agata78 said:
If I use the same equation for example for my initial problem I am not getting the magic anwser for question (ii) 1.727 Np.
You should be getting the 1.727 Np. Show your calculation and maybe we can figure out what's going wrong.
Does anyone know the equation for current ratio expressed in Nepers for the following problem:
Current Output = 250 mA
Current Input = 1.406 A
I think you have those swapped (according to the original post). I think you mean
Current Input = 250 mA
Current Output = 1.406 A
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In the mean time, allow me to make a few comments on decibels.
- Decibels are commonly used by many people in many fields.
- Decibels are defined in terms of power ratios -- not current ratios, not voltage ratios, but power ratios. And there is an assumption that whenever you use decibels, the ratio involved is a ratio of powers.
- With decibels, logarithms are base 10.
- The definition for value in units of decibels (I'll just call it L_{\mathrm{dB}}) is L_{\mathrm{dB}} = 10 \ \log_{10} \left( \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}} \right). (although technically, P_{\mathrm{out}} and P_{\mathrm{in}} don't have to be output and input powers, they can be any powers. The important point is that they be powers. Defining it this way, as input and output, is how one would calculate the gain of a circuit.)
So where does changing the "10" to "20" come from when working with currents and voltages? It starts with an assumption: The input resistance of the circuit is the same as the output resistance of the circuit. This is a
big assumption. But it is also a common one. Normally you want the input resistance of one stage in a circuit to equal the output resistance of the previous stage for maximum power transfer.* So usually, the input and output resistances of all stages are the same, typically something like 50 Ohms (some systems use 75 Ohms like cable TV).
*(Technically you want the impedance of one to the complex conjugate of the impedance of the other, but I'm getting more complicated than I need to right now.)
So what is the power to voltage relationship between input and output voltages and powers? P_{\mathrm{in}} = \frac{V_{\mathrm{in}}^2}{R_{\mathrm{in}}} and P_{\mathrm{out}} = \frac{V_{\mathrm{out}}^2}{R_{\mathrm{out}}}.
But since the input and output resistances are equal, P_{\mathrm{in}} = \frac{V_{\mathrm{in}}^2}{R} and P_{\mathrm{out}} = \frac{V_{\mathrm{out}}^2}{R}.
Plugging that into our decibel equation gives
L_{\mathrm{dB}} = 10 \ \log_{10} \left( \frac{\frac{V_{\mathrm{out}}^2}{R}}{\frac{V_{\mathrm{in}}^2}{R}} \right) = 10 \ \log_{10} \left( \frac{V_{\mathrm{out}}^2}{V_{\mathrm{in}}^2} \right) = 10 \ \log_{10} \left[ \left( \frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} \right)^2 \right] = 20 \ \log_{10} \left( \frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} \right)
You can do the same thing with current ratios, realizing that P = I^2R, and following the same idea. And that's where the "20" comes from.
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Now a few comments on Nepers.
- Nepers are not as commonly used, at least from my experience (I could be wrong, but today is the first I ever remember hearing of them.)
- Nepers are not defined in terms of powers, so you can simply plug your voltages or currents right in, without having to worry about changing "10" to "20" or anything like that, when working with voltages and currents.
- With Nepers, logarithms are base e.
- The definition value in units of Nepers (I'll just call it L_{\mathrm{Np}}) is L_{\mathrm{Np}} = \ln \left( \frac{X_{\mathrm{out}}}{X_{\mathrm{in}}} \right), where X can be a voltage or a current.
