Waveguides and Transmission LinesGeneral Wave PropagationMaxwell's EquationsCurl of emph {z}-ComponentCurl of EThe Field Component EquationsThe Curl of HHx = f(non-Transverse Components)Hx = f(z)(cont)Hy = f(z)E = f(z)TEM, TE, and TM WavesTEM Waves (again)TEM Helmholtz EquationTEM has Static Transverse FieldsStatic E FieldAmp`{e}re's Law (I)Faraday's LawTEM Wave ImpedanceTEM Fields (last)TEM General SolutionTEM Waves in Hollow Waveguides?EECS 117Lecture 25: Field Theory of T-Lines and WaveguidesProf. NiknejadUniversity of California, BerkeleyUniversity of California, Berkeley EECS 117 Lecture 25 – p. 1/23Waveguides and Transmission LinesWe started this course by studying transmission linesby using the concept of distributed circuits. Now we’dlike to develop a field theory based approach toanalyzing transmission lines.Transmission lines have one or more disconnectedconductors. Waveguides, though, can consist of asingle conductor. We’d also like to analyze waveguidestructures, such as a hollow metal pipe, or a hollowrectangular structure. These are known as waveguides.These structures have a uniform cross-sectional area.We shall show that these structures can support wavepropagation in the axial direction.University of California, Berkeley EECS 117 Lecture 25 – p. 2/23General Wave PropagationWe shall assume that waves in the guide take thefollowing formE(x, y, z) = [e(x, y) + ˆzez(x, y)] e−jβz= Ee−jβzH(x, y, z) = [h(x, y) + ˆzhz(x, y)] e−jβzIt’s important to note that we have broken the wave intotwo components, a part in the plane of thecross-section, or the transverse component e(x, y), andcomponent in the direction of wave propagation, anaxial component, ez(x, y).Recall that TEM plane waves have no components inthe direction of propagation.University of California, Berkeley EECS 117 Lecture 25 – p. 3/23Maxwell’s EquationsNaturally, the fields in the waveguide or T-line have tosatisfy Maxwell’s equations. In particular∇ × E = −jωµHRecall that ∇ × (Ff) = ∇f × F + f∇ × F∇ × E = −jβe−jβzˆz × E + e−jβz∇ × ENote that ∇ × (ˆzez(x, y)) does not have a ˆz-componentwhereas ∇ × e has only a ˆz-component∇ × e =ˆx ˆy ˆz∂∂x∂∂y∂∂zExEy0= −ˆx∂Ey∂z+ˆy∂Ex∂z+ˆz(∂Ey∂x−∂Ex∂y)University of California, Berkeley EECS 117 Lecture 25 – p. 4/23Curl of z-ComponentSince Exand Eyhave only (x, y) dependence∇ × e = ˆz(∂Ey∂x−∂Ex∂y)Taking the curl of the ˆz-component generates only atransverse component∇ ׈zez(x, y) =ˆx ˆy ˆz∂∂x∂∂y∂∂z0 0 ez= ˆx∂ez∂y− ˆy∂ez∂xUniversity of California, Berkeley EECS 117 Lecture 25 – p. 5/23Curl of ECollecting terms we see that the curl of E has twoterms, an axial term and a transverse term∇ × E = −jωµH = (−jβe−jβz) (ˆz × e)|{z }t−plane+e−jβzˆz|{z}axial(∂Ey∂x−∂Ex∂y) + ˆx∂ez∂y− ˆy∂ez∂x|{z }t−planeUniversity of California, Berkeley EECS 117 Lecture 25 – p. 6/23The Field Component EquationsNote that ˆz × (Exˆx + Eyˆy) = Exˆy − Eyˆx, so thex-component of the curl equation givesjβEy+∂ez∂y= −jωµHx(1)and the y-component givesjβEx+∂ez∂x= jωµHy(2)The z-component defines the third of our importantequations∂Ey∂x−∂Ex∂y= −jωµhz(x, y)(3)University of California, Berkeley EECS 117 Lecture 25 – p. 7/23The Curl of HNote that ∇ × H = jωǫE, and so a set of similarequations can be derived without any extra mathjβHy+∂hz∂y= jωǫEx(4)jβHx+∂hz∂x= −jωǫEy(5)∂Hy∂x−∂Hx∂y= jωǫez(x, y)(6)University of California, Berkeley EECS 117 Lecture 25 – p. 8/23Hx = f(non-Transverse Components)We can now reduce these (6) equations into (4)equations if we take ezand hzas known components.SincejβHx= −∂hz∂x− jωǫEyandEy=−∂ez∂y− jωµHx1jβsubstituting Eyinto the above equationjβHx= −∂hz∂x−jωǫjβ−∂ez∂y− jωµHxjβHx= −∂hz∂x+ωǫβ∂ez∂y+ jω2µǫβHxUniversity of California, Berkeley EECS 117 Lecture 25 – p. 9/23Hx = f(z) (cont)Collecting terms we have and k2= ω2µǫjβ − jk2βHx=−∂hz∂x+ωǫβ∂ez∂yLet k2c= k2− β2and simplifyHx=jk2cωǫ∂ez∂y− β∂hz∂x(7)In the above eq. we have found the transversecomponent x in terms of the the axial components ofthe fieldsUniversity of California, Berkeley EECS 117 Lecture 25 – p. 10/23Hy = f(z)We can also solve for Hyin terms of ezand hzjβHy= jωǫEx−∂hz∂yjEx= −1β∂ez∂x+jωµβHyjβHy= ωǫ−1β∂ez∂x+jωµβHy−∂hz∂yCollecting termsjβ2− jk2βHy= −ωǫβ∂ez∂x− β∂hz∂yHy=−jk2cωǫ∂ez∂x− β∂hz∂y(8)University of California, Berkeley EECS 117 Lecture 25 – p. 11/23E = f(z)In a similar fashion, we can also derive the followingequationsEx=−jk2cβ∂ez∂x+ ωµ∂hz∂y(9)Ey=jk2c−β∂ez∂y+ ωµ∂hz∂x(10)Notice that we now have found a functional relationbetween all the transverse fileds in terms of the axialcomponents of the fieldsUniversity of California, Berkeley EECS 117 Lecture 25 – p. 12/23TEM, TE, and TM WavesWe can classify all solutions for the field componentsinto 3 classes of waves.TEM waves, which we have already studied, have noz-component. In other words ez= 0 and hz= 0TE waves, or transverse electric waves, has atransverse electric field, so while ez= 0, hz6= 0 (alsoknown as magnetic waves)TM waves, or transverse magnetic waves, has atransverse magnetic field, so while hz= 0, ez6= 0 (alsoknown as electric waves)University of California, Berkeley EECS 117 Lecture 25 – p. 13/23TEM Waves (again)From our equations (7) - (10), we see that if ezand hzare zero, then all the fields are zero unless kc= 0This can be seen by working directly with equations (1)and (5)jβEy= −jωµHxjβHx= −jωǫEyjβEy=−ωµβ(−jωǫEy)β2Ey= ω2µǫEy= k2EyThus β2= k2, or kc= 0University of California, Berkeley EECS 117 Lecture 25 – p. 14/23TEM Helmholtz EquationThe Helmholtz Eq. (∇2+ k2)E simplifies for the TEMcase. Take the x-component∂2∂x2+∂2∂y2+∂2∂z2+ k2Ex= 0Since the z-component of the field is a complexexponential∂2∂x2+∂2∂y2− β2+ k2Ex= 0Since k2= β2∂2∂x2+∂2∂y2Ex= 0University of California, Berkeley EECS 117 Lecture 25 – p. 15/23TEM has Static Transverse