Difference Threshold Test for M -ary Orthogonal FSK Signaling in Rayleigh Fading Channels
Young Gil Kim
Sang Wu Kim
Dept. of Electrical Engineering
Dept. of Electrical Engineering
University of Seoul
Korea Advanced Institute of Science and Technology
ygkim@ieee.org
swkim@san.kaist.ac.kr
Abstract-We present the difference threshold test (DTT) for
M -ary orthogonal frequency shift keying (FSK) signaling in
Rayleigh flat fading channels. The proposed DTT declares an
erasure whenever the difference between the largest and the
second largest energy detector outputs does not exceed a fixed
threshold. We show that the DTT is an approximation of the
Bayesian erasure test, but provides almost the same performance as the Bayesian erasure test. For (7, 3) and (31, 15)
Reed-Solomon (RS)-coded systems, the DTT provides power
gains of 0.8 dB and 0.4 dB over the ratio threshold test (RTT),
respectively. We prove that the limiting performance of the
DTT approaches that of the RTT for very large M .
I. I NTRODUCTION
The use of side information permits identification and erasure of symbols that have been impaired by channel effects
such as fading, jamming, background noise, etc. Since more
erasures can be corrected than errors, it is advantageous to determine the reliability of the received symbols and erase unreliable symbols prior to the decoding process. There are a number of methods for generating side information, and their performances have been analyzed in [1]-[7]. Particularly in [7],
the difference threshold test (DTT) for binary orthogonal signaling was proposed.
In this paper, we present the DTT for M -ary orthogonal frequency shift keying (FSK) signaling in Rayleigh fading channels. The DTT declares an erasure whenever the difference
between the largest and the second largest energy detector outputs does not exceed a fixed threshold, γ. Hard decision corresponds to a special case of the DTT with γ set to zero. We
show that the DTT is an approximation of the Bayesian erasure test, but provides almost the same performance as the
Bayesian erasure test. A similar reliability test, called ratio
threshold test (RTT), generates side information about the reliability of the received symbol by comparing the ratio of the
largest to the second largest decision statistic [1],[8]. We show
that the DTT provides power gains of 0.8 dB and 0.4 dB over
the RTT for (7, 3) and (31, 15) Reed-Solomon (RS)-coded
systems, respectively, in Rayleigh flat fading channels. We
prove that the limiting performance of the DTT approaches
that of the RTT for very large M .
This paper consists of six sections. In Section II, we describe the system model. In Section III, we show that the DTT
is an approximation of the Bayesian erasure test. In Section
IV, we derive the probabilities of symbol error, symbol erasure, and decoding error probability. Also, the limiting performance of the DTT for very large M is analyzed. In Section
V, numerical results are presented. Finally, conclusions are
provided in Section VI.
II. S YSTEM M ODEL
Let
si (t) = Acos(2πfi t),
i = 1, 2, ..., M,
0≤t≤T
(1)
be the transmitted signal, where A is the signal amplitude and
fi is the ith tone frequency. We assume that the tone frequencies {fi } are chosen such that the signals {si (t), i =
1, 2, ..., M } are noncoherently orthogonal, i.e., |fi+1 − fi | =
1/T . The received signal r(t) at the input of the energy detector given that s1 (t) is transmitted is
r(t) = gAcos(2πf1 t + θ) + n(t),
0≤t≤T
(2)
where g is Rayleigh distributed, θ is uniformly distributed over
[0, 2π), and n(t) is the white Gaussian noise of one-sided
spectral density N0 . We assume that g and θ are independent,
and are also independent of n(t). Then, the noncoherent mth
energy detector output Zm , given that s1 (t) is transmitted, is
Zm
2
2 T
=
r(t) cos(2πfm t)dt
T 0
2
2 T
+
r(t) sin(2πfm t)dt
(3)
T 0
 2

A2 T

g
cos
θ
+
n

c,1
2


2
(4)
=
A2 T

+
, m=1

2 g sin θ + ns,1



m = 1,
(nc,m )2 + (ns,m )2 ,
where
nc,m
=
ns,m
=
2
T
2
T
T
0
0
n(t) cos(2πfm t)dt,
m = 1, 2, ..., M
n(t) sin(2πfm t)dt,
m = 1, 2, ..., M
T
are independent Gaussian random variables each with mean
zero and variance N0 /2. Since g is Rayleigh distributed and
θ is uniformly distributed over [0, 2π], g cos θ is a Gaussian
random variable with mean zero and variance E[g 2 ]/2. Thus,
0-7803-7802-4/03/$17.00 © 2003 IEEE
2743
the conditional probability density function (pdf) PZm (z|1) of
Zm , given that s1 (t) is transmitted, is
1 −z/2σ2
1,
e
m=1
2σ12
(5)
PZm (z|1) =
1 −z/2σ02
e
, m = 1
2σ 2
0
where
2σi2
=
Ēs + N0 , i = 1
i=0
N0 ,
III. D IFFERENCE T HRESHOLD T EST
In this section, we derive the DTT for M -ary orthogonal
signaling from the Bayesian erasure test. With the Bayesian
erasure test, the received signal is erased if the a posteriori
probability of correct decision is less than a given threshold
φ [2]:
(7)
Pr(s = sd |Z) ≤ φ
where s ∈ {s1 , s2 , ..., sM } is the transmitted signal
when represented in an orthogonal expansion form, Z =
(Z1 , Z2 , ..., ZM ), and
max
m=1,2,...,M
(8)
Zm .
We define log-likelihood ratio (LLR) Λm as
Λm = ln
Pr(s = sd |Z)
.
Pr(s = sm |Z)
(9)
Then
e−Λm Pr(s = sd |Z)
M
e−Λm Pr(s = sd |Z)
=
=
m=1
Pr(s = sm |Z)
M
M
e−Λm
=
Assuming that each symbol is transmitted with equal probability, the LLR Λm in (9) can be expressed as
Pr(Z|s = sd )
.
Pr(Z|s = sm )
(12)
Since the conditional joint pdf of Z is
P (Z|s = sm ) =

M −1
 Z
1
1
1
m
· 2 exp − 2 − 2
2
 2σ1
2σ0
2σ1
2σ0
M
i=1,i=m
the LLR Λm in (12) is given by

M
1  Zm
1 Zd
1
Λm =
+
Zi − 2 − 2
2
2
2 σ1
σ0
σ1
σ0
i=1,i=m
1 1
1
=
− 2 (Zd − Zm ).
2 σ02
σ1
− 12
Pr(s = sd |Z) ≈ 1/ 1 + e
where
d = arg
1
σ2
0
−
1
σ2
1
max
m=1,2,...,M,m=d
(Zd −Zd )
Zm
Zd − Zd ≤ γ ,
.
(16)
(17)
(18)
where Zd − Zd is the difference between the largest and the
second largest energy detector outputs.
IV. P ERFORMANCE A NALYSIS
The probability of not correctly decoding PE for (N, K)
RS-coded system with errors-and-erasures decoding is
N −K−e
N
−K 2
n
[per (γ)]e [pe (γ)]t [pc (γ)]n−e−t
e,
t
e=0
t=0
(19)
where per (γ) is the probability of symbol erasure, pe (γ) is the
probability of symbol error, pc (γ) is the probability of correct
decision, and N = M − 1. It follows from (18) that the DTT
generates an erasure if
PE = 1−
max Zk ≤ Zm < max Zk + γ
k=m
k=m
Zm ≥ max Zk + γ
k=m
(20)
(21)
for some m = 1, when s1 (t) is transmitted. Notice that
conventional hard decision corresponds to the special case of
γ = 0. Because of the symmetry of the signal set, the probability of symbol error is independent of the transmitted signal if
each signal is transmitted equally likely. Thus, for our analysis
we assume that s1 (t) is transmitted without loss of generality.
Then, the probability of symbol error pe (γ) is
(13)


Zi ,

pe (γ) = Pr(∪M
m=2 Zm ≥ max Zk + γ|1)
k=m
∞
− z−γ
2
= (M − 1)
(1 − e 2σ1 )

Zi 
i=1,i=d
(14)
2744
(22)
γ
− z−γ
2
M
is the argument of the second largest decision statistic.
Since (16) is a monotonically increasing function of Zd − Zd ,
the DTT is to erase the symbols if
(11)
1.
m=1
Λm = ln
for some parameter γ ≥ 0 and each m ∈ {1, 2, ..., M }, and
produces an error if
m=1
Pr(s = sd |Z)
Thus,
(10)
Pr(s = sm |Z)
(15)
m=1
(6)
where Ēs = E[g 2 ]A2 T /2 is the average received code symbol energy. If an RS code of rate r is used, then the average
received information bit energy Ēb is Ēs /(r log2 M ).
d = arg
From (11) and (14), we get
M
− 12 σ12 − σ12 (Zd −Zm )
0
1
e
.
Pr(s = sd |Z) = 1/
1 − 2σz2
e 0 dz
2σ02
M
−2 M −2
= (M − 1)e−γ/N0
k
·(1 − e
·
2σ
0
)M −2
(23)
k=0
k
(−1)
.
(k + 1){(k + 1)(Ēs /N0 + 1) + 1}
(24)
Note that if we let γ = 0 in (24), then we get the wellknown probability of symbol error formula with hard decision
in Rayleigh fading channels [9]. The probability of correct
decision pc (γ) is
Pr(Z1 ≥ max Zm + γ|1)
m=1
∞
− z−γ
1 − z2
2
(1 − e 2σ0 )M −1 2 e 2σ1 dz
=
2σ1
γ
M
−1
M − 1
γ
− E¯ +N
= e s 0
k
pc (γ)
=
(25)
(26)
k=0
(−1)k
·
1 + k(Ēs /N0 + 1)
The right-hand side of (33) is the maximum possible code rate
for error-free communication. Following the same argument
in [8], the optimum code rate ropt that minimizes (Ēb /N0 )min
can be found from (34) as
−1
e , γ>0
ropt =
(36)
0.46, γ = 0.
(27)
For values of r given in (36), the (Ēb /N0 )min is
e ln 2(= 2.75 dB), DTT
(Ēb /N0 )min =
4.79(= 6.80 dB), hard decision.
(37)
This is equivalent to the minimum required Ēb /N0 for errorfree communication with the RTT [8].
whereas the probability of symbol erasure per (γ) is
per (γ) = 1 − pc (γ) − pe (γ).
In Appendix A, we derive the probability of symbol error
pe (γ) and the probability of correct decision pc (γ) in detail.
Limit Analysis
In this subsection, we show that the performance of the
DTT is the same as that of the RTT as M goes to infinity.
The following propositions are proved in Appendix B.
Proposition 1:
lim pc (γ) = 2−(γ
+1)N0 /r E¯b
M →∞
where γ =
V. N UMERICAL R ESULTS
(28)
(29)
γ
N0 ln M .
Figure 1 is a plot of the probability of not correctly decoding PE versus Ēb /N0 for a (7, 3) RS code with M = 8 and
(31, 15) RS code with M = 32 in a Rayleigh fading channel.
The probability of not correctly decoding with the Bayesian
erasure test is evaluated by computer simulation. The thresholds of the DTT and the RTT are optimally chosen for each
Ēb /N0 . We find that the DTT provides power gains of 0.8 dB
and 0.4 dB over the RTT for (7, 3) and (31, 15) RS codes, respectively, when the probability of not correctly decoding PE
is 10−3 . Also, the DTT provides almost the same performance
as the Bayesian erasure test. The power gain provided by the
DTT over the RTT diminishes with increasing M (or N ), and
in the limit of large M both the DTT and the RTT provides the
same performance in accordance with the limit analysis.
Proposition 2:
lim pe (γ) =
M →∞
VI. C ONCLUSION
0,
γ>0
N
− rE¯0
1−2
b
,
γ = 0.
(30)
Notice that γ = 0 and r = 1 in (30) yields the same result
given in [10]. The asymptotic probability of symbol erasure
can be obtained by applying Propositions 1 and 2 in (28),
which results in
¯
1 − 2−(γ +1)N0 /rEb , γ > 0
lim pe (γ) =
(31)
0
γ = 0.
M →∞
Thus, for large M , all errors becomes erasures if γ > 0. Following the analysis in [8], error free communication is possible with the DTT and errors-and-erasures decoding provided
r
1 − per (γ) − 2pe (γ)
−(γ +1)N /rE¯
0
b
2
, γ>0
=
−N0 /r E¯b
− 1 γ = 0.
2·2
<
or equivalently
Ēb
N0
>
γ +1
r log2 (1/r)
1
r{1−log2 (r+1)}
= (Ēb /N0 )min .
γ>0
γ = 0.
We presented the DTT for M -ary orthogonal FSK signaling
in Rayleigh flat fading channels. The proposed DTT declares
an erasure whenever the difference between the largest and the
second largest energy detector outputs does not exceed a fixed
threshold. We have shown that the DTT is an approximation
of the Bayesian erasure test, but provides almost the same performance as the Bayesian erasure test. For (7, 3) and (31, 15)
RS-coded systems, the DTT provides power gains of 0.8 dB
and 0.4 dB over the RTT, respectively. The power gain provided by the DTT over the RTT diminishes with increasing M
(or N ), and in the limit of large M both the DTT and the RTT
provides the same performance.
(32)
(33)
(34)
(35)
R EFERENCES
[1] A. J. Viterbi, “A robust ratio-threshold technique to mitigate
tone and partial band jamming in coded MFSK systems,” IEEE
MILCOM, pp. 22.4.1-22.4.5, Oct. 1982.
[2] C. W. Baum and M. B. Pursley, “Bayesian methods for erasure
insertion in frequency-hop communication systems with partialband interference,” IEEE Trans. Commun., vol. 40, No. 7, pp.
1231-1238, Jul. 1992.
[3] C. M. Keller and M. B. Pursley, “Diversity combining for channels with fading and partial-band interference,” IEEE J. Select.
Areas Commun., vol. SAC-5, No. 2, pp. 248-260, Feb. 1987.
2745
[4] L. Yang and L. Hanzo, “Performance analysis of coded M ary orthogonal signaling errors-and-erasures decoding over
frequency-selective fading channels,” IEEE J. Select. Areas
Commun., vol. 19, No. 2, pp. 211-221, Feb. 2001.
[5] L. Yang, K. Yen and L. Hanzo, “A Reed-Solomon coded DSCDMA system using noncoherent M -ary orthogonal modulation over multipath fading channels,” IEEE J. Select. Areas
Commun., vol. 18, No. 11, pp. 2240-2251, Nov. 2000.
[6] L. Yang and L. Hanzo, “Low-complexity erasure insertion in
frequency-hopping spread-spectrum communications subject to
fading and partial-band interference,” IEEE Globecom, Nov. 2529, 2001, San Antonio, Texas, USA, pp. 796-800.
[7] I. Su and J. Wu, “Difference threshold test for asynchronous
BFSK frequency-hopped multiple access systems over Rician
channels,” Electronic Letters, vol. 35, No. 18, pp. 1512-1513,
Sept. 1999.
[8] S. W. Kim and W. Stark, “Performance limits of Reed-Solomon
coded CDMA with orthogonal signaling in a Rayleigh-fading
channel,” IEEE Trans. Commun., vol. 46, No. 9, pp. 1125-1134,
Sep. 1998.
[9] J. G. Proakis, Digital Communications. pp. 834, 4th Ed.,
McGraw-Hill, 2001.
[10] M. V. Hegde and W. Stark, “Asymptotic performance of M -ary
orthogonal signals in worst case partial-band and Rayleigh fading,” IEEE Trans. Commun., vol. 36, pp. 989-992, Aug. 1988.
A PPENDIX B
We prove the two propositions in Section 5. The derivation
follows the same mathematical techniques in [8].
Proposition 1:
lim pc (γ) = 2−(γ
(38)
= (M − 1) · Pr(Z2 ≥ max Zk + γ|1)
(39)
k=2
∞
1 − z2
= (M − 1)
Pr(max Zk ≤ z − γ|1) 2 e 2σ0 dz
k=2
2σ0
0
∞
− z−γ
2
= (M − 1)
(1 − e 2σ1 )
where
·(1 − e
2σ
0
)M −2
−
1
e
2σ02
= (M − 1)e−γ/N0
z
2σ 2
0
M
−2 =
→
k=0
k
(49)
ln L = (M − 1) ln(1 − M −au )
1
1
= (M − 1)(−M −au − M −2au − M −3au − · · ·)
2
3
(50)
≈ −M (1−au) , for large M
implying
0,
1,
lim L =
M →∞
u < 1/a
u > 1/a.
(51)
Therefore, we get
lim pc (γ)
∞
=
M →∞
1/a
−
2
1 − u −γ/2σ12
e 2 ·e
du
2
(γ +1)N0
r E¯b
(52)
(53)
.
Proposition 2:
(41)
lim pe (γ) =
0,
γ>0
N
− rE¯0
1−2
M →∞
, γ = 0.
b
(54)
Proof: Letting u = (z − γ)/σ12 in (23) gives
The probability of correct decision pc (γ) for the DTT is
(42)
pc (γ) = Pr(Z1 ≥ max Zm + γ|1)
m=1
∞
1 − z2
=
Pr(max Zm ≤ z − γ|1) 2 e 2σ1 dz (43)
m=1
2σ1
0
∞
z
− z−γ
−
1
2
2
=
(1 − e 2σ0 )M −1 2 e 2σ1 dz
(44)
2σ1
γ
M
−1 γ
M −1
−
= e E¯s +N0
k
lim pe (γ) =
∞
(M − 1)M −bu (1 − e−u/2 )
lim b
M →∞
M →∞
0
·(1 − M
−bu M −2
)
−
ln M · e
γ
2σ 2
0
du
(55)
where
k=0
k
(−1)
.
1 + k(Ēs /N0 + 1)
(48)
(40)
dz
(−1)
·
.
(k + 1){(k + 1)(Ēs /N0 + 1) + 1}
·
1
2 ln 2
as M → ∞. If we let L = (1 − M −au )M −1 , then
=
M −2
k
rĒb
1
+
N0
log2 M
rĒb 1
N0 2 ln 2
a
γ
− z−γ
2
(46)
We derive (24) and (27). The probability of symbol error
pe (γ) for the DTT is
k=m
+1)N0 /r E¯b
γ
where γ = N0 ln
M.
Proof: Letting u = (z − γ)/σ12 in (26) gives
∞
1 u − γ2
lim pc (γ) = lim
(1 − M −au ) e− 2 e 2σ1 du (47)
M →∞
M →∞ 0
2
A PPENDIX A
pe (γ) = Pr(∪M
m=2 Zm ≥ max Zk + γ|1)
M →∞
(45)
2746
b
=
→
σ12
ln M
rĒb
.
2N0 ln 2
2σ02
(56)
(57)
as M → ∞. If we let
L = (M − 1)M −bu (1 − M −bu )M −2 ln M · M −γ
−
where M −γ = e
γ
2σ 2
0
(58)
, then for large M
ln L ≈ (1 − bu − γ ) ln M
+ ln(ln M ) + (M − 2) ln(1 − M −bu )
= ln(ln M ) + (1 − bu − γ ) ln M +
(M − 2)
1
1
·(−M −bu − M −2bu − M −3bu − · · ·).
2
3
(59)
(60)
Since −(M − 2)M −bu term in (60) predominates over all others,
if 1 − bu > 0
(61)
lim ln L = −∞,
M →∞
implying
lim L = 0,
(62)
if u < 1/b.
M →∞
Thus, we can write
lim pe (γ) =
∞
(M − 1)M −bu−γ
lim b
M →∞
M →∞
1/b
−u/2
·(1 − e
(63)
) ln M du.
Note that the term (1 − M −bu )M −2 in (55) has in effect been
eliminated from the integrand, but has been compensated for
by increasing the lower limit from zero to 1/b. The compensation is exact as M approaches infinity. Performing integration
in (63) gives
lim pe (γ)
M →∞
=
=
−
N0
lim M −γ (1 − 2 rE¯b )
M →∞
0,
γ>0
N
− rE¯0
1−2
b
, γ = 0.
(64)
(65)
Fig. 1. The probabilities of not correctly decoding for the hard decision, the
RTT, the DTT, and the Bayesian erasure test for (7, 3) and (31, 15) RS-codes
2747