The propagation constantk, can be expressedas-（分）f>f.A--()--()-11if f > f., k, is a real number, and the factore-jk. stands for thewavepropagating along the positive z-direction.LIf f < fo, k, is an imaginary number, then e-jk.= =ewhich states that this time-varying electromagnetic field is nottransmitted,butis anevanescentfieldFor a given mode andin a given size waveguide, f. is the lowestfrequency of the mode to be transmitted. In view of this, the waveguideacts likea high-pass filterUV
The propagation constant kz can be expressed as − − − = = − j 1, 1 , 1 c 2 c c 2 c 2 c f f f f k f f f f k f f k k z if , is a real number, and the factor stands for the wave propagating along the positive z-direction. c f f z k k zz j e − If f f c , kz is an imaginarynumber, then 1 j 2 c e e − − − = f f kz k zz which states that this time-varying electromagnetic field is not transmitted, but is an evanescent field. For a given mode and in a given size waveguide, is the lowest frequency of the mode to be transmitted. In view of this, the waveguide acts like a high-pass filter. c f
2元Fromkwe can find the cutoff wavelengtha correspondingthe cutoff propagation constantk,as2元2入。 =kmCThe cutoff frequency or the cutoff wavelength is related to thedimensions of the waveguide a, b and the integers m, n . For a givensize of waveguide, different modes have different cutoff frequenciesand cutoff wavelengths. A mode of higher order has a higher cutofffrequency,orashortercutoffwavelengthThe cutoff wavelength of the TE.oTEo1TE10wave is 2a, and that of TE2o wave is a.TE20The left figure gives the distribution ofTMuthe cutoffwavelengthfora waveguide11a2awith a=2b .u7
0 a 2a c From , we can find the cutoff wavelength corresponding the cutoff propagation constant as 2π k = c k c 2 2 c c 2π 2 + = = b n a k m The cutoff frequency or the cutoff wavelength is related to the dimensions of the waveguide a, b and the integers m, n . For a given size of waveguide, different modes have different cutoff frequencies and cutoff wavelengths. A mode of higher order has a higher cutoff frequency, or a shorter cutoff wavelength. The cutoff wavelength of the TE10 wave is 2a, and that of TE20 wave is a. The left figure gives the distribution of the cutoff wavelength for a waveguide with a = 2b . TM11 TE01 TE20 TE10
If a > a. , then the corresponding mode will be cut off. From thefigure we see that if a > 2a , all modes will be cut off.If a< a<2a, then only TEio waveexists,while allothermodes are cutTEolTE10off. If a<a,then the other modesTE20will be supported.TMuHence, if the operating wavelength1o元satisfiesthe inequalitya2aa<<2aThen the transmission of a single mode is realized, and the TEjo wave isthe single mode to be transmitted.The transmission of a singlemode wave is necessaryin practicesince it is helpful for coupling energy into or out of the waveguideTEro wave is usually used, and it is called the dominant mode ofthe rectangularwaveguideuV
TM11 TE01 TE20 TE10 0 a 2a c TE10 wave is usually used, and it is called the dominant mode of the rectangular waveguide. If , then the corresponding mode will be cut off. From the figure we see that if , all modes will be cut off. c 2a If , then only TE10 wave exists, while all other modes are cut off . a 2a If , then the other modes will be supported. a Hence, if the operating wavelength satisfies the inequality a 2a Then the transmission of a single mode is realized, and the TE10 wave is the single mode to be transmitted. The transmission of a single mode wave is necessary in practice since it is helpful for coupling energy into or out of the waveguide. Cutoff area
In practice, we usually take a>2b to realize the transmission of thesingle mode TEjo in the frequency band a< a<2a .To support the TEjo mode the sizes of the rectangular waveguideshould satisfythe followinginequality元23a<2The lower limit forthe narrow side depends on the transmittedpower, the allowable attenuation, and the weight per unit lengthIn practice, we usually take a= 0.7a and b = (0.4 ~ 0.5)a or (0.1 ~ 0.2)aAs the wavelengthisincreasedthe sizes ofthe waveguidemustbeincreased proportionallyto ensurethe dominantmodeisabovecutoffIf the frequencyis very low, the wavelength will be very long so thatitmay not be convenientfor use.Consequently, metal waveguides are used for microwave bandsabove3GHzuV
The lower limit for the narrow side depends on the transmitted power, the allowable attenuation, and the weight per unit length. As the wavelength is increased, the sizes of the waveguide must be increased proportionally to ensure the dominant mode is above cutoff. If the frequency is very low, the wavelength will be very long so that it may not be convenientfor use. In practice, we usually take to realize the transmission of the single mode TE10 in the frequency band . a 2b a 2a a 2 2 b In practice, we usually take and or . a = 0.7 b = (0.4 ~ 0.5)a (0.1~ 0.2)a To support the TE10 mode the sizes of the rectangular waveguide should satisfy the following inequality Consequently, metal waveguides are used for microwave bands above 3GHz
The phase velocity y,can be found from the phase constantas01k.（（园）Wherev:Ifthe inside ofthewaveguideis vacuum,thenVueVu.8Since the operating frequency f > f. and the operating wavelength< 2, we have y, > c for a vacuum waveguide. Hence, the phase velocitydoes not represent the energy velocity in a waveguide.The phase velocity depends on not only the sizes of the waveguide.the modes, and the properties of the media within the waveguide, butalso the frequency. Hence, an electromagnetic wave will also experiencedispersionin a waveguideUV
The phase velocity can be found from the phase constant as p v v v f f v k v z − = − = = 2 c 2 c p 1 1 Where . If the inside of the waveguide is vacuum, then 1 v = v = = c 0 0 1 The phase velocity depends on not only the sizes of the waveguide, the modes, and the properties of the media within the waveguide, but also the frequency. Hence, an electromagnetic wave will also experience dispersion in a waveguide. Since the operating frequency and the operatingwavelength , we have for a vacuum waveguide. Hence, the phase velocity does not representthe energy velocity in a waveguide. c f f c v c p