for more technical articles.

**Links to individual sections below****GO TO**Introduction**GO TO**Integrating The Plate Current Pulse**GO TO**Fourier's Analysis of the Plate Current Pulse**GO TO**Making the Graphs**GO TO**Formulas Used in Class AB Power Amplifier Calculations**GO TO**An Example 30,000 watt RF Amplifier**GO TO**Another Example: Same Amplifier, Different Power Output**GO TO**Applying the Calculations to Amateur Radio**GO TO**Disclaimer

**This page is under construction!**

.

There are a few reasons one might want to know the operating parameters
of the commonly used class AB power amplifiers:

Adjustment of loading on the plate circuit for most efficient operation,

Adjustment of plate voltage for most efficient operation,

Estimate the linearity and accuracy of in-line wattmeters,

and calculate resonant load resistance to calculate tank circuit 'Q',

and judge alignment of double tuned plate tanks.

Calculate the harmonic content which must be suppressed by the tank circuit.

.

I am assuming the reader has some technical knowledge of RF Amplifiers. The class AB amplifier has a conduction angle of greater than 180 degrees but less than 360 degrees. In practice, the conduction angle of plate current pulses are usually between 185 degrees and 200 degrees at maximum power output. Assume stiff power supplies for, plate, screen grid, and bias.

The theory behind the calculations is fairly simple.

First, find the conduction angle of the plate current pulse. This is done
graphically because the integral for analyzing the plate current pulse can
not be worked backwards by any mathematical formulas. We can integrate
the plate current pulse for trial values of conduction angle and plot
the results on a graph. We know the idle current of the amplifier and
the average plate current during operation. The conduction angle is
read off the graph opposite the ratio of plate current to idle current.
The peak value of the plate current pulse can be calculated from the
conduction angle.

Secondly, use Fourier's integral to analyze the plate current pulse for fundamental frequency content, along with the lower harmonics. Formulas can be used to calculate Fourier coefficients, once conduction angle is known. Trial values of conduction angle can also be used here to produce a graph of the Fourier coefficients for the fundamental frequency and lower harmonics. The Fourier fundamental frequency coefficient for a given conduction angle is multiplied by the peak value of the plate current pulse to find the fundamental RF frequency current contained in that conduction angle. Once one has the fundamental RF current, the fundamental RF voltage and resonant load resistance can be calculated from the amplifier power output.

Integration to find the area under the curve can be simplified by using a
cosine wave, which is symmetrical about the zero 'Y' axis, and multiplying
the result by 2. That gives us the area under the curve and dividing by 2
&pi gives the area average for the full 360 degree cycle.

**Note:**
that &theta equals the conduction angle divided by 2, as we are using the
cosine wave. Also note that radian measure is used!

On the left,
I_{idle}, and I_{b}, which is the plate current with a
black picture (no setup). On the right, are the values in trigonometry.
The idle current 'a' equals ' -(cos &theta)'. Note that (cos &theta) is
negative and -(cos &theta) results in a positive value. The plate current
pulse is described by: y = (cos &theta + a) and the integral is
&int (cos &theta + a). This is a simple integration
from a book of integrals. Integration yields: A = (sin &theta + a&theta)
and dividing by 2&pi yields: A_{avg} = 1/&pi (sin &theta + a&theta)
for the complete 360 degree cycle.

Inspection of the plate current pulse gives us some proportions:

I_{b} / I_{idle} = A_{avg} / - cos&theta,

I_{b max} / I_{idle} = (1 - cos&theta) / - cos&theta, and

I_{b max} / I_{b} = (1 - cos&theta) / A_{avg}

'N' below is the number of the harmonic content contained in the cosine wave. N = 1, is the fundamental component and N = 2, is the second harmonic, etc.

a_{n} = 2/&pi&int f(&theta) cos N&theta
d&theta, and from the integral sign to the right

we have
&int (cos&theta = a) cos N&theta d&theta =

&int u dv = uv - &int v du

where u = cos (&theta + a), dv = cos N&theta, du = - sin &theta,

and v = &int cos N&theta = (sin N&theta) / N

After integration, the formula for the fundamental Fourier coefficient is:

a_{1} = 2/&pi [&theta/2 - (sin 2&theta)/4],

and the harmonic coefficients can be found by:

a_{n} = 2/&pi N [sin(1-N)&theta / 2(1-N) - sin(1-N)&theta / 2(1+N)]

The peak value of the fundamental RF component of the plate current pulse is
i_{1}, and

i_{1} = a_{1} B / (1 - cos &theta), where 'B' is the peak
value of the plate current pulse, I _{b max}, and I_{b
max} = I_{idle}(1 - cos&theta) / - cos&theta

Combining the above two formulas and simplifying yields:

i_{1} = a_{1} I_{idle} / - cos&theta

I

Using a computer program, one can evaluate the expressions:

A_{avg} = 1/&pi (sin &theta + a&theta),

I_{b} / I_{idle} = A_{avg} / - cos&theta,

a_{1} = 2/&pi [&theta/2 - (sin 2&theta)/4], and

i_{1} = a_{1} I_{idle} / - cos&theta

in increments of 0.05 degrees from 185 degrees to 270 degrees and print the results to a table in a text file. Remember to convert degrees into radians for these formulas.

It is not even necessary to make a graph of these two functions. Average
plate current is read from the transmitter plate current ammeter and the
ratio of I_{b} / I_{idle} can be read directly from the
table in the text file. For example, an average plate current of 3.80
amperes and an idle current of 0.8 amperes yields an I_{b} /
I_{idle} of 4.7500. The closest I_{b} / I_{idle}
in the chart excerpt below is 4.757275 in the third line and the conduction
angle is 188.60 degrees and A_{avg} is 0.356694.

Ib/Iidle 4.806988, CA 188.50, A-avg 0.356239, a1 0.547136, i1 7.382902

Ib/Iidle 4.781985, CA 188.55, A-avg 0.356466, a1 0.547412, i1 7.343512

Ib/Iidle 4.757275, CA 188.60, A-avg 0.356694, a1 0.547688, i1 7.304581

Ib/Iidle 4.732851, CA 188.65, A-avg 0.356922, a1 0.547964, i1 7.266100

Ib/Iidle 4.708709, CA 188.70, A-avg 0.357150, a1 0.548241, i1 7.228062

The column 'a1' is the Fourier coefficient for the conduction angle and the
column 'i1' is the fundamental current component contained in the plate
current pulse.

**Note:** The column 'i1' is for an I_{idle} of '1' or unity.
This 'i1' must be normalized to a different I_{idle} by multiplying
this number by the different I_{idle}.

Note that milliamperes works just as well for I_{idle} and
I_{b} but do not mix milliamperes and amperes. If you start out
with one unit of measure, use them for all calculations!

If one knows I_{b}, I_{idle}, and the conduction angle, the
other important parameters needed to start calculations are: plate voltage,
E_{b}; screen grid voltage, E_{g2}; and the average
amplifier output power, P_{o amp}.

**Note:** If anyone wants the complete table above, send an E-Mail to
the address at the end of this page and I will send the table as an
attachment to an E-Mail back to you.

**Formula one**: Power delivered by the plate circuit, P_{plate} = P
_{o amp} / N_{cir}, where P_{o amp} is the amplifier
output power delivered to a load, and N_{cir} is the efficiency
factor for the plate tank circuit.

**Formula two**: Peak voltage of plate circuit RF,
e_{1} = 2 P_{plate} / i_{1}, where
i_{1} is the fundamental RF current component of the plate current
pulse.

**Formula three**: Resonant load resistance, R_{L} =
e_{1} / i_{1}

**Formula four**: Plate voltage efficiency factor, N_{e} =
e_{1} / E_{b}, where E_{b} is the plate DC voltage.

**Formula five**: Conduction angle efficiency factor, N_{&theta} =
a_{1} / 2 A_{avg}, where a_{1} is the Fourier
coefficient from tables as is A_{avg}.

**Formula six**: Amplifier efficiency factor, N_{amp} =
N_{e} N_{&theta} N_{cir}

**Formula seven**: Also, amplifier efficiency factor, N_{amp} =
P_{o amp} / E_{b} I_{b}, where E_{b} and
I_{b} are the DC plate voltage and average current respectively.

**Formula eight**: Power, plate dissipation, P_{p} = (1 -
N_{p}) E_{b} I_{b}, where N_{p} =
plate circuit efficiency factor, N_{e} N_{&theta}.

**Formula nine**: 'Q' or 'quality factor' = R_{L} /
X_{c out}, where X_{c out} is the reactance of the output
capacitance of the tube, from the tube data sheet.

**Formula ten**: peak RF voltage at peak sinc, e_{pk sinc} =
e^{2} = 2 P_{plate pk sinc} R_{L}, where
P_{plate pk sinc} is the power delivered by the plate circuit at
peak sinc levels, or P_{plate} * 1.68.

**Formula eleven**: E_{b min} = E_{b} - e_{pk sinc}

**Formula twelve**: Voltage amplification factor, A_{v} =
e_{1} / e_{1 in}, where e_{1 in} is the drive
voltage input to the cathode.

**Formula thirteen**: Current amplification factor, A_{i} =
i_{1} / i_{1 in}, where i_{1 in} is the drive
current input to the cathode.

**Formula fourteen**: Cathode input resistance, R_{in} = e_{1
in} / i_{1}, - or - in the case of the grounded grid amplifier,
R_{in} = R_{L} / A_{v}.

**Formula fifteen**: Power gain, G_{p} = A_{v}^{
2} R_{in} / R_{L}, - or - A_{i}^{ 2}
R_{L} / R _{in}, - or - in the case of the grounded grid
amplifier, A_{v}, - or - R_{L} / R_{in}.

The example television transmitter is a General Electric TT-B510 and B530 final visual RF amplifier combination. The B530 amplifier runs a pair of 4CX15000 tetrode tubes. These tubes are feed 90 degrees out of phase through a hybrid splitter and the two amplifier outputs are combined by a hybrid combiner. This hybrid combiner sums the RF back in-phase for feeding the transmission line to a second hybrid combiner. This second hybrid combiner sums the outputs of the visual and aural final amplifiers together and splits the combined output into two components which are 90 degrees out of phase. One output feeds the transmission line to the antenna for the North-South direction, and the other output feeds the transmission line to the antenna for the East-West direction.

The 4CX15000 RF amplifiers are of the grounded grid configuration where
both the control and screen grids are grounded for RF and the tube is fed
via cathode. The plate is also grounded for RF, for those who are familiar
with TE mode aperture coupling where the electron stream within the tube is
coupled to the tank cavity through the ceramic section of the tube. The
standard grounded grid circuit model is used for calculations. The plate
tank is a double tuned circuit. The primary circuit is a cavity which
contains the amplifier tube, and the secondary circuit is a coupling link
to the primary, with series and shunt air variable tuning capacitors. The
bandwidth of the tank circuit is approximately 4.2 Mhz. The efficiency
factor, N_{cir} is estimated to be 0.9 for this tank circuit and
0.95 for the input matching circuit feeding the amplifier.

Typical operation:

Frequency of operation: channel 13, with visual carrier at 211.125 Mhz.

Total power output = 30,000 watts at peak sinc levels.

Plate voltage, E_{b} = 6000 volts DC.

Plate current, I_{b} = 3.80 amperes average at black picture for
each tube.

Plate current at idle, I_{idle} = 0.8 amperes.

Screen grid voltage, E_{g2} = 750 volts DC.

Screen grid current, I_{g2} = 100 milliamperes DC.

Each amplifier delivers 15,000 watts at peak sinc level.

The wattmeter measures average transmitter power with a black picture, and
the ratio of peak sinc to average level is 1.68 - so P_{o amp} = 8929
watts average with a black picture.

**Formula one**: Power delivered by the plate circuit, P_{plate}
= P_{o amp} / N_{cir}, where P_{o amp} is the
amplifier output power delivered to a load, and N_{cir} is the
efficiency factor for the plate tank circuit. P_{plate} = 8929 /
0.9 = 9921 watts.

**From the table excerpt above:**

I_{b} / I_{idle} = 4.7500 and

the conduction angle = 188.6 degrees and i_{1} = 7.304581 amperes
peak which is the value for an I_{idle} of 1 ampere. In this case,
7.304581 * 0.8 = 5.844 amps, which is the peak value of the fundamental RF
current component of the plate current pulse.

**Formula two**: Peak voltage of plate circuit RF, e_{1} = 2
P_{plate} / i_{1}, where i_{1} is the fundamental
RF current component of the plate current pulse. e_{1} = 2 * 9921
/ 5.844 = 3396 volts peak of fundamental RF component.

**Formula three**: Resonant load resistance, R_{L} =
e_{1} / i_{1} = 3396 / 5.844 = 581 ohms.

**Formula four**: Plate voltage efficiency factor, N_{e} =
e_{1} / E_{b}, where E_{b} is the plate DC voltage.
N_{e} = 3396 / 6000 = 0.5660

**Formula five**: Conduction angle efficiency factor, N_{&theta}
= a_{1} / 2 A_{avg}, from the table excerpt above, and

0.547688 / 2 * 0.356694 = N_{&theta} = 0.7677

**Formula six**: Amplifier efficiency factor, N_{amp} =
N_{e} N_{&theta} N_{cir}, and

0.5660 * 0.7677 * 0.9 = 39.11 percent.

**Formula seven**: Also, amplifier efficiency factor, N_{amp} =
P_{o amp} / E_{b} I_{b}, and

8929 / 6000 * 3.80 = 39.16 percent.

The results of Formula six and Formula seven are very close together in value. This is the acid test of the accuracy of the integration of the plate current pulse, the solution to Fourier's integral, and the calculations up to this point.

**Formula eight**: Power, plate dissipation, P_{p} = (1 -
N_{p}) E_{b} I_{b}, where N_{p} =
plate circuit efficiency factor, N_{e} N_{&theta}, and

(1 - 0.5660 * 0.7677) * 6000 * 3.80 = 12,893 watts, which is below the
maximum of 15,000 watts plate dissipation for the 4CX15000 tube.

**Formula nine**: 'Q' or 'quality factor' = R_{L} / X_{c
out}, where X_{c out} is the reactance of the output
capacitance of the tube, where 'c out' = 24.5 pf, from the tube data sheet
and 'Q' = 581 / 30.769 = 18.88.

The maximum allowable 'Q' is probably around 20. A higher operating 'Q'
results in higher circulating currents in the double tuned tank circuit
causing excessive heating and possible destruction, as witnessed in a few
cases.

**Formula ten**: peak RF voltage at peak sinc, e_{pk sinc} =
e^{2} = 2 P_{plate pk sinc} R_{L}, and
(2 * 9921 * 1.68 * 581)^{1/2} = 4401 volts at peak sinc level.

**Formula eleven**: E_{b min} = E_{b} - e_{pk
sinc}, and 6000 - 4401 = 1599 volts. This is the absolute value of
the negative excursion of the RF at peak sinc levels. If the RF voltage
swing becomes greater, this value approaches E_{g2} = 750 volts,
and I_{g2} increases. If I_{g2} increases substantially
above the 100 ma of typical operation, the amplifier becomes non-linear.
This condition may also indicate error in the alignment of the double tuned
tank circuit. A double humped swept frequency response, which has higher
amplitude at the visual carrier frequency, will have higher than normal
resonant load resistance, R_{L}.

It is the RF voltage at peak sinc modulation levels which limit the
amplifier efficiency at average power.

**Formula twelve**: Voltage amplification factor, A_{v} =
e_{1} / e_{1 in}, where e_{1 in} is the drive
voltage input to the cathode. From the tube data sheet, approximate drive
voltage is 345 volts at peak sinc, or 266 volts at average power, and

the voltage amplification factor is 3396 / 266 = 12.767

**Formula thirteen**: Current amplification factor, A_{i} =
i_{1} / i_{1 in}, where i_{1 in} is the drive
current input to the cathode.

The grounded grid amplifier has a current amplification factor of 1 (one).

**Formula fourteen**: Cathode input resistance, R_{in} =
e_{1 in} / i_{1}, or in the case of the grounded grid
amplifier, R_{in} = R_{L} / A_{v}.

266 / 5.844 = 45.57 ohms, - or - 581 / 12.767 = 45.51 ohms approximate cathode
input resistance. This is close to the nominal 50 ohm impedance of
transmission line and a simple low 'Q' broadband &pi or T network will
suffice for matching to the preceding driver amplifier.

**Formula fifteen**: Power gain, G_{p} = A_{v}^{
2} R_{in} / R_{L}, - or - A_{i}^{ 2}
R_{L} / R _{in}, - or - in the case of the grounded grid
amplifier, A_{v}, - or - R_{L} / R_{in}.

In the case of the grounded grid amplifier, G_{p} = A_{v},
or 12.767 approximately.

This example calculates the operating parameters for a different power output from the same amplifier.

From above, the average power delivered by the plate circuit, P_{plate}
is 9921 watts with a black picture at 100 percent power and, R_{L}
is 581 ohms.

What are the amplifier operating parameters at 50 percent power?

9921 * 0.5 = 4960.5 watts delivered by the plate circuit.
i_{1}^{2} = 2 P_{plate} / R_{L}, and
(2 * 4960.5 / 581)^{1/2} = i_{1} = 4.1323 amperes.

**Note**: this is at an I_{idle} of 0.8 amperes DC. Divide the
i_{1} by 0.8 amperes to get the i_{1} at an idle current of
1.0 amperes: 4.1323 / 0.8 = i_{1} = 5.1653

The 5.1653 figure is only for use with the computer generated tables.
i_{1} = 4.1323 amperes for 50 percent amplifier power output.

From the computer generated tables, the closest i_{1} is 5.156047
in the middle line below in the far right column. Also, conduction angle =
192.70 degrees, I_{b} / I_{idle} = 3.395605. Since
I_{idle} = 0.8, I_{b} = 2.716 amperes DC.

Ib/Iidle 3.418215, CA 192.60, A-avg 0.375095, a1 0.569719, i1 5.191800

Ib/Iidle 3.406865, CA 192.65, A-avg 0.375328, a1 0.569993, i1 5.173853

Ib/Iidle 3.395605, CA 192.70, A-avg 0.375560, a1 0.570267, i1 5.156047

Ib/Iidle 3.384435, CA 192.75, A-avg 0.375792, a1 0.570542, i1 5.138381

Ib/Iidle 3.373352, CA 192.80, A-avg 0.376024, a1 0.570816, i1 5.120854

**Formula ten**: peak RF voltage, e_{1} =
e^{2} = 2 P_{plate} R_{L}, and
(2 * 4960.5 * 581)^{1/2} = e_{1} =2401 volts.

**Formula four**: Plate voltage efficiency factor, N_{e} =
2401 / 6000 = N_{e} = 0.40012

**Formula five**: Conduction angle efficiency factor, N_{&theta}
= a_{1} / 2 A_{avg}, from the table excerpt above, and

0.570267 / 2 * 0.375560 = N_{&theta} = 0.75921

**Formula six**: Amplifier efficiency factor, N_{amp} =
N_{e} N_{&theta} N_{cir}, and

0.40012 * 0.75921 * 0.9 = 27.34 percent.

**Formula seven**: Also, amplifier efficiency factor, N_{amp} =
P_{o amp} / E_{b} I_{b}, where P_{o amp} =
P_{plate} N_{cir}, and

4960.5 * 0.9 / 6000 * 2.716 = 27.40 percent which compares well with the
27.34 percent above.

**Formula eight**: Power, plate dissipation, P_{p} = (1 -
N_{p}) E_{b} I_{b}, where N_{p} = plate
circuit efficiency factor, N_{e} N_{&theta}, and

(1 -
0.40012 * 0.75921) * 6000 * 2.716 = P_{p} = 11,346 watts plate
dissipation.

Note: At 100 percent power, P_{p} = 12,893 watts. Reducing the
output power by 50 percent only reduced the plate dissipation by 12
percent.

.

The engineer in charge of maintenance on the above television transmitter
told me that he was getting an efficiency of 55 percent. I asked if he
thought that might be a red flag as I knew that was not possible. He said
that class AB amplifiers were capable of that efficiency. It took some
convincing but I finally got him to borrow a different thru-line wattmeter
from one of our other transmitter sites. I heard later that when he tried
to do a transmitter power meter calibration with the borrowed wattmeter,
the screen grid current, I_{g2}, became very high at only 50
percent of licensed power and sinc pulses were very compressed and the
chroma levels were quite low, which were signs of a badly mistuned plate
tank circuit. The transmitter had only ran another day when the secondary
tuning sections of both of the plate tank circuits had burned up.

This transmitter had a built in sweep generator and diode detectors on the
amplifier output for use in alignment, which was done at zero power with no
DC plate voltage applied. The engineer had been doing the alignment like
on our other transmitters, by applying video sweep into the visual modulator
and connecting a spectrum analyzer to directional couplers in the
transmission lines at the amplifier outputs. The diode detectors are not
directional couplers, and the limiting action of the diodes made the
frequency response of the plate tank circuit look good even though it was
peaked up high over the visual carrier frequency. The resulting high
resonant load resistance, R_{L}, and much higher circuit 'Q',
caused the secondary sections of both tank circuits to overheat.

The example amplifiers above use the grounded grid configuration but the calculations involving the plate circuit are the same for the grounded cathode circuit as well. The difference in formulas is related to the amplifier input, and are included in the formula section above in Formulas Thirteen, Fourteen, and Fifteen.

The frequency of operation will have an effect on the estimated value of
the tuned circuit efficiency factor, N_{cir}, for the plate tank
circuit and the matching network to the amplifier inputs. The estimate of
N_{cir} = 0.9 for the plate tank circuit, and 0.95 for the input
matching network is for a frequency of around 200 Mhz. and comes from the
transmitter manufacturer. Higher operating frequencies will cause less
efficiencies and lower operating frequencies will cause higher
efficiencies. How much lower or higher? Use your best guess, unless
someone actually knows. In the HF range, the heating loss in those
circuits is probably only a few percent and the error in the estimate
should not cause significant error in the calculated results.

Inaccurate wattmeters are the most common source of error. An accurate power measurement is essential! A careful power measurement using a calorimeter is preferred.

Note that Formulas nine and twelve require values from the tube data sheet.

**Linear Amplifiers for Single Sideband**

The amplifier must be operated within its linear limits. If the amplifier
input is fed with RF modulated with steady state two tone audio, the plate
current ammeter will read the average plate current, I_{b}. Read
the ratio of I_{b} / I_{idle} in the tables (or graph) to
find conduction angle, A_{avg}, a_{1}, and i_{1}.
Multiply i_{1} by the I_{idle} current and proceed with
calculations starting with Formula one.

The ratio of the power at the peak of the modulation envelope to average
power is a factor of 2.0. When you get to Formula ten, substitute 2.0 in
place of the 1.68 used for video, to find e_{1} at the peak of the
modulation envelope.

**Linear Amplifiers for FM**

Again, the amplifier must be operated within its linear limits. There is
no amplitude modulation envelope, so average power output and peak power
output are the same! Read across from I_{b} / I_{idle} in
the tables, multiply i_{1} by the I_{idle} current and
proceed with calculations starting with Formula one. Skip Formula ten, and
use e_{1} in Formula eleven.

**Comments on the Calculations**

* You get to Formula eleven and find that E_{b min} is less than
E_{g2} and I_{g2} is not excessive, or in the case of a
triode tube or transistor, that E_{b min} is less than zero, and
there is no distortion on the amplifier output.

Wattmeters seem to err on the safe side. This is an indication that the
amplifier output is somewhat lower than the wattmeter indicates.
Calculations result in a resonant load resistance which is falsely too
high, and swing of e_{1} which is falsely too high.

* How accurate and how linear is your in-line or other wattmeter?

Take wattmeter, P_{o amp}, and amplifier meter readings,
I_{b}, at 25, 50, 75, and 100 percent power and do the calculations
for each as far as Formula three for resonant load resistance,
R_{L}. The resonant load resistance should calculate to nearly the
same in each case, allowing for a couple percent error for meter reading
error. If the resonant load resistance steadily increases or steadily
decreases, the wattmeter is non-linear for power, and any readings from it
are suspect.

* What is the harmonic power which must be suppressed by the tank circuit?

Calculate a_{n} = 2/&pi N [sin(1-N)&theta / 2(1-N) - sin(1-N)&theta
/ 2(1+N)], for your conduction angle at average power. 'N' is the number
of the harmonic: N=2 is second harmonic, N=3 is third, etc.

Plug the a_{n} into the formula: i_{n} = a_{n}
I_{idle} / - cos&theta, to find the amplitude of the harmonic
current, and the power is: i_{n}^{2} R_{L} / 2.

Remember, &theta is conduction angle / 2, and use radian measure in the
above formulas!

This article was written using notes I made while doing research back in the late 1980s and early 1990s. I did not keep track of the references. Some of the formulas came from a 1950s McGraw-Hill 'Radio Handbook' and some were derived by me.

I was on my own for the sections on integrating the plate current pulse, and the related proportions, and the solution to Fourier's integral. It was not treated in enough depth in college, and I could find almost nothing with enough depth in our local libraries, or in my own reference collection. It took me a while to get it right! I am sure others before me have done this. I was just not lucky enough to discover their work.

.

- Please send E-Mail to: mathison (aatt) sdf-eu.org
- .
- Last revision 2013-01-28