ELEC 456 - Mobile Communications
Assignment #1 - Due Feb. 12, 1999

  1. Is cellular radio system capacity limited by noise or by interference?
    Explain why this is the case.

    2. The system is limited by interference from other users of the cellular system. This interference comes from users of the same channel in co-cells (co-channel interference or CCI) as well as adjacent-channel interference (ACI) from users of nearby frequencies. The cellular system is designed so that the CCI and ACI are reduced to acceptable levels. Due to the power levels used in the cellular system, these levels exceed that of the noise from other sources.

  2. Which of TDMA, CDMA, and GSM cellular systems is inherently suited to data services? Why?

    3. GSM was designed from the ground up to be compatible with the ISDN data communications standard.

  3. Briefly describe the major sources of CCI and ACI.

    4. As mentioned in #1, the main source of CCI is from users of the same channel in co-cells. The main source of ACI is from users of nearby channels in the same cell or adjacent cells.

  4. What is frequency re-use and what factors limit re-use in a practical cellular system?

    5. Frequency re-use is where the radio channels that are used in one cell are used again to serve other users in some nearby cell. This technique is what gives the cellular system such a large capacity. The limitation on re-use is co-channel interference: the channel cannot be re-used in a cell that is close enough to cause significant CCI.

  5. Assuming a regular hexagonal tessellation, what are the first five cluster sizes and the corresponding distances between cocells?

    6. i j Size Distance
    0 1 1 1
    1 1 3 1.732
    0 2 4 2
    1 2 7 2.646
    0 3 9 3
    2 2 12 3.464

  6. Find the re-use distance D/R if the required CCI is 18 dB with a = 3.5. What is the corresponding minimum cluster size?

    7. D/R = (M*C/Ic)^(1/a) = (6 * 63.1)^(1/3.5) = 5.45

    N = (D/R)^2 / 3 = 9.9 Next valid cluster size is 12.

     

  7. Consider an AMPS-like system using a multiple-channel bandwidth scheme to increase capacity. If we have a total bandwidth of 12 MHz and we split this bandwidth equally between 20 kHz and 30 kHz channels, what is the total number of channels? For a seven-cell cluster, what is the radius of the inner cell (as a fraction of the radius of the outer cell)? The minimum C/Ic for the 20 and 30 kHz bands is 22 dB and 18 dB, respectively. Use a = 4.

    8. 12 MHz / 50 kHz = 240. Total of 480 channels.
    D/R0 = (6 * 63.1)^(1/4) = 4.411
    D/R1 = (6 * 158.5)^(1/4) = 5.553
    R0/R1 = 0.79

  8. Use Okumura’s model to find the propagation loss in dB for a suburban area with:
    f = 1 GHz, ht = 20 m, Pt = 5W, Gt = 4, Gr = 2, hr = 2 m, d = 10 km.
    Check the result using Hata’s formula.

    9. Okumura:
    step 1: Free-space loss = -20 log (l/(4pd)) = 112.45 dB
    step 2: From fig. 3.14, A(f,d) = 30 dB
    step 3: From fig. 3.15, suburban correction = 9 dB
    step 4: a0 = 20 log (ht/200) + 10 log (hr/3) = -21.76 dB
    Total = L0 + A(f,d) - Garea - a0 = 155 dB
    Hata:
    L = 69.55 + 26.16 log f -13.82 log ht - A(hr) + (44.9 - 6.55 log ht) log d
    A(hr) = (1.1 log f - 0.7) hr - (1.56 log f - 0.8) = 1.32 dB
    L = 165 dB

  9. Using sx2 = 1, plot the Rican PDF p(r) for a = 0, a = 1 and for a = 2. As shown in figure on page 72 of the course notes, be sure to normalize a for the computation and normalize r for the plot axis.

    10. See below

  10. Find the shadow loss when the wavelength is l = 25 cm, the distance from the obstruction to the mobile is d1 = 1 km, the distance from the base to the obstruction is d2 = 6 km, and the height of the obstruction is h = 10 m. Repeat with d1 = 2 km and d2 = 5 km. Explain the difference between the answers in view of the fact that in both cases the total distance is 7 km.

Since d1, d2 >> h >> l, use the approximation.
x1 = -h * sqrt(2/ l * (1/ 1000 + 1/ 6000)) = -0.966, from graph, loss = 14 dB
x2 = -h * sqrt(2/ l * (1/ 2000 + 1/ 5000)) = -0.748, from graph, loss = 12 dB

In the first case the shadow loss is greater because the mobile is closer to the obstruction.
From the geometry, the mobile is farther below the horizon.

 

r=0:0.1:5;
sig2 = 1;
a=[0; 1; 2] / sqrt(sig2);
for i=1:length(a)
p(i,:)=r/(sig2).*exp(-(r.^2 + a(i)^2)/(2*sig2)).*besseli(0,a(i)*r/sig2);
end
figure(1); clf;
plot(r/sqrt(2*sig2),p(1,:)); hold on;
plot(r/sqrt(2*sig2),p(2,:),'--');
plot(r/sqrt(2*sig2),p(3,:),'-.'); hold off;
legend('a=0','a=1','a=2'); grid on;
title('Ricean distribution for unit variance');