Research ArticleHigh Order Differential Frequency HoppingDesign and Analysis

Yong Li12 Fuqiang Yao2 and Ba Xu2

1 Institute of Communications Engineering PLA University of Science and Technology Nanjing 210007 China2Nanjing Telecommunication Technology Institute Nanjing 210007 China

Correspondence should be addressed to Yong Li liy771121163com

Received 28 November 2014 Accepted 9 June 2015

Academic Editor Marco Mussetta

Copyright copy 2015 Yong Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

This paper considers spectrally efficient differential frequency hopping (DFH) system design Relying on time-frequency diversityover large spectrum and high speed frequency hopping DFH systems are robust against hostile jamming interference Howeverthe spectral efficiency of conventional DFH systems is very low due to only using the frequency of each channel To improve thesystem capacity in this paper we propose an innovative high order differential frequency hopping (HODFH) scheme Unlike intraditional DFH where the message is carried by the frequency relationship between the adjacent hops using one order differentialcoding in HODFH the message is carried by the frequency and phase relationship using two-order or higher order differentialcoding As a result system efficiency is increased significantly since the additional information transmission is achieved by thehigher order differential coding at no extra cost on either bandwidth or power Quantitative performance analysis on the proposedscheme demonstrates that transmission through the frequency and phase relationship using two-order or higher order differentialcoding essentially introduces another dimension to the signal space and the corresponding coding gain can increase the systemefficiency

1 Introduction

As one of the two basicmodulation techniques used in spreadspectrum communications frequency hopping (FH) wasoriginally designed to be inherently secure and reliable underadverse battle conditions for military applications Howeverthere are three major bottlenecks low data rate narrowbandwidth frequency hopping and a low hopping rate whenusing frequency hopping in the high frequency (HF) bandCurrently the data rate of the existing HF frequency hoppingradio is limited to 2400 bps or less the reliable data rateis 1200 bps the highest hopping rate is only tens of hopsper sec and the antijamming capability is weak Undercurrent conditions it is difficult to achieve high-capacitycommunications with HF communications technology whenthere are harsh electromagnetic environments

In the 1990s the Sanders Company from the UnitedStates of America pioneered an enhanced hopping spreadspectrum radio (CorrelatedHopping Enhanced Spread Spec-trum) referred to as CHESS Radio [1 2] It was based on

the differential frequency hopping system and was a goodsolution to improve the data rate and anti-tracking-jammingabilities among other issues Its hopping rate is 5000 hopsper sec and the frequency hopping bandwidth is 2MHzthese were breakthroughs compared to conventional HFcommunication systems However the conventional DFHsystem has low spectral efficiency over large bandwidthTypically conventional DFH systems require large band-width which is proportional to the requirement on thefrequency number Along with the ever increasing demandon inherently secure high data rate wireless communicationsnew techniques that are more efficient and reliable have tobe developed In literature [3] a FH system called modulateddifferential frequency hopping was proposed in this systemsome information was used to produce a pseudo-randomcarrier frequency slot (as in the DFH technology) and otherswere changed into narrowband modulated waveforms (asin the conventional FH technology) The message-drivenfrequency hopping (MDFH) scheme proposed in literature[4] possesses higher spectral efficiency than conventional FH

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 586080 9 pageshttpdxdoiorg1011552015586080

2 Mathematical Problems in Engineering

communication system In literature [5] based on the idea ofMDFH a new scheme was proposed to improve the abilityof anti-fixed-frequency interference In literature [6 7] theantijamming performance of the MDFH was analyzed indepth

In this paper we propose an innovative high orderdifferential frequency hopping (HODFH) scheme The basicidea is that the frequencies and the phases of the signalare numbered and then using the two-order or higherorder differential coding to code the message and the signalnumber The message is parsed into several parts each partwill be transmitted by different order differential codingIn other words the serial message stream is converted intoparallel data stream transmitted by each order differentialcodingAt the receiver the transmitting frequency is capturedusing a filter bank as in the FSK receiver rather than usingthe frequency synthesizer As a result the frequency can beblindly detected at each hop This relaxes the burden of strictfrequency synchronization at the receiver And the phase isestimated as in the PSK receiver Then the frequency and thephase are used to decode the message The system spectralefficiency is significantly improved because the high orderdifferential coding and the phases do not require extra coston bandwidth and power

In this paper in-depth performance analysis is con-ducted based on both theoretical derivation and simulationexamples Our analysis demonstrates that phases essentiallyadd another dimension to the signal space and the highorder differential coding can increase the system spectralefficiency

2 The Concept of High Order DifferentialFrequency Hopping

21 Transmitter Design Unlike the conventional DFH sys-tems in which only the frequencies are coded by the one-order coding the HODFH systems code the frequenciesand their phases by high order coding That is to say inthe conventional DFH system only the carrier frequenciescarry the message however in the HODFH system not onlyfrequencies but also the phases carry the information In aHODFH system we add several phases to the frequenciestherefore the signal number is significantly increased almostwithout increase in bandwidth

Let 119873119888

be the total number of frequencies with1198911 1198912 119891119873

119888

being the set of all frequencies Let119872119896be the

number of initial phases where the frequency is 119891119896in each

hop with 1206011 1206012 120601119872119896

being the set of all phases Define119904(119891

119899119894 120601

119899119895) as the combinated signal of these frequencies and

phases in the 119899th hop it is clear that there are 119873119904= sum

119873119888

119899=1 119872119896

signals The set S with 1199041 1199042 119904119873119904

being its elements isthe set of all the combinated signals

Let 119873119900be the order of the HODFH system we start by

reshaping the one-dimensional set 119878 into119873119900-dimensional set

S1198631times1198632timessdotsdotsdottimes119863119873119900

where the subscript denotes element numberof each dimension and 1198631 times 1198632 times sdot sdot sdot times 119863

119873119900

le 119873119904

denote 1199041198891 1198892 119889119873119900

as the element of the set S1198631times1198632timessdotsdotsdottimes119863119873119900

We assume that 1198631 times 1198632 times sdot sdot sdot times 119863

119873119900

= 119873119904 Divide the

1st order bits 2nd order bits middot middot middot

middot middot middotXn1 Xn2

Xn

XnN119900

Noth order bits

Figure 1 The 119899th block of the information data

data stream into blocks of length 119871 ≜ sum119873119900

119894=1 119861119894 where 119861119894(le

log2119863119894) is the number of bits that is transmitted through the

119894th order differential coding Denote the 119899th block by 119883119899

The block 119883119899can be grouped into 119873

119900vectors denoted by

[1198831198991 1198831198992 119883119899119873

119900

] where 119883119899119894

denotes the 119894th order bitsas shown in Figure 1 We will transmit119883

119899within one hop

The transmitter block diagram of the proposed HODFHscheme is illustrated in Figure 2 Each input data block119883119899 is fed into a serial-to-parallel (SP) converter where

the data bits are split into 119873119900parallel data streams Denote

the signals in the 119899th and (119899 minus 1)th hop by 119904119899

119889119899

1 119889119899

2 119889119899

119873119900

and119904119899minus1119889119899minus11 119889119899minus12 119889

119899minus1119873119900

The index of the 119904119899

119889119899

1 119889119899

2 119889119899

119873119900

is obtained by 119873119900-

order differential coding the index of the 119904119899minus1119889119899minus11 119889119899minus12 119889

119899minus1119873119900

andinformation data block 119883

119899 The differential processing is

called a function119866 and expressed by the following equations

119889119899

1 = 1198661 (119889119899minus11 119889

119899minus12 119889

119899minus1119873119900

1198831198991)

119889119899

119898= 119866

119898(119889

119899

119898minus1 119883119899119898) when 119898 = 2 119873

119900

(1)

Using the index vectors to select the combinated signalthe signal transmitted over the channel is 119904119899

119889119899

1 119889119899

2 119889119899

119873119900

Recall that 119904(119891

119899119894 120601

119899119895) is the combination of the frequency

and phase The transmitted signal in the 119899th hop can beexpressed as

119904119899

119889119899

1 119889119899

2 119889119899

119873119900

= 119904 (119891119899119894 120601

119899119895) = 119860 cos (2120587119891

119899119894119905 + 120601

119899119895)

0 le 119905 lt 119879ℎ

(2)

where119860 is the signal amplitude119891119899119894is the center frequency of

the 119899th hop120601119899119895

is the possible phases of the 119899th hop and119879ℎis

the hop duration The passband waveform of the transmittedsignal may be expressed as

119904 (119905) = radic2119879ℎ

Reinfin

sum

119899=minusinfin

119890minus1198952120587(119891

119899119894+120601119899119895)119905

119892 (119905 minus 119899119879ℎ) (3)

where 119892(119905) is the signal pulse-shaping filter

22 Receiver Design The receiver block diagram of the pro-posed HODFH scheme is illustrated in Figure 3 Recall thatthere are 119873

119888frequencies 1198911 1198912 119891119873

119888

that are extendedout 119873

119904signals First we must detect which frequency is

transmitted For this purpose a bank of 119873119888bandpass filters

(BPF) each centered at 119891119894(119894 = 1 2 119873

119888) with the same

channel bandwidth as the transmitter is deployed at thereceiver front end Since only one frequency band is occupiedat any given hop we simply measure the outputs of bandpassfilters at each possible signaling frequency The actual carrier

Mathematical Problems in Engineering 3

SPconverter

Table look-up

and signal

generation

Delay

s(t)

nth data blockXn

XnN119900

Xn2

Xn1

dnN119900minus1

dn1dn1

dn2

dnminus11 dnminus12 dnminus1N119900

middot middot middot

dnminus1N119900

dnN119900= GN119900

(dnN119900minus1 XnN119900

)

dn2 = G2(dn1 Xn2)

dn1 = G1(dnminus11 dnminus12 dnminus1N119900 Xn1)

Figure 2 The transmitter block diagram of the proposed HODFH scheme

frequency at a certain hopping period can be detected byselecting the one that captures the strongest signal As a resultblind detection of the carrier frequency is achieved at thereceiver

More specifically the received signal is a convolution ofthe transmitted signal 119904(119905)with the channel impulse responseand it can be expressed as

119903 (119905) = int

+infin

minusinfin

ℎ (120591) 119904 (119905 minus 120591) 119889120591 + 119899 (119905) (4)

where ℎ(120591) is the channel impulse response and 119899(119905) denotesadditive white Gaussian noise (AWGN) Accordingly theoutputs of bandpass filters are given by

119911119894(119905) = int

+infin

minusinfin

119902119894(120591) 119903 (119905 minus 120591) 119889120591 for 119894 = 1 119873

119888 (5)

where 119902119894(119905) is the ideal bandpass filter centered at frequency

119891119894 If the channel is ideal that is ℎ(119905) = 120575(119905) then

119911119894(119905) = int

+infin

minusinfin

119902119894(120591) 119904 (119905 minus 120591) 119889120591 + 119906

119894(119905)

for 119894 = 1 119873119888

(6)

where 119906119894(119905) = int

+infin

minusinfin

119902119894(120591)119899(119905minus120591) is the filtered version ofWGN

If the signal-to-noise ratio is sufficiently high and there is nostrong jamming as in most communication systems thereis one and only one significantly strong signal among theoutputs of the filter bank Suppose that the 119897th filter capturesthis distinctive signal during the 119899th hop then the estimatedhopping frequency 119891

119899119894= 119891

119897 The same procedures can be

carried out to determine the carrier frequency at each hopNext the estimated hopping frequency119891

119899119894is used to con-

trol NCOThe output signal multiplied by the received signaland NCO is used to estimate the phase of the transmittedsignal Denote the estimated phase as 120601

119899119895

Then the estimated hopping frequency 119891119899i and the esti-

mated phase 120601119899119895

are used to estimate the transmitted signalDenote the estimated signal as 119904

119899

119889119899

1 119889119899

2 119889119899

119873119900

The estimated

index 119889119899

1 119889119899

119873119900

at the 119899th hop and the estimated index119889119899minus11 119889

119899minus1119873119900

at the (119899 minus 1)th hop are fed into the function119866minus1119898

(1 le 119898 le 119873119900) the inverse function of 119866

119898 to recover the

information data blocksThis processing can be expressed bythe following equations

1198831198991 = 119866

minus11 (119889

119899minus11 119889

119899minus12 119889

119899minus1119873119900

119889119899

1)

119883119899119898

= 119866minus1119898

(119889119899

119898minus1 119889119899

119898) when 119898 = 2 119873

119900

(7)

It then follows that the estimate of the 119899th block 119883119899can

be obtained as119883119899= [119883

1198991 1198831198992 119883119899119873119900

]

3 Bit Error Probability for HODFH

In HODFH system the input bit stream is carried by hoppingfrequency and its phase So firstly we analyze hoppingfrequency symbol error probability and phase symbol errorprobability respectively and then deduce the bit error rate(BER) of the HODFH

31 Symbol Error Probability Firstly we analyze the hoppingfrequency symbol error probability Based on the receiverdesign in HODFH analysis of the hopping frequency symbolerror probability is analogous to that of noncoherent FSKdemodulation For noncoherent detection of 119872

119865-ary FSK

signals the probability of symbol error is given by [8 eqn(5-4-46) page 310]

119901119904FSK (

120576119904

1198730)

=

119872119865minus1

sum

119899=1(minus1)119899+1 (

119872119865minus 1

119899)

1119899 + 1

exp [minus119899120576119904

(119899 + 1)1198730]

(8)

where 1205761199041198730 is the SNR per symbol For a HODFH system

with 119873119888frequencies 119872

119865= 119873

119888 Let 119901

(119865)

119904HODFH denote theprobability of the hopping frequency symbol detection error

4 Mathematical Problems in Engineering

Delay

Selectlargest

Transmitted signal

estimateand table

look-up

PSconverter

BPF f2

BPF f1

r(t)

fni

njdn

1 dn

2 dn

N119900

dnminus1

1 dnminus1

2 dnminus1

N119900

Xn1

Xn2Xn

XnN119900

BPF fN119888

Xn1 = Gminus11 (dnminus11 d

nminus1

2 dnminus1

N119900 d

n

1)

Xn2 = Gminus12 (dn1 d

n

2)

XnN119900= Gminus1

N119900dn

N119900minus1 d

n

N119900

EnergydetectionEnergy

detection

Energydetection

Phas

ede

tect

ion

Figure 3 The receiver block diagram of the proposed HODFH scheme

(corresponding to the symbol error in FSK) in HODFH thenwe have

119901(119865)

119904HODFH (120576119865

119904

1198730)

=

119873119888minus1

sum

119899=1(minus1)119899+1 (

119873119888minus 1119899

)1

119899 + 1exp[minus

119899120576119865

119904

(119899 + 1)1198730]

(9)

where 120576119865

1199041198730 is the SNR of hopping frequency per symbol

Next we focus on the phase symbol error probabilityThephase symbol error probability can be calculated in a similarmanner as that of the 119872

119875-ary PSK The probability of phase

symbol error probability is given by [8 eqn (5-2-56) amp (5-2-55) page 268]

119901119904PSK (

120576119904

1198730) = 1minusint

120587119872119875

minus120587119872119875

119901Θ119903

(Θ119903) 119889Θ

119903 (10)

119901Θ119903

(Θ119903) =

12120587

exp(minus120576119904

1198730sin2Θ

119903)

sdot int

infin

0119881 exp

[[[

[

minus

(119881 minus radic2 (1205761199041198730) cosΘ119903

)

2

2]]]

]

119889119881

(11)

For a HODFH system signal phase detection is imple-mented at the condition that the hopping frequency isdetected correctly for which the probability is (1minus119901

(119865)

119904HODFH)the probability of phase symbol error can be calculated as thatof the119872

119875-ary PSK Let 119901(119875)

119904HODFH denote the probability of thephase symbol detection error (corresponding to the symbolerror in PSK) Recall that there are119872

119896possible phases in each

hopping frequency we have

119901(119875)

119904HODFH (120576119875

119904

1198730) = 1minusint

120587119872119896

minus120587119872119896

119901Θ119903

(Θ119903) 119889Θ

119903 (12)

where 120576119875

1199041198730 is the SNR of phase per symbol and it will be

substituted in (11) when we calculate 119901(119875)

119904HODFHOnly when the hopping frequency and the phase are

detected correctly the transmitted signal can be estimatedcorrectly Since in a HODFH system the transmitted signalcarries the frequency and phase information the SNR of the

hopping frequency per symbol is the same as the SNR ofthe phase that is 120576119865

1199041198730 = 120576

119875

1199041198730 We denote the SNR per

symbol by 1205761199041198730 The overall symbol error probability of the

HODFH scheme is calculated as the combination of 119901(119865)119904HODFH

and 119901(119875)

119904HODFH defined as 119901119904HODFH which can be found in

119901119904HODFH (

120576119904

1198730)

= 1

minus(1minus119901(119865)

119904HODFH (120576119904

1198730))(1minus119901

(119875)

119904HODFH (120576119904

1198730))

= 119901(119865)

119904HODFH (120576119904

1198730)+119901

(119875)

119904HODFH (120576119904

1198730)

minus119901(119865)

119904HODFH (120576119904

1198730) sdot 119901

(119875)

119904HODFH (120576119904

1198730)

(13)

32 Bit Error Probability It is reasonable to assume that theprobability of the index 119889

119899

1198941 le 119894 le 119873

119900 estimated incorrectly

is inversely proportionate to the number of elements at the 119894thdimensionality when the symbol error occurs Let 119901Index119894 bethe error probability of the 119894th dimensional index we have

119901Index119894 (120576119904

1198730) =

119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730) (14)

For the first order bits 1198831198991 it is only decided by

119889119899minus11 119889

119899minus1119873119900

and 119889119899

1 The correct probability of119889119899minus11 119889

119899minus1119873119900

is equal to the symbol correct probability(1 minus 119901

119904HODFH) Recall that there are 1198611 bits at the first orderThe BER of the first order bits denoted as 119901

1198871HODFH isobtained

1199011198871HODFH (

120576119904

1198730) =

21198611minus1

21198611 minus 1(1

minus(1minus119901Index1 (120576119904

1198730))(1minus119901

119904HODFH (120576119904

1198730)))

=21198611minus1

21198611 minus 1(21198631 minus 1

1198631119901119904HODFH (

120576119904

1198730)

minus1198631 minus 11198631

(119901119904HODFH (

120576119904

1198730))

2)

(15)

Mathematical Problems in Engineering 5

For the119898th order bits119883119899119898

(2 le 119898 le 119873119900) it is decided by

119889119899

119898minus1 and 119889119899

119898 Only the 119889

119899

119898minus1 and 119889119899

119898are estimated correctly

the information bits can be recovered correctly Assume thatthe error probability of 119889

119899

119898minus1 and 119889119899

119898is independent This

is insured by the processing of reshaping the 1-dimensionalset 119878 into 119873

119900-dimensional set S

1198631times1198632timessdotsdotsdottimes119863119873119900 Let 119901119894119895Index be the

correct probability of the 119894th and the 119895th dimensional indexsimultaneity we have

119901119894119895

Index (120576119904

1198730) = (1minus119901Index119894 (

120576119904

1198730))

sdot (1minus119901Index119895 (120576119904

1198730))

= (1minus119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730))

sdot (1minus

119863119895minus 1

119863119895

119901119904HODFH (

120576119904

1198730)) = 1minus

119863119895minus 1

119863119895

sdot 119901119904HODFH (

120576119904

1198730)minus

119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730)

+

(119863119894minus 1) (119863

119895minus 1)

119863119894119863119895

(119901119904HODFH (

120576119904

1198730))

2

(16)

Recall that there are 119861119898bits at the119898th order The BER of

the119898th order bits denoted as 119901119887119898HODFH is obtained

119901119887119898HODFH =

2119861119898minus1

2119861119898 minus 1(1minus119901

119898minus1119898Index )

=2119861119898minus1

2119861119898 minus 1(119863119898minus1 minus 1119863119898minus1

119901119904HODFH (

120576119904

1198730)

+119863119898

minus 1119863119898

119901119904HODFH (

120576119904

1198730)

minus(119863

119898minus1 minus 1) (119863119898

minus 1)119863119898minus1119863119898

(119901119904HODFH (

120576119904

1198730))

2)

(17)

Therefore the total BER of a HODFH system denoted as119901119887HODFH can be obtained as

119901119887HODFH (

120576119904

1198730)

= (1198611

sum119873119900

119899=1 119861119899) sdot 119901

1198871HODFH (120576119904

1198730)

+

119873119900

sum

119898=2(

119861119898

sum119873119900

119899=1 119861119899) sdot 119901

119887119898HODFH (120576119904

1198730)

(18)

Since sum119873119900

119899=1 119861119899 = 119871 (18) can be transformed as

119901119887HODFH (

120576119904

1198730) =

1119871

(11986111199011198871HODFH (120576119904

1198730)

+

119873119900

sum

119898=2119861119898119901119887119898HODFH (

120576119904

1198730))

(19)

We assume that one symbol not correctly detected willincur the 119873

119900dimensional index being not correctly esti-

mated That is to say 119901Index119894 and 119901119894119895

Index will satisfy thefollowing equations

119901Index119894 (120576119904

1198730) = 119901

119904HODFH (120576119904

1198730)

119901119894119895

Index (120576119904

1198730)

= (1minus119901119904HODFH (

120576119904

1198730))(1minus119901

119904HODFH (120576119904

1198730))

= 1minus 2119901119904HODFH (

120576119904

1198730)+(119901

119904HODFH (120576119904

1198730))

2

(20)

Substitute (20) into (15) and (17) the uniformof the upperbound of BER for the first and the119898th order bits denoted as119901119880

119887119898HODFH can be obtained as

119901119880

119887119898HODFH =2119861119898minus1

2119861119898 minus 1(2119901

119904HODFH (120576119904

1198730)

minus(119901119904HODFH (

120576119904

1198730))

2) 1 le 119898 le 119873

119900

(21)

When the symbol error probability 119901119904HODFH is small

enough the bit error probability 119901119880

119887119898HODFH can be approx-imated as

119901119880

119887119898HODFH (120576119904

1198730) =

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

1 le 119898 le 119873119900

(22)

Substitute (22) into (19) the upper bound of total BER forthe HODFH system is obtained

119901119880

119887HODFH (120576119904

1198730) =

1119871

119873119900

sum

119898=1119861119898119901119880

119887119898HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(23)

The lower bound of total BER denoted as 119901119871

119887HODFHcan be obtained by assuming that the symbol error onlyincurs bits error in that hopping The error probability ofthe 119894th dimensional index is expressed as (14) and the

6 Mathematical Problems in Engineering

119894th dimensional index error only incurs the 119894th order bitsTherefore the lower bound of total BER is

119901119871

119887HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

(24)

To summarize our discussions above we have the follow-ing

Proposition In HODFH the total BER 119901119887HODFH(1205761199041198730) is

bounded by

1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

le 119901119887HODFH (

120576119904

1198730)

le1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(25)

33 Further Discussions about Bit Error Probability Recallthat there are 119873

119888available frequencies 1198911 1198912 119891119873

119888

forthe HODFH system and they distribute in broad band Itonly occupies narrow band centered at 119891

119894(119894 = 1 2 119873

119888)

available frequencies at any given moment The SNR usuallyis unequal at each narrow band In this occasion the symbolerror probability of the frequency and phase cannot be simplycalculated by (9) and (12) The symbol correct decisionprobability119901

119888119891119894

of the frequency119891119894can be deduced by [8 eqn

(5-4-41) page 309] Because the random variables 119877119894(119894 =

1 2 119873119888) are statistically independent and identically dis-

tributed and we assume that the signals centered at the 119873119888

available frequencies 1198911 1198912 119891119873119888

have equal energy anddifferent noise the probability 119901

119888119891119894

can be expressed as

119901119888119891119894

= int

infin

0

119873119888

prod

119898=1119898 =119894

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)119901

119877119894

(119909) 119889119909 (26)

where 1205902119898

= 121198730119898 is the variance of the AWGN atfrequency 119891

119894and

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)

= int

120590119894119909120590119898

0119901119877119898

(119903119898) 119889119903

119898= 1minus 119890

minus1205902119894119909221205902119898

= 1minus 119890minus1198730119894119909

221198730119898

(27)

Substitute (27) into (26) we obtain the general probability

119901119888119891119894

= int

infin

0

119873119888

prod

119898=2119898 =119894

(1minus 119890minus1198730119894119909

221198730119898) 119901

119877119894

(119909) 119889119909 (28)

We have known that in a HODFH system signal phasedetection is implemented at the condition that the hoppingfrequency centred at 119891

119894is detected correctly for which the

probability is 119901119888119891119894

the probability of phase symbol error canbe calculated as (10) and (11) but the SNR 120576

119875

1199041198730 of phase

per symbol will be substituted by 1205761198730119894 which is the SNRat frequency 119891

119894 Let 119901

(119875119894)

119904HODFH denote the probability of thephase symbol detection error at frequency119891

119894 In this occasion

the overall symbol error probability of the HODFH schemeis calculated by

1199011015840

119904HODFH = 1minus

119873119888

sum

119894=1119901119894sdot 119901

119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

= 1minus1119873119888

119873119888

sum

119894=1119901119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

(29)

where the 119901119894(119894 = 1 2 119873

119888) is the probability that the

signals centered at frequency 119891119894and the 1199011 1199012 119901119873

119888

allequal 1119873

119888 Substitute the probability 119901

119904HODFH by 1199011015840

119904HODFHin (16) and (17) we can calculate the total BER of a HODFHsystem by (18) when it works in the occasion that the119873119888available frequencies 1198911 1198912 119891119873

119888

distribute in broadband and the SNR is not equal at each narrow band centeredat 119891

119894

4 Simulation Results

Without loss of generality in this paper the HODFH systemin the AWGN channel with equal SNR at anywhere wassimulated and its bit error rate (BER) performance wasanalyzed

Example 1 BER performance of the HODFH system inAWGN channel the performance of BER at the informa-tion rates of 9600 bps 19200 bps and 28800 bps in AWGNchannel was compared For the information rate of 9600 bps119873119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and 119861

119896= 1 (119896 =

1 119873119900) For the information rate of 19200 bps 119873

119900= 2

119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 2 (119896 = 1 119873

119900)

For the information rate of 28800 bps 119873119900

= 2 119873119888

= 64119872

119896= 4 (119896 = 1 119873

119888) 1198611 = 4 and 1198612 = 2

Figure 4 is the BER performance of the HODFH systemin AWGN channel The requirement for the energy-per-bitnoise rate (119864

1198871198730) becomes lower when the information

rate increases In practice it is usually possible that onlythe energy-per-symbol noise rate (119864

1199041198730) can be measured

According to formula 1198641199041198730 = 119864

1198871198730 + 10 log(119896) (119896 is the

number of bits per symbol) convert 1198641198871198730 to 119864

1199041198730 and

then the higher 1198641199041198730 is required for the higher information

rate which is reasonable

Example 2 The BER performance of information rate28800 bps at different number of phases assume 119873

119900= 2 and

119861119896= 3 (119896 = 1 119873

119900) The numbers of available frequencies

119873119888are 64 32 and 16 the numbers of phases119872

119896are 4 8 and

16 respectively

Mathematical Problems in Engineering 7

0 1 2 3 4 5 6 7 8 9 10 11

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps No = 2 Nc = 64 Mk = 4 B1 = 4 B2 = 2

Figure 4 The BER performance at information rates of 9600 bps19200 bps and 28800 bps in AWGN channel

0 2 4 6 8 10 12 14 16

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

No = 2 Nc = 64 Mk = 4 Bk = 3

No = 2 Nc = 32 Mk = 8 Bk = 3

No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 5 The BER performance of information rate 28800 bps atdifferent number of phases

As seen in Figure 5 when more phrases are used whilekeeping the information rate the same there is poorerperformance for the HODFH system It will spend about6 dB extra 119864

1198871198730 to obtain the same BER performance

of 10minus3 when the number of phases increases from 8 to16 Therefore the number of phases should be kept to aminimum if the number of transmission signals has reachedthe requirements

Example 3 TheBERperformance of theHODFH system andthe conventional DFH (CDFH) system for the informationrate of 9600 bps the HODFH system has the followingparameters 119873

119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and

0 2 4 6 8 10 12

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

9600bps CDFH Nc = 128 BPH = 2

19200 bps CDFH Nc = 128 BPH = 4

Figure 6 The BER performance of the HODFH system and theCDFH system

119861119896= 1 (119896 = 1 119873

119900) for the information rate of 19200 bps

it has parameters 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) The CDFH has the following

parameters for information rate of 9600 bps 119873119888= 128 and

BPH = 2 for information rate of 19200 bps 119873119888= 128 and

BPH = 4

Figure 6 shows the BER performance comparison ofthe HODFH system and the conventional DFH system forinformation rates of 9600 bps and 19200 bps At the sameinformation rate the HODFH system has a better BERperformance than the conventional DFH system The higherinformation rate is the higher gain is obtained by theHODFH system when compared to the conventional DFHsystem

Example 4 Compare the BER performances of the HODFHsystem at three information rates (9600 bps 19200 bps and28800 bps) with the theoretical BER performance of QPSK8PSK 16PSK 32PSK and 64PSK Assume the HODFHsystem has the following parameters respectively For theinformation rate of 9600 bps119873

119900= 2119873

119888= 128119872

119896= 2 (119896 =

1 119873119888) and 119861

119896= 1 (119896 = 1 119873

119900) For the information

rate of 19200 bps 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) For the information rate of

28800 bps we have three combinations the first is 119873119900

= 2119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

the second is 119873119900= 2 119873

119888= 32 119872

119896= 8 (119896 = 1 119873

119888) and

119861119896

= 3 (119896 = 1 119873119900) and the third is 119873

119900= 2 119873

119888= 16

119872119896= 16 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

Figure 7 presents the BER performance of the HODFHsystem and the theoretical BER performance of QPSK 8PSK16PSK 32PSK and 64PSK When the information rate is9600 bps the HODFH system carries two bits of information

8 Mathematical Problems in Engineering

0 5 10 15 20 25 30

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

QPSK theory8 PSK theory16 PSK theor32 PSK theor64 PSK theor9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 3

28800 bps HODFH No = 2 Nc = 64 Mk = 8 Bk = 3

28800 bps HODFH No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 7 The BER performance of the HODFH system andtheoretical BER performance of the M-PSK

per hop (symbol) which is the same as the QPSK the BERperformance of the HODFH system is worse than the QPSKWhen the information rate is 19200 bps the HODFH systemcarries four bits of information per hop (symbol) which is thesame as the 16PSK but it has better BERperformance than the16PSK When the information rate is 28800 bps the HODFHsystemcarries six bits of information per hop (symbol) whichis the same as the 64PSK but its performance is much betterthan the 64PSK no matter how many phases are used bythe HODFH system The BER performance of the HODFHsystem is better than theM-PSK when the information rate ishighThe carrier frequencies of theHODFH systems are usedto transmit information the HODFH systems have higherinformation rate without extra cost on power and do not needextra cost on power when the numbers of phases adopted bythe HODFH systems and the M-PSK are equal The HODFHsystems have better BER performance at the same energy-per-bit noise rate compared to the PSK

Example 5 Compare the BER performance of informationrate 19200 bps at the occasion that the SNR is equal andunequal at each narrow band Assume that the SNRs at119873

1198882

narrow bands are different with the SNRs at the other 1198731198882

narrow bands Use SNR119889to denote the difference among

the 119873119888SNRs The SNR

119889increases from 2 to 8 by increment

2 Use the average SNR in the broad band as the symbolsignal-to-noise ratio The HODFH system has the followingparameters 119873

119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888) and

119861119896= 2 (119896 = 1 119873

119900)

0 2 4 6 8 10 12 14

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

Equal SNRUnequal SNR SNRd = 2

Unequal SNR SNRd = 4

Unequal SNR SNRd = 6

Unequal SNR SNRd = 8

Figure 8 The BER performance of information rate 19200 bps atSNR

119889= 2 4 6 and 8

Example 6 Compare the BER performance of informationrate 28800 bps at different number of phases when the SNRdifference SNR

119889= 4 and 6 The SNR is similar as Example 5

The HODFH system has the following parameters 119873119900= 2

119873119888= 64 119861

119896= 3 (119896 = 1 119873

119900) and 119872

119896=4 8 and 16 (119896 =

1 119873119888)

As seen in Figure 8 the BER performance is worse whenthe SNR is unequal at each narrow band despite the averageSNR of the unequal signal-to-noise ratio broad band beingthe same as the SNR of the equal signal-to-noise ratio Asthe difference of the SNRs is increasing the performance isdegraded

Figure 9 shows that the BER performance is degradedwhen the number of phases used by the HODFH systemincreases This recalls the conclusion in Example 2 that thenumber of phases should be kept to a minimum if the num-ber of transmission signals has reached the requirements

5 Summary

In the future communication will have higher informationrate requirements and this paper proposed a solution forthis with a novel differential frequency hopping high orderdifferential frequency hopping method This was done byconumbering the frequencies and their phases so that thesystem information rate was improved without having toincrease the frequency resources The BER performance ofthe HODFH system is better than the conventional DFHsystem and M-PSK in the AWGN channel In this workthe number of phases can be adjusted to match the channelconditions to obtain the optimal performance

Although using phases to expand the signal set canimprove the bandwidth efficiency their introduction of the

Mathematical Problems in Engineering 9

0 2 4 6 8 10 12 14 16 18

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus7

10minus6

EbN0 (dB)

SNRd = 4 Mk = 4

SNRd = 4 Mk = 8

SNRd = 4 Mk = 16

SNRd = 6 Mk = 4

SNRd = 6 Mk = 8

SNRd = 6 Mk = 16

Figure 9 BER performance of information rate 28800 bps atdifferent number of phases when SNR

119889= 4 and 6

phases makes the HODFH system more sensitive to fre-quency offset and timing errors That is to say implementa-tion of this system is more complex but it is a price worthpaying for the information rate improvement and increasedperformance

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] D L Herrick and P K Lee ldquoCHESS a new reliable high speedHF radiordquo in Proceedings of the IEEE Military CommunicationsConference (MILCOM rsquo96) pp 684ndash690 McLean Va USA1996

[2] D L Herrick P K Lee and L L Ledlow Jr ldquoCorrelated fre-quency hopping an improved approach to HF spread spectrumcommunicationsrdquo in Proceedings of the Tactical Communica-tions Conference pp 319ndash324 IEEE Fort Wayne Ind USAApril-May 1996

[3] B Yang and Y Shen ldquoModulated (soft-decision) differentialfrequency hopping systemrdquo in Proceedings of the InternationalConference on Information and Communication Technology(ICICT rsquo07) pp 209ndash212 March 2007

[4] Q Ling and T Li ldquoMessage-driven frequency hopping designand analysisrdquo IEEE Transactions on Wireless Communicationsvol 8 no 4 pp 1773ndash1782 2009

[5] D Wang H Zhao and Z Fan ldquoA new scheme for message-driven FH systemrdquo in Proceedings of the International Con-ference on Future Information Technology and ManagementEngineering (FITME rsquo10) pp 395ndash398 October 2010

[6] L Zhang H Wang and T Li ldquoAnti-jamming message-drivenfrequency hopping Part I System designrdquo IEEE Transactionson Wireless Communications vol 12 no 1 pp 70ndash79 2013

[7] L Zhang and T Li ldquoAnti-jamming message-driven frequencyhopping-Part II capacity analysis under disguised jammingrdquoIEEE Transactions on Wireless Communications vol 12 no 1pp P80ndashP88 2013

[8] J G Proakis Digital Communications McGraw-Hill 4th edi-tion 2001

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Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

communication system In literature [5] based on the idea ofMDFH a new scheme was proposed to improve the abilityof anti-fixed-frequency interference In literature [6 7] theantijamming performance of the MDFH was analyzed indepth

In this paper we propose an innovative high orderdifferential frequency hopping (HODFH) scheme The basicidea is that the frequencies and the phases of the signalare numbered and then using the two-order or higherorder differential coding to code the message and the signalnumber The message is parsed into several parts each partwill be transmitted by different order differential codingIn other words the serial message stream is converted intoparallel data stream transmitted by each order differentialcodingAt the receiver the transmitting frequency is capturedusing a filter bank as in the FSK receiver rather than usingthe frequency synthesizer As a result the frequency can beblindly detected at each hop This relaxes the burden of strictfrequency synchronization at the receiver And the phase isestimated as in the PSK receiver Then the frequency and thephase are used to decode the message The system spectralefficiency is significantly improved because the high orderdifferential coding and the phases do not require extra coston bandwidth and power

In this paper in-depth performance analysis is con-ducted based on both theoretical derivation and simulationexamples Our analysis demonstrates that phases essentiallyadd another dimension to the signal space and the highorder differential coding can increase the system spectralefficiency

2 The Concept of High Order DifferentialFrequency Hopping

21 Transmitter Design Unlike the conventional DFH sys-tems in which only the frequencies are coded by the one-order coding the HODFH systems code the frequenciesand their phases by high order coding That is to say inthe conventional DFH system only the carrier frequenciescarry the message however in the HODFH system not onlyfrequencies but also the phases carry the information In aHODFH system we add several phases to the frequenciestherefore the signal number is significantly increased almostwithout increase in bandwidth

Let 119873119888

be the total number of frequencies with1198911 1198912 119891119873

119888

being the set of all frequencies Let119872119896be the

number of initial phases where the frequency is 119891119896in each

hop with 1206011 1206012 120601119872119896

being the set of all phases Define119904(119891

119899119894 120601

119899119895) as the combinated signal of these frequencies and

phases in the 119899th hop it is clear that there are 119873119904= sum

119873119888

119899=1 119872119896

signals The set S with 1199041 1199042 119904119873119904

being its elements isthe set of all the combinated signals

Let 119873119900be the order of the HODFH system we start by

reshaping the one-dimensional set 119878 into119873119900-dimensional set

S1198631times1198632timessdotsdotsdottimes119863119873119900

where the subscript denotes element numberof each dimension and 1198631 times 1198632 times sdot sdot sdot times 119863

119873119900

le 119873119904

denote 1199041198891 1198892 119889119873119900

as the element of the set S1198631times1198632timessdotsdotsdottimes119863119873119900

We assume that 1198631 times 1198632 times sdot sdot sdot times 119863

119873119900

= 119873119904 Divide the

1st order bits 2nd order bits middot middot middot

middot middot middotXn1 Xn2

Xn

XnN119900

Noth order bits

Figure 1 The 119899th block of the information data

data stream into blocks of length 119871 ≜ sum119873119900

119894=1 119861119894 where 119861119894(le

log2119863119894) is the number of bits that is transmitted through the

119894th order differential coding Denote the 119899th block by 119883119899

The block 119883119899can be grouped into 119873

119900vectors denoted by

[1198831198991 1198831198992 119883119899119873

119900

] where 119883119899119894

denotes the 119894th order bitsas shown in Figure 1 We will transmit119883

119899within one hop

The transmitter block diagram of the proposed HODFHscheme is illustrated in Figure 2 Each input data block119883119899 is fed into a serial-to-parallel (SP) converter where

the data bits are split into 119873119900parallel data streams Denote

the signals in the 119899th and (119899 minus 1)th hop by 119904119899

119889119899

1 119889119899

2 119889119899

119873119900

and119904119899minus1119889119899minus11 119889119899minus12 119889

119899minus1119873119900

The index of the 119904119899

119889119899

1 119889119899

2 119889119899

119873119900

is obtained by 119873119900-

order differential coding the index of the 119904119899minus1119889119899minus11 119889119899minus12 119889

119899minus1119873119900

andinformation data block 119883

119899 The differential processing is

called a function119866 and expressed by the following equations

119889119899

1 = 1198661 (119889119899minus11 119889

119899minus12 119889

119899minus1119873119900

1198831198991)

119889119899

119898= 119866

119898(119889

119899

119898minus1 119883119899119898) when 119898 = 2 119873

119900

(1)

Using the index vectors to select the combinated signalthe signal transmitted over the channel is 119904119899

119889119899

1 119889119899

2 119889119899

119873119900

Recall that 119904(119891

119899119894 120601

119899119895) is the combination of the frequency

and phase The transmitted signal in the 119899th hop can beexpressed as

119904119899

119889119899

1 119889119899

2 119889119899

119873119900

= 119904 (119891119899119894 120601

119899119895) = 119860 cos (2120587119891

119899119894119905 + 120601

119899119895)

0 le 119905 lt 119879ℎ

(2)

where119860 is the signal amplitude119891119899119894is the center frequency of

the 119899th hop120601119899119895

is the possible phases of the 119899th hop and119879ℎis

the hop duration The passband waveform of the transmittedsignal may be expressed as

119904 (119905) = radic2119879ℎ

Reinfin

sum

119899=minusinfin

119890minus1198952120587(119891

119899119894+120601119899119895)119905

119892 (119905 minus 119899119879ℎ) (3)

where 119892(119905) is the signal pulse-shaping filter

22 Receiver Design The receiver block diagram of the pro-posed HODFH scheme is illustrated in Figure 3 Recall thatthere are 119873

119888frequencies 1198911 1198912 119891119873

119888

that are extendedout 119873

119904signals First we must detect which frequency is

transmitted For this purpose a bank of 119873119888bandpass filters

(BPF) each centered at 119891119894(119894 = 1 2 119873

119888) with the same

channel bandwidth as the transmitter is deployed at thereceiver front end Since only one frequency band is occupiedat any given hop we simply measure the outputs of bandpassfilters at each possible signaling frequency The actual carrier

Mathematical Problems in Engineering 3

SPconverter

Table look-up

and signal

generation

Delay

s(t)

nth data blockXn

XnN119900

Xn2

Xn1

dnN119900minus1

dn1dn1

dn2

dnminus11 dnminus12 dnminus1N119900

middot middot middot

dnminus1N119900

dnN119900= GN119900

(dnN119900minus1 XnN119900

)

dn2 = G2(dn1 Xn2)

dn1 = G1(dnminus11 dnminus12 dnminus1N119900 Xn1)

Figure 2 The transmitter block diagram of the proposed HODFH scheme

frequency at a certain hopping period can be detected byselecting the one that captures the strongest signal As a resultblind detection of the carrier frequency is achieved at thereceiver

More specifically the received signal is a convolution ofthe transmitted signal 119904(119905)with the channel impulse responseand it can be expressed as

119903 (119905) = int

+infin

minusinfin

ℎ (120591) 119904 (119905 minus 120591) 119889120591 + 119899 (119905) (4)

where ℎ(120591) is the channel impulse response and 119899(119905) denotesadditive white Gaussian noise (AWGN) Accordingly theoutputs of bandpass filters are given by

119911119894(119905) = int

+infin

minusinfin

119902119894(120591) 119903 (119905 minus 120591) 119889120591 for 119894 = 1 119873

119888 (5)

where 119902119894(119905) is the ideal bandpass filter centered at frequency

119891119894 If the channel is ideal that is ℎ(119905) = 120575(119905) then

119911119894(119905) = int

+infin

minusinfin

119902119894(120591) 119904 (119905 minus 120591) 119889120591 + 119906

119894(119905)

for 119894 = 1 119873119888

(6)

where 119906119894(119905) = int

+infin

minusinfin

119902119894(120591)119899(119905minus120591) is the filtered version ofWGN

If the signal-to-noise ratio is sufficiently high and there is nostrong jamming as in most communication systems thereis one and only one significantly strong signal among theoutputs of the filter bank Suppose that the 119897th filter capturesthis distinctive signal during the 119899th hop then the estimatedhopping frequency 119891

119899119894= 119891

119897 The same procedures can be

carried out to determine the carrier frequency at each hopNext the estimated hopping frequency119891

119899119894is used to con-

trol NCOThe output signal multiplied by the received signaland NCO is used to estimate the phase of the transmittedsignal Denote the estimated phase as 120601

119899119895

Then the estimated hopping frequency 119891119899i and the esti-

mated phase 120601119899119895

are used to estimate the transmitted signalDenote the estimated signal as 119904

119899

119889119899

1 119889119899

2 119889119899

119873119900

The estimated

index 119889119899

1 119889119899

119873119900

at the 119899th hop and the estimated index119889119899minus11 119889

119899minus1119873119900

at the (119899 minus 1)th hop are fed into the function119866minus1119898

(1 le 119898 le 119873119900) the inverse function of 119866

119898 to recover the

information data blocksThis processing can be expressed bythe following equations

1198831198991 = 119866

minus11 (119889

119899minus11 119889

119899minus12 119889

119899minus1119873119900

119889119899

1)

119883119899119898

= 119866minus1119898

(119889119899

119898minus1 119889119899

119898) when 119898 = 2 119873

119900

(7)

It then follows that the estimate of the 119899th block 119883119899can

be obtained as119883119899= [119883

1198991 1198831198992 119883119899119873119900

]

3 Bit Error Probability for HODFH

In HODFH system the input bit stream is carried by hoppingfrequency and its phase So firstly we analyze hoppingfrequency symbol error probability and phase symbol errorprobability respectively and then deduce the bit error rate(BER) of the HODFH

31 Symbol Error Probability Firstly we analyze the hoppingfrequency symbol error probability Based on the receiverdesign in HODFH analysis of the hopping frequency symbolerror probability is analogous to that of noncoherent FSKdemodulation For noncoherent detection of 119872

119865-ary FSK

signals the probability of symbol error is given by [8 eqn(5-4-46) page 310]

119901119904FSK (

120576119904

1198730)

=

119872119865minus1

sum

119899=1(minus1)119899+1 (

119872119865minus 1

119899)

1119899 + 1

exp [minus119899120576119904

(119899 + 1)1198730]

(8)

where 1205761199041198730 is the SNR per symbol For a HODFH system

with 119873119888frequencies 119872

119865= 119873

119888 Let 119901

(119865)

119904HODFH denote theprobability of the hopping frequency symbol detection error

4 Mathematical Problems in Engineering

Delay

Selectlargest

Transmitted signal

estimateand table

look-up

PSconverter

BPF f2

BPF f1

r(t)

fni

njdn

1 dn

2 dn

N119900

dnminus1

1 dnminus1

2 dnminus1

N119900

Xn1

Xn2Xn

XnN119900

BPF fN119888

Xn1 = Gminus11 (dnminus11 d

nminus1

2 dnminus1

N119900 d

n

1)

Xn2 = Gminus12 (dn1 d

n

2)

XnN119900= Gminus1

N119900dn

N119900minus1 d

n

N119900

EnergydetectionEnergy

detection

Energydetection

Phas

ede

tect

ion

Figure 3 The receiver block diagram of the proposed HODFH scheme

(corresponding to the symbol error in FSK) in HODFH thenwe have

119901(119865)

119904HODFH (120576119865

119904

1198730)

=

119873119888minus1

sum

119899=1(minus1)119899+1 (

119873119888minus 1119899

)1

119899 + 1exp[minus

119899120576119865

119904

(119899 + 1)1198730]

(9)

where 120576119865

1199041198730 is the SNR of hopping frequency per symbol

Next we focus on the phase symbol error probabilityThephase symbol error probability can be calculated in a similarmanner as that of the 119872

119875-ary PSK The probability of phase

symbol error probability is given by [8 eqn (5-2-56) amp (5-2-55) page 268]

119901119904PSK (

120576119904

1198730) = 1minusint

120587119872119875

minus120587119872119875

119901Θ119903

(Θ119903) 119889Θ

119903 (10)

119901Θ119903

(Θ119903) =

12120587

exp(minus120576119904

1198730sin2Θ

119903)

sdot int

infin

0119881 exp

[[[

[

minus

(119881 minus radic2 (1205761199041198730) cosΘ119903

)

2

2]]]

]

119889119881

(11)

For a HODFH system signal phase detection is imple-mented at the condition that the hopping frequency isdetected correctly for which the probability is (1minus119901

(119865)

119904HODFH)the probability of phase symbol error can be calculated as thatof the119872

119875-ary PSK Let 119901(119875)

119904HODFH denote the probability of thephase symbol detection error (corresponding to the symbolerror in PSK) Recall that there are119872

119896possible phases in each

hopping frequency we have

119901(119875)

119904HODFH (120576119875

119904

1198730) = 1minusint

120587119872119896

minus120587119872119896

119901Θ119903

(Θ119903) 119889Θ

119903 (12)

where 120576119875

1199041198730 is the SNR of phase per symbol and it will be

substituted in (11) when we calculate 119901(119875)

119904HODFHOnly when the hopping frequency and the phase are

detected correctly the transmitted signal can be estimatedcorrectly Since in a HODFH system the transmitted signalcarries the frequency and phase information the SNR of the

hopping frequency per symbol is the same as the SNR ofthe phase that is 120576119865

1199041198730 = 120576

119875

1199041198730 We denote the SNR per

symbol by 1205761199041198730 The overall symbol error probability of the

HODFH scheme is calculated as the combination of 119901(119865)119904HODFH

and 119901(119875)

119904HODFH defined as 119901119904HODFH which can be found in

119901119904HODFH (

120576119904

1198730)

= 1

minus(1minus119901(119865)

119904HODFH (120576119904

1198730))(1minus119901

(119875)

119904HODFH (120576119904

1198730))

= 119901(119865)

119904HODFH (120576119904

1198730)+119901

(119875)

119904HODFH (120576119904

1198730)

minus119901(119865)

119904HODFH (120576119904

1198730) sdot 119901

(119875)

119904HODFH (120576119904

1198730)

(13)

32 Bit Error Probability It is reasonable to assume that theprobability of the index 119889

119899

1198941 le 119894 le 119873

119900 estimated incorrectly

is inversely proportionate to the number of elements at the 119894thdimensionality when the symbol error occurs Let 119901Index119894 bethe error probability of the 119894th dimensional index we have

119901Index119894 (120576119904

1198730) =

119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730) (14)

For the first order bits 1198831198991 it is only decided by

119889119899minus11 119889

119899minus1119873119900

and 119889119899

1 The correct probability of119889119899minus11 119889

119899minus1119873119900

is equal to the symbol correct probability(1 minus 119901

119904HODFH) Recall that there are 1198611 bits at the first orderThe BER of the first order bits denoted as 119901

1198871HODFH isobtained

1199011198871HODFH (

120576119904

1198730) =

21198611minus1

21198611 minus 1(1

minus(1minus119901Index1 (120576119904

1198730))(1minus119901

119904HODFH (120576119904

1198730)))

=21198611minus1

21198611 minus 1(21198631 minus 1

1198631119901119904HODFH (

120576119904

1198730)

minus1198631 minus 11198631

(119901119904HODFH (

120576119904

1198730))

2)

(15)

Mathematical Problems in Engineering 5

For the119898th order bits119883119899119898

(2 le 119898 le 119873119900) it is decided by

119889119899

119898minus1 and 119889119899

119898 Only the 119889

119899

119898minus1 and 119889119899

119898are estimated correctly

the information bits can be recovered correctly Assume thatthe error probability of 119889

119899

119898minus1 and 119889119899

119898is independent This

is insured by the processing of reshaping the 1-dimensionalset 119878 into 119873

119900-dimensional set S

1198631times1198632timessdotsdotsdottimes119863119873119900 Let 119901119894119895Index be the

correct probability of the 119894th and the 119895th dimensional indexsimultaneity we have

119901119894119895

Index (120576119904

1198730) = (1minus119901Index119894 (

120576119904

1198730))

sdot (1minus119901Index119895 (120576119904

1198730))

= (1minus119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730))

sdot (1minus

119863119895minus 1

119863119895

119901119904HODFH (

120576119904

1198730)) = 1minus

119863119895minus 1

119863119895

sdot 119901119904HODFH (

120576119904

1198730)minus

119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730)

+

(119863119894minus 1) (119863

119895minus 1)

119863119894119863119895

(119901119904HODFH (

120576119904

1198730))

2

(16)

Recall that there are 119861119898bits at the119898th order The BER of

the119898th order bits denoted as 119901119887119898HODFH is obtained

119901119887119898HODFH =

2119861119898minus1

2119861119898 minus 1(1minus119901

119898minus1119898Index )

=2119861119898minus1

2119861119898 minus 1(119863119898minus1 minus 1119863119898minus1

119901119904HODFH (

120576119904

1198730)

+119863119898

minus 1119863119898

119901119904HODFH (

120576119904

1198730)

minus(119863

119898minus1 minus 1) (119863119898

minus 1)119863119898minus1119863119898

(119901119904HODFH (

120576119904

1198730))

2)

(17)

Therefore the total BER of a HODFH system denoted as119901119887HODFH can be obtained as

119901119887HODFH (

120576119904

1198730)

= (1198611

sum119873119900

119899=1 119861119899) sdot 119901

1198871HODFH (120576119904

1198730)

+

119873119900

sum

119898=2(

119861119898

sum119873119900

119899=1 119861119899) sdot 119901

119887119898HODFH (120576119904

1198730)

(18)

Since sum119873119900

119899=1 119861119899 = 119871 (18) can be transformed as

119901119887HODFH (

120576119904

1198730) =

1119871

(11986111199011198871HODFH (120576119904

1198730)

+

119873119900

sum

119898=2119861119898119901119887119898HODFH (

120576119904

1198730))

(19)

We assume that one symbol not correctly detected willincur the 119873

119900dimensional index being not correctly esti-

mated That is to say 119901Index119894 and 119901119894119895

Index will satisfy thefollowing equations

119901Index119894 (120576119904

1198730) = 119901

119904HODFH (120576119904

1198730)

119901119894119895

Index (120576119904

1198730)

= (1minus119901119904HODFH (

120576119904

1198730))(1minus119901

119904HODFH (120576119904

1198730))

= 1minus 2119901119904HODFH (

120576119904

1198730)+(119901

119904HODFH (120576119904

1198730))

2

(20)

Substitute (20) into (15) and (17) the uniformof the upperbound of BER for the first and the119898th order bits denoted as119901119880

119887119898HODFH can be obtained as

119901119880

119887119898HODFH =2119861119898minus1

2119861119898 minus 1(2119901

119904HODFH (120576119904

1198730)

minus(119901119904HODFH (

120576119904

1198730))

2) 1 le 119898 le 119873

119900

(21)

When the symbol error probability 119901119904HODFH is small

enough the bit error probability 119901119880

119887119898HODFH can be approx-imated as

119901119880

119887119898HODFH (120576119904

1198730) =

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

1 le 119898 le 119873119900

(22)

Substitute (22) into (19) the upper bound of total BER forthe HODFH system is obtained

119901119880

119887HODFH (120576119904

1198730) =

1119871

119873119900

sum

119898=1119861119898119901119880

119887119898HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(23)

The lower bound of total BER denoted as 119901119871

119887HODFHcan be obtained by assuming that the symbol error onlyincurs bits error in that hopping The error probability ofthe 119894th dimensional index is expressed as (14) and the

6 Mathematical Problems in Engineering

119894th dimensional index error only incurs the 119894th order bitsTherefore the lower bound of total BER is

119901119871

119887HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

(24)

To summarize our discussions above we have the follow-ing

Proposition In HODFH the total BER 119901119887HODFH(1205761199041198730) is

bounded by

1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

le 119901119887HODFH (

120576119904

1198730)

le1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(25)

33 Further Discussions about Bit Error Probability Recallthat there are 119873

119888available frequencies 1198911 1198912 119891119873

119888

forthe HODFH system and they distribute in broad band Itonly occupies narrow band centered at 119891

119894(119894 = 1 2 119873

119888)

available frequencies at any given moment The SNR usuallyis unequal at each narrow band In this occasion the symbolerror probability of the frequency and phase cannot be simplycalculated by (9) and (12) The symbol correct decisionprobability119901

119888119891119894

of the frequency119891119894can be deduced by [8 eqn

(5-4-41) page 309] Because the random variables 119877119894(119894 =

1 2 119873119888) are statistically independent and identically dis-

tributed and we assume that the signals centered at the 119873119888

available frequencies 1198911 1198912 119891119873119888

have equal energy anddifferent noise the probability 119901

119888119891119894

can be expressed as

119901119888119891119894

= int

infin

0

119873119888

prod

119898=1119898 =119894

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)119901

119877119894

(119909) 119889119909 (26)

where 1205902119898

= 121198730119898 is the variance of the AWGN atfrequency 119891

119894and

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)

= int

120590119894119909120590119898

0119901119877119898

(119903119898) 119889119903

119898= 1minus 119890

minus1205902119894119909221205902119898

= 1minus 119890minus1198730119894119909

221198730119898

(27)

Substitute (27) into (26) we obtain the general probability

119901119888119891119894

= int

infin

0

119873119888

prod

119898=2119898 =119894

(1minus 119890minus1198730119894119909

221198730119898) 119901

119877119894

(119909) 119889119909 (28)

We have known that in a HODFH system signal phasedetection is implemented at the condition that the hoppingfrequency centred at 119891

119894is detected correctly for which the

probability is 119901119888119891119894

the probability of phase symbol error canbe calculated as (10) and (11) but the SNR 120576

119875

1199041198730 of phase

per symbol will be substituted by 1205761198730119894 which is the SNRat frequency 119891

119894 Let 119901

(119875119894)

119904HODFH denote the probability of thephase symbol detection error at frequency119891

119894 In this occasion

the overall symbol error probability of the HODFH schemeis calculated by

1199011015840

119904HODFH = 1minus

119873119888

sum

119894=1119901119894sdot 119901

119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

= 1minus1119873119888

119873119888

sum

119894=1119901119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

(29)

where the 119901119894(119894 = 1 2 119873

119888) is the probability that the

signals centered at frequency 119891119894and the 1199011 1199012 119901119873

119888

allequal 1119873

119888 Substitute the probability 119901

119904HODFH by 1199011015840

119904HODFHin (16) and (17) we can calculate the total BER of a HODFHsystem by (18) when it works in the occasion that the119873119888available frequencies 1198911 1198912 119891119873

119888

distribute in broadband and the SNR is not equal at each narrow band centeredat 119891

119894

4 Simulation Results

Without loss of generality in this paper the HODFH systemin the AWGN channel with equal SNR at anywhere wassimulated and its bit error rate (BER) performance wasanalyzed

Example 1 BER performance of the HODFH system inAWGN channel the performance of BER at the informa-tion rates of 9600 bps 19200 bps and 28800 bps in AWGNchannel was compared For the information rate of 9600 bps119873119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and 119861

119896= 1 (119896 =

1 119873119900) For the information rate of 19200 bps 119873

119900= 2

119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 2 (119896 = 1 119873

119900)

For the information rate of 28800 bps 119873119900

= 2 119873119888

= 64119872

119896= 4 (119896 = 1 119873

119888) 1198611 = 4 and 1198612 = 2

Figure 4 is the BER performance of the HODFH systemin AWGN channel The requirement for the energy-per-bitnoise rate (119864

1198871198730) becomes lower when the information

rate increases In practice it is usually possible that onlythe energy-per-symbol noise rate (119864

1199041198730) can be measured

According to formula 1198641199041198730 = 119864

1198871198730 + 10 log(119896) (119896 is the

number of bits per symbol) convert 1198641198871198730 to 119864

1199041198730 and

then the higher 1198641199041198730 is required for the higher information

rate which is reasonable

Example 2 The BER performance of information rate28800 bps at different number of phases assume 119873

119900= 2 and

119861119896= 3 (119896 = 1 119873

119900) The numbers of available frequencies

119873119888are 64 32 and 16 the numbers of phases119872

119896are 4 8 and

16 respectively

Mathematical Problems in Engineering 7

0 1 2 3 4 5 6 7 8 9 10 11

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps No = 2 Nc = 64 Mk = 4 B1 = 4 B2 = 2

Figure 4 The BER performance at information rates of 9600 bps19200 bps and 28800 bps in AWGN channel

0 2 4 6 8 10 12 14 16

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

No = 2 Nc = 64 Mk = 4 Bk = 3

No = 2 Nc = 32 Mk = 8 Bk = 3

No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 5 The BER performance of information rate 28800 bps atdifferent number of phases

As seen in Figure 5 when more phrases are used whilekeeping the information rate the same there is poorerperformance for the HODFH system It will spend about6 dB extra 119864

1198871198730 to obtain the same BER performance

of 10minus3 when the number of phases increases from 8 to16 Therefore the number of phases should be kept to aminimum if the number of transmission signals has reachedthe requirements

Example 3 TheBERperformance of theHODFH system andthe conventional DFH (CDFH) system for the informationrate of 9600 bps the HODFH system has the followingparameters 119873

119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and

0 2 4 6 8 10 12

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

9600bps CDFH Nc = 128 BPH = 2

19200 bps CDFH Nc = 128 BPH = 4

Figure 6 The BER performance of the HODFH system and theCDFH system

119861119896= 1 (119896 = 1 119873

119900) for the information rate of 19200 bps

it has parameters 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) The CDFH has the following

parameters for information rate of 9600 bps 119873119888= 128 and

BPH = 2 for information rate of 19200 bps 119873119888= 128 and

BPH = 4

Figure 6 shows the BER performance comparison ofthe HODFH system and the conventional DFH system forinformation rates of 9600 bps and 19200 bps At the sameinformation rate the HODFH system has a better BERperformance than the conventional DFH system The higherinformation rate is the higher gain is obtained by theHODFH system when compared to the conventional DFHsystem

Example 4 Compare the BER performances of the HODFHsystem at three information rates (9600 bps 19200 bps and28800 bps) with the theoretical BER performance of QPSK8PSK 16PSK 32PSK and 64PSK Assume the HODFHsystem has the following parameters respectively For theinformation rate of 9600 bps119873

119900= 2119873

119888= 128119872

119896= 2 (119896 =

1 119873119888) and 119861

119896= 1 (119896 = 1 119873

119900) For the information

rate of 19200 bps 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) For the information rate of

28800 bps we have three combinations the first is 119873119900

= 2119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

the second is 119873119900= 2 119873

119888= 32 119872

119896= 8 (119896 = 1 119873

119888) and

119861119896

= 3 (119896 = 1 119873119900) and the third is 119873

119900= 2 119873

119888= 16

119872119896= 16 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

Figure 7 presents the BER performance of the HODFHsystem and the theoretical BER performance of QPSK 8PSK16PSK 32PSK and 64PSK When the information rate is9600 bps the HODFH system carries two bits of information

8 Mathematical Problems in Engineering

0 5 10 15 20 25 30

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

QPSK theory8 PSK theory16 PSK theor32 PSK theor64 PSK theor9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 3

28800 bps HODFH No = 2 Nc = 64 Mk = 8 Bk = 3

28800 bps HODFH No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 7 The BER performance of the HODFH system andtheoretical BER performance of the M-PSK

per hop (symbol) which is the same as the QPSK the BERperformance of the HODFH system is worse than the QPSKWhen the information rate is 19200 bps the HODFH systemcarries four bits of information per hop (symbol) which is thesame as the 16PSK but it has better BERperformance than the16PSK When the information rate is 28800 bps the HODFHsystemcarries six bits of information per hop (symbol) whichis the same as the 64PSK but its performance is much betterthan the 64PSK no matter how many phases are used bythe HODFH system The BER performance of the HODFHsystem is better than theM-PSK when the information rate ishighThe carrier frequencies of theHODFH systems are usedto transmit information the HODFH systems have higherinformation rate without extra cost on power and do not needextra cost on power when the numbers of phases adopted bythe HODFH systems and the M-PSK are equal The HODFHsystems have better BER performance at the same energy-per-bit noise rate compared to the PSK

Example 5 Compare the BER performance of informationrate 19200 bps at the occasion that the SNR is equal andunequal at each narrow band Assume that the SNRs at119873

1198882

narrow bands are different with the SNRs at the other 1198731198882

narrow bands Use SNR119889to denote the difference among

the 119873119888SNRs The SNR

119889increases from 2 to 8 by increment

2 Use the average SNR in the broad band as the symbolsignal-to-noise ratio The HODFH system has the followingparameters 119873

119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888) and

119861119896= 2 (119896 = 1 119873

119900)

0 2 4 6 8 10 12 14

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

Equal SNRUnequal SNR SNRd = 2

Unequal SNR SNRd = 4

Unequal SNR SNRd = 6

Unequal SNR SNRd = 8

Figure 8 The BER performance of information rate 19200 bps atSNR

119889= 2 4 6 and 8

Example 6 Compare the BER performance of informationrate 28800 bps at different number of phases when the SNRdifference SNR

119889= 4 and 6 The SNR is similar as Example 5

The HODFH system has the following parameters 119873119900= 2

119873119888= 64 119861

119896= 3 (119896 = 1 119873

119900) and 119872

119896=4 8 and 16 (119896 =

1 119873119888)

As seen in Figure 8 the BER performance is worse whenthe SNR is unequal at each narrow band despite the averageSNR of the unequal signal-to-noise ratio broad band beingthe same as the SNR of the equal signal-to-noise ratio Asthe difference of the SNRs is increasing the performance isdegraded

Figure 9 shows that the BER performance is degradedwhen the number of phases used by the HODFH systemincreases This recalls the conclusion in Example 2 that thenumber of phases should be kept to a minimum if the num-ber of transmission signals has reached the requirements

5 Summary

In the future communication will have higher informationrate requirements and this paper proposed a solution forthis with a novel differential frequency hopping high orderdifferential frequency hopping method This was done byconumbering the frequencies and their phases so that thesystem information rate was improved without having toincrease the frequency resources The BER performance ofthe HODFH system is better than the conventional DFHsystem and M-PSK in the AWGN channel In this workthe number of phases can be adjusted to match the channelconditions to obtain the optimal performance

Although using phases to expand the signal set canimprove the bandwidth efficiency their introduction of the

Mathematical Problems in Engineering 9

0 2 4 6 8 10 12 14 16 18

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus7

10minus6

EbN0 (dB)

SNRd = 4 Mk = 4

SNRd = 4 Mk = 8

SNRd = 4 Mk = 16

SNRd = 6 Mk = 4

SNRd = 6 Mk = 8

SNRd = 6 Mk = 16

Figure 9 BER performance of information rate 28800 bps atdifferent number of phases when SNR

119889= 4 and 6

phases makes the HODFH system more sensitive to fre-quency offset and timing errors That is to say implementa-tion of this system is more complex but it is a price worthpaying for the information rate improvement and increasedperformance

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] D L Herrick and P K Lee ldquoCHESS a new reliable high speedHF radiordquo in Proceedings of the IEEE Military CommunicationsConference (MILCOM rsquo96) pp 684ndash690 McLean Va USA1996

[2] D L Herrick P K Lee and L L Ledlow Jr ldquoCorrelated fre-quency hopping an improved approach to HF spread spectrumcommunicationsrdquo in Proceedings of the Tactical Communica-tions Conference pp 319ndash324 IEEE Fort Wayne Ind USAApril-May 1996

[3] B Yang and Y Shen ldquoModulated (soft-decision) differentialfrequency hopping systemrdquo in Proceedings of the InternationalConference on Information and Communication Technology(ICICT rsquo07) pp 209ndash212 March 2007

[4] Q Ling and T Li ldquoMessage-driven frequency hopping designand analysisrdquo IEEE Transactions on Wireless Communicationsvol 8 no 4 pp 1773ndash1782 2009

[5] D Wang H Zhao and Z Fan ldquoA new scheme for message-driven FH systemrdquo in Proceedings of the International Con-ference on Future Information Technology and ManagementEngineering (FITME rsquo10) pp 395ndash398 October 2010

[6] L Zhang H Wang and T Li ldquoAnti-jamming message-drivenfrequency hopping Part I System designrdquo IEEE Transactionson Wireless Communications vol 12 no 1 pp 70ndash79 2013

[7] L Zhang and T Li ldquoAnti-jamming message-driven frequencyhopping-Part II capacity analysis under disguised jammingrdquoIEEE Transactions on Wireless Communications vol 12 no 1pp P80ndashP88 2013

[8] J G Proakis Digital Communications McGraw-Hill 4th edi-tion 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

SPconverter

Table look-up

and signal

generation

Delay

s(t)

nth data blockXn

XnN119900

Xn2

Xn1

dnN119900minus1

dn1dn1

dn2

dnminus11 dnminus12 dnminus1N119900

middot middot middot

dnminus1N119900

dnN119900= GN119900

(dnN119900minus1 XnN119900

)

dn2 = G2(dn1 Xn2)

dn1 = G1(dnminus11 dnminus12 dnminus1N119900 Xn1)

Figure 2 The transmitter block diagram of the proposed HODFH scheme

frequency at a certain hopping period can be detected byselecting the one that captures the strongest signal As a resultblind detection of the carrier frequency is achieved at thereceiver

More specifically the received signal is a convolution ofthe transmitted signal 119904(119905)with the channel impulse responseand it can be expressed as

119903 (119905) = int

+infin

minusinfin

ℎ (120591) 119904 (119905 minus 120591) 119889120591 + 119899 (119905) (4)

where ℎ(120591) is the channel impulse response and 119899(119905) denotesadditive white Gaussian noise (AWGN) Accordingly theoutputs of bandpass filters are given by

119911119894(119905) = int

+infin

minusinfin

119902119894(120591) 119903 (119905 minus 120591) 119889120591 for 119894 = 1 119873

119888 (5)

where 119902119894(119905) is the ideal bandpass filter centered at frequency

119891119894 If the channel is ideal that is ℎ(119905) = 120575(119905) then

119911119894(119905) = int

+infin

minusinfin

119902119894(120591) 119904 (119905 minus 120591) 119889120591 + 119906

119894(119905)

for 119894 = 1 119873119888

(6)

where 119906119894(119905) = int

+infin

minusinfin

119902119894(120591)119899(119905minus120591) is the filtered version ofWGN

If the signal-to-noise ratio is sufficiently high and there is nostrong jamming as in most communication systems thereis one and only one significantly strong signal among theoutputs of the filter bank Suppose that the 119897th filter capturesthis distinctive signal during the 119899th hop then the estimatedhopping frequency 119891

119899119894= 119891

119897 The same procedures can be

carried out to determine the carrier frequency at each hopNext the estimated hopping frequency119891

119899119894is used to con-

trol NCOThe output signal multiplied by the received signaland NCO is used to estimate the phase of the transmittedsignal Denote the estimated phase as 120601

119899119895

Then the estimated hopping frequency 119891119899i and the esti-

mated phase 120601119899119895

are used to estimate the transmitted signalDenote the estimated signal as 119904

119899

119889119899

1 119889119899

2 119889119899

119873119900

The estimated

index 119889119899

1 119889119899

119873119900

at the 119899th hop and the estimated index119889119899minus11 119889

119899minus1119873119900

at the (119899 minus 1)th hop are fed into the function119866minus1119898

(1 le 119898 le 119873119900) the inverse function of 119866

119898 to recover the

information data blocksThis processing can be expressed bythe following equations

1198831198991 = 119866

minus11 (119889

119899minus11 119889

119899minus12 119889

119899minus1119873119900

119889119899

1)

119883119899119898

= 119866minus1119898

(119889119899

119898minus1 119889119899

119898) when 119898 = 2 119873

119900

(7)

It then follows that the estimate of the 119899th block 119883119899can

be obtained as119883119899= [119883

1198991 1198831198992 119883119899119873119900

]

3 Bit Error Probability for HODFH

In HODFH system the input bit stream is carried by hoppingfrequency and its phase So firstly we analyze hoppingfrequency symbol error probability and phase symbol errorprobability respectively and then deduce the bit error rate(BER) of the HODFH

31 Symbol Error Probability Firstly we analyze the hoppingfrequency symbol error probability Based on the receiverdesign in HODFH analysis of the hopping frequency symbolerror probability is analogous to that of noncoherent FSKdemodulation For noncoherent detection of 119872

119865-ary FSK

signals the probability of symbol error is given by [8 eqn(5-4-46) page 310]

119901119904FSK (

120576119904

1198730)

=

119872119865minus1

sum

119899=1(minus1)119899+1 (

119872119865minus 1

119899)

1119899 + 1

exp [minus119899120576119904

(119899 + 1)1198730]

(8)

where 1205761199041198730 is the SNR per symbol For a HODFH system

with 119873119888frequencies 119872

119865= 119873

119888 Let 119901

(119865)

119904HODFH denote theprobability of the hopping frequency symbol detection error

4 Mathematical Problems in Engineering

Delay

Selectlargest

Transmitted signal

estimateand table

look-up

PSconverter

BPF f2

BPF f1

r(t)

fni

njdn

1 dn

2 dn

N119900

dnminus1

1 dnminus1

2 dnminus1

N119900

Xn1

Xn2Xn

XnN119900

BPF fN119888

Xn1 = Gminus11 (dnminus11 d

nminus1

2 dnminus1

N119900 d

n

1)

Xn2 = Gminus12 (dn1 d

n

2)

XnN119900= Gminus1

N119900dn

N119900minus1 d

n

N119900

EnergydetectionEnergy

detection

Energydetection

Phas

ede

tect

ion

Figure 3 The receiver block diagram of the proposed HODFH scheme

(corresponding to the symbol error in FSK) in HODFH thenwe have

119901(119865)

119904HODFH (120576119865

119904

1198730)

=

119873119888minus1

sum

119899=1(minus1)119899+1 (

119873119888minus 1119899

)1

119899 + 1exp[minus

119899120576119865

119904

(119899 + 1)1198730]

(9)

where 120576119865

1199041198730 is the SNR of hopping frequency per symbol

Next we focus on the phase symbol error probabilityThephase symbol error probability can be calculated in a similarmanner as that of the 119872

119875-ary PSK The probability of phase

symbol error probability is given by [8 eqn (5-2-56) amp (5-2-55) page 268]

119901119904PSK (

120576119904

1198730) = 1minusint

120587119872119875

minus120587119872119875

119901Θ119903

(Θ119903) 119889Θ

119903 (10)

119901Θ119903

(Θ119903) =

12120587

exp(minus120576119904

1198730sin2Θ

119903)

sdot int

infin

0119881 exp

[[[

[

minus

(119881 minus radic2 (1205761199041198730) cosΘ119903

)

2

2]]]

]

119889119881

(11)

For a HODFH system signal phase detection is imple-mented at the condition that the hopping frequency isdetected correctly for which the probability is (1minus119901

(119865)

119904HODFH)the probability of phase symbol error can be calculated as thatof the119872

119875-ary PSK Let 119901(119875)

119904HODFH denote the probability of thephase symbol detection error (corresponding to the symbolerror in PSK) Recall that there are119872

119896possible phases in each

hopping frequency we have

119901(119875)

119904HODFH (120576119875

119904

1198730) = 1minusint

120587119872119896

minus120587119872119896

119901Θ119903

(Θ119903) 119889Θ

119903 (12)

where 120576119875

1199041198730 is the SNR of phase per symbol and it will be

substituted in (11) when we calculate 119901(119875)

119904HODFHOnly when the hopping frequency and the phase are

detected correctly the transmitted signal can be estimatedcorrectly Since in a HODFH system the transmitted signalcarries the frequency and phase information the SNR of the

hopping frequency per symbol is the same as the SNR ofthe phase that is 120576119865

1199041198730 = 120576

119875

1199041198730 We denote the SNR per

symbol by 1205761199041198730 The overall symbol error probability of the

HODFH scheme is calculated as the combination of 119901(119865)119904HODFH

and 119901(119875)

119904HODFH defined as 119901119904HODFH which can be found in

119901119904HODFH (

120576119904

1198730)

= 1

minus(1minus119901(119865)

119904HODFH (120576119904

1198730))(1minus119901

(119875)

119904HODFH (120576119904

1198730))

= 119901(119865)

119904HODFH (120576119904

1198730)+119901

(119875)

119904HODFH (120576119904

1198730)

minus119901(119865)

119904HODFH (120576119904

1198730) sdot 119901

(119875)

119904HODFH (120576119904

1198730)

(13)

32 Bit Error Probability It is reasonable to assume that theprobability of the index 119889

119899

1198941 le 119894 le 119873

119900 estimated incorrectly

is inversely proportionate to the number of elements at the 119894thdimensionality when the symbol error occurs Let 119901Index119894 bethe error probability of the 119894th dimensional index we have

119901Index119894 (120576119904

1198730) =

119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730) (14)

For the first order bits 1198831198991 it is only decided by

119889119899minus11 119889

119899minus1119873119900

and 119889119899

1 The correct probability of119889119899minus11 119889

119899minus1119873119900

is equal to the symbol correct probability(1 minus 119901

119904HODFH) Recall that there are 1198611 bits at the first orderThe BER of the first order bits denoted as 119901

1198871HODFH isobtained

1199011198871HODFH (

120576119904

1198730) =

21198611minus1

21198611 minus 1(1

minus(1minus119901Index1 (120576119904

1198730))(1minus119901

119904HODFH (120576119904

1198730)))

=21198611minus1

21198611 minus 1(21198631 minus 1

1198631119901119904HODFH (

120576119904

1198730)

minus1198631 minus 11198631

(119901119904HODFH (

120576119904

1198730))

2)

(15)

Mathematical Problems in Engineering 5

For the119898th order bits119883119899119898

(2 le 119898 le 119873119900) it is decided by

119889119899

119898minus1 and 119889119899

119898 Only the 119889

119899

119898minus1 and 119889119899

119898are estimated correctly

the information bits can be recovered correctly Assume thatthe error probability of 119889

119899

119898minus1 and 119889119899

119898is independent This

is insured by the processing of reshaping the 1-dimensionalset 119878 into 119873

119900-dimensional set S

1198631times1198632timessdotsdotsdottimes119863119873119900 Let 119901119894119895Index be the

correct probability of the 119894th and the 119895th dimensional indexsimultaneity we have

119901119894119895

Index (120576119904

1198730) = (1minus119901Index119894 (

120576119904

1198730))

sdot (1minus119901Index119895 (120576119904

1198730))

= (1minus119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730))

sdot (1minus

119863119895minus 1

119863119895

119901119904HODFH (

120576119904

1198730)) = 1minus

119863119895minus 1

119863119895

sdot 119901119904HODFH (

120576119904

1198730)minus

119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730)

+

(119863119894minus 1) (119863

119895minus 1)

119863119894119863119895

(119901119904HODFH (

120576119904

1198730))

2

(16)

Recall that there are 119861119898bits at the119898th order The BER of

the119898th order bits denoted as 119901119887119898HODFH is obtained

119901119887119898HODFH =

2119861119898minus1

2119861119898 minus 1(1minus119901

119898minus1119898Index )

=2119861119898minus1

2119861119898 minus 1(119863119898minus1 minus 1119863119898minus1

119901119904HODFH (

120576119904

1198730)

+119863119898

minus 1119863119898

119901119904HODFH (

120576119904

1198730)

minus(119863

119898minus1 minus 1) (119863119898

minus 1)119863119898minus1119863119898

(119901119904HODFH (

120576119904

1198730))

2)

(17)

Therefore the total BER of a HODFH system denoted as119901119887HODFH can be obtained as

119901119887HODFH (

120576119904

1198730)

= (1198611

sum119873119900

119899=1 119861119899) sdot 119901

1198871HODFH (120576119904

1198730)

+

119873119900

sum

119898=2(

119861119898

sum119873119900

119899=1 119861119899) sdot 119901

119887119898HODFH (120576119904

1198730)

(18)

Since sum119873119900

119899=1 119861119899 = 119871 (18) can be transformed as

119901119887HODFH (

120576119904

1198730) =

1119871

(11986111199011198871HODFH (120576119904

1198730)

+

119873119900

sum

119898=2119861119898119901119887119898HODFH (

120576119904

1198730))

(19)

We assume that one symbol not correctly detected willincur the 119873

119900dimensional index being not correctly esti-

mated That is to say 119901Index119894 and 119901119894119895

Index will satisfy thefollowing equations

119901Index119894 (120576119904

1198730) = 119901

119904HODFH (120576119904

1198730)

119901119894119895

Index (120576119904

1198730)

= (1minus119901119904HODFH (

120576119904

1198730))(1minus119901

119904HODFH (120576119904

1198730))

= 1minus 2119901119904HODFH (

120576119904

1198730)+(119901

119904HODFH (120576119904

1198730))

2

(20)

Substitute (20) into (15) and (17) the uniformof the upperbound of BER for the first and the119898th order bits denoted as119901119880

119887119898HODFH can be obtained as

119901119880

119887119898HODFH =2119861119898minus1

2119861119898 minus 1(2119901

119904HODFH (120576119904

1198730)

minus(119901119904HODFH (

120576119904

1198730))

2) 1 le 119898 le 119873

119900

(21)

When the symbol error probability 119901119904HODFH is small

enough the bit error probability 119901119880

119887119898HODFH can be approx-imated as

119901119880

119887119898HODFH (120576119904

1198730) =

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

1 le 119898 le 119873119900

(22)

Substitute (22) into (19) the upper bound of total BER forthe HODFH system is obtained

119901119880

119887HODFH (120576119904

1198730) =

1119871

119873119900

sum

119898=1119861119898119901119880

119887119898HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(23)

The lower bound of total BER denoted as 119901119871

119887HODFHcan be obtained by assuming that the symbol error onlyincurs bits error in that hopping The error probability ofthe 119894th dimensional index is expressed as (14) and the

6 Mathematical Problems in Engineering

119894th dimensional index error only incurs the 119894th order bitsTherefore the lower bound of total BER is

119901119871

119887HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

(24)

To summarize our discussions above we have the follow-ing

Proposition In HODFH the total BER 119901119887HODFH(1205761199041198730) is

bounded by

1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

le 119901119887HODFH (

120576119904

1198730)

le1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(25)

33 Further Discussions about Bit Error Probability Recallthat there are 119873

119888available frequencies 1198911 1198912 119891119873

119888

forthe HODFH system and they distribute in broad band Itonly occupies narrow band centered at 119891

119894(119894 = 1 2 119873

119888)

available frequencies at any given moment The SNR usuallyis unequal at each narrow band In this occasion the symbolerror probability of the frequency and phase cannot be simplycalculated by (9) and (12) The symbol correct decisionprobability119901

119888119891119894

of the frequency119891119894can be deduced by [8 eqn

(5-4-41) page 309] Because the random variables 119877119894(119894 =

1 2 119873119888) are statistically independent and identically dis-

tributed and we assume that the signals centered at the 119873119888

available frequencies 1198911 1198912 119891119873119888

have equal energy anddifferent noise the probability 119901

119888119891119894

can be expressed as

119901119888119891119894

= int

infin

0

119873119888

prod

119898=1119898 =119894

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)119901

119877119894

(119909) 119889119909 (26)

where 1205902119898

= 121198730119898 is the variance of the AWGN atfrequency 119891

119894and

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)

= int

120590119894119909120590119898

0119901119877119898

(119903119898) 119889119903

119898= 1minus 119890

minus1205902119894119909221205902119898

= 1minus 119890minus1198730119894119909

221198730119898

(27)

Substitute (27) into (26) we obtain the general probability

119901119888119891119894

= int

infin

0

119873119888

prod

119898=2119898 =119894

(1minus 119890minus1198730119894119909

221198730119898) 119901

119877119894

(119909) 119889119909 (28)

We have known that in a HODFH system signal phasedetection is implemented at the condition that the hoppingfrequency centred at 119891

119894is detected correctly for which the

probability is 119901119888119891119894

the probability of phase symbol error canbe calculated as (10) and (11) but the SNR 120576

119875

1199041198730 of phase

per symbol will be substituted by 1205761198730119894 which is the SNRat frequency 119891

119894 Let 119901

(119875119894)

119904HODFH denote the probability of thephase symbol detection error at frequency119891

119894 In this occasion

the overall symbol error probability of the HODFH schemeis calculated by

1199011015840

119904HODFH = 1minus

119873119888

sum

119894=1119901119894sdot 119901

119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

= 1minus1119873119888

119873119888

sum

119894=1119901119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

(29)

where the 119901119894(119894 = 1 2 119873

119888) is the probability that the

signals centered at frequency 119891119894and the 1199011 1199012 119901119873

119888

allequal 1119873

119888 Substitute the probability 119901

119904HODFH by 1199011015840

119904HODFHin (16) and (17) we can calculate the total BER of a HODFHsystem by (18) when it works in the occasion that the119873119888available frequencies 1198911 1198912 119891119873

119888

distribute in broadband and the SNR is not equal at each narrow band centeredat 119891

119894

4 Simulation Results

Without loss of generality in this paper the HODFH systemin the AWGN channel with equal SNR at anywhere wassimulated and its bit error rate (BER) performance wasanalyzed

Example 1 BER performance of the HODFH system inAWGN channel the performance of BER at the informa-tion rates of 9600 bps 19200 bps and 28800 bps in AWGNchannel was compared For the information rate of 9600 bps119873119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and 119861

119896= 1 (119896 =

1 119873119900) For the information rate of 19200 bps 119873

119900= 2

119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 2 (119896 = 1 119873

119900)

For the information rate of 28800 bps 119873119900

= 2 119873119888

= 64119872

119896= 4 (119896 = 1 119873

119888) 1198611 = 4 and 1198612 = 2

Figure 4 is the BER performance of the HODFH systemin AWGN channel The requirement for the energy-per-bitnoise rate (119864

1198871198730) becomes lower when the information

rate increases In practice it is usually possible that onlythe energy-per-symbol noise rate (119864

1199041198730) can be measured

According to formula 1198641199041198730 = 119864

1198871198730 + 10 log(119896) (119896 is the

number of bits per symbol) convert 1198641198871198730 to 119864

1199041198730 and

then the higher 1198641199041198730 is required for the higher information

rate which is reasonable

Example 2 The BER performance of information rate28800 bps at different number of phases assume 119873

119900= 2 and

119861119896= 3 (119896 = 1 119873

119900) The numbers of available frequencies

119873119888are 64 32 and 16 the numbers of phases119872

119896are 4 8 and

16 respectively

Mathematical Problems in Engineering 7

0 1 2 3 4 5 6 7 8 9 10 11

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps No = 2 Nc = 64 Mk = 4 B1 = 4 B2 = 2

Figure 4 The BER performance at information rates of 9600 bps19200 bps and 28800 bps in AWGN channel

0 2 4 6 8 10 12 14 16

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

No = 2 Nc = 64 Mk = 4 Bk = 3

No = 2 Nc = 32 Mk = 8 Bk = 3

No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 5 The BER performance of information rate 28800 bps atdifferent number of phases

As seen in Figure 5 when more phrases are used whilekeeping the information rate the same there is poorerperformance for the HODFH system It will spend about6 dB extra 119864

1198871198730 to obtain the same BER performance

of 10minus3 when the number of phases increases from 8 to16 Therefore the number of phases should be kept to aminimum if the number of transmission signals has reachedthe requirements

Example 3 TheBERperformance of theHODFH system andthe conventional DFH (CDFH) system for the informationrate of 9600 bps the HODFH system has the followingparameters 119873

119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and

0 2 4 6 8 10 12

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

9600bps CDFH Nc = 128 BPH = 2

19200 bps CDFH Nc = 128 BPH = 4

Figure 6 The BER performance of the HODFH system and theCDFH system

119861119896= 1 (119896 = 1 119873

119900) for the information rate of 19200 bps

it has parameters 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) The CDFH has the following

parameters for information rate of 9600 bps 119873119888= 128 and

BPH = 2 for information rate of 19200 bps 119873119888= 128 and

BPH = 4

Figure 6 shows the BER performance comparison ofthe HODFH system and the conventional DFH system forinformation rates of 9600 bps and 19200 bps At the sameinformation rate the HODFH system has a better BERperformance than the conventional DFH system The higherinformation rate is the higher gain is obtained by theHODFH system when compared to the conventional DFHsystem

Example 4 Compare the BER performances of the HODFHsystem at three information rates (9600 bps 19200 bps and28800 bps) with the theoretical BER performance of QPSK8PSK 16PSK 32PSK and 64PSK Assume the HODFHsystem has the following parameters respectively For theinformation rate of 9600 bps119873

119900= 2119873

119888= 128119872

119896= 2 (119896 =

1 119873119888) and 119861

119896= 1 (119896 = 1 119873

119900) For the information

rate of 19200 bps 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) For the information rate of

28800 bps we have three combinations the first is 119873119900

= 2119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

the second is 119873119900= 2 119873

119888= 32 119872

119896= 8 (119896 = 1 119873

119888) and

119861119896

= 3 (119896 = 1 119873119900) and the third is 119873

119900= 2 119873

119888= 16

119872119896= 16 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

Figure 7 presents the BER performance of the HODFHsystem and the theoretical BER performance of QPSK 8PSK16PSK 32PSK and 64PSK When the information rate is9600 bps the HODFH system carries two bits of information

8 Mathematical Problems in Engineering

0 5 10 15 20 25 30

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

QPSK theory8 PSK theory16 PSK theor32 PSK theor64 PSK theor9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 3

28800 bps HODFH No = 2 Nc = 64 Mk = 8 Bk = 3

28800 bps HODFH No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 7 The BER performance of the HODFH system andtheoretical BER performance of the M-PSK

per hop (symbol) which is the same as the QPSK the BERperformance of the HODFH system is worse than the QPSKWhen the information rate is 19200 bps the HODFH systemcarries four bits of information per hop (symbol) which is thesame as the 16PSK but it has better BERperformance than the16PSK When the information rate is 28800 bps the HODFHsystemcarries six bits of information per hop (symbol) whichis the same as the 64PSK but its performance is much betterthan the 64PSK no matter how many phases are used bythe HODFH system The BER performance of the HODFHsystem is better than theM-PSK when the information rate ishighThe carrier frequencies of theHODFH systems are usedto transmit information the HODFH systems have higherinformation rate without extra cost on power and do not needextra cost on power when the numbers of phases adopted bythe HODFH systems and the M-PSK are equal The HODFHsystems have better BER performance at the same energy-per-bit noise rate compared to the PSK

Example 5 Compare the BER performance of informationrate 19200 bps at the occasion that the SNR is equal andunequal at each narrow band Assume that the SNRs at119873

1198882

narrow bands are different with the SNRs at the other 1198731198882

narrow bands Use SNR119889to denote the difference among

the 119873119888SNRs The SNR

119889increases from 2 to 8 by increment

2 Use the average SNR in the broad band as the symbolsignal-to-noise ratio The HODFH system has the followingparameters 119873

119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888) and

119861119896= 2 (119896 = 1 119873

119900)

0 2 4 6 8 10 12 14

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

Equal SNRUnequal SNR SNRd = 2

Unequal SNR SNRd = 4

Unequal SNR SNRd = 6

Unequal SNR SNRd = 8

Figure 8 The BER performance of information rate 19200 bps atSNR

119889= 2 4 6 and 8

Example 6 Compare the BER performance of informationrate 28800 bps at different number of phases when the SNRdifference SNR

119889= 4 and 6 The SNR is similar as Example 5

The HODFH system has the following parameters 119873119900= 2

119873119888= 64 119861

119896= 3 (119896 = 1 119873

119900) and 119872

119896=4 8 and 16 (119896 =

1 119873119888)

As seen in Figure 8 the BER performance is worse whenthe SNR is unequal at each narrow band despite the averageSNR of the unequal signal-to-noise ratio broad band beingthe same as the SNR of the equal signal-to-noise ratio Asthe difference of the SNRs is increasing the performance isdegraded

Figure 9 shows that the BER performance is degradedwhen the number of phases used by the HODFH systemincreases This recalls the conclusion in Example 2 that thenumber of phases should be kept to a minimum if the num-ber of transmission signals has reached the requirements

5 Summary

In the future communication will have higher informationrate requirements and this paper proposed a solution forthis with a novel differential frequency hopping high orderdifferential frequency hopping method This was done byconumbering the frequencies and their phases so that thesystem information rate was improved without having toincrease the frequency resources The BER performance ofthe HODFH system is better than the conventional DFHsystem and M-PSK in the AWGN channel In this workthe number of phases can be adjusted to match the channelconditions to obtain the optimal performance

Although using phases to expand the signal set canimprove the bandwidth efficiency their introduction of the

Mathematical Problems in Engineering 9

0 2 4 6 8 10 12 14 16 18

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus7

10minus6

EbN0 (dB)

SNRd = 4 Mk = 4

SNRd = 4 Mk = 8

SNRd = 4 Mk = 16

SNRd = 6 Mk = 4

SNRd = 6 Mk = 8

SNRd = 6 Mk = 16

Figure 9 BER performance of information rate 28800 bps atdifferent number of phases when SNR

119889= 4 and 6

phases makes the HODFH system more sensitive to fre-quency offset and timing errors That is to say implementa-tion of this system is more complex but it is a price worthpaying for the information rate improvement and increasedperformance

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] D L Herrick and P K Lee ldquoCHESS a new reliable high speedHF radiordquo in Proceedings of the IEEE Military CommunicationsConference (MILCOM rsquo96) pp 684ndash690 McLean Va USA1996

[2] D L Herrick P K Lee and L L Ledlow Jr ldquoCorrelated fre-quency hopping an improved approach to HF spread spectrumcommunicationsrdquo in Proceedings of the Tactical Communica-tions Conference pp 319ndash324 IEEE Fort Wayne Ind USAApril-May 1996

[3] B Yang and Y Shen ldquoModulated (soft-decision) differentialfrequency hopping systemrdquo in Proceedings of the InternationalConference on Information and Communication Technology(ICICT rsquo07) pp 209ndash212 March 2007

[4] Q Ling and T Li ldquoMessage-driven frequency hopping designand analysisrdquo IEEE Transactions on Wireless Communicationsvol 8 no 4 pp 1773ndash1782 2009

[5] D Wang H Zhao and Z Fan ldquoA new scheme for message-driven FH systemrdquo in Proceedings of the International Con-ference on Future Information Technology and ManagementEngineering (FITME rsquo10) pp 395ndash398 October 2010

[6] L Zhang H Wang and T Li ldquoAnti-jamming message-drivenfrequency hopping Part I System designrdquo IEEE Transactionson Wireless Communications vol 12 no 1 pp 70ndash79 2013

[7] L Zhang and T Li ldquoAnti-jamming message-driven frequencyhopping-Part II capacity analysis under disguised jammingrdquoIEEE Transactions on Wireless Communications vol 12 no 1pp P80ndashP88 2013

[8] J G Proakis Digital Communications McGraw-Hill 4th edi-tion 2001

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

Delay

Selectlargest

Transmitted signal

estimateand table

look-up

PSconverter

BPF f2

BPF f1

r(t)

fni

njdn

1 dn

2 dn

N119900

dnminus1

1 dnminus1

2 dnminus1

N119900

Xn1

Xn2Xn

XnN119900

BPF fN119888

Xn1 = Gminus11 (dnminus11 d

nminus1

2 dnminus1

N119900 d

n

1)

Xn2 = Gminus12 (dn1 d

n

2)

XnN119900= Gminus1

N119900dn

N119900minus1 d

n

N119900

EnergydetectionEnergy

detection

Energydetection

Phas

ede

tect

ion

Figure 3 The receiver block diagram of the proposed HODFH scheme

(corresponding to the symbol error in FSK) in HODFH thenwe have

119901(119865)

119904HODFH (120576119865

119904

1198730)

=

119873119888minus1

sum

119899=1(minus1)119899+1 (

119873119888minus 1119899

)1

119899 + 1exp[minus

119899120576119865

119904

(119899 + 1)1198730]

(9)

where 120576119865

1199041198730 is the SNR of hopping frequency per symbol

Next we focus on the phase symbol error probabilityThephase symbol error probability can be calculated in a similarmanner as that of the 119872

119875-ary PSK The probability of phase

symbol error probability is given by [8 eqn (5-2-56) amp (5-2-55) page 268]

119901119904PSK (

120576119904

1198730) = 1minusint

120587119872119875

minus120587119872119875

119901Θ119903

(Θ119903) 119889Θ

119903 (10)

119901Θ119903

(Θ119903) =

12120587

exp(minus120576119904

1198730sin2Θ

119903)

sdot int

infin

0119881 exp

[[[

[

minus

(119881 minus radic2 (1205761199041198730) cosΘ119903

)

2

2]]]

]

119889119881

(11)

For a HODFH system signal phase detection is imple-mented at the condition that the hopping frequency isdetected correctly for which the probability is (1minus119901

(119865)

119904HODFH)the probability of phase symbol error can be calculated as thatof the119872

119875-ary PSK Let 119901(119875)

119904HODFH denote the probability of thephase symbol detection error (corresponding to the symbolerror in PSK) Recall that there are119872

119896possible phases in each

hopping frequency we have

119901(119875)

119904HODFH (120576119875

119904

1198730) = 1minusint

120587119872119896

minus120587119872119896

119901Θ119903

(Θ119903) 119889Θ

119903 (12)

where 120576119875

1199041198730 is the SNR of phase per symbol and it will be

substituted in (11) when we calculate 119901(119875)

119904HODFHOnly when the hopping frequency and the phase are

detected correctly the transmitted signal can be estimatedcorrectly Since in a HODFH system the transmitted signalcarries the frequency and phase information the SNR of the

hopping frequency per symbol is the same as the SNR ofthe phase that is 120576119865

1199041198730 = 120576

119875

1199041198730 We denote the SNR per

symbol by 1205761199041198730 The overall symbol error probability of the

HODFH scheme is calculated as the combination of 119901(119865)119904HODFH

and 119901(119875)

119904HODFH defined as 119901119904HODFH which can be found in

119901119904HODFH (

120576119904

1198730)

= 1

minus(1minus119901(119865)

119904HODFH (120576119904

1198730))(1minus119901

(119875)

119904HODFH (120576119904

1198730))

= 119901(119865)

119904HODFH (120576119904

1198730)+119901

(119875)

119904HODFH (120576119904

1198730)

minus119901(119865)

119904HODFH (120576119904

1198730) sdot 119901

(119875)

119904HODFH (120576119904

1198730)

(13)

32 Bit Error Probability It is reasonable to assume that theprobability of the index 119889

119899

1198941 le 119894 le 119873

119900 estimated incorrectly

is inversely proportionate to the number of elements at the 119894thdimensionality when the symbol error occurs Let 119901Index119894 bethe error probability of the 119894th dimensional index we have

119901Index119894 (120576119904

1198730) =

119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730) (14)

For the first order bits 1198831198991 it is only decided by

119889119899minus11 119889

119899minus1119873119900

and 119889119899

1 The correct probability of119889119899minus11 119889

119899minus1119873119900

is equal to the symbol correct probability(1 minus 119901

119904HODFH) Recall that there are 1198611 bits at the first orderThe BER of the first order bits denoted as 119901

1198871HODFH isobtained

1199011198871HODFH (

120576119904

1198730) =

21198611minus1

21198611 minus 1(1

minus(1minus119901Index1 (120576119904

1198730))(1minus119901

119904HODFH (120576119904

1198730)))

=21198611minus1

21198611 minus 1(21198631 minus 1

1198631119901119904HODFH (

120576119904

1198730)

minus1198631 minus 11198631

(119901119904HODFH (

120576119904

1198730))

2)

(15)

Mathematical Problems in Engineering 5

For the119898th order bits119883119899119898

(2 le 119898 le 119873119900) it is decided by

119889119899

119898minus1 and 119889119899

119898 Only the 119889

119899

119898minus1 and 119889119899

119898are estimated correctly

the information bits can be recovered correctly Assume thatthe error probability of 119889

119899

119898minus1 and 119889119899

119898is independent This

is insured by the processing of reshaping the 1-dimensionalset 119878 into 119873

119900-dimensional set S

1198631times1198632timessdotsdotsdottimes119863119873119900 Let 119901119894119895Index be the

correct probability of the 119894th and the 119895th dimensional indexsimultaneity we have

119901119894119895

Index (120576119904

1198730) = (1minus119901Index119894 (

120576119904

1198730))

sdot (1minus119901Index119895 (120576119904

1198730))

= (1minus119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730))

sdot (1minus

119863119895minus 1

119863119895

119901119904HODFH (

120576119904

1198730)) = 1minus

119863119895minus 1

119863119895

sdot 119901119904HODFH (

120576119904

1198730)minus

119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730)

+

(119863119894minus 1) (119863

119895minus 1)

119863119894119863119895

(119901119904HODFH (

120576119904

1198730))

2

(16)

Recall that there are 119861119898bits at the119898th order The BER of

the119898th order bits denoted as 119901119887119898HODFH is obtained

119901119887119898HODFH =

2119861119898minus1

2119861119898 minus 1(1minus119901

119898minus1119898Index )

=2119861119898minus1

2119861119898 minus 1(119863119898minus1 minus 1119863119898minus1

119901119904HODFH (

120576119904

1198730)

+119863119898

minus 1119863119898

119901119904HODFH (

120576119904

1198730)

minus(119863

119898minus1 minus 1) (119863119898

minus 1)119863119898minus1119863119898

(119901119904HODFH (

120576119904

1198730))

2)

(17)

Therefore the total BER of a HODFH system denoted as119901119887HODFH can be obtained as

119901119887HODFH (

120576119904

1198730)

= (1198611

sum119873119900

119899=1 119861119899) sdot 119901

1198871HODFH (120576119904

1198730)

+

119873119900

sum

119898=2(

119861119898

sum119873119900

119899=1 119861119899) sdot 119901

119887119898HODFH (120576119904

1198730)

(18)

Since sum119873119900

119899=1 119861119899 = 119871 (18) can be transformed as

119901119887HODFH (

120576119904

1198730) =

1119871

(11986111199011198871HODFH (120576119904

1198730)

+

119873119900

sum

119898=2119861119898119901119887119898HODFH (

120576119904

1198730))

(19)

We assume that one symbol not correctly detected willincur the 119873

119900dimensional index being not correctly esti-

mated That is to say 119901Index119894 and 119901119894119895

Index will satisfy thefollowing equations

119901Index119894 (120576119904

1198730) = 119901

119904HODFH (120576119904

1198730)

119901119894119895

Index (120576119904

1198730)

= (1minus119901119904HODFH (

120576119904

1198730))(1minus119901

119904HODFH (120576119904

1198730))

= 1minus 2119901119904HODFH (

120576119904

1198730)+(119901

119904HODFH (120576119904

1198730))

2

(20)

Substitute (20) into (15) and (17) the uniformof the upperbound of BER for the first and the119898th order bits denoted as119901119880

119887119898HODFH can be obtained as

119901119880

119887119898HODFH =2119861119898minus1

2119861119898 minus 1(2119901

119904HODFH (120576119904

1198730)

minus(119901119904HODFH (

120576119904

1198730))

2) 1 le 119898 le 119873

119900

(21)

When the symbol error probability 119901119904HODFH is small

enough the bit error probability 119901119880

119887119898HODFH can be approx-imated as

119901119880

119887119898HODFH (120576119904

1198730) =

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

1 le 119898 le 119873119900

(22)

Substitute (22) into (19) the upper bound of total BER forthe HODFH system is obtained

119901119880

119887HODFH (120576119904

1198730) =

1119871

119873119900

sum

119898=1119861119898119901119880

119887119898HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(23)

The lower bound of total BER denoted as 119901119871

119887HODFHcan be obtained by assuming that the symbol error onlyincurs bits error in that hopping The error probability ofthe 119894th dimensional index is expressed as (14) and the

6 Mathematical Problems in Engineering

119894th dimensional index error only incurs the 119894th order bitsTherefore the lower bound of total BER is

119901119871

119887HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

(24)

To summarize our discussions above we have the follow-ing

Proposition In HODFH the total BER 119901119887HODFH(1205761199041198730) is

bounded by

1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

le 119901119887HODFH (

120576119904

1198730)

le1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(25)

33 Further Discussions about Bit Error Probability Recallthat there are 119873

119888available frequencies 1198911 1198912 119891119873

119888

forthe HODFH system and they distribute in broad band Itonly occupies narrow band centered at 119891

119894(119894 = 1 2 119873

119888)

available frequencies at any given moment The SNR usuallyis unequal at each narrow band In this occasion the symbolerror probability of the frequency and phase cannot be simplycalculated by (9) and (12) The symbol correct decisionprobability119901

119888119891119894

of the frequency119891119894can be deduced by [8 eqn

(5-4-41) page 309] Because the random variables 119877119894(119894 =

1 2 119873119888) are statistically independent and identically dis-

tributed and we assume that the signals centered at the 119873119888

available frequencies 1198911 1198912 119891119873119888

have equal energy anddifferent noise the probability 119901

119888119891119894

can be expressed as

119901119888119891119894

= int

infin

0

119873119888

prod

119898=1119898 =119894

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)119901

119877119894

(119909) 119889119909 (26)

where 1205902119898

= 121198730119898 is the variance of the AWGN atfrequency 119891

119894and

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)

= int

120590119894119909120590119898

0119901119877119898

(119903119898) 119889119903

119898= 1minus 119890

minus1205902119894119909221205902119898

= 1minus 119890minus1198730119894119909

221198730119898

(27)

Substitute (27) into (26) we obtain the general probability

119901119888119891119894

= int

infin

0

119873119888

prod

119898=2119898 =119894

(1minus 119890minus1198730119894119909

221198730119898) 119901

119877119894

(119909) 119889119909 (28)

We have known that in a HODFH system signal phasedetection is implemented at the condition that the hoppingfrequency centred at 119891

119894is detected correctly for which the

probability is 119901119888119891119894

the probability of phase symbol error canbe calculated as (10) and (11) but the SNR 120576

119875

1199041198730 of phase

per symbol will be substituted by 1205761198730119894 which is the SNRat frequency 119891

119894 Let 119901

(119875119894)

119904HODFH denote the probability of thephase symbol detection error at frequency119891

119894 In this occasion

the overall symbol error probability of the HODFH schemeis calculated by

1199011015840

119904HODFH = 1minus

119873119888

sum

119894=1119901119894sdot 119901

119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

= 1minus1119873119888

119873119888

sum

119894=1119901119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

(29)

where the 119901119894(119894 = 1 2 119873

119888) is the probability that the

signals centered at frequency 119891119894and the 1199011 1199012 119901119873

119888

allequal 1119873

119888 Substitute the probability 119901

119904HODFH by 1199011015840

119904HODFHin (16) and (17) we can calculate the total BER of a HODFHsystem by (18) when it works in the occasion that the119873119888available frequencies 1198911 1198912 119891119873

119888

distribute in broadband and the SNR is not equal at each narrow band centeredat 119891

119894

4 Simulation Results

Without loss of generality in this paper the HODFH systemin the AWGN channel with equal SNR at anywhere wassimulated and its bit error rate (BER) performance wasanalyzed

Example 1 BER performance of the HODFH system inAWGN channel the performance of BER at the informa-tion rates of 9600 bps 19200 bps and 28800 bps in AWGNchannel was compared For the information rate of 9600 bps119873119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and 119861

119896= 1 (119896 =

1 119873119900) For the information rate of 19200 bps 119873

119900= 2

119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 2 (119896 = 1 119873

119900)

For the information rate of 28800 bps 119873119900

= 2 119873119888

= 64119872

119896= 4 (119896 = 1 119873

119888) 1198611 = 4 and 1198612 = 2

Figure 4 is the BER performance of the HODFH systemin AWGN channel The requirement for the energy-per-bitnoise rate (119864

1198871198730) becomes lower when the information

rate increases In practice it is usually possible that onlythe energy-per-symbol noise rate (119864

1199041198730) can be measured

According to formula 1198641199041198730 = 119864

1198871198730 + 10 log(119896) (119896 is the

number of bits per symbol) convert 1198641198871198730 to 119864

1199041198730 and

then the higher 1198641199041198730 is required for the higher information

rate which is reasonable

Example 2 The BER performance of information rate28800 bps at different number of phases assume 119873

119900= 2 and

119861119896= 3 (119896 = 1 119873

119900) The numbers of available frequencies

119873119888are 64 32 and 16 the numbers of phases119872

119896are 4 8 and

16 respectively

Mathematical Problems in Engineering 7

0 1 2 3 4 5 6 7 8 9 10 11

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps No = 2 Nc = 64 Mk = 4 B1 = 4 B2 = 2

Figure 4 The BER performance at information rates of 9600 bps19200 bps and 28800 bps in AWGN channel

0 2 4 6 8 10 12 14 16

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

No = 2 Nc = 64 Mk = 4 Bk = 3

No = 2 Nc = 32 Mk = 8 Bk = 3

No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 5 The BER performance of information rate 28800 bps atdifferent number of phases

As seen in Figure 5 when more phrases are used whilekeeping the information rate the same there is poorerperformance for the HODFH system It will spend about6 dB extra 119864

1198871198730 to obtain the same BER performance

of 10minus3 when the number of phases increases from 8 to16 Therefore the number of phases should be kept to aminimum if the number of transmission signals has reachedthe requirements

Example 3 TheBERperformance of theHODFH system andthe conventional DFH (CDFH) system for the informationrate of 9600 bps the HODFH system has the followingparameters 119873

119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and

0 2 4 6 8 10 12

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

9600bps CDFH Nc = 128 BPH = 2

19200 bps CDFH Nc = 128 BPH = 4

Figure 6 The BER performance of the HODFH system and theCDFH system

119861119896= 1 (119896 = 1 119873

119900) for the information rate of 19200 bps

it has parameters 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) The CDFH has the following

parameters for information rate of 9600 bps 119873119888= 128 and

BPH = 2 for information rate of 19200 bps 119873119888= 128 and

BPH = 4

Figure 6 shows the BER performance comparison ofthe HODFH system and the conventional DFH system forinformation rates of 9600 bps and 19200 bps At the sameinformation rate the HODFH system has a better BERperformance than the conventional DFH system The higherinformation rate is the higher gain is obtained by theHODFH system when compared to the conventional DFHsystem

Example 4 Compare the BER performances of the HODFHsystem at three information rates (9600 bps 19200 bps and28800 bps) with the theoretical BER performance of QPSK8PSK 16PSK 32PSK and 64PSK Assume the HODFHsystem has the following parameters respectively For theinformation rate of 9600 bps119873

119900= 2119873

119888= 128119872

119896= 2 (119896 =

1 119873119888) and 119861

119896= 1 (119896 = 1 119873

119900) For the information

rate of 19200 bps 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) For the information rate of

28800 bps we have three combinations the first is 119873119900

= 2119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

the second is 119873119900= 2 119873

119888= 32 119872

119896= 8 (119896 = 1 119873

119888) and

119861119896

= 3 (119896 = 1 119873119900) and the third is 119873

119900= 2 119873

119888= 16

119872119896= 16 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

Figure 7 presents the BER performance of the HODFHsystem and the theoretical BER performance of QPSK 8PSK16PSK 32PSK and 64PSK When the information rate is9600 bps the HODFH system carries two bits of information

8 Mathematical Problems in Engineering

0 5 10 15 20 25 30

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

QPSK theory8 PSK theory16 PSK theor32 PSK theor64 PSK theor9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 3

28800 bps HODFH No = 2 Nc = 64 Mk = 8 Bk = 3

28800 bps HODFH No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 7 The BER performance of the HODFH system andtheoretical BER performance of the M-PSK

per hop (symbol) which is the same as the QPSK the BERperformance of the HODFH system is worse than the QPSKWhen the information rate is 19200 bps the HODFH systemcarries four bits of information per hop (symbol) which is thesame as the 16PSK but it has better BERperformance than the16PSK When the information rate is 28800 bps the HODFHsystemcarries six bits of information per hop (symbol) whichis the same as the 64PSK but its performance is much betterthan the 64PSK no matter how many phases are used bythe HODFH system The BER performance of the HODFHsystem is better than theM-PSK when the information rate ishighThe carrier frequencies of theHODFH systems are usedto transmit information the HODFH systems have higherinformation rate without extra cost on power and do not needextra cost on power when the numbers of phases adopted bythe HODFH systems and the M-PSK are equal The HODFHsystems have better BER performance at the same energy-per-bit noise rate compared to the PSK

Example 5 Compare the BER performance of informationrate 19200 bps at the occasion that the SNR is equal andunequal at each narrow band Assume that the SNRs at119873

1198882

narrow bands are different with the SNRs at the other 1198731198882

narrow bands Use SNR119889to denote the difference among

the 119873119888SNRs The SNR

119889increases from 2 to 8 by increment

2 Use the average SNR in the broad band as the symbolsignal-to-noise ratio The HODFH system has the followingparameters 119873

119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888) and

119861119896= 2 (119896 = 1 119873

119900)

0 2 4 6 8 10 12 14

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

Equal SNRUnequal SNR SNRd = 2

Unequal SNR SNRd = 4

Unequal SNR SNRd = 6

Unequal SNR SNRd = 8

Figure 8 The BER performance of information rate 19200 bps atSNR

119889= 2 4 6 and 8

Example 6 Compare the BER performance of informationrate 28800 bps at different number of phases when the SNRdifference SNR

119889= 4 and 6 The SNR is similar as Example 5

The HODFH system has the following parameters 119873119900= 2

119873119888= 64 119861

119896= 3 (119896 = 1 119873

119900) and 119872

119896=4 8 and 16 (119896 =

1 119873119888)

As seen in Figure 8 the BER performance is worse whenthe SNR is unequal at each narrow band despite the averageSNR of the unequal signal-to-noise ratio broad band beingthe same as the SNR of the equal signal-to-noise ratio Asthe difference of the SNRs is increasing the performance isdegraded

Figure 9 shows that the BER performance is degradedwhen the number of phases used by the HODFH systemincreases This recalls the conclusion in Example 2 that thenumber of phases should be kept to a minimum if the num-ber of transmission signals has reached the requirements

5 Summary

In the future communication will have higher informationrate requirements and this paper proposed a solution forthis with a novel differential frequency hopping high orderdifferential frequency hopping method This was done byconumbering the frequencies and their phases so that thesystem information rate was improved without having toincrease the frequency resources The BER performance ofthe HODFH system is better than the conventional DFHsystem and M-PSK in the AWGN channel In this workthe number of phases can be adjusted to match the channelconditions to obtain the optimal performance

Although using phases to expand the signal set canimprove the bandwidth efficiency their introduction of the

Mathematical Problems in Engineering 9

0 2 4 6 8 10 12 14 16 18

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus7

10minus6

EbN0 (dB)

SNRd = 4 Mk = 4

SNRd = 4 Mk = 8

SNRd = 4 Mk = 16

SNRd = 6 Mk = 4

SNRd = 6 Mk = 8

SNRd = 6 Mk = 16

Figure 9 BER performance of information rate 28800 bps atdifferent number of phases when SNR

119889= 4 and 6

phases makes the HODFH system more sensitive to fre-quency offset and timing errors That is to say implementa-tion of this system is more complex but it is a price worthpaying for the information rate improvement and increasedperformance

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] D L Herrick and P K Lee ldquoCHESS a new reliable high speedHF radiordquo in Proceedings of the IEEE Military CommunicationsConference (MILCOM rsquo96) pp 684ndash690 McLean Va USA1996

[2] D L Herrick P K Lee and L L Ledlow Jr ldquoCorrelated fre-quency hopping an improved approach to HF spread spectrumcommunicationsrdquo in Proceedings of the Tactical Communica-tions Conference pp 319ndash324 IEEE Fort Wayne Ind USAApril-May 1996

[3] B Yang and Y Shen ldquoModulated (soft-decision) differentialfrequency hopping systemrdquo in Proceedings of the InternationalConference on Information and Communication Technology(ICICT rsquo07) pp 209ndash212 March 2007

[4] Q Ling and T Li ldquoMessage-driven frequency hopping designand analysisrdquo IEEE Transactions on Wireless Communicationsvol 8 no 4 pp 1773ndash1782 2009

[5] D Wang H Zhao and Z Fan ldquoA new scheme for message-driven FH systemrdquo in Proceedings of the International Con-ference on Future Information Technology and ManagementEngineering (FITME rsquo10) pp 395ndash398 October 2010

[6] L Zhang H Wang and T Li ldquoAnti-jamming message-drivenfrequency hopping Part I System designrdquo IEEE Transactionson Wireless Communications vol 12 no 1 pp 70ndash79 2013

[7] L Zhang and T Li ldquoAnti-jamming message-driven frequencyhopping-Part II capacity analysis under disguised jammingrdquoIEEE Transactions on Wireless Communications vol 12 no 1pp P80ndashP88 2013

[8] J G Proakis Digital Communications McGraw-Hill 4th edi-tion 2001

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

For the119898th order bits119883119899119898

(2 le 119898 le 119873119900) it is decided by

119889119899

119898minus1 and 119889119899

119898 Only the 119889

119899

119898minus1 and 119889119899

119898are estimated correctly

the information bits can be recovered correctly Assume thatthe error probability of 119889

119899

119898minus1 and 119889119899

119898is independent This

is insured by the processing of reshaping the 1-dimensionalset 119878 into 119873

119900-dimensional set S

1198631times1198632timessdotsdotsdottimes119863119873119900 Let 119901119894119895Index be the

correct probability of the 119894th and the 119895th dimensional indexsimultaneity we have

119901119894119895

Index (120576119904

1198730) = (1minus119901Index119894 (

120576119904

1198730))

sdot (1minus119901Index119895 (120576119904

1198730))

= (1minus119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730))

sdot (1minus

119863119895minus 1

119863119895

119901119904HODFH (

120576119904

1198730)) = 1minus

119863119895minus 1

119863119895

sdot 119901119904HODFH (

120576119904

1198730)minus

119863119894minus 1

119863119894

119901119904HODFH (

120576119904

1198730)

+

(119863119894minus 1) (119863

119895minus 1)

119863119894119863119895

(119901119904HODFH (

120576119904

1198730))

2

(16)

Recall that there are 119861119898bits at the119898th order The BER of

the119898th order bits denoted as 119901119887119898HODFH is obtained

119901119887119898HODFH =

2119861119898minus1

2119861119898 minus 1(1minus119901

119898minus1119898Index )

=2119861119898minus1

2119861119898 minus 1(119863119898minus1 minus 1119863119898minus1

119901119904HODFH (

120576119904

1198730)

+119863119898

minus 1119863119898

119901119904HODFH (

120576119904

1198730)

minus(119863

119898minus1 minus 1) (119863119898

minus 1)119863119898minus1119863119898

(119901119904HODFH (

120576119904

1198730))

2)

(17)

Therefore the total BER of a HODFH system denoted as119901119887HODFH can be obtained as

119901119887HODFH (

120576119904

1198730)

= (1198611

sum119873119900

119899=1 119861119899) sdot 119901

1198871HODFH (120576119904

1198730)

+

119873119900

sum

119898=2(

119861119898

sum119873119900

119899=1 119861119899) sdot 119901

119887119898HODFH (120576119904

1198730)

(18)

Since sum119873119900

119899=1 119861119899 = 119871 (18) can be transformed as

119901119887HODFH (

120576119904

1198730) =

1119871

(11986111199011198871HODFH (120576119904

1198730)

+

119873119900

sum

119898=2119861119898119901119887119898HODFH (

120576119904

1198730))

(19)

We assume that one symbol not correctly detected willincur the 119873

119900dimensional index being not correctly esti-

mated That is to say 119901Index119894 and 119901119894119895

Index will satisfy thefollowing equations

119901Index119894 (120576119904

1198730) = 119901

119904HODFH (120576119904

1198730)

119901119894119895

Index (120576119904

1198730)

= (1minus119901119904HODFH (

120576119904

1198730))(1minus119901

119904HODFH (120576119904

1198730))

= 1minus 2119901119904HODFH (

120576119904

1198730)+(119901

119904HODFH (120576119904

1198730))

2

(20)

Substitute (20) into (15) and (17) the uniformof the upperbound of BER for the first and the119898th order bits denoted as119901119880

119887119898HODFH can be obtained as

119901119880

119887119898HODFH =2119861119898minus1

2119861119898 minus 1(2119901

119904HODFH (120576119904

1198730)

minus(119901119904HODFH (

120576119904

1198730))

2) 1 le 119898 le 119873

119900

(21)

When the symbol error probability 119901119904HODFH is small

enough the bit error probability 119901119880

119887119898HODFH can be approx-imated as

119901119880

119887119898HODFH (120576119904

1198730) =

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

1 le 119898 le 119873119900

(22)

Substitute (22) into (19) the upper bound of total BER forthe HODFH system is obtained

119901119880

119887HODFH (120576119904

1198730) =

1119871

119873119900

sum

119898=1119861119898119901119880

119887119898HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(23)

The lower bound of total BER denoted as 119901119871

119887HODFHcan be obtained by assuming that the symbol error onlyincurs bits error in that hopping The error probability ofthe 119894th dimensional index is expressed as (14) and the

6 Mathematical Problems in Engineering

119894th dimensional index error only incurs the 119894th order bitsTherefore the lower bound of total BER is

119901119871

119887HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

(24)

To summarize our discussions above we have the follow-ing

Proposition In HODFH the total BER 119901119887HODFH(1205761199041198730) is

bounded by

1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

le 119901119887HODFH (

120576119904

1198730)

le1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(25)

33 Further Discussions about Bit Error Probability Recallthat there are 119873

119888available frequencies 1198911 1198912 119891119873

119888

forthe HODFH system and they distribute in broad band Itonly occupies narrow band centered at 119891

119894(119894 = 1 2 119873

119888)

available frequencies at any given moment The SNR usuallyis unequal at each narrow band In this occasion the symbolerror probability of the frequency and phase cannot be simplycalculated by (9) and (12) The symbol correct decisionprobability119901

119888119891119894

of the frequency119891119894can be deduced by [8 eqn

(5-4-41) page 309] Because the random variables 119877119894(119894 =

1 2 119873119888) are statistically independent and identically dis-

tributed and we assume that the signals centered at the 119873119888

available frequencies 1198911 1198912 119891119873119888

have equal energy anddifferent noise the probability 119901

119888119891119894

can be expressed as

119901119888119891119894

= int

infin

0

119873119888

prod

119898=1119898 =119894

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)119901

119877119894

(119909) 119889119909 (26)

where 1205902119898

= 121198730119898 is the variance of the AWGN atfrequency 119891

119894and

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)

= int

120590119894119909120590119898

0119901119877119898

(119903119898) 119889119903

119898= 1minus 119890

minus1205902119894119909221205902119898

= 1minus 119890minus1198730119894119909

221198730119898

(27)

Substitute (27) into (26) we obtain the general probability

119901119888119891119894

= int

infin

0

119873119888

prod

119898=2119898 =119894

(1minus 119890minus1198730119894119909

221198730119898) 119901

119877119894

(119909) 119889119909 (28)

We have known that in a HODFH system signal phasedetection is implemented at the condition that the hoppingfrequency centred at 119891

119894is detected correctly for which the

probability is 119901119888119891119894

the probability of phase symbol error canbe calculated as (10) and (11) but the SNR 120576

119875

1199041198730 of phase

per symbol will be substituted by 1205761198730119894 which is the SNRat frequency 119891

119894 Let 119901

(119875119894)

119904HODFH denote the probability of thephase symbol detection error at frequency119891

119894 In this occasion

the overall symbol error probability of the HODFH schemeis calculated by

1199011015840

119904HODFH = 1minus

119873119888

sum

119894=1119901119894sdot 119901

119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

= 1minus1119873119888

119873119888

sum

119894=1119901119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

(29)

where the 119901119894(119894 = 1 2 119873

119888) is the probability that the

signals centered at frequency 119891119894and the 1199011 1199012 119901119873

119888

allequal 1119873

119888 Substitute the probability 119901

119904HODFH by 1199011015840

119904HODFHin (16) and (17) we can calculate the total BER of a HODFHsystem by (18) when it works in the occasion that the119873119888available frequencies 1198911 1198912 119891119873

119888

distribute in broadband and the SNR is not equal at each narrow band centeredat 119891

119894

4 Simulation Results

Without loss of generality in this paper the HODFH systemin the AWGN channel with equal SNR at anywhere wassimulated and its bit error rate (BER) performance wasanalyzed

Example 1 BER performance of the HODFH system inAWGN channel the performance of BER at the informa-tion rates of 9600 bps 19200 bps and 28800 bps in AWGNchannel was compared For the information rate of 9600 bps119873119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and 119861

119896= 1 (119896 =

1 119873119900) For the information rate of 19200 bps 119873

119900= 2

119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 2 (119896 = 1 119873

119900)

For the information rate of 28800 bps 119873119900

= 2 119873119888

= 64119872

119896= 4 (119896 = 1 119873

119888) 1198611 = 4 and 1198612 = 2

Figure 4 is the BER performance of the HODFH systemin AWGN channel The requirement for the energy-per-bitnoise rate (119864

1198871198730) becomes lower when the information

rate increases In practice it is usually possible that onlythe energy-per-symbol noise rate (119864

1199041198730) can be measured

According to formula 1198641199041198730 = 119864

1198871198730 + 10 log(119896) (119896 is the

number of bits per symbol) convert 1198641198871198730 to 119864

1199041198730 and

then the higher 1198641199041198730 is required for the higher information

rate which is reasonable

Example 2 The BER performance of information rate28800 bps at different number of phases assume 119873

119900= 2 and

119861119896= 3 (119896 = 1 119873

119900) The numbers of available frequencies

119873119888are 64 32 and 16 the numbers of phases119872

119896are 4 8 and

16 respectively

Mathematical Problems in Engineering 7

0 1 2 3 4 5 6 7 8 9 10 11

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps No = 2 Nc = 64 Mk = 4 B1 = 4 B2 = 2

Figure 4 The BER performance at information rates of 9600 bps19200 bps and 28800 bps in AWGN channel

0 2 4 6 8 10 12 14 16

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

No = 2 Nc = 64 Mk = 4 Bk = 3

No = 2 Nc = 32 Mk = 8 Bk = 3

No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 5 The BER performance of information rate 28800 bps atdifferent number of phases

As seen in Figure 5 when more phrases are used whilekeeping the information rate the same there is poorerperformance for the HODFH system It will spend about6 dB extra 119864

1198871198730 to obtain the same BER performance

of 10minus3 when the number of phases increases from 8 to16 Therefore the number of phases should be kept to aminimum if the number of transmission signals has reachedthe requirements

Example 3 TheBERperformance of theHODFH system andthe conventional DFH (CDFH) system for the informationrate of 9600 bps the HODFH system has the followingparameters 119873

119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and

0 2 4 6 8 10 12

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

9600bps CDFH Nc = 128 BPH = 2

19200 bps CDFH Nc = 128 BPH = 4

Figure 6 The BER performance of the HODFH system and theCDFH system

119861119896= 1 (119896 = 1 119873

119900) for the information rate of 19200 bps

it has parameters 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) The CDFH has the following

parameters for information rate of 9600 bps 119873119888= 128 and

BPH = 2 for information rate of 19200 bps 119873119888= 128 and

BPH = 4

Figure 6 shows the BER performance comparison ofthe HODFH system and the conventional DFH system forinformation rates of 9600 bps and 19200 bps At the sameinformation rate the HODFH system has a better BERperformance than the conventional DFH system The higherinformation rate is the higher gain is obtained by theHODFH system when compared to the conventional DFHsystem

Example 4 Compare the BER performances of the HODFHsystem at three information rates (9600 bps 19200 bps and28800 bps) with the theoretical BER performance of QPSK8PSK 16PSK 32PSK and 64PSK Assume the HODFHsystem has the following parameters respectively For theinformation rate of 9600 bps119873

119900= 2119873

119888= 128119872

119896= 2 (119896 =

1 119873119888) and 119861

119896= 1 (119896 = 1 119873

119900) For the information

rate of 19200 bps 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) For the information rate of

28800 bps we have three combinations the first is 119873119900

= 2119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

the second is 119873119900= 2 119873

119888= 32 119872

119896= 8 (119896 = 1 119873

119888) and

119861119896

= 3 (119896 = 1 119873119900) and the third is 119873

119900= 2 119873

119888= 16

119872119896= 16 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

Figure 7 presents the BER performance of the HODFHsystem and the theoretical BER performance of QPSK 8PSK16PSK 32PSK and 64PSK When the information rate is9600 bps the HODFH system carries two bits of information

8 Mathematical Problems in Engineering

0 5 10 15 20 25 30

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

QPSK theory8 PSK theory16 PSK theor32 PSK theor64 PSK theor9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 3

28800 bps HODFH No = 2 Nc = 64 Mk = 8 Bk = 3

28800 bps HODFH No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 7 The BER performance of the HODFH system andtheoretical BER performance of the M-PSK

per hop (symbol) which is the same as the QPSK the BERperformance of the HODFH system is worse than the QPSKWhen the information rate is 19200 bps the HODFH systemcarries four bits of information per hop (symbol) which is thesame as the 16PSK but it has better BERperformance than the16PSK When the information rate is 28800 bps the HODFHsystemcarries six bits of information per hop (symbol) whichis the same as the 64PSK but its performance is much betterthan the 64PSK no matter how many phases are used bythe HODFH system The BER performance of the HODFHsystem is better than theM-PSK when the information rate ishighThe carrier frequencies of theHODFH systems are usedto transmit information the HODFH systems have higherinformation rate without extra cost on power and do not needextra cost on power when the numbers of phases adopted bythe HODFH systems and the M-PSK are equal The HODFHsystems have better BER performance at the same energy-per-bit noise rate compared to the PSK

Example 5 Compare the BER performance of informationrate 19200 bps at the occasion that the SNR is equal andunequal at each narrow band Assume that the SNRs at119873

1198882

narrow bands are different with the SNRs at the other 1198731198882

narrow bands Use SNR119889to denote the difference among

the 119873119888SNRs The SNR

119889increases from 2 to 8 by increment

2 Use the average SNR in the broad band as the symbolsignal-to-noise ratio The HODFH system has the followingparameters 119873

119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888) and

119861119896= 2 (119896 = 1 119873

119900)

0 2 4 6 8 10 12 14

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

Equal SNRUnequal SNR SNRd = 2

Unequal SNR SNRd = 4

Unequal SNR SNRd = 6

Unequal SNR SNRd = 8

Figure 8 The BER performance of information rate 19200 bps atSNR

119889= 2 4 6 and 8

Example 6 Compare the BER performance of informationrate 28800 bps at different number of phases when the SNRdifference SNR

119889= 4 and 6 The SNR is similar as Example 5

The HODFH system has the following parameters 119873119900= 2

119873119888= 64 119861

119896= 3 (119896 = 1 119873

119900) and 119872

119896=4 8 and 16 (119896 =

1 119873119888)

As seen in Figure 8 the BER performance is worse whenthe SNR is unequal at each narrow band despite the averageSNR of the unequal signal-to-noise ratio broad band beingthe same as the SNR of the equal signal-to-noise ratio Asthe difference of the SNRs is increasing the performance isdegraded

Figure 9 shows that the BER performance is degradedwhen the number of phases used by the HODFH systemincreases This recalls the conclusion in Example 2 that thenumber of phases should be kept to a minimum if the num-ber of transmission signals has reached the requirements

5 Summary

In the future communication will have higher informationrate requirements and this paper proposed a solution forthis with a novel differential frequency hopping high orderdifferential frequency hopping method This was done byconumbering the frequencies and their phases so that thesystem information rate was improved without having toincrease the frequency resources The BER performance ofthe HODFH system is better than the conventional DFHsystem and M-PSK in the AWGN channel In this workthe number of phases can be adjusted to match the channelconditions to obtain the optimal performance

Although using phases to expand the signal set canimprove the bandwidth efficiency their introduction of the

Mathematical Problems in Engineering 9

0 2 4 6 8 10 12 14 16 18

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus7

10minus6

EbN0 (dB)

SNRd = 4 Mk = 4

SNRd = 4 Mk = 8

SNRd = 4 Mk = 16

SNRd = 6 Mk = 4

SNRd = 6 Mk = 8

SNRd = 6 Mk = 16

Figure 9 BER performance of information rate 28800 bps atdifferent number of phases when SNR

119889= 4 and 6

phases makes the HODFH system more sensitive to fre-quency offset and timing errors That is to say implementa-tion of this system is more complex but it is a price worthpaying for the information rate improvement and increasedperformance

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] D L Herrick and P K Lee ldquoCHESS a new reliable high speedHF radiordquo in Proceedings of the IEEE Military CommunicationsConference (MILCOM rsquo96) pp 684ndash690 McLean Va USA1996

[2] D L Herrick P K Lee and L L Ledlow Jr ldquoCorrelated fre-quency hopping an improved approach to HF spread spectrumcommunicationsrdquo in Proceedings of the Tactical Communica-tions Conference pp 319ndash324 IEEE Fort Wayne Ind USAApril-May 1996

[3] B Yang and Y Shen ldquoModulated (soft-decision) differentialfrequency hopping systemrdquo in Proceedings of the InternationalConference on Information and Communication Technology(ICICT rsquo07) pp 209ndash212 March 2007

[4] Q Ling and T Li ldquoMessage-driven frequency hopping designand analysisrdquo IEEE Transactions on Wireless Communicationsvol 8 no 4 pp 1773ndash1782 2009

[5] D Wang H Zhao and Z Fan ldquoA new scheme for message-driven FH systemrdquo in Proceedings of the International Con-ference on Future Information Technology and ManagementEngineering (FITME rsquo10) pp 395ndash398 October 2010

[6] L Zhang H Wang and T Li ldquoAnti-jamming message-drivenfrequency hopping Part I System designrdquo IEEE Transactionson Wireless Communications vol 12 no 1 pp 70ndash79 2013

[7] L Zhang and T Li ldquoAnti-jamming message-driven frequencyhopping-Part II capacity analysis under disguised jammingrdquoIEEE Transactions on Wireless Communications vol 12 no 1pp P80ndashP88 2013

[8] J G Proakis Digital Communications McGraw-Hill 4th edi-tion 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

119894th dimensional index error only incurs the 119894th order bitsTherefore the lower bound of total BER is

119901119871

119887HODFH (120576119904

1198730)

=1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

(24)

To summarize our discussions above we have the follow-ing

Proposition In HODFH the total BER 119901119887HODFH(1205761199041198730) is

bounded by

1119871

119873119900

sum

119898=1119861119898

2119861119898minus1 (119863119898

minus 1)(2119861119898 minus 1)119863

119898

119901119904HODFH (

120576119904

1198730)

le 119901119887HODFH (

120576119904

1198730)

le1119871

119873119900

sum

119898=1119861119898

21198611198982119861119898 minus 1

119901119904HODFH (

120576119904

1198730)

(25)

33 Further Discussions about Bit Error Probability Recallthat there are 119873

119888available frequencies 1198911 1198912 119891119873

119888

forthe HODFH system and they distribute in broad band Itonly occupies narrow band centered at 119891

119894(119894 = 1 2 119873

119888)

available frequencies at any given moment The SNR usuallyis unequal at each narrow band In this occasion the symbolerror probability of the frequency and phase cannot be simplycalculated by (9) and (12) The symbol correct decisionprobability119901

119888119891119894

of the frequency119891119894can be deduced by [8 eqn

(5-4-41) page 309] Because the random variables 119877119894(119894 =

1 2 119873119888) are statistically independent and identically dis-

tributed and we assume that the signals centered at the 119873119888

available frequencies 1198911 1198912 119891119873119888

have equal energy anddifferent noise the probability 119901

119888119891119894

can be expressed as

119901119888119891119894

= int

infin

0

119873119888

prod

119898=1119898 =119894

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)119901

119877119894

(119909) 119889119909 (26)

where 1205902119898

= 121198730119898 is the variance of the AWGN atfrequency 119891

119894and

119875(119877119898lt

120590119894

120590119898

119877119894| 119877

119894=119909)

= int

120590119894119909120590119898

0119901119877119898

(119903119898) 119889119903

119898= 1minus 119890

minus1205902119894119909221205902119898

= 1minus 119890minus1198730119894119909

221198730119898

(27)

Substitute (27) into (26) we obtain the general probability

119901119888119891119894

= int

infin

0

119873119888

prod

119898=2119898 =119894

(1minus 119890minus1198730119894119909

221198730119898) 119901

119877119894

(119909) 119889119909 (28)

We have known that in a HODFH system signal phasedetection is implemented at the condition that the hoppingfrequency centred at 119891

119894is detected correctly for which the

probability is 119901119888119891119894

the probability of phase symbol error canbe calculated as (10) and (11) but the SNR 120576

119875

1199041198730 of phase

per symbol will be substituted by 1205761198730119894 which is the SNRat frequency 119891

119894 Let 119901

(119875119894)

119904HODFH denote the probability of thephase symbol detection error at frequency119891

119894 In this occasion

the overall symbol error probability of the HODFH schemeis calculated by

1199011015840

119904HODFH = 1minus

119873119888

sum

119894=1119901119894sdot 119901

119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

= 1minus1119873119888

119873119888

sum

119894=1119901119888119891119894

sdot (1minus119901(119875119894)

119904HODFH)

(29)

where the 119901119894(119894 = 1 2 119873

119888) is the probability that the

signals centered at frequency 119891119894and the 1199011 1199012 119901119873

119888

allequal 1119873

119888 Substitute the probability 119901

119904HODFH by 1199011015840

119904HODFHin (16) and (17) we can calculate the total BER of a HODFHsystem by (18) when it works in the occasion that the119873119888available frequencies 1198911 1198912 119891119873

119888

distribute in broadband and the SNR is not equal at each narrow band centeredat 119891

119894

4 Simulation Results

Without loss of generality in this paper the HODFH systemin the AWGN channel with equal SNR at anywhere wassimulated and its bit error rate (BER) performance wasanalyzed

Example 1 BER performance of the HODFH system inAWGN channel the performance of BER at the informa-tion rates of 9600 bps 19200 bps and 28800 bps in AWGNchannel was compared For the information rate of 9600 bps119873119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and 119861

119896= 1 (119896 =

1 119873119900) For the information rate of 19200 bps 119873

119900= 2

119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 2 (119896 = 1 119873

119900)

For the information rate of 28800 bps 119873119900

= 2 119873119888

= 64119872

119896= 4 (119896 = 1 119873

119888) 1198611 = 4 and 1198612 = 2

Figure 4 is the BER performance of the HODFH systemin AWGN channel The requirement for the energy-per-bitnoise rate (119864

1198871198730) becomes lower when the information

rate increases In practice it is usually possible that onlythe energy-per-symbol noise rate (119864

1199041198730) can be measured

According to formula 1198641199041198730 = 119864

1198871198730 + 10 log(119896) (119896 is the

number of bits per symbol) convert 1198641198871198730 to 119864

1199041198730 and

then the higher 1198641199041198730 is required for the higher information

rate which is reasonable

Example 2 The BER performance of information rate28800 bps at different number of phases assume 119873

119900= 2 and

119861119896= 3 (119896 = 1 119873

119900) The numbers of available frequencies

119873119888are 64 32 and 16 the numbers of phases119872

119896are 4 8 and

16 respectively

Mathematical Problems in Engineering 7

0 1 2 3 4 5 6 7 8 9 10 11

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps No = 2 Nc = 64 Mk = 4 B1 = 4 B2 = 2

Figure 4 The BER performance at information rates of 9600 bps19200 bps and 28800 bps in AWGN channel

0 2 4 6 8 10 12 14 16

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

No = 2 Nc = 64 Mk = 4 Bk = 3

No = 2 Nc = 32 Mk = 8 Bk = 3

No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 5 The BER performance of information rate 28800 bps atdifferent number of phases

As seen in Figure 5 when more phrases are used whilekeeping the information rate the same there is poorerperformance for the HODFH system It will spend about6 dB extra 119864

1198871198730 to obtain the same BER performance

of 10minus3 when the number of phases increases from 8 to16 Therefore the number of phases should be kept to aminimum if the number of transmission signals has reachedthe requirements

Example 3 TheBERperformance of theHODFH system andthe conventional DFH (CDFH) system for the informationrate of 9600 bps the HODFH system has the followingparameters 119873

119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and

0 2 4 6 8 10 12

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

9600bps CDFH Nc = 128 BPH = 2

19200 bps CDFH Nc = 128 BPH = 4

Figure 6 The BER performance of the HODFH system and theCDFH system

119861119896= 1 (119896 = 1 119873

119900) for the information rate of 19200 bps

it has parameters 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) The CDFH has the following

parameters for information rate of 9600 bps 119873119888= 128 and

BPH = 2 for information rate of 19200 bps 119873119888= 128 and

BPH = 4

Figure 6 shows the BER performance comparison ofthe HODFH system and the conventional DFH system forinformation rates of 9600 bps and 19200 bps At the sameinformation rate the HODFH system has a better BERperformance than the conventional DFH system The higherinformation rate is the higher gain is obtained by theHODFH system when compared to the conventional DFHsystem

Example 4 Compare the BER performances of the HODFHsystem at three information rates (9600 bps 19200 bps and28800 bps) with the theoretical BER performance of QPSK8PSK 16PSK 32PSK and 64PSK Assume the HODFHsystem has the following parameters respectively For theinformation rate of 9600 bps119873

119900= 2119873

119888= 128119872

119896= 2 (119896 =

1 119873119888) and 119861

119896= 1 (119896 = 1 119873

119900) For the information

rate of 19200 bps 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) For the information rate of

28800 bps we have three combinations the first is 119873119900

= 2119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

the second is 119873119900= 2 119873

119888= 32 119872

119896= 8 (119896 = 1 119873

119888) and

119861119896

= 3 (119896 = 1 119873119900) and the third is 119873

119900= 2 119873

119888= 16

119872119896= 16 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

Figure 7 presents the BER performance of the HODFHsystem and the theoretical BER performance of QPSK 8PSK16PSK 32PSK and 64PSK When the information rate is9600 bps the HODFH system carries two bits of information

8 Mathematical Problems in Engineering

0 5 10 15 20 25 30

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

QPSK theory8 PSK theory16 PSK theor32 PSK theor64 PSK theor9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 3

28800 bps HODFH No = 2 Nc = 64 Mk = 8 Bk = 3

28800 bps HODFH No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 7 The BER performance of the HODFH system andtheoretical BER performance of the M-PSK

per hop (symbol) which is the same as the QPSK the BERperformance of the HODFH system is worse than the QPSKWhen the information rate is 19200 bps the HODFH systemcarries four bits of information per hop (symbol) which is thesame as the 16PSK but it has better BERperformance than the16PSK When the information rate is 28800 bps the HODFHsystemcarries six bits of information per hop (symbol) whichis the same as the 64PSK but its performance is much betterthan the 64PSK no matter how many phases are used bythe HODFH system The BER performance of the HODFHsystem is better than theM-PSK when the information rate ishighThe carrier frequencies of theHODFH systems are usedto transmit information the HODFH systems have higherinformation rate without extra cost on power and do not needextra cost on power when the numbers of phases adopted bythe HODFH systems and the M-PSK are equal The HODFHsystems have better BER performance at the same energy-per-bit noise rate compared to the PSK

Example 5 Compare the BER performance of informationrate 19200 bps at the occasion that the SNR is equal andunequal at each narrow band Assume that the SNRs at119873

1198882

narrow bands are different with the SNRs at the other 1198731198882

narrow bands Use SNR119889to denote the difference among

the 119873119888SNRs The SNR

119889increases from 2 to 8 by increment

2 Use the average SNR in the broad band as the symbolsignal-to-noise ratio The HODFH system has the followingparameters 119873

119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888) and

119861119896= 2 (119896 = 1 119873

119900)

0 2 4 6 8 10 12 14

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

Equal SNRUnequal SNR SNRd = 2

Unequal SNR SNRd = 4

Unequal SNR SNRd = 6

Unequal SNR SNRd = 8

Figure 8 The BER performance of information rate 19200 bps atSNR

119889= 2 4 6 and 8

Example 6 Compare the BER performance of informationrate 28800 bps at different number of phases when the SNRdifference SNR

119889= 4 and 6 The SNR is similar as Example 5

The HODFH system has the following parameters 119873119900= 2

119873119888= 64 119861

119896= 3 (119896 = 1 119873

119900) and 119872

119896=4 8 and 16 (119896 =

1 119873119888)

As seen in Figure 8 the BER performance is worse whenthe SNR is unequal at each narrow band despite the averageSNR of the unequal signal-to-noise ratio broad band beingthe same as the SNR of the equal signal-to-noise ratio Asthe difference of the SNRs is increasing the performance isdegraded

Figure 9 shows that the BER performance is degradedwhen the number of phases used by the HODFH systemincreases This recalls the conclusion in Example 2 that thenumber of phases should be kept to a minimum if the num-ber of transmission signals has reached the requirements

5 Summary

In the future communication will have higher informationrate requirements and this paper proposed a solution forthis with a novel differential frequency hopping high orderdifferential frequency hopping method This was done byconumbering the frequencies and their phases so that thesystem information rate was improved without having toincrease the frequency resources The BER performance ofthe HODFH system is better than the conventional DFHsystem and M-PSK in the AWGN channel In this workthe number of phases can be adjusted to match the channelconditions to obtain the optimal performance

Although using phases to expand the signal set canimprove the bandwidth efficiency their introduction of the

Mathematical Problems in Engineering 9

0 2 4 6 8 10 12 14 16 18

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus7

10minus6

EbN0 (dB)

SNRd = 4 Mk = 4

SNRd = 4 Mk = 8

SNRd = 4 Mk = 16

SNRd = 6 Mk = 4

SNRd = 6 Mk = 8

SNRd = 6 Mk = 16

Figure 9 BER performance of information rate 28800 bps atdifferent number of phases when SNR

119889= 4 and 6

phases makes the HODFH system more sensitive to fre-quency offset and timing errors That is to say implementa-tion of this system is more complex but it is a price worthpaying for the information rate improvement and increasedperformance

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] D L Herrick and P K Lee ldquoCHESS a new reliable high speedHF radiordquo in Proceedings of the IEEE Military CommunicationsConference (MILCOM rsquo96) pp 684ndash690 McLean Va USA1996

[2] D L Herrick P K Lee and L L Ledlow Jr ldquoCorrelated fre-quency hopping an improved approach to HF spread spectrumcommunicationsrdquo in Proceedings of the Tactical Communica-tions Conference pp 319ndash324 IEEE Fort Wayne Ind USAApril-May 1996

[3] B Yang and Y Shen ldquoModulated (soft-decision) differentialfrequency hopping systemrdquo in Proceedings of the InternationalConference on Information and Communication Technology(ICICT rsquo07) pp 209ndash212 March 2007

[4] Q Ling and T Li ldquoMessage-driven frequency hopping designand analysisrdquo IEEE Transactions on Wireless Communicationsvol 8 no 4 pp 1773ndash1782 2009

[5] D Wang H Zhao and Z Fan ldquoA new scheme for message-driven FH systemrdquo in Proceedings of the International Con-ference on Future Information Technology and ManagementEngineering (FITME rsquo10) pp 395ndash398 October 2010

[6] L Zhang H Wang and T Li ldquoAnti-jamming message-drivenfrequency hopping Part I System designrdquo IEEE Transactionson Wireless Communications vol 12 no 1 pp 70ndash79 2013

[7] L Zhang and T Li ldquoAnti-jamming message-driven frequencyhopping-Part II capacity analysis under disguised jammingrdquoIEEE Transactions on Wireless Communications vol 12 no 1pp P80ndashP88 2013

[8] J G Proakis Digital Communications McGraw-Hill 4th edi-tion 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

0 1 2 3 4 5 6 7 8 9 10 11

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps No = 2 Nc = 64 Mk = 4 B1 = 4 B2 = 2

Figure 4 The BER performance at information rates of 9600 bps19200 bps and 28800 bps in AWGN channel

0 2 4 6 8 10 12 14 16

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

No = 2 Nc = 64 Mk = 4 Bk = 3

No = 2 Nc = 32 Mk = 8 Bk = 3

No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 5 The BER performance of information rate 28800 bps atdifferent number of phases

As seen in Figure 5 when more phrases are used whilekeeping the information rate the same there is poorerperformance for the HODFH system It will spend about6 dB extra 119864

1198871198730 to obtain the same BER performance

of 10minus3 when the number of phases increases from 8 to16 Therefore the number of phases should be kept to aminimum if the number of transmission signals has reachedthe requirements

Example 3 TheBERperformance of theHODFH system andthe conventional DFH (CDFH) system for the informationrate of 9600 bps the HODFH system has the followingparameters 119873

119900= 2 119873

119888= 128 119872

119896= 2 (119896 = 1 119873

119888) and

0 2 4 6 8 10 12

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

9600bps CDFH Nc = 128 BPH = 2

19200 bps CDFH Nc = 128 BPH = 4

Figure 6 The BER performance of the HODFH system and theCDFH system

119861119896= 1 (119896 = 1 119873

119900) for the information rate of 19200 bps

it has parameters 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) The CDFH has the following

parameters for information rate of 9600 bps 119873119888= 128 and

BPH = 2 for information rate of 19200 bps 119873119888= 128 and

BPH = 4

Figure 6 shows the BER performance comparison ofthe HODFH system and the conventional DFH system forinformation rates of 9600 bps and 19200 bps At the sameinformation rate the HODFH system has a better BERperformance than the conventional DFH system The higherinformation rate is the higher gain is obtained by theHODFH system when compared to the conventional DFHsystem

Example 4 Compare the BER performances of the HODFHsystem at three information rates (9600 bps 19200 bps and28800 bps) with the theoretical BER performance of QPSK8PSK 16PSK 32PSK and 64PSK Assume the HODFHsystem has the following parameters respectively For theinformation rate of 9600 bps119873

119900= 2119873

119888= 128119872

119896= 2 (119896 =

1 119873119888) and 119861

119896= 1 (119896 = 1 119873

119900) For the information

rate of 19200 bps 119873119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888)

and 119861119896

= 2 (119896 = 1 119873119900) For the information rate of

28800 bps we have three combinations the first is 119873119900

= 2119873119888= 64119872

119896= 4 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

the second is 119873119900= 2 119873

119888= 32 119872

119896= 8 (119896 = 1 119873

119888) and

119861119896

= 3 (119896 = 1 119873119900) and the third is 119873

119900= 2 119873

119888= 16

119872119896= 16 (119896 = 1 119873

119888) and 119861

119896= 3 (119896 = 1 119873

119900)

Figure 7 presents the BER performance of the HODFHsystem and the theoretical BER performance of QPSK 8PSK16PSK 32PSK and 64PSK When the information rate is9600 bps the HODFH system carries two bits of information

8 Mathematical Problems in Engineering

0 5 10 15 20 25 30

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

QPSK theory8 PSK theory16 PSK theor32 PSK theor64 PSK theor9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 3

28800 bps HODFH No = 2 Nc = 64 Mk = 8 Bk = 3

28800 bps HODFH No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 7 The BER performance of the HODFH system andtheoretical BER performance of the M-PSK

per hop (symbol) which is the same as the QPSK the BERperformance of the HODFH system is worse than the QPSKWhen the information rate is 19200 bps the HODFH systemcarries four bits of information per hop (symbol) which is thesame as the 16PSK but it has better BERperformance than the16PSK When the information rate is 28800 bps the HODFHsystemcarries six bits of information per hop (symbol) whichis the same as the 64PSK but its performance is much betterthan the 64PSK no matter how many phases are used bythe HODFH system The BER performance of the HODFHsystem is better than theM-PSK when the information rate ishighThe carrier frequencies of theHODFH systems are usedto transmit information the HODFH systems have higherinformation rate without extra cost on power and do not needextra cost on power when the numbers of phases adopted bythe HODFH systems and the M-PSK are equal The HODFHsystems have better BER performance at the same energy-per-bit noise rate compared to the PSK

Example 5 Compare the BER performance of informationrate 19200 bps at the occasion that the SNR is equal andunequal at each narrow band Assume that the SNRs at119873

1198882

narrow bands are different with the SNRs at the other 1198731198882

narrow bands Use SNR119889to denote the difference among

the 119873119888SNRs The SNR

119889increases from 2 to 8 by increment

2 Use the average SNR in the broad band as the symbolsignal-to-noise ratio The HODFH system has the followingparameters 119873

119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888) and

119861119896= 2 (119896 = 1 119873

119900)

0 2 4 6 8 10 12 14

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

Equal SNRUnequal SNR SNRd = 2

Unequal SNR SNRd = 4

Unequal SNR SNRd = 6

Unequal SNR SNRd = 8

Figure 8 The BER performance of information rate 19200 bps atSNR

119889= 2 4 6 and 8

Example 6 Compare the BER performance of informationrate 28800 bps at different number of phases when the SNRdifference SNR

119889= 4 and 6 The SNR is similar as Example 5

The HODFH system has the following parameters 119873119900= 2

119873119888= 64 119861

119896= 3 (119896 = 1 119873

119900) and 119872

119896=4 8 and 16 (119896 =

1 119873119888)

As seen in Figure 8 the BER performance is worse whenthe SNR is unequal at each narrow band despite the averageSNR of the unequal signal-to-noise ratio broad band beingthe same as the SNR of the equal signal-to-noise ratio Asthe difference of the SNRs is increasing the performance isdegraded

Figure 9 shows that the BER performance is degradedwhen the number of phases used by the HODFH systemincreases This recalls the conclusion in Example 2 that thenumber of phases should be kept to a minimum if the num-ber of transmission signals has reached the requirements

5 Summary

In the future communication will have higher informationrate requirements and this paper proposed a solution forthis with a novel differential frequency hopping high orderdifferential frequency hopping method This was done byconumbering the frequencies and their phases so that thesystem information rate was improved without having toincrease the frequency resources The BER performance ofthe HODFH system is better than the conventional DFHsystem and M-PSK in the AWGN channel In this workthe number of phases can be adjusted to match the channelconditions to obtain the optimal performance

Although using phases to expand the signal set canimprove the bandwidth efficiency their introduction of the

Mathematical Problems in Engineering 9

0 2 4 6 8 10 12 14 16 18

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus7

10minus6

EbN0 (dB)

SNRd = 4 Mk = 4

SNRd = 4 Mk = 8

SNRd = 4 Mk = 16

SNRd = 6 Mk = 4

SNRd = 6 Mk = 8

SNRd = 6 Mk = 16

Figure 9 BER performance of information rate 28800 bps atdifferent number of phases when SNR

119889= 4 and 6

phases makes the HODFH system more sensitive to fre-quency offset and timing errors That is to say implementa-tion of this system is more complex but it is a price worthpaying for the information rate improvement and increasedperformance

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] D L Herrick and P K Lee ldquoCHESS a new reliable high speedHF radiordquo in Proceedings of the IEEE Military CommunicationsConference (MILCOM rsquo96) pp 684ndash690 McLean Va USA1996

[2] D L Herrick P K Lee and L L Ledlow Jr ldquoCorrelated fre-quency hopping an improved approach to HF spread spectrumcommunicationsrdquo in Proceedings of the Tactical Communica-tions Conference pp 319ndash324 IEEE Fort Wayne Ind USAApril-May 1996

[3] B Yang and Y Shen ldquoModulated (soft-decision) differentialfrequency hopping systemrdquo in Proceedings of the InternationalConference on Information and Communication Technology(ICICT rsquo07) pp 209ndash212 March 2007

[4] Q Ling and T Li ldquoMessage-driven frequency hopping designand analysisrdquo IEEE Transactions on Wireless Communicationsvol 8 no 4 pp 1773ndash1782 2009

[5] D Wang H Zhao and Z Fan ldquoA new scheme for message-driven FH systemrdquo in Proceedings of the International Con-ference on Future Information Technology and ManagementEngineering (FITME rsquo10) pp 395ndash398 October 2010

[6] L Zhang H Wang and T Li ldquoAnti-jamming message-drivenfrequency hopping Part I System designrdquo IEEE Transactionson Wireless Communications vol 12 no 1 pp 70ndash79 2013

[7] L Zhang and T Li ldquoAnti-jamming message-driven frequencyhopping-Part II capacity analysis under disguised jammingrdquoIEEE Transactions on Wireless Communications vol 12 no 1pp P80ndashP88 2013

[8] J G Proakis Digital Communications McGraw-Hill 4th edi-tion 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

0 5 10 15 20 25 30

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

QPSK theory8 PSK theory16 PSK theor32 PSK theor64 PSK theor9600bps HODFH No = 2 Nc = 128 Mk = 2 Bk = 1

19200 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 2

28800 bps HODFH No = 2 Nc = 64 Mk = 4 Bk = 3

28800 bps HODFH No = 2 Nc = 64 Mk = 8 Bk = 3

28800 bps HODFH No = 2 Nc = 16 Mk = 16 Bk = 3

Figure 7 The BER performance of the HODFH system andtheoretical BER performance of the M-PSK

per hop (symbol) which is the same as the QPSK the BERperformance of the HODFH system is worse than the QPSKWhen the information rate is 19200 bps the HODFH systemcarries four bits of information per hop (symbol) which is thesame as the 16PSK but it has better BERperformance than the16PSK When the information rate is 28800 bps the HODFHsystemcarries six bits of information per hop (symbol) whichis the same as the 64PSK but its performance is much betterthan the 64PSK no matter how many phases are used bythe HODFH system The BER performance of the HODFHsystem is better than theM-PSK when the information rate ishighThe carrier frequencies of theHODFH systems are usedto transmit information the HODFH systems have higherinformation rate without extra cost on power and do not needextra cost on power when the numbers of phases adopted bythe HODFH systems and the M-PSK are equal The HODFHsystems have better BER performance at the same energy-per-bit noise rate compared to the PSK

Example 5 Compare the BER performance of informationrate 19200 bps at the occasion that the SNR is equal andunequal at each narrow band Assume that the SNRs at119873

1198882

narrow bands are different with the SNRs at the other 1198731198882

narrow bands Use SNR119889to denote the difference among

the 119873119888SNRs The SNR

119889increases from 2 to 8 by increment

2 Use the average SNR in the broad band as the symbolsignal-to-noise ratio The HODFH system has the followingparameters 119873

119900= 2 119873

119888= 64 119872

119896= 4 (119896 = 1 119873

119888) and

119861119896= 2 (119896 = 1 119873

119900)

0 2 4 6 8 10 12 14

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

EbN0 (dB)

Equal SNRUnequal SNR SNRd = 2

Unequal SNR SNRd = 4

Unequal SNR SNRd = 6

Unequal SNR SNRd = 8

Figure 8 The BER performance of information rate 19200 bps atSNR

119889= 2 4 6 and 8

Example 6 Compare the BER performance of informationrate 28800 bps at different number of phases when the SNRdifference SNR

119889= 4 and 6 The SNR is similar as Example 5

The HODFH system has the following parameters 119873119900= 2

119873119888= 64 119861

119896= 3 (119896 = 1 119873

119900) and 119872

119896=4 8 and 16 (119896 =

1 119873119888)

As seen in Figure 8 the BER performance is worse whenthe SNR is unequal at each narrow band despite the averageSNR of the unequal signal-to-noise ratio broad band beingthe same as the SNR of the equal signal-to-noise ratio Asthe difference of the SNRs is increasing the performance isdegraded

Figure 9 shows that the BER performance is degradedwhen the number of phases used by the HODFH systemincreases This recalls the conclusion in Example 2 that thenumber of phases should be kept to a minimum if the num-ber of transmission signals has reached the requirements

5 Summary

In the future communication will have higher informationrate requirements and this paper proposed a solution forthis with a novel differential frequency hopping high orderdifferential frequency hopping method This was done byconumbering the frequencies and their phases so that thesystem information rate was improved without having toincrease the frequency resources The BER performance ofthe HODFH system is better than the conventional DFHsystem and M-PSK in the AWGN channel In this workthe number of phases can be adjusted to match the channelconditions to obtain the optimal performance

Although using phases to expand the signal set canimprove the bandwidth efficiency their introduction of the

Mathematical Problems in Engineering 9

0 2 4 6 8 10 12 14 16 18

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus7

10minus6

EbN0 (dB)

SNRd = 4 Mk = 4

SNRd = 4 Mk = 8

SNRd = 4 Mk = 16

SNRd = 6 Mk = 4

SNRd = 6 Mk = 8

SNRd = 6 Mk = 16

Figure 9 BER performance of information rate 28800 bps atdifferent number of phases when SNR

119889= 4 and 6

phases makes the HODFH system more sensitive to fre-quency offset and timing errors That is to say implementa-tion of this system is more complex but it is a price worthpaying for the information rate improvement and increasedperformance

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] D L Herrick and P K Lee ldquoCHESS a new reliable high speedHF radiordquo in Proceedings of the IEEE Military CommunicationsConference (MILCOM rsquo96) pp 684ndash690 McLean Va USA1996

[2] D L Herrick P K Lee and L L Ledlow Jr ldquoCorrelated fre-quency hopping an improved approach to HF spread spectrumcommunicationsrdquo in Proceedings of the Tactical Communica-tions Conference pp 319ndash324 IEEE Fort Wayne Ind USAApril-May 1996

[3] B Yang and Y Shen ldquoModulated (soft-decision) differentialfrequency hopping systemrdquo in Proceedings of the InternationalConference on Information and Communication Technology(ICICT rsquo07) pp 209ndash212 March 2007

[4] Q Ling and T Li ldquoMessage-driven frequency hopping designand analysisrdquo IEEE Transactions on Wireless Communicationsvol 8 no 4 pp 1773ndash1782 2009

[5] D Wang H Zhao and Z Fan ldquoA new scheme for message-driven FH systemrdquo in Proceedings of the International Con-ference on Future Information Technology and ManagementEngineering (FITME rsquo10) pp 395ndash398 October 2010

[6] L Zhang H Wang and T Li ldquoAnti-jamming message-drivenfrequency hopping Part I System designrdquo IEEE Transactionson Wireless Communications vol 12 no 1 pp 70ndash79 2013

[7] L Zhang and T Li ldquoAnti-jamming message-driven frequencyhopping-Part II capacity analysis under disguised jammingrdquoIEEE Transactions on Wireless Communications vol 12 no 1pp P80ndashP88 2013

[8] J G Proakis Digital Communications McGraw-Hill 4th edi-tion 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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International Journal of

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

0 2 4 6 8 10 12 14 16 18

BER

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus7

10minus6

EbN0 (dB)

SNRd = 4 Mk = 4

SNRd = 4 Mk = 8

SNRd = 4 Mk = 16

SNRd = 6 Mk = 4

SNRd = 6 Mk = 8

SNRd = 6 Mk = 16

Figure 9 BER performance of information rate 28800 bps atdifferent number of phases when SNR

119889= 4 and 6

phases makes the HODFH system more sensitive to fre-quency offset and timing errors That is to say implementa-tion of this system is more complex but it is a price worthpaying for the information rate improvement and increasedperformance

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] D L Herrick and P K Lee ldquoCHESS a new reliable high speedHF radiordquo in Proceedings of the IEEE Military CommunicationsConference (MILCOM rsquo96) pp 684ndash690 McLean Va USA1996

[2] D L Herrick P K Lee and L L Ledlow Jr ldquoCorrelated fre-quency hopping an improved approach to HF spread spectrumcommunicationsrdquo in Proceedings of the Tactical Communica-tions Conference pp 319ndash324 IEEE Fort Wayne Ind USAApril-May 1996

[3] B Yang and Y Shen ldquoModulated (soft-decision) differentialfrequency hopping systemrdquo in Proceedings of the InternationalConference on Information and Communication Technology(ICICT rsquo07) pp 209ndash212 March 2007

[4] Q Ling and T Li ldquoMessage-driven frequency hopping designand analysisrdquo IEEE Transactions on Wireless Communicationsvol 8 no 4 pp 1773ndash1782 2009

[5] D Wang H Zhao and Z Fan ldquoA new scheme for message-driven FH systemrdquo in Proceedings of the International Con-ference on Future Information Technology and ManagementEngineering (FITME rsquo10) pp 395ndash398 October 2010

[6] L Zhang H Wang and T Li ldquoAnti-jamming message-drivenfrequency hopping Part I System designrdquo IEEE Transactionson Wireless Communications vol 12 no 1 pp 70ndash79 2013

[7] L Zhang and T Li ldquoAnti-jamming message-driven frequencyhopping-Part II capacity analysis under disguised jammingrdquoIEEE Transactions on Wireless Communications vol 12 no 1pp P80ndashP88 2013

[8] J G Proakis Digital Communications McGraw-Hill 4th edi-tion 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of