Your company has just been awarded a license to operate a digital Personal Commu

Question

Your company has just been awarded a license to operate a digital Personal Communication Network (PC'N) using GSM 1800 to meet the following requirements as shown in Table 2. Suggest a suitable cellular topology and draw the frequency plan to meet 0/1 requirement. Give your reason. Calculate the minimum power received. P_r(min) and estimate the cell radius. R. Can your system accommodate and support the total subscribers? Assume that one traffic channel (TCH) can accommodate 24 subscribers and for practical implementation at least one carrier frequency be allocated for control channel (CCH) in each sector.

Explanation / Answer

Hi below i have given the solution for all your question for your reference,

Suggest a suitable cellular topology and draw the frequency plan to meet C/I requirement. Give your reason.

N

1

3

4

7

9

12

13

CI_MEAN_OMNI

0.568109

8.917731

11.10416

15.35733

17.26735

19.45378

20.06212

CI_WC_OMNI

-8.00049

6.286211

9.148933

14.2535

16.41193

18.81425

19.47222

CI_MEAN_3SEC

7.194082

14.9967

17.05269

21.0759

22.89372

24.98318

25.56615

CI_WC_3SEC

5.339322

13.68894

15.87537

20.12854

22.03857

24.22499

24.83333

CI_MEAN_6SEC

13.50904

19.88706

21.68325

25.2996

26.9699

28.91262

29.45852

CI_WC_6SEC

8.349622

16.69924

18.88567

23.13884

25.04887

27.23529

27.84363

Since the C/I(th) = 16.6 dB, the suitable cluster for each system if (C/I)threshold(mean) ≥ 16.6 dB are:

SECTOR

N

OMNI

12

3 X 120Ëš

7

6 X 60Ëš

3

Thus, the best option would be N = 3 with 6 X 60Ëš sectoring to reduce trunk and also fulfill the worst C/I condition, i.e. N = 3/18.

Frequency plan:

Total Available Channel = (1856.10MHz - 1832.50MHz) / 20000 = 118

ARFCN

Start

512

End

629

N

3/18

ARFCN

A1

B1

C1

A2

B2

C2

A3

B3

C3

A4

B4

C4

A5

B5

C5

A6

B6

C6

TRx1

512

513

514

515

516

517

518

519

520

521

522

523

524

525

526

527

528

529

TRx2

530

531

532

533

534

535

536

537

538

539

540

541

542

543

544

545

546

547

TRx3

548

549

550

551

552

553

554

555

556

557

558

559

560

561

562

563

564

565

TRx4

566

567

568

569

570

571

572

573

574

575

576

577

578

579

580

581

582

583

TRx5

584

585

586

587

588

589

590

591

592

593

594

595

596

597

598

599

600

601

TRx6

602

603

604

605

606

607

608

609

610

611

612

613

614

615

616

617

618

619

TRx7

620

621

622

623

624

625

626

627

628

629

Calculate the minimum power received, Pr(min) and estimate the cell radius, R.

EIRP = 43dBmW = 43-30=13dBW

BGSM = 200K

Nf = 5 dB

Thus, 10 log10 Nf = 5

Nf = 100.5 = 3.16227766

Boltzmann's constant, k = 1.3803 x 10-23 J/0K

T0 = 2900K

Pn = kTeB = k(Tant + Te + Tf1), & Te = T0 (Nf - 1)

Thus, Pn = k (T0 (Nf - 1) + T0) BGSM

Pn = (1.3803 x 10-23) (290 (3.16227766 - 1) + 290) (200K)

Pn = 2.5316 x 10-15 W

SNRMIN = 20 dB = Pr / Pn

Thus, 10 log10 (Pr / Pn) = 20

Pr = Pn (102) = (2.5316 x 10-15) (102) = -125.97 dB

FSPL = 32.44 + 20 log10d (km) + 20 log10f (MHz)

FSPL = 32.44 + 20 log101+ 20 log101800 = 97.5455 dB

Pr = EIRP + Gr + GD - PL - Lother - Lfade

-125.97 = 13 + 3 + 5 - PL - 3.3 - 6

PL = 125.97 + 13 + 3 + 5 - 3.3 - 6 = 137.67 dB

PL (dB) = FSPL + 10γ log10(d/d0)

PL (dB) = FSPL + 10γ log10(d/d0)

137.67 = 97.5455 + 10 (3.5) log10(d/1)

137.67 = 97.5455 + 35 log10d

d = 10 ((137.67 - 97.5455) / 35) = 14 km

Thus, Pr(min) = -125.97 & cell radius, R = 14km

Can your system accommodate and support the total subscribers? Assume that one traffic channel (TCH) can accommodate 24 subscribers and for practical implementation at least one carrier frequency be allocated for control channel (CCH) in each sector.

Total Available Channel = (1856.10MHz - 1832.50MHz) / 20000 = 118

Traffic channel (TCH) = 118 - 18 = 100

No of base station = Total area covered / 2.6R2 = 9185 / 2.6(14)2 = 18

No of cluster = 18/3 = 6

Total carrier for TCH = 100 x 8 = 800 carrier for GSM

Total no of subscribers that can be supported (based on 24 subscriber / channel)

= (Total no of TCH) (no of subscribers/channel) (no of clusters)

= (800) (24) (6) = 115200

Thus, the system may accommodate and support the total subscribers as the value is less than the minimum no of subscriber, which are 25000.

A cellular system with cluster size, N = 4/12 is shown in Figure Q5 with the traffic intensity in each cell is as shown in Table Q5. The average call holding time is 180 seconds. There are 528 traffic channels (duplex) available and ready to be assigned. The expected Grade of Service (GOS) is 2%.

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