Manfred research group presents new wireless communication method FBS. Prof. Michael Bank M.Sc. Boris Hill Dr. Miriam Bank

[email protected]

1

Content Slide 1. What is the problem?

3

2. FBS sources

17

3. FBS – I

19

4. FBS - II

25

5. Simulation

35

6. Simulation results

46

7. Instead of conclusion

53

2

1. What is the problem? Today there are three multiplexing methods: (TS - symbol duration, τ - reflected signal delay.) TS ≈ τ

TS >> τ

TDMA + Equalizing

OFDM

CDMA TS

TS

τ

TS << τ

TS

τ τ

K

E

CDMA f

1

ES = 1 N0

Inf.

DEMOD

ES

PRBS

K

N 0 + ∑ Ec K =1

3

In tough mobile communication conditions: • Speed more than 100km/h • Frequency more than 1 GHz • Reflection signals large number each method has serious problems: TDMA needs high order often removable transversal filter. In CDMA signal to noise ratio is increased noticeable. OFDMA was conception β of UMTS and there is opinion that next generation for mobile/wireless will be: OFDMA-based systems. OFDMA is transmitting method in DAB, DVB-T, DVB-H, WiMax, Wi-Fi Nevertheless OFDMA has a serious problems too.

4

Conditions are changing during one symbol time • Usually investigators assume that Doppler Shift does not change during one symbol. But symbol duration today achieves 1ms (for example in DVB-T and in DAB systems). Let central frequency of OFDMA signal be 1GHz. Assume that there is Tx or Rx moving with speed 120 km/h. During one symbol, car moves by 1/3 cm. In comparison with λ = 30 cm it is noticeable conditions variation.

The main problem is orthogonality destruction • There are orthogonality problems in a receiver. Receiver synchronization system is not able to correct current symbol throughout its duration. It has a delay. Moreover, in the case of OFDMA systems, frequencies of different carriers can be shifted in different directions. In the case of Doppler effect and in absence of fast response synchronization correction system, each subsequent symbol will arrive later or earlier. That means that each following symbol has additional phase shift. This fact is of the main importance in Doppler Effect problems.

Receiving spectrum does not correspond to FFT parameters • The Automatic Frequency Control system in a receiver cannot correct decoding FFT parameters instantly. So we will be able to detect a signal on the transmitted frequencies only, but not on the frequencies received after Doppler Effect influence. 5

Inter Carrier Interference • Due to orthogonality deterioration, Inter Carrier Interference (ICI) appears.

Pilot signals don't reflect channel conditions • Due to ICI Pilot signal phase depend not only on channel conditions, but also on neighboring carriers phases.

There are Doppler Shift simulation problems • An OFDMA signal Spectrum does not include intermediate points between spectral components. So it is impossible to simulate the components frequencies deviation accordingly to Doppler Shift. Usually, this simulation is done by use of the “Doppler Filter”, which can change signal phases. But the main problem – orthogonality deterioration – this method cannot simulate.

6

Experimental assessment of DVD-T signal receiving quality during a moving. Experiments were performed in collaboration with the Measurement Laboratory of the Israeli Communication co. - Bezeq. The Bezeq mobile laboratory allows measurement electromagnetic field density and Bit Error Rate (BER) after Viterbi decoder. In addition, the equipment enables the observation of image quality including synchronization disturbances.

7

The section of a road chosen for the experiments, is quite a straight path . Traffic during the experiments was not heavy.

Ashdod

Receiving place

Tx

Ashqelon

Israel

8

Two sets of measurements were accomplished for two speeds: V = 40km/h , V = 120km/h

The number of synchronization losses increases proportionally with the vehicle speed because of Doppler effect.

- The Digital Audio Broadcasting (DAB) system may represent a possible solution of this problem by means of differential modulation. Nevertheless, it is impossible in the case of QAM modulation. - In the OFDM systems we may use the pilot signals on the same additional carriers like in Terrestrial Digital Video Broadcasting (DVB-T) system, but this does not work in the case of OFDMA system, where various sources use various carriers. Symbol i Multiplier

LPF

Reference Symbol

This solution requires noticeable redundancy increasing. If time conditions changing is comparable with symbol duration, this redundancy increasing can be 100% and more. 10

Explanation

Doppler in OFDM Distance increasing or decreasing causes the change in the receiving symbol duration by kd. Symbol period number does not depend on moving Tx or Rx , therefore the frequency changes by kd as well. There are frequency and time changing. Without Doppler. With Doppler.

f ⋅ k d and T / k d where k d = 1 +

V cos ϕ C

( f k +i +1 − f k +i )T = 1 [( f i +1 − f i ) ⋅ k d ] ⋅ [T / k d ] = 1

So Doppler shift does not effect orthogonally condition, but… 1. The receiver does not change symbol time instantly. 2.

The receiver does not change FFT parameters instantly.

It is possible to assume, that in Doppler shift situation T is constant due to symbol synchronization system in receiver. In this case we get: a – additional phase shift ∆ωt, b – orthogonallity disturbance (especially important in the case of OFDMA, when carrirs can move in different ways), c – additional phase shift due to incompatibility carrier frequencies and spectral components after FFT.

11

Explanation

Doppler in OFDM In the case of constant symbol duration (IFFT interval) T:

∆fi = ∆fi+ n

1 1 1 ( k + i ) k d − ( k + i ) = ( k + i )( k d − 1) T T T 1 1 1 = ( k + i + n )( k d − 1) = k ( k d − 1) + ( i + n )( k d − 1) T T T

In the case of Doppler, the frequency shift consists of two components:

1 k (k d − 1) - constant component: T

1 (i + n)(k d − 1) - small variable component, which proportional to carrier number: T = 100 µ s = 10 , ∆ f = 1 = 10 , f = 1GHzT, Example. −4

FFT is made on receiving frequencies:

k = 10

9

10

4

T

= 10 5 , k d , cos ϕ =1 = 1 +

4

1 / 30 = 1 + 10 − 7 300000

∆ f 1 = 10 4 ⋅ 10 5 ⋅ 10 − 7 = 100 . 000 Hz ∆ f 2 = 10 4 ⋅ (10 5 + 1) ⋅ 10 − 7 = 100 . 001 Hz

Conclusion: frequency shift on all carriers and corresponding Phase Shift (∆D) are approximately the same. However, because of constant T, sampling frequency is not changing. So there is incompatibility between receiving carrier frequencies and spectral components after FFT. It causes orthogonality deterioration. 12

Explanation

Short Delay in OFDM

In case of mobile transmitter the following symbol can be received slightly earlier or later. This delay depends on its speed and direction. In this case ∆t << T.

Example: T = 0.1 ms, f1 = 1GHz = 1000 MHz, V = 120 km/h. In this case, on first carriers we have phase shifts due to delay equal 3.90.

In case of FBS-8 with ∆f = 10 kHz the eighth frequency (f8) will be 1000.07 MHz. Eighth carrier will have phase shift equals 3.90010.

120000 300 = 0.3m, V = = 33.3...m/ s, 1000 60⋅ 60 LT =100µs = 33.3...m/ s ⋅10−4 s = 0.3...cm

λ1 =

ϕ = 2π ⋅ 0.3...30 = 3.90

Conclusion: phase shifts coursed by moving transmitter distance changing are approximately the same on all carriers. 13

Explanation

MPP in OFDM Multipath Propagation (MPP) causes the phase shift which depends on k, i, ∆t, A/AR. The use of Guard Interval prevents the orthogonality deterioration caused by MPP. The reflection signal is one more signal with the same frequency and with two additional phase shift components. First phase shift component does not depend on sub-carrier number and second one is proportional to sub-carrier number. 1 Direct signal: Reflection signal:

si (t ) = A sin[2π (k + i )

si , R (t ) = AR sin[ 2π (k + i )

T

t + ϕi ]

1 1 1 ⎧ ⎫ (t + ∆t ) + ϕi ] = AR sin ⎨[2π ⋅ k (t + ∆t ) + ϕi ] + 2π ⋅ i (t + ∆t )⎬ T T T ⎩ ⎭

1 ∆t First phase shift component is equal on all frequencies (∆φconst ) T 1 Second phase shift component is proportional to carrier number 2π ⋅ i ∆t T (∆φvar) One can see that MPP phase shift does not depend on modulation phase. Moreover it is possible to assume that A/AR does not change essentially during one symbol. 2π ⋅ k

So MPP phase shift is a serious problem, but amplitude shift can be compensated by well known Amplitude Gain Control (AGC) methods. 14

Explanation

Pilot deterioration caused by neighboring signals •Due to ICI Pilot signal phase depend not only on channel conditions, but also on neighboring carriers phases. BER of OFDM QPSK 4 signals with doppler 3% -3% 3% -3%

0

10

theor doppler correction with pilots correction with "ideal" pilots

-1

10

Doppler shift = 3%

-2

BER

10

-3

10

-4

10

-5

10

-6

10

0

2

4

6 Eb/No

8

10

12

15

Anyway, OFDMA problems : Multipath

Propagation, Doppler shift and delays influences, Pilot signals necessity, Orthogonality disturbing. New method :

Combining OFDMA and CDMA principles names Frequency Bank Signal (FBS), which allows above problems solving without additional signals, like a pilots or test for equalizing process.

16

This is one of simulation results, correspond to realistic, not extreme situation.

OFDMA and FBS systems under conditions of Doppler Effect (2%), MPP, and receiver phase noise and jitter influence (5 – 100) . Pilot/direct signal amplitude ratio in OFDMA is 2. 17

2. FBS sources There are three sources of FBS method: OFDM method, Phase

distortions compensation in PAL TV and Walsh function.

Phase Distortions Compensation in PAL There is an assumption, that color of two neighboring pixels does not change and corresponds to transmitting phase φ. Before transmitting color signal its phase is changed every second pixel from φ to – φ. In decoder second pixel phase sign is returned from – φ to φ. As a result , phase deviations in channel are compensated, because phases half-sum does not depend on equal phase shifts.

φ

φ+∆

Line n -φ

φ+∆ Returning φ-∆

-φ+∆

φ

Summing φ-∆ Line n + 1

18

Walsh Functions

1 1 1 1 1 1 1 1

1 1 1 1 0 0 0 0

1 1 0 0 1 1 0 0

1 1 0 0 0 0 1 1

1 0 1 0 1 0 1 0

1 0 1 0 0 1 0 1

1 0 0 1 1 0 0 1

1 0 0 1 0 1 1 0

These are orthogonal signals, which means that any signal is not sum or difference of other signals. 19

3. FBS – I using MPSK modulation The first version of the FBS method (FBS-1) can be illustrated by transforming an OFDMA system with four one-carrier MPSK modulation signals to FBS-1 system with the same four signals on the same four carriers . Let N = 4, each symbol of each signal is transmitted four times on four carriers. Sign of phase ϕi (+ or -) corresponds to one of Walsh functions. Symbol of a single signal is transmitted four times (on four frequencies), but on the same frequencies we can transmit four signals. We can recover the useful signal from four signals sum , due to Walsh functions orthogonality. 1111 1100 1010 1001

or

+ ++ + ++ - +-++--+

ϕ 1 ϕ1 ϕ2

ϕ1

ϕ1

ϕ2 - ϕ2 - ϕ2

ϕ 3 - ϕ 3 ϕ 3 - ϕ3 ϕ4 - ϕ4 - ϕ4 ϕ4 20

From OFDMA to FBS (N = 8) The same frequency band, bit rate and power. φ1 Rb/N

1

Present OFDM

Proposed invention FBS +/-φ1

φ2

1

Rb/N φ N -1 Rb/N

N1

f

N1

Rb/N

+/-φ2 R /N b +/-φN - 1 Rb/N The phase sign corresponds to Walsh function of order N

All phase shifts due to channel influence are compensated after opposite phase sign changing in receiver .

21

Three Signals in FBS S1 Receiver (N = 4)

S1 S2 S3

**

** Receiver 1

S1 S2 S3

Receiver 1 changes the phase sign in frequencies 3 and 4. For signal S1 we get all equal phases (without inversion). For signals S2 and S3 we can get phases compensation, meaning that the S2 and S3 do not effect the receiving of the S1.

22

FBS signal N −1

s ( kl ), FBS = E l ∑ e

j [ 2 π f k t + ( − 1 )W kl (θ l + β l )]

,

k =0

Wkl

Where El - component magnitude, l = 0,…, N -1 θl - initial phase, chosen for certain signal. For example, it is 450 or another, βl - information symbol of the l st FBS signal fk = f0+k∆f - FBS carrier frequencies, k = 0,…, N -1 , Wkl - sequence of phases of the l st FBS carrier pattern.

l/k 0

1

2

3

4

5

6

7

0

2

2

2

2

2

2

2

2

1

2

2

2

2

1

1

1

1

2

2

2

1

1

2

2

1

1

3

2

2

1

1

1

1

2

2

4

2

1

2

1

2

1

2

1

5

2

1

2

1

1

2

1

2

6

2

1

1

2

2

1

1

2

7

2

1

1

2

1

2

2

1

23

FBS TX and RX

(N = 4)

24

Problems solutions Phase shift influence in FBS-I In the case of non-symmetrical Walsh function (for example 1 1 -1 -1 1 1 -1 -1 ) we have additional phase shift, which does not depend on phase information. In the case of symmetrical Walsh function (for example 1 1 -1 -1 -1 -1 1 1 ) Doppler Shift and Delay do not have any influence on FBS receiving. Doppler Shift (∆ ) D

Before phase changing in receiver After phase changing in receiver

Before phase changing in receiver

After phase changing in receiver

ϕ + ∆D − ϕ + ∆D − ϕ + ∆D ϕ + ∆D ϕ + ∆D − ϕ + ∆D − ϕ + ∆D ϕ + ∆D

ϕ + ∆D ϕ − ∆D

ϕ − ∆D ϕ + ∆D ϕ + ∆D ϕ − ∆D ϕ − ∆D

ϕ + ∆D

∑ ϕ + 8ϕ

Delay (i∆d)

ϕ −ϕ + ∆d −ϕ + 2∆d ϕ + 3∆d ϕ + 4∆d −ϕ + 5∆d −ϕ + 6∆d ϕ + 7∆d

ϕ ϕ − ∆d

ϕ − 2∆d ϕ + 3∆d ϕ + 4∆d

ϕ − 5∆d

ϕ − 6∆d ϕ + 7∆d

∑ϕ = 8ϕ

25

4. FBS-II (FBS) using MPSK or MQAM Main idea: Instead of two carriers in OFDMA with QPSK (4 bits) we use I signal (2 bits) and Q signal (2 bits) and transmit they on N carriers with sign changing, correspond to individual Walsh function. FBS with 16QAM OFDM with QPSK φN

φ1 φ2

f S1

S2

S3 S4

I(S1) × 1 -1 -1 1 -1 1 1 -1 Q(S1) × 1 -1 1 -1 1 -1 1 -1

FFT-1

I(S4) × 1 -1 -1 1 -1 1 1 -1 Q(S4) × 1 -1 1 -1 1 -1 1 -1

FFT-1

It is known that a change from QPSK to 16QAM must correspond with an increase of power by 4dB. Subsequently, the frequency band becomes twice as narrow. In our case, the frequency band remains at the same frequency since we have made use of the same sub-carriers. The signal to noise ratio also does not change. Actually, the same information is transmitted on N sub-carriers. Essential information (I and Q) is summed using an arithmetical method. However, noise components are summed using a root-mean-square method. So we have got, that different between OFDMA and FBS by Eb/N0 is smaller than 4dB. 26

FBS Principle

I and Q transmitting values

I and Q after sign changing which correspond to the Walsh function

Channel influence

I and Q values after sign opposite changing

mean values of I and Q

I

Q

27

Short Selective Fading in FBS We transmit four FBS signals S4, S5, S6 and S7.

This is transmitted signal.

We have deep fading on frequency number 4. This fading results in receiving signal phase shift. A phase shift dependence on the transmitted phases has the normal distribution. The mean value of each signal has following magnitudes:

S 4 ⇒ ∆ϕ = 9.88 0 S 5 ⇒ ∆ϕ = 8.14 0 S 6 ⇒ ∆ϕ = 14 .38 0 S 7 ⇒ ∆ϕ = 14 .7 0

In the case of OFDMA the deep fading results in total signal loss!

28

FBS Signal Mathematical Presentation

( )k

k The sign of I or Q we can write as I ⋅ − 1 l or Q ⋅ (− 1) m , where k is 0 or 1 Let us transmit ith symbol with amplitude Ai,j and phase φi,j of the jth signal, where j is from 1 up to N/2. This symbol can be presented as a sum of two following orthogonal components: I i , j = Ai , j sin(ϕ i , j + θ j )

Qi , j = Ai , j cos(ϕ i , j + θ j ), where θj is the initial phase, chosen for the jth signal. Using the line l of the Walsh-Hadamard matrix for Ij and the line m of the WalshHadamard matrix for Qj, we can present FBS signal for transmitting this symbol as follows: N

{

S i , j = ∑ I i , j (− 1) k =1

Wl ,k

cos 2πf k t − Qi , j (− 1)

Wm , k

sin 2πf k t

}

where Wl,k is the kth value in line l of the Walsh-Hadamard matrix, Wm,k is the kth value in line m of the Walsh-Hadamard matrix. 29

FBS transmitter block diagram for transmitting four signals

S1 four bits

A1and φ1 of 16QAM

I1and Q1

I1 -I1 -I1 I1 -I1 I1 I1 -I1 Q1 Q1 Q1 Q1 -Q1 -Q1 -Q1 -Q1

A1(f1) A1(f2) A1(f3) A1(f4) A1(f5) A1(f6) A1(f7) A1(f8) φ1(f1) φ1(f2) φ1(f3) φ1(f4) φ1(f5) φ1(f6) φ1(f7) φ1(f8 )

S4 four bits

A4 and φ4 of 16QAM

I4 and Q4

FFT-1

I4 -I4 -I4 I4 I4 -I4 -I4 I4 Q4 Q 4 Q 4 Q 4 Q 4 Q 4 Q 4 Q 4

A4(f1) A4(f2) A4(f3) A4(f4) A4(f5) A4(f6) A4(f7) A4(f8) φ4(f1) φ4(f2) φ4(f3) φ4(f4) φ4(f5) φ4(f6) φ4(f7) φ4(f8 )

FFT-1

30

FBS receiver block diagram (one out of 4 users)

FFT

A∑(f1) A∑(f2) A∑(f3) A∑(f4) A∑(f5) A∑(f6) A∑(f7) A∑(f8) φ∑(f1) φ∑(f2) φ∑(f3) φ∑(f4) φ∑(f5) φ∑(f6) φ∑(f7) φ∑(f8 )

I∑(f1) I∑(f2) I∑(f3) I∑(f4) I∑(f5) I∑(f6) I∑(f7) I∑(f8) Q∑(f1) Q∑(f2) Q∑(f3) Q∑(f4) Q∑(f5) Q∑(f6) Q∑(f7) Q∑(f8 )

I∑(f1) -I∑(f2) -I∑(f3) I∑(f4) -I∑(f5) I∑(f6) I∑(f7) -I∑(f8) Q∑(f1) Q∑(f2) Q∑(f3) Q∑(f4) -Q∑(f5) -Q∑(f6) -Q∑(f7) -Q∑(f8 )

∑I/8 ∑Q/8

A1and φ1 of 16QAM

S1 four bits

31

Comparison between OFDMA and FBS with respect to required Eb/N0 There are four signals in both system S1, S2, S3 and S4.. All signals are transmitted in both system on N = 8 carriers. We can think that in FBS we transmit eight signals four I and four Q. Each signal I or Q transmit two bits. So we transmit in both systems during a one symbol (T) 16 bits.

FBS with 16QAM

OFDM with QPSK

I(S1) × 1 -1 -1 1 -1 1 1 -1 Q(S1) × 1 -1 1 -1 1 -1 1 -1

φN

φ1 φ2

f S1

S2

S3 S4

Correspond to this constellation the mean value of I or Q power is 22 = 4. For transmitting two bits in FBS we need eight components, so summing power will be 32. For using the same power we have to take each component in OFDMA with power 32 or amplitude 5.66. Let us noise power spectral density in both system will be N0 and T = 1.

I(S4) × 1 -1 -1 1 -1 1 1 -1 Q(S4) × 1 -1 1 -1 1 -1 1 -1 I 1011

1010

1110

1111

1001

1000

1100

1101

1

3 Q

0001

0000

0100

0101

0011

0010

0110

0111

32

Let us compare Eb/N0. In OFDMA Eb/N0 will be (32/2)/N0 = 16/N0. In FBS after multiplying by Walsh function in decoder we are getting eight equals amplitudes (in this case we must sum these amplitudes arithmetically). So summing amplitude of one I or Q component will be 2*8 = 16. Summing power will be 162 = 256. Noise on all

carrier will be different, so we must sum powers (not amplitudes) arithmetically. In this case we will get: Eb/N0 = (256/2)/8/N0 = 16/N0. In case of FBS we have to increase Eb/N0 for getting BER like in OFDMA because different between distances in both constellations. In case of OFDMA d = 7.99. with noise N0 In case of FBS we have to take amplitudes multiplying by 8 and noise by root of 8, so we have d = 5.66. For getting the same BER we have to give in FBS power or Eb/N0 by 20log(7.99/5.66) = 3 dB more. I I

d

5.66

3⋅ 8

1⋅ 8 Q

d

Q

33

Actually different in Eb/N0 little smaller approximately 2,7 dB. Because for getting two

errors in case of OFDMA distance increase by root of 2. In case of FBS distance increase by 2.

Possible 16 QAM constellation

We can decrease difference by another IdB, if we implement not rectangle constellation, but sphere one, for example 1,5,10. The gain of 16QAM Sphere Mapping is 1 dB. (See: B. Sklar. Digital Communications. Pearson Education, Inc 2005).

34

Eb/N0 in OFDMA with pilot signals In real systems the pilot signals level is grater than information signals level. In case of equal levels value of Eb/N0 increases, because pilot noise influences on detecting result. At that, this increasing does not depend on pilots rate. As usual pilot signal level is grater than information level by 3 – 9 dB. In following simulations we use 6 dB difference. For correct comparison we take all the signal of OFDMA power (including pilots) and FBS is the same. If pilot rate is Rp , we have to increase Eb/N0 by:

1 −1) + 4 Rp dB 10 log 1 Rp (

For Rp = 1/10 we have to move BER receiving results by 1dB Eb/N0 – direction. For Rp = 1/2 we have to move BER receiving results by 4dB Eb/N0 – direction.

35

5. Simulation It is well-known, that the channel model for digital signal includes Modem. Therefore it is impossible to

It is well-known, that the channel model for digital signal includes Modem. Therefore it is impossible to compare different modulation systems using one of known channel model. A typical situation involving a cellular system was implemented in order to make a comparison between OFDMA and the FBS systems. Receiver is situated in the center of 10 km diameter zone (cell). Receiver acquires signals from four transmitters located on the zone border. All transmitters can move with speed 120 km per our. Suppose that reflected signal with maximal delay has 5 km additional way, that is maximal delay equals T = 10 µs. Let us choose symbol duration 100 µs. So frequency difference between sub carriers will be ∆f = 1/T = 10 kHz. Transmitter moving causes phase shift and Doppler shift. Let central carrier frequency will be 10 9 Hz. So the phase shift per one symbol time due to delay will be 3.90. Doppler shift by these conditions can be 1% of ∆f. FBS modulation is 16QAM.

Tx S1

Tx S2 Tx S3

Rx

5 km

S4 T x 36

ANALYTICAL SIMULATION OF DOPPLER EFFECT Discrete Fourier transform is defined by N −1

X k = X (ωk ) = ∑ x(t n )e − jωk tn ,

direct transform :

k = 0,..., N − 1

n =0

1 xn = x (tn ) = N

inverse transform :

tn =

where

T n, N

ωk =

2π , T

N −1

∑

k =0

X (ω k )e jω k t n ,

x n = x (t n ).

Due to the Doppler shift, the new frequencies are: where

∆ω =

2π δ( T

x n

ω k = ωk + ∆ω ,

δ denotes relative shift).

Then the new (shifted) signal is

1 = x ( t n ) = N

n = 0, ..., N − 1

N −1

∑

k =0

X (ω k )e

j ω k t n

1 = N

N −1

∑

k =0

j ω + ∆ ω )t n X (ω k )e ( k 37

Results: Spectrum of the sifted signal (explicit expression): (a) Exact formula:

(b) Linear Approximation.

X k

1 − e 2π = N

jδ

N −1

X

∑

i=0

1− e

k −i 2π j (δ − i ) N

N − 1 ⎞ 2π jδ ⎛ δ ⎟+ X k = X k ⎜1 + π j N N ⎝ ⎠

N −1

X k −i

i =1

−

∑

1− e

2π j i N

Conclusion: for known ∆ω or δ , we can compute all components amplitudes and phases after Doppler Effect influence. Therefore, in the contrary to Matlab Doppler Filter method, orthogonality is not preserved. It is very important for pilot signal efficiency estimation.

38

Simulation of the Doppler effect, Stuffing Method FFT stuffing means to add S -1 'zeros' (s = 2n) between each carrier. Let us conduct an examination based on the following example. We would like to transmit an OFDM signal with eight carriers. We can decide the real spectrum in FFT dimension (N/2) to be 16 bits, where eight bits are used for transmitting information bits and pilot signals. Now we have to transform this spectrum to a complex form. For the purpose of frequency changing, we make a stuffing. That is, S-1 points are placed between all carriers .

39

Assumed characteristics of simulated system It is well-known, that the channel model for digital signal includes Modem. Therefore it is impossible to compare different modulation systems using one of known channel model. The following principal characteristics where choose for simulation : Frequency band, signal power and bit rate must be the same in both systems. All conditions must be the same in both systems. Conditions must be very difficult, but real. It is possible to put different parameters separately or together • Signals are transmitted by means of frames • Each frame includes 100 - 1000 symbols • Each frame starts with one synchronizing symbol • The main problem is phase variations from symbol to symbol. These changes are accumulated. • Receiver synchronization system is not able to make a correction of symbol duration and frequency changing, during several symbols. • Phase variations are different in all signals. • Reflection signals contain three phase shift components: • ∆φconst any phase shift, the same on all carriers, • ∆φvar phase shift, proportional to carrier number, • ∆φN accidental phase shift depends on propagation conditions diversity on different carriers (noise behavior)*. • The accidental phase shift ∆φN is inherent in direct signal too. 40

Simulation Conditions One can divide the transmitting and receiving conditions on three cases: • Conditions does not changing during a frame. That is all cars moving with constant speed and road is a straight line –”Const. Cond.” • Speed and directions are changing during a frame “Tx Ripple” • As a result of a variable Doppler Shift and MPP the receiver synchronization is not ideal. There are Jitter and Wander – “Rx Ripple”

“Const. Cond.” • Guard interval duration, • Each signal amplitude, • Constant Doppler shift, • Constant phase shift, • Reflection signal amplitude, • Reflection signal Doppler shift, • Reflection signal constant phase shift due to delay, • Reflection signal phase shift due to delay, proportional to carrier number. 41

Explanation

In real cellular systems attenuation of reflection signals with delay more than 8µs more than 20 dB.

42

Explanation

Doppler shift and Short Delay in Direction signal and Reflection signal.

1 d ⋅ 2 cos α

x α

x

d

d 1 ⋅ 2 cos α

Reflected signal length d/Cosα

α

Absolute distance changing x is the same. So difference of delay from beginning to end of symbol will be the same in direct and reflected signals. So we have to give the same condition on Doppler Shift and Short Delay to direct signal and to reflection signal.

43

Simulation Functional Diagram

Const. Cond.

Tx Ripple

Noise

addition Signal construction

Desired signal extraction Rx Ripple

BER calculation

PRBS Generation

44

Signal spectrum examples Eb/N0 = 4.3 dB ( BER ≈ 10-2 )

45

For presented simulations we have choused following parameters: For central carrier frequencies 1GHz : Symbol duration 100 µs, so ∆f = 10 kHz Maximal Doppler Shift (ϕ = 900) is 100 Hz, so ∆fd/∆f is 1% Short Delay of each symbol will be 3.90 , For central carrier frequencies 2.4GHz : Symbol duration 50 µs, so ∆f = 20 kHz Maximal Doppler Shift (ϕ = 900) is 240 Hz, so ∆fd/∆f is 1.2% Short Delay of each symbol will be 4.20 . In both cases guard interval was more than maximal reflection signal delay. Receiver has a AGC system with length 9 symbol samples

Redundancy is ratio of pilots number to information carriers number in percents. FBS

Redundancy

0%

OFDMA

OFDMA

OFDMA

With pilot rate 1/10

With pilot rate 1/5

With pilot rate 1/2

11.1%

25%

100% 46

6. Simulation Results

OFDMA and FBS systems under conditions of Doppler Effect (1%) and MPP. Pilot/direct signal amplitude ratio in OFDMA is 2. 47

OFDMA and FBS systems under conditions of Doppler Effect (1%), MPP, and jitter (mean value is 7.50) in receiver influence. Pilot/direct signal amplitude ratio in OFDMA is 2. 48

OFDMA and FBS systems under conditions of Doppler Effect (1%), MPP and phase noise (the standard deviation of random phase variation was 50) in receiver influences. Pilot/direct signal amplitude ratio in OFDMA is 2. 49

OFDMA and FBS systems under conditions of Doppler Effect (2%), MPP, jitter and receiver phase noise influence (mean value is 7.50) . Pilot/direct signal amplitude ratio in OFDMA is 2. 50

OFDMA and FBS system under conditions of Doppler Effect (1%), MPP, jitter and receiver phase noise (mean value is 7.50) influence. Pilot/direct signal amplitude ratio in OFDMA is 1. 51

OFDMA and FBS systems under conditions of Doppler Effect (4%), MPP, and jitter and receiver phase noise (mean value is 7.50) influence Pilot/direct signal amplitude ratio in OFDMA is 2. 52

OFDMA and FBS systems under conditions of Doppler Effect (2%), MPP, jitter and receiver phase noise (mean value is 150) influence. Pilot/direct signal amplitude ratio in OFDMA is 2. 53

7. Instead of Conclusion FBS in 3G FBS system can be implemented in all wireless communications including broadcasting systems, personal telephone systems, military systems and so on. Let us illustrate this first of all in the case of one of third generation (3G) cellular telephone systems: UTRA TDD. In this method signal after multiplexer is spread into 16 times with the help Pseudorandom Sequence (PRS) and input to modulator QPS

3G

3G+

UTRA TDD

UTRA TDD FBS

× 16 MUX

QPSK × 16 PRS MUX

QPSK

In the case of UTRA TDD FBS signal after multiplexer is input to modulator QPSK and after this it is spread into 16 times with the help of Walsh functions for N = 16. The changeover from UTRA TDD to UTRA TDD FBS allows getting either frequency band to twice narrower, or user number twice bigger. Like in usual CDMA system case, we can divide users from one UTRA TDD FBS group between several neighboring cells. Like in CDMA system case, there is soft handover possibility in FBS Method. 54

Anyway you can get: ‰ Considerable Decreasing Doppler shift, delays, reflections and fading influences, ‰ No need of any test or pilot signals or equalizing process ‰ Eliminating the long time selective fading ‰ Reduction of output power density, ‰ Decreasing narrow band interference influence, ‰ Using soft hand over.

You are getting all these without frequency band and power increasing. 55

7. Bibliography: 1. M. Bank “On increasing OFDM method frequency efficiency opportunity”, IEEE Transactions on Broadcasting, 50(2), 2004 (165-171) 2. Michel Bank, Miriam Bank, Boris Hill, Hanit Selecter Hill Patent PCT/Il N 2006000926: “A Wireless Mobile Communication System without Pilot Signals” Patent PCT International Application N0 PCT/IL 2006000926 3. M. Bank, B. Hill, J. Gavan. ”New multiple access method for wireless mobile systems”, Proc. XXVIIIth General Assembly of International Union Radio Science (URSI), New Delhi, India, 23-29 Oct. 2005 4. M. Bank, M. Haridim, B. Hill, Miriam Bank, “Wireless mobile communication systems without pilot signals”, WSEAS Trans. on Signal Processing, 2(12), 2006 (1592-1595) 5. M. Bank “System free of channel problems inherent in changing mobile communication systems”, IEEE, Electronics Letters -- 29 March 2007 -- Volume 43, Issue 7, p. 401-402.

56

57