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Unformatted text preview: CHAPTER 14 14.1. A parallelplate waveguide is known to have a cutoff wavelength for the m = 1 TE and TM modes of λ c 1 = . 4 cm. The guide is operated at wavelength λ = 1 mm. How many modes propagate? The cutoff wavelength for mode m is λ cm = 2 nd/m , where n is the refractive index of the guide interior. For the first mode, we are given λ c 1 = 2 nd 1 = . 4 cm ⇒ d = . 4 2 n = . 2 n cm Now, for mode m to propagate, we require λ ≤ 2 nd m = . 4 m ⇒ m ≤ . 4 λ = . 4 . 1 = 4 So, accounting for 2 modes (TE and TM) for each value of m , and the single TEM mode, we will have a total of 9 modes. 14.2. A parallelplate guide is to be constructed for operation in the TEM mode only over the frequency range < f < 3 GHz. The dielectric between plates is to be teflon ( ± R = 2 . 1). Determine the maximum allowable plate separation, d : We require that f < f c 1 , which, using (7), becomes f < c 2 nd ⇒ d max = c 2 nf max = 3 × 10 8 2 √ 2 . 1 ( 3 × 10 9 ) = 3 . 45 cm 14.3. A lossless parallelplate waveguide is known to propagate the m = 2 TE and TM modes at frequencies as low as 10GHz. If the plate separation is 1 cm, determine the dielectric constant of the medium between plates: Use f c 2 = c nd = 3 × 10 10 n( 1 ) = 10 10 ⇒ n = 3 or ± R = 9 14.4. A d = 1 cm parallelplate guide is made with glass ( n = 1 . 45) between plates. If the operating frequency is 32 GHz, which modes will propagate? For a propagating mode, we require f > f cm Using (7) and the given values, we write f > mc 2 nd ⇒ m < 2 f nd c = 2 ( 32 × 10 9 )( 1 . 45 )(. 01 ) 3 × 10 8 = 3 . 09 The maximum allowed m in this case is thus 3, and the propagating modes will be TM 1 , TE 1 , TM 2 , TE 2 , TM 3 , and TE 3 . 14.5. For the guide of Problem 14.4, and at the 32 GHz frequency, determine the difference between the group delays of the highest order mode (TE or TM) and the TEM mode. Assume a propagation distance of 10 cm: From Problem 14.4, we found m max = 3. The group velocity of a TE or TM mode for m = 3 is v g 3 = c n s 1 − ± f c 3 f ² 2 where f c 3 = 3 ( 3 × 10 10 ) 2 ( 1 . 45 )( 1 ) = 3 . 1 × 10 10 = 31 GHz 249 14.5. (continued) Thus v g 3 = 3 × 10 10 1 . 45 s 1 − ± 31 32 ² 2 = 5 . 13 × 10 9 cm / s For the TEM mode (assuming no material dispersion) v g,T EM = c/n = 3 × 10 10 / 1 . 45 = 2 . 07 × 10 10 cm/s. The group delay difference is now 1t g = z ± 1 v g 3 − 1 v g,T EM ² = 10 ± 1 5 . 13 × 10 9 − 1 2 . 07 × 10 10 ² = 1 . 5 ns 14.6. The cutoff frequency of the m = 1 TE and TM modes in a parallelplate guide is known to be f c 1 = 7 . 5 GHz. The guide is used at wavelength λ = 1 . 5 cm. Find the group velocity of the m = 2 TE and TM modes. First we know that f c 2 = 2 f c 1 = 15 GHz. Then f = c/λ = 3 × 10 8 /. 015 = 20 GHz. Now, using (23), v g 2 = c n s 1 − ± f c 2 f ² 2 = c n s 1 − ± 15 20 ² 2 = 2 × 10 8 /n m / s n was not specified in the problem. 14.7. A parallelplate guide is partially filled with two lossless dielectrics (Fig. 14.23) where ± R 1 = 4 . 0, ± R 2 = 2 . 1, and d = 1 cm. At a certain frequency, it is found that the TM 1 mode propagates through the guide without suffering any reflective loss at the dielectric interface....
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This note was uploaded on 06/10/2008 for the course EE 325 taught by Professor Brown during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Brown
 Electromagnet, Gate

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