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Some general questions on capacity in FDMA, TDMA and CDMA
", "advice": "Part A:
\n$\\frac{S}{I}_{linear}=10^{\\frac{27}{10}}=501.19$
\n$C_{FDMA}=\\frac{B_T}{\\frac{B_c}{3}\\left(M\\left(\\frac{S}{I}_{min}\\right)\\right)^{\\frac{2}{v}}}$
\n$C_{FDMA}=\\frac{B_T}{\\frac{B_c}{3}\\left(M\\left(\\frac{S}{I}_{min}\\right)\\right)^{\\frac{2}{v}}}=\\frac{20,000,000}{\\frac{10,000}{3}\\left(M\\left(6\\cdot501.19\\right)\\right)^{0.5}}=109.41$ channels/cell
\n109.41 channels is not possible, therefore $C_{FDMA}=109$channels/cell
\n\nPart B:
\n$C_{FDMA}=\\frac{B_T}{\\frac{B_c}{3}\\left(M\\left(\\frac{S}{I}_{min}\\right)\\right)^{\\frac{2}{v}}}$
\nRearrange for $\\frac{S}{I}_{min}$:
\n$\\frac{S}{I}_{min}=\\frac{1}{M}\\left(\\frac{B_T}{\\frac{B_c}{3}C}\\right)^\\frac{v}{2}=31.56$
\n$\\frac{S}{I}_{dB}=10\\log_{10}\\left(31.56\\right)\\approx15$ dB
\n\nPart C:
\n$\\frac{S}{I}=\\frac{S}{I_s+I_a}=\\frac{S}{\\left(M-1\\right)S+I)a}=\\frac{1}{(M-1)+\\frac{I_a}{S}}$
\n$I_a$ is negligible, therefore $I=I_s$:
\n$\\frac{S}{I}=\\frac{1}{(M-1)}$
\n$M=C_{CDMA}=\\frac{I}{S}+1$
\n\nPart D:
\nData rate: $R_b=1$ Mb/s
\nBit period: $T_b=1/R_b=1\\mu s$
\nOne slot duration: $1500\\cdot T_b=1500\\cdot 10^{-6}
\nTotal slot time = $16\\cdot 0.0015=24\\mu s$
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\n$C_{FDMA}=\\frac{B_T}{\\frac{B_c}{3}\\left(M\\left(\\frac{S}{I}_{min}\\right)\\right)^{\\frac{2}{v}}}$
\nwhere $B_T$ 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.
\n\nA 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.
\n$C_{FDMA}=$[[0]] channels/cell
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\n$\\frac{S}{I}_{dB}=$[[0]] dB
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\n\nIf 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.
\n$C_{CDMA}=$[[0]]
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\nDetermine the duration of the frame: [[0]] ms
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