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Chapter 9 Guided Electromagnetic Waves导行电磁波 Several wave guiding systems, Electromagnetic waves in rectangular and circular waveguides Coaxial line,Cavity resonator 1. TEM Wave, TE Wave, and TM Wave 2. Equations for Electromagnetic Waves in Rectangular Waveguides 3. Characterization of Electromagnetic Waves in Rectangular Waveguides 4. TE10 Wave in Rectangular Waveguides 5. Group Velocity 6. Circular Waveguides 7. Transmitted Power and Loss in Waveguides 8. Resonant Cavity 9. Coaxial Lines The electromagnetic waves to be transmitted along a confined path are called guided electromagnetic waves, and the systems to transmit the guided electromagnetic waves are called the wave guiding systems. Two-wire line, coaxial line, strip line, microstrip, and metal waveguides are often used in practice. we will discuss the metal waveguides and the coaxial line only. Two-wire line Strip line Coaxial line Rectangular waveguide Microstrip line Circular waveguide Dielectric waveguide, Fiber optic 1. TEM Wave, TE Wave, and TM Wave E E es H es H TEM wave E TE wave es H TM wave The wave guiding systems in which an electrostatic field can exist must be able to transmit TEM wave. From Maxwell’s equations we can prove that the metal waveguide cannot transmit TEM wave. The main properties of several wave guiding systems Systems Wave types EM Wave band shielding Two-wire line TEM wav Poor > 3m Coaxial line TEM wave Good > 10cm Strip line TEM wave Poor Centimeter Microstrip line Quasi-TEM wave Poor Centimeter Rectangular waveguide TE or TM wave Good Centimeter Millimeter Circular waveguide TE or TM wave Good Centimeter Millimeter Fiber optic TE or TM wave Poor Optical wave The general approach to study the wave guiding systems Suppose the wave guiding system is infinitely long, and let it be placed along the z-axis and the propagating direction be along the positive z-direction. Then the electric and the magnetic field intensities can be expressed as E ( x, y, z ) E0 ( x, y )e jk z z H ( x, y, z ) H 0 ( x, y )e jk z z where kz is the propagation constant in the z-direction, and they satisfy the following vector Helmholtz equation: 2 E 2 E 2 E 2 2 2 2 k E 0 y z x 2 2 2 H H H k 2 H 0 x 2 y 2 z 2 The above equation includes six components, E x , E y , E z and H x , H y , H z , in rectangular coordinate system, and they satisfy the scalar Helmhotz equation. Based on the boundary conditions of the wave guiding system and by using the method of separation of variables, we can find the solutions for these equations. From Maxwell’s equations, we can find the relationships between the x-component or the y-component and the z-component as Ex 1 Ez H z j k j z 2 kc x y Ey 1 Ez H z j k j z 2 kc y x 1 E z H z j j k z 2 kc y x 1 E H z H y 2 j z jk z kc x y Hx Where kc2 k 2 k z2 . These relationships are called the representation of the transverse垂直 components by the longitudinal纵向 components. We only need to solve the scalar Helmholtz equation for the longitudinal components, and then from the relationships between the transverse components and the longitudinal components all transverse components can be derived. In the same way, in cylindrical coordinates the z-component can be expressed in terms of the r-component and –component as Er 1 E z H z j k j z 2 kc r r E 1 k z Ez H z j j 2 kc r r Hr 1 Ez H z j j k z 2 kc r r H 1 E z k z H z j j 2 kc r r 2. Equations for Electromagnetic Waves in Rectangular Waveguides Select the rectangular coordinate system and let the broad side be placed along the x-axis, the narrow side along the y-axis, and the propagating direction be along the z-axis. For TM waves, Hz = 0 , and according y to the method of longitudinal fields, the component b a z Ez should first be solved, and from , x which the other components can be derived. The z-component of the electric field intensity can be written as E z E z 0 ( x , y ) e jk z z It satisfies the following scalar Helmholtz equation, i.e. 2 Ez 2 Ez 2 k Ez 0 c 2 2 x y And the amplitude is found to satisfy the same scalar Helmholtz equation, given by 2 Ez 0 2 Ez 0 kc2 Ez 0 0 2 2 x y In order to solve the above equation, the method of separation of variables is used. Let E z 0 ( x、y) X ( x)Y ( y) We obtain X Y kc2 X Y where X" denotes the second derivative of X with respect to x, and Y" denotes the second derivative of Y with respect to y. X Y kc2 X Y The second term on the left side of the above equation is a function of y only, while the right side is a constant. The only way the equation can be satisfied is that both terms on the left side are constants. Now let X k x2 X Y k y2 Y where k x and k y are called the separation constants, and they can be found by using the boundary conditions. Obviously kc2 k x2 k y2 The two equations are second order ordinary differential equations, and the general solutions, are respectively X C1 cos k x x C 2 sin k x x Y C 3 cos k y y C 4 sin k y y where all the constants C1 , C2 , C3 , C4 , and k x , k y , depend on the boundary conditions. Since the component Ez is parallel to the walls, we have Ez = 0 at the boundaries x = 0, a and y = 0, b . Using these results we find mπ nπ kx , m 1,2,3, ky , n 1,2,3, b b And all the field components are E z E0 sin mπ nπ jk z z x sin ye a b Ex j k z E0 mπ mπ nπ x sin cos 2 kc a a b y e jk z z Ey j k z E0 nπ mπ nπ x cos sin 2 kc b a b y e jk z z Hx j E0 nπ Hy j kc2 mπ nπ sin x cos b a b E0 mπ kc2 y e jk z z mπ nπ cos x sin a a b y e jk z z E z E0 sin mπ nπ jk z z x sin ye a b Ex j k z E0 mπ mπ nπ x sin cos 2 kc a a b y e jk z z Ey j k z E0 nπ mπ nπ x cos sin 2 kc b a b y e jk z z Hx j E0 nπ Hy j kc2 mπ nπ sin x cos b a b E0 mπ kc2 y e jk z z mπ nπ cos x sin a a b y e jk z z (c) If mmodes ormnand iswith zero, then for TM wave), and all Ez and 0 ( Hare 0called z the (e) (d) Since larger are multi-valued, m pattern the of modes the ofishigher has (b) The plane zof= the 0n is a wave front.nBecause therelated amplitude related (a) The phase electromagnetic wave is tofield the variable components willalso bemodes, zero. Thus the mwith andless n are non-zero integrals, and order multiple or the forms, higher called and multiple that modes. A m pair and of n m are and called n lead the to modes a ztoonly, the variables x and y. Hence, a traveling x andwhile y, thethe TMamplitude wave is a to non-uniform plane wave. they have clear physical meanings. The value ofinstance, m stands for the number of mode, lower and it isor denoted the the modes. TMmn Since mode. m and n are TM not11is zero, denotes andthe wave isorder formed in lower theasz-direction, andboth aFor standing wave in the x-the of half-cycle variations ofis=the fields along thewave broadwith side,this while n denotes lowest pattern mode of the of field TM wave for m TM 1 , n = in 1 the , and rectangular the waveguide. character is 11 direction and y-direction. that for the narrow side. called TM 11 wave or mode. Similarly, we can derive all the components of a TE wave in the rectangular waveguide, as given by mπ nπ H z H 0 cos x cos a b y e jk z z Hx j k z H 0 mπ mπ nπ jk z z x cos y e sin kc2 a a b Hy j k z H 0 nπ mπ nπ jk z z x sin y e cos kc2 b a b Ex j H 0 nπ Ey j kc2 mπ nπ jk z z x sin y e cos b a b H 0 mπ kc2 mπ nπ jk z z x cos y e sin a a b where m, n 0, 1, 2, , but both should not be zero at the same time. TE wave has the multi-mode characteristics as the TM wave. The lowest order mode of TE wave is the TE01 or TE10 wave. 3. Characterization of Electromagnetic Waves in Rectangular Waveguides Since kc2 k 2 k z2, or k z2 k 2 kc2, if k kc , then k z 0 . This means that the propagation of the wave is cut off, and kc is called the cutoff propagation constant. 2 kc2 k x2 k y2 mπ n π k a b 2 2 c From k 2πf , we can find the cutoff frequency f c corresponding to the cutoff propagation constant kc , as given by 2 kc 1 m n fc 2π 2 a b 2 The propagation constant kz can be expressed as f k z k 1 c f 2 2 f k 1 c , f 2 fc jk 1, f f fc f fc if f f c , k z is a real number, and the factor e jk z z stands for the wave propagating along the positive z-direction. 2 If f f c , k z is an imaginary number, then e jk z z e f kz c 1 f 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,f c is the lowest frequency of the mode to be transmitted. In view of this, the waveguide acts like a high-pass filter. From k 2π , we can find the cutoff wavelength c corresponding the cutoff propagation constant kc as c 2π 2 2 2 kc m n a b 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截止频率, or a shorter cutoff wavelength截止波长. TE01 The cutoff wavelength of the TE10 wave is 2a, and that of TE20 wave is a. TE10 TE20 TM11 0 a 2a c The left figure gives the distribution of the cutoff wavelength 截 止 波 长 for a waveguide with a 2b . TE01 TE10 TE20 TM11 0 a 2a Cutoff area If c , then the corresponding mode will be cut off. From the figure we see that if 2a , all modes will be cut off. If a 2a , then only TE10 wave exists, while all other modes are cut off . If a , then the other modes will be supported. c Hence, if the operating wavelength 工作波长 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单模传输 wave is necessary in practice since it is helpful for coupling energy into or out of the waveguide. TE10 wave is usually used, and it is called the dominant mode主模 of the rectangular waveguide矩形波导. In practice, we usually take a 2b to realize the transmission of the single mode TE10 in the frequency band频带 a 2a . To support the TE10 mode the sizes of the rectangular waveguide should satisfy the following inequality 2 a b 2 The lower limit for the narrow side depends on the transmitted power, the allowable attenuation衰减, and the weight per unit length. In practice, we usually take a 0.7 and b (0.4 ~ 0.5)a or (0.1 ~ 0.2)a . As the wavelength is increased, the sizes of the waveguide must be increased proportionally to ensure the dominant mode主模 is above cutoff. If the frequency is very low, the wavelength will be very long so that it may not be convenient for use. Consequently, metal waveguides are used for microwave bands above 3GHz. The phase velocity vp can be found from the phase constant as vp Where v 1 kz v f 1 c f 2 v 1 c 2 v . If the inside of the waveguide is vacuum, then v 1 0 0 c Since the operating frequency f f c and the operating wavelength c , we have vp c for a vacuum waveguide. Hence, the phase velocity does not represent the energy velocity 能速in a waveguide. The phase velocity 相 速 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色散 in a waveguide. Based on the relationship between the wavelength and the phase constant, we find the wavelength of the electromagnetic wave in a waveguide, g , as 2π g 2 2 kz f 1 c 1 f c where is the operating wavelength工作波长. The quantity g is called the guide wavelength波导波长. Due to f f c , c ,thus g . The ratio of the transverse electric to the transverse magnetic field intensities as the waveguide impedance of the waveguide. For a TM wave the waveguide impedance is Z TM Ey Ex Hy Hx 2 Z TM f Z 1 c Z 1 f c 2 Z In the same way, we find the waveguide impedance阻抗 of a TE wave as Z TE Z f 1 c f 2 Z 1 c 2 If f f c , c , then Z TM and Z TE are both imaginary numbers. This means that the transverse横向 electric field and the transverse π 横向magnetic field have a phase difference of . Hence, there is no 2 energy flow in the z-direction, and it indicates that the propagation of the electromagnetic wave is cut off. Example. The inside of a rectangular metal waveguide is vacuum, and the cross-section截面 is 25mm10mm. What modes can be transmitted if an electromagnetic wave of frequency f 10 4 MHz enters the waveguide? Will the modes be changed if the waveguide is filled with a perfect dielectric of relative permittivity介电常数 r 4 ? Solution: Due to the inside is vacuum, the operating wavelength is c 30 mm f and the cutoff wavelength is c 2 2 m n a b 2 50 m 2 6.25n 2 Then the cutoff wavelength of TE10 wave is c 50mm , that of TE20 wave is c 25mm , and that of TE01 wave is λc 20mm . The cutoff wavelength of the higher modes will be even shorter. In view of this, only TE10 wave can be transmitted in this waveguide. Example. The inside of a rectangular metal waveguide is vacuum, and the cross-section截面 is 25mm10mm. What modes can be transmitted if an electromagnetic wave of frequency f 10 4 MHz enters the waveguide? Will the modes be changed if the waveguide is filled with a perfect dielectric of relative permittivity介电常数 r 4 ? If the waveguide is filled with a perfect dielectric of r 4 , then the operating wavelength is 15mm r Hence, TE10 and TE20 waves can be transmitted, and some other modes TE01,TE30,TE11,TM11,TE21,TM21 can exist. 4. TE10 Wave in Rectangular Waveguides Let m 1, n 0 , we find Ey j Hx j H 0 π kc2 π jk z sin x e z a a k z H 0 π π jk z z sin x e 2 kc a a π H z H 0 cos x e jk z z a And H y E x E z 0 . The corresponding instantaneous瞬时 values are E y (r , t ) H x (r , t ) 2H 0 π π π sin x sin( t k z ) z 2 kc 2 a a 2k z H 0 π π π sin x sin( t k z ) z 2 kc 2 a a π H z (r , t ) 2 H 0 cos x sin( t k z z ) a The above equations are simplified as π π E y (r , t ) A sin x sin( t k z z ) 2 a π π H x (r , t ) B sin x sin( t k z z ) 2 a π H z (r , t ) C cos x sin( t k z z ) a Where A, B, C are positive real numbers. Ey Hz z Hx Hz g Hx The right figure gives the distributions x of the TE10 wave along the z-direction and Ey x-direction at t = 0 . a A standing wave驻波 is found in the x -direction, while a traveling wave is seen in the z -direction. The amplitude of Hz follows a cosine function, while the amplitudes of Hx and Ez depend on x with the sine function. But all of them are independent of the variable y. The electric and magnetic field lines and the currents of TE10 wave . z x y a Electric field lines Magnetic field lines g x z b y y The electric currents on the inner walls z x The modes with higher orders TE10 TE11 TE20 TE21 TM11 TM21 Electric field lines Magnetic field lines Let m = 1, n = 0, we find the cutoff wavelength of TE10 mode as c 2a It means that the cutoff wavelength of the TE10 wave is independent of the narrow side. The phase velocity相速 and the guide wavelength波导波长 can be v found as vp g 2 2 1 2a 1 2a To visualize the physical meaning 物理含义 of the phase velocity, the energy velocity, as well as the guide wavelength for the TE10 wave, the expression of electric field intensity Ey is rewritten as E y E0 (e π j x a e π j x a ) e jk z z π 1 j aπ x - j aπ x sin x (e e ) a 2j Furthermore, we have Ey E0e jk ( x cos z sin ) E0e jk ( x cos z sin ) cos 2a c which states that a TE10 wave can be considered as the resultant wave comprising two uniform plane waves均匀平面波 with the same propagation constant k . The propagating directions of the two plane waves are laid on the xzz plane. They are parallel to the broad a ② ① wall, and the two plane waves are combined into a plane wave taking a zigzag path between the two narrow x walls. If c , then 0 . The plane wave will be reflected vertically between two narrow walls. Hence it cannot propagate in the z-direction and is cut off. when the wave loops of the two plane waves meet, a wave loop of the resultant wave is formed. A wave node of the resultant wave is formed when the wave nodes of the two plane waves meet. z B a x ② A D ① C The bold lines denote the wave loops of plane wave ①, and the dashed lines denote that of plane wave ②. Obviously, the length of the line AB is equal to the guide wavelength , and the length of the line AC is equal to the operating wavelength . If the inside of the waveguide is vacuum, then the length of the line AC is equal to the wavelength in vacuum. From the figure, we find g 2 g 1 sin 1 cos 2 c The space phase of plane wave ① is changed by 2 from A to C, while that of the resultant wave is changed by 2 over the distance AB. In view of this, the phase velocity of the resultant wave is greater than that of the uniform plane wave v, z B a ② A D ① v vp sin v vp 1 c C x 2 From the point of the view of energy, when the energy carried by plane wave ① arrives at C from A, the movement in z-direction is just over the distance AD. Hence, the energy velocity is less than the energy velocity of the uniform plane wave v. From the figure, we find the energy velocity as ve v sin v vp ve v 1 c 2 Example. The broad side of a rectangular waveguide filled with air satisfies the condition a 2 , and the operating frequency is 3GHz. If the operating frequency is required to be higher than the cutoff frequency of the TE10 wave by 20% and less than the cutoff frequency of the TE01 by 20%. Find: (a) The sizes for a and b. (b) The operating wavelength, the phase velocity, the guide wavelength, and the wave impedance for the designed waveguide. Solution: (a) The cutoff wavelength of the TE10 wave is c 2a , and c c the cutoff frequency is f c . The cutoff wavelength of TE01 wave c 2a c is c 2b , and the cutoff frequency is f c . According to the given 2b condition, we have c c 3 10 9 1.2 3 109 0.8 2b 2a We find a 0.06m , b 0.04m . Take a 0.06m , b 0.04m. (b) The operating wavelength, the phase velocity, the guide wavelength, and the wave impedance: vp g c 0.1m f c 1 2a 2 1 2a Z TE 1 0 2 5.42 103 m/s 0.182m Z 1 2a 2 682Ω 5. Group Velocity When the phase velocity is frequency dependent, a single phase velocity alone cannot account for the speed at which a wave consisting of multiple frequency components propagates. As an example, we consider an amplitude-modulated wave to illustrate the concept of the group velocity. Suppose an electromagnetic wave propagating in the z-direction has two components with frequencies close to each other as given by A1 ( z, t ) A0 cos(1t k1 z ) A2 ( z, t ) A0 cos( 2 t k 2 z ) with the resultant signal A A1 A2 2 A0 cos(Δ t Δkz) cos(0t k0 z ) where 1 0 2 (1 2 ) Δ 1 ( ) 1 2 1 k 0 2 (k1 k 2 ) Δk 1 (k k ) 0 2 1 A A1 A2 2 A0 cos(Δ t Δkz) cos(0t k0 z ) Since 1 ~ 2 , and Δ 0 . Therefore, in a very short time interval, the first cosine function show little change, but the second cosine function has large variations. So 0represents the carrier frequency while Δ is the frequency of the envelope or the modulating frequency. This is an amplitude-modulated signal with a slower variation in the amplitude. If the medium is non-dispersive, the envelope of the amplitude is moving together with the carrier, both maintaining the sinusoidal behavior in the movement. Therefore, by the locus of a stationary point on the envelope, we can find the velocity of the envelope, and this velocity is called the group velocity, denoted as vg . Let Δt Δkz constant , we find vg dz Δ dt Δk For non-dispersive media, the relationship between k and is linear, and Δ d . We obtain Δk dk vg d dk Let 0t k0 z constant , and we find the phase velocity of the carrier as vp 0 k0 Since k in non-dispersive media, we find the group velocity as 1 vg d dk dk d 1 vp In non-dispersive media, the group velocity is equal to the phase velocity. For dispersive media, the relationship between k and is nonlinear. In this case, for a given operating frequency , can be 0 k (ω) expanded by Taylor series 0 around as dk k ( ) k 0 ( 0 ) d 0 1 d 2k 2 ( 0 ) 2 2 d 0 For a narrow band signal, take only the first two terms as approximation, so that dk k ( ) k 0 ( 0 ) d 0 Δ d Consider vg , we have Δk dk 1 dk d vg d 0 dk 0 With a nonlinear relation between the propagation constant k and the frequency for a dispersive medium, the phase velocity is frequency dependent and it is not the same as the group velocity. Carrier Envelope Carrier It gives the waveforms of the above narrow band signal at three different moments for the case of vp 2vg .P is a stationary point on the envelope, and P is that for the carrier. When the displacement of the point P is d, the point P is moved only byd , (d d ) because the velocity of the envelope is less. The actual signal waveform will be modified as it propagates. dk d ω 1 ω dvp 1 ω dvp 1 2 Consider , we find dω dω vp vp vp dω vp vp dω vp vg dv p 1 v p d For the narrow band signal, the above equation becomes vp vg dv 1 p vp d 0 If the phase velocity vp is independent of frequency, vg v p If If dvp d dvp d dvp d 0 , then 0 , then vg vp , and the dispersion is called normal dispersion. 0 , then vg vp , and it is called abnormal dispersion. dv For a rectangular waveguide, p 0 , it is normal dispersive d and the group velocity is 2 2 f vg v 1 c v 1 ve f c The group velocity is equal to the energy velocity in the rectangular waveguide, which is the same behavior for all normal dispersive media. The phase velocity vp and the group velocity vg in a waveguide satisfy the following equation vp vg v 2 When an electromagnetic wave is propagating in a conductive medium, abnormal dispersion is observed. In this case, the group velocity is not equal to the energy velocity, and the above equation is not valid for this case. 6. Circular Waveguides The inner radius a is the only dimension to be specified. Select the cylindrical coordinate system, and let the z-axis be the axis of the cylinder. Similar to the rectangular waveguide, the longitudinal components Ez y or Hz is first obtained, from which the transverse components Er , E , Hr , H a x can be derived. , z The field intensities in waveguide can be written as E ( r , , z ) E 0 ( r , ) e jk z z the H (r , , z ) H 0 (r , )e jk z z The corresponding longitudinal components are, respectively E z ( r , , z ) E z 0 ( r , ) e jk z z H z (r , , z ) H z 0 (r , )e jk z z For a TM wave, Hz = 0 . In a source-free region, Ez satisfies the scalar Helmholtz equation given by 2 Ez k 2 Ez 0 Expanding this equation in cylindrical coordinate system, we have 2 Ez 0 1 E20 1 2 Ez 0 2 k Ez 0 0 c 2 2 2 r r r r Using the method of separation of variables is used, and let E z 0 (r, ) R(r ) ( ) Substituting it into the above equation gives r 2 R rR 2 2 kc r R R where R and R are the second and the first derivatives of the function R with respect to r, respectively, and is the second derivative of the function with respect to . Using the same derivation as before,we obtain the equation for the function as m 2 0 The general solution is A1 cos m A2 sin m Since the period of variation of the field with the angle is 2 . Hence m must be integers so that m 0, 1, 2 The circular waveguide is symmetrical with respect to the zaxis; thus the plane 0 can be chosen arbitrarily. In this way, we can always select the plane properly so that the first term cos m or the second term sin m vanishes. Therefore, the solution of can be expressed as cos m sin m A We find d2 R dR 2 2 2 r r ( k r m )R 0 c 2 dr dr 2 Let kc r x , then the above equation becomes the standard Bessel equation d2 R dR 2 2 x x ( x m )R 0 2 dx dx 2 The general solution is R BJ m ( x) CN m ( x) whereJ m ( x) is the first kind of Bessel function of order m, and N m ( x) is the second kind of Bessel function of order m. If r 0 , x 0 , N m (0) , But the field should be finite in the waveguide. Hence the constant C 0 . The solution should then be R BJ m (kc r ) Consider all results above, we find the general solution of Ez as cos m jk z z E z E0 J m (kc r ) e sin m And the transverse components are Er j E j cos m jk z z k z E0 Jm (kc r ) e kc sin m sin m jk z z k z mE0 J ( k r ) e m c 2 kc r cos m sin m jk z z J ( k r ) e m c 2 kc r cos m cos m jk z z E0 H j Jm (kc r ) e kc sin m Hr j mE0 where Jm (kc r ) is the first derivative of Bessel function J m (kc r ) . The constant kc depends on the boundary condition. The components Ez and E are tangential to the inner wall of the circular waveguide; hence, E z E 0 at r a . 2 P kc2 mn We find a Pmn is the n -th root of the first kind of Bessel function of order m. The values of Pmn n 1 2 3 4 0 2.405 5.520 8.654 11.79 1 3.832 7.016 10.17 13.32 2 5.136 8.417 11.62 14.80 m A pair of m and n corresponds to aPmn , and that corresponds to a kind of field distribution or a mode. Hence, the electromagnetic waves have multiple modes in a circular waveguide also. For the TE wave, Ez = 0. We can use the same approach to find the component Hz first, and then the other transverse components can be determined. TE wave: cos m jk z z H z H 0 J m (kc r ) e sin m cos m jk z z kH H r j z 0 Jm (kc r ) e kc sin m sin m jk z z k z mH 0 H j 2 J m (kc r ) e kc r cos m Er j mH 0 E j kc2 r H 0 kc sin m jk z z J m (kc r ) e cos m cos m jk z z Jm (kc r ) e sin m Based on the boundary conditions, we find Pmn 2 kc a .2 is the root of the first derivative of Bessel function. Where Pmn The values of Pmn n 1 2 3 4 0 3.832 7.016 10.17 13.32 1 1.841 5.332 8.526 11.71 2 3.054 6.705 9.965 13.17 m As with the rectangular waveguide, if k kc , then the propagation constant k z 0 , meaning that the wave is cut off, so that propagation ceases. From k c 2 πf c 2π For TM wave, we have c fc For TE wave, we have fc Pmn ; 2π a c 2π a Pmn Pmn ; 2π a c 2π a Pmn Cutoff area The following figure gives the cutoff wavelength of several modes in a circular waveguide. The TE11 wave has the longest TE11 cutoff wavelength, and the next one TM01 TE21 is the TM01 wave. TE01 0 a 2a 3a 4a c The cutoff wavelengths of the TE11 and TM01 waves, respectively, as TE 11 : c 3.41a, TM 01 : c 2.62a If the operating wavelength satisfies the following inequality 2.62a 3.41a The transmission of a single mode (TE11 wave) can be realized, and the TE11 wave is the dominant mode for the circular waveguide. If the operating wavelength is given, to realize the transmission of only the TE11 wave, the radius a must satisfy the following inequality: a 3.41 2.62 From the cutoff frequencies or the cutoff wavelengths, the phase velocity, the group velocity, the guide wavelength and the wave impedance of each mode can be found using the same equations as those for the rectangular waveguide. TE11 TE01 TM01 Electric field lines Magnetic field lines Example. A circular waveguide of radius a = 5mm is filled with a perfect dielectric of relative permittivity r = 9 . If it is to be operated in the dominant TE11 mode, find the permissible frequency range. Solution: To operate on the mode TE11, the operating wavelength must satisfy the following inequality 2.62a 3.41a Hence max 3.41 5mm 17.1mm min 2.62 5mm 13.1mm The corresponding range of the operating frequency is f max f min v min v max 1 min 0 1 max 0 7634MHz 5848MHz 7. Transmitted Power and Loss in Waveguides The longitudinal component of the complex energy flow density vector is given by the cross product of the transverse components of the electric and magnetic fields. The integration of the real part over the cross-sectional area of the waveguide gives the transmitted power. Take the TE10 wave as an example, we find the transmitted power as abE02 P 2Z TE Where E 0 is the amplitude of the electric field in the middle of the broad side If the dielectric strength of the filling dielectric is Eb , then the maximum transmitted power of the rectangular waveguide is abEb2 Pb 4Z TE In practice, the transmitted power is limited as P 1 ~ 1 Pb for 3 5 safety purpose. The two major mechanisms for energy loss are imperfect dielectric and finite conductivity of the waveguide walls. The effect of the dielectric can be accounted for by introducing the equivalent permittivity to replace the original one, i.e. e j A vigorous analysis of the waveguide walls is very complicated. An approximation that retains the magnetic field that would have existed if the walls were perfectly conducting may be employed. If we assume the attenuation constant is k , then the amplitude of the electric field intensity propagating along the positive zdirection has the form E E0 e k z The transmitted power can be expressed as P P0 e 2 k z P P0 e 2 k z Take the derivative of the above equation with respect to z, we find the power attenuation per unit length as P 2k P z Obviously, that is the power loss per unit length, so that Pl1 2k P Hence, the attenuation constant k is obtained as k Pl1 2P To calculate the loss of the walls, we consider a piece of conductor making up the broad side wall, with unit width and length and a thickness equal to d . The current in the conductor is flowing y in the z-direction. The resistance of the piece 1 1 of conductor is given by d l 1 πf R S 1 S d z x where is the conductivity of the wall. RS is called the surface resistivity. The surface resistivities of three types of metal Metals RS Silver 2.52 10 7 Copper f 2.61 10 7 Aluminum f 3.26 10 7 f The surface current density is the current per unit width. Hence the power loss per unit length and width of the waveguide wall PlS is PlS J S2 R S where the surface current J S en H S , and H S is the magnetic field intensity on the surface of the wall. Taking the integration of PlS over the inner wall for a section of the waveguide of unit length, the power loss per unit length of the wall Pl1 can be obtained. TM1 1 For a given size of rectangular waveguide, the loss of the TE10 wave is minimum. For a given width, the smaller is the narrow wall, the larger will be the attenuation constant. The loss of the TE01 wave in a circular waveguide is minimum in the higher frequency range. 。 The cutoff wavelength of the TE01 wave is not the largest. In order to realize the transmission of the single mode TE01, the modes TM01, TE21 and TE11 have to be suppressed. r For the same cross-section, the perimeter of a rectangle is larger than that of a circle, and the loss in a circular waveguide is less than that of the rectangular waveguide. However, when a TE11 wave is propagating in a circular waveguide, the fields could be rotated. An elliptical waveguide does not result in the rotation of the fields, and the loss is less E also. In addition, in order to reduce the wall losses, the inner surface should be polished, and plated with silver or gold. To prevent oxidation of the surface, the waveguide may be filed with inert gas. Example. Calculate the attenuation caused by finite conductivity of the wall when a TE10 wave is propagating in a rectangular waveguide. Solution: When a TE10 wave is propagating in a rectangular waveguide, there are x-component and z-component of the surface currents on the broad sides, while there is only the y-component on the narrow sides. y Therefore, the power loss per unit length of the broad wall is z x a a 2 Pla 2 J Sz RS dx J Sx2 RS dx 0 0 Where J Sz e y H x , J Sx e y H z . The power loss per unit length of the narrow wall is b Plb 2 J S y RS dy 2 0 Where J Sy e x H z . Then the total power loss per unit length is Pl1 Pla Plb Based on the transmitted power P and the total power loss per unit length Pl1 , we find the attenuation constant as P k l1 2P 1 2 2 2 b a 2a 1 2a RS 8. Resonant Cavity In microwave band, the lumped LC tank circuits cannot be used, we usually employ a transmission line to construct a resonant device, and it is called cavity resonator. with the increase of the resonant frequency the inductance and the capacitance must be reduced. However, for small L and C, distributed effects cannot be neglected. The inductance of the lead wires of capacitors, the distributed capacitances among the coils or the devices have to be considered. This means that a pure capacitor or a pure inductor is very difficult to be made at microwave frequencies. Furthermore, with the increase in frequency, the radiation effect of the circuits becomes significant, and the power loss in the dielectric of the capacitor is more severe as well. All of these will result in the decrease of the quality factor Q of the lumped tank circuit. When a metal plate is placed at the end of a waveguide, the electromagnetic wave will be completely reflected, leading to a standing wave. For a rectangular waveguide operating in the dominant mode, the closed end corresponds to a wave node for the electric field since it is tangential to the metal plate. Another wave node for the electric λ field appears at a distance g from the closed end. If one more metal 2 plate is placed there, the boundary condition is still satisfied, y d b z a g /2 x In this way, there is a standing wave in the cavity formed by the waveguide walls and the end plates. Based on the field intensity and the boundary condition, we find the equations for the standing waves in the cavity as follows: π H z H 0 (e e ) cos x a k aH π H x j z 0 (e jk z z e jk z z ) sin x π a aH 0 jk z z jk z z π Ey j (e e ) sin x π a jk z z jk z z π H z 2 jH 0 sin( k z z ) cos x a k aH π H x 2 j z 0 cos( k z z ) sin x π a 2aH 0 π Ey sin( k z z ) sin x π a There are standing waves of the electric and the magnetic fields along both the x-direction and the z-direction, but they are out of the time phase by π . When the electric energy is maximum, the magnetic 2 energy is zero. Conversely, when the magnetic energy is maximum, the electric energy is zero. The electromagnetic energy is exchanged between the electric field and the magnetic field, and this phenomenon is called resonance. So the cavity is called a resonant cavity, and it is used as a resonant device in microwave circuits. For a given cavity, resonance occurs only at certain frequencies. The particular frequency is called resonant frequency, and the corresponding wavelength is called the resonant wavelength. g d l , 2 If the length of the cavity is l 1,2,3, Then the boundary condition can be satisfied, and the resonance will happen. Hence, the resonant frequency or wavelength of a resonant cavity is multi-valued, and it is called multiple resonance. since the guide wavelength is related to the mode, different modes have different resonant frequencies. 2 2 m π n π Consider k z2 k 2 , when d l g , k z d lπ , k z lπ , 2 d a b and we find 2 2 mπ nπ lπ k a b d 2 From k 2π 2πf , we find mnl f mnl 2 2 2 m n l a b d 1 2 2 2 2 m n l a b d 2 The resonant wavelengths and frequencies depend on not only the sizes of the cavity, but also the mode. A set of mnl corresponds to a mode. For instance, TE101 stands for that a rectangular waveguide cavity operating on a TE10 wave, and the length of the cavity is half of the guide wavelength. In order to properly design the coupling and the tuning devices of the cavity, knowledge about the distribution of the fields in the cavity is required. The figure gives the distribution of the fields in a rectangular cavity operating at the TE101 mode. d As with other resonant devices, y z a real cavity always has some loss. a z Electric field lines In order to assess the loss of a resonant device, the quality factor Q is usually employed, and its definition is 0W Magnetic field lines Pl x x b y Q where 0 is the resonant angular frequency, W is the total energy, Pl is the power loss in the cavity. The maximum value of the energy density of the electric field in the cavity is a 3bd02 2 | H 0 |2 W 2π 2 In order to calculate the power loss of the cavity wall, the same method for waveguide analysis may be applied. We find the power loss of a rectangular cavity operating with the TE101 mode as 2a 3b a 3 d ad 3 2d 3b 2 Pl 2 R | H | S 0 d2 and the Q value is 03 2a 3bd 3 Q 2 4π RS (2a 3b a 3d ad 3 2d 3b) The resonant angle frequency for the TE101 mode is 101 2πf101 π 2 1 1 a d 2 Therefore, the Q value for the TE101 mode can be expressed as πZb (a 2 d 2 )3 Q101 4 RS (2a 3b a 3d ad 3 2d 3b) where Z . Since the circular waveguide has less loss, the Q value of the cylindrical cavity is higher, and it is more popular than the rectangular cavity. The method for calculating the resonant frequency and the Q value of a cylindrical cavity is the same as that above. TM wave: f TM TE wave: f TE QTE 1 2π 1 2π d 2 Pmn lπ a d 2 lπ Pmn a d 2 m 2 1 Pmn QTM 2 lπa P d d 2a 2 π 1 d 2 mn lπa ) 2 ( Pmn 2 3 2 2 2 a l π a 2 a ml π a 2 ) 2π ( Pmn 1 d d d d Pmn 2 TE01l modes have higher Q values, and the maximum Q value of the TE011 mode occurs around d 2a. If = 3cm, then Q value will be 104~4104. 。 The approach to increase the Q value is the same as that of decreasing the loss of waveguide wall. In addition, the volume of the cavity should be as larger as possible to increase the stored energy, while the area of the walls should be as small as possible to decrease the loss. Example. Show that for any mode the resonant wavelength r of the cavity can be expressed as c r l 1,2,3, 2 l 1 c 2d where c is the cutoff wavelength, and d is the length of the cavity. Consider k z2 k 2 kc2 Solution: If the length d l g 2 , then k z d lπand k z lπ . Resonance will occur. d 2π And due to k 2π , k c , we find c r 2 π 2π 2π l d r c 2 2 2 1 1 l 2 2 λr λc 2d Rewriting the equation gives the general formula. 9. Coaxial Lines A coaxial line is shown in the figure, with an inner radius a and an outer radius b. The electromagnetic wave propagates in the region between the two conductors, which may be filled with air or a dielectric. The coaxial line is a good transmission y line in microwave band. It possesses the electromagnetic shielding function as the waveguides, but it has a wider frequency b x range of operation. z a Electric field lines Magnetic field lines The coaxial line is a typical TEM wave transmission line a coaxial line, and the electric field lines are along with the radial direction , while the magnetic field lines are a set of circles. A coaxial line can also be considered as a circular waveguide that supports TE and TM waves, besides the TEM wave. However, if we properly design the dimensions according to the operating frequency, these non-TEM waves can be restrain . The method for analyzing non-TEM waves in a coaxial line is similar to that for a circular waveguide. However, the coaxial line has an internal conductor, the range of the variable r is a r b , andr 0 . Hence, the second kind of Bessel function with the singularity at r = 0 should be the solution of Bessel equation as well, i.e. R BJ m ( x) CN m ( x) TE11 TM01 TE10 0 (b - a) (a + b) c For TM and TE waves,based on the boundary conditions the cutoff propagation constant are found first, then the cutoff wavelengths can be obtained. TM01 TE10 0 TEM wave TE11 (b - a) (a + b) c The TE11 wave has the longest cutoff wavelength, and it is π(a b) . To restrain the non-TEM wave, the operating wavelength must satisfy the following inequality π ( a b) In other words, the dimensions of the coaxial line should satisfy the following inequality ab π 3 Hence, in order to eliminate the higher order modes in a coaxial line, the dimensions have to be decreased as the frequency increases. But small sizes will result in the increase of loss and the restriction of the transmitted power. For this reason, the coaxial line is usually used for the frequencies below 3GHz. However, the operating frequency has no lower limit, and the coaxial line can also be used to construct a cavity.