determine the probability that Z1 � Z2.
P�Z1 � Z2 � �
∞
z2� 0
P�Z1 � z2 � fZ2
�z2 � dz2
�
∞
z2� 0
FZ1
�z2 � fZ2
�z2 � dz2
�
∞
z2� 0 1 � exp
� � z22
2σ21
� L 1
∑l � 0
�z2
2 � � 2σ21 ��� l
l! � fZ2
�z2 � dz2
� ∞
z2� 0 1 � exp
� � z22
2σ21
� L 1
∑l � 0
�z2
2 � � 2σ21 ��� l
l! �zL
2
σ22Λ
�L 1 � � 2 exp
� � z22 � Λ2σ2
2
� IL 1� z2 � Λ
σ22
� dz2
� 1 �L 1
∑l � 0
1�2σ2
1 � l l!exp
� � � Λ2σ2
2
� � ∞
z � 0exp
� � z2
2σ21
� z2
2σ22
�zL � 2l
σ22Λ
�L 1 � � 2 IL 1
� z � Λσ2
2
� dz
� 1 �L 1
∑l � 0
1�2σ2
1 � l l!σ22Λ
�L 1 � � 2 exp
� � � Λ2σ2
2
� � ∞
z � 0e α2z2
zL � 2lIL 1�γz � dz
where
α2 � 1
2σ21� 1
2σ22
γ � � Λ � σ22
The integral may be evaluated as (see Lindsey, Watson)
∞
z � 0e α2z2
zL � 2lIL 1�γz � dz � l!γL 1
2Lα2�L � l � eγ2 � � 4α2 � l
∑k � 0
� l � L � 1l � k
� γ2 � � 4α2 � � kk!
Thus
P�Z1 � Z2 � �
L 1
∑l � 0
1�2σ2
1 � l l!σ22Λ
�L 1 � � 2 exp
� � � Λ2σ2
2
� � l!γL 1
2Lα2�L � l � eγ2 � � 4α2 �
5-16 CHAPTER 5. NONCOHERENT RECEIVERS
l
∑k � 0
� l � L � 1l � k
� γ2 � � 4α2 � � kk!
�L 1
∑l � 0
1�2σ2
1 � lσ22Λ
�L 1 � � 2 exp
� � � Λ2σ2
2
� � l!γL 1
2Lα2�L � l � eγ2 � � 4α2 �
l
∑k � 0
� l � L � 1l � k
� γ2 � � 4α2 � � kk!
�L 1
∑l � 0
1�2σ2
1 � lσ22Λ
�L 1 � � 2 exp
�γ2 � � 4α2 � � Λ
2σ22
� γL 1
2Lα2�L � l �
l
∑k � 0
� l � L � 1l � k
� γ2 � � 4α2 � � kk!
α2 � σ21 � σ2
2
2σ21σ2
2
γ2 � � 4α2 � ��Λ � σ4
2 � 2σ21σ2
2
σ21 � σ2
2
� Λσ21
2σ22
�σ2
1 � σ22 �
γ2 � � 4α2 � � Λ2σ2
2
��Λ � σ4
2 � 2σ21σ2
2
σ21 � σ2
2
� Λ2σ2
2
� Λ2σ2
2
�σ2
1
σ21 � σ2
2
� 1 �� Λ
2�σ2
1 � σ22 �
P�Z1 � Z2 � � exp
� � Λ2�σ2
1 � σ22 �
� L 1
∑l � 0
1�2σ2
1 � lσ22Λ
�L 1 � � 2 γL 1
2Lα2�L � l �
l
∑k � 0
� l � L � 1l � k
� γ2 � � 4α2 � � kk!
� e Λ2 � σ
21 σ2
2 � σ2
1
σ21 � σ2
2
� L L 1
∑l � 0
�σ2
2
σ21 � σ2
2
� l l
∑k � 0
� l � L � 1l � k
� γ2 � � 4α2 � � kk!
� e Λ2 � σ
21 σ2
2 � σ2
1
σ21 � σ2
2� L L 1
∑l � 0
�σ2
2
σ21 � σ2
2� l l
∑k � 0
� l � L � 1l � k
� 1k!
�Λσ2
1
2σ22
�σ2
1 � σ22 �� k
6. Frequency Shift Keying (FSK)
Frequency shift keying communicates information by transmitting different frequencies. It can be demodulatednoncoherently (by measuring the received energy at the different frequencies). It performance is worse than coher-ently demodulated signals but may be simpler.
b�t � �
∞
∑l � ∞
bl pT�t � lT �� bl �
� � 1 � � 1 �
6. 5-17
�b�t �
VCOs�t � ��� �
� �
n�t �
��r
�t �
Figure 5.3: FSK Modulator
s�t � � � 2P
∞
∑l � ∞
cos�2π�fc � b
�t � ∆ f � t � θ � pT
�t � lT �
where ∆ f is half the difference between the two transmitted frequencies and θ is an unknown (to the receiver)phase. We let f0 � f � ∆ f and f1 � f � ∆ f . When bi � � 1 then a signal at frequency f1 is transmitted. Whenbi � � 1 then a signal at frequency f0 is transmitted. The two frequencies f0 and f1 are separated far enough tomake the two signals orthogonal. (Minimum shift keying has the minimum separation in order to make the signalsorthogonal).
The receiver decides signal � 1 was transmitted if�Y 1
� � �Y1�and otherwise decides signal 1. The random
variables at the output of the low pass filters are
Xc � 1 � iT � � � Eδ�bi 1 � 1 � cos
�θ ��� ηc � 1
Xs � 1 � iT � � � Eδ�bi 1 � 1 � sin
�θ ��� ηs � 1
Xc � 1�iT � � � Eδ
�bi 1 � � 1 � cos
�θ ��� ηc � 1
Xs � 1�iT � � � Eδ
�bi 1 � � 1 � sin
�θ ��� ηs � 1
where δ�a � b � � 1 if a � b and is zero otherwise. In the absence of noise (ηx � i � 0) it is easy to see that when
bi 1 ��� 1 that Y1 � � E and Y 1 � 0. The error probability of binary FSK is
Pe � b � 12
e Eb� 2N0 �
5-18 CHAPTER 5. NONCOHERENT RECEIVERS
Noncoherent Demodulator
r�t �
�
�
����
����
�
�
�2 � T cos
�2π�fc � ∆ f � t �
�2 � T sin
�2π�fc � ∆ f � t �
�
�
LPF
LPF
�
�
�
�
��
��
t � iT
t � iT
�
�
Xs � 1
Xc � 1
� � 2
� � 2
�
�
�+
�Y 1
� 2
�
�
����
����
�
�
�2 � T cos
�2π�fc � ∆ f � t �
�2 � T sin
�2π�fc � ∆ f � t �
�
�
LPF
LPF
�
�
�
�
��
��
t � iT
t � iT
�
�
Xs � 1
Xc � 1
� � 2
� � 2
�
�
�+
�Y1� 2
Figure 5.4: Noncoherent Demodulator
7. 5-19
Figure 5.5: Output Densities For Noncoherent Receivers.
Figure 5.6: Density for Y1 � Y 1 given +1 Transmitted.
7. Differential Phase Shift Keying (DPSK)
�b � t � DifferentialEncoder
a � t � � ����
�
�2Pcos � 2π fct � θ �
�s � t �
n � t �
��r � t �
Modulator
b�t � �
∞
∑l � ∞
bl pT�t � lT �� bl �
� � 1 � � 1 � �a�t � �
∞
∑l � ∞
al pT�t � lT �� al �
� � 1 � � 1 � �Differential Encoder is such that
bl � 1 � al � al 1
bl � � 1 � al � � al 1 �
For example
l � � � � 2 � 1 0 1 2 3 ��� �bl � 1 1 1 � 1 1 � 1al � 1 1 1 1 � 1 � 1 1
s�t � � � 2Pa
�t � cos
�2π fct � θ ��
Optimum Demodulator
Figure 5.7: Error Probability of FSK with Noncoherent Detection.
5-20 CHAPTER 5. NONCOHERENT RECEIVERS
Figure 5.8: Error Probability for Differential Phase Shift Keying
r�t �
�
�
����
����
�
�
�2 � T cos
�2π fct �
�2 � T sin
�2π fct �
�
�
LPF
LPF
��� �
��� �
�
�
�
�
t � iT Xs�iT �
t � iT Xc�iT �
DelayT
DelayT
�
�
����
����
�
�Zi
� � 0 dec bi 1 � � 1� 0 dec bi 1 � � 1
Xc�iT ��� � Eai 1 cosθ � ηc � i �
Xs�iT ��� � Eai 1 sinθ � ηs � i �
The random variables ηc � i and ηs � i are independent identically distributed Gaussian random variables with mean 0and variance N0 � 2. Thus
Zi � Xc�iT � Xc
���i � 1 � T ��� Xs
�iT � Xs
� �i � 1 � T �
Zi � Re W �iT � W � ���
i � 1 � T � �where W
�iT ��� Xc
�iT � � jXs
�iT � . The error probability for DPSK is
Pe � b � 12
e E � N0 �Thus differential phase shift keying is 3dB better than FSK with noncoherent detection. However, errors tend tooccur in pairs.
To derive the above expression for DPSK consider the low pas filter with impulse response h�t ��� pT
�t � . The
output of the lowpass filters can be expressed as
Xc�t � �
∞ ∞
�2 � T cosωcτh
�t � τ � r � τ � dτ
7. 5-21
Xc�iT � �
∞ ∞
�2 � T cosωcτpT
�iT � τ � r � τ � dτ
�
iT
�i 1 � T
�2 � T cosωcτ ∞
∑l � ∞
� 2Pal cos�ωcτ � θ � pT
�τ � lT ��� n
�τ � � dτ
�
iT
�i 1 � T � 2P
�2 � Tai 1 cosωcτcos
�ωcτ � θ � dτ � ηc � i
nc � i is Gaussian random variable, mean 0 variance N0 � 2. Assuming ωcT � 2πn
Xc�iT ��� � Eai 1 cosθ � ηc � i
Similarly
Xs�iT ��� � Eai 1 sinθ � nc � i
ThusZi � Xc
�iT � Xc
���i � 1 � T ��� Xs
�iT � Xs
� �i � 1 � T �
Note that if we write W�iT � � Xc
�iT � � jXs
�iT � that Zi � Re W �
iT � W � ���i � 1 � T � � . It is clear that this represents
the phase difference between two consecutive symbols.Let
U1 � Xc�iT ��� Xc
���i � 1 � T �
2
U2 � Xs�iT ��� Xs
� �i � 1 � T �
2
U3 � Xc�iT � � Xc
���i � 1 � T �
2
U4 � Xs�iT � � Xs
� �i � 1 � T �
2
Zi � U21 � U2
2 � �U2
3 � U24 �
Assume bi 1 � � 1 so that ai 1 � ai 2 then
Pe � 0 � P�Z � 0
�ai 1 � ai 2 �
� P � U21 � U2
2 � U23 � U2
4�
U1 � N�µ1 � σ2 �
U2 � N�µ2 � σ2 �
µ1 � 12� 2P
�ai 1T cosθ � ai 2T cosθ �
� 12� 2P
�ai 1 � ai 2 � T cosθ
µ2 � 12� 2P
�ai 1 � ai 2 � T sinθ
σ2 � 14 N0T � N0T �
� 12
N0T
U3 � N�µ3 � σ2 � U4 � N
�µ4 � σ2 �
µ3 � 0 � µ4 � 0 �
E U1U2 � � E
�Xc�iT ��� Xc
���i � 1 � T �
2� � Xs
�iT ��� Xs
���i � 1 � T �
2�
5-22 CHAPTER 5. NONCOHERENT RECEIVERS
� 14
E Xc�iT � Xs
�iT ��� Xc
�iT � Xs
� �i � 1 � T �
� Xc� �
i � 1 � T � Xs�iT ��� Xc
� �i � 1 � T � Xs
� �i � 1 � T � �
E Xc�iT � Xs
�jT � � � E Xc
�iT � � E Xs
�jT � � since independent
E U1U2 � � � E Xc�iT � � � E Xc
� �i � 1 � T � �
2
� � E Xs�iT � ��� E Xs
���i � 1 � T � �
2
�� E U1 � E U2 � � U1 � U2 independent
Similarly�U1 � U3 � independent
�U2 � U3 � independent
�U3 � U4 � independent
�U1 � U4 � independent U2 � U4 � inde-
pendentThus U2
1 � U22 is independent of U2
3 � U24 . From the results derived for noncoherent FSK it is easy to show that
P�U2
1 � U22 � U2
3 � U24 � � 1
2e E � N0 �
Thus differential phase shift keying is 3dB better than FSK with noncoherent detection. However, errors tend tooccur in pairs.
8. Problems
1. A noncoherent communication system employs the signals
si�t ��� Asin
�π
tT� pT
�t � cos
�2π fit � θi � i � 0 � 1
where 2π fiT is an integer multiple of 2π and θi is unknown. Determine the optimal receiver for this system.Determine the error probability. (Assume additive white Gaussian noise).
2. Consider the following DPSK (differential phase shift keying) communication system. The information tobe transmitted is given by the following data waveform
b�t ���
∞
∑l � ∞
bl pT�t � lT �
where bl is a sequence of i.i.d. random variables with P�bl � � 1 � � P
�bl � � 1 � � 1 � 2. The differential
encoder produces another data stream
a�t ���
∞
∑l � ∞
al pT�t � lT �
where al � al 1 if bl � 1 otherwise al �� al 1. The transmitted waveform is
s�t ���
∞
∑l � ∞
� 2Pal pT�t � lT � cos
�2π fct � θ �
where θ is unknown to the receiver and P is the power. Consider the receiver shown below. Determinethe error probability for deciding on bl. Express your answer in terms of the signal energy and the noise(additive white Gaussian) spectral density N0 � 2 Hint: Use the following transformation of variables
U1 � Xc�iT ��� Xc
� �i � 1 � T �
2
U2 � Xs�iT ��� Xs
� �i � 1 � T �
2
U3 � Xc�iT � � Xc
� �i � 1 � T �
2
U4 � Xs�iT � � Xs
� �i � 1 � T �
2
then write the error probability in terms of these variables.
8. 5-23
r�t �
�
�
����
����
�
�
�2 � T cos
�2π fct �
�2 � T sin
�2π fct �
�
�
LPF
LPF
� ���
����
�
�
�
�
t � iT Xs�iT �
t � iT Xc�iT �
DelayT
DelayT
�
�
����
����
�
�Zi
� � 0 dec bi 1 ��� 1� 0 dec bi 1 � � 1
3. Determine the loss in signal-to-noise ratio for using hard decisions (versus soft decisions) on an additivewhite Gaussian noise channel using orthogonal signals and noncoherent detection (based on the cutoff rate)at low signal-to-noise ratios.
4. (a) Consider a communication system using DPSK. Show that over anytime duration of 2T seconds thetwo possible transmitted signals are orthogonal. That is, given an arbitrary reference signal (phase) in thetime period
� �l � 2 � T � � l � 1 � T � , show that the two waveforms corresponding to bl � � 1 and bl � � 1 are
orthogonal.(b) Using the above or otherwise, derive the optimal receiver for detecting data bit bl .(c) Show that the optimum receiver is identical to that derived in class (if it is not already in that form).
5. (a) Consider a communication system with modulation and demodulation producing a discrete time (mem-oryless) channel with transition probabilities p
�b�a � , a � A � b � B. If the decoder knows these transition
probabilities then we showed in class (and in the homework) that codes of rate R exist to transmit informa-tion with error probability upper bounded by
Pe � 2 N�R0 R �
where R0 is called the cutoff rate. In this problem consider a communication system with a receiver that doesnot know exactly the transition probabilities. The receiver uses transition probability p
� �b�a � as if they were
the actual transition probabilities. (The channel statistics and the receiver statistics are memoryless). Showthat there exist codes of rate R such that
Pe � 2N�R �0 R �
whereR�0 � � lnJ
�0 �
5-24 CHAPTER 5. NONCOHERENT RECEIVERS
J�0 � min
λ � 0minp�x � E J �λ � X1 � X2 � �
and
J�λ�a1 � a2 ��� ∑
b � B
p�b�a1 ��p� �
b�a2 �
p� �
b�a1 � � λ �
(b) Show that when p� �
b�a � � p
�b�a � the optimizing value for λ is 1/2 so the result reduces to the normal
cutoff rate.(Hint: Use Cauchy inequality:
� ∑a � A
p�a � w � a � u � a ��� 2
� � ∑a � A
p�a � w2 � a � � � ∑
a � A
p�a � u2 � a � �
with equality if w�a � � cu
�a � for some positive constant c and use w
�a � � �
p�b�a ��� � 1 λ � � 2 and u
�a � ��
p�b�a ��� λ � 2).
(c) Show that for a noncoherent detection of binary (orthogonal) FSK the exponential density functions (usedin the receiver) results in square law combining (as opposed to maximum likelihood combining) and that thecutoff rate has the form
R�0 � 1 � log2
�1 � D
� ��Determine an explicit form for D
�.
6. Let H0 and H1 be two events. Let pi�x1 � ��� � � xn � be the conditional density of X1 � ��� � � Xn given Hi occurred.
(a) Assume that given Hi,�X j � n
j � 1 is a sequence of i.i.d. Gaussian random variables with mean� � 1 � i � E
and variance N0 � 2. Use the Chernoff bound to show that
P�
p1�X � � p0
�X � �H0 � � e nE � N0
(b) Now let X1 ��� ����� Xn be independent discrete random variables taking values +1 and � 1 with
P�Xi ��� 1
�H0 � � p �
P�Xi � � 1
�H0 � � 1 � p �
P�Xi � � 1
�H1 � � p �
P�Xi ��� 1
�H1 � � 1 � p �
Thus if the number of components of x equal to +1 is d then p0�x � � pd � 1 � p � n d and p1
�x � � pn d � 1 � p � d .
Use the Chernoff bound to show that
P�p1�X � � p0
�X � �H0 � � e n
� ln � 4p�1 p � �
(c) Again let X1 � ��� � � Xn be independent discrete random variables with Xi taking nonnegative integer valuesonly. Let
p0�x1 � � ����� xn ���
n
∏i � 1
λxi0 e λ0
xi!� xi � 0 � 1 � i � n
and
p1�x1 ��� ����� xn ���
n
∏i � 1
λxi1 e λ1
xi!
If λ1 � λ0 find the best Chernoff bound on
Pe � 0 � P�
p1�X � � p0
�X � �H0 �
andPe � 1 � P
�p0�X � � p1
�X � �H1 ���
8. 5-25
Let s�i be the optimal value of s for minimizing the bound to Pe � i. Show s
�1 � 1 � s
�0.
(d) Again let X1 ��� ����� X2n be independent discrete random variables with Xi taking nonnegative integer valuesonly. Let
p0�x1 � � ��� � x2n ���
n
∏i � 1
λxi1 e λ1
xi!
2n
∏i � n � 1
λxi0 e λ0
xi!� xi � 0 � 1 � i � 2n
and
p1�x1 ��� ����� x2n ���
n
∏i � 1
λxi0 e λ0
xi!
2n
∏i � n � 1
λxi1 e λ1
xi!
If λ1 � λ0 find the best Chernoff bound on
Pe � 0 � P�
p1�X � � p0
�X � �H0 �
andPe � 1 � P
�p0�X � � p1
�X � �H1 ���
7. (a) Show that ∑∞i � 1 φi
�t � φ �i � s � � δ
�t � s � for any complete orthonormal set of functions φi
�t � . ( Hint: Show
that the above function satisfies the definition of a delta function, that ∞ ∞
f�t � δ � t � s � dt � f
�s �
for all functions f�t � ).
(b) Let K�s � t � be a positive covariance function of the zero mean Gaussian random process X
�t � . Let K 1 � 2
be the negative square root of this function as defined in class. Show that the process
Y�s � �
K 1 � 2 � s � t � X �
t � dt
is a white Gaussian noise process. (You need to show that E Y � s � Y � � t � ��� δ�t � s � ).
8. Let K�s � t � be a real covariance matrix of a random process with eigenvalues λi and (real) eigenfunctions φi.
Define (as in class) K2 � s � t � as
K2 � s � t ���
K�s � u � K �
u � t � du
and Kn � s � t � as
Kn � s � t ���
Kn 1 � s � u � K �u � t � du �
Also define eK � s � t � as
eK�s � t � � ∞
∑n � 0
Kn � s � t �n!
�Show that
eK�s � t � � ∞
∑i � 1
eλiφi�s � φi
�t ��