Propagation of Stochastic Signals in a RandomMedium

Manoel J.L. Alves and Marcelo Sampaio de Alencar∗Institute of Advanced Studies in Communications

Institute of Superior Studies of the AmazonFederal University of Campina Grande

Campina Grande PB, BrazilE-mails: [malencar,jacinto]@dee.ufcg.edu.br

Abstract— The fluctuation in the received signal level,caused by diverse factors, that affect the electromagnetic wavepropagation in a channel is studied. The signal behavior isanalyzed for a channel that presents flat or frequency selectivefading. Numerical simulations were performed to verify themultipath effect between the transmitter and receiver. A MatLabprogram was written, which describes the influence that thecharacteristics of the channel can have on the transmitted signal.

Keywords: Propagation. Fading. Simulation.

I. INTRODUCTION

An important question in radio communication systemsis to understand the aspects that involve the behavior ofelectromagnetic waves in a random medium of propagation.However, a previous knowledge of the channel can be decisivefor the choice of modulation and or coding technique used fortransmission [1]-[3].

In the past few years, progress has been obtained on theknowledge of propagation modes and on the effects that theenvironment can have on the electromagnetic waves. One canmention, for example, the influence of physical structures,such as buildings and geographical accidents, changes in theelectromagnetic properties of the medium, meteorological andatmospheric alterations [4],[5].

The propagation of the electromagnetic wave is affectedby obstacles that cause reflection, refraction and scattering.The transmitted signal arrives at the receiver by several paths,known as multipath propagation [6]. Those components thatarrive at the receiver can diminish the performance of thechannel, producing the fading phenomenon. This effect iscommon and produces strong random fluctuations on thereceived signal level. In the case of mobile systems, it isassociated with the relative movement between the transmitterand receiver [7].

For the deployment of a radio communication system adetailed study of phenomena that influence its performanceis needed. Mathematical models are used to obtain a model ofthe behavior of the signal in the fading environment [8]-[10].

The remaining of the paper is organized as follows. SectionII presents the receiver of a communication system thatoperates in an environment with multipath. The characteristicsof flat or frequency selective fading channels necessary to

Fig. 1. Representation of pulsed signals transmitted through two differentpaths.

the implementation of a radio communication channel areanalyzed in Section III. A MatLab program was developedto quantify the several parameters that are relevant to theinterpretation of the results shown in the last section, alongwith the conclusions.

II. MATHEMATICAL MODEL

The receiver of a radio communication system in an en-vironment with multipath must be capable to operate witha disturbed signal. Each field component that arrives has acertain power value and delay [11]. Between the componentsexist, therefore, a difference of arrival instants, that will bereferred as power delay from the channel. Fig. 1 shows apulse train with duration Tbb, a repetition period Trep and themultipath delay τmax.

In this analysis it is assumed that the repetition period ofthe pulse is superior to the delay originated by the some paths.For two distinct signals with different bandwidths, in the sameenvironment with multipath, the transmitted pulse is of theform

x(t) = �e{p(t) exp(j2πfct)}, (1)

where p(t) defines the format of a repetitive baseband pulsetrain. Ideally, for its duration the amplitude of each pulseis constant. It is possible to consider an arbitrary value A.

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However, to simplify the final expressions, the following pulseformat [11] will be used,

p(t) =

⎧⎪⎨⎪⎩

2√

τmax

Tbb, for 0 ≤ t ≤ Tbb

0, otherwise

. (2)

When a pulse is transmitted over a channel with low-pass filtertransfer function its output includes a carrier shift and delay.Considering the composition of the signals for N paths, theresult becomes

r(t) =12

N−1∑i=0

ai(exp(jθi)) · p(t − τi). (3)

The factor 2 in (2) simplifies the expression, resulting in

r(t) =N−1∑i=0

ai(exp(jθi)) ·√

τmax

Tbbrect

[t − τi

Tbb− 1

2

], (4)

where N is the number de multipath components and the rectindicates a rectangular pulse [11].

The resultant field for an instant t◦ is the superposition ofseveral components that lead to a received power proportionalto the square of its module. However, considering (3), thepower is proportional to the product of the function by itsconjugate

v(t) = r(t) × r∗(t)

v(t) =14

N−1∑j=0

N−1∑i=0

aj(t)[p(t − τj) exp(jθj)]

· ai(t)[p(t − τi) exp(−jθi)]. (5)

Finally, to deduce the final expression obtained from (5),the first two terms of the series are admitted initially, withi = 0, j = 1 and i = 1, j = 0. Thus, with the expansion forthese two terms, one has

v1(t) =14[2a1(t)a◦(t)p(t − τ1)p(t − τ◦)2 cos(θ◦ − θ1)]. (6)

As said previously the delays of the components i and j arelarger than Tbb and their values are very close. Therefore, fori = j = k, one obtains

vk(t) = a2k(t)p2(t − τk), (7)

whose average value for t = t◦ is obtained with the integrationin the interval [0, τmax]. Therefore,

P (t◦) =| r(t◦) |2= 1τmax

∫ τmax

0

r(t) × r∗(t)dt

=1

τmax

∫ τmax

0

(N−1∑k=0

vk(t)

)dt.

Substituting the expression for r(t),

P (t◦) =1

τmax

N−1∑k=0

a2k(t◦)

·∫ τmax

0

{√τmax

Tbbrect

[t − τi

Tbb− 1

2

]}2

dt. (8)

The result of the integration gives

P (t◦) =N−1∑k=0

a2k(t◦). (9)

For the baseband signal p(t), Tbb is much smaller than thedelay between the channel components and Equation (9) showsthat the total received power P (t◦) is found as the sum of theindividual powers of each multipath component [11], [12].

III. MODEL FOR MULTIPATH FADING CHANNELS

When transmitting a pulse of small duration and limitedamplitude, the received signal in a channel with multipathwill be a pulse train with a difference of amplitude, time, andformat, as shown in Fig.1. Also, it must be considered that thecharacteristics of time variation in a channel with multipathobey a statistical process. Therefore, assuming a carrier withslow variation in the amplitude, the transmitted signal s(t) inthe channel can be represented as

s(t) = �e [f(t) exp(jωct)] , (10)

where f(t) describes the format of the modulation signal. Inthis analysis the received signal will be represented u(t).

There are several propagation paths, and each one of them isassociated with a delay and an attenuation factor. Thus, the re-ceived signal will be the superposition of several components,of the form

u(t) =∫ ∞

−∞α(τ, t)s(t − τ)dτ, (11)

where α(τ, t) is the attenuation of the signal component withdelay τ at time instant t. Substituting s(t) into the Equation(11), one obtains the resultant received signal

u(t) = �e

{[∫ ∞

−∞α(τ, t) exp(−jωcτ)f(t − τ)dτ

]ejωct

}.

(12)Calling d(t) the integral in (12), one has

d(t) =∫ ∞

−∞α(τ, t) exp(−jωcτ)f(t − τ)dτ, (13)

where d(t) is interpreted as the convolution in time of thesignal f(t) with the time varying attenuation.

A.1 Flat FadingFor flat fading it is interesting to consider modulation with

pulses of short duration, in order to produce a large spectralwidth in the frequency domain. A pulse f(t) of duration τ =1.1364 nanoseconds was considered, as shown in Fig. 2.

The mathematical representation of the modulation pulse is

f(t) =

⎧⎨⎩

A, for −τ/2 ≤ t ≤ τ/2

0, otherwise, (14)

where A is the amplitude of the signal. The Fourier transformof the function f(t), which gives the representation in thefrequency domain, is

F (ω) =12π

∫ ∞

−∞Ae−jωtdt =

(τA

2π

)sin(ωτ/2)(ωτ/2)

, (15)

304

where ω represents the angular frequency [13],[14]. A non-periodic function produces a continuous spectrum, as shownin Fig. 3.

Consider a channel with first order transfer function H(ω),

H(ω) =1

1 + jωT, (16)

whereωc =

1T

= 2πfc (17)

defines the cutoff frequency for the channel. The behavior ofthe modulus and argument of this function are shown in Fig. 4,respectively. Cutoff frequencies of 1900 MHz and 892 MHzwere used.

The signal at the output of the propagation channel, inthe frequency domain, can be found multiplying the signaltransform by the transfer function of the channel, of the form

Fs(ω) = F (ω)H(ω) =(

τA

2π

)sin(ωτ/2)

(ωτ/2)(1 + jωT. (18)

Fig. 5 shows the modulus and argument of Equation (18),assuming a pulse with duration of τ = 1.1364 nanosecondsand cutoff frequencies of 1900 MHz and 892 MHz. However,from the figures is observed that a difference exists among theinput pulse transform modulus and the channel response. Thetransform modulus of the signal is symmetrical and presents aconstant amplitude, but the channel response lacks symmetryalso differs in amplitude.

The Fourier inverse transform of Fs(ω) produces the signalin the time domain [14]. This gives

fs(t) =12π

(τA

2π

)∫ ∞

−∞

sin(ωτ/2)ejωt

(ωτ/2)(1 + jωT )dω, (19)

which is the signal at the input of the receiver, as shownin Fig. 6. This figure shows that with the largest cutofffrequency the channel becomes flatter and the format of thepulse approaches that of the transmitted signal. This impliesin a smaller bit error rate (BER). The given response assumescutoff frequencies of 1900 MHz and 892 MHz.

A.2 Frequency Selective FadingChannels with frequency selective fading are more difficult

to model than channels with flat fading. In this case, thechannel is interpreted as narrowband linear filter and eachpath must be modeled separately [11]. Therefore, to exemplifythis fading type, the same pulse of flat fading was consideredwith duration of τ = 1.1364 nanoseconds and channel cutofffrequencies equal to 450 MHz and 340 MHz.

In this case, the transfer function response introduces largerinfluences in the amplitude and phase of components at higherfrequencies, as shown in Fig. 7.

Fig. 8 shows the frequency response modulus and argument,respectively. Comparing the modulus of Fig. 8 and Fig. 5 it isverified that the amplitude difference is more accentuated forthe components of higher frequencies. Fig. 9 represents thetime domain response compared with Fig. 6, that representsthe flat fading response. One notices that there is a largerdistortion of the transmitted pulse.

−6 −4 −2 0 2 4 6

x 10−10

0

0.2

0.4

0.6

0.8

1

1.2

Pulse duration (in nanoseconds)

Pul

se a

mpl

itude

Fig. 2. Modulation pulse in a random medium.

−8 −6 −4 −2 0 2 4 6 8

x 1010

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency (in hertz)

Tra

nmitt

ed p

ulse

Fou

rier

tran

sfor

m m

odul

e

Cutoff frequency = 1900 MHzCutoff frequency = 892 MHz

Fig. 3. Continuous spectrum of a pulse transmitted in the frequency domain.

108

109

1010

1011

−10

−8

−6

−4

−2

0

Frequency (in hertz)

Tra

nsfe

r fu

nctio

n m

odul

e (d

B)

108

109

1010

1011

−80

−60

−40

−20

0

Frequency (in hertz)

Tra

nsfe

r fu

nctio

n ar

gum

ent

Cutoff frequency = 1900 MHzCutoff frequency = 892 MHz

Fig. 4. Transfer function response modulus and argument with flat fading.

−8 −6 −4 −2 0 2 4 6 8

x 1010

0

0.2

0.4

0.6

0.8

1

Frequency (in hertz)

Cha

nnel

res

pons

e m

odul

e

−8 −6 −4 −2 0 2 4 6 8

x 1010

−100

0

100

200

Frequency (in hertz)

Cha

nnel

res

pons

e ar

gum

ent

Cutoff freq. = 1900 MHzCutoff freq. = 892 MHz

Fig. 5. Frequency domain channel response modulus and argument of apulse with duration of τ = 1.1364 nanoseconds and flat fading.

305

−5 0 5 10 15 20

x 10−10

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (in seconds)

Pul

se r

espo

nse

with

flat

fadi

ng

Cutoff frequency = 1900 MHzCutoff frequency = 892 MHz

Fig. 6. Time domain response of transmitted pulse for a channel with flatfading.

108

109

1010

−10

−8

−6

−4

−2

0

Frequency (in hertz)

Tra

nsfe

r fu

nctio

n m

odul

e (d

B)

108

109

1010

−80

−60

−40

−20

0

Frequency (in hertz)

Tra

nsfe

r fu

nctio

n ar

gum

ent

Cutoff frequency = 450 MHzCutoff frequency = 340 MHz

Fig. 7. Transfer function response modulus and argument for frequencyselective fading.

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

x 1010

0

0.2

0.4

0.6

0.8

1

Frequency (in hertz)

Cha

nnel

res

pons

e m

odul

e

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

x 1010

−100

0

100

200

Frequency (in hertz)

Cha

nnel

res

pons

e ar

gum

ent

Cutoff freq. = 450 MHz

Cutoff freq. = 340 MHz

Fig. 8. Frequency domain channel response modulus and argument of apulse with duration of τ = 1.1364 nanoseconds subject to frequency selectivefading.

−5 0 5 10 15 20

x 10−10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (in seconds)

Pul

se r

espo

nse

with

flat

fadi

ng

Cutoff frequency = 450 MHzCutoff frequency = 340 MHz

Fig. 9. Time domain response of the transmitted pulse for a channel withfrequency selective fading.

IV. CONCLUSION

This paper analyzes the influence that the characteristicsof the channel exert on the transmitted signal, with the aidof a MatLab program. Thus, the use of simulation permittedthe comparison of signals submitted to flat and frequencyselective fading. In this way, for the channels with flat fading,the amplitude and phase of the several components do notsuffer severe alterations. Regarding the selective fading, theconsequence is that the components of the received signaldo not keep the same relations of amplitude and phase and,moreover, comparing the transmitted signal with the received,a noticeable distortion in the waveform at the receiver isverified.

ACKNOWLEDGMENT

The authors thank the support of the Brazilian Councilfor Scientific and Technological Development (CNPq), ofthe Institute of Superior Studies of the Amazon (IESAM)and of the Institute of Advanced Studies in Communications(IECOM).

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