I do not intuitively understand why max power is transferred when the characteristic impedance of a transmission line is equivalent to the impedance of a load.

A voltage wave going through the transmission line is already traveling through a certain impedance. When it encounters the load, the load's impedance is exactly the same, so the voltage wave should pass through it in the same way it has been passing through the previous part of the transmission line. The load should act as just another part of the transmission line because there are no differences between the impedance.


Many answers aren't addressing my question about why the wave would be absorbed by the load rather than pass right through it.

  • 1
    \$\begingroup\$ You may start simple by analysing simple DC circuit. Using ohms law and power formula, you will notice maximum power will dissipate in the load when its resistance equals to the battery's internal resistance. \$\endgroup\$ Commented Jan 20, 2017 at 14:39
  • \$\begingroup\$ You're mixing up the line impedance in a wrong way. The line impedance has nothing to do with actual power dissipation in this case. It just gives you the relationship between I and V on the line. When encountering a real load, eg resistor, power is dissipated according to Ohm's law. V and I on the point where the line goes into the load have to be the same as in the line, but Ohm's law U = R*I of the load also applies. If the load is different than the line imped., the only way to acchieve a valid solution of I and V at this point is a backwards traveling wave, which takes some of the power. \$\endgroup\$
    – GNA
    Commented Oct 30, 2019 at 22:20

5 Answers 5


If you don't match the impedance, parts of the energy flowing through the transmission line is reflected at the point where the impedance changes.

The reflection factor for the wave is:

\$ \Gamma = \frac{Z_{new}-Z_0}{Z_{new}+Z_0} \$

The power transmission has its maximum when no wave is reflected. This is when \$ Z_{new} = Z_0 \$ (matched impedance).

If you leave the end of the line open. You have a reflection of "1". The wave is coming back as it was sent. If you short the end, you get a factor of "-1". The wave is reflected inverted. For any resistance other than these three cases, some energy is reflected and some not.

More detailed explaination:

Typical RF circuit

In an RF system you have three impedances: The source impedance \$ Z_i\$, load impedance \$Z_L\$ and line impedance \$Z_0\$.

Maximum Power transfer occurs if the input impedance \$Z_{in}\$, seen by the source, equals the complex conjugate of \$Z_{i}\$. this is basic knowledge in AC analysis for every frequency range.

$$ Z_{in} = \overline{Z_{i}} $$

The line impedance does not have to be equal to the source impedance. Only the resulting input impedance (which depends on the line impedance) has to.

The input impedance \$Z_{in}\$ can be calculated or evaluated graphically using a smith chart.

$$Z_{in} = Z_0 \cdot \frac{Z_L+ Z_0 \tanh \gamma l}{Z_0+Z_L \tanh \gamma l}$$ For a detailed explaination of this formula you can look into various RF books or Wikipedia.

Example 1: If you have a source and load impedance of both 50 Ohms, a valid solution is a 50 Ohm line with any length, as long as it is considered lossless. The input impedance seen into the line equals 50 Ohms and therefore the above condition for maximum power transfer is fullfilled.

Example 2: If you take a 100 Ohm load and want to connect it to a 50 Ohm source, a valid way would be:

Take a line with an characteristic impedance of \$Z_0=70.7\Omega (=\sqrt{50 \Omega \cdot 100 \Omega})\$ and a length of a quarter wavelength. The input impedance \$Z_{in}\$ is equal to \$50 \Omega\$ in this case. Therefore, maximum power transfer occurs.

You see, that the line impedance does not necessarily have to be the same as the source/load impedance. However, an easy way to achieve a valid solution is to set all impedances equal, as shown in the first example.

I hope this explains it a little better.


First start by reading here and and in particular the section "Single-source transmission line driving a load"

Why is max power is transferred when the characteristic impedance of a transmission line is equivalent to the impedance of a load.

Well that is not exactly true. You should say "equivalent to the Real part of the impedance of the load."

You should know that there are 3 passive elements: Resistors, Capacitors and Inductors.

Of those only the Resistor can dissipate power because it has a Real value impedance.

$$ Z_R = R $$

Capacitors and Inductors are reactive components and cannot dissipate power (we're talking about ideal components here). Their impedance only has an imaginary part

$$ Z_C = 1/jwC $$ $$ Z_L = jwL $$

That j makes these imaginary.

These reactive componets can only influence the amplitude and phase relation of a signal. Since they cannot dissipate power no power is lost in these components.

An (ideal) transmission line can be seen as a distributed network of Capacitors and Inductors, so no resistors ! The characteristic impedance of a transmission line tells us something about the relations between amplitude, phase, currents and voltages of the waves traveling through it.

In the middle of a transmission line the wave traveling through it "sees" the same characteristic impedance in front and behind. It cannot dissipate into these impedances as they are reactive, they cannot dissipate power.

However at the end of the transmission line at the load, the characteristic impedance ends and turns into a real impedance. The amplitude and phase relations are not changed when the load impedance has the same value as the characteristic impedance of the transmission line. So the wave travels into the load as if nothing has changed. If there was a difference, then part of the wave would reflect.

In the load the wave cannot travel further but since the impedance is real it is dissipated and turned into heat.

  • \$\begingroup\$ Your said "It cannot dissipate into these impedances as they are reactive, they cannot dissipate power." Maybe you can explain (further elaborate) to OP that the "characteristic impedance" is just a number that characterizes the transport media, an invariant, and does not represent the "real active" load to the wave when it travels along? The wikipedia omits this important distinction, and creates a confusion in unsophisticated minds, although it mentions later a link to "wave impedance", which is the one that the EM wave actually interacts with. \$\endgroup\$ Commented Jan 20, 2017 at 23:29
  • \$\begingroup\$ But the characteristic impedance of a transmission line is real? \$\endgroup\$ Commented Feb 2, 2017 at 0:43
  • \$\begingroup\$ @Goldname If you mean, does the characteristic impedance of a transmission line have a real part ? No, it does not so Re(Ztl) = 0. That does not mean it is a short, it means Ztl has a purely imaginary value. A TL can be seen as distributed Capacitors and Inductors. There are no resistors (assuming an ideal TL) so the real part is not present. \$\endgroup\$ Commented Feb 2, 2017 at 6:45
  • \$\begingroup\$ Yes you're right that there are no real impedance components, but the imaginary parts cancel out to become real, which I'm confused about. Z0 = sqrt(jwL/jwC) or something like that, to give a purely real characteristic impedance, right? \$\endgroup\$ Commented Feb 2, 2017 at 16:39
  • \$\begingroup\$ but the imaginary parts cancel out to become real How ?? If there would be a real part then a (ideal) transmission line would be able to dissipate power. And that is not possible. The imaginary part does not become real, that is simply impossible.. So if your Zo formula become real somehow, you made a mistake. \$\endgroup\$ Commented Feb 2, 2017 at 20:46

Probably you know that the refleted wave is zero if the matching is perfect. But why the reflection occurs if there is a wrong load - I suppose that this is the interesting thing.

Explanation: In the very beginning you should note that the characteristic line impedance is not something resistorlike that the wave goes through. It describes the wave - whatlike waves are possible on that line geometry and materials.

assume the the line have a charasteristic impedance Zc. It means the ratio between voltage and current in one and strictly one directionally travelling wave. Up/Ip = Zc = Ur/Ir at every moment and at every point of the line.Here p refers towards the load propagating wave and r refers the reflected wave.

These equations are the mathemathical fact for waves along a loseless two wire line. They are true for momentary voltage and current values as well as for rms values.

For load (this time a resistor =RL) there is another equally valid fact: Ohm's Law. Load's total voltage UL and total current IL obeys Ohm's Law. UL/IL = RL.

Total voltage and total current are the superpositions of the voltages and currents of the different waves. UL = Up + Ur. IL=Ip-Ir. Minus sign is caused by the fact that the reflected wave carries energy towards the beginning of the line.

We have now the equations Up/Ip = Zc = Ur/Ir and (Up+Ur)/(Ip-Ir) = RL.

A short calculation shows that the reflected wave is zero i.e Ur=0 and Ir=0 if RL=Zc. The total load voltage UL is in that case =Up, the coming wave voltage.

Because we assumed the line to be loseless and the signal source can't get anything from the load without the reflected wave, the power transmission must be at it's maximum when RL = Zc. Of course this can be derived purely by calculations, too.

Without calculations we can say, that if RL =Zc, the coming wave directly fits to the load (= same U/I) and no reflected wave is needed to get all the basic U/I laws satisfied.

A little trickier mathematics is needed if we want to prove that nonlinear or reactive additions to RL do not ingrease the transmitted power.


You almost answer your own question but miss a few details. Indeed one of the reasons max power transfer occurs in a matched line is that the wave traveling down a line with some characteristic impedance "feels no difference" in travelling down a load of the same imepdence, and hence no reflections. However, note that your load is typically a resistor, and the impedance is purely real. This means that once the wave travels through the load it is dissipated as heat and no longer exists on the line. On other other hand because a tranmission line (loss less) is modelled with reactive components only, it cannot dissipiate any energy and the wave is just propogated from one portion to another.


Intuitively it looks very simple. It is known that if the load impedance does not match the impedance of transmission line, the incoming wave is partially reflected. It occurs in both cases of impedance mismatch, just the reflected phase is different. In both cases the reflected wave carries energy back into transmission line, so the power dissipated in load is less. So it looks obvious that the power will be at maximum if no reflection occurs, which happens only if the impedance of load matches the line.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Dave Tweed
    Commented Jan 21, 2017 at 2:15

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