Question E4D12:
What is the link margin in a system with a transmit power level of 10 W (+40 dBm), a system antenna gain of 10 dBi, a cable loss of 3 dB, a path loss of 136 dB, a receiver minimum discernable signal of -103 dBm, and a required signal-to-noise ratio of 6 dB?  

Choices:  
A: -8dB  
B: -14dB  
C: +8dB  
D: +14dB    

Correct Answer is  C: +8dB  

To find link margin, do a link budget to get received power, then compare it to the required minimum receive power.

Step 1: EIRP (include gains, subtract losses)
$$\mathrm{EIRP} = P_t + G_{\text{ant}} - L_{\text{cable}} = 40 + 10 - 3 = 47\ \text{dBm}$$

Step 2: Received power after path loss
$$P_r = \mathrm{EIRP} - L_{\text{path}} = 47 - 136 = -89\ \text{dBm}$$

Step 3: Required receive power (MDS plus required SNR)
$$P_{\text{req}} = \mathrm{MDS} + \mathrm{SNR}_{\text{req}} = -103 + 6 = -97\ \text{dBm}$$

Step 4: Link margin
$$\text{Margin} = P_r - P_{\text{req}} = (-89) - (-97) = +8\ \text{dB}$$

Therefore, the correct answer is C: +8 dB.

Why the other choices are incorrect:
- A: -8 dB — this comes from reversing the subtraction: computing $-97 - (-89)$ instead of $-89 - (-97)$.
- B: -14 dB — a result of compounding sign/arithmetic errors (e.g., subtracting antenna gain or wrong dBm conversion). It does not match the correct link budget.
- D: +14 dB — this is what you get if you compare to MDS only and ignore the required SNR: $-89 - (-103) = +14$ dB.

Useful formulas and tips:
- Link budget (dB units): $$P_r = P_t + G - L - L_{\text{path}}$$
- Required threshold: $$P_{\text{req}} = \mathrm{MDS} + \mathrm{SNR}_{\text{req}}$$
- Link margin: $$\text{Margin} = P_r - P_{\text{req}}$$
- In dB arithmetic: add gains, subtract losses. Positive margin means the link meets the requirement.
- dBm conversion: $10\ \text{W} = 40\ \text{dBm}$ (since $1\ \text{W} = 30\ \text{dBm}$; +10 dB per 10× power).