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In a communication system, -110 dBm of signal is passed through a low noise ampl...
shrutihire

shrutihire

@copilot In a communication system, -110 dBm of signal is passed through a low noise amplifier with 45dB gain, after which it is passed through a fixed amplifier with 35 dB gain. What is the resultant signal strength?
Answer:
1 watt
1 milliwatts
1 nanowatt
1 microwatt
1y

copilot

To find the resultant signal strength, we need to calculate the power after amplification by the low noise amplifier and then by the fixed amplifier. Let's break it down.
  1. Initial Signal Power:
    • The initial signal power is -110 dBm.
  2. Gain of Low Noise Amplifier:
    • Gain = 45 dB
  3. Gain of Fixed Amplifier:
    • Gain = 35 dB
First, calculate the power after the low noise amplifier:
  • Power after low noise amplifier = Initial power + Gain of low noise amplifier
  • Power = -110 dBm + 45 dB = -65 dBm
Then, calculate the power after the fixed amplifier:
  • Power after fixed amplifier = Power after low noise amplifier + Gain of fixed amplifier
  • Power = -65 dBm + 35 dB = -30 dBm
Now, convert the resultant signal strength from dBm to watts:
  • The formula to convert from dBm to watts is: $ P(\text{watts}) = 10^{\left(\frac{P(\text{dBm}) - 30}{10}\right)} $
Let's calculate the power in watts:
P=10(−30−3010)=10−6 watts=1 microwattP = 10^{\left(\frac{-30 - 30}{10}\right)} = 10^{-6} \text{ watts} = 1 \text{ microwatt}
Thus, the resultant signal strength is 1 microwatt.
1y
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