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EEE8120 Formative Assessment

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In NUMBAS, to insert the equation $A = BC + \\frac{\\sqrt{D}}{E^2}$, the correct format is as follows: A = B*C + sqrt(D)/E^2. Please make a note of this.

A cellular system that comprises of hexagonal cells with radius is shown in Figure 1.

", "advice": "

Part A:

The adjacent side of the right-angled triangle has a length of $i_hd\\sqrt{3}+j_hd\\sqrt{3}\\cos{(60^\\circ)}$ and the opposite side has a length of $j_hd\\sqrt{3}\\sin{(60^\\circ)}$.

Therefore:

$U=\\sqrt{(i_hd\\sqrt{3}+j_hd\\sqrt{3}\\cos{(60^\\circ)})^2+(j_hd\\sqrt{3}\\sin{(60^\\circ)})^2}$

$U=\\sqrt{3i_h^2d^2+3i_hj_hd^2+3j_h^2d^2(\\cos^2{(60^\\circ)}+\\sin^2{(60^\\circ)})}$

$U=\\sqrt{3}\\sqrt{i_h^2+i_hj_h+j_h^2}$

Part B:

If $K=i_h^2+i_hj_h+j_h^2$ and $U=3\\sqrt{d}\\sqrt{i_h^2+i_hj_h+j_h^2}$

$U=d\\sqrt{3K}$

$U/d=\\sqrt{3K}$

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Starting from the original cell, the destination cell is reached by moving $i_h$ cells vertically, rotating $60^\\circ$, and then moving $j_h$ cells in this direction

If the distance between neighbouring cell centres is $d\\sqrt{3}$, derive the distance $U$ between the source and destination cell in terms of $i_h$ and $j_h$.

Answer here: [[0]]

$\"Figure1\"/$

Given that the number of cells in a cluster, $K$, is defined as $K=i_h^2+i_hj_h+j_h^2$, derive an expression for the co-channel reuse distance $U/d$ in terms of $K$.

Answer here: [[0]]

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The path-loss model for the received power at a mobile user in a cell is given by:

$P_{RX}=P_{TX}d^{-v}$

where $P_{RX}$ is received power, $P_{TX}$ is the transmitted power from all base stations, $d$ is the distance from the transmitter to the receiver and $v$ is the path-loss exponent.

", "advice": "

Part A:

Total signal power:

$S=P_{TX}d^{-v}$

Total received power:

$I=\\sum^{M}_{i=1}P_{TX}d_i^{-v}=MP_{TX}X^{-v}$

Therefore $S/I$:

$\\frac{S}{I}=\\frac{P_{TX}d^{-v}}{MP_{TX}X^{-v}}=\\frac{1}{M}\\left(\\frac{X}{d}\\right)^{-4}=\\frac{1}{M}\\left(\\frac{X}{d}\\right)^{-4}$

Part B:

$\\frac{S}{I}_{linear}=10^{\\frac{SNR_{dB}}{10}}=57.54$

$\\frac{S}{I}=\\frac{1}{M}\\left(\\frac{X}{d}\\right)^{-4}$

$\\frac{X}{d}\\longrightarrow$ co-channel re-use distance $\\longrightarrow\\frac{X}{d}=\\sqrt{3K}$

$\\frac{S}{I}=\\frac{1}{M}\\left(\\sqrt{3K}\\right)^v=\\frac{1}{M}\\left(3K\\right)^{\\frac{2}{v}}$

Rearranging for $K$:

$K=\\frac{1}{3}\\left(\\frac{MS}{I}\\right)^{\\frac{2}{v}}=\\frac{1}{3}\\left(6\\cdot57.54\\right)^{0.5}=6.19$ clusters, therefore minimum cluster size, $K=7$.

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If a mobile user also receives signals from $M$ interferers that are all at a distance $X$ from the mobile user, derive the signal-to-interference ratio ($\\frac{S}{I}$) of the mobile user, given a distance $d$ and that all interferers have the same transmission power, $P_{TX}$, and path-loss exponent, $v$.

SIR=[[0]]

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For satisfactory performance of an arbitrary cellular system, a signal-to-interference ratio of $17.6$ dB is required. Find the minimum cluster size, $K$, needed to achieve this, considering $v=4$ and there are $M=6$ equidistant interferers.

K=[[0]]

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Some general questions on capacity in FDMA, TDMA and CDMA

", "advice": "

Part A:

$\\frac{S}{I}_{linear}=10^{\\frac{27}{10}}=501.19$

$C_{FDMA}=\\frac{B_T}{\\frac{B_c}{3}\\left(M\\left(\\frac{S}{I}_{min}\\right)\\right)^{\\frac{2}{v}}}$

$C_{FDMA}=\\frac{B_T}{\\frac{B_c}{3}\\left(M\\left(\\frac{S}{I}_{min}\\right)\\right)^{\\frac{2}{v}}}=\\frac{20,000,000}{\\frac{10,000}{3}\\left(M\\left(6\\cdot501.19\\right)\\right)^{0.5}}=109.41$ channels/cell

109.41 channels is not possible, therefore $C_{FDMA}=109$channels/cell

Part B:

$C_{FDMA}=\\frac{B_T}{\\frac{B_c}{3}\\left(M\\left(\\frac{S}{I}_{min}\\right)\\right)^{\\frac{2}{v}}}$

Rearrange for $\\frac{S}{I}_{min}$:

$\\frac{S}{I}_{min}=\\frac{1}{M}\\left(\\frac{B_T}{\\frac{B_c}{3}C}\\right)^\\frac{v}{2}=31.56$

$\\frac{S}{I}_{dB}=10\\log_{10}\\left(31.56\\right)\\approx15$ dB

Part C:

$\\frac{S}{I}=\\frac{S}{I_s+I_a}=\\frac{S}{\\left(M-1\\right)S+I)a}=\\frac{1}{(M-1)+\\frac{I_a}{S}}$

$I_a$ is negligible, therefore $I=I_s$:

$\\frac{S}{I}=\\frac{1}{(M-1)}$

$M=C_{CDMA}=\\frac{I}{S}+1$

Part D:

Data rate: $R_b=1$ Mb/s

Bit period: $T_b=1/R_b=1\\mu s$

One slot duration: $1500\\cdot T_b=1500\\cdot 10^{-6}

Total slot time = $16\\cdot 0.0015=24\\mu s$

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The capacity of a cellular system employing FDMA is defined as

$C_{FDMA}=\\frac{B_T}{\\frac{B_c}{3}\\left(M\\left(\\frac{S}{I}_{min}\\right)\\right)^{\\frac{2}{v}}}$

where $B_T$ is the total bandwidth, $B_c$ is the channel bandwidth, $M$ is the number of interferers, is the minimum signal-to-interference ratio and $v=4$ is the path-loss exponent.

A cellular system has a total bandwidth of 20 MHz and each channel has a bandwidth of 10 kHz. If the mobile user can tolerate a minimum signal-to-interference ratio of 27 dB and there are 6 interferers effecting the mobile user, determine the capacity of the system.

$C_{FDMA}=$[[0]] channels/cell

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An equivalent cellular system employing FDMA occupies the same total bandwidth as the cellular system in part a), but each channel has a bandwidth of 40 kHz. Determine the minimum signal-to-interference ratio that is required for this system to achieve the same capacity as the system in part a).

$\\frac{S}{I}_{dB}=$[[0]] dB

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The capacity of a CDMA cellular system, $C_{CDMA}$ is dependent on the number of users, $M$, in each cell.

If each user in the cell is transmitting at $S$ Watts and intercell interference $I_a$ is assumed to be neglible, then derive the overall capacity of the system.

$C_{CDMA}=$[[0]]

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A TDMA frame comprises 16 time slots, where each time slot contains 1500 bits. The data is transmitted as a rate of 1 Mbps.

Determine the duration of the frame: [[0]] ms

Enter N/A if unable to answer

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Timeout

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Five minutes to go

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The formative assessment to EEE8120. You will have 120 minutes to attempt all questions. Once the timer expires, questions will no longer be answerable.

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