# What happens to the reflected wave when it gets back to the generator? (1 Viewer)

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#### iVenky Basically, we have a transmission line not matched to load R. So there is a forward wave and a reflected wave. Now when the reflected wave reaches back to source 'S', does it get absorbed by the source? Does that mean the source 'S' gets back some of the power (equal to the reflected power) that is initially given to the forward wave?

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#### Baluncore

The reflected wave will be absorbed by the 50 ohm series matching resistor in/at the source.
That source to line matching resistor is there to prevent secondary reflections.
Reflected energy will appear as heat and not be recycled.

#### alan123hk

Basically, we have a transmission line not matched to load R. So there is a forward wave and a reflected wave. Now when the reflected wave reaches back to source 'S', does it get absorbed by the source? Does that mean the source 'S' gets back some of the power (equal to the reflected power) that is initially given to the forward wave?
The logic formula is : Absorbed Power = Incident Power - Reflected Power.

Since the source impedance is 50 ohm, which is matched with the 50 ohm impedance of the transmission line, therefore, no reflected power from the source occurs again, namely Absorbed Power = Incident Power, all the power back from the load is entirely absorbed by the source.

There is a possible circumstance in which the net transmit power of the source may momentarily become negative. That is, when the reflected power is large enough to generate a transient voltage higher than the original source voltage. At this point, the current actually flows back to the positive terminal of the source, so the source absorbs purely rather than providing power at this moment, although this does not happen in the configuration shown in your diagram.

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#### Rive

Does that mean the source 'S' gets back some of the power (equal to the reflected power) that is initially given to the forward wave?
Usually driver circuits has some kind of protection against this kind of interference from the driven line. It is difficult enough to keep the power voltages 'clean', sudden 'boosts' are not needed. So even if the reflection not gets absorbed by the (hopefully: matching) source impedance, it will not be recovered either.

#### Baluncore

It is important to model a source as a voltage source with zero impedance, or as a current source with an infinite impedance. A series or parallel resistor can then be employed to match the source to a transmission line.
There is a possible circumstance in which the net transmit power of the source may momentarily become negative.
That is not really true. Energy is transmitted, the rate of energy transmission is power. The reflected wave is reflected energy that will all be dissipated as heat in the resistor, so there is no negative energy, or energy flow back into the source.

The OP shows an ideal voltage source with an output impedance of zero. It has a perfect 50 ohm series matching resistor, into a line with a 50 ohm characteristic impedance. The OP also shows a sinewave as the signal, not a step. The length of the line is not specified.

Depending on line length and mismatch, the return of the reflected sinewave may cause the line to appear reactive. That may phase shift the current relative to the source voltage, but the transmitted energy will never be negative, because the reflected wave will always have less energy than the transmitted wave.

The length of a perfectly matched line is irrelevant. If you consider the source as providing a voltage step to the line through the resistor, you will see that the reflected step appears across the resistor and cannot influence the voltage of the zero impedance voltage source.

#### alan123hk

That is not really true. Energy is transmitted, the rate of energy transmission is power. The reflected wave is reflected energy that will all be dissipated as heat in the resistor, so there is no negative energy, or energy flow back into the source.
I agree that negative power output doesn't happen in the circuit diagram shown by OP as I said "although this does not happen in the configuration shown in your diagram", indeed, the negative power doesn't occur if the source impedance (no matter put it inside or outside the source) is matched with the transmission line.

However, momentary negative power is possible if both the source impedance and load impedance are not matched with the transmission line.
Of course, this phenomenon only lasts for a very short time, it will disappear after the system is stabilized.

#### Baluncore

However, momentary negative power is possible if both the source impedance and load impedance are not matched with the transmission line.
Of course, this phenomenon only lasts for a very short time, it will disappear after the system is stabilized.
You are suggesting that reflected energy can flow back into a voltage source with a zero impedance. But if any reflected energy got past the matching network to reach the source it would all be reflected from the source by the infinite impedance mismatch.

#### alan123hk

it would all be reflected from the source by the infinite impedance mismatch.
Yes, the idea voltage source reflects all reflected energy from load back to the source terminal of the transmission line.
However, I think that transient negative power (power is absorbed by the source) is still possible occur under certain circumstances, the two situations can happen simultaneously.

Let's assume a transmission line is terminated in an open circuit termination and fulfills the other conditions to behave as a pure capacitor, then drive it with an ideal AC voltage source at the source terminal, and then the phase shift between the voltage and the current become 90 degrees, that means the voltage source gives both positive and negative power, although the voltage source has zero impedance and indeed reflects all input power from the transmission line back to the transmission line.

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#### Rive

@alan123hk , if the question is about the presented model, then you don't have space to maneuver at all, the source side has matching impedance (50Ohm resistor and 0Ohm voltage source): if the question is an engineering question presented with just a general model, then you don't have enough data to maneuver and also, you have to work with the usual components/problems (what means you should expect at least some cutting diodes and secondary/thertiary reflections along the whole line due the unavoidable impedance mismatch common in any existing setup).

#### Baluncore

The reactive current that flows on a transmission line is VAR and does not represent real energy, so is not relevant to power.

You should compute power by considering the total energy flow over a full cycle, then dividing by the period of one cycle.

Yes, the idea voltage source reflects all reflected energy from load back to the source terminal of the transmission line.
You appear to be arguing that “momentary negative power” can happen with a non-ideal source.
Anyhow, take that non-ideal source, with a series terminating resistor. Factor out the non-ideal impedance component of the source, combine it with the series resistor. That leaves an ideal source with a more complex matching impedance. There is now an ideal source so there can be no “momentary negative power”.

First - I want to say, in my understanding - ANY non-resistive LOAD will result in Instantaneous (more concise than momantary) Negative Power flow INTO the source.

The Ideal V source has Zero Impedance, so current can and will flow so that Power Flows into the source.

When we measure (and discuss) power - we typically do this over full cycles, true, but there are definite times when we need to consider that within a cycle power flow is both directions - in an ideal or pretty much any real source. An ideal source connected to an ideal capacitive load - is zero (net) but 50% in and 50% out power flow relative to both the source and the load.

As an example - the need for freewheeling diodes or snubbers on switching power supply ( sources), even in the ideal model - these are needed to ensure the calculations are solvable.

As far what happen "in" the source - where this energy goes at the instantaneous timeframe, is not part of the definition of an ideal source model.

#### Baluncore

First - I want to say, in my understanding - ANY non-resistive LOAD will result in Instantaneous (more concise than momantary) Negative Power flow INTO the source.
Instantaneous power involves division by zero and so must remain undefined. You probably mean energy.

You misunderstand reactive current. It does not represent real power because it does not come from the source voltage, it is reflected by the source, so it circulates on the line.
Volts * amps = power. Sine * cosine = zero.
The only real power involved in reactive current is the resistive voltage drop along the wire, (or in the termination), due to the reactive current.

As an example - the need for freewheeling diodes or snubbers on switching power supply ( sources), even in the ideal model - these are needed to ensure the calculations are solvable.
No. Flyback diodes provide a path for inductive energy to reach a supply reservoir capacitor when a switch is turned off and the magnetic field collapses. Snubbers are used to reduce dv/dt or di/dt to values that will not destroy real semiconductors. They have nothing to do with transmission lines or making calculations possible.

You have lost me as to how instantaneous power requires division by zero. At any given instant P = V * I. Take the extreme - but ideal case of a Voltage source connected to a capacitor ( or dummy cap load ) . For one quarter of the cycle from Vs=V p to +Vpeak, current flows into the capacitor, and charges it, so power is flowing from the source to the capacitor, the capacitor is now charged, it has stored REAL energy. As the Vs falls from +Vp the capacitor discharges and current flow back into the source. When we get to Vs = 0 Vcap = 0 .. there is no energy in the CAP - the REAL Energy moved from the cap into the source. The average power over time is zero - but the instantaneous is not...

The relationship of an ideal source and load - I do not see how the application has much to do with it, there are more elements. But if the load on the source is not perfectly resistive ( perfectly tuned) the source will see instantaneous power flow into it. Isn't this whole idea of the antenna tuner being as close as possible to the transmitter - most other systems can deal with this energy flow. A transmitter typically can not but this is no where near an ideal source; a poorly tuned Transmitter +antenna set up will kill many transmitters, because they can not absorb the reflected energy..

The point of comment regarding flyback and using ideals - if we have current flowing in an inductor, and attempt to stop the current, an IDEAL source in series with an ideal diode and inductor) - we are storing real energy in the inductor, in this configuration the source can not absorb the energy as the diode blocks the current. This is a case where the diode + source (or any switch in series with the current flow) create a system that can not absorb the energy back from the inductor.

As for the calculations - the energy has to go somewhere... if we stop the current in an inductor, the dI/dt is very high, if the switch is "ideal" you can not calculate the result - the "dt" is zero...put a flyback or snubber in your circuit you provide some where for the energy to go, and you can calculate a solution, or in the real world, you do not break things, not just the semiconductors.

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#### alan123hk

if the question is about the presented model, then you don't have space to maneuver at all,
I am not trying to maneuver anything, all I want is to express something I believe to be true with my limited English ability. I am also not sure if there is any inaccuracy or error in my wording

Take the extreme - but ideal case of a Voltage source connected to a capacitor ( or dummy cap load ) . For one quarter of the cycle from Vs=V p to +Vpeak, current flows into the capacitor, and charges it, so power is flowing from the source to the capacitor, the capacitor is now charged, it has stored REAL energy. As the Vs falls from +Vp the capacitor discharges and current flow back into the source. When we get to Vs = 0 Vcap = 0 .. there is no energy in the CAP - the REAL Energy moved from the cap into the source. The average power over time is zero - but the instantaneous is not...
I totally agree with you.
Your description is accurate and detailed.

The following link is about ideal voltage source, just want to share the interesting property of it.
https://en.wikipedia.org/wiki/Voltage_source

"The internal resistance of an ideal voltage source is zero; it is able to supply or absorb any amount of current. The current through an ideal voltage source is completely determined by the external circuit. When connected to an open circuit, there is zero current and thus zero power. When connected to a load resistance, the current through the source approaches infinity as the load resistance approaches zero (a short circuit). Thus, an ideal voltage source can supply unlimited power.
No real voltage source is ideal; all have a non-zero effective internal resistance, and none can supply unlimited current. However, the internal resistance of a real voltage source is effectively modeled in linear circuit analysis by combining a non-zero resistance in series with an ideal voltage source (a Thévenin equivalent circuit)"

My personal thoughts is that if current flows into the positive voltage terminal and flows out from the negative voltage terminal at any time, no matter how short or how long, the power of the voltage source can be considered negative.

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#### Tom.G

Two approaches:

1)
Power (in Watts) = I2 x R
or Power = Current2 x Resistance
or Watts = Amps2 x Ohms
"The internal resistance of an ideal voltage source is zero...
Power = I2 x 0
Power = ?

Wikipedia strikes again!

2) Two AC generators with identical voltage and frequency are connected in parallel but with a phase difference. Which way does the instantaneous power flow.

• #### Kiran sahu

we found that a quarter-wavelength of transmission line turned a short circuit into an open circuit. Indeed, with an appropriate length (or driven at an appropriate frequency), the shorted line could have an inductive or a capacities reactant. In general, the impedance observed at the terminals of a transmission line has a more complicated dependence on the termination.

Typical microwave measurements are made with a length of transmission line between the observation point and the terminals of the device under study, whether that be an antenna or a transistor. In this section, the objective is a way of visualising the relation between the impedance at the "generator" terminals and the impedance of the "load." We will find that a representation of the variables in the reflection coefficient plane is valuable both conceptually and practically.

To Toms point in 1) Power = I2 x 0... this is the energy dissipated in the source, just as an ideal source would supply infinite current into a 0 Ohm load. The ideal source model does not account for energy or power. If we look at the circuit we have V and I --- it is an apparent paradox, but this is where the ideal model does not hold up.
If we are looking at reactive circuit elements in real applications - IMO - we should always consider they are storing energy, but we do not typically consider energy in circuit analysis.

As for 2) Without discussing real impedance - ... any V difference is undefined amount of current in one direction and then the other. Power flows one direction and then the other- Ideal model, this is irrelevant, in the real world very relevant.

#### alan123hk

An ideal source is a two-terminal device which can maintain a fixed voltage independent of the load resistance or the output current. It is often used as a mathematical abstraction that simplifies the analysis of real electric circuits. We can't or don't need to put a resistance inside because resistance implies converting electrical energy to thermal energy and causing change of voltage when the current flows, it then leads to a violation of the fixed voltage definition.

For an ideal voltage source, there is no definition of where internal energy comes from and where it goes. When using the ideal voltage source to simplify the analysis of real electric circuits, we don't need to know where the energy goes while it absorbing power, just like we don't ask where the energy comes from while it providing power.

The definition of electric power is : P = V*I
https://en.wikipedia.org/wiki/Electric_power

In the case of resistive (Ohmic, or linear) loads, Joule's law can be combined with Ohm's law (V = I·R) to produce alternative expressions for the amount of power that is dissipated : P=I^2*R

For a two-port network, if there is a voltage (V) across it, the current (I) flows in to its positive terminal and flows out from its negative terminal, then it absorbs power = V*I.

In real application, if the absorbed electric energy will be totally dissipated as heat in the source, then we can either assume it just dissipates in the ideal voltage source, or to use a whatever complex model to describe the energy conversion process in detail, including other possible forms of energy conversions likes chemical energy in rechargeable battery, mechanical energy..etc.
We can choose any method to analyze and design the system.

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#### Baluncore

This topic has become confused by the many off-topic examples that involve currents flowing in and out. Power is the rate of flow of energy, and yes, for beginners, power = voltage * current.

For transmission lines I see the propagation of energy to be the key consideration, but that energy does NOT flow in the wires of the transmission line as current. Energy propagates along the line through the insulation as an EM wave. The line conductors guide the electric field, while the magnetic field induces currents in the guiding conductors. Energy flow is the cross product of the electric and the magnetic fields, which is called the Poynting vector. Named after John Henry Poynting; https://en.wikipedia.org/wiki/Poynting_vector Now, the current in the wire is a proxy for the magnetic field, and the voltage between the wires is a proxy for the electric field. That is why for wire circuits, power = volts * amps.

Turning left twice sends you back the way you came. Likewise, when an EM wave encounters a mirror it is reflected. A mirror is a conductive sheet. The incident magnetic field induces a perpendicular current to flow in the surface of the sheet, which generates another perpendicular, hence equal and opposite magnetic field in the sheet. That cancels the wave into the sheet, so the energy must be reflected.

Notice that a mirror does not require a power supply to function.

Likewise, when you terminate a transmission line in an ideal voltage source, with zero impedance, you are providing a mirror that will reflect the incident line energy. Again, it does not need a power supply to function.

By placing a matching network between the ideal source and the line you can change the incident EM energy into heat, thereby eliminating the reflection. You can do the same to a vanity mirror by painting it black.

• iVenky and Klystron

#### iVenky

This topic has become confused by the many off-topic examples that involve currents flowing in and out. Power is the rate of flow of energy, and yes, for beginners, power = voltage * current.

For transmission lines I see the propagation of energy to be the key consideration, but that energy does NOT flow in the wires of the transmission line as current. Energy propagates along the line through the insulation as an EM wave. The line conductors guide the electric field, while the magnetic field induces currents in the guiding conductors. Energy flow is the cross product of the electric and the magnetic fields, which is called the Poynting vector. Named after John Henry Poynting; https://en.wikipedia.org/wiki/Poynting_vector Now, the current in the wire is a proxy for the magnetic field, and the voltage between the wires is a proxy for the electric field. That is why for wire circuits, power = volts * amps.

Turning left twice sends you back the way you came. Likewise, when an EM wave encounters a mirror it is reflected. A mirror is a conductive sheet. The incident magnetic field induces a perpendicular current to flow in the surface of the sheet, which generates another perpendicular, hence equal and opposite magnetic field in the sheet. That cancels the wave into the sheet, so the energy must be reflected.

Notice that a mirror does not require a power supply to function.

Likewise, when you terminate a transmission line in an ideal voltage source, with zero impedance, you are providing a mirror that will reflect the incident line energy. Again, it does not need a power supply to function.

By placing a matching network between the ideal source and the line you can change the incident EM energy into heat, thereby eliminating the reflection. You can do the same to a vanity mirror by painting it black.
That's a great explanation. When I look at this in terms of V and I, I also felt that the source gets some of its power back since the source sees a Z=50 (& not Z=60) looking into the transmission line (hence power is higher to begin with) and the reflected wave has an opposite direction of current that gets absorbed onto the source, but now when I think about in terms of E and H, the power is actually Poynting vector and if that's the case the EM field is getting dissipated as heat in the resistor. Is that what you are saying? For that, I may have to understand how this Poynting vector is calculated for a transmission line and an unmatched load I suppose...

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