# Why does "impedance matching" in the context of reflections demand that resistive impedance and reactive impedance are treated separately?

I have assumed that for reflections, Ohm had to be matched between source and load, but resistive and reactive impedance did not have to be matched separately. I was told this is wrong, and that reactance and resistance have to be treated separately.

For example, R_source = sqrt(2), R_load=1, Z_load =j, Z_source = 0j, matches impedance, sqrt(2) Ohm on both source and load, but, reactance and resistance are not matched separately (i.e., reactance is not cancelled out and then resistance matched, the "complex conjugate". )

Since resistance affects the impedance from reactive components, assuming source & load resistance behaves as series resistance, then R_load = 1/sqrt(2), R_source = sqrt(2)-1, Z_load = j, Z_source = 0j, matches the Ohm for Z_source & Z_load. So it still seems like Ohm can be matched, without matching reactance and resistance separately.

Is there any way to pin point to exactly why reflections would still be a problem, if the impedance is matched in Ohm, but not necessarily matching the reactance and resistance separately?

Edit: I assume the answer is that the "magnitude", or actual ability to resist current flow, of the reactive elements is phase dependent. Thus, when the source signal encounters the load, the actual ability to resist (or conduct) the signal, may be out of phase if the phase is not matched, and lower or higher than it would be once the reactive element has "caught up" (after the phase shift has been accounted for). Therefore, there will still be reflections if the phase is not matched, even if the magnitude is matched.

• When you write "Ohm", do you mean resistance or impedance? Please edit accordingly. Commented May 4, 2023 at 9:25
• I explained this in an answer to your previous question. What was it that you didn't understand? Commented May 4, 2023 at 9:59
• Your 2nd and 3rd paragraph above does not read at all well. Please try and fix this. Commented May 4, 2023 at 10:04
• You use "reflection" multiple times but make not even a single reference to Γ. Similarly you use "match" but do not define it. As a result, your post reads as unclear and confused. Your ongoing (considering several recent posts) confusion would seem to be better resolved by what providing background information you took in recently which prompted these questions. Discussing those materials would seem to get at the root of your confusion much better than asking questions resulting from that confusion. Commented May 5, 2023 at 10:10
• @BipedalJoe if you are done with this question you should follow this guidance: electronics.stackexchange.com/help/someone-answers Commented Jul 21, 2023 at 8:36

Is there any way to pin point to exactly why reflections would still be a problem, if the impedance is matched in Ohm, but not necessarily matching the reactance and resistance separately?

You implied earlier (in a comment) that when you use the term "Ohm" you refer to the complex impedance formed by a resistance and a reactance. If this is incorrect then please say so. Anyway...

If a cable has a complex characteristic impedance then, to prevent reflections, you must use a complex terminator hence you must match both resistive and reactive parts.

But, at RF (greater than around 100 kHz), cables don't have a complex characteristic and, you would terminate a cable in a resistance that matches the cable.

Why am I focussing on cables you might ask; cables have significant length and, are therefore, the exclusive cause of all significant signal reflections.

To understand reflections (and reflection coefficient) we can set up a thought experiment: -

• Assume a cable (transmission-line) of characteristic impedance $$\Z_0\$$
• At some distance away $$\Z_L\$$ terminates the cable
• Assume a voltage ($$\V_F\$$) applied at the non-terminated end of the line
• The current ($$\I_F\$$) that initially flows into the t-line equals $$\V_F\$$ divided by $$\Z_0\$$

When the voltage and accompanying current reach the end of the t-line and meet $$\Z_L\$$ there will be a potential violation of ohm's law if $$\Z_0\$$ does not equal $$\Z_L\$$. We have to fix this violation. The mathematical process of fixing this violation uncovers the meaning of the reflection coefficient.

At no point am I assuming anything about the complex nature of these impedances.

For instance, if $$\Z_L\$$ > $$\Z_0\$$ we have to consider a mechanism that prevents the potential violation of ohm's law. The options are: -

• Somehow make the voltage arriving at $$\Z_L\$$ a bit bigger and, at the same time
• Somehow make the current arriving at $$\Z_L\$$ a bit smaller
• So, we "adjust" voltage and current in such a way so as to produce a ratio of $$\Z_L\$$
• Those "adjustments" have to "go somewhere" and, indeed they form a reflection

Algebraically we could say: -

$$\dfrac{V_F + \delta V_F}{I_F - \delta I_F} = Z_L$$

In effect I've added a "bit" of voltage and, subtracted a "bit" of current. It drills-down like this: -

$$\dfrac{V_F}{I_F}\cdot \dfrac{1 + \delta}{1 - \delta} = Z_L\longrightarrow Z_0\cdot \dfrac{1 + \delta}{1 - \delta} = Z_L$$

$$\text{Hence,}\hspace{1cm}\delta Z_0 +\delta Z_L = Z_L - Z_0$$

$$\text{And,}\hspace{1cm}\delta = \dfrac{Z_L-Z_0}{Z_L+Z_0}$$

But, of course, we call $$\\delta\$$ by it's usual name (reflection coefficient) $$\\Gamma\$$. The symbol $$\\delta\$$ is just a device I invented to get through the thought experiment.

However, the important subtlety that prevents an ohm's law violation is the "bit" we add to voltage and the "bit" we subtract from current ($$\\delta V_F\$$ and $$\\delta I_F\$$).

If we examined their ratio we would find it is $$\Z_0\$$. This means that they can naturally flow (together) back into the transmission line because they have the perfectly correct ratio to do so.

That is called a reflection and travels from load to source.

Clearly, if $$\V_F\$$ and $$\I_F\$$ were originally of a ratio that matched the load ($$\Z_0\$$) impedance (right from the start), we wouldn't need to set up the algebra that figured out how to deal with the unwanted signals and, there would be no thoughts of violating ohm's law nor talk of reflections.

The reflection coefficient $$\\Gamma\$$ is defined as: $$\Gamma=\frac{Z_L-Z_S}{Z_L+Z_S}$$

$$\\Gamma=0\$$ to have no reflected power, Setting the numerator to zero and expanding the impedance, $$R_L+jX_L=R_S+jX_S\tag{equ 1}$$

The real and imaginary parts are independent quantities and so must be equal separately for there to be no reflection.

I think “matched by Ohm” means the magnitude of Z, |Z|. Then equ 1 can be written $$|Z_L|\angle Z_L=|Z_S|\angle Z_S$$

Both the magnitude and the angle must be separately equal for no reflection.

• Hi that's great, I know those equations too. What I asked about is why. But, as I mentioned in the edit, I think the reason is that for reactance, how much current is actually "resisted" is varying over the wave. And that is why reflections become a problem, when magnitude is matched but phase isn't. That's the best "pin point" to why I can think of. Commented May 4, 2023 at 17:27

It's like the size of my shoes should be 42. I bought a pair of size 42 items and they do not at all fit my feet. What's wrong? The items happen to be hats, but why don't they fit? Their size is 42 and 42 is good for my feet!

The pure number of ohms presents a resistance or the absolute value of a impedance (=complex number) or the value of the real part of the impedance or the value of the imaginary part of the impedance. 50 ohms as impedance is not uniquely defined before one says also one of the next:

• the 50 ohms is resistive (that's common assumption when one says #it's 50 ohms)
• the 50 ohms has the phase angle xxx degrees, where xxx is a number.

Resistive impedance means xxx= zero.

Matching problems cannot be solved right if impedance phase angle is not taken into the account. Splitting an impedance to real and imaginary parts is in math the same as knowing the absolute value and the phase angle.

• The shoe reference is probably not helpful Commented May 4, 2023 at 14:35
• @kazooplay9 doesn't really address why phase is important. As I mentioned in my edit, my best guess is because how much the current is resisted varies over the wave. And that's why the "magnitude" is not matched, if phase isn't matched, even though it is matched in terms of total current. Commented May 4, 2023 at 17:29
• As for shoe reference, I welcome all references and answers. People are helpful in their own way, and that's great. Commented May 4, 2023 at 17:49

Much talk of electricity facts that the questioner probably already has heard but cannot understand intuitively like he can understand practical facts which contain only visible and touchable things.

The question is about reflection and preventing it by having matched impedances. There exists other impedance matching cases, too, for ex. matching for best noise performance. But lets talk about matching a load to a transmission line.

When we have a load connected to the end of a transmission line the electromagnetic wave coming along the line sinks to the load without generating a reflected wave if the load is matched i.e. it has the same resistance as the transmission line impedance - that's 50 ohm resistive load to a 50 ohm coaxial cable.

These 50 ohms mean different things. In resistor it's the ratio of the voltage over the resistor and the current through the resistor - the Ohm's law. In transmission line the 50 ohm is the ratio of the voltage and current which belong to the same wave which propagates to one direction and are measured in the same point on the line. The electromagnetic wave which propagates along the line is actually outside the metal and it contains electric and magnetic fields, but we can follow the wave in simple regular 2 conductor transmission lines by following the voltage of the electric field between the conductors and the induced current at the surface of the metal.

The arriving wave has a certain ratio between its voltage and current - the line impedance - which in low loss lines is a real number even when we make wave calculations with phasors for sinusoidal waves. The voltage and current components have the same time waveform (no phase shift) if the line doesn't have remerkable losses.

The load impedance can be resistive or with sinusoidal waves we coveniently also use complex load impedances. If the load impedance happens to be something else than a resistance which is equal with the line impedance, a reflection occurs.

The reflection coefficient which is presented in an older answer, is the ratio how strong is the reflected wave when compared to the arriving wave. The strength of the reflected wave can be derived in math easily. Only such waves can exist which have a certain line type and material specific propagation velocity and which have a certain line type and material specific ratio (=the line impedance) between its voltage and current components. At the end of the line must that relation be valid for the wave. The same voltage and current must also obey ohms law caused by the load. That cannot happen with 60 ohm resistor connected to a 50 ohm line.

To unviolate those untouchable boundary conditions we must assume that a reflected wave is born at the end of the line. By having at the same time 2 waves which propagate to opposite directions we can satisfy the wave voltage/current impedance -rule and Ohm's law for the load. In Ohm's law we use the net voltage (=sum) and net current (=difference) of the 2 waves. Making this calculation gives the reflection coefficient. The calculation is presented in every transmission line theory book.

Why a 50 ohm non-resistive load is not matched to a 50 ohm transmission line? Answer: The momentary values of the wave voltage and current do not fit. If the load voltage and current have phase difference their momentary value ratio is not 50 ohm, no matter their peak or r.m.s-values have ratio =50 ohm. No reflection -case needs that the current-voltage ratios fit as momentary values, i.e. in every moment of the time.

When we have connected to the load some other signal source than a transmission line, the source can also have internal impedance. Impedance matching problem is in many cases power transmission maximizing. That's achieved by having a complex conjugate load for certain signal source impedance. That's because the imaginary parts cancel each other when in series. Cancelling the imaginary parts reduce the total absolute value of the series impedance, it makes possible more current with certain source voltage. Equal real parts are easy to be proved the biggest transferred power case. Proving it, of course, needs ability to handle find the maximum -problems. It's well presented in circuit theory textbooks.

Often - especially in radio circuits - engineers use wave formulation for all signal paths, no matter is it a transmission line or 2 interconnected subcircuits in so small circuit board (compared to the wavelength) that all wave phenomenons could be skipped and perfect results would be possible with Ohm's and Kirchoff's laws only. That unified approach occurs as scattering parameters in measurements and calculations. The method unification makes calculations and measurements easier when there are mixed cases in the same system. Some parts may need using waves for realistic results. When every signal is considered as a wave a single method covers all signals in the circuit.

But, as said above, there are other impedance matching problems, for ex. noise optimization. I halt my writing in this phase.

I think it is that reactive impedance is time-varying. Reflections in mechanical systems happen when the termination is either "open" or "closed", or somewhere in between, up until it is matched. The reactive impedance is matched with resistive and current through resistive, but, not at the same time. It is matched with either a slight delay, or, slightly ahead of time. So when the signal has passed through the transmission line and meets the "load" (that includes reactance in the context of this question and is matched in terms of combined Ohm but not in phase), it is not at that time the same "Ohm" or ability to resist/conduct current. Therefore, there will be a tendency for reflections, the reflection coefficient will not be 0 when considering time variation of phase.

• Reactive impedance is not time-varying in a linear network as implied by your question. So, this answer seems wrong. Commented May 8, 2023 at 23:40
• It's not time-varying. The buck stops there. Commented May 8, 2023 at 23:48
• That's not what time-varying means. Commented May 8, 2023 at 23:53
• Stop shifting goal posts. I was not talking about some other answers, I am talking about yours. And the impedance is not time-varying, that's the whole point of specifying an impedance. It describes the behaviour of the circuit in face of harmonically changing, including DC, signals, without depending on time. This is literally first week, first lecture knowledge, and we shouldn't have to discuss it. Commented May 8, 2023 at 23:56
• You're really squandering your chance to learn something here! As said, getting things wrong happens. It's not a big deal. Insisting that one's wrong model of the world be left uncommented is a bit of a bummer, and I invite you to do better in communicating with the people trying to explain things to you in their spare time. I wish you a nice day now. Commented May 9, 2023 at 1:00