# Attenuation of a signal

Suppose that I have a wire between two terminals and I want to send a high frequency ($$f$$) signal from one to another.

Is the inductance and capacitance per unit of distance enough to calculate the attenuation of the signal as a function of the wire length?

If so, how can I deduce it?

• Note that for real cables, both impedance and loss are specified. (Remember that an ideal LC circuit has no loss.) – Jon Custer Feb 20 '20 at 13:39

A perfect coaxial cable (pictured as a capacitance $$\Gamma$$ per unit distance and an inductance per unit distance $$\Lambda$$) would not dissipate any signal. It can be described by the D'Alembert equation:

$$\frac{\partial^2 V}{\partial z^2} = \Gamma \Lambda \frac{\partial^2 V}{\partial t^2} = \frac{1}{c^2} \frac{\partial^2 V}{\partial t^2}.$$

If you add a resistance per unit length $$\rho$$, the previous equation is modified as follows:

$$\frac{\partial^2 V}{\partial z^2} = \Gamma \Lambda \frac{\partial^2 V}{\partial t^2} + \Gamma \rho \frac{\partial V}{\partial t}= \frac{1}{c^2} \frac{\partial^2 V}{\partial t^2} + \frac{\kappa}{c} \frac{\partial V}{\partial t},$$

where $$\kappa = \rho \sqrt{\frac{\Gamma}{\Lambda}}$$ is an inverse attenuation length. Indeed, considering a complex wavevector $$k$$ such as $$V(x,t) = V_0 \exp(i(\omega t - kx))$$, we find the following characteristic equation on $$k$$:

$$k^2 = \frac{\omega^2}{c^2} - i \frac{\omega}{c} \kappa$$

which can be solved easily for $$\kappa \ll \omega/c$$ by removing $$\kappa^2/4$$ and noticing that $$(\omega/c - i \kappa/2)^2 = \omega^2/c^2 - i \omega/c \kappa - \kappa^2/4 \approx k^2$$. Thus $$k \approx \omega/c - i\kappa/2$$, and finally:

$$V(x,t) = V_0 \exp(i(\omega(t-x/c))) \exp(-\kappa x/2)$$

The power attenuation coefficient can thus be expressed as $$G_{\mathrm{dB}}(L) = 10 \log_{10} \left(\left| \frac{V(L)}{V(0)} \right|^2 \right) = - 10 \frac{\kappa L}{\ln(10)}$$.

• This only accounts for one kind of loss in the wire (resistance of the wire). It's also possible to have losses due to conductance of the dielectric medium between the wire and the return path, and due to radiation. – The Photon Feb 20 '20 at 2:15
• That is true, I did not include the conductance in my model for simplicity but it also induce losses. I didn't think about radiation as a source of attenuation though. I guess it will still be present even if you assume no resistivity in the wire and no conductance between the wire and around, wouldn't it? – QuantumApple Feb 20 '20 at 10:59

Suppose that I have a wire between two terminals and I want to send a high frequency (f) signal from one to another.

Just a comment here: in any kind of signal integrity problem, you want to consider the return path for your signal as nearly as important as the geometry of the wire that's nominally carrying the signal.

So don't just think of "a wire between two terminals". Also think about where the return or reference current is flowing for signal currents on the wire. Is it in another wire in a ribbon cable? Or in a ground plane below the wire? Or ground planes above and below the wire? Are there any irregularities or disruptions in the return connection?

Is the inductance and capacitance per unit of distance enough to calculate the attenuation of the signal as a function of the wire length?

There are three main sources of attenuation

• Resistive loss in the wire (and return path). You need to know the resistance of your wire (per unit length) to calculate this effect. The value can also increase at high frequencies due to the skin effect.

• Conductive loss in the dielectric. This is important when current can flow from the wire to the return conductor. Again this is measured per unit length of line. This is likely negligible if you're talking about a wire suspended in air above a ground plane. But if you try to transmit a high frequency signal on a circuit board made with a dielectric not designed for high frequencies it can become a problem.

• Radiation. If your wire is too far separated from the return path, and you try to transmit too high a signal frequency, the wire can become an antenna and radiate signal energy away into space. Analyzing this effect generally requires considering the complete geometry of the wire and return conductor, usually using a numerical method rather than an analytical one.

So none of these effects are really described by the inductance and capacitance per unit length of the wire, although the resistive and conductive effects can be described by extending the transmission line element model to include a resistor (R) in series with the inductive element, and a conductor (G) in parallel with the capacitive element: