Waveguides Free Space. Modal Excitation. Daniel S. Weile. Department of Electrical and Computer Engineering University of Delaware

Transcription

1 Modl Excittion Dniel S. Weile Deprtment of Electricl nd Computer Engineering University of Delwre ELEG 648 Modl Excittion in Crtesin Coordintes

2 Outline 1 Aperture Excittion Current Excittion

3 Outline 1 Aperture Excittion Current Excittion 2 Apertures in Ground Plne Plne Current Sheets

4 Outline Aperture Excittion Current Excittion 1 Aperture Excittion Current Excittion 2 Apertures in Ground Plne Plne Current Sheets

5 Excittion of Aperture Excittion Current Excittion Our discussion so fr hs ignored how energy might be coupled into guides. After ll mode is by definition solution tht cn exist without source. How re modes excited? Let us consider our stndrd guide. Over the z = plne, let us ssume E x = E y = f (x, y) We further ssume the wveguide is mtched somewhere for z > nd seek the solution for the modes lunched.

6 Modl Expnsion Aperture Excittion Current Excittion Since E x = we look for TE x solutions. We cn thus write Modl Expnsion for ψ ψ = m=1 n= A mn sin mπx To find the A mn we must impose f (x, y) = E y z= = m=1 n= How do we find the coefficients? nπy cos b e γmnz γ mn A mn sin mπx nπy cos b

7 Aside: Useful Nottion Aperture Excittion Current Excittion Two functions tht often rise in mthemticl physics re the Kronecker Delt δ mn = { 1 if m = n otherwise nd the ever present nd strngely nmed Neumnn s Number ɛ n = { 1 if n = 2 if n

8 Aside: Orthogonl Functions Aperture Excittion Current Excittion Note the Orthogonlity Reltions sin mπx cos mπx sin mπx Know these! nπx sin dx = 2 δ mn for m, n = 1, 2,... nπx cos dx = δ mn for m, n =, 1, 2,... ɛ m nπx cos dx = for m, n =, 1, 2,...

9 Finding the Coefficients Aperture Excittion Current Excittion To find the coefficients, we multiply nd integrte: = b Thus m=1 n= f (x, y) sin m πx γ mn A mn = b m=1 n= A mn = 2ɛ n bγ mn b cos n πy b dxdy sin mπx γ mn A mn b 2ɛ n nπy cos b sin m πx f (x, y) sin mπx cos n πy b dxdy δ mm δ nn = γ m n A b m n 2ɛ n nπy cos b dxdy

10 Observtions Aperture Excittion Current Excittion A similr procedure cn be followed if the mgnetic field is given. Excittion by field with both x- nd y-components cn be solved by superposition. Note tht A finite number of modes propgtes. An infinite number of modes evnesces. How cn we excite given mode? Cn we void exciting some modes entirely? How?

11 Power Flow Aperture Excittion Current Excittion Becuse our modes re TE x, we cn compute the power flow in the guide using Plugging in P = dx b dy P = [ m=1 n= b p=1 q= [E y H x ] z= dxdy. E mn sin mπx ] nπy cos b k 2 ( pπ ) 2 jωµγpq Epq sin pπx cos qπy b Why do we use different vribles in the two summtions?

12 Power Flow Aperture Excittion Current Excittion Thus P = m=1 n= p=1 q= = k 2 ( pπ ) 2 jωµγpq b m=1 n= dy cos nπy b E pqe mn qπy cos b b k 2 ( mπ ) 2 2ɛ n jωµγmn = where (Y ) mn is the modl dmittnce. Thus, ech mode crries its own power. E mn 2 m=1 n= dx sin mπx pπx sin b 2ɛ n (Y ) mn E mn 2

13 Wveguide Junction Aperture Excittion Current Excittion Consider the junction between the wveguides below. Their width is. y c b z We ssume the field in the junction is { sin πx E y z= y < c y > c

14 Wveguide Junction Aperture Excittion Current Excittion Now, we hve seen tht the field in the junction is given by f (x, y) = E y z= = m=1 n= γ mn A mn sin mπx nd tht the coefficients cn be computed from γ mn A mn = E mn = 2ɛ n b b f (x, y) sin mπx Immeditely, we see the only excited modes hve by orthogonlity. m = 1 nπy cos b nπy cos b dxdy

15 Computing the Coefficients Aperture Excittion Current Excittion Plugging in we find nd E 1 = 2 c sin 2 πx b dxdy = c b, E 1n = 4 c sin 2 πx nπy cos b b dxdy = 2 nπc sin nπ b. Most of these modes re of course evnescent.

16 Power Flow nd Admittnce Aperture Excittion Current Excittion Applying our power formul P = c2 [ ( 2b (Y ) sin nπ )] 2 (Y ) b 1n nπc n=1 We cn lso define n perture dmittnce by defining the perture voltge s the line integrl of E cross the center of the perture. Then Y = 2P V 2 b

17 Outline Aperture Excittion Current Excittion 1 Aperture Excittion Current Excittion 2 Apertures in Ground Plne Plne Current Sheets

18 Aperture Excittion Current Excittion Current Excittion of In perture excittion, we mtch mode series to given field in the junction. In current excittion, we re given current. The electric field is continuous t the current. The mgnetic field is discontinuous t the current. Note tht in this wy, perture excittion cn be thought of s mgnetic current excittion. Suppose for simplicity we ssume current sheet in the z = plne with J s = u x f (x, y) nd tht the guide is mtched in both directions. We cn ssume the wves generted re TM x. (Why?)

19 Aperture Excittion Current Excittion Current Excittion of The most generl solution for TM wves cn be written s ψ + = ψ = m= n=1 m= n=1 B + mn cos mπx B mn cos mπx nπy sin b e γmnz z > nπy sin b e+γmnz z < Continuity of the tngentil E-field (which is proportionl to ψ nd its trnsverse derivtives) requires B mn = B + mn = B mn

20 Aperture Excittion Current Excittion Current Excittion of The mgnetic field hs to be discontinuous by the current, i.e. Thus, u z ( H + H ) = J s f (x, y) = [ H y H y + ]z= = m= n=1 2γ mn B mn cos mπx nπy sin b nd we hve the Field Coefficients 2γ mn B mn = J mn = 2ɛ m b dx b dy f (x, y) cos mπx nπy sin b

21 Power Flow Aperture Excittion Current Excittion The power supplied cn be computed from wht we know of the Poynting Theorem: P = dx b dy E x Jx z= After substituting the Fourier series for E x nd J x nd ppeling to orthogonlity, we find n expression for Power Flow P = m= n=1 b 4ɛ m (Z ) mn J mn 2

22 An Exmple Aperture Excittion Current Excittion Consider coxilly fed wveguide with the probe ffixed to the opposite wll: y c b x The probe cts s n ntenn; by imge theory, we expect the current to be pproximtely Probe Current J x = I cos[k( x)]δ(y c)

23 An Exmple Aperture Excittion Current Excittion J mn = 2ɛ m b dx b = 2I ɛ m b dy f (x, y) cos mπx nπc sin b nπy sin b dx cos[k( x)] cos mπx = 2I ɛ m k sin k sin nπc b b[(k) 2 (mπ) 2 ] The impednce seen by the probe cn be computed in the usul wy: where I i is the feed current Z i = 2P I i 2 I i = I cos k

24 Outline Apertures in Ground Plne Plne Current Sheets 1 Aperture Excittion Current Excittion 2 Apertures in Ground Plne Plne Current Sheets

25 Hve We Seen This Before? Apertures in Ground Plne Plne Current Sheets We lredy know one wy to compute the field rdited from n perture, but consider: All methods must give the sme result becuse of uniqueness. This method is spectrl method. Spectrl methods converge fstest when sptil methods converge slowest. Why?

26 Hve We Seen This Before? Apertures in Ground Plne Plne Current Sheets We lredy know one wy to compute the field rdited from n perture, but consider: All methods must give the sme result becuse of uniqueness. This method is spectrl method. Spectrl methods converge fstest when sptil methods converge slowest. Why? This method elucidtes different physics. Thus, suppose we re given the perture field E x (x,, z) nd we wnt the field for y >. How should we proceed?

27 Modl Expnsion Apertures in Ground Plne Plne Current Sheets As before, we cn expnd the field in modes. Now, however, we hve continuum of modes, rther thn discrete set. Therefore, we choose the Wve Function ψ(x, y, z) = 1 4π 2 dk x dk z f (k x, k z )e jkx x e jky y e jkzz Note tht this is subject s usul to the usul seprtion or dispersion reltion k 2 x + k 2 y + k 2 z = k 2 Our unknown function is the Fourier Trnsform of ψ: ψ(k x, k z ) = f (k x, k z )e jky y

28 Modl Expnsion Apertures in Ground Plne Plne Current Sheets Given E x in constnt y plne, we choose TE z modes. From before, these modes re relted to the wve function with E x = ψ y H x = 1 E y = ψ x H y = 1 E z = H z = 1 jωµ We cn esily compute the 2 ψ jωµ x z 2 ψ jωµ y z ( 2 z 2 + k 2 ) ψ Spectrl Domin TE z Equtions Ẽ x = jk y ψ Ẽ y = jk x ψ Ẽ z = Hx = Hy = kx kz jωµ ky kz jωµ Hz = k 2 k 2 z jωµ ψ ψ ψ

29 The Solution Apertures in Ground Plne Plne Current Sheets Since we know the electric field in the perture, we cn write Ẽ x (k x, k z ) = jk y f (k x, k z ) y= or f (k x, k z ) = 1 Ẽ x (k x, k z ) jk y y= Of course, cre must be tken in the computtion of k y : { j kx k y = 2 + kz 2 k 2 for k < kx 2 + kz 2 k 2 kx 2 kz 2 for k > kx 2 + kz 2 How re the signs chosen? Wht does this sy bout determining the current fr wy from it?

30 An Exmple: The Infinite Slot Apertures in Ground Plne Plne Current Sheets To simplify the exmple, we use 2D problem independent of z. Consider the infinite slot shown below: x y If we consider the TEM mode, H is z directed everywhere. We therefore use H z s our wve function: H z = 1 2π f (k x )e jkx x e jky y dk x

31 The Infinite Slot Apertures in Ground Plne Plne Current Sheets From this expression, we see immeditely H z = f (k x )e jky y To get the electric field, we use Ampere s Lw: H = jωɛe Now, in the spectrl domin, this becomes k H = jωɛẽ From symmetry, this eqution becomes k x H z = jωɛe y k y H z = jωɛe x

32 The Infinite Slot Apertures in Ground Plne Plne Current Sheets Now, ssume we know the electric field in the slot. We hve just shown Ẽ x y= = k y ωɛ f (k x) Now, given the incident wve nd the symmetry in the z-direction, we cn pproximte Therefore Ẽ x y= = E E x y= { E x < 2 x > e jkx x dx = 2E k x sin k x 2

33 The Infinite Slot Apertures in Ground Plne Plne Current Sheets From this we find tht f (k x ) = ωɛ 2E sin k x k y k x 2 The remining field cn be found from this vi the formuls on the previous pges. As usul, the vlue of k y is dictted by ensure tht the wve Evnesces (nd does not explode) in the direction of propgtion, or, Trvels wy from (nd not towrds) the perture. This implies k y = { j k 2 x k 2 for k < k x k 2 k 2 x for k > k x

34 Outline Apertures in Ground Plne Plne Current Sheets 1 Aperture Excittion Current Excittion 2 Apertures in Ground Plne Plne Current Sheets

35 Plne Current Sheets Apertures in Ground Plne Plne Current Sheets Plne current sheets excittion of free spce cn be computed in the sme wy. Assuming the sheet is in the y = plne, nd A = µψu z, we cn write the Fourier trnsformed (with respect to x nd z) equtions for TM z propgtion: H x = jk y ψ H y = jk x ψ H z = E x = kx kz jωɛ ky kz Ey = jωɛ E z = k 2 k 2 z jωɛ ψ ψ ψ

36 Plne Current Sheets Apertures in Ground Plne Plne Current Sheets We need to expnd ψ on either side of the sheet: ψ + = f + (k x, k z )e jk + y y ψ = f (k x, k z )e jk y y for y > for y < Of course, k + y = k y = { j kx 2 + kz 2 k 2 k 2 kx 2 kz 2 for k < kx 2 + kz 2 for k > kx 2 + kz 2

37 Boundry Condition Apertures in Ground Plne Plne Current Sheets At the sheet E x nd E z hve to be continuous, so f + (k x, k z ) = f (k x, k z ) = f (k x, k z ) We lso need u y [ H + H ] = J jk y + ψ jky ψ = J x (k x, k z ) y= where J x (k x, k z ) = 2jk + y f (k x, k z ) = J x (k x, k z ) f (k x, k z ) = 1 2jk + y J x (k x, k z ) J x (k x, k z )e jkx x e jkzz dxdz

38 An Importnt Exmple Apertures in Ground Plne Plne Current Sheets Consider current sheet with J z = Ilδ(x)δ(z). We lredy know this cretes mgnetic vector potentil with A z = µψ nd e jk x 2 +y 2 +z 2 ψ = Il 4π x 2 + y 2 + z 2 On the other hnd, we cn tret it s current sheet. The trnsform is J z = 1 4π 2 J x (k x, k z )e jkx x e jkzz dxdz = Il 4π 2

39 An Importnt Exmple Apertures in Ground Plne Plne Current Sheets Given this so tht, for y >, f (k x, k z ) = 1 2jk + y Il 4π 2 ψ(k x, y, k z ) = jil 8π 2 k y e jky y where k y = k + y = { j kx 2 + kz 2 k 2 k 2 kx 2 kz 2 for k < kx 2 + kz 2 for k > kx 2 + kz 2

40 An Importnt Exmple Apertures in Ground Plne Plne Current Sheets Therefore, we cn find ψ from n inverse trnsform: For y > ψ(x, y, z) = jil 8π 2 1 k y e jkx x e jky y e jkzz dxdz For y <, both y nd k y chnge sign, so we cn write ψ(x, y, z) = jil 1 8π 2 e jkx x e jky y e jkzz dxdz k y without cvet, with the understnding tht k y chnges sign t the y = plne.

41 The Weyl Identity Apertures in Ground Plne Plne Current Sheets Invoking equivlence (how?) we cn equte the two different expressions for ψ. This gives rise to the Weyl Identity e jkr r = 1 1 e jkx x e jky y e jkzz dxdz 2πj k y Wht does this formul men?