Question 20.16: A signal of 1.0 mW passes through an amplifier of gain 30 dB...
A signal of 1.0 mW passes through an amplifier of gain 30 dB and then along a cable where the attenuation is 18 dB.
a What is the overall gain of the signal in dB?
b What is the output power at the end of the cable?
a Given that gain and attenuation expressed in decibels are logarithmic, we can say that overall gain
= gain in amplifier + gain in cable
= gain in amplifier – attenuation in cable
= 30 – 18 = 12 dB
b Substitute values in gain = 10 lg\left(\frac{P_{2}}{P_{1}}\right)
So 12 = 10 lg\left(\frac{P_{2}}{1.0 \times 10^{-3}}\right),
therefore lg\left(\frac{P_{2}}{1.0 \times 10^{-3}}\right) = 1.2
Take inverse logs to give output power
P_{2} = 10^{1.2} \times 1.0 \times 10^{-3} = 15.8 mW \approx 16 mWRelated Answered Questions
Question: 20.WE.5
Verified Answer:
Step 1 The signal-to-noise ratio must be at least ...
Question: 20.17
Verified Answer:
a Coaxial cable contains a central copper core sur...
Question: 20.15
Verified Answer:
Signal-to-noise ratio = 10 lg(1.0) = 0.0 dB
Question: 20.14
Verified Answer:
Attenuation = 10 lg\left(\frac{P_{1}}{P_{...
Question: 20.13
Verified Answer:
Attenuation = 10 lg \left( \frac{P_{1}}{P...
Question: 20.WE.4
Verified Answer:
Step 1 Apply the decibel equation to each cable in...
Question: 20.12
Verified Answer:
a The value on the vertical axis (the voltage) has...
Question: 20.11
Verified Answer:
a Binary 1111 is (1 × 8) + (1 × 4) + (1 × 2) + (1 ...
Question: 20.10
Verified Answer:
a 14 = (1 × 8) + (1 × 4) + (1 × 2); in binary, 111...
Question: 20.9
Verified Answer:
AM covers a larger area than FM. FM can reach all ...