2.1 Calculating levels

1. The determination of the level L in Np relies on a voltage or current ratio A2/A1. The choice of a logarithmic measure with base e corresponds to the exponential decrease of voltage, current and power for energy transmission on electric lines. 

Especially for the voltage the following from Figure 2-2 holds:

To illustrate the equations there are two parallel cables with length l showing a line with attenuation coefficient alpha. On the left their beginnings are connected by an arrow indicating U1, on the right their endings are connected with an arrow indicating U2 = U1 times e to the power of – alpha times l 
Figure 2-2: Voltage definition of a transmission line.
Voltage ratio: \( \frac{U_2}{U_1} = e^{-\alpha \cdot l } \)     (2-1)

Voltage level in Np: \( L_u = \text{ln} \frac{U_2}{U_1} = \text{ln}(e^{- \alpha \cdot l }) = - \alpha \cdot l \)     (2-2)

In case of a power ratio \( \frac{P_2}{P_1} = \frac{U_2^2/R_2}{U_1^2/R_1} \) from

\( L_u = \text{ln} \frac{U_2}{U_1} = \text{ln} \sqrt{\frac{U_2^2/R_2}{U_1^2/R1} \cdot \frac{R_2}{R_1} } = \frac{1}{2} \text{ln} (\frac{U_2^2/R_2}{U_1^2/R_1} \cdot \frac{R_2}{R_1} ) = \underbrace{\frac{1}{2} \text{ln} \frac{P_2}{P_1}}_{L_p \text{ in }N_p} + \underbrace{\frac{1}{2} \text{ln} \frac{R_2}{R_1}}_{\text{correction factor}} \) in Np     (2-3)

The following power level results:

\( L_p = \frac{1}{2} \text{ln} \frac{P_2}{P_1} \) for R2 = R1     (2-4)

Only if the same resistance occurs at both points of the transmission system, that are to be compared (P = I²·R = U²/R), P ~ A² holds for the relationship between amplitude and power ratio. If this is not the case, a correction factor (resistance ratio) applies for the transfer from voltage to current or power ratio. Then, the values of voltage, current, and power level will differ a lot.

2. The level in bel or decibel is directly based on the power ratio P2/P1.The following power level results:
\( L_p = \text{lg} \frac{P_2}{P_1} \) in B  or \( L_p = 10 \text{lg} \frac{P_2}{P_1} \) in dB,     (2-5)
and by conversion 
\( L_p = 10 \text{lg} \frac{P_2}{P_1} = 10 \text{lg}\frac{U_2^2/R_2}{U_1^2/R_1} = \underbrace{20 \text{lg}\frac{U_2}{U_1}}_{L_u \text{ in dB}} + \underbrace{10 \text{lg}\frac{R_1}{R_2}}_{\text{correction factor}} \) in dB 
the corresponding voltage level:
\( L_u = 20\text{lg} \frac{U_2}{U_1} \) for R2 = R1     (2-6)
The following logarithm laws have been used:
\( log(a \cdot b ) = log(a) + log(b) \),   \( log(a^b) = b \cdot log(a) \)
In practice, the power ratio most often is given in decibel.

3. The relation between the corresponding values in Neper and decibel results from
 \( L_u [\text{Np}] = \text{ln} \frac{U_2}{U_1} = \underbrace{\text{ln} 10}_{2,3}\text{lg}\frac{U_2}{U_1} = \underbrace{\frac{\text{ln} 10}{20}}_{0,1151} \underbrace{ \cdot 20\text{lg} \frac{U_2}{U_1}}_{L_u \text{ in dB}} \)     and     (2-7)

  \( L_p [\text{Np}] = \frac{1}{2} \text{ln} \frac{P_2}{P_1} = \frac{\text{ln} 10}{2} \cdot \text{lg} \frac{P_2}{P_1} = \underbrace{\frac{\text{ln} 10}{20}}_{0,1151}\cdot \underbrace{10 \text{lg} \frac{P_2}{P_1}}_{L_p \text{ in dB}} \)     (2-8)
The comparison of coefficients gives the conversions
10 dB ≈ 1,151 Np and     (2-9)
1 Np ≈ 8.686 dB.     (2-10)
The abbreviations Np and B or dB can be converted into one another in the same way as units of quantities of the same kind.