Let's solve each part step by step using **Friis’ formula** for noise figure: ### Friis’ Formula for Noise Figure For a cascade of three components: \[ F_{total} = F_1 + \frac{F_2 - 1}{G_1} + \frac{F_3 - 1}{G_1 G_2} \] where: - \( F = \) Noise Factor (linear, not dB) - \( G = \) Gain (linear, not dB) Noise Figure (NF) in dB: \( NF = 10 \log_{10}(F) \) Gain (G) in dB: \( G_{dB} = 10\log_{10}(G) \Rightarrow G = 10^{G_{dB}/10} \) --- ## (a) **What is the overall noise figure?** **Order:** Antenna → LNA (NF = 2 dB, Gain = 25 dB) → Receiver (NF = 6 dB) - LNA: \( NF_1 = 2\,dB \) → \( F_1 = 10^{2/10} = 1.585 \), \( G_1 = 10^{25/10} = 316.23 \) - Receiver: \( NF_2 = 6\,dB \) → \( F_2 = 10^{6/10} = 3.981 \) #### Applying Friis’ Formula (only two stages since the antenna is noiseless): \[ F_{total} = F_1 + \frac{F_2 - 1}{G_1} \] \[ F_{total} = 1.585 + \frac{3.981 - 1}{316.23} \] \[ F_{total} = 1.585 + \frac{2.981}{316.23} \] \[ F_{total} = 1.585 + 0.0094 = 1.5944 \] **Noise Figure (dB):** \[ NF_{total} = 10 \log_{10}(1.5944) \approx 2.02\, \text{dB} \] --- ## (b) **Antenna → LNA → Cable (20 dB loss) → Receiver** - Cable: **Loss** \( = 20\,dB \) → \( G_2 = 10^{-20/10} = 0.01 \), \( F_2 = 1/G_2 = 100 \) - Add receiver after cable: \( NF_3 = 6\,dB \rightarrow F_3 = 3.981 \) Order: **LNA (F1, G1) → Cable (F2, G2) → Receiver (F3)** \[ F_{total} = F_1 + \frac{F_2 - 1}{G_1} + \frac{F_3 - 1}{G_1 G_2} \] \[ F_{total} = 1.585 + \frac{100 - 1}{316.23} + \frac{3.981 - 1}{316.23 \times 0.01} \] \[ F_{total} = 1.585 + \frac{99}{316.23} + \frac{2.981}{3.1623} \] \[ = 1.585 + 0.313 + 0.943 \approx 2.841 \] **Noise Figure (dB):** \[ NF_{total} = 10 \log_{10}(2.841) \approx 4.53\, \text{dB} \] --- ## (c) **Antenna → Cable (20 dB loss) → LNA → Receiver** - Cable now comes before LNA. - LNA: \( F_2 = 1.585, G_2 = 316.23 \) - Cable: \( F_1 = 100, G_1 = 0.01 \) - Receiver: \( F_3 = 3.981 \) \[ F_{total} = F_1 + \frac{F_2 - 1}{G_1} + \frac{F_3 - 1}{G_1 G_2} \] \[ F_{total} = 100 + \frac{1.585 - 1}{0.01} + \frac{3.981 - 1}{0.01 \times 316.23} \] \[ = 100 + \frac{0.585}{0.01} + \frac{2.981}{3.1623} \] \[ = 100 + 58.5 + 0.943 = 159.443 \] **Noise Figure (dB):** \[ NF_{total} = 10 \log_{10}(159.443) \approx 22.0\, \text{dB} \] --- ## (d) **Conclusion** - **Placing the LNA close to the antenna minimizes the overall noise figure, even with lossy cable after the LNA.** - **If the cable comes before the LNA, the noise contribution of the cable is greatly amplified and dominates the overall system noise.** - **Always place the LNA as close as possible to the antenna to preserve signal quality.** --- ## **Summary Table** | Configuration | Noise Figure (dB) | |------------------------------- |:----------------:| | Antenna-LNA-Receiver | 2.02 | | Antenna-LNA-Cable-Receiver | 4.53 | | Antenna-Cable-LNA-Receiver | 22.0 | --- ### **Key Takeaway** **Always put the LNA before any lossy cable to achieve the best noise performance.**