Adapted from notes by ECE 5317-6351 Prof. Jeffery T. Williams Microwave Engineering
Fall 2019 Prof. David R. Jackson Dept. of ECE
Notes 9 Waveguiding Structures Part 4: Rectangular and Circular Waveguide
1 Rectangular Waveguide
. One of the earliest waveguides. . Still common for high power and low- loss microwave / millimeter-wave applications.
. It is essentially an electromagnetic pipe with a rectangular cross-section.
Single conductor ⇒ No TEM mode
For convenience:
. a ≥ b (the long dimension lies along x).
2 Rectangular Waveguide (cont.)
y Two types of modes: PEC TEz , TMz
221/2 b ε,, µσ kzc=( kk − ) x z a k=ω µεcr = kj0 ε(1 − tan δ d)
σ εε= − j The cutoff wavenumber kc is real. c ω σ We need to solve for kc. =−−εε′jj ′′ ω
=εεcc′ − j ′′ ffk>=−: kk22 cz c εc′′ =εc′ 1 − j 22 εc′ f< fcz: k =−− jk c k =εδcd′ (1 − j tan )
=εε0 rd(1 − j tan δ ) 3 TEz Modes
For +z propagation: y
− jkz z Hzz( xyz,,) = h( xye ,) PEC
where b ε,, µσ ∂∂22 ++2 = x 22kcz h( xy,0) ∂∂xy z a 221/2 kcz≡−( kk)
From previous field table: Subject to B.C.’s:
− j ∂∂EHzz = +ωµ ∂H z Ekxz2 =⇒= kx∂∂ y Ex 00@yb= 0, c ∂y j ∂∂EH Ek=−+zzωµ yz2 ∂H z kyxc ∂∂ E =⇒=00@xa= 0, y ∂x
4 TEz Modes (cont.)
∂∂22 +=−2 22hz( xy,,) kcz h( xy) (eigenvalue problem) ∂∂xy
Using separation of variables, let hz ( xy, ) = X( xY) ( y)
d22 X dY ⇒Y +=− X k2 XY dx22 dy c Must be a constant
This is the 22 11d X dY 2 (If we take one term across the equal sign, we “separation ⇒22 +=−kc have a function of x equal to a function of y.) equation”. X dx Y dy
11d22 X dY ⇒=−kk22and =− X dx22xyY dy
22 2 where kkkxyc+= “separation equation” 5 TEz Modes (cont.) Hence, Xx() Yy() hxyz( ,) =++ (cos A kxB xx sin kxC )(cos kyD y sin ky y )
∂h z = 0 @yb= 0, A ∂y Boundary Conditions: ∂h z = 0 @xa= 0, B ∂x nπ A ⇒=D0 and kn = =0,1,2,... y b mπ B ⇒=B0 and km= =0,1,2,... x a
22 mxπ ny π 2 mππ n hz ( xy,) = Amn cos cos and kc = + a b ab
6 TEz Modes (cont.)
Therefore, 1/2 k= kk22 − ππ zc( ) mn − jkz z Hz ( xyz, ,) = Amn cos x cos y e 221/2 mnππ ab =−−2 k ab
22 From the field table, we obtain the following: mnππ kc = + ab
jnωµ π m π n π − = jkz z Ex 2 Amn cos x sin ye Note: kbc a b m = 0,1,2, jmωµ π m π n π − E= − Asin x cos yejkz z y 2 mn n = 0,1,2, kac a b
jk mπ mnππ − = z jkz z But m = n = 0 Hx 2 Amn sin x cos ye kac a b is not allowed!
jk nπ mnππ − = z jkz z Hy 2 Amn cos x sin ye (non-physical solution) kbc a b − jkz H= zAˆ 00 e;0 ∇⋅ H ≠
7 TEz Modes (cont.)
Reason for non-physical solution
Start with the vector wave equation:
2 ∇×( ∇×H) − kH =0 Vector wave equation: from Maxwell’s equations.
Take divergence of both sides.
2 ∇⋅( ∇×( ∇×HkH)) − ∇⋅ =0 The divergence of a curl is zero.
∇⋅H =0 Magnetic Gauss law
8 TEz Modes (cont.)
Reason for non-physical solution
Revisit how we obtained the vector Helmholtz equation:
2 ∇×( ∇×H) − kH =0 Vector wave equation: from Maxwell’s equations.
22 ∇( ∇⋅H) −∇ H − kH =0 From definition of vector Laplacian ∇×∇×H
Now use:
∇⋅H =0 Magnetic Gauss law A needed assumption!
22 ∇+H kH =0 Vector Helmholtz equation (what we have solved) 9 TEz Modes (cont.)
Reason for non-physical solution
Vector wave equation ⇒ magnetic Gauss law
Vector Helmholtz equation ⇒ magnetic Gauss law
The vector Helmholtz equation does not guarantee that the magnetic Gauss law is satisfied. In the mathematical derivation, we need to assume the magnetic Gauss law in order to arrive at the vector Helmholtz equation.
All of the modes that we get by solving the Helmholtz equation should be checked to make sure that they do satisfy the magnetic Gauss law.
Note: The TE00 mode is the only one that violates the magnetic Gauss law.
10 TEz Modes (cont.)
Lossless case (εc = εε = ′)
221/2 2 1/2 mnππ kmn =−=−− kk22mn k zc( ( ) ) ab
⇒ = mn mn TEmn mode is at cutoff when kkc (k = ωc µε )
22 mn 1 mn fc = + 2 µε ab
Lowest cutoff frequency is for TE10 mode (a > b)
We will 10 1 fc = Dominant TE mode revisit this 2a µε (lowest f ) mode later. c
11 TEz Modes (cont.)
At the cutoff frequency of the TE10 mode (lossless waveguide):
cc c λ =dd = = d = d 10 2a ffc 1 2a µε
so
a = λd /2 ff= c
12 TEz Modes (cont.)
To have propagation:
ff> c
so 1 f > 2a µε or 11c λ a >==dd 2 f µε 22f
or λ a > d 2
Example: Air-filled waveguide, f = 10 GHz. We have that a > 3.0 cm / 2 = 1.5 cm.
13 TMz Modes
Recall: y PEC − jkz z Ezz( xyz,,) = e( xye ,) b ε,, µσ x where z a ∂∂22 +=−2 22ez( xy,,) kecz( xy) (eigenvalue problem) ∂∂xy 221/2 kcz=( kk − )
Subject to B.C.’s: E z = 0 @xa= 0,
@yb= 0,
Thus, following same procedure as before, we have the following result:
14 TMz Modes (cont.) Xx() Yy() exyz( ,) =++ (cos A kxB xx sin kxC )(cos kyD y sin ky y )
Boundary Conditions: ez = 0 @yb= 0, A
@xa= 0, B
nπ A ⇒=C0 and kn= =0,1,2,... y b mπ B ⇒=A0 and kmx = =0,1,2,... a
22 mπ n π 2 mnππ eBz= mn sin x sin y and kc = + a b ab
15 TMz Modes (cont.)
1/2 =22 − Therefore kzc( kk) 221/2 mnππ − mnππ jkz z =−−k 2 Ez ( xyz, ,) = Bmn sin x sin y e ab ab
22 mnππ kc = + From the field table, we obtain the following: ab
jnωεc π mnππ − jk z = = z m 1, 2, 3 Hx 2 Bmn sin x cos ye kbc a b n =1, 2, 3
jmωε π mnππ − = − c jkz z Hy 2 Bmn cos x sin ye kac a b Note: If either m or n is zero, the jk mπ mnππ − = − z jkz z field becomes a trivial one in Ex 2 Bmn cos x sin ye kac a b the TMz case.
jk nπ mnππ − = z jkz z Ey 2 Bmn sin x cos ye kbc a b
16 TMz Modes (cont.)
′ Lossless case (εc = εε = ) (same as for 22 mn 22mn 2 mnππ TE modes) kzc=−=−− kk( ) k ab
22 mn 1 mn fc = + 2 µε ab
The lowest cutoff frequency is obtained for the TM11 mode
22 11 111 fc = + 2 µε ab Dominant TM mode (lowest fc)
17 Mode Chart
′ y Lossless case (εc = εε = ) PEC Two cases are considered: b ε,, µσ b < a / 2 x Single mode operation z a f TE10 TE20 TE01 TE11 TM11 The maximum bandwidth for single-mode operation is 67%. ff21− BW ≡ ba≤ /2 f0 ( )
b > a / 2 f0 =center frequency Single mode operation
f 22 TE TE TE mn 1 mn 10 01 20 TE fc = + 11 2 µε ab TM11 18 Dominant Mode: TE10 Mode For this mode we have y
10 π PEC mn=1, = 0, k = c a
b ε,, µσ Hence we have x π − jkz z Hz = A10 cos xe z a a 1/2 2 ka π − 10 2 π z jkz z kk= = k − Hx = j A10 sin xe zz π a a ωµ π π ja − jkz z − jkz z Ey = − A10 sin xe Ey = E10 sin xe π a a E10 −π AE≡ 10 jaωµ 10 EEHxz= = y = 0
19 Dominant Mode: TE10 Mode (cont.)
The fields can be put in terms of E : 10 y PEC π − jkz z Ey = E10 sin xe a b ε,, µσ x 1 π − H= − Esin xejkz z x 10 a ZaTE z
−ππ − jkz z Hz = E10 cos xe 2 1/2 jaωµ a π kk=10 = k 2 − zz a ωµ Z = EEHxz= = y = 0 TE kz
20 Dispersion Diagram for TE10 Mode
Lossless case (εc = εε = ′) 2 π kk=β =2 − > z ω ffc a k 22= ω µε vg = slope
ω10 1 c (TEM mode, or “Light line”) µε z β=k = ω µε
vp = slope
β
ω Phase velocity: v = p β Velocities are slopes on dω the dispersion plot. Group velocity: v = g dβ 21 Field Plots for TE10 Mode y PEC Top view
b ε,, µσ x z a
x E z a H y y
b b x a z End view Side view 22 Field Plots for TE10 Mode (cont.) y
Top view PEC
b ε,, µσ x z a x z a J s H y y b b x a z End view Side view
Note: One can cut a narrow z-directed slot in the center of the top wall without disturbing the current. 23 Power Flow for TE10 Mode Time-average power flow in the z direction for +z mode:
ab Note: + 1 * P10 =Re (E ×⋅ H) zˆ dydx ab ∫∫ π 2 00 2 x ab ∫∫sin dydx = ab a 2 1 00 = − * Re ∫∫ Eyx H dydx 2 00 π − E= Esin xejkz z y 10 1 ab2 kz −2αz a = ReEe10 1 π − 22 ωµ jkz z Hx = − E10 sin xe ZaTE ωµ Simplifying, we have ZTE = kz +−ab 2 2αz P10 = Re{kEz } 10 e 4ωµ
At breakdown: Note: EE= For a given maximum electric field level (e.g., the 10 c breakdown field), the power is increased by increasing the cross-sectional area (ab). 24 Dielectric Attenuation for TE10 Mode
From Notes 7 we have:
ff> c
22 kzd=−=−βα j kk c
22 β =Re kk − c
22 αdc=−−Im kk
22 β≈−kk0 µεrr c k 2µεtan δ α ≈ 0 rr d d 2β
π k=−= k′ jk ′′ k µε1 −j tan δ k = 0 rr d c a
25 Conductor Attenuation for TE10 Mode
PP= + P (0) 0 10 z=0 (calculated on previous slide) Recall α = l c y 2P0
Rs
R 2 = s b Lossless Pls(0) ∫ Jd 2 C x z a Js = nHˆ × on conductor
CC=+left C right ++ C bot C top
26 Conductor Attenuation for TE10 Mode
y
Side walls Rs
− =left =×=−=−ˆ ˆˆjkz z @x 0: Jsz x Hx=0 yH yA10 e Lossless − b @:x= a Jright =−× xˆ H = yH ˆˆ =− yA e jkz z szxa= 10 x z a
Hence: π − jkz z Hz = A10 cos xe a left right − jkz z Jsy= J sy = − Ae10 π kaz − jkz z Hx = j A10 sin xe π a
27 Conductor Attenuation for TE10 Mode (cont.)
Top and bottom walls
bot y @y= 0: J = yHˆ × s y =0 Rs @:y= b Jtop =−× yHˆ s yb= JJtop= − bot ss b Lossless (The fields of this mode are independent of y.) x z a Hence:
bot π − jk z z π − J= Acos xe jkz z sx 10 Hz = A10 cos xe a a π π bot kaz − jkz z kaz − jkz z Jsz = − j A10 sin xe Hx = j A10 sin xe π a π a
28 Conductor Attenuation for TE10 Mode (cont.) y
Rs We then have:
ba b Lossless RR22 = ssleft + bot Plss(0) 2∫∫J dy J dx x 2200 ba 2 22 z a =R Jleft dy ++ R Jbot J bot dx s∫∫ sy s( sx sz ) 00 ba22 2 ππkaz =Rss − A10 dy + R A10 cos x+− j A10 sin x dx ∫∫π 00aa ba 2 a 2 22ππkaz =++Rs A10 dycos x dx sin x dx ∫∫aaπ ∫ 00 0 2 2 2 aaka =RA b ++z s 10 2 22π
29 Attenuation for TE10 Mode (cont.)
Assume f > fc y R k ≈ β (The wavenumber is taken as that s z of a guide with perfect walls.)
b Lossless 23 2 aaβ Pls(0) = RA10 b ++ 2 x 22π z a −π AE= ab 2 10 jaωµ 10 PE0 = β 10 4ωµ 2 22 10 π Simplify, using β =kk − kc = c a
P (0) α = l c Final result: 2P0 Rs 2 32 απc = (2b+ ak ) [np/m] ab3 βη( k )
30 Attenuation for TE10 Mode (cont.) y R Final Formulas s
b Lossless Two alternative forms for the x final result: z a
Rs 2 32 απc = (2b+ ak ) [np/m] ab3 βη( k )
2 Rfsc12b αc = 1+ [np/m] bη 2 af 1/− ( ffc )
31 Attenuation for TE10 Mode (cont.)
Brass X-band air-filled waveguide (σ ≈×2.6 107 [S/m])
Xband : 8− 12 [GHz] (See the table on the next slide.)
a = 2.0 cm
(from the Pozar book)
32 Attenuation for TE10 Mode (cont.)
Microwave Frequency Bands Letter Designation Frequency range L band 1 to 2 GHz S band 2 to 4 GHz C band 4 to 8 GHz X band 8 to 12 GHz Ku band 12 to 18 GHz K band 18 to 26.5 GHz Ka band 26.5 to 40 GHz Q band 33 to 50 GHz U band 40 to 60 GHz V band 50 to 75 GHz E band 60 to 90 GHz W band 75 to 110 GHz F band 90 to 140 GHz D band 110 to 170 GHz
(from Wikipedia) 33 Modes in an X-Band Waveguide
a = 2.29cm (0.90in) “Standard X-band waveguide” (WR90) b =1.02cm (0.40in)
Xband : 8− 12 [ GHz ] Mode fc [GHz]
TE10 6.55 1"
TE20 13.10 b TE01 14.71 0.5" a TE11 16.10
TM11 16.10 50 mil (0.05”) thickness TE30 19.65
TE21 19.69
TM21 19.69 34 Example: X-Band Waveguide
a = 2.29cm Determine β, α, and λg (as appropriate) at 10 GHz and 6 GHz for the TE mode in a b = 1.02cm 10 εµ00, lossless air-filled X-band waveguide.
@ 10 GHz
222 ππ2 1010 π β= ω2 µε −= − 8 a 2.99792458× 10 0.0229 β =158.25 [rad/m]
22ππ λ = = = 0.0397 g β 158.25
λg = 3.97 [cm]
222π π πωf Lossless :2kf= = = = =ω µε = π µε λddcf/ c d c d 35 Example: X-Band Waveguide (cont.)
@6 GHz
1/2 1/2 229 2 2 ππ2 6× 10 π kz =−=ω µε − × 8 a 2.99792458 10 0.0229
2 2 ππ2 6× 109 =−− j 8 0.0229 2.99792458× 10 = − j 55.04 [1/m]
α = 55.04 [np/m] = 478.08 [dB/m] 2π λ = g β
Evanescent mode: β = 0; λg is not defined!
36 Fields of a Guided Wave
Fields Equations in Cylindrical Coordinates
These are useful for a circular waveguide. j ωε ∂∂EH = c zz Hkρ 2 z kc ρφ∂∂ ρ These equations give the transverse field components in terms of the − j ∂∂E kH longitudinal components, Ez and Hz. = ωε z± zz Hφ 2 c kyc ∂∂ρρ Fz( ) = e jkz z
− j ∂∂EHzzωµ Ek=±+ 22 ρ 2 z k = ω µε kc ∂∂ρ ρφ c
22 kcz= kk − j kE∂∂ H = zz+ωµ z Eφ 2 kc ρφ∂∂ ρ
37 Circular Waveguide
a ε,, µσ TMz mode: 22 ∇=−ez(ρφ,,) kecz( ρφ)
PEC (eigenvalue problem)
2 22 z kkkzc= −
This means any combination of The solution in cylindrical coordinates is: these two functions.
Jknc()ρ sin(nφ ) ez (ρφ, ) = Yknc(ρφ ) cos( n )
Note: The value n must be an integer to have unique fields.
38 References for Bessel Functions
Jxn ()= Bessel function of the first kind of order n
Yxn ()= Bessel function of the second kind of order n
References:
. M. R. Spiegel, Schaum’s Outline Mathematical Handbook, McGraw-Hill, 1968.
. M. Abramowitz and I. E. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Government Printing Office, Tenth Printing, 1972.
. N. N. Lebedev, Special Functions & Their Applications, Dover Publications, New York, 1972.
39 Plot of Bessel Functions
1 1
0.8 n = 0 J n (0) is finite 0.6 n = 1 0.4 J (xJ0)(x) n = 2 n J1(x) 0.2 Jn(2, x)
0
0.2
0.4
− 0.403 0.6 0 1 2 3 4 5 6 7 8 9 10 0 x 10 x
2 nππ n 1 Jxn ( ) ~ cos x− −, x →∞ Jxn ( ) ~ x n= 0,1,2,...., x→ 0 π x 24 2!n n
40 Plot of Bessel Functions (cont.)
1 0.521 n = 0 0 n = 1 1 n = 2
2
Y0(x) Yn (x) Yn (0) is infinite Y1(x) 3 Yn(2, x)
4
5
6
− 6.206 7 0 1 2 3 4 5 6 7 8 9 10 0 x x 10 2 x Yx0 ( ) ~ ln+=γγ , 0.5772156,x → 0 π 2 2 nππ n Yxn ( ) ~ sin x− −, x →∞ π x 24 12 Yxn ( ) ~−− ( n 1)! , n = 1,2,3,....., x → 0 π x 41 Circular Waveguide (cont.)
Choose (somewhat arbitrarily) cos(nφ )
Jknc()ρ enz (ρφ, ) = cos(φ ) Yknc()ρ
The field should be finite on the z axis.
Yknc()ρ is not allowed
Hence, we have ez(ρφ,) = cos( n φ ) Jknc ( ρ )
− jkz z Ez(ρφ, , z) = cos( n φ ) Jknc ( ρ ) e
42 Circular Waveguide (cont.)
B.C.’s: Eaz ( ,,φ z) = 0 Hence Jnc( ka )0=
Sketch for a typical value of n (n ≠ 0).
Jxn () Note: Pozar uses the notation pmn.
xn3 x
xn1 xn2
x ka= x k = np c np c a
Note: The value xn0 = 0 is not included ρ (This is true unless n = 0, in which Jxnn0 = J n(00) = since this would yield a trivial solution: a case we cannot have p = 0.)
43 Circular Waveguide (cont.)
TMnp mode:
ρ − jkz z Ez (ρφ, , z) = cos( nJx φ ) n np e n= 0,1, 2 a
2 2 xnp kkz =−= p1,2,3,...... a
44 Cutoff Frequency: TMz
2 22 kkkzc= − Assume k is real here.
xnp At f = fc : kz = 0 kk= = c a x 2πf TM µε = np c a
TM c c fx= d c ≡ c np d ε 2π a r
45 Cutoff Frequency: TMz (cont.)
xnp values p \ n 0 1 2 3 4 5
1 2.405 3.832 5.136 6.380 7.588 8.771
2 5.520 7.016 8.417 9.761 11.065 12.339
3 8.654 10.173 11.620 13.015 14.372
4 11.792 13.324 14.796
TM01, TM11, TM21, TM02, ……..
46 TEz Modes
Proceeding as before, we now have that
− jkz z Hz(ρφ, , z) = cos( n φ ) Jknc ( ρ ) e
Set Eaφ ( ,,φ z) = 0 H ρ = 0 ( ρ =a ) ∂H ∂ 1 ρ H z (From Ampere’s law) Eφ = − jzωεc ∂∂ ρ
∂H ⇒=z 0 ∂ρ ρ=a
′ The prime denotes derivative Hence Jnc( ka )0= with respect to the argument.
47 TEz Modes (cont.)
Jnc′( ka )0=
Jxn′() Sketch for a typical value of n (n ≠ 1).
xn′3 x ′ xn1 xn′ 2
kac= x np′
xnp′ kpc = =1,2,3,..... We don’t need to consider p = 0; a this is explained on the next slide.
48 TEz Modes (cont.)
ρ − jkz z Hz (ρφ, , z) = cos( nJx φ ) n np′ e p= 1, 2, a
Note: If p = 0, then xnp′ = 0
We then have, for p = 0: ρ ′ n ≠ 0 Jxn np = Jn (00) = (trivial solution) a ρ ′ n = 0 Jx00np = J(01) = a
−−jkzz z jk z − jkz ⇒Hz = e ⇒= H zeˆˆ ⇒= H ze (nonphysical solution) (violates the magnetic Gauss law)
The TE00 mode is not physical. 49 Circular Waveguide (cont.)
TEnp mode:
ρ − ′ jkz z Hz (ρφ, , z) = cos( nJx φ ) n np e n= 0,1, 2 a
′ 2 2 xnp kkz =−= p1,2,3,...... a
50 Cutoff Frequency: TEz
2 22 kkkzc= − Assume k is real here.
x′ k = 0 kk= = np z c a
x′ 2πf TE µε = np c a
Hence TE c c fx= d ′ c = c np d ε 2π a r
51 Cutoff Frequency: TEz
x´np values p \ n 0 1 2 3 4 5
1 3.832 1.841 3.054 4.201 5.317 5.416
2 7.016 5.331 6.706 8.015 9.282 10.520
3 10.173 8.536 9.969 11.346 12.682 13.987
4 13.324 11.706 13.170
TE11, TE21, TE01, TE31, ……..
52 TE11 Mode
The dominant mode of circular waveguide is the TE11 mode.
Electric field Magnetic field
(From Wikipedia)
TE10 mode of TE11 mode of rectangular waveguide circular waveguide
The TE11 mode can be thought of as an evolution of the TE10 mode of rectangular waveguide as the boundary changes shape.
53 TE11 Mode (cont.)
The attenuation due to conductor loss for the TE11 mode is:
2 Rfsc11 αc = + 2 2 aη fx11′ −1 1/− ( ffc )
x11′ = 1.841 x′ k = 11 c a
The derivation is in the Pozar book (see Eq. 3.133).
54 TE01 Mode
The TE01 mode of circular waveguide has the unusual property that the conductor attenuation decreases with frequency. (With most waveguide modes, the conductor attenuation increases with frequency.)
2 R ( ff/ ) α TE01 = s c c aη 2 1/− ( ffc )
Reason: This mode has current only in the φ direction, and this component of current (corresponding to Hz) decreases as the frequency increases (for a fixed power flow down the guide, i.e., a fixed Eφ). (Please see the equations on the next slide.)
The TE01 mode was studied extensively as a candidate for long-range communications – but eventually fiber-optic cables became available with even lower loss. It is still useful for some high-power applications.
Note: This mode is not the dominant mode! 55 TE01 Mode (cont.)
The fields of the TE01 mode are:
ρ − ′ jkz z Hz = Jx0 01 e a
x′ 1 ρ − = ωµ 01 ′′ jkz z Eφ j2 Jx0 01 e akc a (0,1) Hρφ= − EZ/ TE
ωµ (0,1) = ZTE (0,1) kz
56 TE01 Mode (cont.) P (0) TE TM l 11 01 TE21 αc = 2P0 αc TM11
Note: P0 = 0 at cutoff
TE01
f
TE11 TM01 TE21 TE01 fc fc fc fc
Note: The attenuation increases at high frequency for all other modes, due to Rs. 57 TE01 Mode (cont.)
Practical Note:
The TE01 mode has only an azimuthal (φ - directed) surface current on the wall of the waveguide. Therefore, it can be supported by a set of conducting rings, while the lower modes (TE11 ,TM01, TE21, TM11) will not propagate on such a structure.
(A helical spring will also work fine.)
E TE mode: 01 H
58 TE01 Mode (cont.) VertexRSI's Torrance Facility is a leading supplier of antenna feed components for the various commercial and military bands. A patented circular polarized 4-port diplexer meeting all Intelsat specifications leads a full array of products.
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59 TE01 Mode (cont.)
From the beginning, the most obvious application of waveguides had been as a communications medium. It had been determined by both Schelkunoff and Mead, independently, in July 1933, that an axially symmetric electric wave (TE01) in circular waveguide would have an attenuation factor that decreased with increasing frequency [44]. This unique characteristic was believed to offer a great potential for wide-band, multichannel systems, and for many years to come the development of such a system was a major focus of work within the waveguide group at BTL. It is important to note, however, that the use of waveguide as a long transmission line never did prove to be practical, and Southworth eventually began to realize that the role of waveguide would be somewhat different than originally expected. In a memorandum dated October 23, 1939, he concluded that microwave radio with highly directive antennas was to be preferred to long transmission lines. "Thus," he wrote, “we come to the conclusion that the hollow, cylindrical conductor is to be valued primarily as a new circuit element, but not yet as a new type of toll cable” [45]. It was as a circuit element in military radar that waveguide technology was to find its first major application and to receive an enormous stimulus to both practical and theoretical advance.
K. S. Packard, “The origins of waveguide: A case of multiple rediscovery,” IEEE Trans. Microwave Theory and Techniques, pp. 961-969, Sept. 1984.
60 TE01 Mode (cont.)
“In a memorandum dated October 23, 1939, he concluded that microwave radio with highly directive antennas was to be preferred to long transmission lines.”
Recall the comparison of dB attenuation:
Waveguiding system: dB= 8.686(αz)
λ Wireless system: 0 dB=− 10log10 (GGtr) −+ 20log10 20log10 (r) 4π
61