RRM Lecture 2

8/3/2019 RRM Lecture 2

http://slidepdf.com/reader/full/rrm-lecture-2 1/45

1

Radio Resource Management Mohammed Elmusrati – Vaasa University 1

Radio Resource Management

1- Mobile Wireless Channels

2- OFDM3- Performance Measures

4- Diversity Techniques

PART II Wireless Channels and Techniques


Introduction

• Before we start the detailed analysis of RRM algorithms weneed to introduce the main concepts of mobile channels.

• This is very important topic to see the effects of channels on

transmitted signals.

• We need also to introduce the performance measure tools.

This is important because it gives the possibility to evaluate

the received signals if it is excellent, good, poor, very

poor, ..etc.

• The above two topics will be covered in this lecture.



2


Mobile Channel Characteristics

• In wireless communication, the information source or thetransmitter modulate the information signals (analog or digital)with a higher frequency carrier.

• The total modulated signal is fed to the transmitter antenna,which propagates this signal into space with a speed close tothe speed of light.

• The transmitted signal should have some characteristicsenabling the intended receiver(s) to select it, and treat the other signals as (co or cross) channel interference.

• The receiver ’s antenna will transform these electromagneticfluctuations into (very weak) electric current.



• This electrical signal is first filtered to permit only the required signal

bandwidth to pass. Then it is amplified, demodulated, …, until we get a

signal very similar to the transmitted one.

• The received signal will face a lot of degradations such as the propagationloss which reduces the transmitted signal power in order of (d 2 up to d 5)

depends on the propagation environment, where d is the distance between

the transmitter and the receiver, another types of degradation called fading,co/cross channel interference, thermal and other types of noise sources,

Intersymbol interferences (ISI), and others.

• These types of degradation which occurred on the received signals should

be studied and understood in order to design a successful and reliable

communication.





4


Example

-10

-8

-6

-4

-2

0

2

4

time

P o w e r d B w

It is clear the largefluctations of the

received power because of the

second randomdelay path


Example

The Matlab Code:

f0=1.8e9; % carrier frequency

tao=1e-6*rand(200,1); %generation of 200 random delay

Pr=(2+1.8*cos(2*pi*f0*tao))/2;% instantaneous power

Pr_db=10*log10(Pr); % expressing the power in dBw

plot(Pr_db) %plot the power

• More details about channel behaviors is shown in the next

slide.



5



Fading channel manifestations

Large-scale fading due to

motion over large areas

Small-scale fading due to

small changes in position

Variations

about the

mean

Mean

signal

attenuation

Time

variance of

the channel

Time

spreading of

the signal

Time-delay

Domain description

FFFSF

Frequency

Domain description

FFFSF

Time

Domain description

SFFaF

Doppler-shift

Domain description

SFFaF

FTFT

1

2 3

4

5 6

7 10

8 9 11 12

13

1415

16

1718

FF: Flat FadingFSF: Frequency selective fading

FaF: Fast Fading

SF: Slow Fading

FT: Fourier Transform



• The mobile channel is characterized by two fading effects :Large-scale fading and Small scale fading .

• Large scale fading represents the average signal power

attenuation or the path loss due to motion over large areas.

• In slide 9, the large scale fading is shown in blocks 1,2, and 3.

This phenomenon is affected by prominent terrain contours

(e.g., hills, forests, clumps of buildings, etc.) between the

transmitter and receiver. The receiver is often said to beshadowed by such prominences.



6



• The measurements and analysis show that the large scalefading can be described mathematically as a mean-path loss(nth-power law) and a log-normally distributed variations aboutthe mean.

• The received power in the presence of the large scale fadingcan be represented as:

• Where P t = transmitted power,P r =received power,d =distance between the transmitter and receiver, n= losspower (=2 in free space), it has been shown that n=4 givesgood model for mobile channels, and χ is a random variablewith log-normal distribution.

r t n P P

d

χ =



• From the previous slide we can see the received power interms of large scale fading.

• We may represent the received signal in dB as

• Where χσ denotes a zero mean, Gaussian random variable (indecibels) with standard deviation σ (also in decibels). χσ is site

and distance dependent. The variance σ2 can be between 5

up to 8 dB or even more.

( ) ( )10 _ _ 10 logr t

P dB P dB n d dBσ

χ = − +



7



• Small scale fading (blocks 4-18) refers to the dramaticchanges in signal amplitude and phase due to small changes

(as small as half-wavelength) in the spatial positioning

between a receiver and transmitter.

• As indicated in slide 9, the small scale fading manifests itself in two mechanisms:

– Time spreading of the signal (or signal dispersion )

– Time-variant behavior of the channel.

• For mobile radio applications, the channel is time-variantbecause motion between the transmitter and receiver results

in propagation path changes.



• Small scale fading is a random phenomenon and can bemodeled using different probability density functions depends

on the scenario. The most popular models are

– Rayleigh fading: in this model the received signal amplitude is

represented as random variable with Rayleigh distribution. This model

is accurate when represents multiple reflective paths that are large innumber, and if there is no line-of-sight signal component.

– Rician fading: When there is a dominant non-fading signal component

present, such as line-of-sight propagation path, the small-scale fading

envelope is described by a Rician probability density function.



8


Mobile Channel Characteristics• There are three basic mechanisms that impact signal propagation

in a mobile communication system – Reflection occurs when a propagating electromagnetic wave impinges upon

a smooth surface with very large dimension relative to the RF signalwavelength (λ).

– Diffraction occurs when the propagation path between the transmitter andreceiver is obstructed by a dense body with dimensions that are largerelative to λ, causing secondary waves to be formed behind the obstructedbody. It is often termed shadowing because the diffracted field can reachthe receiver even when shadowed by an impenetrable obstruction.

– Scattering occurs when a radio wave impinges on either a large, roughsurface or any surface whose dimensions are on order of λ or less, causingthe energy to be spread out (scattered) or reflected in all directions. Typicalscatterers are lampposts, street signs, and foliage.



Diffraction

L i n e o

f S i g h

t

reflections

scattering



9


Multipath Scenario

• Multipath causes two different kinds of

problems

– Fading (due to the phase differences) in the

band pass, i.e., before demodulation

– Inter-Symbol-Interference (ISI) in the

baseband domain



• The effects of multi-path could be

illustrated using the following

basic figure

t1 t2

transmitter

t1+ τ

receiver

t2+ τ

Multi-pathcomponents of

first pulse

First received pathlast received path

d τ

When > 1/BW then we have large time spreading in the received signal which canlead to considerable signal distortion. BW is the signal bandwidth.

d τ

t t



10



• Using complex notation, a transmitted signal is written as:

• Where Re{.} denotes the real part of {.}, and f c is the carrier frequency.

• The base-band g (t ) is called the complex envelop of s(t ) and

can be expressed as

• Where R(t ) is the envelop magnitude, and Φ (t ) is its phase.

For phase- or frequency modulation, R(t ) is almost constant.

( ) ( ){ }2Re c j f t

s t g t eπ

=

( ) ( )( )

( )( ) j t j t

g t g t e R t eφ φ

= =



• Now we discuss the small scale fading in more details.Assume that the received signal is arrived from different paths

where each path has time delay τn then the received signal

can be represented as

• Where αn(t) is the gain of path n at time t, m is the number of

paths, and s(t ) is the transmitted signal.

• Usually it is easier to work with base band signal, thensubstitute for s(t) (see the previous slide)

( ) ( ) ( )( )1

m

n n

n

r t t s t t α τ

=

= −∑



11



• We obtain

• From above the equivalent received baseband signal is

• The above relation is the discrete convolution between thetransmitted signal and the channel impulse response.

r t ( ) =Re α n

t ( ) g t −τ n

t ( )( )n=1

m

∑ e j2π f

ct −τ

nt ( )#

$%&

'

())

*

+,,=Re α

nt ( )e

j2π f cτ

nt ( ) g t −τ

nt ( )( )

n=1

m

∑-./

0/

12/

3/e

j2π f ct

'

())

*

+,,

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )2

1 1

c n n

m m j f t j t

n n n n

n n

z t t e g t t t e g t t π τ θ

α τ α τ − −

= =

= − = −∑ ∑



• It is clear that so that it is enough for thedelay to be 0.5/f c to get the phase of the received signal be

changed by π (180 degree).

• For example, for a cellular radio operating at f c =1800 MHz,the delay which makes the phase rotate by π is 0.5/f c =0.277

nanosecond which corresponds to a change in propagationdistance of 8.33 cm !!

• Since the phases of the received paths can be summedconstructively or destructively then the total amplitude is

dramatically changed.

( ) ( )2n c nt f t θ π τ =



12



• The received signal strengths are different for each path.Because the signal experiences different obstacles which

have different parameters such as sizes, reflection coefficient,

and so on.

• The path’s parameters can be described in terms of orthogonal components x n(t) and y n(t), where

• If the number of such stochastic components is large, then the

orthogonal components will be random variables with

Gaussian distribution.

( ) ( ) ( )( )n j t

n n n x t jy t t eθ

α

−

+ =



• These orthogonal components yield the small-scale (fading)magnitude r 0(t ), where

• In case of non dominant component (i.e. no line-of-sightcomponent or in other words xr and yr are zero mean

Gaussian distribution), it is not difficult to prove that r 0(t ) has

Rayleigh distribution such as

• is the predetection meanpower of the multi-path signal

( ) ( ) ( ) ( ) ( ) ( ) ( )2 2

0, ,

r r r n r n

n n

r t x t y t x t x t y t y t = + = =∑ ∑

( )

2

0 002 2

0

exp for 02

0 otherwise

r r r

f r σ σ

⎧ ⎛ ⎞− ≥⎪ ⎜ ⎟

= ⎨ ⎝ ⎠⎪⎩

2σ



13



• If there is dominant component or line-of-sight componentthen r 0(t ) has Rician distribution such as

• Where A denotes the peak magnitude of the non-faded signalcomponent (called the specular component), and I 0(.) is themodified Bessel function of the first kind and zero order.The Rician distribution is usually described in terms of parameter K , where

( )

2 2

00 00 02 2 2

0

exp for 0, 02

0 otherwise

r Ar r A I r A

f r σ σ σ

⎧ ⎛ ⎞⎡ ⎤+ ⎛ ⎞⎣ ⎦⎪ ⎜ ⎟− ≥ ≥⎪ ⎜ ⎟⎜ ⎟= ⎝ ⎠⎨ ⎝ ⎠⎪⎪⎩

( )2 22 K A σ =



100 200 300 400 500 600 700 800 900 1000-150

-140

-130

-120

-110

-100

-90

-80

-70

distance

r e c e i v e d p

o w e r ( d B )

Large scale fading

Small scale fading



14



• The channel characteristics can be analyzed by sending very wide band signalsuch as impulses or spread spectrum signal then testing the received signal.

• For example, if an impulse has been sent, then if the channel is ideal we willreceive just one impulse after the propagation delay. But because of the pathsthen we will receive a signal with multiple peaks, this signal represents theimpulse response of the channel as shown in the next slide.

• Different information can be extracted from such a signal as the maximumexcess delay, this signal is called the multi-path intensity profile.

• In the measurements, the time is measured from first arrived peak up to some

threshold (can be 10-20 db less than the strongest component).• Practically the received signal is never be zero because of the background

noise.



0 Tm τ

Fourier Transform

f 0=1/Tm Coherence bandwidth

Spaced-frequency

correlationfunction

Multi-pathintensity

profile

Dualfunctions

f cf c-f d f c+f d

Δf 0

Δt

Dualfunctions

Fourier Transform

T0=1/f d Coherence time

Doppler power

spectrum

Spaced-timecorrelation

function

a

b

d

c

|R(Δt )|

( )S v( )S τ

|R(Δf )|



15



• Let us call the time duration of transmitted symbol as Ts, if Tm >Ts then we

have frequency selective fading

• This condition occurs whenever the received multi-path components of asymbol extend beyond the symbol’s time duration.

• Such multi-path dispersion of the signal yields the same kind of ISI

distortion that is cased by an electronic filter.

• In this case, one can mitigate the fading because many of the path’s

components are resolvable.

• If Tm <Ts then we have frequency nonselective or flat fading. In this caseall received multi-path components of a symbol arrive within the symbol

time duration; hence the components are not resolvable.



• If we take the Fourier transform of the multi-path intensity profile we get

the spaced-frequency correlation function as shown in figure (b) in slide

28.

• One can say that it represents the channel’s frequency transfer function.

• The coherence bandwidth f 0 is a statistical measure of the range of

frequencies over which the channel passes all spectral components with

approximately equal gain and linear phase.

• Tm is not always the best parameter to express the channel excess delay.

A more useful parameter is the delay spread, most often characterized in

terms of its root-mean squared (rms) value, called the rms delay spread

as:

[ ]( )2

2 E E

τ σ τ τ ⎡ ⎤= −⎣ ⎦



16



• There are many approximations for the relation between thechannel coherence bandwidth and the rms delay spread.

• For example if the coherence bandwidth is defined as the

frequency interval over which the channel’s complexfrequency transfer function has a correlation of at least 0.9,

the coherence bandwidth is approximately

• If the correlation is at least 0.5 then the relation is 0

1

50 f τ

σ

≅

0

1

5 f

τ σ

≅



• The flat fading or frequency selective fading can be described in frequencydomain as well.

• We saw that the channel is frequency selective if Tm >Ts otherwise it is flatfading channel.

• We can say also that the channel is frequency selective if f 0< W, where Wis the signal bandwidth, in PSK we can approximate W=1/ Ts.

• The above statement means that, the frequency selective fading distortionoccurs whenever a signal’s spectral components are not all affectedequally by the channel.

• Some of the signal’s spectral components falling outside the coherencebandwidth will be affected differently (independently), compared with thosecomponents contained within the coherence bandwidth.



17



frequenct

s p e c t r a l d e n s i t y

f 0

Transmitted signal

bandwidth

Channelfrequency

transfer function

Typicalfrequency

selectivefading case

(f 0<W)



frequency

s p e c t r a l

d e n s i t y

f 0

Transmitted signal

bandwidth

Channel

frequencytransfer function

Typical flatfading case

(f 0<W)



18



• Signal dispersion and coherence bandwidth discussed before explains

channel’s time-spreading properties but it does not explain the

characteristics of the time-varying nature of the mobile channel.

• The mobile channel is time varying because of free motion of mobile usersas well as the movements of objects (such as cars) within the channel.

• Figure (c) in slide 28 shows the function R(Δt), designated the space-time

correlation function; it is the autocorrelation function of the channel’s

response to a sinusoid.

• This function specifies the extent to which there is correlation between thechannel’s response to a sinusoid sent at time t1 and response to a similar

sinusoid sent at time t2, where Δt= t2 - t1 .



• The coherence time T 0 is a measure of the expected timeduration over which the channel’s response is essentially

invariant.

• To measure the time variant nature of the channel, we usenarrow band signal, for example single frequency (sinusoidal

signal). The same signal is transmitted at different timeintervals (Δt ) then we compute the cross-correlation function

between the transmitted and received signal (as shown in

Figure (c) in slide 28).

• The function R(Δt ) and the parameter T 0 provide knowledge

about the fading rapidity of the channel.



19



• It can be shown that, for a constant mobile velocity V and anunmodulated CW signal having wavelength λ, the normalizedR(Δt) may be described as

• Where is the zero-order Bessel function of the first kind,V Δt is distance traveled, and k=2π/ λ is the free-space phase

constant (transforming distance to radians of phase).• It has been shown experimentally that the received signalsare statistically uncorrelated if distance between the receivingantennas is just 0.4 λ !

( ) ( )0 R t J kV t Δ = Δ

( )0. J



• The time-variant nature or fading rapidity mechanism of the channel can

be viewed in terms of two degradation categories as shown in slide 9: fast

fading and slow fading .

• The term fast fading is used for describing channels in which T0<Ts,where T0 is the channel coherence time, and Ts is the time duration of the

transmission symbol.

• Fast fading describes a condition where the time duration in which the

channel behaves in a correlated manner is short compared with the time

duration of a symbol.

• Therefore, it can be expected that the fading character of the channel will

change several times during the time span of a symbol, leading to

distortion of the baseband pulse shape.

• A channel is generally reffered to as introducing slow fading if T0>Ts



20



• Taking the Fourier transform of the spaced-time correlation function we

get the frequency domain also called Doppler shift domain (see figure (d)

in slide 28).

• It can be shown that for outdoor channels, vertical receive antenna withconstant azimuthal gain, uniform angle of arrival (0,2π), and unmodulated

CW signal, the signal spectrum at the antenna terminal is

( )2

1

1

0 else where

d c d c

cd

d

f f v f f

v f S v f

f π

⎧− + ≤ ≤ +

⎪ ⎛ ⎞⎪ −= −⎨ ⎜ ⎟

⎝ ⎠⎪⎪⎩



• The largest magnitude of S(v) occurs when the scatterer isdirectly a head of the moving antenna platform or directly

behind it.

• In that case, the magnitude of the frequency shift is given by

• When the transmitter and receiver move toward each other, f d

is positive, and when they move away from each other f d isnegative.

• For scatterers directly broadside of the moving platform, the

magnitude of the frequency shift is zero.

d

V f

λ =

where V is the relative velocity and λ isthe signal wavelength.



21



• The Doppler spread f d is regarded as the typical fading rate of the channel. T0 was described as the expected time duration

over which the channel’s response to a sinusoid is essentially

invariant. And the relation was given as T0 =1/ f d.

• When T0 is defined more precisely as the time duration over which the channel’s response to a sinusoid yields a

correlation between them at least 0.5, the relationship

between f d and T0 is approximately

0

9

16 d

T f π

=



• In Doppler domain, a channel is said to be fast fading if the symbol rate 1/

Ts (=W) is less than the fading rate 1/T0(=f d); that is, fast fading is

characterized by

•

Otherwise the channel is slow fading.• From previously, one can see that from signal dispersion point of view, the

coherence bandwidth f 0 sets up upper limit on the signalling rate (or

bandwidth) that can be used without suffering frequency-selective

distortion, and from Doppler spreading the channel fading rate f d sets a

lower limit on the signalling rate that can be used without suffering a fast

fading distortion.

0or d sW f T T < >



22



• From previous discussions we conclude that the transmitted signalarrives to the receiver ’s antenna in form of paths, each path has itsown delay and gain, if the maximum delay difference is less thanthe symbol duration, the channel is called flat fading channel, in thiscase using equalizers may not “considerably” enhance thereception quality.

• If the maximum delay difference is greater than the symbol durationthen it is called frequency selective channel, in this case equalizersor Rake receivers (for CDMA) could be used to mitigate the fading.However, more complicated receiver would be needed.

• How fast the channel characteristics will change? This could bedescribed by studying the time-varying nature of the mobilechannel. The rate of change is related to the Doppler shift of thechannel.



• The channel impulse response can be represented such as aFIR filter:

( ) ( ) ( )

( ) ( ) ( )( )

2

1

, ,

;

dk

k

m j f t

k k

k

j

k k n

k

h t t e t

t t e t d t

π

ϕ

τ α δ τ

χ α ρ ρ

=

= −

= =

∑

Where ρ k is the absolute channel gain for path k , f dk is the Doppler frequency for

path k , ϕk is the phase offset for path k , χ is random number with lognormal

distribution, and d k(t) is the distance of the path k between transmitter and

receiver at time t .



23



• The channel gain may includes transmitter and receiver antenna gains aswell.

• If the transmitted signal is s(t), then the received signal will be theconvolution between s(t) and the channel impulse response such as

• If the symbol duration greater than the maximum delay difference (FlatFading) then the above equation can be simplified as

• In this case the channel can be represented by a multiplicative operation(with a complex number) rather than convolution.

( ) ( ) ( ) ( ) ( )2

1

, ,dk

m j f t

k k

k

r t s t h t t e s t π

τ α τ

=

= ∗ = −∑

( ) ( ) ( ) ( ) ( ) ( ) ( )2 2

0 0

1 1

;dk dk

m m j f t j f t

k k

k k

r t s t t e b t s t where b t t eπ π

τ α τ α

= =

= − = − =∑ ∑


Mitigating ISI

• In many situations the problem of ISI is more stringent than the problem of the additivenoise. The limitations of noise can be mitigated through several ways such as increasing thetransmission power, improving the front-end receiver amplifier, using channel coding, higher antenna gains, and so on.

• However, the problem of ISI cannot be mitigated by just increasing the transmission power or improving the SNR. It can be mitigated by using equalizers which reduce the ISI problem.High quality equalizers which can work at high data rate are very complex. Moreover,equalizers need training sequence to be transmitted periodically to track channel variations.This reduces overall system capacity and efficiency.

• There is another way to mitigate the ISI by using CDMA, where the autocorrelation is verysmall. Therefore, when a copy of the transmitted code arrives after a certain delay (> chipduration), it will have very small contribution because of low correlation. Moreover, it canenhance the communication quality with Rake receiver (exploit the time diversity behavior).

• In wireless channels if the delay spread of the channel is 1 ms, then the symbol rate will belimited to be less than 1 kSymbol/s!! Using equalizers this can be relaxed to few kSymbols/s, but still not enough for high data rate.

• One clever way to mitigate the ISI problem for high data rate systems is using OrthogonalFrequency Division Multiplexing (OFDM).



24



ISI Limitations

• To understand the main concept of OFDM, let’s take this simpleexample.

• Example: For certain wireless channel, the significant delay spread is0.1 ms, find the symbol rate (roughly) in order to have free ISIreception without using equalizers.

• If it is possible to have 1024 parallel orthogonal transmission

simultaneously, what is the achievable symbol rate in this case(without equalizers).

• Solution• In the first case, the symbol rate should be around 5 kSymbol/s.

• In the second case, now we have 1024 parallel orthogonaltransmission, where every branch may send at 5 kSymbol/s, i.e., thetotal symbol rate becomes: 5.12 MSymbol/s



25


ISI Limitations

• From previous we conclude that: Wide bandwidth signals (i.e.,signal with high data rate), will suffer from the ISI problem

according to the channel dispersion. One very effective way to

solve this problem is to divide this high bandwidth to many

smaller bandwidth signals and send every one over

orthogonal carriers.

• One intuitive way to do that is by reducing the transmitted

data rate through serial to parallel converter and then usingsinusoidal carriers to transmit the parallel resultant signal.

• Next slide shows simple realization of the multi-carrier (MC)

transmission for three channels.


MC-FDMA Realization

Serial to

Parallel(S/P)Encoder

a0a1a2a3a4a5

a0a3

a1a4

a2a5

Rate=3f s

f s

f s

f s

( )1 1cos 2 f t π θ +

( )2 2cos 2 f t π θ +

( )3 3cos 2 f t π θ +

C h a n n e l

MF

MF

MF

P/S

( )1 1cos 2 f t π θ +

( )2 2cos 2 f t π θ +

( )3 3cos 2 f t π θ +

( ) s t

( )1 s t

( )2 s t

( )3 s t

( ) x t ( ) y t

MF=Matched Filter Multiplier (Mixer)

sum



26


MC-FDMA Realization

• What is the minimum frequency separation between sine waves toguarantee orthogonality between different channels so that it would be

easy to demodulate them without interfere each other?

• Let’s assume that the symbol duration after serial to parallel encoder is Ts seconds (it means that the symbol duration was Ts/3 before S/P

encoder, or generally Ts/N if we use N parallel channels). Thefrequency separations must satisfies the following condition:

• When A=1, it is called orthonormal.

( ) ( )0

0,1cos 2 cos 2

0,

sT

i i k k

s

i k f t f t dt

A i k T π θ π θ

∀ ≠⎧+ + = ⎨

≠ =⎩ ∫


MC-FDMA Realization

• To find the separation condition necessary for orthogonality,

• We may guarantee that the above is zero when

( ) ( )

( )( ) ( )( )

( )( )( )

( )( )( )

( )( )( ) ( )

0

0

0

1cos 2 cos 2

1cos 2 cos 2

2

sin 2 sin 21

2 2 2

1sin 2 sin

4

s

s

s

T

i i k k

s

T

i k i k i k i k

s

T

i k i k i k i k

s i k i k

i k s i k i k

i k s

f t f t dt T

f f t f f t dt T

f f t f f t

T f f f f

f f T f f T

π θ π θ

π θ θ π θ θ

π θ θ π θ θ

π π

π θ θ θ θ π

+ + =

⎡ ⎤− + − + + + +⎣ ⎦

⎡ ⎤− + − + + +

= + ≅⎢ ⎥− +⎢ ⎥⎣ ⎦

⎡ ⎤− + − − −⎣ ⎦−

∫

∫ ≅0

( ) ( ) ( )1

1, any integer 0

i k s i k i i

s s

n f f T n n f f f f

T T −

− = = ≠ ⇒ − = ⇒ − =

Minimum frequency separation



27


MC-FDMA Realization

• If the original (broadband) signal s(t) has bandwidth Bt asshown in figure, then after dividing to the total band into three

sub-bands we will have the following system

3

2

s f

2

s f

S/P

( )2

S f

( )2

1S f

( )2

2S f ( )

2

3S f

3

2

s f −

2

s f 2

s f

2

s f −

2

s f −

2

s f −

2

s f

2

s f −

s 2 s f 5

2

s f

The required bandwidth with using thistechnique is higher than the original

bandwidth


MC-FDMA Realization

• If we applied this technique in practice we would need evenhigher bandwidth, because we should keep guard band

between different signals’ bands to handle any frequency

changes according to channel or local oscillator drifts. And also

because the bands are not perfectly limited (otherwise it would

need ∞ time duration)!

• This is shown below. This explains why the idea of multi-carrier

parallel transmission was not applied before. It considerablyreduce the spectrum efficiency.



28


OFDM

• When VLSI implementation of DSPs becomes very efficient and atreasonable price, it becomes possible to handle the multicarrier

orthogonality by using Fast Fourier Transform (FFT) and its spouse

Inverse Fast Fourier Transform (IFFT).

• Because the modulation process is done jointly with IFFT, we may

not consider the phase shifts.

• In this case the minimum shift between carriers to handle the

orthogonality becomes: , hence, the spectrum can begreatly reduced (it will be same as the original spectrum as shown

in the figure below)

• Prove this result using slide 52.

• Now we do not have spectrum loss.

1

2 2

s

s

f

T

=

3

2

s f 3

2

s f −


Simple OFDM Transmitter

• Simple description of OFDM transmitter is shown next.

• It should be noted that, different other realizations are also

possible.

S/PEncoder

Bit streamwith rate Rb

Adaptive

M-level

Modulation

e.g., 8-ary

PSK, QPSK,

16QAM, ..

+

Coding and

Power

Allocation

Multicarrier orthogonalmodulation

usingIFFT

Cyclicprefix

(to

reduce

the ISI)

P/S

RFModulator

Information via feedback channel (e.g.,received SNR per sub-channel)



29



• We can see from the previous figure that we use serial to parallel encoder to divide the data rate on each subchannel to Rb/N, where Rb is theoriginal data rate and N is the number of subchannels.

• Next block we have baseband modulator. In this part, the bits are mappedinto symbols. Each symbol may carry more than 1 bit. For example, wemay use PAM with M- possible levels, which carry log2(M) bits. Another possible way, is to use M-ary PSK modulation. Of course, higher modulation level means higher rate on the subchannel. The cost is lessperformance (i.e., higher BER). To optimize the transmission we need toknow the received signal quality of the subchannels. This information isobtained from the receiver through the feedback channel.

• Sending information about every subchannel is not efficient, because largefeedback bandwidth would be needed. There are several methods tooptimize the transmission of the channel information.



• Power allocation and coding are used to optimize the transmission(highest possible throughput and in the same time achieving theQoS requirements such as BER and latency constraints per singleor group of subchannels).

• The optimization between power allocation per subchannel,channel coding (technique and rate), and multi-level modulation is

very challenging problem. It is known as NP- complex problem,however, there are so many suboptimal solutions in the literature.

• Next to that we observe the multi-carrier modulator which is theheart of the OFDM system. It has been realized with the InverseFast Fourier Transform (IFFT) which can be done efficiently withreasonable price VLSI circuits.



30



• Let’s see how the IFFT modulator work. The IFFT performsexactly the same function of Inverse DFT but in much more

efficient way. The IDFT is given by:

( ) ( )1

2

0

N j mk N

m

x k X m eπ

−

=

=∑

IFFT

( )0 x

( )1 x

( )1 x N −

( )0 X

( )1 X

( )1 X N −

Note that is discrete representationof the complex sinusoidal signal. The

orthogonality concept is applied here, i.e.,

2 j mk N e π

( )

( )

1 122 2

0 0

1,

0,elsewhere

N N j m b k N j mk N j bk N

k k

e e e

m bm b

π π π

δ

− −−−

= =

=

=⎧= − = ⎨

⎩

∑ ∑


Simple OFDM Transmitter • The principle of OFDM transmitter can be described as follows.

• The S/P encoder, with N branches, will divide the bandwidth of the

signal into N equally spaced bands where every band is Bt/N, where Bt is the total signal bandwidth.

• Think about the input of the IFFT as frequency domain input, then the

output will be time samples separated by 1/B t seconds.

• Actually the IFFT will maintain the orthogonality between the different

bands of its input. However, observe that the output of the IFFT is

mixed from all inputs. For example

• In the receiver, it is possible to retrieve the original signal again by

using FFT operation as explained in the next slide.

x 0( ) = X m( )m=0

N −1

∑ , x 1( ) = X m( )e j2π m N

m=0

N −1

∑ ,



31



• When the output signal of the IFFT is applied as an input to FFT

system, the output will be the same original signal applied to the

IFFT.

• We can prove this easily as, since the DFT (or its efficient

version FFT) operation is

• So that FFT has identical structure as IFFT.

• If the output of the IFFT applied as input to FFT we obtain

( ) ( )1

2

0

N j bk N

k

Z b x k e π

−−

=

=∑

Z b( ) = 1

N x k ( )e− j2π bk N

k =0

N −1

∑ =1

N X m( )e j2π mk N

m=0

N −1

∑"#$ %

&'e− j2π bk N

k =0

N −1

∑ =1

N X m( )e

j2π m−

b( )k N

m=0

N −1

∑k =0

N −1

∑

⇒ Z b( ) =1

N X m( )e

j2π m−b( )k N

k =0

N −1

∑ =

m=0

N −1

∑1

N X m( ) e

j2π m−b( )k N

k =0

N −1

∑m=0

N −1

∑ ,

from slide 59, ej2π m−b( )k N

k =0

N −1

∑ =δ m−b( )⇒ Z b( ) =1

N X b( )

m=0

N −1

∑ = X b( )


• The process of IFFT in frequency domain can be described in this Figure.

IFFT

S/PEncoder

P/SDecoder

Note: Theoretically

any orthogonal

transformation could

be used instead of

IFFT/FFT. However,

practically IFFT/FFT is

preferred because of

its feature in removing

ISI as will be

explained



32



• Next we have the cyclic prefix block. The main motivation of OFDMmodulation is to handle the ISI problem which is very acute for very shortduration symbols.

• Very short duration symbols means high data rate (large bandwidth).When the delay spread of the channel (e.g., maximum delay withconsiderable power) more than the symbol duration then ISI will occur.

• In OFDM the symbol duration considerably increased, however, some ISIproblem could still occur. In practice, the ISI problem can not be eliminated100% (bandwidth-time duration dilemma). Moreover, the channel

dispersion leads also to interference between subchannels (ICI) which is amore serious problem.

• This small ISI problem can be effectively removed (actually reduced) byusing cyclic prefix and simple equalizer.

• Next we will talk about the cyclic prefix concepts.


Cyclic Prefix• The principle of cyclic prefix is to extend the transmitted signal

beyond the nominal symbol period into a guard interval, so as

to provide a cyclic signal.

• The receiver correlator, however, only integrate over thenominal symbol period. Thus any signals delayed by channel

dispersion remain orthogonal, eliminating both ISI and ICI

provided the delay is less than the guard interval.

• The guard interval is therefore always implemented in OFDMsystems operating on a dispersive channels.

TTgCorrelator window



33


Cyclic Prefix

• Cyclic prefix is a crucial feature of OFDM used to combat the inter-symbol-interference (ISI) and inter-channel-interference (ICI)

introduced by the multi-path channel through which the signal is

propagated.

• The basic idea is to replicate part of the OFDM time-domain

waveform from the back to the front to create a guard period. Theduration of the guard period Tg should be longer than the delay

spread of the target multi-path environment

prefix

TTg

Note that, the prefix duration is added toprevent ISI.Adding part of original symbol in this prefix

duration will maintain the orthogonality betweensubchannels so that we mitigate inter channel

interference (ICI).


Simple OFDM Receiver

• Next slide shows simple OFDM receiver. The blocksrepresent the inverse actions done at the transmitter.

• Because of the added noise and interferences we need to add

some redundancy bits to increase the robustness of thetransmitted symbols.

• This is known as forward error correction codes. The channel

coding enable the receiver to correct certain number of errors

when happened.

• Sometimes this is known as coded OFDM (COFDM).

However, it is natural to add channel coding to all OFDM

systems.



34


Simple OFDM Receiver

RFDemodulator

Channel state information sent throughfeedback channel

S/P

Multicarrier demodulation

usingFFT

Cyclicprefix

remove

BasebandDemodulation

+

Equalization+

Subband

channel

estimation

P/S Bitstream


Interleaving

• The impulsive noise or the deep fading problem may causepacket loss, where most of the packet bits arrive in error, or sometimes, the packet could not be decoded.

• One solution to mitigate this problem is to distribute thepacket bits over several subchannels. For example if each

packet consists of 8 bits. And we have 8 subchannels so thatwe send one bit from each packet over one subchannel. If one subchannel lost because of fading or noise, then we lossonly one bit from each packet. The FEC will be able to correctthis bit in each packet.

• This technique will give some kind of diversity. Next slideshows simple realization of the interleaving with OFDM



35


Interleaving - Deinterleaving

S/PEncoder

Multicarrier orthogonalmodulation

usingIFFT

I n t e r l e a v i n g

Bin=b0b1b2b3b4b5b6b7b8b9bAbBbCbDbEbF

b0 b5 bA

b1 b6 bB

b2 b7 bC

b3 b8 bD

b4 b9 bE

BinBout

Bout=b0b5bAb1b6bBb2b7bCb3b8bDb4b9bE

d e i n t e r l e a v i n g

B’outB’in

B’out=b0b5bAb1b6bBb2b7bCb3b8bDb4b9bE

b0 b1 b2 b3 b4

B5 b6 b7 b8 b9

bA bB bC bD bE

B’in=b0b1b2b3b4b5b6b7b8b9bAbBbCbDbEbF




36


Performance Measure in WirelessCommunication

• In the previous lecture we show one performance measure for CDMA

systems which is known the signal to noise ratio or bit energy to noise

spectral density ratio.

• In next few slides we will explain this in more details. Now we will showsome very common ways to measure the performance in wireless

communication systems.

• Measuring the performance is very important to evaluate or assess the

quality of communication.

• Each operator should achieve the necessary services and performance for their costumers.

• This is known the Quality of Services (QoS) which should be guaranteed by

the operator.


The Quality of Service (QoS)

• Generally the QoS denotes the communication properties such as:

– Communication reliability

– Security

– Probability of Outage

– Communication performance

• Throughput

• Packet loss

• Delay

• Delay Spread (Jitter)

• Bit error rate

– Communication cost

– Priority



37


The Quality of Service (II)

• The QoS level depends on the type of communication.

• It can be divided to

– Real time data streams

• Audio samples

• Video frames

• Sensor inputs

• Actuator commands

– Non-real time data streams

• Internet exploring

• File DOWN/UP-loads


Audio Streams

• The un-coded data rate depends on the sound quality.

– Telephone quality needs (with PCM) 64 Kbit/s

– CD quality needs 1.4 Mb/s (two-channel-stereo stream)

• The compression methods can dramatically reduce the required data rate

at small degradation in the quality

• Recommended maximum delay for real time audio streams is 32 ms• In mobile communication, some audio compression methods are used to

reduce the required bandwidth without much of scarifying the voice quality.

Usually the Linear Predictive Code (LPC) is used for this purpose.

• With LPC the required bit rate for voice is reduced from 64 Kb/s to only 13

Kb/s!



38


Throughput

• The throughput of a stream defines the amount of user data transferred ina certain time unit between source and sink.

• Because of possible errors during the transmission there is possibility of retransmission requirement will reduce the actual throughput.Furthermore, the data do not contain only payload data but also other datalike coding bits. This is also reduce the throughput.

• Generally throughput ≠ data rate [data]=[Address ; Payload ; Coding Bits ; ..]

• Throughput ≤ Data Rate

• Some applications need high throughput such as video transmission, highquality sound, and download of large files.

• Speech signals and text email transmission are examples of lowthroughput applications.


Delay

• Transferring data packets between stream source and sinkconsumes time spent for processing, and queuing in the

nodes and for transmission in the network.

• An upper bound on the delay Dmax denotes the maximum timeany packet of a stream will need to be transferred.

• For real time applications the maximum allowed delay is

restricted. More relaxed delay is allowed for non-real time

applications.

• The allowed delay of packets determine the priority of

transmission.



39


Jitter

• Data packets of the same stream may experience differenttransfer delays.

• The delay variation is usually called jitter.

• Jitter can be represented as the difference between themaximum delay and the minimum delay of the packets.

• The problem of jitter may cause the packets do not arrive inthe correct order. For example if the delay of the first packet is

much more than the delay of the second one, this may causethe second packet to arrive first! This problem can be solvedby numbering the packets, however, this has other problemsuch as reducing the efficiency and increasing the systemcomplexity.


Bit Error Rate

• The bit error rate is an important parameter of thecommunication QoS.

• Relatively high BER can be allowed for speech

communication (in worst case ~10-2).

• Low BER should be achieved for Data communication (10-6

for certain applications).

• Usually we monitor the received Eb/No to estimate the BER.The relation is well defined for additive white noise channel.

But the relation can be very complex for fading channel, in

this case a lot of assumptions are used to derive tractable

mapping.



40


Bit Error Rate• Generally speaking the BER can be expressed as

• Where f(.) is the mapping between the BER and the SINR.

• This mapping depends on:

– Modulation type (FSK, PSK, 8PSK, 16QAM,..etc.)

– Channel type (additive noise, interference structure,..)

– Fading type (flat fading, frequency selective channel, ..etc)

• For simple additive noise channel and BPSK modulation

the relation is given by

( )0b BER f E N =

0

2BER=

b E Q

N

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

( ) 0 ,2

1 22

≥= ∫ ∞

− xdt e xQ

x

t

π

where


Signal to Interference and Noise Ratio

(SINR)

-10 -5 0 5 10 1510

-7

10-6

10-5

10-4

10-3

10-2

10-1

10

Eb/N

0(dB)

P r o b a b i l i t y

o f B i t E r r o r

Coherent PSK

Coherent FSK

Non-Coherent FSK



41


SNR versus BER in Fading Channels

• In reality the received signal magnitude increase anddecrease randomly even when the transmitted power is fixed.

This amplitude fluctuation is due the multi-path characteristic

and the time varying nature of mobile channels as shown in

previously.

• Consequently, the received SNR increase and decrease in arandom manner, i.e. the SNR becomes a random process.

• In this case it is not accurate to use the previous mappingbetween the SNR and the BER because it gives only the

instantaneous BER.


SNR versus BER in Fading Channels

• If the received energy bit to noise spectral density (Γ=Eb/No)

is a R.V. with the probability density function

• The average BER is given by

• Where g(Γ) is the instantaneous mapping between SNR andthe BER

( ) f ΓΓ

( ) ( )0

B P g f d ∞

Γ= Γ Γ Γ ∫



42


Packet Loss

• Packets may be lost due to congestions, deep fading,impulsive noises, buffer overflow, hardware errors etc.

• The probability of packet loss is one of the QoS parameters in

communication systems.


Diversity Techniques

• The problem of fading is very serious in wireless communication systems.

There is always probability for the signal amplitude to be less than the

minimum required to detect it. This causes loss of packets.

• There are some solutions to mitigate this problems (it is not possible to avoidit 100%). Some solution are expensive and difficult and some others are

much cheaper and easier.

• Some of the expensive solution is to use very high sophisticated equalizers to

mitigate the ISI problem, joint equalizers and smart antennas, Rake

receivers, MIMO systems,..

• The most common simple solution is to use diversity.

• The diversity can be utilized on different dimensions such as frequency, time,

spatial, polarization, ..etc.



43



• The frequency diversity is utilized by sending the informationwith different carriers. The separation between carriers shouldbe enough to experience independent (or at leastuncorrelated) channels.

• If the probability of one carrier to be faded is p then theprobability of two distinct carriers to be faded in the same timeis p2.

• The frequency diversity is used for example in GSM wherethey use slow frequency hopping (217 hops/s) to compensatefor those cases where the mobile unit is moving very slowly(or not at all) and experiencing deep fading.



• Time Diversity: can be provided by transmitting the signal on L differenttime slots separation of at least T0 (see slide# 28).

• Interleaving, when used along with error-correction coding is a form of time diversity.

• Spatial diversity is one simple and efficient solution for fading problems. Itis based on simple idea is to use more than one antenna to receive the

signal. The antennas should be separated enough to be at leastuncorrelated channels (for example 10 wavelengths).

• It is possible to combine the signals from different antennas by differenttechniques. The simplest way is by just add all signals together. Another method by selecting the best antenna and use its signal for decoding. Bestperformance is obtained by coherently adding all branches’ signalstogether. Several other combining algorithms were discussed in DigitalCommunication course.



44


Diversity Analysis

• From slide 82 we presented that the average BER over fadingchannel is given by

• Where Γ=Eb/No.

• Since the received magnitude has a Rayleigh distribution then

Γ will have exponential distribution such as

( ) ( )0

B P g f d ∞

Γ= Γ Γ Γ ∫

( ) [ ]1

exp , where f E Γ

Γ⎛ ⎞Γ = − Γ = Γ⎜ ⎟Γ Γ⎝ ⎠


Diversity Analysis

• If each diversity (signal) branch, i=1,2,…,M, has aninstantaneous E b /No= Γ i and we assume that each branch

has the same average Eb/No, then

• The probability that a single branch has Eb/No to be less than

some threshold Γ t is

( ) [ ]1

exp , where 1,..,ii i f E i M Γ

Γ⎛ ⎞Γ = − Γ = Γ ∀ =⎜ ⎟Γ Γ⎝ ⎠

( ) ( )0 0

1exp 1 exp

t t t

t ir i i i i P f d d

Γ Γ

Γ

⎛ ⎞Γ Γ⎛ ⎞Γ ≤ Γ = Γ Γ = − Γ = − −⎜ ⎟ ⎜ ⎟

Γ Γ Γ⎝ ⎠ ⎝ ⎠ ∫ ∫




Diversity Analysis

• The probability that all M independent signal diversitybranches are received simultaneously with an Eb/No lessthan some threshold value Γ t is

• Example:

• For certain mobile communication system the channel can be

expressed as Rayleigh channel with an average Eb/No =5 dB.The packet will be lost if the received Eb/No < 4 dB. Find theprobability of packet lost if using single antenna then if using 4antennas with large separation.

( )1 2, , , 1 exp

M t

t t t

r M P

⎡ ⎤⎛ ⎞ΓΓ ≤ Γ Γ ≤ Γ Γ ≤ Γ = − −⎢ ⎥⎜ ⎟Γ⎝ ⎠⎣ ⎦

L


Diversity Analysis

• Solution

• When using 4 uncorrelated antennas the probability of successful reception increases to

( )

0.4

0.5

4 10 2.51

5 10 3.16

2.51

1 exp 0.54!3.16

The probability of successful reception is only: 1-0.54=0.46

t t

dB

t

r i

dB

dB

P

Γ = ⇒ Γ = =

Γ = ⇒ Γ = =

⎛ ⎞

Γ ≤ Γ=

− −=

⎜ ⎟⎝ ⎠

4

2.511 1 exp 0.91

3.16

⎡ ⎤⎛ ⎞− − − =⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦

Documents

RRM Lecture 2