Question

Question: In audio and communications systems, the gain, \(\beta \) in decibals is defined for an amplifer as,

\(\beta = {\bf{10log}}\left( {\frac{{{P_{{\bf{out}}}}}}{{{P_{{\bf{in}}}}}}} \right)\)

Where \({P_{{\bf{in}}}}\) is the power input to the system and \({P_{{\bf{out}}}}\) is the power output. (a) A particular amplifer puts out 135 W of power for and input of 1.0 mW. What is its gain in dB? (b) if a signal to noise ratio of 93 dB is specified, what is the noise power if the output signal is 10 W?

Short answer

(a) The gain due to increase in power is 51 dB.

(b) The noise power is \(5 \times {10^{ - 9}}\;{\rm{W}}\).

Step-by-step solution

Determination of noise level

The sound level is determined by the logarithmic power output and input ratio.

The formula of the sound level also determines the noise power in terms of the logarithmic ratio of output power and input power.

Given information

Given data:

The output of the amplifer is \({P_{out}} = 135\;{\rm{W}}\).

The input of the amplifer is \({P_{{\rm{in}}}} = 1.0\;{\rm{mW}}\).

Find the gain in sound level

(a)

The gain in sound level can be calculated as:

\(\begin{array}{c}\beta = 10\log \left( {\frac{{{P_{{\rm{out}}}}}}{{{P_{{\rm{in}}}}}}} \right)\\ = 10\log \left( {\frac{{135\;{\rm{W}}}}{{\left( {1\;{\rm{mW}}} \right)\left( {\frac{{{{10}^{ - 3}}\;{\rm{W}}}}{{1\;{\rm{mW}}}}} \right)}}} \right)\\ = 10\log \left( {135 \times {{10}^3}} \right)\;{\rm{dB}}\\ = 51\;{\rm{dB}}\end{array}\)

Thus, the gain due to increase in power is 51 dB.

Find the noise power

(b)

Given data:

The sound level is \(\beta = 93\).

The output signal power is \({P_{{\rm{signal}}}} = 10\;{\rm{W}}\).

The noise power can be calcaulted as:

\(\begin{array}{c}\beta = 10\log \left( {\frac{{{P_{{\rm{signal}}}}}}{{{P_{{\rm{noise}}}}}}} \right)\\93 = 10\log \left( {\frac{{10\;{\rm{W}}}}{{{P_{{\rm{noise}}}}}}} \right)\\{10^{9.3}} = \frac{{\left( {10\;{\rm{W}}} \right)}}{{{P_{{\rm{noise}}}}}}\\{P_{{\rm{noise}}}} = 5 \times {10^{ - 9}}\;{\rm{W}}\end{array}\)

Thus, the noise power is \(5 \times {10^{ - 9}}\;{\rm{W}}\).