Respuesta :
Answer:
P₂ = 0.5 W
Explanation:
It is given that,
The attenuation of a signal is -10 dB.
We need to find the final signal power if it was originally 5 W.
Using attenuation for signal :
[tex]dB=10\text{log}_{10}(\dfrac{P_2}{P_1})[/tex]
Put dB = -10, P₁ = 5 W
Put all the values,
[tex]-10=10\text{log}_{10}(\dfrac{P_2}{5})\\\\-1=\text{log}_{10}(\dfrac{P_2}{5})\\\\0.1=\dfrac{P_2}{5}\\\\P_2=5\times 0.1\\\\P_2=0.5\ W[/tex]
So, the final signal power is 0.5 W.
The final signal power if it was originally 5W is; 0.5 W
Signal Power
We are given;
- Attenuation signal; dB = -10 dB
- Original signal power; P1 = 5 W
Formula for attenuation signal is;
dB = 10log_10_ (P2/P1)
Where P2 is final signal power
Thus;
-10 = 10log_10_(P2/5)
-10/-10 = log_10_(P2/5)
-1 = log_10_(P2/5)
0.1 = P2/5
P2 = 0.1 × 5
P2 = 0.5 W
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