Question E4D13:
What is the received signal level with a transmit power of 10 W (+40 dBm), a transmit antenna gain of 6 dBi, a receive antenna gain of 3 dBi, and a path loss of 100 dB?  

Choices:  
A: -51 dBm  
B: -54 dBm  
C: -57 dBm  
D: -60 dBm    

Correct Answer is  A: -51 dBm  

To find received signal level, use the link budget in dB:

$$P_R(\text{dBm}) = P_T(\text{dBm}) + G_T(\text{dBi}) + G_R(\text{dBi}) - L_{\text{path}}(\text{dB})$$

Compute:
- Given: $P_T = +40\ \text{dBm}$ (10 W), $G_T = +6\ \text{dBi}$, $G_R = +3\ \text{dBi}$, $L_{\text{path}} = 100\ \text{dB}$
- $$P_R = 40 + 6 + 3 - 100 = -51\ \text{dBm}$$

Why the correct answer is correct:
- The link budget adds gains and subtracts losses, all in dB units. The arithmetic yields $-51$ dBm, so choice A is correct.

Why the other choices are incorrect:
- B: −54 dBm results from forgetting the receive antenna gain (+3 dBi): $40 + 6 - 100 = -54$.
- C: −57 dBm results from forgetting the transmit antenna gain (+6 dBi): $40 + 3 - 100 = -57$.
- D: −60 dBm results from omitting both antenna gains: $40 - 100 = -60$.

Useful formulas and rules of thumb:
- Link budget (dB): $$P_R = P_T + G_T + G_R - L_{\text{path}} - L_{\text{other}}$$
- Power conversion: $$P(\text{dBm}) = 10\log_{10}\big(P(\text{mW})\big)$$
  - $1\ \text{W} = 30\ \text{dBm}$, $10\ \text{W} = 40\ \text{dBm}$
- Add dB gains, subtract dB losses. A 3 dB change ≈ factor of 2 in power; 10 dB ≈ factor of 10.