Transmission lines Part II

Transmission lines Part II

We were left with this sentence:

For now I'll stop here, I want to give a chance to assimilate the concepts accurately, as soon as I integrate the topic of transmission lines, my intent is to arrive at last describe the waveguides and resonant cavities.

Now it starts!!

Transmission lines “long” e “cut”



In DC circuits and low frequency AC using coaxial cables to protect the weak signals from the degradation caused by “induced noise” produced by magnetic or electric stray fields neglecting the characteristic impedance. This is due to the relatively short time in which the reflections take place in line, compared to the period of the waveforms or pulses of significant signals in a circle. As we saw earlier, if a transmission line is connected to a DC voltage source, It will behave as a resistor of equal value to the characteristic impedance of the line only for the time it takes for the impulse to reach the opposite end and return as a reflected pulse. After that time (short 10 mS for the coaxial cable along 1 Km last example), source “sees” only the terminating impedance, whatever it is.

If the circuit is low frequency, so short delays introduced by a transmission line between when the AC source emits a voltage spike and when the source “sees” that peak loaded impedance termination are of little importance.

Although we know that the signal magnitudes along the length of the line are not equal at a given time due to the propagation of the signal to (quasi) light speed, the actual phase difference between the signals of start and end of line is negligible.

In these cases, we can say that the transmission lines in question are electrically court, because their propagation effects are much faster periods ducts signals.

Unlike, a line electrically long It is one in which the propagation time is a large fraction, or even a multiple of the signal period. A line “long” is generally regarded as one in which the waveform of the full source signal at least a quarter of a cycle (90of “rotation”) before the incident signal reaches the end of the line.

To put this into perspective, we have to express the distance traveled by a voltage or current signal along a transmission line in relation to its frequency. A signal with a frequency of 50 Hz completes a cycle in 20 ms. At the speed of light (300.000 Km / s), equivalent to a distance of 6000 Km that a voltage or current signal will propagate in that period. If the transmission line speed factor is less than 1, the propagation velocity will be less than 300.000 Km per second and the distance will be lower by the same factor. But even if we used the velocity factor of the coaxial cable from the last example (0.66), the distance is still 3960 Km ! Whatever distance we calculate for a given frequency is called length d’then signal.

A simple formula to calculate the wavelength is as follows:

The Greek letter “lambda” tiny (l) It represents the wavelength, in any length unit used in the figure of speed (If Km per second, then wavelength in Km, if meters per second, then wavelength in meters). The propagation speed is usually the speed of light when calculating the wavelength of the signal wave in the open air or in a vacuum, but it will be lower if the transmission line has a speed factor of less than 1.

Is a line “long” Long is considered at least 1/4 of length, to an AC power system 50 Hz, power lines must pass 990 Km long before the effects of the propagation time become significant. The cables connecting an audio amplifier to the speakers should be long over 4,95 km before the reflections of the line significantly affect an audio signal from 10 kHz!

Although I am convinced that some esoteric audio somewhere in the world will have pulled off this argument to extort money by offering some weird contraption that removes the effects of these spectral reflections!!!

When it comes to radio frequency systems, however, the length of the transmission line is far from trivial. Consider a radio signal 100 MHz: its wavelength is only 3 meters, even at full speed of propagation of light (300.000 Km / s). A transmission line that carries this signal should not be longer than about 75 cm to be considered “long!” With a cable velocity factor of 0,66, this critical length is reduced to 50 cm.

When an electric source is connected to a load through a transmission line “short”, the load impedance dominates the circuit. That means, when the line is short, its characteristic impedance is of little importance for the behavior of the circuit. We see it testing a coaxial cable with an ohmmeter: cable law “open” from the central conductor to the outer conductor if the cable end is left unfinished.

Even if the line acts as a resistance for a short period of time after which the instrument is connected (circa 50 Ω for a RG-58 cable / The), immediately he behaves like a simple “open circuit:” the final impedance of the line is open circuit.

Since that time the combined response of an ohm meter and the human being who uses greatly exceeds the propagation time of the round-trip up and down the cable and only record the terminating impedance (load). It is the extreme speed of the propagated signal that makes us unable to detect the impedance of the transient 50 Ω cable with an ohmmeter.

When a source is connected to a load through a transmission line “long”, the characteristic impedance of the line dominates the own load impedance in determining the behavior of the circuit.

In other words, an electrically line “long” It serves as the main component of the circuit, with its features which obscure the load. With a source connected to one end of the cable and a load to another, the current drawn from the source is primarily a function of the line and not the load. This is increasingly true the longer the transmission line.

Consider our hypothetical cable 50 Ω of infinite length, surely the ultimate example of a transmission line “long”: regardless of the type of load connected to one end of this line, source (connected at the other end) will only see 50 Ω impedance, since the endless length of the line prevents the signal from reaching the end never where the load is connected. In this scenario, the line impedance exclusively defines the behavior of the circuit, making the completely irrelevant load.

The most effective way to minimize the impact of the length of the transmission line on the behavior of the circuit is to match the characteristic impedance of the line load impedance. If the load impedance is equal to the impedance of the line, any signal source connected to the other end of the line “will see” exactly the same impedance and will have exactly the same amount of current drawn from it, regardless of the length of the line.

In this condition of perfect impedance matching, the line length affects only the amount of time delay from the starting signal to the source to report the arrival of the load. However, the perfect combination of line and load impedances is not always practical or possible.

standing waves and resonance

Every time that occurs a mismatch between the impedance of the transmission line and the load, there will be some reflections. If the incident signal is an AC waveform continues, these reflections will be summed with more incoming wavelength incident forms to produce stationary waveforms calls standing wave .

The following illustration shows how an incident signal in a triangle shape (I used this waveform for simplicity of design) is transformed into a reflection of mirror image to the achievement of the end of the line is not terminated. The transmission line in this illustrative sequence is shown as a single thick line, rather than a pair of wires, for reasons of simplicity. The incident wave is shown traveling from left to right, while the reflected wave traveling from right to left:

The incident wave reflects the end of the transmission line is not terminated.

If we add the two waveforms together, we discover that a third standing wave form is created along the length of the line

The sum of the incident and reflected waves is a standing wave.

This third wave in red represents the only voltage along the line, being the sum is representative of the incident and reflected voltage waves. Swings in instantaneous amplitude, but does not propagate along the length of the cable.

Standing waves are quite abundant in the real world. Consider a rope agitated from one end and linked to each other (only a means of hand movement cycle shown, moving verso ithe low):

Standing waves on a rope.

Both nodes (points of little or no vibration) and antinodes (maximum vibration points) I remain fixed along the length of the rope. The effect is more pronounced when the free end is shaken at the right frequency. The plucked strings exhibit the same behavior “stationary”, with “knots” maximum and minimum along their length vibration. The main difference between a plucked string and a shock cord is that the plucked string provides the natural frequency of vibration “correct” to maximize the effect of standing waves:

Standing waves on a plucked string.

Even the wind that blows through an open tube produces standing waves; this time, the waves are vibrations of air molecules (sound) inside the tube rather than vibration of a solid object. If the stationary wave ends in a knot (minimum width) or an antinode (maximum amplitude) It depends on the fact that the other end of the tube is open or closed:

Stationary waves scores in open tubes.

One end of the closed tube must be a wave node, while one end of the open tube must be an antinode. By analogy, the end anchored to a vibrating string must be a node, while the free end (is present) It must be an antinode.

Note how there is more of a wavelength suitable to produce stationary waves of vibrating air inside a tube which correspond exactly to the end points of the tube. This is true for all systems in standing waves: standing waves resonate with the system for any frequency (wavelength) related to node points / antinodes of the system. Another way of saying this is that there are more resonant frequencies for any system that supports standing waves.

All higher frequencies are integer multiples of the minimum frequency (fundamental) for the system. The sequential progression of harmonics from one to the next resonant frequency defines the frequencies of tone for the system:

Harmonics in open-ended tubes

Actual frequencies (measured in Hertz) for each of these harmonics or overtones depend on the physical length of the tube and the speed of wave propagation, which it is the speed of sound in air.

Since the transmission lines support standing waves and force these waves to possess nodes and antinodes according to the type of end termination impedance of the load, also they possess the resonance at certain frequencies from the physical length and the velocity of propagation. The resonance of the transmission line, however, is a bit 'more complex of string resonance or air in the pipes, because we have to consider both the voltage so that the current wave.

This complexity is made easier to understand through computer simulation. To start, examine a source, a transmission line and a load perfectly matched. All components have an impedance of 75 Oh:

Transmission line perfectly matched.

Usando LTspice (free) to simulate the circuit, We will specify the transmission line (
t1) with a characteristic impedance of 75 Oh (z0 = 75) and a propagation delay of 1 microsecond (tr = 1u). This is a convenient method of expressing the physical length of a transmission line: the amount of time it takes for a wave to propagate along its entire length. If you were a real cable 75 Oh, perhaps a kind of coaxial cable “RG-59B / The”, the type commonly used for television distribution cable, with a velocity factor of 0,66 It would be around long 198 meters. As 1 mS is the period of a signal to 1 MHz, I will choose to shift the frequency of the AC source from zero (quasi) 1MHz, to see how the system reacts when exposed to signals ranging from DC to 1 wavelength.

By performing this simulation and tracing the voltage drops in the circuit, we see that the source voltage on the graph diagram records a 1 volt constant, while every other point has one 0,5 volt constant:

In a system in which all impedances are perfectly balanced, there can be no standing waves, and then there are “peaks” O “valli” resonant in the Bode graph.

Now, We change the load impedance to 999 MΩ, to simulate an open transmission line. We should definitely see some reflections on the line now that the frequency is moved by 1 mHz a 1 MHz:

Line open transmission

Resonances on the open transmission line.

The voltages rise and the scanning interval are lowered at different frequencies along by 1 mHz a 1 Mhz. There are five points of interest along the horizontal axis analysis: 0 Hz, 250 kHz, 500 kHz, 750 kHz e 1 Mhz. We will test each with regard to the voltage and current at different points of the circuit.

A 0 Hz (in reality 1 mHz), the signal is practically DC, and the circuit behaves in much the same way as would a DC battery source 1 volt. There is no loop current

A 250 kHz, we see the zero voltage and the maximum current at the end of the transmission line, but still at full voltage at the end of the load:

f = 250 KHz:

entrance: V = 0 — I = 13,33 mA;

fine: V = 1 — I = 0.

You're probably wondering, how can it be?

How can we get a full tension to the line open while there is zero voltage input?

The answer lies in the paradox of the standing wave. With a source frequency of 250 kHz, the length of the line is exactly right for 1/4 wavelength to fit from one end. With the end of the open-circuit of the line load, there can be no current, but there will be tension. Therefore, the load terminal of an open circuit transmission line is a current node (zero) and a voltage antinode (maximum amplitude):

A 500 kHz, exactly half of a standing wave is based on the transmission line, and here we see another point analysis in which the current of the source drops to zero and the voltage source at the end of the transmission line rises again to the full voltage

Wave full of half transmission line open half wave.

A 750 kHz, The situation seems very similar to 250 kHz: zero voltage source end and maximum current. This is due to 3/4 suspended of a wave along the transmission line, with the result that the source “sees” a short circuit where it connects to the transmission line, even if the other end of the line is open circuit

When the supply rate reaches 1 MHz, on the transmission line is present permanent wave. At this point, the source-end of the line undergoes the same voltage amplitudes and the end of the load current: full voltage and current zero. Basically, the source “sees” an open circuit at the point where it connects to the transmission line.

Double standing waves on the transmission line open to full wave.

Similarly, a short-circuited transmission line generates standing waves, although the node and antinode of the assignments for the voltage and current are reversed. What follows is the SPICE simulation in which the shorting jumper is simulated by a load impedance of 1 MD

short-circuited transmission line

transmission line short-circuited in detail

Needless to repeat all the drawings invertedo the input with the output behavior.

In both of these examples of open-circuit line and line short-circuited, the reflection of energy is total: the 100% of the incident wave that reaches the end of the line is reflected back toward the source.

HE, however, the transmission line is terminated with impedances other than open or a short, reflections will be less intense, as will be the difference between the minimum and maximum values ​​of voltage and current along the line.

Suppose we need to end our sample line with a resistor 100 Ω instead with a resistor 75 Oh.

weak resonances of a transmission line mismatch

weak resonances Details

At all frequencies, the source voltage remains stable at 1 volt, as it should. Even the load voltage remains stable, but at a lower voltage: 0,5714 volt. However, both the line input voltage that the voltage drop across the impedance of 75 Ω of the source vary with frequency.

A 0 Hz : entrance: V = 0,5714, I = 5.715 mA
A 250 KHz: entrance: V = 0,4286, I = 7,619 mA
A 500 Khz: entrance: V = 0.5714, I = 5.715 mA
A 750 KHz: entrance: V = 0,4286, I = 7,619 mA
A 1 MHz: entrance: V = 0.5714, I = 5.715 mA

In odd harmonics of the fundamental frequency (250 kHz e 750 kHz) see different voltage levels at each end of the transmission line, because at those frequencies standing waves and terminating at one end in a knot and the other end in a antinodes.

Unlike the examples of open-circuit and short-circuited transmission lines, the maximum and minimum voltage levels along this transmission line do not reach the same extreme values ​​of voltage source 0% e 100%, but we still points “minimum” e “maximum” voltage. The same applies for the current

One way to express the severity of the standing waves is a relationship between the maximum amplitude (antinode) and the minimum amplitude (node), for voltage or current. When a line is terminated by an open or a short, this standing wave ratio , O SWR It is valued at infinity, because the minimum amplitude will be equal to zero, and any finite value divided by zero results in an infinite . In this example, with a line from 75 Ω terminated impedance 100 Oh, the SWR will be finished: 1.333, calculated by taking the maximum line voltage 250 kHz o 750 kHz (0,5714 volt) and dividing by the minimum line voltage ( 0,4286 volt).

The standing wave ratio can also be calculated by taking the terminating impedance of the line and the characteristic impedance of the line and by dividing the larger of the two values ​​for the smallest. In this example, the terminating impedance of 100 Ω divided by the characteristic impedance 75 Ω produces a quotient of exactly 1,33, in a way very similar to the previous calculation.

A perfectly terminated transmission line will have a SWR 1, since the voltage at any position along the length of the line will be the same, and likewise for the current. Again, This is usually considered ideal, not only because the reflected waves are not energy supplied to the load, but because of the high voltage and current values ​​created by the antinodes of the standing waves can overload the isolation of the transmission line (high voltage) and conductors (high current), respectively.

Furthermore, a transmission line with a high SWR tends to act as an antenna, radiating electromagnetic energy away from the line, rather than channeling all the load. This is usually undesirable, since the radiated energy could “mate” with neighboring conductors, producing signal interference.

An interesting note at this point is that the antennas which typically resemble open or short-circuited transmission lines are often designed to operate at high standing wave ratios, for the reason to maximize the transmission and reception signals..

This second article the term here, too thinly then the risk of confusing the reader.

I'm meditating on the third and final chapter with real examples of impedance transformer and continue immediately after the article with the waveguides and resonant cavities.

As you may have noticed I have tried to eliminate the mathematical treatises replacing them with drawings and practical examples.

With the LT spice simulator, while being limited in the functions, still manages to give raw data and interpreted in the right way, can be compared to that which would explicitly prohibitively expensive professional simulators.

A heartfelt thank you to the public Livio for its significant help with LT spice.


1 reply
  1. theremino
    theremino says:

    As the previous, this is also a very clear article. I read it but I reread tonight I want to better understand how they add up the waves in the various cases of different lengths.


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