Photograph of the rectangular waveguide components

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Angel Willis
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1 Waveguides

2 Photograph of the rectangular waveguide components

3 BACKGROUND A transmission line can be used to guide EM energy from one point (generator) to another (load). A transmission line can support only a transverse electromagnetic (TEM) wave at microwave frequencies (roughly GHz), transmission lines become inefficient due to skin effect and dielectric losses; A transmission line may operate from dc ( / = 0) to a high frequency; A waveguide is another means of achieving the same goal. A waveguide can support many possible field configurations i.e. TE, TM, TEM. Waveguides are used at that range of frequencies to obtain larger bandwidth and lower signal attenuation. Waveguide can operate only above a certain frequency called the cutoff frequency and therefore acts as a high-pass filter.

4 Waveguides Transmission lines that consists of two or more conductor may support transverse electromagnetic (TEM) waves It is characterized by the lack of longitudinal field components TEM waves have a uniquely defined voltages, current and characteristics impedance Waveguides, often consisting of a single conductor, support transverse electric (TE) and/or transverse magnetic (TM) waves It is characterized by the presence of longitudinal magnetic or electric, respectively, field components.

5 Electromagnetic wave (1/) x E x Direction of Propagation k z z y B y H y An electromagnetic wave is a travelling wave which has time varying electric and magnetic fields which are perpendicular to each other and the direction of propagation, z S.O. Kasap, Optoelectronics (Prentice Hall)

11 Non-TEM mode waveguide structures (a) rectangular waveguide (c) dielectric slab waveguide (b) circular waveguide, (d) fiber optic waveguide.

12 TERMINOLOGY Kc Cut off wave number: How many wave passes per meter? Cut off frequecy: A waveguide can operate a certain frequency is called cutoff frequency. It is act as a high pass filter. Below this frequency attenuation occurs and above which propagation takes place. Each set of integers m and n gives a different field pattern or mode, referred to as TMmn mode in the wave guide. Integer m equals the number of half cycle variations in the x direction and n in the y direction. The dominant mode is the mode with lowest cut off frequency. Mode: There are a number of ways in which electrical energy can propagate along a wave guide. All these modes must satisfy certain boundary conditions. TEM=Waves in free space known as transverse electromgnetic (TEM). Transmission lines that consist of two or more conductors may support TEM waves, characterize by lack of longitudinal field components. Waveguides, often consisting of a single conductor support transverse electric (TE) and/or TM waves, characterized by the presence of longitudinal magnetic or electric field component.

13 General solutions for TEM, TE, AND TM Waves General solutions to Maxwell equation for the specific case of TEM, TE and TM wave propagation in waveguides. Conductor boundaries that are parallel to the z-axis. The wavenumber, k k k=is the unbounded propagation wave number Cutoff wavenumber, k c k k If dielectric loss is present, c Where, β = phase constant) o r ( 1 j tan )

14 TEM Waves Transverse electromagnetic (TEM) waves are characterized by E z = H z = 0. The wave impedance of a TEM mode can be found as the ratio of the transverse electric and magnetic fields: Z TEM E H x y The other pair of transverse field components Z TEM H E x y

15 The procedure for analyzing a TEM line can be summarized as: Solve Laplace s equation which contain several unknown constants Find these constants by applying the boundary conditions for the known voltages on the conductors Compute e, E,h,H Compute V from I The characteristic impedance is given by Z o = V/I

16 TE Waves The starting point will be the homogenous wave equation. Transverse electric (TE) waves, (also referred to as H-waves) E z =0 and H z 0 The TE wave impedance can be found as Z TE E H x y H E x y k Which is seen to be frequency dependent TE waves can be supported inside closed conductors, as well as between two or more conductors.

17 TM Waves TM waves (transverse magnetic) also referred to as E-waves are characterized E z 0 and H z =0 The TM wave impedance can be found as Z TM E H x y H E x y k Which is frequency dependent TM waves can be supported inside closed conductors, as well as between two or more conductors.

18 Attenuation Due to Dielectric Loss Attenuation in a transmission line or waveguide can be caused by either dielectric loss or conductor loss. If d is the attenuation constant due to dielectric loss, and c is the attenuation constant due to conductor loss, then the total attenuation constant is = d + c. If the line or guide is completely filled with a homogeneous dielectric, the attenuation due to lossy dielectric can be calculated from the propagation constant, and this result will apply to any guide or line with a homogenous dielectric filling.

19 In practice, most dielectric materials have a very small loss (tan << 1) so these expression can be reduces to: ) tan (1 j k k k j r o o c c d j k k k jk k k jk k k c c c tan tan tan Using the complex dielectric constant allows the complex propagation constant to be written as

20 The real wave number in the absence of loss is o o r k c k j (phase constant) The result applies to any TE or TM wave; If the guide is completely filled with the dielectric, it can also be used for TEM (k c = 0, = k) kc d d k k tan tan Np/m (TEM waves)

21 RECTANGULAR WAVEGUIDE Rectangular waveguides were one of the earliest types of transmission lines used to transport microwave signals and are still used today for many applications. The hollow rectangular waveguide can propagate TM and TE modes, but not TEM waves, since only one conductor is present. We will see that the TE and TM modes of a rectangular waveguide have cutoff frequencies below which propagation is not possible, similar to the TM and TE modes of the parallel plate guide.

22 Photograph of Ka-band (WR-8) rectangular waveguide components.

25 TE Mode It is characterized by fields E z =

26 TE Mode It is characterized by fields E z =0. The guide is filled with a material of permittivity and permeability. Waveguide along the x-axis, so that a>b. Fig: Geometry of a rectangular waveguide

27 The propagation constant; m n k kc k a b Cutoff frequency, f cmn f c mn k c 1 m a n b Or f c u ' m n a b Where, u p = u = 1 με lossless = phase velocity of the uniform plane wave in the dielectric medium (σ = 0, μ, ε) filling the waveguide. The cutoff wavelength λ c is given by λ c =u /fc.

28 The mode with the lowest cutoff frequency is called the dominant mode. Since we have assumed a>b, the lowest f cmn occurs for the TE 10 (m=1, n=0). for TE, (m,n) may be (0,1) or (1,0) but not (0,0). Both m and n cannot be zero at the same time because this will force the filed component to vanish.

29 From the homogenous wave equation E E x = jωμ nπ k c b H mπx 0 Cos Sin( nπy a b ) e jβz This equation consists of cosine and sine component. So all the fields will not vanish if m and n assume a value of 0. Thus the TE 10 mode is the dominant. There is no TE 00 mode. 1 f c 10 a At a given operation f, only those modes having f c <f will propagate. Modes with f c >f will lead to an imaginary β, meaning that all field component will decay exponentially away from the source of excitation. Such modes are referred to as cutoff, or evanescent modes.

30 If more than one mode is propagating, the waveguide is said to be overmoded. Wave impedance the ration can be constituted the wave impedance in the guide ; Z TE E H x y H E x y k Type equation here. Z TE = η 1 (fmn/f) = η r 1 (fmn/f) η= μ/ε ηo = μ o ε o 10π = 377 η=the intrinsic impedance of the dielectric material filling the guide η 0 =the intrinsic impedance of the free space

31 Guide wavelength g k g is thus greater than. the wavelength of a plane wave in the filling medium. Phase velocity u p k 1/ This shows u p u (un bounded medium). If u =c, then u p is greater than the speed of light in vacuum. The group velocity u g is the velocity with which the resultant repeated reflected waves are traveling down the guide and is given u g The group velocity in the guide is always less than or equal to u. So it is evident that U p u g =u Which is greater than the speed of light (plane wave) in the filling medium. In the vast majority of applications the operating frequency and guide dimensions are chosen so that only the dominant TE 10 mode will propagate. fc u' 1 f

32 Which is greater than the speed of light (plane wave) in the filling medium. In the vast majority of applications the operating frequency and guide dimensions are chosen so that only the dominant TE 10 mode will propagate.

33 The attenuation due to conductor loss for the TE 10 mode Np/m ) ( ) / / ( k a b k b a R b a a a b R P P s s l c

34 TM Modes Characterized by fields with H z =???????

35 TM Modes Characterized by fields with H z =0 while E z must satisfy the reduced wave equation. Propagation constant c From the homophonous wave equation E k k k Ez = e z e jβz = E 0 Sin mπx Sin( nπy a b ) e jβz As the equation consists of sine component only so a value of zero for m or n will become zero and all other field component will vanish as well. Cutoff frequency is same as that of the TE mn mode. The lowest order TM mn to propagete is the TM 11 mode. There is no TM 00, TM 01, or TM 10 modes. (as the filed expression for E and H are identically zero if either m or n is zero) m a n b

36 Having a cutoff frequency for TM 11 f c 1 11 a b Which is seen to be larger than f c10 forte 10 mode Wave impedance can be computed in the same way as for the TE modes. Z TM E H x y H E x y k Figure summarizes results for TE and TM wave propagation in rectangular waveguide.

37 Figure : Field lines for some of the lower order modes of a rectangular waveguide.

39 Problem 1: A standard air-filled rectangular waveguide with dimensions a = cm, b = cm is fed by a 4-GHz carrier from a coaxial cable. Determine if a TE 10 mode will be propagated. If so, calculate the phase velocity and the group velocity.

40 Problem (a). Show that a rectangular waveguide does not support TM 10 and TM 01 modes. (b) Explain the difference TE mn and TM mn

41 Show that a rectangular waveguide does not support TM10 and TM01 modes. (b) Explain the difference TE mn and TM mn

42 Problem : A tunnel is modeled as an air-filled metallic rectangular waveguide with dimensions a = 8 m and b = 16 m. Determine whether the tunnel will pass: (a) a 1.5-MHz AM broadcast signal, (b) a 10-MHz FM broadcast signal.

43 Problem 3: Design an air-filled rectangular waveguide with the cut-off frequency of a TE 10 mode, f cte10 = 5 GHz whereas the TE 01 mode, f cte01 = 1 GHz.

44 Problem 4: A X3 cm waveguide is filled with a dielectric material with ε r =5. If the waveguide operates at 0 GHz with TE 11 mode, find the a) The cut off frequency

45 Problem 5: A waveguide, with dimensions a = 1 cm and b =0.7 cm, is to be used at 0 GHz. Determine the wave impedance for the dominant mode when i) the guide is empty, and ii) the guide is filled with polyethylene (whose ε r =:5).

46 Problem 6: A hollow rectangular waveguide is used to transmit signals at a carrier frequency of 6 GHz. Design the waveguide so that the cut-off frequency of the dominant TE mode is lower than the carrier by 5% and the next mode is at least 5% higher than the carrier frequency.

47 A hollow rectangular waveguide is used to transmit signals at a carrier frequency of 6 GHz. Design the waveguide so that the cut-off frequency of the dominant TE mode is lower than the carrier by 5% and the next mode is at least 5% higher than the carrier frequency.

48 THANK YOU

Waveguides GATE Problems

Waveguides GATE Problems One Mark Questions. The interior of a 20 20 cm cm rectangular waveguide is completely 3 4 filled with a dielectric of r 4. Waves of free space wave length shorter than..can be