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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition

Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition simplifies even the toughest textbook questions with step-by-step solutions and easy explanations.

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    Solution Manual for Engineering Electromagnetics and Waves: Advanced Topics Umran S. Inan, Aziz S. Inan, Ryan K. Said

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    Contents A Transmission Lines: Advanced Topics 1 C Cylindrical Waveguides 9 D Cavity Resonators 37 E Field-Matter Interactions: Advanced Topics 45 F Electromagnetic Radiation and Elementary Antennas 58

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    Solutions to Addendum A Transmission Lines: Advanced Topics A.1 Nonlinear termination. We start by plotting (see Figure A.1) the current-voltage characteristics of the source and load ends of the 50 Ω line given by 5 = 10 ) s + 9 s and ) L = 0 : 35 ( 1 e 9 L = 2 ) At t = 0, a voltage of amplitude 9 + 1 = Z 0 R s + Z 0 V 0 = 50 10 + 50 ( 5 V ) ≃ 4 : 17 V and an accompanying current ) + 1 = 9 + 1 = Z 0 ≃ 0 : 0833 A are launched at the source end of the line, as indicated by point A which is the intersection of the line 9 + 1 = Z 0 ) + 1 (i.e., line OA) with the source-end line, defined by 5 = 10 ) s + 9 s . Next, we draw a straight line (line AB) with slope [ Z 0 ] 1 = [ 50 ] 1 passing through point A and find its intersection with the current-voltage characteristics of the nonlinear load which is denoted as point B, as shown. The coordinates of point B which correspond to the values of 9 L and ) L at t = t + d = 2 + ns are approximately given by 9 L ≃ 1 : 072 V and ) L ≃ 0 : 145 A respectively. Using the value of 9 L , the amplitudes of the reflected voltage and current can be determined as 9 1 = 9 L 9 + 1 ≃ 1 : 072 4 : 167 = 3 : 095 V and ) 1 = 9 1 = Z 0 ≃ + 0 : 0619 A respectively. We then draw a straight line (line BC) with slope +[ Z 0 ] 1 and passing through point B and find its intersection with the source-end line indicated as point C, as shown in Figure A.1. This process continues on until steady-state is reached at approximately point G or at t ≃ 8 t d = 16 ns. Note that at steady-state, the source- and load-end voltages and currents become 9 s = 9 L ≃ 2 : 50 V and ) s = ) L ≃ 0 : 25 A which correspond to the intersection point of the source- and load-end voltage-current characteristics as indicated in Figure A.1.

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    2 Addendum A/ Transmission Lines: Advanced Topics Figure A.1 Figure for Problem A.1. Application of the graphical Bergeron tech- nique to Problem A.1. The points A,B,C,D,E,F, and G correspond respectively to times t d ; 2 t d ; 3 t d ; 4 t d ; 5 t d ; 6 t d , and 7 t d . A.2 Nonlinear source. (a) We start by plotting the current-voltage characteristics of the source and load ends of the 50 Ω line given by ) s = 8 ( 5 9 s ) ( 5 9 s ) 2 and ) L = R 1 L 9 L ≃ 0 Note that the current-voltage characteristics of the load end of the line is approximately the horizontal voltage axis since R L ≃ 10 M Ω ≃ 1 . At t = 0, an incident voltage of amplitude 9 + 1 ≃ 0 : 798 V and its current counterpart of amplitude ) + 1 = Z 0 1 9 + 1 ≃ 16 : 0 mA (which are the coordinates of point A, the intersection point between line OA and the nonlinear source-end current-voltage characteristics as shown) are launched at the source end of the 50 Ω line. Next, we draw a straight line (line AB) with slope [ Z 0 ] 1 = [ 50 ] 1 passing through point A and find its intersection with the load-end current-voltage characteristics which is denoted as point B on the horizontal voltage axis as shown. The coordinates of point B ( ) B = 0 mA and 9 B ≃ 1 : 60 V) correspond to the values of ) L and 9 L at t = 2 + ns from which the amplitudes of the reflected current and voltage from the load end can be determined as ) 1 = ) L ) + 1 = 0 16 = 16 mA

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    3 and 9 1 = 9 L 9 + 1 ≃ 1 : 6 0 : 798 ≃ 0 : 798 V respectively. We then draw another straight line (line BC) with slope +[ Z 0 ] 1 passing through point B and find its intersection with the nonlinear source-end current-voltage characteristics indicated as point C as shown in Figure A.2. Using these coordinates ( 9 C ≃ 2 : 31 V ; ) C ≃ 14 : 3 mA), the amplitudes of the current and voltage reflected from the source end can be determined as ) + 2 = ) C ) + 1 ) 1 ≃ 14 : 3 16 + 16 = 14 : 3 mA and 9 + 2 = 9 C 9 + 1 9 1 ≃ 2 : 31 2  0 : 798 ≃ 0 : 714 V respectively. This process continues on. Using the voltage coordinates of points B, D, F, etc., the load-end voltage can be sketched with respect to time as shown. (b) It can be approximately observed from the Bergeron diagram in Figure A.2 that the steady state condition is reached in t ≃ 12 t d = 24 ns, or at point M. Figure A.2 Figure for Problem A.2. Application of the graphical Bergeron technique and 9 L versus t . A.3 Effects of source risetime. The one-way time delay along the trace is given by t d = ( 80 ps-cm 1 )( 25 cm ) = 2 ns. At t = 0, an incident voltage of amplitude

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    4 Addendum A/ Transmission Lines: Advanced Topics 9 + 1 ( t ) = 8 > < > : Z 0 V 0 ( R s + Z 0 ) t r t = ( 4 t r ) t t  t r Z 0 V 0 R s + Z 0 = 4 V t  t r is launched from the source end of the trace. This voltage reaches the short-circuited end of the trace at t = t d , when a voltage of amplitude 9 1 ( t ) = 8 > < > : L 9 + 1 ( t t d ) = ( 4 t r ) ( t t d ) t d < t  t d + t r 4 V t  t d + t r (where t d = 2 ns) reflects back towards the source since L = 1. This reflected voltage arrives at the source end of the line at t = 2 t d and a new voltage of amplitude 9 + 2 ( t ) = 8 > < > : s 9 1 ( t t d ) = ( 2 : 4 t r ) ( t 2 t d ) 2 t d < t  2 t d + t r 2 : 4 V t  2 t d + t r is launched back towards the load since s = 12 : 5 50 12 : 5 + 50 = 0 : 6 At t = 3 t d , a new reflected voltage 9 2 ( t ) of amplitude 9 2 ( t ) = 8 > < > : L 9 + 2 ( t t d ) = ( 2 : 4 t r ) ( t 3 t d ) 3 t d < t  3 t d + t r 2 : 4 V t  3 t d + t r is launched towards the source. This process continues indefinitely, with the voltages and currents along the trace gradually reaching their steady-state values. The source-end voltage 9 s ( t ) can be written in terms of the above voltages as 9 s ( t ) = ( 4 t r ) [ tu ( t ) ( t t r ) u ( t t r )] (A.1) ( 4 t r ) [( t 2 t d ) u ( t 2 t d ) ( t 2 t d t r ) u ( t 2 t d t r )] (A.2) + ( 2 : 4 t r ) [( t 2 t d ) u ( t 2 t d ) ( t 2 t d t r ) u ( t 2 t d t r )] (A.3) ( 2 : 4 t r ) [( t 4 t d ) u ( t 4 t d ) ( t 4 t d t r ) u ( t 4 t d t r )] (A.4) + ( 1 : 44 t r ) [( t 4 t d ) u ( t 4 t d ) ( t 4 t d t r ) u ( t 4 t d t r )] (A.5) + : : : (A.6)

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    5 Figure A.3 Figure for Problem A.3. 9 s versus t for t r = 1 ns and t r = 0 : 25 ns. (a) For t r = 1 ns, the source-end voltage 9 s ( t ) given by the above expression is sketched as a function of t in Figure A.3. (b) For t r = 250 ps, the source-end voltage 9 s ( t ) is sketched in Figure A.3. We see that all transi- tions (e.g., those starting at 0, 4, or 8 seconds) are sharper for t r = 250 ps. A.4 Effects of source risetime. The phase velocity along the line can be found as v p = [ Z 0 C ] 1 = [( 50 Ω)( 1 pF-cm 1 )] 1 = 20 cm-ns 1 . At t = 0, a voltage of amplitude 9 + 1 ( t ) = 8 > < > : Z 0 V 0 ( R s + Z 0 ) t r t = ( 2 t r ) t 0 < t  t r Z 0 V 0 R s + Z 0 = 2 V t  t r is applied from the source end of the line with source rise time t r = 1 ns. This voltage reaches the load end of the line t = t d , when a voltage of amplitude 9 1 ( t ) = 8 > < > : L 9 + 1 ( t t d ) = ( 2 t r ) ( t t d ) t d < t  t d + t r 2 V t  t d + t r reflects back towards the source since L ≃ + 1. This voltage arrives at the source end at t = 2 t d when a new voltage of amplitude 9 + 2 ( t ) = 8 > < > : s 9 1 ( t t d ) = ( 2 3 t r ) ( t 2 t d ) 2 t d < t  2 t d + t r 2 3 V t  2 t d + t r is launched back towards the load since the source reflection coefficient is s = 25 50 25 + 50 = 1 3 At t = 3 t d , a new voltage of amplitude

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    6 Addendum A/ Transmission Lines: Advanced Topics 9 2 ( t ) = 8 > < > : ( 2 3 t r ) ( t 3 t d ) 3 t d < t  3 t d + t r 2 3 V t  2 t d + t r is launched back towards the source. This process continues on indefinitely. The source- and the load-end voltage 9 s ( t ) and 9 L ( t ) can be written in terms of the above voltages as 9 s ( t ) = ( 2 t r ) [ tu ( t ) ( t t r ) u ( t t r )] (A.7) + ( 2 t r ) [( t 2 t d ) u ( t 2 t d ) ( t 2 t d t r ) u ( t 2 t d t r )] (A.8) ( 2 3 t r ) [( t 2 t d ) u ( t 2 t d ) ( t 2 t d t r ) u ( t 2 t d t r )] (A.9) ( 2 3 t r ) [( t 4 t d ) u ( t 4 t d ) ( t 4 t d t r ) u ( t 4 t d t r )] (A.10) + ( 2 9 t r ) [( t 4 t d ) u ( t 4 t d ) ( t 4 t d t r ) u ( t 4 t d t r )] (A.11) + : : : (A.12) and 9 L ( t ) = ( 2 t r ) [( t t d ) u ( t t d ) ( t t d t r ) u ( t t d t r )] (A.13) + ( 2 t r ) [( t t d ) u ( t t d ) ( t t d t r ) u ( t t d t r )] (A.14) ( 2 3 t r ) [( t 3 t d ) u ( t 3 t d ) ( t 3 t d t r ) u ( t 3 t d t r )] (A.15) ( 2 3 t r ) [( t 3 t d ) u ( t 3 t d ) ( t 3 t d t r ) u ( t 3 t d t r )] (A.16) + ( 2 9 t r ) [( t 5 t d ) u ( t 5 t d ) ( t 5 t d t r ) u ( t 5 t d t r )] (A.17) + : : : (A.18) (a) For l = 5 cm, the one-way time delay is t d = l = v p = ( 5 cm ) = ( 20 cm-ns 1 ) = 0 : 25 ns. Substituting t d = 0 : 25 ns and t r = 1 ns in the above expressions yield the source- and the load-end voltages 9 s ( t ) and 9 L ( t ) which are both sketched as a function of t in Figure A.4. (b) For l = 50 cm, the one-way time delay is t d = 2 : 5 ns. Again substituting t d = 2 : 5 ns and t r = 1 ns in the above expressions yield the source-end and load-end voltages 9 s ( t ) and 9 L ( t ) which are both sketched as a function of t in Figure A.4. We note that, as expected, the steady-state is achieved much quicker for t d = 0 : 25, and with much less ringing.

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    7 Figure A.4 Figure for Problem A.4. 9 s ( t ) and 9 rmL ( t ) versus t for t d = 0 : 25 ns and t d = 2 : 5 ns. A.5 Half-wave coaxial line resonator. (a) Using Table 2.2 of the text with a = 1 mm, b = 4 mm, and f = 3 GHz, the line parameters of the air-filled coaxial resonator can be evaluated as L = 0 : 2ln4 ≃ 0 : 277  H-m 1 C = 55 : 6 ln4 ≃ 40 : 1 pF-m 1 R = 4 : 15  10 8  5  10 3 p 3  10 9 1  4  10 6 ≃ 2 : 84 Ω -m 1 and G = 0. Since ! L ≃ 5226 ≫ 2 : 84 ≃ R , low-loss approximation is valid and the real and imaginary parts of the propagation constant can be calculated using (3.73) as ≃ R 2 √ C L ≃ 0 : 0171 np-m 1 ≃ ! p LC ≃ 62 : 9 rad-m 1 Using these values, the quality factor follows as Q air ≃ 2 ≃ 1839

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    8 Addendum A/ Transmission Lines: Advanced Topics (b) For teflon-filled coaxial line, the line parameters C and G have new values given by C teflon ) c 2 v 2 p C air ≃ ( 30 20 : 7 ) 2 ( 40 : 1 ) ≃ 84 : 2 pF-m 1 G ≃ 3 : 3  10 4 ln b a ≃ 2 : 38  10 4 S-m 1 Again low-loss approximation is valid since ! L ≫ R and ! C ≫ G . Using (3.73), we find ≃ 0 : 0632 and ≃ 91 : 1 from which the Q factor follows as Q teflon ≃ 1442. (c) Using the line parameters in part (a) along with (3.72), the characteristic impedance of the air- filled coaxial resonator can be calculated as Z 0 ≃ 83 : 14 j 0 : 0226 ≃ 83 : 1 Ω . From Section A.2.2, the element values of the equivalent series RLC circuit can be computed as R eq = Z 0  2 ≃ 0 : 0710 Ω L eq =  Z 0 2 ! ≃ 6 : 93 nH C eq = 2 Z 0 ! ≃ 0 : 406 pF respectively.

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    Solutions to Addendum C Cylindrical Waveguides C.1 Frequency range of the dominant mode. (a) The cutoff frequency of the dominant TE 10 mode in an air-filled hollow WR-10 rectangular waveguide with a = 0 : 254 cm and b = 0 : 127 cm is given by f c 10 = c 2   a ≃ 3  10 10 2 ( 0 : 254 ) ≃ 59 : 1 GHz (b) Since a = 2 b , the next higher-order modes are TE 11 , TE 20 , and TM 11 . The cutoff frequency of these three modes is f c 11 = f c 20 = 2 f c 10 ≃ 118 GHz (c) Therefore, the frequency band over which only the dominant TE 10 mode can propagate is f c 20 f c 10 ≃ 59 : 1 GHz. C.2 AM waves through a tunnel? The minimum tunnel dimension a min needed for the dominant TE 10 mode to propagate at 1 MHz can be calculated as f c 10 ≃ 3  10 8 2 a min = 10 6 ! a min = 150 m ! As long as a > a min = 150 m, TE 10 will propagate regardless of the value of the other dimension b . In practice, since there is no tunnel which has such a large dimension, AM transmission through a tunnel is not feasible.

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    10 Addendum C/ Cylindrical Waveguides C.3 Propagation experiment in a road tunnel. (a) For the rectangular railway tunnel under consideration ( a = 17 m and b = 4 : 9 m), the cutoff frequencies of the TE 10 and TE 01 modes are given by f c 10 = c 2 a ≃ 3  10 8 2 ( 17 ) ≃ 8 : 82 MHz f c 01 = c 2 b ≃ 3  10 8 2 ( 4 : 9 ) ≃ 30 : 6 MHz (b) Using (C.17), the lowest-order modes that do not propagate at the 100 MHz FM frequency are TE 92 and TM 92 since the cutoff frequency of these two modes is given by f c 92 ≃ 3  10 8 2  √ ( 2  4 : 9 ) 2 + ( 9  17 ) 2 ≃ 100 : 3 MHz C.4 Radio-wave propagation in railway tunnels. For tunnel 1 ( a = 8 m and b = 4 m), the total number of propagating modes at 100 MHz is 26. These modes are TE 10 , TE 01 , (TE and TM) 02 , (TE and TM) 11 , (TE and TM) 12 , (TE and TM) 20 , (TE and TM) 21 , (TE and TM) 22 , (TE and TM) 30 , (TE and TM) 31 , (TE and TM) 32 , (TE and TM) 40 , (TE and TM) 41 , and (TE and TM) 50 . For tunnel 2 ( a = 6 m and b = 4 m), the total number of propagating modes at 100 MHz is 18. Unlike tunnel 1, (TE and TM) 32 , (TE and TM) 40 , (TE and TM) 41 , and (TE and TM) 50 modes do not propagate. C.5 Rectangular waveguide modes. (a) For the WR-137 rectangular air waveguide ( a = 3 : 484 cm, b = 1 : 58 cm), the cutoff frequencies of the four lowest-order modes TE 10 , TE 20 (and TM 20 ), TE 01 , and TE 11 (and TM 11 ) are respectively calculated from (C.17) as f c 10 ≃ 4 : 31 GHz, f c 20 ≃ 8 : 61 GHz f c 01 ≃ 9 : 49 GHz, and f c 11 ≃ 10 : 4 GHz. (b) For the dominant TE 10 mode, the phase velocity  v p and the guide wavelength   at the two end frequencies (i.e., 5.85 GHz and 8.20 GHz) of the specified frequency range can be calculated using (C.20) and (C.21) as  v p 10 j f = 5 : 85 GHz ≃ 3  10 8 √ 1 ( 4 : 31 5 : 85 ) 2 ≃ 4 : 43  10 8 m-s 1  v p 10 j f = 8 : 20 GHz ≃ 3  10 8 √ 1 ( 4 : 31 8 : 20 ) 2 ≃ 3 : 52  10 8 m-s 1   10 f = 5 : 85 GHz ≃ 3  10 8 = ( 5 : 85  10 9 ) √ 1 ( 4 : 31 5 : 85 ) 2 ≃ 7 : 57 cm   10 f = 8 : 20 GHz ≃ 3  10 8 = ( 8 : 20  10 9 ) √ 1 ( 4 : 31 8 : 20 ) 2 ≃ 4 : 30 cm

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    11 C.6 Dominant mode in a rectangular waveguide. (a) The cutoff frequency of the dominant TE 10 mode in the WR-137 rectangular air waveguide ( a = 3 : 484 cm, b = 1 : 58 cm) can be calculated from (C.17) as f c 10 = c 2 a ≃ 3  10 10 2 ( 3 : 484 ) ≃ 4 : 31 GHz Using this value, the propagation constant  10 and wave impedance Z TE 10 at 3.5 GHz can be found using (C.16) and (C.28) as  10 =  10 = √ ( f c 10 f ) 2 1 ≃ 2  ( 3 : 5  10 9 ) 3  10 8 √ ( 4 : 31 3 : 5 ) 2 1 ≃ 52 : 5 np-m 1 Z TE 10 =  air √ 1 ( f c 10 = f ) 2 ≃ 377 √ 1 ( 4 : 31 = 3 : 5 ) 2 ≃ j 526 : 3 Ω Note that the dominant TE 10 mode is not propagating at 3.5 GHz as expected since f c 10 ≃ 4 : 31 GHz > f = 3 : 5 GHz. (b) Repeating the same calculations as in part (a) at 7 GHz, we have  10 = j √ 1 ( f c 10 f ) 2 ≃ j 2  ( 7  10 9 ) 3  10 8 √ 1 ( 4 : 31 7 ) 2 ≃ j 115 : 6 rad-m 1 Z TE 10 ≃ 377 √ 1 ( 4 : 31 = 7 ) 2 ≃ 478 Ω Note that the dominant TE 10 mode is a propagating mode at 7 GHz since  4 : 31 GHz < 7 GHz. C.7 Waveguide wavelength. (a) Using (C.21), the guide wavelength of the dominant TE 10 mode propagating in a rectangular air waveguide with a = 2 b = 7 : 214 cm can be written as   10 = c = f √ 1 ( f c 10 = f ) 2 ≃ 30 √ [ f ( GHz )] 2 ( 2 : 08 ) 2 cm where c ≃ 3  10 10 cm-s 1 and f c 10 = c = ( 2 a ) ≃ 2 : 08 GHz are used. Note that the expression found for   10 above is only valid for f >  2 : 08 GHz (i.e., propagating case). This expression (i.e.,   10 ) is plotted as a function of frequency f as shown in Figure C.1. (b) If the same waveguide is filled with water (assume ϵ r ≃ 81), then, f c 10 ≃ 2 : 08 = p 81 ≃ 0 : 231 GHz. Therefore, the guide wavelength for this case is given by   10 ≃ 30 = p 81 √ [ f ( GHz )] 2 ( 0 : 231 ) 2 cm

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    12 Addendum C/ Cylindrical Waveguides 2 4 6 8 5 10 15 20 25 f (GHz) λ (cm) TE 10 TE 20 0.2 0.4 0.6 0.8 5 10 15 20 25 λ (cm) f (GHz) TE 10 Air-filled waveguide Water-filled waveguide Figure C.1 Figure for Problem C.7. Right panel:   10 and   20 versus f for air-filled waveg- uide. Left panel:   10 versus f for water-filled waveguide (i.e., ϵ r ≃ 81). which is valid for f >  0 : 231 GHz. This expression is plotted in Figure C.1 as a function of the frequency f . (c) For the propagating TE 20 mode in the air waveguide of part (a), the   20 expression is the same as the   10 expression except 2.08 should be replaced by 4.16 (i.e., f c 20 = 2 f c 10 ). This expression is valid for f >  4 : 16 GHz and is plotted as a function of f in Figure C.1. For the propagating TE 20 mode in the water-filled waveguide of part (b),   20 expression is the same as the   10 expression found in part (b) except that 0.231 should be replaced by 0.462. This expression is valid for f >  0 : 462 GHz and is plotted in Figure C.1. C.8 Rectangular waveguide modes. (a) For the WR-42 rectangular air waveguide ( a = 1 : 067 cm, b = 0 : 432 cm), the cutoff frequencies of the TE10, TE 01 , TE 11 , TE 20 , and TE 02 modes are given by f c 10 ≃ 3  10 10 2 ( 1 : 067 ) ≃ 14 : 06 GHz < 40 GHz f c 01 ≃ 3  10 10 2 ( 0 : 432 ) ≃ 34 : 72 GHz < 40 GHz f c 11 ≃ 3  10 10 2 √ ( 1 : 067 ) 2 + ( 0 : 432 ) 2 ≃ 37 : 46 GHz < 40 GHz f c 20 ≃ 3  10 10 ( 1 : 607 ) ≃ 28 : 12 GHz < 40 GHz f c 02 ≃ 3  10 10 ( 0 : 432 ) ≃ 69 : 44 GHz > 40 GHz Note that all these modes except the TE 02 mode are propagating at 40 GHz. The propagation constants and the wave impedances of these modes can be calculated using (C.18) and (C.19) as  10 ≃ j 2  ( 4  10 10 ) 3  10 8 √ 1 ( 14 : 06 40 ) 2 ≃ j 784 : 3 rad-m 1

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    13 Z TE 10 ≃ 377 √ 1 ( 14 : 06 40 ) 2 ≃ 402 : 7 Ω  01 ≃ j 2  ( 4  10 10 ) 3  10 8 √ 1 ( 34 : 72 40 ) 2 ≃ j 415 : 9 rad-m 1 Z TE 01 ≃ 377 √ 1 ( 34 : 72 40 ) 2 ≃ 759 : 4 Ω  11 ≃ j 2  ( 4  10 10 ) 3  10 8 √ 1 ( 37 : 46 40 ) 2 ≃ j 293 : 8 rad-m 1 Z TE 11 ≃ 377 √ 1 ( 37 : 46 40 ) 2 ≃ 1075 Ω  20 ≃ j 2  ( 4  10 10 ) 3  10 8 √ 1 ( 28 : 1 40 ) 2 ≃ j 595 : 9 rad-m 1 Z TE 20 ≃ 377 √ 1 ( 28 : 1 40 ) 2 ≃ 530 Ω  02 ≃ 2  ( 4  10 10 ) 3  10 8 √ ( 69 : 44 40 ) 2 1 ≃ 1189 np-m 1 Z TE 02 ≃ 377 √ 1 ( 69 : 44 40 ) 2 ≃ j 265 : 65 Ω (b) For the TE 10 mode, we have from (C.20) and (C.21)  v p 10 ≃ 3  10 8 √ 1 ( 14 : 06 40 ) 2 ≃ 3 : 20  10 8 m-s 1   10 ≃ 3  10 10 4  10 10 √ 1 ( 14 : 06 40 ) 2 ≃ 0 : 801 cm Similarly, for the TE 01 mode, we have  v p 01 ≃ 3  10 8 √ 1 ( 34 : 72 40 ) 2 ≃ 6 : 04  10 8 m-s 1   01 ≃ 3  10 10 4  10 10 √ 1 ( 34 : 72 40 ) 2 ≃ 1 : 51 cm In a similar fashion, for the TE 11 mode, we obtain  v p 11 ≃ 8 : 56  10 8 m-s 1 ,   11 ≃ 2 : 14 cm, and for the TE 20 mode, we obtain  v p 20 ≃ 4 : 22  10 8 m-s 1 , and   11 ≃ 1 : 054 cm respectively.

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