Solved problems in waveguides
- 1. Problem Solving Approach in Waveguides Dr.N.Subhashini Assistant Professor Department of ECE SRMValliammai Engineering College 1
- 2. Presentation Outline Introduction to Waveguides Types of Waveguides Parallel plate waveguide Rectangular Waveguide Circular waveguide 2
- 3. Introduction to waveguides A waveguide used for guiding the Electromagnetic energy from one point to another point. Waveguides are used at microwave frequencies (3GHz to 300 GHz) The waveguide can operate only above cutoff frequency and therefore acts as a high pass filter. Conducting walls of waveguide is made up of brass, copper or aluminium. Inside wall is electroplated with silver or gold to improve performance of the waveguide. Confines the electromagnetice wave in the direction defined by the boundaries. 3
- 4. Waveguides (I) Parallel plate waveguide (ii) Rectangular waveguide (iii) Circular waveguide 4
- 5. Types of Waves (i)Transverse Electric (TE) or HWave : The Electric field component is transverse to the direction of the propagation. (ii)Transverse Magnetic (TE) or EWave : The Magnetic field component is transverse to the direction of the propagation. (iii)Transverse Electromagnetic(TEM)Wave: Does not exists in waveguide, both Electric field and the magnetic field component is perpendicular to the direction of propagation. 5
- 6. Modes of propagation The order of the mode refers to the field configuration in the guide and is given by ‘m’ and ‘n’ integer subscripts, as TEmn and TMmn. The ‘m’ subscript corresponds to the number of half wave variations of the field in x direction The ‘n’ subscript corresponds to the number of half wave variations of the field in y direction 6
- 7. Parallel Plate Waveguide • Consider plates in the yz plane at x= 0 & at x= a. • Assume the direction of the propagation is along the z direction • For the TE wave the EZ component is zero. • For the TM wave the Hz component is zero. • The cut off frequency will be decided by the plate separation value ’ a’. 7
- 8. Field Components Transverse Electric Wave 𝐸𝑦 = 𝐶1sin( 𝑚𝜋 𝑎 x)e-jβz 𝐻𝑥 = −𝛽 𝑗𝜔𝜇 𝐶1sin( 𝑚𝜋 𝑎 x)e-jβz 𝐻𝑧 = 𝑗𝑚𝜋 𝜔𝜇𝑎 𝐶1cos( 𝑚𝜋 𝑎 x)e-jβz Transverse Magnetic Wave 𝐻𝑦 = 𝐶1cos( 𝑚𝜋 𝑎 x)e-jβz 𝐸𝑥 = 𝛽 𝜔𝜀 𝐶1cos( 𝑚𝜋 𝑎 x)e-jβz 𝐸𝑧 = 𝑗𝑚𝜋 𝜔𝜀𝑎 𝐶1sin( 𝑚𝜋 𝑎 x)e-jβz 8
- 9. Cut off frequency Propagation constant 𝛾 = 𝛼 + 𝑗𝛽 = ℎ2 − 𝜔 2 𝜇𝜀 = 𝑚𝜋 𝑎 2 − 𝜔2𝜇𝜀 At the cut off frequency f = fc , 𝛾 = 0 Thus, cut off frequency fc = 𝑚 2𝑎 𝜇𝜀 𝛾 = 𝜇𝜀 𝜔𝑐 2 − 𝜔2 9
- 10. Propagation constant Propagation constant 𝛾 = 𝜇𝜀 𝜔𝑐 2 − 𝜔2 Case (i) : f < fc then, the propagation constant is purely real 𝜸 = 𝜶 - Non propagation or Evanescent mode Case (ii) : f > fc then, the propagation constant is purely imaginary 𝜸 = 𝒋𝜷 - propagating mode with zero attenuation 10
- 11. Dominant mode • TE or TM wave that propagates with minimum cut off frequency is called as Dominant mode. • For the parallel plat there is no variations in the y direction thus mode is denoted as TEmo and Tmmo. • TE10 dominant mode of TE wave. 11
- 12. Formulae Propagation constant 𝛾 = 𝛼 + 𝑗𝛽 = ℎ2 − 𝜔 2 𝜇𝜀 = 𝑚𝜋 𝑎 2 − 𝜔2𝜇𝜀 Cut off frequency fc = 𝑚 2𝑎 𝜇𝜀 Cut off wavelength 𝜆𝑐 = 2𝑎 𝑚 Guide wavelength 𝜆𝑔 = 𝜆 1− 𝑓𝑐 𝑓 2 where 𝜆 = 𝑐 𝑓 , c = 3 x108 m/s 12
- 13. Formulae… Phase velocity vp = 𝑐 1− 𝑓𝑐 𝑓 2 Group velocity vg = c 1 − 𝑓𝑐 𝑓 2 Intrinsic impedance 𝜂 = 𝜇 𝜀 In free space 𝜂 = 377 Ω 13
- 14. Formulae… Transverse Electric Wave Wave impedance ZTE = η 1− 𝑓𝑐 𝑓 2 Transverse Magnetic Wave Wave impedance ZTM = η 1 − 𝑓𝑐 𝑓 2 14
- 15. Problem 1 Solution: Cut off frequency fc = 𝑚 2𝑎 𝜇𝜀 Given : m =1, a = 3x 10-2 m f = 1 x 10 9 Hz & Medium is air, εr = 1 & µr = 1 Using the given values fc can be computed Find the cut-off frequency of the TE10 Mode wave operating at 1GHz frequency that propagates in between the parallel plates separated by a distance of 3 cm in air Cut off frequency fc = 5 GHz 15
- 16. Solution: f = 1GHz and fc = 5GHz Condition for Propagation : Wave does not propagates Find the cut-off frequency of the TE10 Mode wave operating at 1GHz frequency, that propagates in between the parallel plates separated by a distance of 3 cm in air . Check whether the wave propagates. Operating frequency ( f ) < Cut off frequency (fc) 16
- 17. Solution: f = 1GHz Cut off frequency fc = 𝑚 2𝑎 𝜇𝜀 = 1 𝑥 3 𝑥108 2𝑥3𝑥10−2𝑥 4𝑥9 = 5 𝑋109 6 = 0.83GHz Wave propagates Find the cut-off frequency of the TE10 Mode wave operating at 1GHz frequency, that propagates in between the parallel plates separated by a distance of 3 cm. Ler εr = 4, µr = 9 . Check whether the wave propagates. Operating frequency ( f ) > Cut off frequency (fc) 17
- 18. Then Propagation constant is 𝛾 = 𝛼 + 𝑗𝛽 = ℎ2 − 𝜔 2 𝜇𝜀 = 𝑚𝜋 𝑎 2 − 𝜔2𝜇𝜀 𝛾 = 𝑚𝜋 𝑎 2 − 𝜔2𝜇𝜀 =102.55 𝛾 is purely real Find the cut-off frequency of the TE10 Mode wave operating at 1GHz frequency, that propagates in between the parallel plates separated by a distance of 3 cm in air . Evaluate the propagation constant. 𝜸 = 𝜶 18 Non propagation or Evanescent mode
- 19. Guide wavelength λg = 𝜆 1− 𝑓𝑐 𝑓 2 Group velocity vg = c 1 − 𝑓𝑐 𝑓 2 Wave impedance ZTE = η 1− 𝑓𝑐 𝑓 2 Phase velocity vp = c 1− fc f 2 Consider the TE10 Mode wave operating at 10GHz frequency and propagating in between the parallel plates separated by a distance of 3 cm in air. Find guide wavelength, group velocity, phase velocity, wave impedance Problem 2 Solution: 19
- 20. Problem 2… Given : m =1, a = 3x 10-2 m f = 1 x 10 9 Hz & Medium is air, εr = 1 & µr = 1 fc = 5 GHz Find 1 − 𝑓𝑐 𝑓 2 = 1 − 5 10 2 = 0.87 Cut off frequency fc = 𝑚 2𝑎 𝜇𝜀 20
- 21. Guide wavelength λg = 𝜆 1− 𝑓𝑐 𝑓 2 𝜆 = 𝑐 𝑓 = 3 𝑥 108 10 𝑥 109 = 0.03 m λg = 𝜆 1− 𝑓𝑐 𝑓 2 = 0.03 0.87 = 0.034 m Problem 2… 21
- 22. Group velocity vg = c 1 − 𝑓𝑐 𝑓 2 vg = 3 x 108 x 0.87 = 2.61 x 108 m/sec Phase velocity vp = c 1− fc f 2 vp = 3 𝑥 108 0.87 = 3.45 x 108 m/sec Problem 2… 22
- 23. Wave impedance ZTE = η 1− 𝑓𝑐 𝑓 2 Intrinsic impedance η = 𝜇 𝜀 = 𝜇𝑜 𝜀𝑜 = 120𝜋 ≅ 377 Ω ZTE = 377 0.87 = 433.33 Ω Problem 2… ZTM = η 1 − 𝑓𝑐 𝑓 2 = 377 x 0.87 = 327.99 Ω 23
- 24. Rectangular Waveguide • A rectangular wave guide is a hollow metallic pipe with rectangular cross section. • Rectangular waveguide is a rigid structure for the propagation of electromagnetic wave without losses. • A rectangular waveguide supports TM and TE modes but not TEM waves. • Rectangular waveguide usually has a cross section with an aspect ratio of 1:2, the width being about twice the height. 24
- 25. Modes of propagation • A direction along the length of the waveguide is called the Longitudinal direction and perpendicular to the wave propagation is called the Transverse direction. • Consider the rectangular coordinate system such that the z- axis is along the longitudinal direction. • TE mode: The electric vector (E) being always perpendicular to the direction of propagation. • TM mode: The magnetic vector (H vector) is always perpendicular to the direction of propagation. • TEM mode: The Transverse electromagnetic wave cannot be propagated within a waveguide, In TEM wave both the electric vector (E vector) and the magnetic vector (H vector) are perpendicular to the direction of propagation. Ez = 0, Hz ≠0 Hz = 0, Ez ≠ 0 Ez = 0, Hz = 0 25
- 26. Field Components Transverse Electric Wave Transverse Magnetic Wave 26
- 27. Formulae Propagation constant 𝛾 = 𝛼 + 𝑗𝛽 = ℎ2 − 𝜔 2 𝜇𝜀 = 𝑚𝜋 𝑎 2 + 𝑛𝜋 𝑏 2 − 𝜔2𝜇𝜀 Cut off frequency fc = 1 2𝜋 𝜇𝜀 𝑚𝜋 𝑎 2 + 𝑛𝜋 𝑏 2 27
- 28. Formulae Cut off wavelength 𝜆𝑐 = 𝑣 𝑓𝑐 Guide wavelength 𝜆𝑔 = 𝜆 1− 𝑓𝑐 𝑓 2 where 𝜆 = 𝑐 𝑓 , c = 3 x108 m/s Phase velocity vp = 𝑐 1− 𝑓𝑐 𝑓 2 Group velocity vg = c 1 − 𝑓𝑐 𝑓 2 28
- 29. Formulae… Transverse Electric Wave Wave impedance ZTE = η 1− 𝑓𝑐 𝑓 2 Dominant Mode : TE10 (m = 1, n = 0) Transverse Magnetic Wave Wave impedance ZTM = η 1 − 𝑓𝑐 𝑓 2 Dominant Mode : TM11 (m = 1, n = 1) 29
- 30. Solution : Given : a = 10 x 10-2 m b = 4 x 10-2 m Minimum frequency is the cut off frequency of the dominant mode , TE10 , thus m=1, n=0 cut off frequency fc = 1 2𝜋 𝜇𝜀 𝑚𝜋 𝑎 2 + 𝑛𝜋 𝑏 2 Problem 3 Determine the minimum frequency which can propagate in an air filled rectangular waveguide with the inner wall dimension as 4 cm x 10 cm. 30
- 31. cut off frequency fc = 3 𝑥 108 2𝜋 1.𝜋 10 𝑥10−2 2 + 0.𝜋 4 𝑥 10−2 2 fc = 1.5 x 109 Minimum frequency fc = 1.5 GHz Problem 3… 31
- 32. Solution : Given : a = 3 x 10-2 m b = 1.2 x 10-2 m The dominant mode of TM wave is TM11 , thus m=1, n=1 cut off frequency fc = 1 2𝜋 𝜇𝜀 𝑚𝜋 𝑎 2 + 𝑛𝜋 𝑏 2 Problem 4 The rectangular metal wave guide filled with a dielectric material of relative permittivity 𝜀𝑟=4 has the inside wall dimension as 3.0 𝑐𝑚 × 1.2 𝑐𝑚. Obtain the cut off frequency for the dominant mode of TM wave 32
- 33. cut off frequency fc = 3 𝑥 108 2𝜋 4 1.𝜋 3 𝑥10−2 2 + 1.𝜋 1.2 𝑥 10−2 2 fc = 6.73 x 109 Cut off frequency fc = 6.73 GHz Problem 4… 33
- 34. Solution : Given : For TE30 mode, fc = 18 GHz, m = 3 , n =0 Broader wall dimension is ‘a’ cut off frequency fc = 1 2𝜋 𝜇𝜀 𝑚𝜋 𝑎 2 + 𝑛𝜋 𝑏 2 Problem 5 Compute the inner broad wall dimension of the rectangular waveguide having TE10 mode as dominant mode and with a cut off frequency of 18 GHz for the TE30 mode. 34
- 35. cut off frequency for TE30 = 18x109 = 3 𝑥 108 2𝜋 3.𝜋 𝑎 2 + 0.𝜋 𝑏 2 18x109 = 3 𝑥 108 2𝜋 3𝜋 𝑎 a = 9 𝑥 108 2𝑥18𝑥109 a = 0.025 m Broader wall dimension ‘a’ = 2.5 cm Problem 5… 35
- 36. Solution : Given : For TE10 mode, m = 1 , n =0 f = 10GHz a = 2.286 cm and b = 1.016 cm cut off frequency fc = 1 2𝜋 𝜇𝜀 𝑚𝜋 𝑎 2 + 𝑛𝜋 𝑏 2 Problem 6 Determine the propagation constant (per meter) for the dominant mode in an air filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm, operating at the frequency of 10 GHz. 36
- 37. cut off frequency for TE10 fc = 3 𝑥 108 2𝜋 1.𝜋 2.286 𝑥10−2 2 + 0.𝜋 1.016 𝑥 10−2 2 fc = 6.56 GHz As f > fc , there is no attenuation to the wave, it propagates with a phase constant β. Propagation constant 𝛾 = 𝛼 + 𝑗𝛽 = ℎ2 − 𝜔 2 𝜇𝜀 = 𝑚𝜋 𝑎 2 + 𝑛𝜋 𝑏 2 − 𝜔2𝜇𝜀 𝛾 = 1.𝜋 2.286 𝑥10−2 2 + 0.𝜋 1.016 𝑥 10−2 2 − 2𝜋𝑥10𝑥109 211.12𝑥10−18 𝛾 = 𝛼 + 𝑗𝛽 = j 204.77 m-1 Problem 6… 37
- 38. Solution : Given : For TE10 mode, m = 1 , n = 0 fc = 6 GHz For TM11 mode, m = 1 , n = 1 fc = 15 GHz cut off frequency fc = 1 2𝜋 𝜇𝜀 𝑚𝜋 𝑎 2 + 𝑛𝜋 𝑏 2 Problem 7 An air filled rectangular waveguide of internal dimension a 𝑐𝑚×𝑏 𝑐𝑚 (𝑎>𝑏) has a cut off frequency of 6 GHz for the dominant TE10 mode and cutoff frequency of 15 GHz for TM11 mode. Find the cut off frequency of the TE01 mode . ____________ 38
- 39. cut off frequency for TE10 fc = 6x109 = 3 𝑥 108 2𝜋 1.𝜋 𝑎𝑥 10−2 2 + 0.𝜋 𝑏 𝑥 10−2 2 𝑎𝑥 10−2 = 0.025 = 2.5 x 10-2 a = 2.5 cm cut off frequency for TM11 fc = 15x109 = 3 𝑥 108 2𝜋 1.𝜋 2.5 𝑥 10−2 2 + 1.𝜋 𝑏 𝑥 10−2 2 Problem 7… 39
- 40. 1.𝜋 2.5 𝑥 10−2 2 + 1.𝜋 𝑏 𝑥 10−2 2 = 100 1 𝑏 𝑥 10−2 2 = (100)2 – 1600 = 8400 𝑏 𝑥 10−2 = (1 / 8400) = 0.0109 𝑏 𝑥 10−2 = 1.09 x 10-2 b = 1.09 cm Problem 7… 40
- 41. cut off frequency for TE01 fc = 3 𝑥 108 2𝜋 0.𝜋 2.5𝑥 10−2 2 + 1.𝜋 1.09 𝑥 10−2 2 fc = 13.76 GHz Problem 7… 41
- 42. • A circular waveguide consists of a hollow metallic cylinder with an inner radius r. • In the inner air-filled volume of the cylinder electromagnetic waves can propagate above mode-specific cut-off frequencies fc,mn. • Solutions of Maxwell's equations can be found using cylindrical coordinates and Bessel functions . • TM waves corrrespond to the infinite number of roots of Jn(xnm) = 0. • TE waves corrrespond to the infinite number of roots of J’n(xnm) = 0. • The dominant mode in a circular waveguide is the TE11 mode, TM01 mode. Circular Waveguide 42
- 43. Circular Waveguide - TM mode 43
- 44. Circular Waveguide - TE mode 44
- 45. Transverse Magnetic Wave Cut off frequency fc = 𝐽𝑛(𝑥𝑛𝑚) 2𝜋𝑎 𝜇𝜀 Transverse Electric Wave Cut off frequency fc = 𝐽𝑛 ′ (𝑥𝑛𝑚) 2𝜋𝑎 𝜇𝜀 Cut off frequency 45
- 46. Solution : Given : a = 1 x 10-2 m Transverse Electric Wave Dominant mode is TE11 Cut off frequency fc = 𝐽𝑛 ′ (𝑥𝑛𝑚) 2𝜋𝑎 𝜇𝜀 fc = 1.841 𝑥 3𝑥108 2𝜋𝑥1𝑥 10−2 = 8.795 GHz Problem 8 An air filled circular waveguide having an inner radius of 1 cm is excited in dominant mode at 10GHz. Find the cut off frequency, guide wavelength, wave impedance 46
- 47. Guide Wavelength 𝜆𝑔= 𝜆 1− ( 𝑓𝑐 𝑓 )2 = 6.3 x 10-2 m Wave Impedance ZTE = η 1− ( 𝑓𝑐 𝑓 )2 = 792 Ω Problem 8… 47
- 48. Solution : Given : a = 4 x 10-2 m , f = 2GHz , εr = 2.25 Transverse Electric Wave Dominant mode is TE11 Cut off frequency fc = 𝐽𝑛 ′ (𝑥𝑛𝑚) 2𝜋𝑎 𝜇𝜀 fc = 1.841 𝑥 3𝑥108 2𝜋 𝑥 2.25 𝑥 4𝑥10−2 = 1.47 GHz Problem 9 A circular waveguide of radius 4 cm is filled with a material of dielectric constant 2.25. The guide is operated at a frequency of 2 GHz. For the dominant TE mode, determine the cutoff frequency, the guide wavelength 48
- 49. Guide Wavelength 𝜆𝑔= 𝜆 1− ( 𝑓𝑐 𝑓 )2 = 0.1 0.678 Where 𝜆 = 𝑣 𝑓 = 3 𝑥 108 2.252𝑥109 = 0.1 Guide Wavelength 𝜆𝑔= 0.147 m Problem 9… 49
- 50. Solution : Given : a = 5 x 10-2 m , f = 3GHz Transverse Electric Wave Dominant mode is TE11 Cut off frequency fc = 𝐽𝑛 ′ (𝑥𝑛𝑚) 2𝜋𝑎 𝜇𝜀 fc = 1.841 𝑥 3𝑥108 2𝜋 𝑥 5 𝑥10−2 = 1.76 GHz Problem 10 An air filled circular waveguide of radius 5 cm with operates at the frequency of 3GHz with TE11 mode propagating through it. Compute the cutoff frequency, the guide wavelength and the propagation constant. 50
- 51. Guide Wavelength 𝜆𝑔= 𝜆 1− ( 𝑓𝑐 𝑓 )2 = 0.1 0.81 Where 𝜆 = 𝑣 𝑓 = 3 𝑥 108 3𝑥109 = 0.1 Guide Wavelength 𝜆𝑔= 0.123m Problem 10… 51
- 52. 𝛾 = 𝛼 + 𝑗𝛽 = ℎ2 − 𝜔2𝜇𝜀 Where h = 𝐽𝑛 ′ (𝑥𝑛𝑚) 𝑎 = 1.841 5 x10−2 =36.82 𝛾 = 𝛼 + 𝑗𝛽 = 36.822 − 2𝑥3.14𝑥3𝑥109 2/(3𝑥108)2 𝛾 = 𝛼 + 𝑗𝛽 = 1.36𝑥103 − 3.95 𝑥 103 =j 50.89 Thus, 𝛂 = 0 (no attenuation) & 𝛃 = 50.89 rad/m (wave propagates) Problem 10… 52
- 53. • Waveguides are used in air borne radar • Used in satellite communication. • Circular waveguides used to receive signal from the antennas which transmit signal around 360o. • They are used Photonic integrated circuits. • Handle high power of energy. • They are used in space crafts. Applications of Waveguide 53
- 54. Thank you 54