EECS 117. Lecture 25: Field Theory of T-Lines and Waveguides. Prof. Niknejad. University of California, Berkeley

Transcription

1 EECS 117 Lecture 25: Field Theory of T-Lines and Waveguides Prof. Niknejad University of California, Berkeley University of California, Berkeley EECS 117 Lecture 25 p. 1/2

2 Waveguides and Transmission Lines We started this course by studying transmission lines by using the concept of distributed circuits. Now we d like to develop a field theory based approach to analyzing transmission lines. Transmission lines have one or more disconnected conductors. Waveguides, though, can consist of a single conductor. We d also like to analyze waveguide structures, such as a hollow metal pipe, or a hollow rectangular structure. These are known as waveguides. These structures have a uniform cross-sectional area. We shall show that these structures can support wave propagation in the axial direction. University of California, Berkeley EECS 117 Lecture 25 p. 2/2

3 General Wave Propagation We shall assume that waves in the guide take the following form E(x,y,z) = [e(x,y) + ẑe z (x,y)] e jβz = Ee jβz H(x,y,z) = [h(x,y) + ẑh z (x,y)] e jβz It s important to note that we have broken the wave into two components, a part in the plane of the cross-section, or the transverse component e(x, y), and component in the direction of wave propagation, an axial component, e z (x,y). Recall that TEM plane waves have no components in the direction of propagation. University of California, Berkeley EECS 117 Lecture 25 p. 3/2

4 Maxwell s Equations Naturally, the fields in the waveguide or T-line have to satisfy Maxwell s equations. In particular E = jωµh Recall that (Ff) = f F + f F E = jβe jβz ẑ E + e jβz E Note that (ẑe z (x,y)) does not have a ẑ-component whereas e has only a ẑ-component ˆx ŷ ẑ e = x y z = ˆx E y z +ŷ E x z +ẑ( E y x E x y ) E x E y 0 University of California, Berkeley EECS 117 Lecture 25 p. 4/2

5 Curl of z-component Since E x and E y have only (x,y) dependence e = ẑ( E y x E x y ) Taking the curl of the ẑ-component generates only a transverse component ˆx ŷ ẑ ẑe z (x,y) = x y z = ˆx e z y ŷ e z x 0 0 e z University of California, Berkeley EECS 117 Lecture 25 p. 5/2

6 Curl of E Collecting terms we see that the curl of E has two terms, an axial term and a transverse term E = jωµh = ( jβe jβz ) (ẑ e) }{{} e jβz t plane }{{} ẑ ( E y x E x y ) + ˆx e z y ŷ e z x axial }{{} t plane + University of California, Berkeley EECS 117 Lecture 25 p. 6/2

7 The Field Component Equations Note that ẑ (E xˆx + E y ŷ) = E x ŷ E yˆx, so the x-component of the curl equation gives (1) jβe y + e z y = jωµh x and the y-component gives (2) jβe x + e z x = jωµh y The z-component defines the third of our important equations E y x E x (3) y = jωµh z(x,y) University of California, Berkeley EECS 117 Lecture 25 p. 7/2

8 The Curl of H Note that H = jωǫe, and so a set of similar equations can be derived without any extra math (4) (5) (6) jβh y + h z y = jωǫe x jβh x + h z x = jωǫe y H y x H x y = jωǫe z(x,y) University of California, Berkeley EECS 117 Lecture 25 p. 8/2

9 Hx = f(non-transverse Components) We can now reduce these (6) equations into (4) equations if we take e z and h z as known components. Since and jβh x = h z x jωǫe y E y = ( e ) z 1 y jωµh x jβ substituting E y into the above equation jβh x = h z x jωǫ jβ jβh x = h z x + ωǫ β ( e ) z y jωµh x e z y + µǫ jω2 β H x University of California, Berkeley EECS 117 Lecture 25 p. 9/2

10 Hx = f(z) (cont) Collecting terms we have and k 2 = ω 2 µǫ ) (jβ j k2 H x = β ( h z x + ωǫ β e z y ) Let kc 2 = k 2 β 2 and simplify H x = j ( kc 2 ωǫ e ) z y β h z (7) x In the above eq. we have found the transverse component x in terms of the the axial components of the fields University of California, Berkeley EECS 117 Lecture 25 p. 10/2

11 Hy = f(z) We can also solve for H y in terms of e z and h z jβh y = jωǫe x h z y ( jβh y = ωǫ 1 β je x = 1 β e z x + jωµ β H y e z x + jωµ β H y ) h z y Collecting terms (8) ) (jβ 2 j k2 β H y = j k 2 c H y = ωǫ β e z x β h z y ( ωǫ e ) z x β h z y University of California, Berkeley EECS 117 Lecture 25 p. 11/2

12 E = f(z) In a similar fashion, we can also derive the following equations E x = j ( kc 2 β e ) z x + ωµ h z (9) y E y = j ( kc 2 β e ) z y + ωµ h z (10) x Notice that we now have found a functional relation between all the transverse fileds in terms of the axial components of the fields University of California, Berkeley EECS 117 Lecture 25 p. 12/2

13 TEM, TE, and TM Waves We can classify all solutions for the field components into 3 classes of waves. TEM waves, which we have already studied, have no z-component. In other words e z = 0 and h z = 0 TE waves, or transverse electric waves, has a transverse electric field, so while e z = 0, h z 0 (also known as magnetic waves) TM waves, or transverse magnetic waves, has a transverse magnetic field, so while h z = 0, e z 0 (also known as electric waves) University of California, Berkeley EECS 117 Lecture 25 p. 13/2

14 TEM Waves (again) From our equations (7) - (10), we see that if e z and h z are zero, then all the fields are zero unless k c = 0 This can be seen by working directly with equations (1) and (5) jβe y = jωµh x Thus β 2 = k 2, or k c = 0 jβh x = jωǫe y jβe y = ωµ β ( jωǫe y) β 2 E y = ω 2 µǫe y = k 2 E y University of California, Berkeley EECS 117 Lecture 25 p. 14/2

15 TEM Helmholtz Equation The Helmholtz Eq. ( 2 + k 2 )E simplifies for the TEM case. Take the x-component ( 2 ) x y z 2 + k2 E x = 0 Since the z-component of the field is a complex exponential ( 2 ) x y 2 β2 + k 2 E x = 0 Since k 2 = β 2 ( 2 ) x y 2 E x = 0 University of California, Berkeley EECS 117 Lecture 25 p. 15/2

16 TEM has Static Transverse Fields The same result applies to E y so that we have where t = ˆx x + ŷ y 2 te(x,y) = 0 This is a two-dimensional Laplace equation. Recall that static fields satisfy Laplace s Eq. So it appears that our wave is a static field in the transverse plane! Therefore, applying our knowledge of electrostatics, we have e(x,y) = t Φ(x,y) Where Φ is a scalar potential. This can also be seen by taking the curl of e. If it s a static field, the curl must be identically zero University of California, Berkeley EECS 117 Lecture 25 p. 16/2

17 Static E Field Notice that ˆx ŷ ẑ t e = x y 0 E x E y 0 = ẑ ( Ey x E ) x y = jωµh z = 0 Since h z = 0, the curl of e is zero and thus the field behaves statically. Since D = t ǫe = 0, we also have 2 tφ(x,y) = 0 And thus we can also define a unique potential function in the transverse plane V 12 = 2 1 e(x,y) dl University of California, Berkeley EECS 117 Lecture 25 p. 17/2

18 Ampère s Law (I) By Ampère s Law (extended to include displacement current) H = jωd + J Or in integral form H dl = C S J ds + jω S D ds But since e z = 0, the surface integral term of displacement current vanishes and we have H dl = J ds = I C Which is an equation satisfied by static magnetic fields. S University of California, Berkeley EECS 117 Lecture 25 p. 18/2

19 Faraday s Law It s easy to show that the electric fields also behave statically by using Faraday s law or C E = jωb E dl = jω S B ds = 0 The RHS is zero since h z = 0 for TEM waves. Thus E dl = 0 and a unique potential can be defined. C University of California, Berkeley EECS 117 Lecture 25 p. 19/2

20 TEM Wave Impedance By equation (4) we have jβh y = jωǫe x Thus the TEM wave impedance can be defined as Z TEM = E x = β H y ωǫ = k ωǫ = ω µǫ ωǫ From equation (5) we have jβh x = jωǫe y = µ ǫ = η Z TEM = E y H x = β ωǫ = µ ǫ = η University of California, Berkeley EECS 117 Lecture 25 p. 20/2

21 TEM Fields (last) Since H x = E y /η and H y = E x /η, we have h(x,y) = ẑ e(x,y) Z TEM Thus we only need to compute the electric field to find all the fields in the problem. This is exactly what we found when we studied uniform plane waves University of California, Berkeley EECS 117 Lecture 25 p. 21/2

22 TEM General Solution 1. Begin by solving Laplace s Eq. in the transverse plane (2D problem) 2. Apply boundary conditions to resolve some of the unknown constants 3. Compute the fields e and h 4. Compute V and I (voltage and currents) 5. The propagation constant γ = jβ = jω µǫ and the impedance is given by Z 0 = V/I University of California, Berkeley EECS 117 Lecture 25 p. 22/2

23 TEM Waves in Hollow Waveguides? What do our equations tell us about wave propagation in a hollow waveguide, such as a metal pipe? If TEM waves travel inside a such a structure, the transverse components must be solutions to the static 2D fields. But if we have a metal conductor surrounding a region, we have already proven in electrostatics that the only solution is a zero field, which are of no interest to us. Thus TEM waves cannot travel in such waveguides! We can see through a metal pipe, so what s going on? There must be other types of waves traveling through it. University of California, Berkeley EECS 117 Lecture 25 p. 23/2