Why does reflection only applies to transmission lines?

Why does the concept of wave reflection seem to only apply to transmission lines? For example, for a simple circuit with two resistances R1 = 50$\Omega$ and R2 = 75$\Omega$, is the voltage wave coming from the first resistance reflected by the amount:

$\Gamma = \dfrac{75-50}{75+50} = 0.2$ ?

Then it would mean a $(0.2)^2 = 0.04 = 4\%$ power reflection and a $1 - 0.04 = 96\%$ power transfer. But then what is the incident power?

I guess you could brush it off as "transmission lines and resistances are different things" but then what IS the fundamental distinction between them? You kind of have a "wave" of electrons "travelling" in a resistance, and I guess that if they hit another resistance with a different ability to let electrons "travel", then they should partially go back, hence be reflected.

• The concept of reflections is important in acoustics as well. Jul 31, 2015 at 1:18

Reflections happen everywhere, not just in transmission lines. Transmission line is a model of the physical situation, which is easy to apply to a pair of conductors whose length is comparable to or larger than the wavelength of the signal, and which is regular in cross section.

What determines whether reflections matter is the frequencies in and the physical size of the circuit. If you have unmatched impedances then you do get reflected waves just as you describe, and either you have to deal with them or they are negligible for some reason. Here are two reasons:

• For exclusively low-frequency circuits, the reflections reflect repeatedly and settle down on a timescale much faster than the signals change. That is, each double reflection is an extra signal which is merely out of phase with the original signal, but as they get more out of phase their amplitude drops quickly enough that they can be neglected. (Even RF circuits can be built this way, as can be seen from a lot of homebuilt HF amateur radio gear.)

As frequency increases, wavelength decreases, and the physical size of your components becomes relatively larger, and you start having to worry about avoiding impedance “bumps”. This is where you start using microstrip design techniques in printed circuits.

• In digital circuits, sharp transitions may have high-frequency components that will reflect but you don't have to worry about this as long as your clock speed is much slower than the length of your traces/wires (there's a conversion via c to make that make sense, of course) because by the time the clock makes its next tick all the signals have settled down to a steady state.

(Note that there are no standing waves here because within the period of a single clock tick the driving signals are steps (high to low or low to high logic levels), not periodic signals.)

As clock speed increases, the settling time available decreases, requiring you to either minimize reflections or minimize signal travel time (so that the settling occurs faster).

• This is starting to make sense, so for example in my theoretical simple circuit, the waves would get reflected by 20% coming from R1 to R2, but they would also be reflected back and forth everywhere in the circuit until Kirchoff's law apply and I don't have to worry about such and such reflection (for low frequencies) is that about right? Jul 30, 2015 at 23:03
• @victorbg Kirchhoff's laws apply no matter what. It's just that you either have the choice of waiting until a steady state and then applying them to your lumped-element model ignoring distances, or applying them to a model that includes transmission lines (or L-C approximations to them). Jul 31, 2015 at 0:06
• Let's suppose that the two resistances are now separated by a long transmission line. From the moment I turn on my generator until it reaches R2, there is no potential difference at R2 right? So that Kirchoff's law for loops doesn't apply yet, it will only apply in the steady state. If you remove the transmission line, it should still apply, only, it will be very (very very) quick. Jul 31, 2015 at 1:06
• KCL still applies. The current loop near the generator is closed by the capacitance of the transmission line near the generator. As that charges, the current there stops, and the capacitance further along the line charges (the loop grows), and so on until the loop encloses R2. Jul 31, 2015 at 1:11

The difference between them is that a transmission line is characterized by both a capacitance and an inductance (and usually some resistance as well). In real life, the transmission of a signal involves both the generation of a magnetic field (since current is flowing) and electric fields (since there is a voltage difference along the conductor). The framework for dealing with these fields are the concepts of inductance and capacitance. A transmission line can be modeled as a distributed inductive/capacitive network, and it is the energy-storage attributes of the transmission line which allow it to produce the effects that it does. So the reason that it behaves differently from an ideal resistor is that it is different. At audio frequencies and short distances these effects really don't matter, but at either high frequencies or long distances they can become important. One of the first applications to demand treatment of this stuff was the transatlantic telegraph cables. Not very high frequencies, but the long lengths caused unexpected problems. You can read here htp://faculty.uml.edu/cbyrne/Cable.pdf for instance, for a discussion.

The electromagnetic effects you are talking about apply to high frequencies.Normally for circuit analysis the frequency is small so reflection and transmission concepts do not apply.

A resistor is a lumped circuit element almost by definition. Transmission lines are used to model situations where the length of the line is close to or greater than the wavelength. If your physical resistor is bigger than the wavelength, you need to model it as something more complex than a simple lumped resistance. One option might be a lossy transmission line.

Transmission line effects occur when the risetime of the driver is faster than the propagation delay of the wire. If this is not the case, the wire typically behaves as a lumped inductance and the load as a lumped capacitance. I have done a lot of modelling using SPICE and measurements of PC boards and that's what I have found.