The solution for the voltage in a transmission line can be written as:

$$V(z) = V_o^+ e^{-jkz} + V_o^- e^{jkz}$$

The voltage $V(z)$ is the difference between the conductors of the line at a certain distance $z$. My question are now:

  • What are exactly $V_o^+$ and $V_o^-$? I know it is the voltage wave travelling forward and backward but exactly what are physically?

  • How can I physically think about a connection between $V_o^+$ and $V_o^-$ and S parameters?


2 Answers 2


A transmission line is linear, which means we can consider the voltage at any point to be the superposition of two other voltages.

Asking what they are "physically" may or may not be a useful question.

I could for example define:

$$ V(z) = Ve^{jkz} + V_{DS}(t) $$

where $V_{DS}(t)$ is a function corresponding to the normalized amplitude of Pink Floyd's Dark Side of the Moon at time t.

What does this mean, "physically"? Nothing at all. It's math, and I can do what I want, as long as I follow the rules. And I have: the superposition principle says that I can take the voltage and define it as the sum of any terms I like.

So perhaps the answer to your question of what these terms are physically is the same: they aren't anything physically. They are mathematical abstractions.

Perhaps the question you should be asking is why this particular superposition of forward and reverse waves is a useful one (unlike my Pink Floyd superposition).

The answer is that by considering the waves divided in this way, we can say the forward and reverse waves independently exhibit the characteristic impedance of the transmission line. For example, if this is a (50+0j) ohm line, then considering just the forward wave, there will always be precisely 50 volts per ampere. The ratio will always be 50 and the two will always be in phase.

This is useful, because it isn't generally true considering a transmission line in whole. If the line isn't terminated by its characteristic impedance, then the ratio of volts to amps may not be 50. And the two may not be in phase, indicating a reactive component to the impedance.

So by defining the voltage to be a superposition of forward and reflected waves, the relationship between current and voltage becomes a simple fixed, real (for idealized lines) ratio. This makes analysis simpler. It also means when the line is terminated in a matched load (or at least, any mismatch is practically negligible), the reflected wave is zero and this cancels a lot of terms in a lot of equations related to transmission lines.

Of course we can't simply ignore reality either, but with this superposition defined that complexity can be factored as the reflection coefficient. And this is also real nice since the reflection coefficient can trivially be calculated if the (possibly mismatched) load impedance is known.


The electromagnetic (EM) wave(s) on an antenna consist of an Electrical component manifested as the electric field associated with voltage and a Magnetic component manifested as the magnetic field associated with current. An EM antenna wave cannot exist without both components. If one divides the above voltage components by the characteristic impedance, Z0, one gets the equation for the current. A very good explanation of s-parameters appears in the following HP Ap Note 95-1.


$b1 = s11(a1) + s12(a2)$

$b2 = s21(a1) + s22(a2)$

The reflected wave(s) on the dipole antenna elements are simply obeying the laws of physics. When the forward wave coming from the antenna feedpoint reaches the open end of a dipole element, there is nothing it can do except undergo wave reflection in the opposite direction. The forward energy wave and the reflected energy wave superpose on the dipole elements to get the standing waves (on the standing wave antenna) but do not affect each other except at an impedance discontinuity and that is where the s-parameter equations can be applied. For a dipole, the discontinuity exists at the feedpoint where the transmission line Z0 may be 50 ohms. The Z0 of an ideal #14 wire horizontal dipole antenna 30 feet in the air is 600 ohms so s11 at the 50 ohm to 600 ohm junction of such an antenna is equal to 0.846. If the feedpoint impedance of the dipole is 50 ohms then $b1$ in the first s-parameter equation is equal to zero, i.e. there are no reflections from the antenna feedpoint back down the 50 ohm transmission line. It follows that $s11(a1)$ has to phasor sum to zero with $s12(a2)$. This can only happen when total destructive interference occurs at the feedpoint. The conservation of energy principle tells us that total destructive interference in one direction at an impedance discontinuity in a typical antenna system must be accompanied by total constructive interference in the opposite direction so all of the available energy winds up in the $b2$ forward wave on the dipole antenna. These same principles apply to any impedance discontinuity in an antenna system including a change in Z0 between two sections of transmission line. Note that all elements of the s-parameter equations are phasors and have an associated phase angle. Luckily, for a resonant dipole, the phase angles are either zero or 180 degrees at the feedpoint so that ordinary math will suffice. More info available in an old magazine article of mine which uses the more common amateur radio terms rather than the s-parameter terms:


Somehow, before my first cup of coffee, I got the idea that the question was about the voltage on an antenna. Mea Culpa. Most of what I said applies to both antennas and transmission lines.


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