Chapter 26, Problem 35A

The signal-to-noise ratio for a tape recorder is listed at 50 dB, meaning that when music is played back, the intensity level of the music is 50 dB greater than that of the noise from tape hiss and so forth. By what factor is the sound intensity of the music greater than that of the noise?

To determine

The factor by which the intensity of sound is greater than the noise generated by the tape recorder is to be determined.

Answer to Problem 35A

The sound intensity by the tape recorder is $\text{100,000}$ times greater than that of the noise.

Explanation of Solution

Given info:

Signal-to-noise ratio, $SNR=50dB$

Formula used:

Signal-to-noise ratio is related to the relative power by the following formula

  $SNR=10Lo{g}_{10}\left(\frac{P_s}{P_n}\right)dB\Rightarrow \frac{P_s}{P_n}={10}^{SNR/10}$ , $P_s$ and $P_n$ are the intensities of signal and noise from the tape recorder respectively.

Calculation:

Put the given value of $SNR$ , we get

  $\frac{P_s}{P_n}={10}^{SNR/10}={10}^{50/10}={10}^5$

Conclusion:

The signal is greater than noise by a factor of $1\times {10}^5$ .