Decibel and Neper

Decibel and Neper

Before dealing with various types of filters, it will be desirable to know the unites in which the attenuation of a network is measured. These unites are (i) the decibel (dB) and (ii) the neper. The origin of the decible (dB) can be traced back to the telephony industry. Our ears respond to sound intensity in a non-linear logartithmic fashion. Since many of the amplifiers wear used to supply an audio-output, the use of a lognarithmic power output scale was deemed reasonable and convenient. The basic unit of power gain (the ratio of power output and the power input to a network) was termed the bel in honour of A.G. Bell. The number to bels by    which the power output P_{out} exceeds the power input P_{in} is defined as

Number of bels =log_{10}\dfrac{p_{out}}{P_{in}}

For example, a power gain of 100 would be equivalent to log_{10} 100 i.e. 2 bels. The bel proved to be too large for most work, so one tenth of a bel the decible (dB), was adopted. In decibels power gain of 100 is equivalent to 10 log_{10} 100=10*2=20 dB.

In short, to convert a power expressed in bels into decibels, we merely multiply the number of bels by 10. Symbolically

Number dB = 10*number of bels

Number of dB 10 log_{10}\dfrac{P_{out}}{P_{in}}

In practice, it is much easier and covenient to measure voltage than to measure power levels. Therefore, a relationship between the voltage gain and the power gain in decibels was developed. It is derived as below

Number of dB = 10 log_{10}\dfrac{V^2_{out}/R_L}{V^2_{in}/R_{in}}=10 log_{10}\left[\dfrac{V_{out}}{V_{in}}\right]^2 \dfrac{R_{in}}{R_L}

=20 log_{10}\dfrac{V_{out}}{V_{in}}

if the power P_{out} and P{in} are associated with equal impedances.

In a similar fashion, the current gain in in decibles can be related to the power gain in decibels and

Number of dB =20 log_{10}\dfrac{I_{out}}{V_{in}}

The nepar is fundamentally a unit of current or voltage ratio.

The attenuation in nepers =log_e\dfrac{I_{out}}{I_{in}}

or =log_e\dfrac{V_{out}}{V_{in}}

The attenuation dB  =20 log_{10}\dfrac{I_{out}}{I_{in}}

=\dfrac{20 log_e\dfrac{I_{out}}{I_{in}}}{log_{10}e} =8.686 log_e\dfrac{I_{out}}{I_{in}}

Thus attenuation in dB = 8.686*attenuation in nepers.

It should be noted that the two units are defined on  alogarithmic base, the decibel being being expressed in logarithms to the base 10 and the neper being expressed in logarithms to the base 10 and the neper being expressed to the base ‘e’. This is since their greatest field of application is in sound transmission, the loudness of which is a logarithmic function of the sound energy.

Example 1. The output and input voltage of a filter network are 10 mV and 20 mV respectively. Determine the attenuation in dB and nepers.

Solution: Input voltage, V_{in}=20 mV

Output voltage, V_{out}=10 mV

Attenuation in dB =20 log_{10}\dfrac{V_{out}}{V_{in}}

=20 log_{10}\dfrac{10}{20}

=-6 dB Ans.

Attenuation in nepers \dfrac{Attenuation\enspace in\enspace dB}{8.686}

=\dfrac{-6}{8.686}

= -0.6908 nepers Ans.