PHILCOTECHREP DIVISION

lume 8 May -June, 1957 No. 3

X4ite....

'editorial 1

.etters to the Editors 2

Simple Lin -Log Curve Check T/Sgt. Lee R. Bishop 3

'What's Your Answer?" 4

leight Finder Alignment Bud M. Compton 5

Determination of the Internal Resistance of a Meter 6 8

Ferrites for Waveguides 9

Design Criteria for Mutual Inductance Transducers 16

volution to March -April "What's Your Answer?" 19

The Polar Impedance Chart and Its Use 20

Radio Location of Underground Cablesand Pipes George W. Spooner 27

quality Control of Vacuum Tubes Gilbert and Rosen 30

Tech Info Mail Bag 32

Technical Sketch of George Boole 33

PHILCO

TECHREP DIVISION

BULLETINPublished bimonthly by

The TechRep Division of Philco CorporationPhiladelphia, Pennsylvania

&Mae John E. Remich

afanagaty Ladilde Francis R. Sherman

&dimrid &-/i/04.4 Robert D. Hunter

Harvey W. Mertz

&liteveialeAre:

Technical Information SectionPhilco TechRep Division22nd St. and Lehigh AvenuePhiladelphia 32, Penna.

If any information contained herein conflicts with a technical order,manual, or other official publication of the U.S. Armed Forces, theinformation in the official publication should be used. Views expressedin this publication are the opinions of the authors and do not necessarilyrepresent those of Philco Corporation. All technical information andcircuits are presented without responsibility as to patent infringement.

Copyright 1957 Philco Corporation

-411110111Cif.. . .

by John E Remich

Manager, Technical Department

EXICON-The New X -Ray Technique

Almost every newspaper and magazine contains references to

the glowing achievements in present-day electronics. New weapons

systems for aircraft, guidance systems for missiles, giant com-puters, Texas Towers, the DEW line-all these are well-knownand much talked about. It is unfortunate that the field of medical

electronics does not receive such attention, for advances in thisspecialized phase of electronics may prove to have far moresignificance in the course of history.

One of the recent significant advances in this field is EXICON-short for Expanded Image Contrastor-developed jointly by Philco

and the Albert Einstein Medical Center. EXICON promises tohave far-reaching effects in the field of radiology. All of you arefamiliar with X-ray photographs, and are aware of their value inthe diagnosis of disease. Although such X-ray pictures are invalu-able to the physician, much of the story is often not apparentbecause of the limitations of human vision. It has been shownthat the average human eye is incapable of detecting a contrastvariation of less than two percent in gray tones. It is, however,precisely these minute variations which may determine the finaldiagnosis, and EXICON makes it possible to see such variations.

Basically, EXICON vastly enhances the contrast of the familiargray X-ray photograph by converting such a photograph intoelectrical signals, which may then be electronically amplified andreconverted to light variations. To further enhance the contrast,the resultant picture is presented in color, with the colors of thespectrum representing different tones of gray. Thus the importantminute variations may be easily seen and a more accuratediagnosis made.

Only the future can determine the true value of this technique,but present indications hold forth the hope that another milestonein medical progress has been passed.

i

LETTERS TO THE EDITORS

"We are experiencing considerable difficulty insecuring parts information on the AN/TRC-24 RadioSet. All technical manuals available omit any listingof Air Force stock number, Signal Corps stock num-ber, manufacturer's part number, of any other infor-mation which might be used in re -ordering parts. Iwould appreciate any information that you cansupply."

Leo J. PinzPhilco TechRep Field Engineer

(You are not alone in wishing to obtain parts in-formation on the AN/TRC-24. All available infor-mation-which is far from complete-is containedin the SIG 7 and 8 manuals. The manuals availableare:

SIG 7 and 8 AN/TRC-24SIG 7 and 8 OA-483/TRCSIG 7 and 8 MK-123/TRCSIG 7 and 8 MK-124/TRC

In these manuals there is some type of stock num-ber for almost every item. Ed.)

"I am interested in obtaining technical data on anoise diode type L-1612 for which the plate current is0.5 ma. This is to be used in conjunction with thedevelopment of a noise generator, to aid in noise fig-ure measurement."

Bennett O'BannonPhilco Team Leader

(The Lansdale Tube Co., Division of Philco Corpo-ration has provided the information that no othercompany, to their knowledge, uses the "L" desig-

nation for tubes, and that no tube with this numberis manufactured by them. However, they do manu-facture a noise diode type L -1262A. This tube alsohas a plate current rating of 0.5 ma and is availablefrom the Lansdale Tube Co. Ed.)

"It is planned to add in the near future courses ontransistors, printed circuits, and single-sideband sys-tems to those subjects currently being given. Any infor-mation pertaining to these subjects would be mostwelcome."

F. C. FarrarPhilco TechRep Field Engineer

(There are a number of excellent texts on transistorsavailable. One of these, Transistors Handbook, byBevitt, is currently being used by the EducationalExtension Program of the TechRep Division. Onthe subject of printed circuits, the three pamphletslisted below are available from the Superintendentof Documents, Government Printing Office, Wash-ington 25, D. C.

"Printed Circuit Techniques"(Catalog No. C13.4:468)

"New Advances in Printed Circuits"(Catalog No. C13.10:192)

"Printed Circuit Techniques, Adhesive TapeResistor System" (Catalog No. C13.4:530)

With regard to single-sideband systems, the Decem-ber, 1956, issue of the Proceedings of the Institute ofRadio Engineers was devoted entirely to SSB. Man-uals on SSB are also available from the publisher ofCQ magazine and from the ARRL. Ed.)

2

A SIMPLE LIN-LOG CURVE CHECKby T/Sgt. Lee R. Bishop

This article describes a simplified test setup for checkingthe characteristic of a lin-log MTI receiver.

AT TIMES IN THE COURSE OF MTItrouble shooting, a case of poor cancel-lation may make a check of the receiverlin-log curve desirable. Since the con-ventional method of checking the curverequires the setting up of special testequipment, such checks are time con-suming and have often been avoided infavor of component substitution. Withthe method described below, the curvemay be easily and rapidly checked withthe same equipment that is used tomake the cancellation measurements.

The importance of a lin-log curvecheck in locating a possible source ofpoor cancellation cannot be overempha-sized. This is particularly true in thecase of a receiver which uses crystaldiodes to supply the lin-log character-istic, because these crystals are extremelyheat sensitive and therefore easily dam-aged.

As previously stated, the followingcheck is made with equipment alreadyon hand. No special connections, net-works, etc., are required. The onlymaterials needed are a synchroscope, apiece of ordinary graph paper, and theMTI Evaluator (UPM-41), which isgenerally near the test bench. Althoughthe check procedure was developed spe-cifically for the CPN-18 receiver, theprinciple may be applied to any receiverof the same type.

To perform the check, proceed asfollows:

1. Calibrate the MTI Evaluator usingthe procedure in the Technical Or-der for the instrument. Trigger theMTI Evaluator with the synchro-scope trigger.

2. Calibrate the synchroscope so thatfull-scale deflection is produced bya signal of 10 volts (r.m.s.).

3. Connect the TARGET OUT jackof the MTI Evaluator to the inputjack of the MTI i-f strip.

4. Remove the first limiter tube fol-lowing the detector (V8 in thecase of the CPN-18). Removal ofthe tube is especially importantbecause it prevents the limiting ac-tion from affecting the accuracy ofthe readings. Connect the VER-TICAL INPUT jack of the syn-chroscope to the first grid pin ofthe limiter tube socket.

5. On the graph paper, let the "X"axis represent decibels with a scaleof 1 db per division, and let the"Y" axis represent voltage with ascale of 5 divisions per volt. It hasbeen found experimentally thatthese proportions offer the bestcomprpmise between reasonablegraph size and optimum curve pre-sentation. The VOLTAGE scaleshould run from 0 to 10 volts. TheDECIBEL scale should start at- 65 db and continue to 0 db. The- 65 db level is in the region ofminimum discernible signal (MDS)for the average receiver, and wellwithin the linear portion of anycurve. Thus it serves as a goodstarting point.

6. Turn the i-f gain control of the re-ceiver under test fully clockwise.Turn the coho off. Grass shouldappear on the synchroscope. Set theMTI Evaluator for the operatepulse condition, and adjust theFIXED TARGET RANGE controlto position the target at a convenientposition on the synchroscope. Donot use the moving target for thischeck.

3

7. Starting with the FIXED TARGETattenuator at -65 db, plot the volt-age amplitude of the fixed targetpulse, as read on the synchroscope,in the correct spot on the graph. Avoltage reading should be taken ateach decibel setting from -65 to 0db.NOTE: Use only the largestswing of the pulse as measuredwith respect to the baseline(positive for the CPN-18). Donot measure peak -to -peak am-plitude.

Theoretically, when a lin-log functionis plotted on a linear graph, the linearpart of the receiver curve should becurved and the log part should bestraight. In practice, however, the curvewill be found to deviate somewhat fromthe theoretical shape since the crystalsonly approximate the desired lin-logcharacteristic. A graph of a typical re-sponse for a lin-log receiver is shown infigure 1.

tu -CO En

Fa -I_I0>>

I0-

-65 -50 -25

DECIBELS

0

Figure 1. Typical Characteristic ofLin -Log Receiver

This check should be made whenevera new set of crystals has been installedor after the strip has satisfactorily passeda subclutter measurement. The graphobtained may then be used for referencein future performance checks and troubleshooting.

- =I 1E11 1E1 NI

"What's Your Answer?"

The problem for this issue was submitted by Art Davidson, PhilcoTechRep Field Engineer assigned as a staff engineer at NAESU.

Given: Twelve batteries. Eleven are 1.5 -volt batteries. The remain-ing battery is higher or lower in voltage than 1.5 volts bya significant amount.

Problem: Using series or parallel battery hookups and a meter, findwhich battery has the odd voltage, and determine whetherthis voltage is higher or lower than 1.5 volts. Only threevoltage measurements may be made with the meter.

4

HEIGHT FINDER ALIGNMENT

by Bud M. ComptonPhilco Site Engineer

This article describes a few of the errors most oftenencountered in height -finding radar equipments, thereasons for these errors, and methods of minimizingthem. Reference material for detailed theoretical studyof these errors is included at the end of the article.

AS THE ART of radar height findingbecomes more sophisticated, and im-provement in equipment increases theattainable accuracy, it becomes worth-while to reconsider the factors whichaffect accuracy. The factors which aremost often encountered include thefollowing:

1. Mechanicala. Antenna levelb. Gyro level (if stabilized)c. Data potentiometers, synchros,

etc., for height and angle infor-mation relative to antenna elec-trical axis

2. Electricala. Power supply voltageb. Linearity of vertical and hori-

zontal sweepsc. Angle marks, range marks,

height lines, etc.d. Gyro calibration

3. Physicala. Earth's curvatureb. Diffractionc. Refractiond. Ground reflection'

4. Equipment design limitationsThis formidable list makes it appar-

ent that academic study of these factorswould become extremely cumbersome.Limitations are therefore arbitrarily im-posed to the degree which is justifiedfor field applications. Accordingly, thereferences at the conclusion of this ar-ticle are given for extended theoreticalresearch.

MECHANICAL FACTORSMechanical adjustments must be

checked periodically because of antennatower settling, wear of gears and shafts,and possible cold flow of metal understress. An antenna out of level by aslittle as 1/10 of a degree introduces aheight error of over 1000 feet at a rangeof 100 nautical miles. Furthermore, thistype of height error is not constantthroughout 360 degrees of azimuth.The variation of error is usually sin-usoidal, thus complicating the problem.With gyro -stabilized equipment there isthe added possibility that the gyro maybe out of level. Antenna level is readilychecked by ase of a spirit level, an itemwhich forms an integral part of manyequipments. If the bubble remains cen-tered during rotation of the antennathrough a full 360°, then the antenna islevel. In the case of gyro -stabilizedequipment, the antenna platform mustfirst be leveled, and for airborne radarsthe aircraft must also be leveled to itsaverage flight attitude.

ELECTRICAL FACTORSElectrical factors are self-evident and

may normally be checked by routineprocedures. As an example, sweep line-arity may be checked on most equip-ments by using the presentation on theRHI. For a check on horizontal linearityof the sweep, it may be possible tocompare the distance between rangemarks at short ranges with the distanceat long ranges by switching betweenranges and making use of scope per -

5

sistence to note superposition. Verticallinearity is usually adjusted by use ofthe height mark. Here the height markcounter can be used to compare the ver-tical distance along, say, the 5 -degreeangle mark. Linearity exists when thisvertical distance is the same from 10 to50 miles as from 50 to 90 miles, orsome other convenient range dependingon the radar being calibrated. Errorsarising from incorrect electrical align-ment can be easily detected by utilizinga list of predetermined check pointswhich gives the proper intersectionpoints of height line with range andangle marks. Naturally, such a checklist is prepared consistent with inherentdesign limitations of the particular radar.

PHYSICAL FACTORSPhysical factors are the most elusive

of all the items affecting field personnel.Diffraction is discussed first, since it isleast amenable to compensation. This isthe term applied to the phenomenon ofthe bending of radio waves as they grazeobstacles. Television and communica-

tions coverage is often increased becauseof this phenomenon, but it always re-sults in errors for height finders. Dif-fraction can be calculated when all thevariables are known. In practice, how-ever, the terrain surrounding a radar isfar too complex to permit numericalevaluation. When a target is picked upunder diffractive conditions, its trueheight will be less than that indicatedby the height finder.

Refraction would cause little troubleif it could be measured readily in the airspace of radar coverage. It is a functionof temperature, moisture, and pressuredistribution in the atmosphere, and istherefore a variable quantity. The varia-tions consist of somewhat typical di-urnal and seasonal changes, togetherwith extreme fluctuations during ab-normal weather. These characteristics ofrefraction are variable with geographicallocation. If height errors cannot be at-tributed to other causes, it is advisableto investigate the possibility of impropercompensation for refraction. In order toaccomplish this, it is imperative that a

Figure I. Refraction Correction Chart for Computation of Height Errors

log of all height accuracy checks bekept. Data from the log can be analyzedby using some system such as the chartillustrated in figure 1.

The log entries should include date,time, range, azimuth, indicated height,actual height, error, type of target, andsome reference to weather. When inter-preting log data it must be rememberedthat errors caused by out -of -level con-ditions are a function of azimuth, andthat electrical misalignment causes er-rors which are independent of azimuth.Diffraction occurs only with grazingpaths, and refraction varies with time,while all of the previously mentionedfactors are independent of time.

The style of the accompanying chartwas chosen to match the presentation ofa particular radar (AN/MPS-14). (Forsets having curved height lines and/orrange marks, the familiar radio horizontype chart may be prepared. On thistype of chart the earth's surface is de-picted as curved rather than flat.) Foreach angle of antenna tilt there are fourcurves, the upper in each group is fork = 1.1, the second for k = 1.2, and soon through k = 1.4. Height ordinateswere computed from the generally ac-cepted equation:

D = V2TCahOn solving for h, this becomes:

2h - 2ka

where D = distance to horizonh = height of targetk = correction for refraction

1

- 1 - a(dn/dh)

= 4/3 for standard atmosphere

a = earth's radius (20.9 x 10" ft)n = refractive index

It should be noted that this relation-ship is not exact, but that the approxi-mation is accurate enough for currentusage with one exception. The equation

is derived for zero antenna tilt and re-mains useful only at small angles. Asthe range of tilt exceeds several degreeswith typical height finders, the refrac-tive correction should be reduced as afunction of the cosine of the nod angle,since there is less bendingeof the radiowaves as the angle of propagation de-parts from being parallel to the planeof constant refractive index. Manufac-turers have considered this problem, andsome equipments utilize circuitry whichvaries the refractive correction as theantenna nods.

In using the chart, lightly mark boththe true and indicated positions of thetarget on the chart. The antenna heightmust be subtracted from each of thesereadings. If the method of checkingheight uses an angle", rather than apoint in space, it is only necessary tonote the angle. Now determine whichantenna angle on the chart comes near-est to the plotted points or angle. Next,step off (from the curve having thesame value of k to which the RHI iscalibrated) from this nearest angle groupa distance equal to the separation be-tween the true and indicated targetpositions. This distance is stepped offdownward if the true position of thetarget was below the indicated position.This procedure normally results in thecorrect value of k. There are, however,times of transitory weather when thevalue of k will be found to lie outsidethe four values given on the chart.Whenever this occurs, it is advisable tore -check other possible factors that mayexplain, or contribute to, indications ofextreme refraction. Repetitive checkingof known targets by such methods willproduce the data required for the deter-mination of typical k values for a givensite.

The chart may he used also to obtaincheck points for R1-11 calibration to anyk value (within its limits). As an illus-tration, assume that the logs indicate anaverage summer value of 1.2 for k,rather than the value 1.1 to which the

RI -II has been previously aligned. Fromthe chart it can be found that for k = 1.2,the earth's curvature calibration at 0°and 150 miles is very nearly 16,600 feet,rather than 18,000 feet for k = 1.1.Check points at other ranges can be de-termined in a similar manner; alwaysremember to add the site elevation toall values taken from the chart.

Conclusion

It would seem that height errorsshould not be tolerated for reasonsother than diffraction, refraction, orworn-out equipment. Complacency isoften a reason for lack of accuracy. Bychecking and re -checking, most align-ment procedures can be improved toyield greater precision.

BIBLIOGRAPHY1. Radio Wave Propagation, Academic Press Inc.,

New York, 1949.2. Propagation of Short Radio Waves, Kerr, D. E.,

Vol. 13, MIT Radiation Laboratory Series,McGraw-Hill Book Co., New York, 1951.

3. Computing the Index of Refraction of the Atmos-phere, Cowan, L W., Hq 3rd Weather Group,Technical Paper No. 1, May, 1952.

4. Forecasting Refractive Index Profiles in the At-mosphere, Cowan, L. W., Hq 3rd Weather Group,Technical Paper No. 2, September, 1953.

5. Interpreting Refractive Index Profiles in Termsof Radar Coverage, Cowan, L. W., Hq 3rdWeather Group, Technical Paper No. 3, Oc-tober, 1953.

6. C 6 E Digest, September, 1956, page 14 (Classi-fied), Hq ADC, Eng Air Force Base, Colorado.

7. "Some Aspects of Radar Coverage," Moss, SamuelM., Philco TechRep Division BULLETIN,Sept. -Oct., 1956, page 18, Philadelphia, Penna.

8. "Split Paints On a Height -Finder Radar," CE Digest, November, 1956, Unclassified, page16, Hq ADC, Eng Air Force Base, Colorado.

9. "Atmospheric Effects on Propagation," A. A. Mc-Kenzie, Electronics, Volume 30, Number 3(March 1, 1957).

Determination of the Internal Resistance of a Meter

The following procedure provides a simple method by which theinternal resistance of a meter can be found.

EmL5V

1. Connect the circuit as shown in the accompanying figure.2. Set R1 for full-scale meter deflection.3. Close Si,4. Set R2 for one-half of full-scale meter deflection.5. Disconnect R2 from the circuit and measure its resistance.

The measured value of R2 is now equal to the meter resistance, sinceone-half of the total current now passes through the meter. Theaccuracy of this procedure may be increased by increasing the valueof both E and Rl. In this case the ratio of R1 to the meter resistancebecomes larger and, insofar as total current is concerned, thus decreasesthe effect of adding R2 in parallel with the meter.

John Krawczyk

8

FERRITES FOR WAVEGUIDES

(Editor's Note: This article, which deals with a recent develop-ment in the field of microwave electronics, originally appeared inthe Digest of U. S. Naval Aviation Electronics and appears herethrough the courtesy of the Naval Aviation Electronics Service Unit.)

INTRODUCTIONRECENT DEVELOPMENTS in microwavetechniques have brought forth a seriesof waveguide fittings that utilize the un-usual properties of some new types ofmagnetic materials. These materials areferrites, a family of iron -containing cer-amic materials with unusual combina-tions of magnetic and electrical proper-ties. The new waveguide devices arealready being applied in modern firecontrol radar where the characteristicsof conventional TR and ATR tubesprevent full use of millimicrosecondpulse widths needed for high definition.

Ferrites are compounds containingoxygen, iron, and other metals such aszinc, manganese, magnesium, or nickel.As an example, the formula for one ofthe common magnesium ferrites is

MgO(Fe203). The ferrites are manufac-tured by mixing oxides of the requiredmaterials in powder form, pressing thepoWder into the desired shape, and fir-ing the shaped mixture at a temperatureof about 2000 degrees Fahrenheit. Thefinished material is a ceramic, and amongits properties is that of high electricalresistance since it contains no metallicparticles. Ferrites are extremely hardand brittle and cannot be machined. Awide variety of ferrites exist, with vary-ing electrical and magnetic properties.

Ferrites have been used commerciallyfor several years. A nickel -zinc ferrite iswidely used in the deflection coils andflyback transformers of television sets,and magnetic recording tape makes useof a .type of ferrite in powder form. Atlow frequencies, ferrites behave like ironalloys, having magnetic permeabilities

of from 4 to 3000. Their major advan-tage is a high electrical resistance whichprevents eddy currents and permits theiruse at frequencies where metallic ironcannot be used.

FERROMAGNETIC RESONANCEThe property of high electrical resis-

tivity permits ferrites to be used at fre-quencies extending into the microwaveregion. At these super -high frequencies,resonance occurs within the iron atomsthemselves, leading to unusual effects.In order to analyze these effects, it isnecessary to delve carefully into thereasons for the existence of the phe-nomenon known as magnetism.

ELECTRON SPINA little known but fundamental prop-

erty of electrons is that, in addition totheir motions around the nucleus of theatoms, they spin on their own axes.This may be compared to the spinningof the earth on its axis as it rotatesaround the sun. The spin causes the

Figure 1. A Commercial Ferrite IsolatorUtilizing the Faraday Rotation Principle.The operation of this model is dia-

grammed in figure 9.

9

Hdc

ANGULARMOMENTUM I

MAGNETIC_,.MOMENT

SPINNING ELECTRONHAVING MASSAND CHARGE

Hdc

hrtDRIVING FORCE

Figure 2. An Electron in a D -C Magnetic Field,with Its Own Magnetic Field Aligned with theD -C Field. The lower circle shows the natural

direction of precession.

electrical charge on the electron to setup a magnetic field, as shown in figure2, with the spinning charge acting likecurrent flowing around a loop. All elec-trons, therefore, behave like very smallmagnets.

Atoms of most elements have as manyelectrons spinning in one direction asthe other, so that the magnetic fields ofthe electrons cancel out or, if there isan uneven number, only one electron'smagnetic field is left uncancelled. Theseatoms do not show any noticeable mag-netic effects. Very careful measurementsby physicists have shown that about halfof all the elements are very slightlymagnetic. Even oxygen, when liquefied,exhibits definite magnetic properties.

FERROMAGNETISMIn the iron atom, an unusual arrange-

ment of electrons leaves four more elec-trons spinning in one direction than inthe other. The uncancelled magnetic fieldof these four electrons leaves the ironatom with a strong magnetism. Eachiron atom is a basic permanent magnet.Since the prefix ferro means iron, amagnetic effect caused by the magnetismof iron atoms is known as ferromag-netism.

GYROSCOPIC ELECTRONSAs a consequence of their spinning

motion, electrons behave like very smallgyroscopes. If a force is applied to theelectron to tilt its spin axis, it will be-have like any other gyroscope, and pre-cess, or wobble. This may be illustratedby experimenting with a gyroscope andstick, as shown in figure 3. Normally,even with the gyroscope spinning, thestick will hang straight down, becauseof gravity. If one tries to move the stickrapidly from side to side, the gyroscopewill force the stick to move, or precess,around in a circle in a direction deter-mined by which way the gyro rotor isspinning.

If the stick is pulled to one side andreleased, it will precess around in a

circle. The direction will be determinedas before by the rotor, and the naturalfrequency by the momentum of thegyroscope and the gravitational field. Ifthe force of gravity could be increased,it would be found that the natural pre-cession frequency would increase pro-portionately. Those who are familiar

X

- DIRECTION OFPRECESSION

Figure 3. A Gyroscope Mounted on the Endof a Freely Pivoted Stick Showing the Direc-tion of Precession When a Force Is Applied

in the X Direction

10

with resonance phenomena will realizethat if a rotational force is applied to thestick at just the natural precession fre-

quency, a very large amplitude of mo-tion (or displacement from the vertical)will result, while a force applied at anyother frequency will produce a muchlower amplitude. Furthermore, if onetries to make the stick move around inthe natural direction, the amplitude willbuild up; but if one tries to make itmove around in the opposite direction,the movement will be resisted.

Electrons behave in a rather similarfashion. Since gravity has little effect onthe electron, a steady d -c magnetic fieldmay be applied to line up the axes oftie spinning electrons. Any precessionquickly dies out when this field is ap-plied. If an alternating field is appliedat right angles to the steady field, theelectrons will wobble, or precess, in thesame fashion as the stick and gyroscopedo when a sideways force is applied.The natural precessional frequency ofan electron in a d -c field is somewherebetween 3 and 9 kilomegacycles, de-pending on the field strength. Thus, ifthe applied a -c magnetic field is at thisnatural frequency, the precessional mo-tion will build up and cause frictionaldamping effects to increase, since theentire iron atom is forced to vibrate.Energy will be extracted from the a -cfield and dissipated by heating theferrites.

FERRITE ATTENUATORIf a piece of ferrite is placed in the

center of a waveguide and a d -c field isapplied, as shown in figure 4, frequen-cies at the resonant frequency of the fer-rite electrons will be attenuated, whileother frequencies will pass with littleattenuation. This resonant attenuationfrequency may he varied over a limitedrange by varying the strength of the d -cfield.

ONE-WAY ISOLATORAn electromagnetic wave traveling

along a waveguide will produce, at apoint off the centerline of the guide, a

Figure 4. A Ferrite Slab Mounted in the Centerof a Waveguide Causes High Attenuation at

the Ferromagnetic Resonant Frequency.

rotating magnetic field. This is illus-trated in figure 5. When the wave ismoving from right to left, the magneticfield at a stationary location, A, will bepointing up, as shown. When point 2arrives at A, the magnetic field will bedirected to the right. Similarly, as points3 and 4 on the wave arrive at A, thefield direction will appear to rotateclockwise. Thus, any point off the cen-ter of the waveguide will see a rotatingmagnetic field as the electromagneticwave goes by.

By applying this same analysis to awave traveling from left to right, it canbe demonstrated that the magnetic fieldat A will rotate counterclockwise.

Assume that a piece of ferrite cannow be placed in the waveguide at A,as shown in figure 6. The d -c magneticfield is applied, and either the magneticfield strength or the microwave fre-quency is adjusted to make the ferrite'selectron resonant frequency equal to themicrowave frequency. Under these con -

f,

j4

i t It-_iJ I___.___J I__

2 3 4

Figure 5. Stationary Point A, Off the Center-line of the Waveguide, Sees a Clockwise Ro-wing Magnetic Field as Points 1, 2, 3, and 4,

on the Moving Wave Pattern, Go Past.

II

Figure 6. A Ferrite Slab, M1lounted at the Off -Center Point A, Will Act as a One -Way Device.

ditions, a wave traveling along the guidefrom left to right will act as a rotatingforce on the electrons in the iron atomsin the direction of natural precession.The precession will build up to a veryhigh amplitude and cause the power ab-sorbed from the electromagnetic waveto be dissipated in the ferrite as heat.

A wave traveling from right to leftwill not suffer much attenuation. Its ro-tating magnetic field will attempt topush the electrons around in the oppo-site direction to their natural precessionand no large movements will occur.Thus a wave traveling from left to rightwill suffer as much as a 10 -decibel at-tenuation, but one traveling from rightto left will lose only about 0.4 decibel.The above is a rather simple isolator,consisting of a short piece of waveguide,a permanent magnet, and a piece offerrite.

FARADAY ROTATIONWhen microwaves are passed through

a piece of ferrite in a magnetic field, an-other effect occurs. If the frequency ofthe microwaves is well above the ferrite'selectron resonant frequency, the planeof polarization of the wave will be ro-tated. This is known as the Faradayrotation effect, and is illustrated in fig-ure 7. A rod of ferrite is placed alongthe axis of the waveguide, and a d -cmagnetic field is set up along the axisby a coil. Suppose a wave entering atthe left end is vertically polarized. As itenters the section containing the ferrite,

it will set up limited precession motionof the electrons. The magnetic fields ofthe wave and the precessing electronsinteract, and the polarization of the waveis rotated. Upon leaving the ferrite, thewave is polarized at a 45 -degree angleif the correct length of ferrite has beenselected.

A more accurate explanation of theFaraday rotation may be given by con-sidering the linearly polarized wave asthe sum of two circularly polarizedwaves rotating in opposite directions.The component wave that is rotating inthe same direction as the natural preces-sion motion of the ferrite electrons willhave some of its energy absorbed, whilethe counter rotating component will notlose energy. This causes the velocity ofpropagation for the former componentto be somewhat less than for the latter,and the plane of polarization is rotatedin the direction of the faster movingwave. For any type of ferrite, the angleof rotation is proportional to both theferrite length and the strength of thed -c magnetic field. It does not dependon frequency.

The most important effect of the Far-aday rotation is that the direction of ro-tation depends only on the electron spinin the ferrite, so that a wave going pastthe ferrite rod in figure 7 will always betwisted as shown, regardless of whichdirection it is traveling. The directionof rotation may be changed by reversingthe d -c magnetic field in the ferrite.

12

WAVEGUIDE

FERRITE

POLARIZATION OFENTERING WAVE

POLARIZATIONOF LEAVING

WAVE

D -C MAGNETICFIELD

a

POLARIZATION OFLEAVING WAVE

POLARIZATIONOF ENTERING

WAVE

b

Figure 7. Faraday Rotation Occurs as the W are Trarels Past the Ferrite Rod. The plane ofrotation is always shifted clockwise as riewed from this angle. regardless of which way

the ware trarels.

FERRITE ISOLATORA practical device utilizing the Fara-

day rotation is shown schematically infigure 8. A plane -polarized wave travel-ing in the guide goes through a rec-tangular -to -round waveguide transition.

As the wave passes the ferrite, its planeof polarization is rotated 45 degreesclockwise, and enters the rectangularoutput waveguide, as shown in part aof figure 8. However, if a wave travelsthrough the guide in the reverse direc-

DIRECTIONOF ROTATION

INPUTENERGY

-or

FERROMAGNETICMATERIAL AND

PERMANENTMAGNET

TRANSFORMATION FROMCYLINDRICAL TO

RECTANGULARWAVEGUIDE

0

OUTPUTENERGY

DIRECTIONOF ROTATION

_

OUTPUT

FERROMAGNETICMATERIAL AND

PERMANENTMAGNET

I

TRANSFORMATION FROMRECTANGULAR

TO CYLINDRICALWAVEGUIDE

b

'NPuTENERGY

Figure 8. Ferrite Isolator. Part a shows the action of a wave traveling in the forwarddirection, while part b shows a ware being attenuated in the reverse direction.

13

REFLECTEDPOWER LOAD

INPUTTERMINAL

LOAD TERMINAL

DIRECTION OFROTATION

OUTPUTTERMINAL

.

FERRITE ROTATOR

REFLECTED POWERDECOUPLING UNIT

(SEPTUM AND PROBE)

RECTANGULAR TOSQUARE WAVEGUIDE TRANSITION

\ s\.\\\\,,,,NZ

s,45 MECHANICALWAVEGUIDE TWIST

SQUARE WAVEGUIDETO RECTANGULARTRANSITION

IPLANE OF POLARIZATIONOF REFLECTED ENERGY

PLANE OF POLARIZATION: OF INCIDENT ENERGY

Figure 9. Ferrite Isolator with Separate, Reflected -Power -Absorbing Load.The ferrite section is set up to twist the wave polarization 45 degrees.

tion, it will also be rotated 45 degreesclockwise, as shown in part b of figure8, and will be at 90 degrees to the planeof the rectangular output waveguide.The waveguide cannot accept this crosspolarized wave, and the energy will bereflected. Properly oriented vane typeabsorbers will absorb this energy with-out affecting waves traveling in the for-ward direction.

In isolators designed to handle highpower, the reflected wave - after itspolarization is rotated-can be coupledout to a separate power -absorbing load.A commercial isolator of this type isshown in figures 1 and 9.

Isolators of the Faraday rotation typeare capable of handling more powerthan the ferromagnetic resonance type,since the reflected energy does not haveto be dissipated in the ferrite. Somemodern isolators of the Faraday rotationtype can provide 30 db attenuation inthe reverse direction for only 0.1 dbloss in the forward direction. Mostcommercial models of this device incor-porate a 221/2 degree step twist at theinput and output waveguide flanges, sothat they may be inserted in a plumb-

ing system without causing unwantedtwists.

One-way isolators ordinarily use per-manent magnets surrounding the centerwaveguide. These provide the d -c mag-netic field along the axis of the ferriteand permit the isolator to be installedas a self-contained unit.

The most important use of the one-way isolator is in keeping reflectedenergy from the magnetron, where fre-quency stability is dependent upon theimpedance into which the magnetronworks. In a typical radar system, a rela-tively large and possibly time -varyingmismatch caused by the scanning an-tenna is connected to the magnetron bymany wavelengths of transmission lines.This condition often causes excessivefrequency pulling and a poor spectrum.In fixed -frequency systems, these effectscan be minimized by judicious selectionof components, and by tuning proce-dures. However, for tunable systemswhere frequency bandwidths of about10 percent are required, these proceduresare not usable. The use of ferrite iso-lators has proved to be one of the mostadvantageous methods of minimizing

14

V ,

' PvAOR

ANC

NT IN4

ROTATEDE. E ='PbC

TRANSITION FROMCIRCULAR TO

RECTANGULAR*A,..FGUIDE

MODULATEDOUTPUTENERGY

Figure W. Ferrite builatur wall fin fief tr ttttt rignet. Per -nutting I ,iriable ttenuation.

the effects of mismatch in magnetroncircuits.

VARIABLE ATTENUATORAn MEL:R:4E111g L .trutton ut the territe

isolator can he made by using an electro-magnet around the guide, as shown in

figure 10. An input wave, entering atthe left side, comes in polarized with itsE lines horizontal, as shown. If there isno coil current, it will pass the ferritewith no rotation, and suffer an attenua-tion of 3 db in entering the rectangularoutput guide that is mounted at a 45 -degree angle. If current flows throughthe magnetizing coil in one direction,the ferrite will twist the wave polariza-tion 45 degrees to the leftand it willenter the output guide, but will be re-flected and absorbed by the cross polar-ization vane attenuator. This permits theattenuation of the device to be variedfrom 0.1 to 30 db merely by changingcoil current.

This device can be used to producean amplitude -modulated microwave out-put merely by feeding the magnetizingcoil with the output of an audio ampli-fier. It permits a wave having a constantfrequency and amplitude to be ampli-tude -modulated with no attendantfrequency modulation.

Another important application is inthe receiver section of a radar set. A va-riable attenuator can be placed immedi-ately before the receiver detector crystals.When the transmitter pulse is about tobe generated, a step current can be ap-

plied to the variable attenuator to causemaximum attenuation. This will protectthe crystals before the transmitter pulsehits them, instead of afterwards, as isthe case when gas type TR tubes arcused. After the pulse, the attenuationcan he reduced slowly, protecting thecrystals from strong reflections fromnearby objects and providing automaticgain control as the range increases. Thistype of attenuator cannot, of course, pro-tect the crystals from any strong pulsesthat may enter the antenna from anothernearby radar set.

MICROWAVE APPLICATION

Ferrite isolators arc a nes% develop-ment and have not as yet appeared inmany equipments. In the next few years,however, they will become familiar de-vice's. Typical applications of these iso-lators are as follows:

1. Load isolation between a micro-wave source and its load, to preventreflections from affecting the generator.

2. Coupled -circuit isolation, couplingseveral microwave circuits to a commonoutput and preventing one source fromaffecting any other.

3. Modulating microwaves with pureamplitude modulation.

4. Fast action radar duplexers.When a new development, like ferro-

magnetic resonators, becomes availableto a rapidly expanding field like micro-wave technology, the many devices thatwill result cannot be foreseen.

I5

DESIGN CRITERIA FOR MUTUALINDUCTANCE TRANSDUCERS

(Editor's Note: This article appears through the courtesy of NBS.)SINCE THE ORIGINAL DEVELOPMENT Ofthe mutual inductance transducer (figure1) by M. L. Greenough, of the NationalBureau of Standards, the device hasfound increasing application at the Bu-reau in electronic distance -measuringinstruments. The transducer detects ex-tremely minute changes in the positionof a nearby conducting plate, and canbe made to record such changes withan accuracy of 5 percent or better. Thisuseful characteristic, together with easeof calibration, has led to its adoption ina number of noncontacting displacementgauges for both static and vibratorytypes of measurements. Examples in-clude an electronic micrometer, a man-ometer, an oil -film thickness indicator,a vibration pickup calibrator, and others.

Until recently, design analysis of thetransducer has been restricted to its im-mediate use in a particular instrument.However, there has been a growingneed for general design criteria for usein future applications. To provide thenecessary data, H. M. Joseph and N.Newman, of the Bureau's electronic in-strumentation laboratory, have made adetailed study of the device's operatingprinciples, with major emphasis ontransducers using highly conductingreference plates. Their investigation wascarried out as part of a program of basicinstrumentation sponsored at the Bu-reau by the Office of Naval Research,the Air Research and DevelopmentCommand, and the Atomic Energy Com-mission. The results include a number

Figure I. Experimental Setup of the NBS Mutual -Inductance Transducer. Plate at left isreference surface; coils are wound on plastic form. Graduated wheel at right is calibrated formeasuring changes in surface -to -coil distance. As this distance changes, mutual inductancebetween the two coils changes, and can be detected through suitable electronic circuitry and

indicated on a calibrated meter.

16

SECONDARY

METALPLATE

Figure 2. Schematic Diagram of a Typical NBS Mutual Inductance Probe. Both prinuir N andsecondary coils are fixed with respect to each other. Metal reference plates may be vibrating,provided that the highest mechanical frequency is less than the electrical frequency of the

current energizing the primary coil.

of design recommendations for obtain-ing optimum combinations of stability,sensitivity, and linearity over the rangein which the instrument is to be used.

A typical mutual -inductance probe, asshown in figure 2, contains two co-planar, coaxial coils wound on a dielec-tric core. An r -f source that is regulatedwith respect to the product of the fre-quency and the current energizes theprimary coil. The a -c voltage inducedin the secondary coil then depends pnthe distance from the probe to the ref-erence plate. Suitable electronic circuitrydetects and amplifies the output voltagefrom the secondary coil, and this volt-age is indicated on a calibrated meterthat gives the change in probe -to -surfacemeasurement in inches or centimeters.The instrument is adaptable to veryrapid displacement changes; detectionof these changes is limited only by theresponse of the amplifier and by the en-ergizing frequency applied to the probe.

Results of the NBS investigation showquantitatively the change in mutual in-ductance between two coils when a con-ducting plate moves toward or awayfrom the coils; see figure 3. The changein mutual inductance can be calculatedby use of the theory of images wherebythe conducting plate is replaced by an"image coil" that has the same effect asthe plate on the mutual inductance be-tween the two coils. Thus a system ofthree coils is substituted for the systemof two coils and plate. The equivalenceof electrical and magnetic effects be-tween the two systems allows the useof the image coil in mathematical analy-sis of the flux -linkage variations. One ofthe assumptions in this analysis is thatthe plate is perfectly conducting; inpractice, however, plate conductivity isfinite. Although detailed studies of con-ducting plates with finite conductivityhave not yet been made, preliminaryinvestigations indicate that if a suffi-

1 7

d+ b

b

4

A

4

PRIMARY SECONDARYCOIL

COIL

4- ISEQUIVALENT

TO --

LARGEMETALPLATE

b

21d +b)

PRIMARYCOIL

2d -l -b

\\SECONDARY

COIL

I:: -

IMAGEOFPRIMARY

COIL

Figure 3. Diagram Demonstrating Theory of Images I sed in the Design Analysis of theNBS Mutual -Inductance Transducers with Highly Conducting Reference Plates. At left isthe system of two coils and plate; at right is the equivalent system of three coils, where

the reference plate is replaced by the image of the primary coil.

ciently high frequency is used, in mostapplications the effect reduces the in-strument's sensitivity by only a neg-ligible amount.

From the analysis made so far, thefollowing design recommendations canbe specified:

1. The reference plate should be asnearly a perfect conductor as practicable.

2. The primary or exciting coil shouldbe larger than the secondary coil, topermit greater heat dissipation.

3. If it is not possible for the primarycoil to be coplanar with the secondarycoil, then the primary coil should bethe farther from the plate.

4. The excitation frequency should beas high as possible, provided that theinstabilities of resonance are avoided.Although it is true that operation atresonance will greatly enhance the sen-sitivity, a high degree of sensitivity is

not usually a ruling consideration inprobe design.

5. The leads to the primary coil andto the secondary coil should be shieldedfrom each other, or should be fixedwith respect to each other.

6. The number of ampere -turns on theprimary should be as large as possible.

7. The number of turns on the sec-ondary should be as large as possible,provided that resonance effects areavoided.

8. The ratio of the secondary coilradius to the primary coil radius shouldbe as large as possible.

9. The primary coil should be as largeas possible. However, for linear responsewith displacement, it should not belarger than the reference plate.

All of these recommendations refer toprobes designed for static measurements.However, no error results from using

18

static considerations when making meas-urements on vibrating reference platesif the highest mechanical frequency is

less than the electrical frequency.Another aspect of the investigation

has been the application of electro-dynamic similitude to reduce the amountof experimental work required for thedesign and construction of probes.Electrodynamic similitude is a method

of changing design parameters of a givendimension (such as length or plate con-ductivity) by a scale factor so that theoriginal output voltage can be main-tained by properly altering all para-meters of other dimensions. Scaling thuspermits a probe of convenient size to beused for experiments, instead of onewhich is too large or too small for easeof handling or fabrication.

Solution to March -April "What's Your Answer?"Consideration of the original figure discloses that the final voltage

across the capacitor will be 20 volts, a value which can be obtainedas follows: the charged capacitor will appear as an open circuit andwill impose no loading on R2 or the source, and the voltage acrossR2 (and, therefore, across C) will be determined by simple voltage -divider action. Thus:

ER,E

100K 20

R, + R, 500K 100

The time required to charge C can be found most easily by trans-forming E, R and R2 of the original circuit into an equivalentcircuit, by means of Thevenin's theorem. Redrawing the original

ER, and R' - ,circuit gives the circuit below, where E' = R, + R, RR

R,, +2

E'

R'VVN.,

T ICThe equivalent voltage is thus seen to be 20 volts (confirming thevalue obtained above), and the equivalent resistance is 80,000 ohms.The charging time constant is, therefore, 8 x 10' x 10 '°, or 8 micro-seconds. Assuming that C will charge fully in 5 time constants gives40 microseconds as the answer to the first question.

19

THE POLAR IMPEDANCECHART AND ITS USE

This article describes the construction of the polar impe-dance chart (Smith Chart) and some typical applicationsof this chart in performing transmission line computa-tions which are of general interest to communicationsand electronics personnel.(Editor's Note: This article is substantially the same as thatappearing in the "Civil Aeronautics Administration CourseMaterial" and is reprinted here by permission of that agency.)

INTRODUCTIONTHE INPUT IMPEDANCE Of a transmissionline is the impedance that would be seenlooking toward the load if the line werecut. The input impedance of a trans-mission line that is not terminated in itscharacteristic impedance varies with linelength. If the line is considered to bewithout loss, the variations in input im-pedance are repeated every half wave-length. Because of the repetitive natureof the input impedance, it can be plottedas a function of line length and charac-teristic impedance. The appearance ofthis plot can be made to take any desiredform, one of the most easily used formsbeing that of the polar impedance chartor the Smith chart (after Philip H. Smithwho first developed this type of display).The polar impedance chart relates, in agraphical manner, the input impedance,characteristic impedance, and SWR of atransmission line. The name polar impe-dance chart is used because the displayis completely contained in a circle.

CHART COORDINATESAn impedance may be written in sev-

eral different forms. To give individualinformation about the resistive and re-active components, -the rectangular formis used. If the impedance value is to berepresented by a point on a two-dimen-sional plot, the rectangular form ofexpressing the impedance suggests onecoordinate for resistive components ofall impedances and another coordinate

for the reactive components of all impe-dances. The polar impedance chart isbasically, therefore, a two -coordinatesystem of perpendicularly intersectingcircles, one system of circles for resis-tive components and the other systemof circles for reactive components of allimpedances.

THE RESISTANCE CIRCLESInspection of figure 1 and the polar

impedance chart shown in figure 10 willreveal a system of circles all of whichare tangent to each other at the bottomof the chart and all of which have theircenters on a vertical line. Any one ofthese circles is the locus of all pointsrepresenting impedances with the same

Figure I. Impedance Chart Showing OnlyResistance Circles

20

Figure 2. Impedance Chart Showing OnlyReactance Circles

resistive components. Any two differentpoints on the same resistance circle rep-resent two impedances with the sameresistive component but with differentreactive components.

THE REACTANCE CIRCLESInspection of figure 2 and the impe-

dance chart shown in figure 10 will re-veal a second system of circles only cer-tain arcs of which are included on thechart. Each of these arcs intersects eachof the resistance circles at right angles.These perpendicularly intersecting arcsof circles are the reactance circles. Thepositive reactance circles are on the righthalf of the chart, and the negative reac-tance circles are on the left half of thechart. Any one reactance circle is thelocus of all points representing impe-dances having the same reactive com-ponent. Any two points on a reactancecircle represent two impedances withthe same reactive component but differ-ent resistive components. Any point onthe chart, therefore, represents one andonly one impedance value.

CHART UNITSIt is desirable to use the chart on a

variety of problems having different ini-tial conditions. One of the initial condi-tidns that may change from problem toproblem is characteristic impedance, Z0.

In order to use the chart in solvingtransmission line problems involvingdifferent values of Zo, the units used inlabeling resistance and reactance circlesare in terms of Zo. Referring to figure 3,consider the following illustration:

If a point on the impedance chart ison a resistance circle of 0.5 and a reac-tance circle of -1.0, as read on thescales of the chart, this point representsan impedance of 0.5-j1.0 in chart unitsor a normalized value. To get the actualvalue of the impedance, the Zo of thetransmission line must be known. Thenthe actual impedance value is found bymultiplying the impedance in chartunits by the characteristic impedance,Zo. Assume Zo in this case to be 50 ohms.The actual impedance is then 50(0.5 -j1.0), or 25 - j50 ohms.

Conversely, if the point representingthe impedance 25 - j50 ohms on a50 -ohm line is desired, it is necessary tochange the actual impedance value intochart units (or normalize it) by divid-ing it by Z0. Thus, in the exampleabove, the chart -unit value of 25 - j50would be 0.5 - j1.0 if Zo is 50 ohms.The point of intersection of the 0.5 re-sistance circle and the - 1.0 reactancecircle would then represent the impe-dance 25 - j50.

Figure 3. Impedance Chart Showing Locationof Impedance Value 0.5-J1.0 (Chart Units)

21

Figure 4. Impedance Chart ShowingStanding -Wave -Ratio Circles

STANDING -WAVE -RATIO CIRCLESThe locus of points representing all

input impedances of a line that willproduce the same SWR on the line isa circle with its center at the exactcenter ofcentric circles on the chart is thereforeSWR circles. Figure 4 shows a fewSWR circles. The impedance of theabove example, 25 - j50, is found to lieon an SWR circle of 4.2. This indicatesthat a 50 -ohm line with an input impe-dance of 25-j50 at any point will havean SWR of 4.2. The point in the exactcenter of the chart represents an SWRof unity, since this point represents animpedance equal to Z0. The circle form-ing the outer boundary of the chart isan SWR circle of infinity, since thesepoints represent purely reactive impe-dances.

WAVE -LENGTH SCALESThe input impedance to a lossless

mismatched transmission line varies withthe line length and is the same at twopoints 180° apart. Also the SWR doesnot vary along the line. Therefore, aphysical movement on the line of thepoint at which input impedance is con-sidered may be represented by a move-

ment of the impedance point on thechart along a constant SWR circle. Ingoing once around the chart on thisconstant SWR circle, the point repre-senting line impedance has passedthrough every possible value of the lineinput impedance, or it has moved theequivalent of 180°. Thus once aroundthe chart represents 180° electricallyeven though it is 360° physically on thechart. Refer to point P on the chart offigure 5. This is the point which repre-sented the input impedance of a 50 -ohmline in the previous example. If the in-put impedance of the same line weredesired at some other point, A, whichis 45° toward the generator from P, thepoint on the chart representing inputimpedance at P should be moved clock-wise (toward the generator) on the con-stant SWR circle of 4.2 for a distance of45 electrical degrees, as indicated in fig-ure 5. Note that this 45° electrical dis-tance on the line corresponds to a 90°physical angle on the chart. Likewise,the input impedance at a point B, whichis 60° toward the load from P, may befound by moving the impedance point

STANDINGWAVES OF VOLTAGE

Figure 5. Movement of Impedance Point onChart to Correspond to Physical Movement of

Point Along Line

22

counterclockwise (toward the load) onthe constant SWR circle of 4.2. Noteagain that this 60° movement on theline is represented by a 120° movementon the chart.

The outer boundary of the chart (fig-ure 10) is calibrated in wave lengths forconvenience in measuring off linelengths pertaining to different problems.One set of wave -length scales reads to-ward the generator, or clockwise, andthe other set reads toward the load, orcounterclockwise. If line length in elec-trical degrees is used instead of wavelengths, a protractor may be used, keep-ing in mind the fact that the chart cir-cumference represents 180°. The scalemarked ANGLE OF REFLECTIONCOEFFICIENT may be adapted to readline length in the absence of aprotractor.

VOLTAGE MAXIMUM ANDMINIMUM POINTS

The input impedance of a line at avoltage maximum is a pure resistanceand is greater than Z,,; actually, Rmax =SZ. The bottom half of the pure resist-ance line (or zero reactance line) repre-sents all such impedances. Thus whena point on the chart representing theinput impedance to a line lies on thishalf of the pure resistance line, a voltagemaximum exists on the transmissionline at this point. The upper half of thepure resistance line, therefore, repre-sents the presence of a voltage minimum,since a voltage maximum and a voltageminimum are always separated by 90°of line length. The input impedance ata voltage minimum is also pure resist-ance (R,,,, ).

1/4,

FINDING INPUT IMPEDANCE ATANY POINT ON THE LINE

"I he input impedance at any point ona transmission line can he found if theinput impedance to the line at someother point is known and if Z,, and theelectrical distance between the knownand unknown input impedance pointsare known. The general procedure, us-ing the chart in figure 10, is to first

obtain the chart -unit value of the knowninput impedance by division by Z.Then enter this point on the impedancechart. Note the SWR as read on theSWR scales of the chart. Move thepoint on the chart on this SWR circlein the appropriate direction and theappropriate amount to locate the pointwhich represents the chart -unit value ofthe input impedance at the point desired.Change the chart -unit value as read fromthe chart to actual impedance by multi-plying by Z. Remember that clockwiserotation on the chart corresponds to aphysical movement toward the gener-ator, and that counterclockwise move-ment corresponds to physical movementtoward the load.

EXAMPLES OF PROBLEM SOLU-TIONS USING THE IMPEDANCE

CHARTEXAMPLE ONE

A 50 -ohm line is terminated in an im-pedance of 30 j40 ohms. If the line is60° long, find the value of input impe-dance.Solution (refer to figure 6):

1. Change the terminating impedanceto chart units by dividing by Z. ThusZR is 0.6 -1)0.8 in chart units.

Zo .50ZR .30 + 40(ACTUAL VALUE)

ZR..6 + j.8(CHART UNITS)

13/fe/

Zr .I.95-)1.31CHART UNITS)Zs .975-)651ACTUAL VALUE)

\\

30.0.

OEN Z0.5011

60'

+1404.

rigore 6. Illustration of the ,Sfi.,, In Solvingthe Problem of EXIImple (Me

23

2. Locate this impedance as a pointon the chart on the SWR circle of 3.

3. Move the point along this SWRcircle toward the generator (clockwise)for an electrical distance of 60°

4. Read the chart -unit input impe-dance value at this point on the line as1.95 - j1.30. Find the actual value ofthe impedance by multiplying by 50,the Zo of the line. Thus Zs at 60° is97.5 - j65.0.EXAMPLE TWO

A 100 -ohm line has a standing -waveratio of 4. A voltage minimum exists130° from the load. Find the value ofload impedance.Solution (refer to figure 7):

1. Locate the point on the chart rep-resenting the input impedance of theline at the voltage minimum. This willbe the intersection of the SWR circle of4 and the top half of the pure resistanceline.

2. Move counterclockwise on thisSWR circle (toward the load) for anelectrical distance of 130°. Read thechart -unit value of impedance at theload of 0.56 + j1.03.

-`1"--°ZeRMIN130°

EMIN

Zs AT 130° FROM LC(Rmits)

SWR°4EMAX

ZL..56 +11.03(CHART UNITS)CKZi_r56 +j10 3

ACTUAL VALUE)Zo

VOLTAGESTANDING WAVES

Zo .100J1 FL]

Figure 7. Illustration of the Steps in Solvingthe Problem of Example Two

24

EitiN

EMAX

Z8.11441.74(CHART UNITS)Zs 59.34138.5

(ACTUAL VALUE)

STANDING WAVESOF VOLTAGE

FROM GEN--- To .52.11

ct.I50°

Ist59.3 4)38.5

-40 LOAD

ZeRMAX

Figure 8. Illustration of the Steps in Solvingthe Problem of Example Three

3. Multiply this value by Zo (100ohms) to get the actual value of the loadimpedance: 56 + j103 ohms.EXAMPLE THREE

A 52 -ohm line has an SWR of 2 anda voltage maximum at a point 75° fromthe load. Find the value of input impe-dance at a point 225° from the load.Solution (refer to figure 8):

1. Find the point on the chart repre-senting the input impedance to the lineat the voltage maximum. This will bethe intersection of the SWR circle of 2and the bottom half of the pure resist-ance line.

2. Move this point clockwise (towardthe generator) on this SWR circle forin electrical distance of 150° (the dif-ference between 225° and 75°).

3. Read the chart -unit value of theinput impedance as 1.14 + j0.74. Multi-plying this value by Zo (52 ohms) gives59.3 + j38.5 ohms as the input impe-dance at the point 225° from the load.USE OF THE IMPEDANCE CHART IN

WORKING WITH ADMITTANCEThe input impedance of a line at a

voltage maximum is called Rmax since itis pure resistance and is greater than Zo.

The admittance of the line at this pointwould therefore be a pure conductanceand is called G,,. Likewise, at a volt-age minimum, the input impedance andinput admittance are called R.,,n andGmax, respectively. When Gmin andGmax are changed to chart units by divi-sion by G., the chakacteristic conduct-ance of the line, the points representingthem on the chart are on the pure re-sistance line (or pure conductance line),but in reverse position with respect tothe voltage maximum and voltage mini-mum parts of this line as was foundwhen working with impedance on thechart. Thus to use the chart in problemsdealing with admittance, the top half ofthe pure conductance line representspoints of input admittances at a voltagemaximum, and the bottom half of thisline represents points of input admit-tances to a line at a voltage minimum.This interchange of the Vmax and Vm,parts of the pure resistance line is theonly change needed when working withadmittance on the chart. Movementalong the line is still represented on thechart as a movement of the admittancepoint on a constant SWR circle. Clock-wise rotation represents movementon the line toward the generator, andcounterclockwise rotation representsmovement on the line toward the load,as was the case when working with im-pedance. The resistance circles used inworking problems in impedance arenow considered to be conductance circleswhen working with admittance, and thereactance circles are now susceptancecircles.

It is sometimes convenient to use bothadmittance and impedance methods inworking a transmission line problem.An example of this is the problem ofstub matching when the value of loadimpedance is known. It is possible todo this on the chart and change over atany point in the solution from impe-dance to admittance, or vice versa, aschart -unit values of input impedanceand the corresponding input admittance

are always on the same SW x arc= antiare always physically 180° apart. Thus ifeither the impedance point or the ad-mittance point is known on the chart,the other can be found by merely ex-tending the line joining the known pointwith the center of the chart until it in-tersects the same SWR circle on theopposite side of the chart. This point ofintersection is then the other point (theimpedance point if the admittance pointis known, or the admittance point if theimpedance point is known).EXAMPLE FOUR:

The SWR on a certain line is 4 and avoltage maximum exists at a point 30°from the load. Find the distance fromthe load to the point nearest the loadwhere the input admittance to the linehas a conductance component equal toGO, the characteristic conductance ofthe line.Solution (refer to figure 9):

1. Find the point on the chart whichrepresents the chart -unit value of inputadmittance of the line at the voltagemaximum. This point will be the inter -

FROMGEN

EMAX

EMIN

Ys.I +11.5(CHART UNITS)Ys .Go +j1.56

(ACTUAL VALUE)

EOUIVALENT_____CIRCUIT - at +j1.5G0

TO LOAD

Figure 9. Illustration of the Steps in Solt insthe Problem of Example Four

25

Figure 10. Complete Smith Chart

section of the SWR circle of 4 and thetop half of the pure conductance line.

2. Move on this SWR circle towardthe generator (clockwise) until inter-secting the unity conductance circle. Atthis point the chart -unit value of theinput admittance is 1.0 + j1.5, or the

conductance component is equal to Goand the susceptance component is equalto 1.5Go. Read the distance traveledalong the SWR circle as 63.3° from thevoltage maximum. Thus the distance tothe load from the point found to havean admittance of Go + j1.5Go is 30° +63.3°, or 93.3°.

26

RADIO LOCATION OFUNDERGROUND CABLES AND PIPES

by George W. SpoonerPhilco TechRep Field Engineer

The task of accurately locating buried pipes or cablesmay be considerably simplified by the application of radiolocation methods and devices. This article describes theprinciples of this method of radio locating and indicatesboth its good and bad features.

IT IS OFTEN necessary that a buried cableor pipe be located with sufficient accu-racy to allow excavation to proceed inthe vicinity without incurring damage tothe cable or pipe.

Any metal object of appreciable sizeburied at depths up to 8 feet or moremay be easily and fairly accurately lo-cated by what is known as the high -frequency (or radio) method of detec-tion. The available commercial detectorsoperate on a principle which is similarto that of military type mine detectors.However, the commercial detectors arespecifically designed for the location ofburied pipes or cables, and are there-fore more efficient for this particularapplication.

The equipment consists of a transmit-ter and receiver which operate in thefrequency range between 130 kc and1 mc. (The equipment used by theauthor for these tests operated on a fre-quency of approximately 1 megacycle.)The signal is radiated from a loop an-tenna which is built around the trans-mitter case, and which is orientedperpendicular to the ground when inuse. The receiver is mechanically at-tached to the transmitter by a barapproximately 3 feet long, and is thusheld rigidly in position with respect tothe transmitter. The receiver antenna isalso a loop which is built around thereceiver case. In operation this loop isoriented parallel to the ground, and isthus in a null position with respect to

the transmitted signal field. Figure 1shows the equipment ready for opera-tion. The equipment operates on theprinciple that any metal in the vicinityof a transmitted signal will distort theradiated field pattern. The receiverloop, which is then no longer in thenull position, will provide a signal out-put. The received signal will be bothaurally indicated in the earphones andvisually indicated by a meter providedon the receiver panel. Figure 2 (left)shows the field without a buried object,and figure 2 (right) shows the distortedfield when a buried cable is encountered.

The effect of ordinary soil withoutburied objects is such that there is al-ways a slight warping of the field. Inorder to cdmpensate for this, the receiverloop antenna is tilted slightly with anadjusting knob to obtain a true nullsetting under particular conditions of

Figure 1. Line Drawing Illustrating RelativeOrientation of Transmitter and Receiver

27

Figure 4. Impedance Chart ShowingStanding -Wave -Ratio Circles

STANDING -WAVE -RATIO CIRCLESThe locus of points representing all

input impedances of a line that willproduce the same SWR on the line isa circle with its center at the exactcenter of the chart. The system of con-centric circles on the chart is thereforeSWR circles. Figure 4 shows a fewSWR circles. The impedance of theabove example, 25 - j50, is found to lieon an SWR circle of 4.2. This indicatesthat a 50 -ohm line with an input impe-dance of 25-j50 at any point will havean SWR of 4.2. T4 point in the exactcenter of the chart represents an SWRof unity, since this point represents animpedance equal to Z0. The circle form-ing the outer boundary of the chart isan SWR circle of infinity, since thesepoints represent purely reactive impe-dances.

WAVE -LENGTH SCALESThe input impedance to a lossless

mismatched transmission line varies withthe line length and is the same at twopoints 180° apart. Also the SWR doesnot vary along the line. Therefore, aphysical movement on the line of thepoint at which input impedance is con-sidered may be represented by a move-

ment of the impedance point on thechart along a constant SWR circle. Ingoing once around the chart on thisconstant SWR circle, the point repre-senting line impedance has passedthrough every possible value of the lineinput impedance, or it has moved theequivalent of 180°. Thus once aroundthe chart represents 180° electricallyeven though it is 360° physically on thechart. Refer to point P on the chart offigure 5. This is the point which repre-sented the input impedance of a 50 -ohmline in the previous example. If the in-put impedance of the same line weredesired at some other point, A, whichis 45° toward the generator from P, thepoint on the chart representing inputimpedance at P should be moved clock-wise (toward the generator) on the con-stant SWR circle of 4.2 for a distance of45 electrical degrees, as indicated in fig-ure 5. Note that this 45° electrical dis-tance on the line corresponds to a 90°physical angle on the chart. Likewise,the input impedance at a point B, whichis 60° toward the load from P, may befound by moving the impedance point

P5 - 11 0)

'1'4

STANDINGWAVES OF VOLTAGE

GEN

A

4 5*--- - 60° -

P 8

LOAD

Figure 5. Movement of Impedance Point onChart to Correspond to Physical Movement of

Point Along Line

22

2. Locate this impedance as a pointon the chart on the SWR circle of 3.

3. Move the point along this SWRcircle toward the generator (clockwise)for an electrical distance of 60°

4. Read the chart -unit input impe-dance value at this point on the line as1.95 - j1.30. Find the actual value ofthe impedance by multiplying by 50,the Zo of the line. Thus Zs at 60° is97.5 - j65.0.EXAMPLE TWO

A 100 -ohm line has a standing -waveratio of 4. A voltage minimum exists130° from the load. Find the value ofload impedance.Solution (refer to figure 7):

1. Locate the point on the chart rep-resenting the input impedance of theline at the voltage minimum. This willbe the intersection of the SWR circle of4 and the top half of the pure resistanceline.

2. Move counterclockwise on thisSWR circle (toward the load) for anelectrical distance of 130°. Read thechart -unit value of impedance at theload of 0.56 + j1.03.

E MIN

Zs AT 130' FROM(RUIN)

SWR=4

ZL: 56 +1103(CHART UNITS)CrIZL.56 +003

ACTUAL VALUE)Zo 1004.

EMAX

VOLTAGESTANDING WAVES

Zo :100A

Zs:Fiss IN

130*

Figure 7. Illustration of the Steps in Solvingthe Problem of Example Two

EOM

Z1=1.14+1.74(CHART UNITS)Zs .59.3+138.5

(ACTUAL VALUE)

STANDING WAVESCF VOLTAGE

FROM GEN-- Zaz52.0, -.TO LOAD

--6-. 150'Zs:59.3+138.5

Ts*

zs RMAX

Figure 8. Illustration of the Steps in Solvingthe Problem of Example Three

3. Multiply this value by Zo (100ohms) to get the actual value of the loadimpedance: 56 + j103 ohms.EXAMPLE THREE

A 52 -ohm line has an SWR of 2 anda voltage maximum at a point 75° fromthe load. Find the value of input impe-dance at a point 225° from the load.Solution (refer to figure 8):

1. Find the point on the chart repre-senting the input impedance to the lineat the voltage maximum. This will bethe intersection of the SWR circle of 2and the bottom half of the pure resist-ance line.

2. Move this point clockwise (towardthe generator) on this SWR circle forin electrical distance of 150° (the dif-ference between 225' and 75°).

3. Read the chart -unit value of theinput impedance as 1.14 + j0.74. Multi-plying this value by Zo (52 ohms) gives59.3 + j38.5 ohms as the input impe-dance at the point 225° from the load.USE OF THE IMPEDANCE CHART IN

WORKING WITH ADMITTANCEThe input impedance of a line at a

voltage maximum is called Rmax since itis pure resistance and is greater than Zo.

24

r -The admittance of the line at this pointwould therefore be a pure conductanceand is called Gm". Likewise, at a volt-age minimum, the input impedance andinput admittance are called 11.,, andGmax, respectively. When G.. andGmax are changed to chart units by divi-sion by G., the chaiacteristic conduct-ance of the line, the points representingthem on the chart are on the pure re-sistance line (or pure conductance line),but in reverse position with respect tothe voltage maximum and voltage mini-mum parts of this line as was foundwhen working with impedance on thechart. Thus to use the chart in problemsdealing with admittance, the top half ofthe pure conductance line representspoints of input admittances at a voltage

n maximum, and the bottom half of thisline represents points of input admit-tances to a line at a voltage minimum.This interchange of the Vmax and V,,

3 parts of the pure resistance line is theonly change needed when working withadmittance on the chart. Movementalong the line is still represented on the

) chart as a movement of the admittancepoint on a constant SWR circle. Clock-wise rotation represents movement

) on the line toward the generator, and) counterclockwise rotation represents

movement on the line toward the load,as was the case when working with im-pedance. The resistance circles used inworking problems in impedance arenow considered to be conductance circleswhen working with admittance, and thereactance circles are now susceptancecircles.

It is sometimes convenient to use bothadmittance and impedance methods inworking a transmission line problem.An example of this is the problem ofstub matching when the value of loadimpedance is known. It is possible todo this on the chart and change over atany point in the solution from impe-dance to admittance, or vice versa, aschart -unit values of input impedanceand the corresponding input admittance

are always on the same SWR circle andare always physically 180° apart. Thus ifeither the impedance point or the ad-mittance point is known on the chart,the other can be found by merely ex-tending the line joining the known pointwith the center of the chart until it in-tersects the same SWR circle on theopposite side of the chart. This point ofintersection is then the other point (theimpedance point if the admittance pointis known, or the admittance point if theimpedance point is known).EXAMPLE FOUR:

The SWR on a certain line is 4 and avoltage maximum exists at a point 30°from the load. Find the distance fromthe load to the point nearest the loadwhere the input admittance to the linehas a conductance component equal toG., the characteristic conductance ofthe line.Solution (refer to figure 9):

1. Find the point on the chart whichrepresents the chart -unit value of inputadmittance of the line at the voltagemaximum. This point will be the inter -

FROMGEN

EMAX

1111,CHYAsR:Ti\+UINI.15TS)EMIN

Y=G°+II5G(ACTUALVALUE)

TO LOAD

i.-63.3°-+-30*-- --NU I VA L ENT 71--ir

CIRCUIT +11.5G0

Figure 9. Illustration of the Steps in Solvingthe Problem of Example Four

25

Mgc

c g-,00. LOS a Si

o e 6

n447led".... 6 6 6 IS a ClC 6 6 Ili

1.1.,0111 2

c s 5;,' 3 5CD 1

a Zs.;

a g

6 a ar% gg, 3'

Figure 10. Complete Smith Chart

section of the SWR circle of 4 and thetop half of the pure conductance line.

2. Move on this SWR circle towardthe generator (clockwise) until inter-secting the unity conductance circle. Atthis point the chart -unit value of theinput admittance is 1.0 + j1.5, or the

conductance component is equal to Goand the susceptance component is equalto 1.5G0. Read the distance traveledalong the SWR circle as 63.3° from thevoltage maximum. Thus the distance tothe load from the point found to havean admittance of Go + j1.5Go is 30° +63.3°, or 93.3°.

26

RADIO LOCATION OFUNDERGROUND CABLES AND PIPES

by George W. SpoonerPhilco TechRep Field Engineer

The task of accurately locating buried pipes or cablesmay be considerably simplified by the application of radiolocation methods and devices. This article describes theprinciples of this method of radio locating and indicatesboth its good and bad features.

IT IS OFTEN necessary that a buried cableor pipe be located with sufficient accu-racy to allow excavation to proceed inthe vicinity without incurring damage tothe cable or pipe.

Any metal object of appreciable sizeburied at depths up to 8 feet or moremay be easily and fairly accurately lo-cated by what is known as the high -frequency (or radio) method of detec-tion. The available commercial detectorsoperate on a principle which is similarto that of military type mine detectors.However, the commercial detectors arespecifically designed for the location ofburied pipes or cables, and are there-fore more efficient for this particularapplication.

The equipment consists of a transmit-ter and receiver which operate in thefrequency range between 130 kc and1 mc. (The equipment used by theauthor for these tests operated on a fre-quency of approximately 1 megacycle.)The signal is radiated from a loop an-tenna which is built around the trans-mitter case, and which is orientedperpendicular to the ground when inuse. The receiver is mechanically at-tached to the transmitter by a barapproximately 3 feet long, and is thusheld rigidly in position with respect tothe transmitter. The receiver antenna isalso a loop which is built around thereceiver case. In operation this loop isoriented parallel to the ground, and isthus in a null position with respect to

the transmitted signal field. Figure 1shows the equipment ready for opera-tion. The equipment operates on theprinciple that any metal in the vicinityof a transmitted signal will distort theradiated field pattern. The receiverloop, which is then no longer in thenull position, will provide a signal out-put. The received signal will be bothaurally indicated in the earphones andvisually indicated by a meter providedon the receiver panel. Figure 2 (left)shows the field without a buried object,and figure 2 (right) shows the distortedfield when a buried cable is encountered.

The effect of ordinary soil withoutburied objects is such that there is al-ways a slight warping of the field. Inorder to compensate for this, the receiverloop antenna is tilted slightly with anadjusting knob to obtain a true nullsetting under particular conditions of

Figure 1. Line Drawing Illustrating RelativeOrientation of Transmitter and Receiver

27

Figure 2. Transmitter Field without (Left) and with (Right) Object Which Causes Distortion

moisture content and type of soil. Inactual operation, the adjustment is thenturned slightly so that a very weak sig-nal is heard. Any buried object willthen be indicated by an increase in sig-nal and a change in pitch. The signallevel will be maximum when the trans-mitter is directly over the buried object,and by decreasing the volume with thevolume control provided, it is possibleto find the exact location of the object.If the buried object is a pipe or cable,the greatest signal strength will occurwhen the transmitter loop is directlyabove the object and is oriented parallelto the run of the cable or pipe. On theother hand, if the object is round orsquare (such as a manhole cover) orien-tation of the transmitter has no effect.

Small buried objects will give only aweak indication which will soon bepassed over, whereas a buried pipe orcable will give a good indication forseveral feet on each side. Figure 3 showsa graph of relative signal strength as aburied cable is approached and passedat right angles to its run. This particularcable was approximately 1 inch in di-ameter, and was buried at a depth ofapproximately 30 inches.

Smaller cables or pipes may be foundat considerable depths, depending uponthe proficiency of the operator. Experi-ence is a necessary factor in the under-standing and interpretation of theindications.

Once a buried cable has been located,it is often desirable to trace its run.There are two methods for doing this inthe field. The first requires that the op-erator find the position and orientationof the equipment which gives maximumsignal. This condition will exist whenthe transmitter is directly over the cableand oriented in a direction parallel toits run. The operator then carries theequipment slowly in the direction of thecable run. As long as the transmitter isdirectly over the cable, maximum signalwill be indicated. It should be noted thatif the cable depth varies, there will be achange in signal strength but the maxi-mum will still be directly over its run.

In the second method, the transmitterand receiver are separated. The trans-mitter is then placed on the ground di-rectly over the cable. It should standupright and at right angles to the ap-parent run of the cable. The receiver isthen held face up and level at waistheight and moved slowly in the appar-ent direction of the cable run. A signal

/10 8 6 4 2 1 2 4 6 8 10

FEET

GROUNDLEVEL

FEET

BURIED5ilre. CABLE

Figure 3. Plot of Relative Meter IndicationVersus Horizontal Distance from Object

28

will be indicated when the receiver isanywhere near directly over the cable.When it is exactly over the cable, therewill be a very sharp null. Using thisnull, the cable may be followed for adistance of 100 to 500 feet withoutmoving the transmitter. The maximumdistance in this case is dependent uponthe depth of the cable.

With this equipment it is possible todetermine the depth of a cable withreasonable accuracy. Using the equip-ment in the same way as in the secondmethod above, the null directly over thecable is found, and this is marked onthe ground. The receiver loop is thenrotated clockwise to a 45 -degree angle.(Some equipments have a depth gaugewhich indicates this angle.) With thereceiver loop held at the 45 -degreeangle, it is moved from side to side atright angles to the run of the cable un-til another null is found. This spot is

also marked on the ground. As can be

seen from figure 4, the distance betweenthe two marks on the ground is thenequal to the depth of the cable.

This type of equipment has beenfound to be highly efficient in the loca-tion of buried cables or pipes, and, withpractice and experience, an operator can

B

i BURIEDCABLE

GROUNDLEVEL

A=B

Figure 4. Method of Determining Depthof Buried Object

become highly proficient in its use. Theequipment will not find a buried coin

or small object, but when used withinits limitations, it produces excellent re-sults. It is battery -operated and usesvacuum tubes of low current drain, toensure long battery life. On some makesof equipment, a switch and meter arealso provided on the panel for checkingthe condition of the batteries wheneverthe equipment is to be used.

During tests, it was noted that over-head power lines have no apparent effect

on the operation of the set, and it canthus be operated successfully in thevicinity of such lines.

29

QUALITY CONTROL OF VACUUM TUBESby Walter Gilbert, Philco Accessories Division,

and Stan Rosen, Philco Service Division

(Editor's Note: This article was written by the authors in con-junction with the Electronics Education Unit of the Philco ServiceDivision, and originally appeared in the February, 1957, issueof the Philco Electronic Supervisor.)

NEW AND MODIFIED type circuits areconstantly being designed and developedin the electronics industry. For thisreason, the type of tube to be used withthese circuits must also be redesignedor improved to meet its new application.Also, the tube manufacturer is alwaysimproving upon the type tube in gen-eral use, as may be seen by lookingthrough a tube manual.

In this effort to produce newer andbetter type tubes, the methods of check-ing the quality and efficiency of thesetubes must also be improved upon. Tohelp in preventing tube failure, practi-cally all tubes are put through static lifeand dynamic life tests as they are pro-duced, in order to check each particulartube under actual working conditions.By the use of these methods, defectiveor faulty tubes are detected, thereby sat-isfactorily controlling the quality of thetubes.

However, in some instances these testscannot fully determine the true efficiencyof certain types of tubes. Furthermore,these tests cannot indicate how wellthese tubes will function after lengthyperiods of operation. For this reason,more extensive tests have been devised.

The "beam power pentode" is a goodexample of a particular type of tubewhich requires more extensive testing.In television, this type is usually em-ployed as the horizontal output tubebecause of its high efficiency. In thepast, however, a major fault with thistype was overheating, which in turncaused short life. Many methods wereinvestigated in an effort to minimize thisheating effect, and the most satisfactory

method to be derived was the placingof "radiators" within the tube to helpin dissipating the heat. Figure 1 illus-trates three types of radiators used.

Through further testing, the cause ofthis overheating was determined. It wasfound that if the grids within the tubeare not properly aligned (screen gridalignment is particularly critical), thetube will draw excessive screen current,thus effectively stealing power from theplate circuit. Such a condition is aggra-vated by the fact that as the screen gridis overheated by the excessive flow ofcurrent, it becomes further misaligned,with complete failure as the end result.

For some time now, a completely newtype of test, developed by Philco, hasbeen used. This test determines thedynamic ratio of the plate current tothe screen current, and is the most ac-curate and effective method yet to bedevised. Figure 2 is a block diagram ofthe test setup for this measurement.

The entire test is comparatively sim-ple, and is accomplished as follows: Thecathode is maintained at ground poten-tial, and a 100 -volt peak-to -peak saw -tooth signal at a frequency of 15,750 cpsis applied to the grid, which is biasedat zero volts. Plate voltage is approxi-mately one and a half to three timeslower than the voltage on the screen.Ammeters are placed in both the plateand screen circuits, to indicate the qual-ity of the tubes. Under test, the tube issubjected to these extreme voltage con-ditions for only 3 minutes. The shorttime is important, since if this conditionwere maintained for any length of time,the tube would definitely burn out.

30

Figure 1. Three Different Types of HeatRadiators Which Minimize Overheating

To obtain the dynamic ratio, the plateand screen currents are checked after 30seconds of operation, and the ratio ofthe two readings is computed. Anotherratio is obtained after 3 minutes of oper-ation and, providing these two ratiosindicate only a slight change, the rubeshould give satisfactory operation.

The results obtained from a dynamicratio test of two "beam pentodes," to-gether with the interpretation of theresults, are as follows: The first tube,checked after 30 seconds, had a plate

current of 252 ma and a screen currentof 18.2 ma, making the first ratio 13.3.This same tube checked again after 3minutes operation had 255 ma in theplate circuit and 19.5 ma in the screencircuit, thus changing the ratio to 13.1.Since the change in ratio is small, thistube will provide normal service. Thesecond rube, when measured after 30seconds, also checked normal with 265ma plate current and 19.5 ma screencurrent, giving a ratio of 13.6. How-ever, after 3 minutes operation, the platecurrent had dropped to 190 ma, whilethe screen current had risen to 28 ma.This big change in plate and screencurrents drastically reduced the ratio to6.8, indicating that this particular tubewas slumping and would not give satis-factory service. Such a tube is thereforerejected.

From the tests mentioned above, itcan be seen that there is a constantstriving to maintain and improve thepresent high quality of vacuum tubes.This fact is evidenced by the presenceof suffix letters in many tube type num-bers, and also by the development ofnew tubes to perform the same functionas older type rubes.

Figure 2. Test Setup Used to Determine Dynamic Ratio of Plate Current to Screen Current

31

TECH INFO MAIL BAG

The following problem was posed by M/Sgt. David A. Berger,and is concerned with a trouble appearing in the AN/APN-9Loran equipment.

"Could you advise me as to how to correct the malfunction shownin the figure below. (Editor's Note: The photograph enclosed with therequest has been replaced by the drawing in the figure below for thesake of clarity.) The equipment is the Receiver -Indicator R-65/APN-9,and the presentation illustrated appears with the controls in thefollowing positions:

Function Switch-set to position 5Coarse Delay-set fully counterclockwiseFine Delay-set fully clockwisePRR (Pulse Repetition Rate)-set to 'High'

I have come across a number of sets in which this same troubleis apparent."

(iglu 111111111111111M

Illustration of Incorrect Indicator Presentation

The information given indicates that the trouble is apparent only onposition 5 of the function switch, and provides no indication ofwhat effect, if any, the delay setting has. It would be of great helpin pinpointing the trouble if more information could be obtained.However, from the information given, it appears that the d -c hori-zontal positioning voltage is not present during the time of theupper trace, and that the vertical separation voltage is insufficient.A check on the waveforms at the deflection plates (both horizontaland vertical) would be logical. The shield grounds on the deflectionplate leads should also be checked. These suggestions merely providea starting point, of course, and do not locate the trouble.

Since the trouble is apparent on a number of equipments, otherreaders may have come across it. If so, let us know, in order that wemay forward the information to Sgt. Berger.

32

TECHNICAL SKETCH OF GEORGE BOOLE (1815-1864)EDUCATION

George Boole was born at Lincoln, England, the son of a poorstorekeeper. Lack of money for an adequate education forced Booleto self -education. In this field he did an admirable job of learningGreek and Latin. Boole's early mathematical training was obtainedfrom his father, and at sixteen Boole took a job as an assistant teacherin a nearby school. Boole continued his studies and was quite success-ful in the field of logic, as evidenced by the following.

MAJOR SCIENTIFIC CONTRIBUTIONBoole's most important scientific contribution was the formulation

of a scheme for the solving of problems in the field of logic by theuse of authentic mathematics. In honor of Boole this inspiredscheme is called Boolean algebra. .

Boolean algebra consists of systematic rules for the handling ofthe connectives or, and, not. Modern applications of this algebra toelectronics fall in the field of data and information handling systems,for example, the design of computer circuits.

Boolean algebra may be used to reduce the number of operationsto accomplish a certain desired result to a minimum. For example, acomplicated relay and switching arrangement having redundantoperations can be redesigned, by application of Boolean algebra, intoan efficient system having a minimum of operations.

In Boolean algebra, the symbols of operation are separated andtreated as objects of calculation. A crude example is given in the fol-

lowing statement, which may be set up in symbolic form.If a certain switch is closed or if a certain bypass contact is not

closed, a given relay will energize.Let the condition if a certain switch is closed = A.Let the condition if a certain bypass contact is closed = B.Let the condition a given relay will energize = C.

Stated in Boolean algebra, the statement then becomes:AU B=C

where U is used to symbolize "or," and the line over B representsnegation of B.

PROFESSIONAL CAREERBoole was a teacher in the equivalent of our public elementary

schools for about four years. At twenty he opened his own schooland further pursued the teaching of elementary students for aboutfourteen years.

Several papers by Boole appeared in the Cambridge MathematicalJournal, and following the publishing of the Mathematical Analysisof Logic (a pamphlet), he was appointed Professor of Mathematics,Queens College, Cork, Ireland.

PUBLISHED WORKS1. Mathematical Analysis of Logic, in 1847.2. An Investigation of the Laws of Thought on Which Are Founded

the Mathematical Theories of Logic and Probabilities, in 1854.(This is Boole's master work.)

3. Treatise on Differential Equations, in 1859.4. Treatise on the Calculus of Finite Differences, in 1860.

33

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