Tuesday, November 25, 2008

Open Mic Night, Bob and Lee, Session 27

Because it is the Wednesday before Thanksgiving we will be doing an informal session this week. There is no prepared topic. Come with your questions, tips, stories, etc.

If you would prefer, you can add your question in the comments section and we will address it.

Wednesday, November 19, 2008

Electrical Resonance

November 19, 2008 Educational Radio Net, PSRG 26th session, Lee Bond N7KC

The impedance series is now history. During the course of that 13 week series we looked at several of the most fundamental ideas in the physics of electrical phenomenon and, hopefully, gained some practical knowledge of how these ideas link together to form a basis for our understanding of all things electrical. Let's exercise some of this earlier impedance series material and see how it can be applied to solve practical problems which are routinely encountered on the bench. The first study examined the potentiometer or "pot" and its behavior when used as a voltage divider. The second study examined how energy is moved from a source to a load and also considered the effect of a transmission line in this process. This third study will look at the phenomenon of resonance in both the mechanical and electrical worlds and extend the idea to antennas.

Lets consider mechanical systems first to get an intuitive feel for resonance.

We have all experienced autos which produce nasty sounds at certain speeds or engines where certain parts tend to vibrate depending on engine rpm. Suppose that we have an engine with some sort of attached bracket and the engine is at idle. If we very slowly advance the engine throttle to increase rpm’s there may be a engine rotational speed where the bracket starts to vibrate very strongly. If we continue to advance the throttle, the bracket vibration diminishes and disappears altogether. The mechanical configuration of the bracket has a "natural" frequency which is the frequency of vibration which develops when excited by the engines complicated sounds.

Another demonstration example is a wine goblet shattering when excited by acoustic energy which matches the natural frequency of the goblet. Finally, lets consider the pendulum in a clock. If the clock is unwound and the pendulum is activated we know that it will oscillate back and forth with diminishing amplitude until it stops. The pendulum has a natural frequency primarily determined by its length. If the clock is wound, however, there is a bit of clock mechanism which "taps" the pendulum very slightly at the correct moment to keep the pendulum swinging at constant amplitude, at the natural frequency, and for as long as the energy to produce the tap is present.

All of these examples show that very small forcing energies at the natural frequency of a mechanical system may cause dramatic vibration amplitudes due to resonance. Resonance frequency is where the forcing frequency matches the natural frequency of a system.

The situation in the electrical world is much the same as in the mechanical. Small signals (forcing energy) can appear dramatically larger due to electrical circuit resonance. Such circuits are always structured with resistance, inductance, and capacitive elements. The resistance element stands alone in being immune to the effects of forcing frequencies since the resistance converts energy directly to heat and stores nothing. In contrast to the resistance element, inductance and capacitive elements do not dissipate energy rather they store energy in the form of electric or magnetic fields during one portion of the cycle and return it to the circuit during the next.

Inductance associated with a forcing frequency creates a reactance product which increases with frequency whereas capacitance associated with a forcing frequency creates a reactance product which decreases with frequency. Therefore, given an assembly of resistance, inductive reactance, and capacitive reactance, there is a possibility that at some specific frequency the reactive components value will be equal and opposite hence cancel since they carry opposite sign. Electrical resonance generally indicates that net reactance is zero at a particular frequency. At this resonant frequency the circuit impedance is purely resistive.

One good example of electrical resonance is given by the tuning circuit in a typical radio receiver. The broadcast band, for instance, contains various amounts of energy from 550 Khz to 1500 Khz. The radio needs to respond to a specific station located in this continuum of signals. Using a parallel resonant circuit which is tunable allows one to slide across the band in search of the desired signal. Very slight amounts of received energy from the antenna will excite the resonant circuit and produce signal levels much higher than the excitation level. It is important to note that incoming signal energy is not increased by resonance rather signal amplitude is increased which is then amplified by a suitable active circuit.

Trapping circuits can be constructed from resistive, capacitive, and inductive elements as well. To facilitate this function the elements should be wired in series. At the resonant frequency the net reactance will be zero leaving only resistance as the circuit element. At frequencies off resonance the circuit impedance will always be larger than at resonance due to the combination of series resistance and predominate reactance.

Another example of impedance changing with frequency is the antenna. Lets consider a simple dipole cut to the center of any band. If you were to connect an antenna analyzer to the dipole and sweep from the lower to the upper band edge you would see the antennas feed point impedance, or combination of resistance and reactance, dip at the cut frequency and show only a resistive component. This is the radiation resistance of the antenna at resonance. Resonance frequency is that frequency where net reactance is zero.

Given that the antenna inductance and capacitance values are fixed, at frequencies above the resonance point the antenna is too long, inductive reactance increases, and the antenna impedance increases. Conversely, at frequencies below the resonance point the antenna is too short, capacitive reactance increases and the antenna impedance increases. Since maximum power transfer occurs when transmission line characteristic impedance matches the radiation resistance of the antenna, the trick is to adjust antenna elements such that, at the desired operating frequency, the net reactance is zero and maximum radio frequency current flows in the antenna elements. Since the dipole has a feed point impedance of about 72 ohms at resonance, driving it directly with 50 ohm coaxial line and a 1:1 balun would yield a VSWR of 72/50 or 1.44:1 minimum. Various matching schemes are available to adjust the feed point impedance to match the transmission line.

In certain circumstances electrical resonance can be a nuisance. For example, consider the guy wires associated with a tower installation. Wires similar in length to the radiating elements can seriously detract from a desired radiation pattern. A careful look may reveal that many compressive egg shell insulators may be used to break up the total length of the guys such that any single guy length cannot produce harmonically related radiation in concert with the actual antenna.

In summary, resonance can be a help or hindrance. Electrical resonance is a fundamental concept of electrical theory and, in practical terms, makes our radio endeavors possible.

This concludes the set up for the discussion of resonance. Are there any questions or comments?

This is N7KC for the Educational Radio Net

Tuesday, November 11, 2008

Log Periodic Dipole Antennas, Bob, Session 25

Tonight we will talk about another class of antennas, log periodic antennas. There are different forms of log periodic antennas but we will talk about the most common one, the Log Periodic Dipole Array (LPDA).

This antenna looks and acts similar to the Yagi but unlike the Yagi it covers a wide range of frequencies. This is the LPDA's defining characteristic. A typical design will cover a range of frequencies where the highest frequency is double the lowest. For example you could have one antenna that covered 14 MHz to 30 MHz with very good gain, front to back, and SWR figures over the entire range. You are not limited to the 2:1 frequency coverage. In fact you are only limited by the ability to physically construct the antenna and use it.

The log periodic antenna looks somewhat like a Yagi but, unlike the Yagi, the length of the parallel elements vary so that the tips form a straight line that gets progressively smaller. If you imagine lines running along the tips of both ends of the elements from the largest element to the smallest and extend the lines beyond the end of the antenna until they meet, they would form an angle with the boom as the bisector. The elements are connected in a criss-cross pattern so that, if you are looking down from the top of the antenna, the smallest element on the left side would be connected to the next larger element on the right side, and vice versa. This crisscrossing continues through all of the elements. The antenna is fed at the small end with a balanced signal.

I found a greatly simplified explanation of how this antenna works at radio-electronics.com. The link is at the bottom of the blog post. Let's say we are feeding our antenna with a signal about in the middle of the range. Because of the crisscross arrangement most of the adjacent elements cancel each other. But at the two elements in the middle of the array, which are closest to resonant length, you also have the width between them such that the wave will be 180 degrees out of phase when it reaches the other element. That combined with the crisscross feed causes the two elements to reinforce each other.

One other point, the smaller elements which don't contribute to the radiation, act like the shorter director elements of a Yagi, while the longer elements act like reflectors. This creates a radiation pattern much like a Yagi.

As you tune up and down the usable frequency range, you find that at the higher frequencies the radiation primarily comes from the smaller elements and at the lower frequencies, the larger elements are the ones that radiate.

Of course it's never quite this simple. Depending on design you may have many of the elements contributing to the radiation.

Because the imaginary line along the tips is straight, and the extended lines on each side form an angle, there are some relationships that have to hold. Hopefully it is obvious to all that if you go twice as far away from where the lines meet (the vertex) then the length of the line going across (the element length) will be twice as much. This leads to the formula that the ratio of the length of successive elements has to equal the ratio of the distance from the vertex. This ratio is given the Greek letter tau. This ratio defines the relative distance between elements. In our example of doubling the distance, tau would equal 0.5. To make an effective LPDA you want to have a tau that is as close to 1.0 as is feasible. You can see that tau of 1 would result in parallel lines which wouldn't work. To cover the range you want of double the initial frequency you need a change of length that is actually more than double. If tau is very close to 1 then you will need many elements and a very long boom to achieve that. These are the trade-offs to building a LPDA.

There are ways to add true parasitic elements to the LPDA to improve performance. This is beyond the scope of this discussion and can be found in the Antenna Book.

As usual, I want to point you to the ARRL Antenna Book for an excellent in-depth discussion of building real world LPDA's. Also, as a bonus, you get a LPDA Design program for the PC when you buy the Antenna Book.

Log Perodic Antennas on radio-electronics.com

Wednesday, November 5, 2008

Impedance matching 101

November 5, 2008 Educational Radio Net, PSRG 24th session, Lee Bond N7KC

The impedance series is now history. During the course of that 13 week series we looked at several of the most fundamental ideas in the physics of electrical phenomenon and, hopefully, gained some practical knowledge of how these ideas link together to form a basis for our understanding of all things electrical. Let's exercise some of this earlier impedance series material and see how it can be applied to solve practical problems which are routinely encountered on the bench. The first study examined the potentiometer or "pot" and its behavior when used as a voltage divider. This second study will firstly examine how energy is moved from a source to a load and secondly, consider the effect of a transmission line in this process.

First, we need to understand a very elementary concept in describing mathematical plots. Imagine that we are walking along a straight path which starts to curve uphill. We notice that the walking is getting tougher as the path curves upward. We might make the observation that this is a steep upward slope. As we continue our walk along the path, it levels out and immediately starts to slope downward and we must hold back to avoid running. We might make the observation that this is a steep downward slope. Looking back on our route we see that the high point on the walk was at the highest point on the hill and, further, that the slope was actually zero at that point. So it is with graphical plots. A maximum point (or minimum for that matter) on a graph always occurs at exactly zero slope. If you are skilled with your math then it is an easy matter to set the slope to zero and determine the conditions which will then cause the maximum or minimum on the plot.

It seems to be common knowledge that one must match the antenna impedance to the transmission line to transfer maximum energy per unit time (power) across the connection. What is not so widely known is that we can use a resistive voltage divider to demonstrate the idea directly and with ease.

Let’s set up a demonstration circuit to test the idea. We will set a powerful oscillator to a frequency to 10 Mhz and adjust the output voltage to 100 volts rms. Consider this to be a "perfect" voltage source with zero internal impedance. This means that our oscillator will stubbornly maintain the 100 vrms at its output without regard to load. Now, let’s convert this oscillator to a real world device by adding 50 ohms to the output. This is the equivalent to your radio transmitter which has a 50 ohm output.

Next, we have a large carbon resistor and we can change the resistance value from zero ohms to 200 ohms by merely turning a calibrated knob. Since we suspect that heating of the resistor might be an interesting thing to watch let’s attach a thermometer to the resistor to see how its temperature changes during the demonstration.

Finally, attach the carbon load resistor to the 50 ohm output of our demonstration oscillator to complete the circuit. Let’s also attach an RF voltmeter to the load resistor so that we can log some numbers during the demo process. (see spreadsheet data and plot at end of this article)

So, we are ready to start the test and take some data. To make our point and to keep this short we will just do 3 measurements so set the load resistor to 30 ohms and we notice that the voltage across the load resistor is 37.5 volts. We know from Joule’s Law that power is just voltage squared divided by the resistance so the power (energy per time) dissipated in the load is 46.88 watts or 46.88 joules per second. Checking the thermometer we see that it has moved upscale from room temperature to level 1.

Next, set the load resistor to 50 ohms and we notice that the voltage across the load resistor is 50 volts. Applying Joule’s Law once again we see the dissipated power to be 50 watts or 50 joules per second. The thermometer is now reading higher than level 1 from the first measurement.

Finally, set the load resistor to 80 ohms and we notice that the voltage across the load resistor is 61.5 volts. Applying Joule’s Law once again we see the dissipated power to be 47.34 watts or 47.34 joules per second. The thermometer is now reading very close to level 1 from the first measurement.

Taking a look at our data we see that the load resistor temperature was highest at 50 ohms and dropped off either side of 50. If we were to take multiple data points and plot them on graph paper, such that the vertical axis, the ordinate, represented power and the horizontal axis, the abscissa, represented values of load resistance then we would show a "hill" much like the hiking hill we traversed earlier. The maximum value would occur at zero slope (top of the hill), when the "source" resistance of the oscillator equaled the load resistance. By extension, resistance can be replaced with impedance and you will obtain exactly the same results.

We have taken the graphical approach here but mathematically this is a very clean problem. One simply writes the equation for power dissipated in the load in terms of the simple voltage division associated with the source and load impedance. Compute the slope, set it to zero, and notice that both source and load impedance must be equal to achieve zero slope.

Fine you say but what happens when I separate the source impedance and load impedance with a transmission line? Now things start to become very interesting. If, instead of RF energy, we had used DC then the transmission line could be a simple wire to complete the circuit. I deliberately used RF in the demonstration example to make a point about transmission lines. Although transmission lines are tagged with an impedance value they are not resistors. If you were to connect an ohmmeter across an open 50 ohm transmission line the meter would indicate an open circuit. If you connected the same meter across a 50 ohm resistor then the meter would read exactly 50 ohms. The solitary mission of a resistor is to convert electrical energy to heat. The mission of a transmission line is to transfer energy from point A to point B with minimum loss of energy. For example, the 50 ohm transmitter provides the energy, the 50 ohm transmission line directs the energy with intended minimum loss, and the 50 ohm load consumes the energy. So, what is going on here when the line is in play?

To answer this question we need to understand that transmission lines, coaxial cables for example, have distributed inductance and capacitance throughout. The, so called, characteristic impedance of a transmission line is defined by the square root of the ratio of the distributed inductance over the distributed capacitance of a tiny cross section of the line and is resistive. Our 50 ohm line simply scales the line current such that the ratio of line voltage to line current is 50 and no energy is lost in the process. However, there are two primary energy losses in a transmission line which we need to address. Skin effect currents flowing in the copper conductors and dielectric absorption losses in the insulating material between conductors produces heat loss. Both losses vary as a function of frequency.

There are three cases which we need to consider for the transmission line between the source and load in the demonstration example.

First, consider the infinitely long 50 ohm transmission line. Energy entering and moving down the line is constantly reduced by the skin effect losses and dielectric losses and, eventually, this energy is reduced to zero. Input energy is totally dispersed nearest the input end of the infinite length line. Our demo system behaves as if the line were a 50 ohm resistor and maximum energy is transferred from the source. No useful work has been done.

Secondly, consider a very much shorter 50 ohm line which is terminated with a 50 ohm resistor. This line is so short that skin effect losses, and others, are small so that almost the entire input energy is converted to heat in the load resistor. The 50 ohm termination matches the line so there is no impedance discontinuity and there is no reflected energy. The line and load are matched and maximum power is transferred by the line to the load. Maximum useful work has been done.

Thirdly, using the same short line as above we arrange for the load to be something other than a 50 ohm resistor. Perhaps a 60 ohm resistor is now the load. When the incident energy first encounters the 50 ohm line the characteristic impedance of the line scales the current appropriately for the 50 ohms. Then, at some later time, the traveling energy encounters the 60 ohm termination. Obviously there is now an impedance mismatch and some small fraction of the incident energy is reflected back toward the generating end. Given these circumstances, with the reflected energy in play, it is clear that maximum energy transfer can never be achieved. Less than maximum useful work has been done.

In summary, one can show either graphically or mathematically that maximum energy is transferred when the source impedance equals, or matches, the load impedance. Transmission lines do not dissipate energy as do resistors. The entire system must be matched to realize maximum energy transfer. Transmission line to load mismatches cause energy reflections which always reduce system throughput and degrade performance.

This concludes the set up for the discussion of impedance matching. Are there any questions or comments?

This image is a scan of spreadsheet data and associated plot for the demonstration circuit described above.

Double click the image to see a larger version.

This is N7KC for the Wednesday night Educational Radio Net.