Chapter 4 Guided Waves Contents Waves between parallel ties of propagation, planes. TE and TM waves and their characteristics. TEM waves, veloc Attenuation in parallel plane guides, wave impedance. POINTS TOREMEMBER 1. Guided 2. The field waves use some components guiding surtaces of wave between are Ex x Ey joxHz for the actual propagation of wavees. parallel planes propagating in positive z-direction H and Ez Hy = joe h2 ax electric wave Transverse 4 (H waves) having Ez = 0. Field components of TE waves are: Ey C sinTX-iB a H and 5 Tra dnsverse C, sin mae-iB a Ccos H2jo ua a -iD magnetic waves are such wave having Hz =0 77 LORD ElectromagneticField Theory 78 6. Field components of TM waves are Hy Ca cos = a E, Cjot cos E Ez and 7. TEM 8. 8. When the waves i a - jo ea C4 sin TXeiP2 = a entirely waves are iB strikes transverse in nature. on the walls of waveguide certain losses are taken place. Power lost per unit length Thus attenuation takes place and attenuation factor is a 9. Wave impedance is the ratio of electric field strength 2 (power transmitted) to the magnetic field strength. QUESTION-ANSWERS Q 1. What are the characteristics of TM waves? Ans. 1. Propagation constant, = 2. Cut off wavelength, A 2a m 3. Wavelength, = 4.Velocity;V=m/s When frequency becomes high enough for air as medium C 3x 108 m/s. E- (PTU, Dec. 2006; May 2006)) Auided Waves 79 a 2. What is wave Ans. The Wave feld strength i.e. impedance? (PTU, Dec. 2006 ; May 2006) impedance is the ratio of transverse components of electric to magneu H Q 3. Define velocity of wave propagation. Ans. The velocity of propagation is times the frequency i.e. (PTU, May 2007) p Af and for free space the velocity of propagation is equal to velocity of light i.e. Vp 3 x 108 m/s. Q 4. Why em waves are also called as guided waves? (PTU, May 2007) Ans. As electromagnetic waves are guided along the conducting or dielectric surfacesS so it is called a guided waves. Q5. What do you mean by TEM waves? (PTU, Dec. 2007) Ans. TEM wave is transverse electromagnetic wave, it having both axial components of electric and magnetic field as zero. Q6. Why em waves are also called as guided waves? (PTU, Dec. 2007) Ans. As em waves propagate in guided media such as parallel conducting planes. Also, the em waves propagate by guiding media along the conducting or dielectric surfaces. So it is called as guided waves. the dissipation factor of the dielectric? Ans. D> Dissipation factor Q7. What is and (PTU, May 2008) D cconduction current density J is displacement current densityand for dielectric D ISvery less than 1. 8/Differentiate between phase velocity and group velocity. (PTU, Dec. 2008) Ans. v=VgVp Where Vggroupvelocity pphasevelocity Vvelocity of light velocity offree space whereas phase velocitv ways greater than velocity of free space. When frequency is cut off frequency group than ane group velocity is always less always the LORD Electromagnetic Field Theory 80 velocity becomes zero and phase velocity infinite so no propagation of energy along the waveguide occurs. (PTU, May 2009) Q9. What are uniform plane waves? Ans. Uniform plane wave have components of E and H at every point in free space are perpendicular to each other and perpendicular to the direction of propagation i.e. these waves are transverse in nature and do not have components in the direction of propagation (i.e. E = H, = 0). Q 10. Define clearly dominant and degenerate modes with examples. (PTU, May 2009, 2005) Ans. Dominant modes are the modes which having highest cut off wavelengths for example:TE10 mode of transverse electric field. Degenerate modes are the modes for which the propagation constant y is same. In these modes the small losses result in strong coupling between modes. a 11. Discuss the properties of uniform plane wave. (PTU, Dec. 2008, 2006) Ans. The properties of uniform plane to () The electric E and magnetic field H, at every point in free space are perpendicular each other and also perpendicular to the direction of propagation. (i) The velocity of propagation of plane wave is C= () The EM wave 1 travels with 3x 108 m/s a velocity of light in the free space for any value of frequency (iv) E and H are in phase and amplitude is 120t. Q 12. A uniform plane wave at 300 MHz is passed through a medium of 4, = 1 and E, 78. Find the wave parameters in that medium assuming a = 0. Ans. AS 27tf = 2x x x 300x 10 as andwave H 1,E 78 velocity C = (PTU, Dec. 2006) 81 A d e dW a v e s 3x 10 78 Y Y and 3x10 30 x10 a 13. Distilled water has the constants o = 0, e, = 81, 4, =1. Find refraction index, shase velocity and group (PTU, May 2007) velocity. Ans. Refractive index 120T as = 1 120T 81 120 40 x22 21 9 42 Phase velocity, V,B VHo So Ho 3x10 m/s 3x 10 x 9 81 do groupvelocity Vp dB Q 14. Derive an m/s expression for the parallel conducting planes. Ans. 3 E and H components for TM attenuation factor for the waves are E, bAs E Ee sin m TM10 Wave between (PTU, May 2007) 82 LORDS Electromagnetic Field Theory Hy = Ex iP A4 cos e A4cos xei Jo E and for TM1o m = 1, n = 0 The current density is equal to the tangential component of Ez at x = 0 and x > a Js IEz nA4 Ea The loss is each plane is power loss per unit length = 2 0 Power transmitted = Ea/ (Ex Hy) BA cosx 20 a E Power between x = 0 and X = a space is BA2 cos 20 E X=0 Power transmitted = PA4a 40 E Attenuation factor Power lost per unit length 2xpower transmitted 2 2 (o ea BAAa 40 E V20 1 Js Rs 83 a d e d Waves 22 2 m Bo ea The attenuation factor of TM10 waves reaches a minimum at a frequency which is 3 fc then increases with frequency. a 15. Discuss the propagation characteristics of TE and TM waves. (PTU, Dec. 2007) Ans. Characteristics of TE and TM waves The characteristics are 1. Propagation constant 2. Wavelength 3. Velocity of propagation as h2 and = h2-oue oue +y2 Y a mT a 16. A 10 GHz plane wave travelling in free space has an amplitude Ey = 1 V/m. Find the phase velocity, wavelength Ans. f and the propagation (PTU, Dec. 2008) 10 GHz, Ex = 1 V/m C f 3x10 0.03 m 10x 109 and v Wavelength, a 3 x 0.03 m C = = = 108 m/s for Propagation constant Y jo (o+jo e and o = 0 for free space ?=-oHe and, Y = jB constant. free space LORDS Electromagnetic Field Theory 84 ory Also, B yoH0 Eo (4,=¬, =1) 2x3.14x 10x10 = o x VHo Eo 209.3 3x10 Q 17. A long straight tubular conductor of circular cross-section with outside. diameter of 5 cm and wall thickness of 0.5 cm carries a direct current of 100 amps. Find H: () just outside the wall of tube. (i) at a point in the tube wall half-way between the inner and outer surface. (PTU, May 2005) Ans. I=100 amp, R2 = H () 5 cm 2TtR2 Ra cm 100x2 636.619 A/m H 2T x5x10-2 (i) H- 2 Ra R = R R-R 2.5x 10-2 m 2.5-0.5 2 cm = 0.02 m +R1 _0025+ 0.02 and 2 0.0255 m H 100 2x0.0225 (O.0225)-(o.02)| (0.025-(0.02) J H 334A/m Q 18. Discuss the propagation characteristics of TE modes. Ans. Take the expression of h2 as h2 Y+ 0 ue (PTU, May 2007) 85 Guided Waves h2- ue and MT h and also, a 2 7 -oe At very high frequencies mr 2 a and becomes pure imaginary. i.e. a and Thus for perfectly conducting planes, direction. Also, HE- (2n1 0 ue=m and m c2a E also .Phase or wave velocity attenuation is zero and wave propagates along z 86 LORD Electromagnetic Field Theor, ( = f and =27 A s frequency increases above critical frequency the v tends to speed of light. Q 19. Derive the expression for the attenuation in the parallel plate guide for TE (PTU, May 2008) mode. Ans. The attenuation in parallel plate guide The E and H components for TE mode between conducting plates as; Ey A sin mXi2 a = H, = A ,sin | HThe e jmnA1 Acos ea @pa magnetic induction of linear current density Js= tangential component of H at x = 0 and x = a i.e (Js) = (H2) mrA ua The current does not flow in the direction of in the each propagation plane of conducting parallel plane isJs2 Rs The power per unit length - maA m oua) 20m Where Hm Rs 26om The power transmitted per unit (EH) -2Ey H area is for these modes and the loss Guided Waves 87 BAL sin?(mXx 20 Sin a Now power transmitted for a conductor between BAT sin? 20 a x =0 and x = a is dx BAa 4 Power lost per unit Now, length 2x power transmitted mA Pm 20m 20a2 2x Aa 40 2m2 oHm V20m 2 3 oua"oyE-mn d of uniform plane wave. Q 20. List the properties of an electromagnetic The electric field intensity Ey E = 0, E Determine = Eo cos w (t- zN) as Ey Ez = in free space is given by H. the components of magnetic intensity for expressions the (PTU, Dec. 2008) Ex Eo cos w (t- zV) The components of H: Ans. wave = 0, = xE aa OH Ho(4,= 1) 88 LORDS Electromagnetic Field Theory ax ay Ox oz Eo Ho H+aHy+ az H2) and 1) and E, Ho 2) Hy PO Q21. Derive the relation between E and H in uniform Define intrinsic plane wave impedance and its Ans. The first give maxwell equation is physical significance. VxE - OB ot i.e. for VxE = Ex Ey Ez uniform plane wave travelling in -(a, B,+a,B +a ot z-direction. = 0 i.e. no x and and y components Ez = 0 ax ay az 0 Ex Ey o2 - la, By +a, By +0] propagation. (PTU, May 2009) B) 89 Guided Waves Byexpanding [a, Ey-ay Exl -ax OBx ay By ot = 0Z Ey OBx B HH We have, Ey-and Ex - oHy From 2nd Maxwell's equation is VxH We have, ax ay Ox ay OD ot lay Dx+ ay Dy +az D,] ot Hx Hy Hz = Now, O and D = 0 OX a So, ay az =lD+a, D, +0 0 H Hy la ODy Hy ay Hd= ax Dx +ay Hence, oty eOEx ot and oHx as, 0Z eOHy ot D =¬E 90 LORDS oHy ot and 0Z We have, Ey f (Z Ct) C where also, =f(Z-c)(Z-Ct) ot = f (Z- ct) -C) Ey =-Cf if we take fh = f (Z- C) from, by putting EyCii ot dx =eC z By integration both sides H=eCdz H =e Cf'+K K is constant of integration, C = VHE H,=e Now, 'Ey-f(Z-c) Electromagnetic Field Theory 91 Guided Waves H and E EHx ly Since, E -E+E2 and H H+H2 Thus, E-H and, Hence, n is intrinsic impedance or characteristic impedance of non-conducting medium. For free space n 4Tx107 36T x 109 Where Thus, H 4Ttx n= n 10-7, 1 e = V4Tx 36 x 100 120 7 36T x 10 LORDS Electromagnetic Field Theory 92 For, 22. T = 3.14 n 120 x 3.14 n 377 Q Derive the expression for attenuation factor for TEM conducting planes. Ans. Consider a transmission line terminated in its waves between characteristics parallel (PTU, May 2009) impedance. The voltage and current expressed as, V Voea7 e-iBz .(1) loea7 elf 2) I and = The average power transmitted is Pav - Re given by .(3) IVI By putting (1) and (2) in (3) We have Re [Vo 1o]e-2z Pav Now, the rate of decrease of power is obtain by differentiating with respect to oPay 0Z and GFav OZ - -2az -2a)Ro and OPav 20 and oFav 20 Pav power lost per unit length = 2a Pav Power lost per unit length and Power transmitted (average 2a. power power Powerlost per unit length 2 (power transmitted This is the attenuation factopr. - 2a z. Guided Waves 93 a 23. Discuss the attenuation of wave in parallel planes. Ans. e-Yz =eAd**c)z-jßz Where hg is the attenuation constant for lossy dielectric A is the power loss in the conducting planes As, time average power transmitted is WT e-2a'z The rate of decrease of transmitted power is dW dz Also, 2a' WT Time rate of transmitted power Time averagetransmitted power 1 ' 2a'WT 2a WT WT Time rate of transmitted power 2 Time averagetransmitted power Q 24. Three regions are shown in figure are lossless and non magnetic medium1>er^ =1 medium 2 er2 =4 medium 3 erg = 9 d= 4 frequency = 2GHz. Find medium 3. (a) Wave impedance in ratio (VSWR) (b) Voltage standing wave medium 3 Ans. (a) Wave impedance in HO = 125.6 2 (b) where and vsWR 1-|Pl VSWR+|p I= n-20 pZin +Zo ZinZa at interface of medium 1 and 2. 4Tx 10 V9x 8.85 x 10-12 94 LORDS Electromagnetic Field Theory 4Tx 10 Where 4 Z and 188.4 Q 8.85x1 0 - 1 2 = 5.42826 Q 125.6 282.6-377 p 0.141 282.6+377 1+0.14 vsWR 1-0.14 = =0.14 1.3 Q 25. Show that a uniform plane wave passing in any direction does not have any component at that direction. (PTU, Dec. 2009) Ans. In uniform plane wave only one component is present among x, y and z directions and it is transverse in nature i.e. uniform plane wave is perpendicular to the direction of propagation. Hence it does not have component among the direction of propagation. a 26. Differentiate between phase velocity and group velocity. (PTU, May 2009) Ans. Phase velocity is the velocity of propagation with which the In a direction parallel to the conducting surface. i.e. wave changes phase. 2tf.a 2T a s B- N Whereas, group velocity is defined as the velocity with which the group of the wave as whole propagation. do i.e. Vp Group velocity is also, Where vo is the SS dß always v less than Vg Vp velocity of velocity of light. frequency phase velocity becomes infinite propagation of energy along the waveguide. At cut off i.e. no light and group Q27. Define the term 'Brewster Angle'. Ans. Brewster angle is the angle at which there is represented as tane 2 O00 velocity becomes zero. (PTU, May 2005) no reflection of wave and it is