Solutions to Practice Problems
Practice Problem 23.1
The input power of an amplifier is 6 W. The power gain is AP = 80. What is the output power?
Pout = Pin x Ap = 6 W (80) = 480 W
Practice Problem 23.2
The input power is 15.5 kW. The power output is 10-15 W. Is this system associated with amplification or attenuation?
What is the gain (or attenuation) of this system?
Ap =Pout / Pin = 10-15 W / 15.5 x 103 W = 6.45 10-20
Since Ap is < 1.0, this is attenuation.
Practice Problem 23.3
Convert these two power gains to decibels (dB).
AP =1000
10 x log10 (1000) = 30 dB
AP =0.0001
10 x log10 (0.0001) = -40 dB
Practice Problem 23.4
Convert the following power gains from decibels to decimal gains.
AP,dB = 25 dB:
AP,dB = -6 dB:
AP =
AP =
Practice Problem 23.5
Express Pin = 2 W in decibels as both dBm and dBW.
Pin,dBm =
Pin,dBW =
Practice Problem 23.6
Express 25 dBm in terms of mW and W.
P(in mW) =
P(in W) =
Practice Problem 23.7
The diagram below represents the first three stages of a typical AM or FM receiver. Find the following quantities.
(a) AT and AT,dB
AT = AP1 x AP2 x AP3 = 100 (0.2) 10,000 = 200,000
AT,dB = 10 log10 (200,000) = 53 dB
(b) AP1,dB, AP2,dB, and AP3,dB.
AP1,dB = 10 log10 (100) = 20 dB, AP2,dB = 10 log10 (0.2) = −7 dB, AP3,dB = 10 log10 (10,000) = 40 dB
(c) P1, P2, and Pout.
P1 = Pin x AP1 = 1 pW (100) = 100 pW,
P2 = P1 x AP2 = 100 pW (0.2) = 20 pW
P3 = P2 x AP3 = 20 pW (10,000) = 200 nW
(d) Pin,dBm, P1,dBm, P2,dBm, and Pout,dBm.
Practice Problem 23.8
The signal power at the input to a receiver is 6.2 nW and the noise power at the input to that receiver is 1.8 nW. Find
SNR and SNRdB.