Table of Contents

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Observing Ocean Waves

Tying Mechanical Waves to Electrical Waves

An Electrical Power Plant

Simulation of a Transmission Line Using Finite Elements

Impedance, the Effect of Resistance and Reactance

The Quarter Wave Match

Polarization

Harmonics

The Mathematical Coincidence of the Chromatic Scale

Communication Bandwidth

Signal to Noise Ratio

Specifying a Wave




Observing Ocean Waves

    We stand at the shore of the Pacific Ocean. Large swells from some far away pacific storm smash against a long concrete wall. When the waves strike the wall the water is propelled into the air then falls back. As the water that was propelled into the air falls back a second wave is created traveling back in the direction from which our waves originated. Looking at the waves far from the wall we see that the high point of the swell is traveling towards the shore while the water in the trough is traveling away from the shore. Next we look at the area near the wall where waves are traveling in both directions. We note that there are certain distances from the wall every 1/2 wave length at which the height of the water does not change. Between these nodes the water is alternately twice as high or twice as low but there is no flow of current. Now lets say we have two large cranes. Between the cables of these cranes we have a section of concrete wall which we maneuver until it is parallel to the wall of the shore. If the wall held by our cranes is lowered until it is just above one of the nodes the waves will continue to pass without interference. However if we maneuver the wall slightly further from shore the waves will strike the wall and absorb some of the energy from the wave. If the wave length of the waves were 1/2 as long there would be twice as many nodes but our movable wall could still hang above a node without interference. Suppose we maneuver the hanging wall to a node further from shore and change the wave length slightly. A small change in wave length will cause the nodes far from shore to move shoreward or seaward a greater distance than the nodes near shore..

Tying Mechanical Waves to Electrical Waves

    Lets say we have a long coil spring or a piece of spiral electrical conduit. Jerking the end of the spring will cause a wave to travel down the spring. The compression of the spring will be like the voltage in an electrical circuit. The velocity of the spring will be like the amps in an electrical circuit. The speed that the wave travels down the spring will depend on the stiffness of the spring.

An Electrical Power Plant

    Our electrical power plant generates alternating current at 60 cycles per second. We will connect to the power plant a 10,000 volt transmission line that is the length of 1/4 wave length at 60 cycles. The electrical waves will travel through the transmission line slightly slower than the speed of light so we will make the transmission line 700 miles long. At the far end of the transmission line we will ( with the power off of course ) connect a big jumper wire that short circuits the transmission line. Turning on the power from the power plant we expect to see the wires from our power plant glow like the elements in a gigantic space heater. To our surprise the amp meter at the power plant reads almost zero and every thing is cool. What happened?  The wave of current from the power plant reached the short circuit in 1/240 second. Returning, the wave of current reaches the power plant after 1/120 second. The power plant is 1/2 way through its 1/60 second cycle and is now generating a current in the opposite direction of the returning current. The two currents meet causing a high voltage with no current. Since the voltage in the transmission line must be lower than the voltage in the power plant if current is to flow from the power plant, the power plant is unable to deliver any energy into the transmission line. Next we remove the jumper at the end of the transmission line. Reconnecting the power plant the portion of the transmission line near the power plant begins to glow orange and the power plant labors to produce enough current to maintain its specified voltage. Well there must be a happy medium here somewhere. Suppose we begin connecting strings of light bulbs to load the end of our transmission line. When we have 1250 100 watt light bulbs connected to the end of the transmission line we finally find the current to be equal at both ends and all along the transmission line. What is the electrical resistance of our 125,000 watt load if the voltage is 10,000 volts? Well, by definition, 'ohms law',  watts equal volts times amps and one volt across a one ohm resistance will allow only one amp of current to flow and produce one watt of heat. Divide 125,000 watts by 10,000 volts and we get 12.5 amps. So how many ohms of resistance are required to allow only 12.5 amps to flow if there are 10,000 volts? One ohm of resistance would allow 10,000 amps to flow.  10,000 ohms would allow one amp to flow. So 800 ohms of resistance allows only 12.5 amps to flow. We will now define 800 ohms as the 'characteristic impedance' of our transmission line. Through experimentation we will find that even if we extend our transmission line to any length or supply it with a different frequency or voltage so long as the line is 'terminated'  by the 800 ohm load no power will be reflected.

Simulation of a Transmission Line Using Finite Elements

    One could simulate a transmission line using inductors and capacitors. By taping iron rods to our transmission line and placing the two wires of our transmission line close together at places one could cause the waves to travel more slowly through the transmission line. The places where the iron rods are taped we will call inductors and the places where the wires are close together we will call capacitors. The next question is how do inductors and capacitors affect the flow of electrons in the transmission line? The inductor gives the electrons the electrical equivalent of mass and momentum. Just as a train or ship cannot be started or stopped quickly so the electrons in an inductor cannot be started or stopped quickly. The electrons do not actually have much mass but what is happening is that the movement of the electrically charged electrons causes a magnetic field and the magnetic field is stored in the iron bar. The flow of the electrons are retarded until the magnetic field in the iron bar has been built. When the supply voltage which has been forcing the electrons to move is disconnected the electrons cannot stop flowing through the inductor until the magnetic field has been dissipated. The voltage where the supply was connected will not just go to zero but will continue to decrease until there is no more magnetic field to push the electrons along. The capacitor acts like a reservoir. Just as a reservoir of larger capacity requires a larger number of acre feet of water to make the level of the water rise one foot so a capacitor of greater capacitance requires a larger number of electrons to cause a change of one volt. We know that electrons are negative and are attracted to places where there are no electrons, positive. If we place two wires or plates of opposite polarity very close together with an insulator or 'dielectric' between the plates to block the flow there will be a strong electric field between the plates causing the electrons to pool at the plate.

Impedance, the Effect of Resistance and Reactance

    To fully describe an electrical load we must take into account both the resistive elements or heat generating elements of the load and the reactive elements of the load. The reactance of a circuit results from its inductance and capacitance. Reactance is measured in mhos ( or ohms of reactance ). When a sine wave voltage is applied to an inductive circuit there is a 90 degree lag between the application of the peak voltage and the time of occurrence of the peak current. When a sine wave voltage is applied to a capacitive circuit the peak current occurs 90 degrees before the peak voltage is reached. If we apply an alternating voltage to an inductor and a resistor in series the current will decrease as the frequency or cycles per second increases because the magnetic field of  the inductor cannot be reversed quickly. If we apply an alternating voltage to a capacitor and a resistor in series the current will increase as the frequency or cycles per second increases. If we apply a continuous or direct current (DC) voltage across the plates of a capacitor current will only flow in the circuit until the capacitor has been charged so for current to flow continuously through a capacitor there must be a constant change of voltage.

Now we want to use the frequency, the inductance and the
capacitance to find the reactance.



Since there is 180 degrees of phase difference between inductive reactance and capacitive reactance one must be subtracted from the other.

By convention reactance = inductive_reactance - capacitive_reactance

Now we want to use the resistance and the reactance to find the impedance of our circuit.

The phase angle of the reactance is 90 degrees from the phase angle of the resistance so to find impedance we must use the Pythagorean theorem.

impedance for series circuits = square_root ( ohm * ohm + mho * mho )

For a more complete explanation refer to 'The Radio Amateur's
Handbook' published by the American Radio Relay League,
Newington, Conn 06111



The Quarter Wave Match

    We have a camera lens that we would like to transmit a maximum amount of light to a film. Is there any way that we can reduce the reflection at the lens? The refractive index of glass is about 1.5 meaning that light travels only .66 the speed of light, c, while traveling through glass. The refractive index of magnesium fluoride is about 1.3. We will coat 1/4 wave of magnesium fluoride onto the lens. The ratio of the refractive index of air which is 1 to the refractive index of magnesium fluoride is about the same as the ratio of the refractive index of magnesium fluoride to the refractive index of glass. Therefore the reflection at the air > MgF  'discontinuity' is about the same as the reflection at the MgF > glass discontinuity. The reflected light from the MgF > glass discontinuity will combine with the light at the air > MgF discontinuity to increase the 'admittance' of the light.

Polarization

    Waves on a pond exists in one plane on a surface. Radio waves exist in three dimensions. Observing the sine wave display of this program the waves have amplitude in the y dimension and travel to the right along the z dimension. If one could imagine a sine wave having amplitude into and out of the display screen and traveling to the right as does the wave on the screen the wave so described would have the opposite polarization. The usual way to demonstrate the polarization of light is to rotate two polarized dark glasses lenses with one lens in front of the other. Polarized light from the first lens will be decreased by the second lens if the angle of rotation of the second lens does not make the axis of polarization in the second lens parallel to the axis of polarization of the first lens. Reflected light from the diagonal in our telescope favors the light of one polarization.

Harmonics

    If one were to speak of a single pure tone one would be speaking of a sine wave of sound at one particular frequency. Any wave shape may be generated by combining sine waves of different frequencies, amplitudes and phase offsets. Two sine waves with frequencies of 1 and 2 or of 1 and 3/2 are said to be harmonics. Putting a pure sine wave through an amplifier which distorts the wave shape of the sine wave will produce a family of harmonics all of which are mathematically related to the 'fundamental' or original sine wave. In birds, wolves and people if two sounds are non harmonic the resulting sound will be discordant and associated with acrimony. Two sound waves related by small rational fractional ratios will be associated with harmony. If people in agreeable surroundings hum together very soon the wave lengths of their hums will become harmonic leading to the concept of harmonic convergence.

The Mathematical Coincidence of the Chromatic Scale

    Early musicians were faced with a daunting task. They knew that for their music to be harmonious the strings on their instruments must have certain mathematical ratios. With the Indian Sitar, a stringed instrument, the solution was to make the frets that stop the strings moveable and to divide the octave into 32 steps. For the Violin family the solution was to have no frets and leave it to the musician to stop the strings properly purely by artistry. For the Harp and Piano and related instruments a more general solution was needed. One octave is a ratio of one to two in frequency. It was necessary for the keyboard to go along for many octaves with the ratio of the frequency of one key to the next being the same and yet still be able to have the harmonic ratios. It was found that by using the twelfth root of two as the ratio between the notes a usable scale could be made. The new scale was known as the Chromatic Scale. To display the ratios of the chromatic scale and find the harmonic notes select Harmonic Ratio Finder from the sine wave menu in the tel.exe DOS program. The selection will also allow you to search for logarithmic scales other than the Chromatic Scale. One alternate scale that fits well is 22 semi tones per octave.

Communication Bandwidth

    For people trying to transmit information from one place to another place harmonics are less important. The objective is to transmit the most possible number of bits per second. One bit per second is like one yes or no answer per second. For those of you who have not played the game 20 questions one bit per second may not sound like much information. In the game twenty questions one person thinks of a noun or of an object. The questioners take turns asking questions which may be answered by yes or no. The first question should attempt to eliminate half the objects in the known universe. The first question might be "Is the object near the Earth or the Moon?". The second question should attempt to eliminate half of all the possibilities that were not eliminated by the first question. The second question might be "Is the object an animal?". The game progresses and if the questions are chosen properly the questioners will have the name of the object before 20 questions have been asked. To transmit more bits per second a greater band width is needed. A telephone is capable of transmitting frequencies from 60 to 5000 cycles per second. It is not possible to transmit a television signal through a telephone line unless the pixels of the picture frames of the television signal are transmitted slowly enough to fit the band width of the telephone line or some way is found to increase the bandwidth of the telephone line. A broadcast television station has a bandwidth of 6,000,000 cycles per second. It would be possible to transmit 1200 telephone calls from a television broadcast station at the same time if the signals were formatted properly. The possible colors of each pixel of our telescope image give each pixel a bandwidth of 300,000,000,000,000 cycles per second. Who could comprehend the amount of information which might be transmitted through our telescope?

Signal to Noise Ratio

    If we desire to hear someone talking a mile away perhaps we could connect several amplifiers together to hear their voice. There are at least two problems with this. First there might be other people talking within the mile causing the voices to be jumbled. Second, if there is any noise in the first amplifier in the string of amplifiers perhaps caused by 60 cycle power or perhaps even due to the flow of electrons, the second amplifier will amplify this noise just as it would amplify the useful signal. Suppose we have a radio receiver in an airplane that is designed to receive coded navigation signals. If the noise in the navigation signal is too great the safety of the people in the aircraft might be jeopardized by an error in navigation. It is necessary therefore to specify the ratio of the signal to the noise to be at least four to one or some other measurable ratio. To determine the signal to noise ratio we must first measure the noise with no signal present then add a measured amount of signal and compare the ratio. The known signal strength is varied until the four to one ratio is achieved. The required signal strength to achieve the four to one 'signal plus noise' to 'noise' ratio is called the 'receiver sensitivity'.
There are many ways to increase signal to noise ratio. We might find that there are some frequency bands which have less noise than other frequency bands. We might build the first amplifier in our string from special components or cool the amplifier in liquid helium to decrease thermal noise. Perhaps all the information that we are interested in is contained in the narrow hydrogen alpha red spectral line of ionized hydrogen. By filtering away the light of all the other colors the noise energy can be reduced dramatically For those of you who want to know more about interference filters refer to 'Amateur Telescope Making Book Three' page 376. 'Building a Birefringence Polarizing Monochromator for Solar Prominence' by Paul. Scientific American Books 1956.  There is also a Yahoo group for the subject.


Specifying a Wave

    To review, here are the things we must know to describe a single frequency wave:
wave length or frequency
amplitude
reactance
polarization
phase angle

To describe a mixture of waves we must know:
The information given above for each of the
wave and be able to describe a wave which is
formed by the sum of the reactive components
of all the waves.

In addition we might want to know the
method used to encode information on
the wave for instance AM, SSB, FM, TV
or some other coding and we probably
also need to know the types of noise and
their frequency bands.

 

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