PROCEEDINGS OF THE I.R.E.

r-=+/2z(z-0.3). The cone, with a total included angleof 109.5 degrees, is spaced 0.15 centimeter from the hy-perboloid and operated at -850 volts.

Fig. 9-Comparison of the axial potential for three differentgeometries.

A similar reflector can be made from the plate andcylinder by choosing the cylinder diameter and voltagesuch that the axial potential matches that given by theequation in the above paragraph at three points:z= 0, 0.031, and 0.062 centimeter. This yields a diameterof 0.220 centimeter and a potential of -945 voltsfor the cylinider, thus giving for the axial potential4 = 1625-25 70 tanh 12z. Similarly, matching the axialpotential of the hemisphere and plane at three pointsgives

b = 1625 + 4110, (0 P097n-=1 I.9

where the hemisphere radius and voltage are 0.097 centi-meter and -430 volts, respectively. Fig. 9 shows theclose agreement of the axial potential distribution forthe three geometries throughout the range traversed bythe electrons.

Properties of Ridge Wave Guide*SEYMOUR B. COHNt, MEMBER, I.R.E.

Summary-Equations and curves giving cutoff frequency andimpedance are presented for rectangular wave guide having a rec-tangular ridge projecting inward from one or both sides. It is shownthat ridge wave guide has a lower cutoff frequency;and impedanceand greater higher-mode separation than a plain rectangular waveguide of the same width and height. The cutoff frequency equationis fairly accurate for any practical cross section. The impedanceequation is strictly accurate only for an extremely thin cross section.Values found by the use of this equation have, however, been foundto check experimental values very closely. A number of uses for thistype of wave guide are suggested.

I. APPLICATIONSr HE CROSS-SECTIONAL shape of ridge wave

guide is shown in Fig. 1. This type of wave guideis briefly described in a text by Ramo and Whin-

nery,' where a simple method of calculating the cutofffrequency is given. That method is used in this paper.The lowered cutoff frequency, lowered impedance,

and wide bandwidth free from high-mode interferenceobtainable with ridge wave guide make it useful inmany ways. A few uses are listed below:

(a) It is useful as transmission wave guide, wherea wide frequency range must be covered, and whereonly the fundamental mode can be tolerated. It will beshown that a frequency range of four to one or more

* Decimal classification: RI 18.2. Original manuscript received bythe Institute, May 9, 1946.

t Cruft Laboratory, Harvard University, Cambridge, Mass. Thework reported in this paper was done at the Radio Research Labora-tory under contract with the Office of Scientific Research and Devel-opment, National Defense Research Committee, Division 15.

1 S. Ramo and J. R. Whinnery, "Fields and Waves in ModernRadio,"n John Wiley and Sons, New York, N. Y.; 1944.

can be easily obtained between the cutoff frequenciesof the TE1o and TE20 modes, and six to one or more be-tween those of the TE,o and TE30 modes. The attenua-tion is several times as great as that for ordinary wave

.~~~~~~ ~ ~~~~~~~~~~~~~~~~~~ II I

1 ~~~~~~~~ai2 -a'

(a)

2b2 2b I

I l_- a2 -,-J

at

(b)Fig. 1-Parameters for single-ridge (a) and double-ridge (b) wave-

guide cross-sections.

guide, but is still much less than for ordinary coaxialcable. The reduced cutoff frequency of ridge wave guidealso permits a compact cross section.

1947 783

PROCEEDINGS OF THE I.R.E.

(b) Ridge wave guide has been used successfullyas matching or transition elements in wave-guide tocoaxial junctions. In one type of junction, a quarter-wavelength section of ridge wave guide serves as amatching transformer from the impedance of the guide(utoll-ticket" wave guide, 21Xi-inch cross section) tothe 50-ohm coaxial cable. In another junction, a taperedlength of ridge wave guide gives a gradual match fromstandard 3Xli-inch rectangular wave guide to a 50-ohm coaxial line.2

(c) Various forms of ridge wave guide are usefulalso as filter elements, cavity elements, cavity termina-tions, etc. Wherever an element of line is needed havingreduced cutoff frequency, reduced impedance, or widemode separation, ridge wave guide provides a simple sol-ution.

(d) The attenuation formula for ridge guide (8)shows that the attenuation may be made very high bymaking a, and Z0O, as small as possible. If the guide, orjust the ridges, are made of steel instead of copper, theattenuation may be made about 1000 times greaterthan that for ordinary copper wave guide withoutridges. H. C. Early of the Radio Research Laboratoryhas made use of a length of such wave guide tapered tostandard 3 X 1i-inch wave guide in the design of abroadband matched load.",4 The total length of the loadand taper is only four feet.

(e) Another application, due to Early, is in a wide-band wattmeter,3 in which a wave guide having nearlyconstant impedance over a wide band is required.

II. DESIGN DATA

The design equations use the notation of Fig. 1. a1,a2, b1, and b2 are inside dimensions in centimeters. 01 and02 are the electrical phase lengths in terms of the cutoffwavelength in free space

eg 02- a/2 X 360)Xc,0

where XO is the wavelength in free space at the ridge-guide cutoff frequency.The cutoff of the TE1o mode occurs when the low-

est root of the following equation is satisfied:Bc

cot 01- -bi Yoly

b2 tan 02(1)'

B, is the equivalent susceptance introduced by the dis-continuities in the cross-section, as explained in Ap-pendix 1.6

lS. B. Cohn, "Design of simple broad-band wave guide-to-coaxial

line junction," to be published in PROC. I.R.E.3 H. C. Early, "A wide-band wattmeter for wave guide," PRoc.

I.R.E., vol. 34, pp. 803-807; October, 1946.' H. C. Early, "A wide-band directional coupler for wave guide,"

PROC. I.R.E., vol. 34, pp. 883-887; November, 1946.' (1), (3), (6), and (8) are derived in the appendix.' B. may be calculated from the curves in a paper by J. R. Whin-

nery and H. W. Jamieson, "Equivalent circuits for discontinuities intransmission lines," PROC. I.R.E., vol. 32, pp. 98-116; February,1944.

Equation (1) is accurate if proximity effects are takenfully into account in calculating B,. In the curves of thispaper, proximity effects are neglected, but the resultsare highly accurate so long as (a,-a2/2) >bi.

In terms of 01 and 02, Xc is given by( 900

x1a= 2)x01 02

(2)

where kc= 2a1 is the cutoff wavelength of the guide with-out the ridge, and where 01 and 02 are values satisfying(1).The TE1i-mode cutoff wavelength is plotted in Figs.

2 and 3 for a wide variety of ridge shapes in guide havingcross-section ratios of b1/a1=0.136 ("toll-ticket" waveguide, 21 Xi inch) and 0.500, respectively. The ordinateX,'/2aj=X,'//X=f,/f/' is the ratio of cutoff wavelengthwith the ridge to that without the ridge. The abscissaa2/a1 is the ratio of ridge width to guide width. Eachsolid curve corresponds to a constant value of b2/bi. Asan example, if a particular ridge wave guide has bi/a1=0.5, a2/a1=0.4, and b2/b1=0.1, then from Fig. 3,)X,'//X,=f,/f/'=2.6. If the cutoff frequency without theridge is 2600 megacycles, the cutoff frequency with theridge will be 1000 megacycles.On comparing Fig. 2 and Fig. 3, it will be seen that

there is not a great deal of difference between the cor-responding constant b2/b, curves. The only reason thereis any difference is the size of the discontinuity suscept-ance term, Bc!Yoi, which is small for b1/ai=0.136, andfairly large for bl/al=0.5. If b1/a1 has a value differentfrom 0.136, or 0.5, Figs. 2 and 3 may still be used withlittle error. Fig. 2 should be used for values of bl/a, be-tween zero and about one-third, and Fig. 3 should beused for values of bi/a, in the vicinity of 0.5.The characteristic impedance at infinite frequency for

the TE1o mode is given by

zooo =120ir2b2

X('{i +b2 0t1X,.'.sin 02 + -COS 02 tan-

(3)4

If Zo0. and the cutoff frequency f' are known, the char-acteristic impedance at any frequency f is obtained bymultiplying Z00o by the right-hand side of (4).

Zo xg- 1

000 x /1 (f)(4)

The guide wavelength is also obtained by multiplyingthe space wavelength at the same frequency by theright-hand side of (4).

Equation (4) is plotted in Fig. 4.Constant ZOow curves are plotted in Figs. 2 and 3 as

dashed lines. In the example cited above, the impedanceof a guide having b1/a, = 0.5, a2/a1= 0.4, b2/b1= 0. 1, andxc'/ = 2.6 would be 47 ohms at infinite frequency. Atone and one-half times the cutoff frequency, the im-

784 A ugust

1947 Cohn: Properties of

pedance is multiplied by the factor 1.34, found fromFig. 4, which gives Zo=47X1.34=63 ohms.Equation (3) does not take the discontinuity suscept-

ance fully into account, and consequently it is truly ac-curate only if bi/a, is small. In addition to this, it has

|5- S zO " IoX . |

4ti-"C.

'111.

%. u

r<u

%. tj-4-.

*411.)

3

2.

Ridge Wave Guide 785

bilal0.5

For example, if bl/al=0.2, b2/bj=0.3, and a2/ai=0.5,Fig. 2 gives Zo=28 ohms for bl/al=0.136. Therefore,

b2 | o ZOOD-a-

$~4

(*2.~~~~~~~~~~.

4P? I

2

'ONX III I I IX>}1 -P 1 1 TiUI Itt I ITs

0 0.2 0.4 0.6 0.8 1.4a2/e4

Fig. 2-Characteristic impedance and cutoff wavelength ofridge wave guide (bil/a = 0.136).

the same restrictions as (1). Experiments have givenexcellent checks of Fig. 2 (b1/a1=0.136), while for b1/a1=0.5 the impedance for the above example was foundexperimentally to be about 35 to 40 ohms. To obtain a50-ohm impedance, b2/b1 has to be increased from 0.1 toabout 0.133 (see Part III, below). But even for bl/a,=0.5 (3) is a useful approximation, and gives a goodstarting point in design work.

If bl/a, is not equal to 0.136 or 0.5, ZOOO may still be de-termined very closely from Figs. 2 -and 3. For values ofbi/a1 between about zero and one-third, multiply valuesof Zo. on Fig. 2 by the scale factor

bl/al0.136

For values of bl/a, between about one-third and two-thirds, multiply values on Fig. 3 by

Fig. 3-Characteristic impedance and cutoff wavelength ofridge wave guide (bl/a, = 0.5).

zo

zo w

14.0--. T,7 I

0

3.5 ____ ___-_ )/ ; _

:________---I -: Z _3.0 _ I

2.5

X-iL 2.0

.5

1.0

1.021.03t.051.071.101.151.201.30I .401.501.702 00

3.003.504.50

1.5 2.0

5.0761.1743.2812.8112.1012.0251.8081.5661.1291.3411.2361.1551.0921.0601.011.026

2. 5

I

I .0

f/ ffc

Fig. 4-Function relating characteristic impedance andguide wavelength to frequency.

3.0

I 1 .1 ^ F"ll

PROCEEDINGS OF THE I.R.E.

for bl/ai =0.2, Zo =28 XO.2/0.136 =41.1 ohms. Thecharacteristic impedance was checked experimentallyfor a cross section having bl/a,= , and was found to be

'@1.... .1 2'S'tAi 'flF1m 7, i.1#1I.

1.1

fC2 0.

0.2

0.4

0.2

when a2/a, is one-half, fa'l/fC3 is a maximum, and thegreatest separation of the TEjo and TE30 cutoff fre-quencies is obtained. It is easily shown that, for a2/a, -fc3'/fc3 increases as b2/b, decreases, and in the limit ap-proaches 4/3.The even TEmo-mode cutoffs are given by solutions of

the following equation in which the discontinuity sus-ceptance term has been neglected:

02 = tan-' (- n tan 01) (6)5

where n=bi/b2. For the TE20 mode, the 02 root lies be-tween 90 and 180 degrees for 0 <01 . 90 degrees, and the02 root between 0 and 90 degrees for 90 iegrees <., < 180degrees. The cutoff frequency is given by

fc21 01+ 02

fc2 180°(7)

0 0.1 U.Z U.3 0.4 U0. 0.0 U.7

I2

1.

1.

I .

~ca0.

fCs0.

0.

0.

v0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.6 0.9

Fig. 5-Cutoff-frequency ratios for the TE20 and TE30modes in ridge wave guide.

I.¢

very close to the value calculated by the foregoingmethod.The higher roots of (1) give the cutoff frequencies of

all the odd TEmo modes. The TE30 mode is of consider-able interest, since it is usually the lowest mode that can

cause trouble in a transmission system having a sym-

metrical cross section in both E and H directions atevery point including the ends. For 0<01,.90 degrees,choose the root of 02 between 180 and 270 degrees. For90 degrees<0, 180 degrees, choose the root of 02 be-tween 90 and 180 degrees. For 180 degrees<.0<270 de-rees, choose the root of 02 between 0 and 90 degrees.Once a pair of values 0, and 02 have been determined, theratiofC3'/fC3 can be determined from the relation

fe3' 01 + 02

fc3 2700

where fa' and fc3 are the cutoff frequenciesfor theTE30 mode with and without the ridge, respectively.fc3'/fc3 is plotted in Fig. 5 as a function of a2/a, forseveral values of b2/b,, with Bc/ Yol neglected. Note that

This is plotted in Fig. 5 as a function of a2/a, for severalvalues of b2/b,. The maximum value off,2'/f,2 occurs be-tween a2/al,= and , depending upon b2lb,. As b2/b, ismade vanishingly small, the maximum value of fc2'/fc2approaches 3/2 at a,/a2 = 3*

Figs. 2, 3, and 5 show that when a wide frequencyband free from TE20 and TE30 modes is desired, theridge width should be between about " and 2 of the totalguide width.The formulas and curves for a single ridge in a wave

guide are directly applicable to the double-ridge crosssection shown in Fig. 1(b). In this case, the total heightof the guide is 2b, and the total spacing is 2b2. Thus, ifthe width is 23 inches and the height is 4 inch, thenbl/a, = 0.136, and the cutoff curves in Fig. 1 applyexactly. The characteristic-impedance curves apply also,but their values must be doubled. Hence, for a doulble-ridge guide in which bl/al = 0.136, a2/al= 0.35, andb2/b,= 0.2, the relative cutoff wavelength is Xc'/X = 1.9,and the infinite-frequency characteristic impedance isZo0=2X26=52 ohms, by Fig. 2.The attenuation constant in decibels per meter for

copper single-ridge wave guide may be calculated fairlyclosely from the following approximate formula:

a = 6.01(10)-'kv'f + }60)2(tf 2 ~zow

decibels per meter (8)

where a, and bi are in centimeters, and f is in cycles persecond. k is a correction constant a little larger thanunity, which takes account of the more crowded currentdistribution in ridge wave guide than in plain waveguide. If a2/a, is larger than about 3, this term is prob-ably not greater than 1.5.

For double-ridge wave guide, bi should be replaced bythe total guide height, 2b,. If any metal other than cop-per is used, a is proportional to /7u/^.

786 August

Cohn: Properties of Ridge Wave Guide

III. EXPERIMENTAL VERIFICATIONA three-foot length of ridge wave guide having the

cross-sectional dimensions shown in Fig. 6(a) has beentested. For this symmetrical cross section, the design

2b2-0.2252 2b1-0.Cv3'

<4a2.o.g146 l

a .2.36'

(a)

0. 178'+

1.3140'

1.000' ..'.-d 2.840'

(b)Fig. 6-Two experimental cross sections discussee in the text.

method of II is applicable. The parameters are bl/a1=0.136, b2/b1=0.35, and a2/a,=0.40. W0Aithout theridges, the cutoff wavelength would be 2X2.36X2.54= 12.0 centimeters, and the cutoff frequency would be30,000/12.0 = 2500 megacycles. Fig. 2 gives fc/fc' = 1.50and Zo0,/2 = 37 ohms. Therefore, f,' = 1670 megacyclesand ZOOO=74 ohms. Fig. 5 gives approximately fc2'/fc2= 1.10 and fc3'/fc = 1.06. Hence,

fc2' = 2 X 2500 X 1.10 = 5500 megacycles, and

fc3= 3 X 2500 X 1.06 = 7950 megacycles.The calculated and measured cutoff frequencies are

tabulated below.MODE CUTOFFS

MODE

TEjoTE20TE30

CALCULATED1670 megacycles5500 megacycles7950 megacycles

MEASURED1675 megacycles5200 megacycles7900 megacycles.

Ridge wave guide has been used for elements in wide-band junctions between wave guide and coaxial line. Inone type of junction designed for "toll-ticket" wave

guide (a1=2.75 inches, b1=0.375 inch, b1/al=0.436) a

quarter-wavelength section of ridge wave guide is usedas a matching transformer between the 103-ohm guideand the 50-ohm line. The experimental results checkedthe ridge wave guide calculated impedance within a fewper cent. In another type of junction for 3X12-inchwave guide, a tapered length of ridge guide is used tomatch the 50-ohm coaxial line. In this case, the imped-ance calculated for the ridge guide proved less accurate,

because the approximations were less valid with thishigher ratio of bi to a1. The impedance in this caseproved, however, to be about 25 per cent lower than thecalculated value, and hence the impedance curveS inFig. 3, though not very accurate, serve as a valuableguide in the preliminary design of a piece of equipment.In a double-ridge type of junction for 3 X 1'-inch waveguide, the impedance curves checked very well. In thiscase the ratio of b1 to a1 was approximately 0.25.A cross section in 3 X12-inch wave guide that has

been found experimentally to have ZOOO approximatelyequal to 50 ohms is shown in Fig. 6(b). Fig. 3 gives avalue of 65 ohms for bl/a1= 0.5, b2/b1=0.133, and a2/a,=0.352. For the above cross section, the impedancemust be scaled by the factor 0.472/0.500, since b1/al isnot quite 0.5. Therefore, the calculated impedance is61.5 ohms, which is 23 per cent greater than the ap-proximate measured impedance.The paper now under preparation on wave-guide to

coaxial-line junctions will give further details.2

APPENDIXI. THE CUTOFF EQUATION

In the cross section of Fig. l(a), the electromagneticfield at the cutoff frequency may be considered as theresultant of a wave traveling from side to side withoutany longitudinal propagation. As pointed out by S.Ramo and J. R. Whinnery,l such a cross section may betreated at cutoff by assuming it to be an infinitely wide,composite, parallel-strip transmission'line short-cir-cuited at two points. The TE1o-mode cutoff occurs atthe frequency at which this strip transmission line hasits lowest-order resonance. All the other TEmo0 cutoffsoccur at the corresponding m-order resonance fre-quencies. For m odd, the resonance must be of a typegiving an infinite impedance at the center of the crosssection. For m even, this impedance must be zero. Aresonance condition may therefore be set up by settingthe input admittance of half the cross section equal tozero or infinity (Fig. 7). The discontinuity susceptanceBC at the change in height must be included in the cal-culation.

If one examines the equivalent circuit, it is seen thatit is a composite, dissipationless, passive line matchedat both ends, and it is, therefore, matched at everypoint within. Hence, the sum of the admittances acrossx - x must equal zero, and the following relation results:

- Yol cot 01 + Bc + Yo2 tan 02 = 0Bc,

cot 01 , -Y02 Zo1 YolYo1 Z02 tan 02

But in a strip transmission line, the characteristic im-pedance is proportional to the height. Therefore,

Becot 01- -

y=1 (9)b2 tan 02

1947 787

PROCEEDINGS OF THE I.R.E.

which is the cutoff condition for the odd TEmo0 modes(1).For the even modes, the equivalent circuit is shown in

Fig. 7(c).

Binsoi O4I~b24B.n.0'Z0B

|+ 02;-P 1 | c

(a) (b)

Bc

(c)

Fig. 7-Development of the equivalent circuit for ridge wave guide

zo =,

o"

f(f 2

(12)

where fc' is the cutoff frequency of the ridge guide.Since the guide has been assumed thin, the voltage

across the step will be continuous. The voltage distribu-tion in the right half of the cross section will therefore bethe same as that along the shorted composite transmis-sion line shown in Fig. 8.

Since the input impedance is infinite at the open end,the voltage across the guide is a maximum at that point.Transmission-line theory shows that the voltage dis-tribution over the 02 range is given by V= Vo cos 0 from0=0 to 0=02. Over the 01 range it is given by

(01 + 02-0) COS 02V = V1 sin = Vo . sin (01+02 -0),

sin 01 sin 01

In this case, - Yo0 cot 01+B--Yo2 cot 02=0, and hence

cot 01 + - cot 02 = --b2 Yol

(10)

Equation (6) follows readily from this.The discontinuity-susceptance term BC/ Yol is obtain-

able from a paper by J. R. Whinnery and H. W.,Jamieson.6

II. IMPEDANCE EQUATIONIn deriving the impedance equation for ridge wave

guide, it will be assumed that b1/a, is small, so that thediscontinuity susceptance at the edges of the ridge maybe neglected.

If the TE1o mode alone is set up in the wave guide, theE field distribution is the same at all frequencies, includ-ing f =fc and f= oo . The E field can be calculated easilyat the cutoff frequency by the approach used in derivingthe cutoff equation. At f= oo, the wave impedance isthat of free space.7 Hence, if the E field is known, theH field is given by H=E/120r. Both E and H are com-pletely transverse atf= oo, and the current on the top orbottom of the wave guide is completely longitudinal.The current per unit width is equal to the H field inten-sity at the surface of the conductor.The guide impedance at infinite frequency will now

be defined as the ratio of voltage across the center of theguide to the total longitudinal current on the top face

Vo b2Eo 120rb2EoI r al/2 r al/2

2 idx 2 J Edx(11)

The impedance at any other frequency is related toZOOO by the expression

from 0-02 to 0=0"+02.

t

2 bII t

02 - C1Ov .Vo V =VoCOS02

x,@

x=a2/2 xsa& /2

Fig. 8-Approximate voltage distribution acrosshalf of the cross section.

The E field is equal to V/b. Therefore,

E = Eo cos 0, 0 0 -< 02 (13)b2 cos 02

E=-Eo . sin (O1 + 02-0), 02<.0 (01+ 02).bi sin 01

The integral in (11) may be evaluated as follows:I al/2

0O

XIt r (01+02)Edx=- I

2ir JOEdO

B¢' [r@2 b2 cos 02 01+02=Eo-J cos0d0+- sin (01+02-0)dO

27r tJo bi sin 01J2

6.2 b2CS0 1+62\=Eo-{ sin 0 + -c0 cos (01+02-0)

2 7r o b1 sin 0, °2ActI b2 COS 02

=Eo-L{sin 02+- ~~(1-CoOSi)}.27r 8 bi sin 01

Xc'{ b2 01=Eo- sin 02+-COS 02 tan

Substituting this relation in (11) gives

zoo0 =7 For background on this derivation, consult J. C. Slater, "Micro-

wave-Transmission," McGraw-Hill Book Co. New York, N. Y., 1942;chap. 4.

120ir2b2

K'{i+b2 01}'XC'slf 02 + -COS02 tan-b1 ~2)

(14)

788 A ugust