Chapter 5 Cylindrical Cavities and Waveguides

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1 Chapter 5 Cylindrical Cavities and Waveguides We shall consider an electromagnetic field propagating inside a hollow (in the present case cylindrical) conductor. There are no sources inside the conductor, but we shall assume the material is isotropic with electric permittivity, and magnetic permeability,. The speed of the propagating wave is 1/ The direction of propagation will be along the cylindrical axis which is the ẑ direction We shall assume that E(r,t) = E(r)e i t and B(r,t) = B(r)e i t. Maxwell s equations give: [ 2 2 ]E(r.t) =0 2 t [ 2 2 ]B(r.t) =0 2 t [ ]B(r) =0 [ ]E(r) =0 Since the wave is propagating along the ẑ direction we shall further assume that: E = i B(r); B = i E(r) Since the wave is propagating along the ẑ direction we shall further assume that: E(r) = E(x,y)e ±ikz 5.7a B(r) = B(x,y)e ±ikz 5.7b Thus Eq. (5.3 and 5.4) become [ 2 t + 2 k 2 ]E(x,y) =0 5.8b [ 2 t + 2 k 2 ]B(x,y) =0 5.8a where

2 t 2 = 2 x y 2 t = x x + ŷ y The expressions in Eqs. 5.5 and 5.6 then become; Then Thus, Also, E =[ẑ z + t ] [ẑe z + E t ]=i B(r) where E t = E ẑe z =(ẑ E) ẑ B t = B ẑb z =(ẑ B) ẑ E= ẑ z E t ẑ t E z + t E t = i (B t + ẑb z ) B= ẑ z B t ẑ t B z + t B t = i (E t + ẑe z ) ẑ z E t ẑ t E z = i ẑ (ẑ B) z E t + i (ẑ B t )= t E z ẑ ( t E t )=i B z z B t i (ẑ E t )= t B z ẑ ( t B t )= i E z t E t + z E z =0 t B t + z B z = a 5.13b Finally, one can solve for E t and B t if E z and B z are known (and not both are zero). ike t = t E z i (ẑ B t ) ikb t = t B z + i (ẑ E t ) ik(ẑ B t )=(ẑ t B z )+i (ẑ (ẑ E t )) and ike t = t E z (i /ik)(ẑ t B z )+( 2 /ik)( E t ) ( 2 k 2 )E t = ik t E z i (ẑ t B z ) E t = i( 2 k 2 ) 1 [k t E z (ẑ t B z )] likewise B t = i( 2 k 2 ) 1 [k t B z + (ẑ t E z )] 5.14a 5.14b For waves in the opposite direction change k to k.

3 Transverse electromagnetic wave (TEM): E z and B z are zero everywhere inside cylinder. For TEM waves E TEM = E t : t E TEM = 0 t E TEM = 0 k = k 0 = B TEM =± (ẑ E TEM ) 5.15a 5.15b 5.15c 5.15d Unfortunately, the TEM wave is not supported by a single hollow cylindrical conductor (with infinite conductivity). The surface must be an equipotential surface and inside such a conductor, the electric field vanishes. One needs two or more cylindrical surfaces (such as a coaxial cable) to support a TEM wave. Boundary conditions at the surface The existence of surface charge densities,, and surface current densities, K, atthe interface provide the following boundary conditions: n (D D c ) S = n (B B c ) S = 0 n (E E c ) S = 0 n (H H c ) S = K In the conductor the electric field, E c, (and for time varying electric fields B c ) is zero. Thus, inside the hollow cylinder the boundary conditions can only be satisfied at the interface when (n E) S = 0 516a n B S = b That is, the component of the electric field tangent to the interface (E z ) must equal zero at the surface: E z S = The corresponding condition on B z is (see Eq. (5.11)): n B z S = t B z S = 0 Since we can not have both E z and B z equal to zero everywhere inside the cylinder, there are two simple cases which satisfy the boundary conditions:

4 Transverse Magnetic (TM) wave: B z = 0 everywhere and E z S = 0 E t = i( 2 k 2 ) 1 k t E z B t = i( 2 k 2 ) 1 (ẑ t E z ) 0 =[ 2 t + 2 k 2 ]E z (x,y) Transverse Electric (TE) wave: E z = 0 everywhere and B z n S = 0 B t = i( 2 k 2 ) 1 k t B z E t = i( 2 k 2 ) 1 [ (ẑ t B z )] 0 =[ 2 t + 2 k 2 ]B z (x,y) The differential equations (5.18d) for E z and (5.19d) for B z and the boundary conditions (5.18) and (5.19) give rise to eigenvalues of k (dependent on ) for which the propagation is allowed. Since the boundary conditions for E z and B z are different, the eigenvalues are also different. The allowed TE and TM waves (and the TEM wave, if it exists) provide a complete set of waves from which one can construct an arbitrary electromagnetic disturbance in the waveguide or cavity b 5.18c 5.18d b 5.19c 5.19d Modes in a rectangular waveguide: We shall determine the TE modes in a rectangular waveguide with dimensions a inxandb in y (with a > b) as shown in Fig Note that this means E z = 0everywhere(Eis transverse). E t will be found from B z.so, first one must solve Eq. 5.8 for B z : [ 2 t + 2 k 2 ]B z (x,y) =0 The general solution is: [ 2 x y 2 + k k ]B z (x,y) =0 B z (x,y) =C 1 e +ik r t +C 2 e ik r t r t = xx + yŷ where

5 The form for B z (x,y) which is non-zero when x = y = 0is: B z (x,y) =B o cos(k x x)cos(k y y) 5.20 The boundary conditions are n B z S = 0 x B o cos(k x x)cos(k y y) x=0,a = 0 y B o cos(k x x)cos(k y y) y=0,b = 0 These give: sin(k x a)=sin(k x 0)=0or 5.21 k x = m /a; m = 0,1,2,... sin(k y b)=sin(k y 0)=0 or 5.21b k y = n /b; n = 0,1,2, B z (x,y) =B o cos(m x/a)cos(n y/b) ( m a )2 +( n b )2 = 2 k 2 ; m,n = 0,1,2,3... k 2 = 2 ( m a )2 +( n b )2 > 0 > 1 ( m a )2 +( n 1/2 b )2 min = a ; m = 1,n = 0 mn = ( m a )2 +( n b )2 1/2 5.22a 5.22d 5.22e 5.22f For a non-trivial solution, m and n can not both be zero. Equation (5.22d) provides a cutoff on the wave vector, k, sincefork 2 < 0 the factor e +ikz becomes e ±kz and the wave would not propagate. The full solution for each TE mn mode is: B t = i( 2 k 2 ) 1 k t B z e i(kz t) ; B t (m,n) = ik ( m a )2 +( n 1 b )2 Bo,m,n [x (m /a)sin(m x/a)cos(n y/b) + ŷ(n /b)cos(m x/a)sin(n y/b)]e i(kz t) E t = k (ẑ B t) k 2 = 2 The solution for m = 1, n = 0is: ( m a )2 +( n b )2 5.23a

6 There is no propagation for B t = i ka B ox sin( x/a)expi(kz t) ẑb z = ẑb o cos( x/a)expi(kz t) E t = i a B oŷsin( x/a)expi(kz t) k = k 10 = 2 1 ( a )2 < 1 ( a )2 Note the 90 degree phase difference between the B x and B z arising from the i = e i /2 factor. The B t and E t are 180 degrees out of phase. 5.24a 5.24b 5.24c 5.24d Usually one designs the wave guide so that the m = 1, n = 0 mode is the dominant TE mode. One can define the general k mn as follows: k mn = 2 2 mn mn = ( m a )2 +( n b )2 1/2 5.25a 5.25b For each mode, the k mn varies with frequency > mn.the mn is the cutoff frequency for the mode. In Fig. 8.4 from Jackson is a plot of k mn /( ) as a function of, wherek =k mn. It is often convenient to choose the dimensions of the waveguide so that at the operating frequency only the lowest mode can occur. Since the wave number, k mn,isalwayslessthan the "free space value,, the wavelength in the waveguide is always larger than the free space wavelength. For the TM modes:

7 E z = E 0 sin(m x/a)sin(n y/b)e i(kz t) E z S = 0atx = 0,a, andy = 0,b E t = i( 2 k 2 ) 1 k t E z = i( 2 k 2 ) 1 ke 0 [x (m /a)cos(m x/a)sin(n y/b) + ŷ(n /b)sin(m x/a)cos(n y/b)]e i(kz t) B t = k k 2 = 2 (ẑ E t ) ( m a )2 +( n b )2 5.26a 5.26b 5.26c IntheTMmodesifn = 0orifm = 0, E z = 0. Hence E t and B t are also zero. The next possible mode is n = m = 1 with 11 = ( 1 a )2 +( 1 b )2 1/ Thus TE 10 min TM = 11 > min TE = mode has the smallest cutoff frequency. a Higher order modes: The following shows the E t for some TE modes (E z = 0). (Taken from N. Stoyanov, Department of Chemistry, MIT, Ph.D. thesis, 2003)

8 Summary of TE and TM The group velocity is given by: TE modes : E z = 0; B z n S = 0 B z (x,y) =B o cos(m x/a)cos(n y/b)e i(kz t) B t = ik ( m a )2 +( n b )2 E t = k (ẑ B t) k 2 = 2 1 t B z ( m a )2 +( n b )2 TM modes: B z = 0; E z S = 0 E z (x,y) =E o sin(m x/a)sin(n y/b)e i(kz t) E t = ik ( m a )2 +( n b )2 B t = (ẑ E t ) k 1 t E z

9 v g = d dk = 1 = k [k2 + 2 mn ] 1/2 d dk [k2 + 2 mn ] 1/ =[ 2 2 mn ] 1/2 1 1 = [1 mn 2 / 2 ] 1/2 An index of refraction can be determined from: k = n( ) c = [ 2 2 mn ] 1/2 n( ) =c [1 2 mn / 2 ] 1/2 = v c p # where the phase velocity, v p, is given by: v p = k = 1 [1 mn 2 / 2 ] 1/ and v g v p = Energy Flow in the Waveguide for TE Modes The time averaged flux of energy is given by the real part of the following expression S = 1 (E 2 H ) = 1 (E 2 t (B t + ẑb z ) ) = 2 k (ẑ B t) (B t + ẑb z ) = 2 k [ẑb t B t B t B z ] ẑ S = 2 k k 2 ( m a ) 2 +( n b )2 2 tb z t B z = C(,m,n) t B z t B z Now we can integrate this over the cross section of the waveguide to find the power: 5.30

10 P TE = ẑ SdA = C(,m,n) tb z t B z da = C(,m,n) [ t (B z t B z ) B z 2 t B z ]da = C(,m,n)[ B z t B z dr B z 2 t B z da] boundary = C(,m,n)[ B z t B z dr B z t 2 B z da] = C(,m,n)[ B z (x x + ŷ y )B z dr B z 2 t B z da] = C(,m,n)[ B z (dxx B z x +dyŷ B z y ) B z 2 t B z da] = 0 + C(,m,n) ( m a )2 +( n b )2 B zb z da = 2 k ( m a ) 2 +( B zb n z da b )2 = 2 = 2 2 The corresponding value for TM modes is P TM = 2 2 mn ( m a ) 2 +( B zb n z da b )2 1 2 mn / 2 ( m a ) 2 +( B zb n z da 5.31 b ) mn / 2 ( m a ) 2 +( E ze n z da 5.32 b )2 Thus at fixed frequency, the power is inversely proportional to the mn 2 and the smallest mode numbers correspond to maximal power.

11 Resonant Cylindrical Cavities In this case, instead of having a propagating wave along the z direction there is a standing wave with z dependence of the form Csin(p z/d) +Dcos(p z/d) 5.33 k = p /d We assume as before that E(r,t) = E(r)e i t and B(r,t) = B(r)e i t. and using cylindrical coordinates one obtains the following. [ ]E z = 0 [ z ]E z = 0 [ (p /d)2 + 2 ]E z = 0 [ m2 + 2 p ]E z = where the k 2 of the wave guide problem becomes: 2 p = 2 (p /d) From Eq. 5.9 we have: z E t + i (ẑ B t )= t E z Whereas before z E t = ike t without altering the e ikz dependence, now E t and E z will have different sin(p z/d) and cos(p z/d) dependence. TM modes: B z = 0; E t = 0atz = 0,d; E z S = 0 E z (,,z,t) =E o J m ( p )e im cos(p z/d)e i t ; p = 0,1,2.. E t = p d sin(p z/d) 2 t(e o J m ( p )e im ) p B t = i p 2 cos(p z/d)(ẑ t (E o J m ( p )e im )) 5.36

12 The J m ( p )e im are solutions to the wave equation in cylindrical coordinates and E z (,,z,t) =0at = R mn = x mn /R; n = 0,1,2... J m (x mn )=0 For each value of p, there is a resonant frequency, mnp = 1 [( xmn R )2 +( p d )2 ] 1/2 5.37