# Step-by-Step Solution ## 1. **Understanding the Problem** - **Given:** - Distance from base station to mobile: \( d = 2 \) km - Received power at \( d \): \( P_r(2\,\text{km}) = -102.82 \) dBm - The channel is **not free space**, so **path loss exponent \( n \) > 2** (due to reflection, diffraction, scattering) - Need to **find the path loss exponent \( n \)** ## 2. **General Path Loss Model** The received power at a distance \( d \) using the **log-distance path loss model**: \[ P_r(d) = P_r(d_) - 10 n \log_{10}\left(\frac{d}{d_}\right) \] where: - \( P_r(d) \): received power at distance \( d \) (in dBm) - \( P_r(d_) \): received power at reference distance \( d_ \) (in dBm) - \( n \): path loss exponent ## 3. **What Is Missing?** To solve for \( n \), we **need \( P_r(d_) \)**, the received power at a reference distance \( d_ \). - Usually, \( d_ = 1 \) m (or another close-in reference distance), and \( P_r(d_) \) is either given or can be calculated from system parameters. - **If not given**, we need to express \( n \) in terms of \( P_r(d_) \). ## 4. **Rearrange the Equation for \( n \)** Solving the path loss equation for \( n \): \[ P_r(d) = P_r(d_) - 10 n \log_{10}\left(\frac{d}{d_}\right) \] \[ 10 n \log_{10}\left(\frac{d}{d_}\right) = P_r(d_) - P_r(d) \] \[ n = \frac{P_r(d_) - P_r(d)}{10 \log_{10}(d/d_)} \] ## 5. **Plug In Values** - \( d = 200 \) m - \( P_r(d) = -102.82 \) dBm - \( d_ = 1 \) m (typical choice) - \( P_r(d_) \) is **unknown** — it's often calculated using the free space formula for the reference distance (if no other information is given). ### **Assume \( P_r(d_) \) is calculated with the free space formula:** \[ P_r(d_) = P_t + G_t + G_r - L_{FS}(d_) \] If \( P_t \), \( G_t \), \( G_r \) are not given, **let's keep \( P_r(d_) \) as a symbol** for now. ## 6. **Compute the Denominator** \[ \log_{10}(d/d_) = \log_{10}(200/1) = \log_{10}(200) \approx 3.301 \] ## 7. **Final Expression for \( n \)** \[ n = \frac{P_r(d_) - (-102.82)}{10 \times 3.301} \] \[ n = \frac{P_r(d_) + 102.82}{33.01} \] --- ## 8. **If You Have \( P_r(d_) \) Value** - **Example:** If \( P_r(d_) = -22.82 \) dBm (hypothetically), then: \[ n = \frac{-22.82 + 102.82}{33.01} = \frac{80}{33.01} \approx 2.42 \] - But you **must use the actual \( P_r(d_) \) for your system**. --- # **Summary Table** | Value | Symbol | Numeric Value | |--------------------------|---------------|------------------| | Distance (mobile) | \( d \) | 200 m | | Reference distance | \( d_ \) | 1 m | | Received power at \( d \)| \( P_r(d) \) | \(-102.82\) dBm | | Received power at \( d_ \) | \( P_r(d_) \) | **unknown** (must be given or computed) | # **Final Formula** \[ \boxed{ n = \frac{P_r(d_) + 102.82}{33.01} } \] - **Insert the correct value for \( P_r(d_) \) to get the path loss exponent.** --- ## **Key Points** - The path loss exponent \( n \) quantifies how quickly signal attenuates with distance. - If \( P_r(d_) \) is known, you can compute \( n \) directly. - For urban or obstructed environments, \( n \) is usually between 2.7 and 4. --- ## **If You Provide \( P_r(d_) \), I Can Give The Numeric Answer.**