Page 199 - Schaum's Outline of Theory and Problems of Applied Physics

184 WAVES AND SOUND [CHAP. 15

SOLVED PROBLEM 15.9
Seventy-six trombones are being played, with each one producing a sound intensity level of 70 dB in the
audience. What is the total sound intensity level?
If the intensity of one trombone is I 1 , its sound intensity level is

I 1
β 1 = 10 log = 70 dB
I 0
The intensity of 76 trombones is 76I 1 . Because log xy = log x + log y and log 76 = 1.9, the sound intensity level
of the trombones is

76 I 1 I 1 I 1
β 76 = 10 log = 10 log 76 + log = 10 log 76 + 10 log
I 0 I 0 I 0
= 10(1.9) dB + β 1 = 19 dB + 70 dB = 89 dB

Increasing the sound intensity by a factor of 76 (7600 percent) leads to only a 27 percent increase in the perceived
loudness. This is the reason why, for example, a solo instrument can be distinguished in a concerto even though a
full orchestra is playing at the same time.

SOLVED PROBLEM 15.10

Find the power gain of an amplifier whose power input is 0.2 W and whose power output is 80 W.

P out 80 W
G dB = 10 log = 10 log = 10 log 400 = 10(2.60) = 26 dB
P in 0.2W
SOLVED PROBLEM 15.11
A record player pickup has an output of 0.002 µW. What is the output power when a 100-dB amplifier is
used with it?
We start with the definition

P out
G dB = 10 log
P in
and divide both sides by 10 to give
G dB P out
= log
10 P in
Now we take the antilogarithm of both sides:

G dB P out
antilog =
10 P in
G dB 100
Hence P out = P in antilog = (2 × 10 −9 W) antilog
10 10
1
10
= (2 × 10 −9 W)(antilog 10) = (2 × 10 −9 W)(10 ) = 2 × 10 W = 20 W
SOLVED PROBLEM 15.12
An RG-58/U coaxial cable has a signal attenuation of 23 dB per 100 m at a frequency of 160 mHz. A
20-m length of this cable is used to couple a 25-W VHF radio transmitter that operates at 160 mHz to an
antenna. How much power reaches the antenna?