Let's answer each part step by step: --- ## Given Data - Number of subcarriers: \( N = 256 \) - Sampling rate: \( F_s = 20\,\text{MHz} \) - Cyclic prefix length: \( N_{CP} = 32 \) - Total RF power at antenna: \( P_T = 10\,\text{dBm} \) - PAPR clipping penalty: \( .8\,\text{dB SNR} \) - Receiver noise power spectral density: \( N_ = -174\,\text{dBm/Hz} \) - Noise figure: \( F = 7\,\text{dB} \) - Available bandwidth: \( B = 15\,\text{MHz} \) (for part d) --- ## a) Useful Symbol Time and CP Overhead ### **1. Useful symbol time \(T_u\)** \[ T_u = \frac{N}{F_s} \] \[ T_u = \frac{256}{20 \times 10^6} = 12.8\,\mu s \] ### **2. Cyclic Prefix time \(T_{CP}\)** \[ T_{CP} = \frac{N_{CP}}{F_s} = \frac{32}{20 \times 10^6} = 1.6\,\mu s \] ### **3. CP overhead factor \( \eta_{CP} \)** \[ \eta_{CP} = \frac{T_u}{T_u + T_{CP}} = \frac{12.8}{12.8 + 1.6} = \frac{12.8}{14.4} \approx .8889 \] --- ## b) Per-Bit Energy at the Receiver \[ E_b = \frac{P_r \cdot \eta_{CP}}{R_b} \] Where, \[ R_b = 2N \cdot \frac{1}{T_u + T_{CP}} \] - \(2\) bits per subcarrier (QPSK) - Number of subcarriers: \(N = 256\) - Symbol time: \(T_u + T_{CP} = 14.4\,\mu s\) \[ R_b = 2 \times 256 \times \frac{1}{14.4 \times 10^{-6}} = 512 \times 69,444.4 = 35,555,555\,\text{bps} \approx 35.56\,\text{Mbps} \] So, \[ E_b = \frac{P_r \cdot .8889}{35,555,555} \] --- ## c) Required \( P_r \) for Target BER **Given:** \[ P_b = 10^{-5} \] For QPSK: \[ P_b = Q\left(\sqrt{2 \frac{E_b}{N_}}\right) \] Find \(E_b/N_\) for \(P_b = 10^{-5}\): From Q-function tables: \[ Q(x) = 10^{-5} \implies x \approx 4.265 \] So, \[ 2 \frac{E_b}{N_} = (4.265)^2 \implies \frac{E_b}{N_} = \frac{18.2}{2} = 9.1 \] So, \[ E_b = 9.1 \cdot N_ \] **Include Noise Figure (NF) and Clipping Penalty:** - \(NF = 7\,\text{dB}\) → Linear: \(10^{7/10} = 5.012\) - Clipping penalty: \(.8\,\text{dB}\) → Linear: \(10^{.8/10} = 1.202\) - Total: \(NF \times \text{Clipping Penalty} = 5.012 \times 1.202 = 6.026\) So, required \(E_b\) increases by \(6.026\) times. **Noise Power:** \[ N_ = -174\,\text{dBm/Hz} = 10^{-17.4}\,\text{W/Hz} = 3.98 \times 10^{-18}\,\text{W/Hz} \] \[ E_b = 9.1 \cdot 3.98 \times 10^{-18} \cdot 6.026 = 2.18 \times 10^{-16}\,\text{J} \] Recall from (b): \[ E_b = \frac{P_r \cdot .8889}{35,555,555} \] So, \[ P_r = \frac{E_b \cdot 35,555,555}{.8889} \] \[ P_r = \frac{2.18 \times 10^{-16} \cdot 35,555,555}{.8889} \] \[ P_r = \frac{7.756 \times 10^{-9}}{.8889} = 8.73 \times 10^{-9}\,\text{W} = 8.73\,\text{nW} \] **Convert to dBm:** \[ P_r (\text{dBm}) = 10 \log_{10}(8.73 \times 10^{-6}) = 10 \left(\log_{10}(8.73) - 6 \right) = 10 \times (-5.058) = -50.58\,\text{dBm} \] --- ## d) Bandwidth Constraint With \( B = 15\,\text{MHz} \): \[ \text{Each subcarrier occupies: } \frac{F_s}{N} \] Total occupied bandwidth: \[ B_{\text{occ}} = \frac{N_a}{N} F_s \] Set \( B_{\text{occ}} \leq 15\,\text{MHz} \): \[ N_a = \frac{B \cdot N}{F_s} = \frac{15\,\text{MHz} \times 256}{20\,\text{MHz}} = 192 \] So, use \( N_a = 192 \) active subcarriers. **Update bitrate:** \[ R_b = 2N_a \cdot \frac{1}{T_u + T_{CP}} = 2 \times 192 \times \frac{1}{14.4 \times 10^{-6}} = 384 \times 69,444.4 = 26,666,666 \approx 26.67\,\text{Mbps} \] **Update \(E_b\):** \[ E_b = \frac{P_r \cdot .8889}{26,666,666} \] **Repeat step (c):** \[ P_r = \frac{E_b \cdot 26,666,666}{.8889} \] \[ P_r = \frac{2.18 \times 10^{-16} \cdot 26,666,666}{.8889} \] \[ P_r = \frac{5.81 \times 10^{-9}}{.8889} = 6.54 \times 10^{-9}\,\text{W} = 6.54\,\text{nW} \] \[ P_r (\text{dBm}) = 10 \log_{10}(6.54 \times 10^{-6}) = 10 \times (\log_{10}(6.54) - 6) = 10 \times (-5.184) = -51.84\,\text{dBm} \] --- # **Summary Table** | Part | Answer | |------|--------| | a) | \( T_u = 12.8\,\mu s \); \( T_{CP} = 1.6\,\mu s \); \( \eta_{CP} = .8889 \) | | b) | \( E_b = \frac{P_r \cdot .8889}{35.56 \times 10^6} \) | | c) | \( P_r \approx 8.73\,\text{nW} = -50.6\,\text{dBm} \) | | d) | \( N_a = 192 \); \( R_b = 26.67\,\text{Mbps} \); \( P_r \approx 6.54\,\text{nW} = -51.8\,\text{dBm} \) | --- **If you need clarification or further breakdown on any step, let me know!**