ChaoticCDMACommunicationsystems
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Chaotic Digital CDMA Communication Systems 2799
6. Channel Capacity of (CD )2MA
For simplicity, in this paper we only study the unsectorized cases. For the sectorized cases all the arguments of the relations between CDMA and (CD)2MA are also valid.
For the Shannon limit, the number of users that we can have in a cell is
encoded signal in CDMA system is given by
x(t) = c(t)d(t) cos !t |
(20) |
and for (CD)2MA, it is given by
x(t) = c(t)d(t) : |
(21) |
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Gp |
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The CDMA receiver multiplies |
x(t) by the PN |
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1:45Gp |
(17) |
waveform to obtain the signal |
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M = Eb=N0 |
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where Eb is the energy per bit and N0 is the noise |
r1(t)=c(t)(x(t)+I(t)) = c2(t)d(t) cos !t+c(t)I(t) |
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power spectral density. Gp is the system processing |
= d(t) cos !t+c(t)I(t) |
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gain which is given by |
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(22) |
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Gp = |
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Bw |
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(18) |
where I(t) denotes the sum of noise and interfer- |
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R |
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ence. For the (CD)2MA receiver, we have |
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where Bw is the bandwidth of the channel and R is |
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the information rate. |
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r2(t) = c(t)(x(t) + I(t)) = d(t) + c(t)I(t) : |
(23) |
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In an actual system, the CDMA cell capac- |
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ity is much lower than the theoretical upper-bound |
Since the frequency of d(t) |
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less |
than |
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value given in Eq. (17). |
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The CDMA cell capac- |
c(t)I(t), from Eq. (23) we know that |
the |
SNR |
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ity is a ected by the receiver modulation perfor- |
of r2(t) can be signi cantly enhanced by using a |
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mance, power control accuracy, interference from |
low-pass lter before we do any further processing. |
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other non-CDMA system sharing the same fre- |
This is not the case when we inspect Eq. (22), in |
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quency band, and other e ects. |
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which d(t) cos !t and c(t)I(t) have similar frequency |
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Every cell in a CDMA system shares the same |
range. We only need a small |
E =N |
CD |
2 |
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b |
0 in ( |
) MA. |
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bandwidth therefore causing the intercell interfer- |
Furthermore, the interference from other users is |
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ence, which we account for by introducing a factor |
also reduced. |
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. The practical range of is 0:5 |
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0:55. The inter- |
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We then used simulation results to show that |
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ference from users in other cells reduces the num- |
(CD) MA has a bigger capacity than CDMA. The |
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ber of users in a cell. The power control accuracy |
following conditions are used. The RF bandwidth |
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is represented by a factor . |
The practical range |
Bw = 1:25 MHz. The chip rate is 1:2288 Mcps, |
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for is 0:5 0:9. The reduction in the interference |
and the data rate is R = 9:6 Kbps. For a CDMA |
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due to the voice activity is represented by which |
system, assume that Eb=E0 = 6 dB, the interfer- |
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has a practical range of 0:45 1. |
Then Eq. (17) |
ence from the neighboring cells = 60%, the voice |
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becomes |
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activity factor = 50%, and the power control |
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Gp |
1 |
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1 |
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accuracy factor = 0:8. Then for a CDMA sys- |
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M = |
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: |
(19) |
tem, the channel capacity is 33 mobile users per |
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Eb=N0 |
1 + |
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cell1. |
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In the rest of this section, |
we give a de- |
The most important benchmark for evaluating |
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sign example to show how (CD)2MA can have a |
the service quality of a digital communication sys- |
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larger capacity than CDMA. In CDMA we use |
tem is the bit-error-rate (BER). In this paper, the |
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phase-shift keying (BPSK) for the data modula- |
desired performance of the (CD)2MA system is cho- |
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tion and quadrature phase-shift keying (QPSK) |
sen to be BER 10−3, which is the same as that |
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for the spreading modulation. |
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However, just for |
used for a CDMA systems. The condition for the |
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the purpose of demonstrating the improvement of |
evaluation is as follows. Suppose that every mo- |
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the (CD)2MA to the channel capacity, we suppose |
bile station has a perfect power control performance |
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that the coherent BPSK is employed for both data |
then the base station receives equal power from |
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modulation and spreading modulation. |
Then the |
each mobile station. We suppose that the delays |
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1Notice that we only study the unsectorized cell here.
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2800 T. Yang & L. O. Chua
due to the multi-paths are distributed uniformly in (0:4 s; 1:2 s), i.e. 1 = 0:4 s and 2 = 1:2 s, which corresponds to the range from 0.5 chip duration to 1.5 chip duration used in CDMA. The simulation results for 60, 90 and 110 users/cell are summarized in Tables 2{4 respectively.
From Table 2 we nd that for (CD)2MA system, Eb=N0 = 4 dB is enough to give a good result under the condition of 60 users/cell. We use simulation to show this result. In our simulation, the RF bandwidth Bw = 1:25 MHz. A 10-shift cipher is used to generate the spread carriers. The data rate is R = 9:6 Kbps. An overhead channel is used to transmit synchronizing impulses for all users. The synchronizing impulses are digitalized into 32-bit floating point numbers. Every user needs to refresh their synchronizing impulses once every second. The bit rate of the overhead channel should be greater than 5:76 Kbps for all 60 users.
The simulation results are shown in Fig. 5. Figure 5(a) shows the mixed signal in the channel which is a mixture of 60 spreading carriers. Figures 5(b){5(d) show the spreading carriers of the users A, B and C, respectively. Figures 5(e){ 5(g) show the message signals (green waveforms) and the recovered signals (red waveforms) for the users A, B and C, respectively. We can see that the digital signals are recovered from the (CD)2MA schemes.
Table |
2. |
The relation |
between |
BER and |
Eb=N0 with |
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60 users/cell. |
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Eb=N0 |
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3 dB |
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4 dB |
5 dB |
6 dB |
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BER |
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3:4 10−3 |
6:8 10−4 |
1:3 10−4 |
3:7 10−5 |
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Table |
3. |
The relation |
between |
BER and |
Eb=N0 with |
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90 users/cell. |
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Eb=N0 |
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3 dB |
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4 dB |
5 dB |
6 dB |
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BER |
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7:5 10−3 |
2:3 10−3 |
7:3 10−4 |
2:9 10−4 |
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Table |
4. |
The relation |
between |
BER and |
Eb=N0 with |
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110 users/cell. |
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Eb=N0 |
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3 dB |
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4 dB |
5 dB |
6 dB |
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BER |
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4:4 10−2 |
3:8 10−2 |
3:1 10−2 |
2:7 10−2 |
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For comparison, we also show the case of Eb=N0 = 4 dB and 110 users/cell. From Table 4 we can see that the interference become so strong that the increase of Eb=N0 cannot decrease the BER signi cantly. The simulation results are shown in Fig. 6. Figures 6(a){6(c) show the recovered signals and the message signals of users A, B and C, respectively. We can see that for user C, there exist
(a)
Fig. 5. Simulation results of the (CD)2MA system with 60 users/cell and Eb=N0 = 4 dB. (a) The mixture of the spreading carriers in the channel. (b) The spread carrier for the user A. (c) The spread carrier for the user B. (d) The spread carrier for the user C. (e) The message signal (green) and the recovered signal (red) for the user A. (f) The message signal (green) and the recovered signal (red) for the user B. (g) The message signal (green) and the recovered signal (red) for the user C.
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Chaotic Digital CDMA Communication Systems 2801
(b)
(c)
(d)
Fig. 5. (Continued)
2802 T. Yang & L. O. Chua
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(e)
(f)
(g)
Fig. 5. (Continued)
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Chaotic Digital CDMA Communication Systems 2803
(a)
(b)
(c)
Fig. 6. Simulation results of the (CD)2MA system with 110 users/cell and Eb=N0 = 4 dB. (a) The message signal (green) and the recovered signal (red) for the user A. (b) The message signal (green) and the recovered signal (red) for the user B.
(c) The message signal (green) and the recovered signal (red) for the user C.
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2804 T. Yang & L. O. Chua
(a)
(b)
(c)
Fig. 7. Simulation results of the (CD)2MA system with 90 users/cell and Eb=N0 = 4 dB. (a) The message signal (green) and the recovered signal (red) for the user A. (b) The message signal (green) and the recovered signal (red) for the user B. (c) The message signal (green) and the recovered signal(red) for the user C.
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very serious bit errors. This is a big bit error probability which may degrade the speech quality very much.
Between 60 users/cell and 110 users/cell, we show the case Eb=N0 = 4 dB and 90 users/cell. From Table 3 we can see that the service quality is decreased but is still usable. We also presented the simulation results in Fig. 7. Figures 7(a){7(c) shows the recovered signals and the message signals of users A, B and C, respectively. We can see that for user C, there exist a few bit errors.
7. Conclusions
In this paper, we present a high capacity chaotic digital CDMA scheme which has twice the channel capacity2 of CDMA. The improvement makes us use almost one-third of the Shannon limit of an ideal channel. We nd that (CD)2MA is a very promising scheme to reduce the subscriber fee of PCS by half, compared to the CDMA system. One of the critical problems in a chaotic CDMA scheme is to maintain the synchronization between two chaotic systems, since the synchronization signals would take a bandwidth of the order of the date rate if the continuous synchronization scheme is used. To overcome this problem we use an impulsive synchronization scheme to synchronize the two chaotic circuits in the linked pair. Thus, we only need a very small bandwidth for transmitting synchronization impulses.
Acknowledgment
This work is supported by the O ce of Naval Research under grant No. N00014-96-1-0753.
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2If the Eb=N0 is increased to 6 dB as that used in CDMA systems, the (CD)2MA system may even triple the channel capacity as shown in Table 3.