Helmholtz Wave Equation TE, TM, and TEM Modes Rect Rectangular Waveguide TE, TM, and TEM Modes Cyl Cylindrical Waveguide.

Transcription

1 Waveguides S. R. Zinka School of Electronics Engineering Vellore Institute of Technology April 26, 2013

2 Outline 1 Helmholtz Wave Equation 2 TE, TM, and TEM Modes Rect 3 Rectangular Waveguide 4 TE, TM, and TEM Modes Cyl 5 Cylindrical Waveguide

3 Outline 1 Helmholtz Wave Equation 2 TE, TM, and TEM Modes Rect 3 Rectangular Waveguide 4 TE, TM, and TEM Modes Cyl 5 Cylindrical Waveguide

4 Helmholtz Wave Equation In a source-less dielectric medium, Taking curl of 2 gives D = 0 B = 0 H = ω D = ωε E 1 E = ω B = ωµ H 2 E = ωµ H E 2 E = ωµ H E 2 E = ωµ ωε E 2 E = E ω 2 µε E 2 E = 0 ω 2 µε E 3 Similarly, it can be proved that 2 H = ω 2 µε H. 4

5 Solution of Helmholtz Equation in Cartesian System Vector Helmholtz equation can be decomposed as shown below: 2 E x + ω 2 µεe x = 0 2 E + ω 2 µε E = 0 2 E y + ω 2 µεe y = 0 2 E y + ω 2 µεe y = 0 Since all the differential equations are similar, let s solve ust one equation using variable-separable method. If E xs can be decomposed into E x = A x B y C z then substituting the above equation into Helmholtz equation gives 2 E x E x 2 2 E x + ω 2 µεe x = E x z 2 + ω 2 µεe x = 0 B y C z 2 A + 2 A x C z 2 B + 2 A x B y 2 C z + 2 ω2 µεa x B y C z = A A x B 2 B y C 2 C z z 2 γ2 = 0

6 Solution of Helmholtz Equation... Contd 1 2 A A x B 2 B y C 2 C z z 2 γ2 = A A x B 2 B y C 2 C z z 2 γ2 x γy γ 2 z 2 = 0 5 The above equation can be decomposed into 3 separate equations: 1 2 A A x 2 γ2 x = B B y 2 γ2 y = C C z z 2 γ2 z = 0 It is sufficient to solve only one of the above equations and it s solution is given as 2 A 2 γ2 xa x = 0 A x = L 1 e γxx + L 2 e γxx = L e γxx + L + e γxx 6

7 Solution of Helmholtz Equation... Contd So, finally E x is given as E x = L e γxx + L + e γxx M e γyy + M + e γyy N e γzz + N + e γzz 7 with the condition that γ 2 x + γ 2 y + γ 2 z = γ 2. 8

8 Outline 1 Helmholtz Wave Equation 2 TE, TM, and TEM Modes Rect 3 Rectangular Waveguide 4 TE, TM, and TEM Modes Cyl 5 Cylindrical Waveguide

9 Maxwell s Equations in Free Space E = ωµ H H = ωε E E z + β ze y = ωµh x β z E x E z E y E x = ωµh y = ωµh z H x = H y = E x = E y = ωε E z ωε E z β z E z β z + β z + ωµ E z β z + ωµ + β zh y = ωεe x β z H x H y H x = ωεe y = ωµεe z The above equations indicate the fact that if longitudinal components E z &H z are known, one can get all the transversal components

10 TE Modes E z = 0 H x = H y = E x = E y = ωε E z ωε E z β z + β z E z β z + ωµ E z β z + ωµ Characteristic impedance of TE waved is given as = = = = β z β z ωµ ωµ Z TE 0 = E x H y = E y H x = ωµ β z

11 TM Modes H z = 0 H x = H y = E x = E y = ωε E z ωε E z β z = + β z E z β z + ωµ E z β z + ωµ Characteristic impedance of TM waved is given as = = = ωε E z ωε E z E z β z β z E z Z TM 0 = E x H y = E y H x = β z ωε

12 TEM Modes E z = 0 & H z = 0 H x = H y = E x = E y = ωε E z ωε E z β z E z β z + β z + ωµ E z β z + ωµ If longitudinal components E z &H z are zeros, all the transversal components become zero... then the wave does not exist at all!!! So, something must be wrong... isn t it?!

13 TEM Modes E z = 0 & H z = 0 H x = H y = E x = E y = ωε E z ωε E z β z E z β z + β z + ωµ E z β z + ωµ Transverse components can exist only when β z = β 0. In other words, β x = 0 and β y = 0.

14 TEM Modes Z TEM 0 Characteristic impedance of TE and TM waves are given as Z TE 0 = E x H y = E y H x = ωµ β z Z TM 0 = E x H y = E y H x = β z ωε Then what is the Characteristic impedance of TEM wave? Since TEM mode is TE as well as TM, we can actually use any one of the above equations... also, for TEM waves β z = β 0, Z TE 0 = ωµ = ωµ = ωµ µ β z β 0 ω µε = ε Z TM 0 = β z ωε = β 0 ωε = ω µε = ωε Have you noticed that the Characteristic impedance of TEM wave is independent of waveguide dimensions?! µ ε

15 Existence of TEM Modes For TEM waves E and H fields exist in transverse plane only since E z = 0 & H z = 0. Also, from the maxwell equation B = 0, we know that magnetic fields are rotational fields. So, line integral of H around any arbitrary closed loop C H dl has to be non-zero. However, from Stokes theorem and Ampere s law, we know that C H dl = H ds = Je ds. Since C H dl = 0, Je has to be a non-zero value. However, we know that Je is conduction current... So, we need at least one conductor inside the waveguide for TEM modes to exist.

16 Outline 1 Helmholtz Wave Equation 2 TE, TM, and TEM Modes Rect 3 Rectangular Waveguide 4 TE, TM, and TEM Modes Cyl 5 Cylindrical Waveguide

17 Longitudinal Component E z From 7, similar to E x, E z also can be written as E z = L e γxx + L + e γxx M e γyy + M + e γyy N e γzz + N + e γzz. 9 If the wave is propagating along +z direction, N = 0. Remaining constants can be derived using boundary conditions. Since E z is tangential to all the four surfaces, boundary conditions are { E z = 0, when x = 0 & x = a E z = 0, when y = 0 & y = b. 10

18 Longitudinal Component E z First let us see the boundary conditions imposed by x = 0 and x = a. Since E z = 0, when x = 0, L + L + = 0 L = L Similarly E z = 0, when x = a, so using the above equation one gets L e γxa + L + e γxa = 0 L + e γxa e γxa = 0 sinh γ x a = 0. Hence, from the boundary conditions, we have γ x a = sinh 1 0 = sin 1 0 = mπ, where m = 0, ±1, ±2,... So, γ x = mπ = 0 + β x. 12 a Similarly, from boundary conditions at y = 0 and y = b, we can derive that M = M + and γ y = nπ b = 0 + β y, where n = 0, ±1, ±2,... 13

19 Longitudinal Component E z Substituting 11, 12, and 13 into 9 gives E z = L e γxx + L + e γxx M e γyy + M + e γyy N e γzz + N + e γzz = L + e γxx + L + e γxx M + e γyy + M + e γyy 0 + N + e γzz = L + M + N + e γxx e γxx e γyy e γyy e γzz = 2L + 2M + N + sinh γ x x sinh γ y y e γzz = 2L + 2M + N + 2 sin β x x sin β y y e γzz, sinh x = sin x }{{} A mπ nπ = A sin a x sin b y e γzz. 14 We also know that β 2 x + β 2 y + β 2 z = β 2 0. So, γ z = β z where mπ 2 nπ β z = β 2 0 β2 x β 2 y = β 2 0 a b 2. 15

20 Longitudinal Component H z When electric field is zero, magnetic field is maximum. So, for magnetic fields, sin function simply becomes cos function. So, we do not have to derive expression for H z once again. So, E z and H z are simply given as mπ E z = Asin a x H z = Bcos mπ a x cos nπ sin b y e γzz 16 nπ b y e γzz 17

21 TE Modes in Rectangular Waveguide E z = 0 Now that we have derived expressions for longitudinal components, we can get transversal components by using the formulas for TE waves: H x = H y = E x = E y = β z β z ωµ ωµ since H z = B cos mπ a x cos nπ b y e γzz. =B β z = B β z = B ωµ =B ωµ [ mπ a [ nπ b [ nπ b [ mπ sin a mπ nπ sin a x cos b y e γzz] mπ nπ cos a x sin b y e γzz] mπ nπ cos a x sin b y e γzz] mπ a x cos nπ b y e γzz]

22 TM Modes in Rectangular Waveguide Similar to TE mode, TM mode s transversal components are given as shown below: H x = H y = E x = E y = ωε E z ωε E z E z β z β z E z since E z = A sin mπ a x sin nπ b y e γzz. = A ωε =A ωε =A β z = A β z [ nπ mπ b sin a x [ mπ mπ cos a a x sin [ mπ mπ cos a a x [ nπ b sin mπ a x cos nπ cos b y e γzz] nπ b y e γzz] nπ sin b y e γzz] nπ b y e γzz]

23 Modes in Rectangular Waveguide Intuitive Explanation If you are fed up with the all the equations given in this PDF file, you may feel free to refer to much simpler explanation given in another PDF file written on rectangular waveguides. There, you can also find formulas for λ g, ω cutoff, etc.

24 Parallel Plate Waveguide All the equations derived for rectangular waveguides can also be used for parallel plate waveguide. After all, parallel plate waveguide is nothing but a rectangular waveguide with a. One maor difference though is, in parallel plate waveguide in addition to TE and TM modes, TEM modes also exist... do you know why?

25 Free Space Waveguide Once again, the equations derived for rectangular waveguides can be used for free space too. It is because, free space waveguide is nothing but a rectangular waveguide with a and b. One maor difference though is, similar to parallel plate waveguide, in free space, TEM modes also exist... do you know why?

26 Outline 1 Helmholtz Wave Equation 2 TE, TM, and TEM Modes Rect 3 Rectangular Waveguide 4 TE, TM, and TEM Modes Cyl 5 Cylindrical Waveguide

27 Maxwell s Equations in Free Space E = ωµ H 1 Ez ρ φ ρ E φ z Eρ z E z ρ 1 ρeφ E ρ ρ ρ φ H = ωε E 1 Hz ρ φ ρ H φ z Hρ z H z ρ ρhφ H ρ ρ φ 1 ρ = ωµh ρ = ωµh φ = ωµh z = ωεe ρ = ωεe φ = ωεe z E ρ = E φ = H ρ = H φ = E z β 2 0 β z β2 z ρ + ωµ ρ φ βz E z ρ φ ωµ H z ρ ωε E z ρ φ β z ρ β 2 0 ωε E z β2 z ρ + β z ρ φ Once again, the above equations indicate the fact that if longitudinal components E z &H z are known, one can get all the transversal components

28 TE Modes E z = 0 E ρ = E z β 2 0 β z β2 z ρ + ωµ ρ E φ = βz E z ρ φ ωµ ρ ωε E z H ρ = ρ φ β z ρ H φ = ωε E z ρ + β z ρ Characteristic impedance of TE waved is given as = φ φ = = = ωµ ρ φ ωµ H z ρ β z ρ βz ρ φ Z TE 0 = E ρ = E φ = ωµ H φ H ρ β z

29 TM Modes H z = 0 E ρ = E z β 2 0 β z β2 z ρ + ωµ E z = ρ φ β 2 0 β z β2 z ρ E φ = βz E z ρ φ ωµ H z = βz E z ρ ρ φ ωε E z H ρ = ρ φ β ωε E z z = ρ ρ φ H φ = β 2 0 ωε E z β2 z ρ + β z = ρ φ β 2 0 ωε E z β2 z ρ Characteristic impedance of TE waved is given as Z TM 0 = E ρ = E φ = β z H φ H ρ ωε

30 TEM Modes Z TEM 0 Characteristic impedance of TE and TM waves are given as Z TE 0 = E ρ = E φ = ωµ H φ H ρ β z Z TM 0 = E ρ = E φ = β z H φ H ρ ωε Then what is the Characteristic impedance of TEM wave? Since TEM mode is TE as well as TM, we can actually use any one of the above equations... also, for TEM waves β z = β 0, Z TE 0 = ωµ = ωµ = ωµ µ β z β 0 ω µε = ε Z TM 0 = β z ωε = β 0 ωε = ω µε = ωε Have you noticed that the Characteristic impedance of TEM wave is independent of waveguide dimensions?! µ ε

31 Outline 1 Helmholtz Wave Equation 2 TE, TM, and TEM Modes Rect 3 Rectangular Waveguide 4 TE, TM, and TEM Modes Cyl 5 Cylindrical Waveguide

32 Helmholtz Wave Equation - Cylindrical Coordinate System Since all the electromagnetic field components have to satisfy the Helmholtz wave equation, one can write the wave equation for E z as [ 1 ρ E z E z ρ ρ ρ ρ 2 φ 2 Solving the above differential equation gives 2 E z + ω 2 µεe z = 0 ] + ω 2 µεe z = E z z 2 [ E z = [AJ n β c ρ + BY n β c ρ] [C sin nφ + D cos nφ] Ee βzz + Fe βzz] 18

33 Longitudinal Components E z & H z In 18, the term Y n β c ρ at ρ = 0; So, this term is physically unacceptable for cylindrical waveguides however, this term is required for coaxial lines... also, if we assume that the wave is propagating in +z direction, then F = 0... So, Similarly, one can write equation for H z as E z = AJ n β c ρ [C sin nφ + D cos nφ] e βzz. 19 H z = BJ n β c ρ [C sin nφ + D cos nφ] e βzz. 20

34 Cylindrical Waveguide TE modes [ ωµ ωµ E ρ = β 2 0 = B [J β2 z ρ φ β 2 0 n β c ρ [nc cos nφ nd sin nφ] e βzz]] β2 z ρ E φ = β 2 0 ωµ H z [ =B β2 z ρ β 2 0 ωµ [β c J β2 n β c ρ [C sin nφ + D cos nφ] e βzz]] z H ρ = β 2 0 β z = B [ β β2 z ρ β 2 0 z [β c J β2 n β c ρ [C sin nφ + D cos nφ] e βzz]] z [ βz βz H φ = β 2 0 = B [J β2 z ρ φ β 2 0 n β c ρ [nc cos nφ nd sin nφ] e βzz]] β2 z ρ where H z = BJ n β c ρ [C sin nφ + D cos nφ] e βzz. Cutoff propagation constant can be derived from the above equations by using boundary conditions. Since E φ is tangential to the boundary of the cylindrical waveguide, boundary conditions are { E φ = 0, when ρ = a. 21 So, boundary condition for TE modes is J n β c a = 0. 22

35 Cylindrical Waveguide TM modes E z E ρ = β 2 0 β z = A β β2 z ρ β 2 0 z [β c J β2 n β c ρ [C sin nφ + D cos nφ] e βzz] z E φ = βz E z β 2 0 = A βz [J β2 z ρ φ β 2 0 n β c ρ [nc cos nφ nd sin nφ] e βzz] β2 z ρ ωε E z ωε H ρ = β 2 0 =A [J β2 z ρ φ β 2 0 n β c ρ [nc cos nφ nd sin nφ] e βzz] β2 z ρ H φ = β 2 0 ωε E z = A β2 z ρ β 2 0 ωε [β c J β2 n β c ρ [C sin nφ + D cos nφ] e βzz] z where E z = AJ n β c ρ [C sin nφ + D cos nφ] e βzz. Cutoff propagation constant can be derived from the above equations by using boundary conditions. Since E φ is tangential to the boundary of the cylindrical waveguide, boundary conditions are { E φ = 0, when ρ = a. 23 So, boundary condition for TM modes is J n β c a = 0. 24