W. Geyi- A Time-Domain Theory of Waveguide

8/3/2019 W. Geyi- A Time-Domain Theory of Waveguide

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Progress In Electromagnetics Research, PIER 59, 267297, 2006

A TIME-DOMAIN THEORY OF WAVEGUIDE

W. Geyi

Research In Motion295 Phillip Street, Waterloo, Ontario, Canada N2L 3W8

AbstractA new time-domain theory for waveguides has been

presented in the paper. The electromagnetic fields are first expandedby using the complete sets of vector modal functions derived fromthe transverse electric field. The expansion coefficients are thendetermined by solving inhomogeneous Klein-Gordon equation interms of retarded Greens function. The theory has been validatedby considering propagation problems excited by various excitationwaveforms, which indicates that the higher order modes play asignificant role in the field distributions excited by a wideband signal.

1. INTRODUCTION

In recent years the major advances made in ultrawideband techniqueshave made time-domain analysis of electromagnetic phenomena animportant research field. A short pulse can be used to obtainhigh resolution and accuracy in radar and to increase informationtransmission rate in communication systems. Another importantfeature of a short pulse is that its rate of energy decay can be sloweddown by decreasing the risetime of the pulse [1].

According to linear system theory and Fourier analysis, theresponse of the system to an arbitrary pulse can be obtained bysuperimposing its responses to all the real frequencies. In other words,the solution to a time domain problem can be expressed in terms of atime harmonic solution through the use of the Fourier transform. Thisprocess can be assisted by the fast Fourier transforms and has beenused extensively in studying the transient response of linear systems.This procedure is, however, not always most effective and is not atrivial exercise since the harmonic problem must be solved for a largerange of frequencies and only an approximate time harmonic solutionvalid over a finite frequency band can be obtained. Another reason isthat the original excitation wave may not be Fourier transformable.


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Thus we are forced to seek a solution in the time domain in somesituations.

In high-speed circuits the signal frequency spectrum of a

picosecond pulse may extend to terahertz regime and signal integrityproblems may occur, which requires a deep understanding of thepropagation characteristics of the transients in a waveguide. Thewaveguide transients have been studied by many authors [224]. Oneof the research topics is to determine the response of the waveguideto an arbitrary input signal. If the input signal of a waveguide in asingle-mode operation is x(t), then after traveling a distance z, theoutput from the waveguide is given by the Fourier integral

y(t) = 12

X()ej[tz ]d (1)

where X() is the Fourier transform of x(t); is the propagationconstant =

2 2c ; and c is the cut-off frequency of the

propagating mode and c = ()1/2 with and being thepermeability and permitivity of the medium filling the waveguiderespectively.

Several methods have been proposed to evaluate the Fourierintegral in (1), such as the method of saddle point integration [2],method of stationary phase [3, 4], contour integration technique [5],and the quadratic approximation of the propagation constant aroundthe carrier frequency [69]. A serious drawback to stationary phaseis that it contradicts the physical realizability [10] as well as causality[13] (i.e., the response appears before the input signal is launched). Amore rigorous approach is based on impulse response function for alossless waveguide, which is defined as the inverse Fourier transformof the transfer function ejz. The impulse response function canbe expressed as an exact closed form and has been applied to studytransient response of waveguide to various input signals [1216].

The response given by (1) is, however, hardly realistic fordescribing the propagation of a very short pulse or an ultrawidebandsignal since it is based on an assumption that the waveguide is in asingle-mode operation. This assumption is reasonable only for a narrowband signal but not valid for a short pulse, which covers a very widerange of frequency spectrum, and will excite a number of higher ordermodes in the waveguide. It should be mentioned that most authorshave discussed transient responses for various input signals, such as astep function, a rectangular pulse or even a impulse on the basis ofan assumption that the waveguide is in a single-mode operation. Thiskind of treatment has oversimplified the problem and the theoretical


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Progress In Electromagnetics Research, PIER 59, 2006 269

results obtained cannot accurately describe the real transient processin the waveguide.

In order to get the real picture of the transient process in an

arbitrary waveguide, we must solve the time-domain Maxwell equationssubject to initial conditions, boundary conditions and excitationconditions, and we should include the higher order mode effects.One such approach is based on the Greens function of Maxwellequations and has been discussed in [20]. Another approach isbased on the field expansions in terms of the eigenfunctions in thewaveguide [2124], which are usually derived from the eigensolutionsof the longitudinal fields. When these field expansions are introducedinto homogeneous Maxwell equations one finds that the expansion

coefficients satisfy the homogeneous Klein-Gordon equations. A wavesplitting technique has been used to solve the homogeneous Klein-Gordon equations by introducing two Greens functions G+(z, t) andG(z, t), which satisfy an integro-differential equation [21], and theexcitation problem in a waveguide has been studied by introducing thetime-domain vector mode functions, each of which is time dependentand satisfies the homogeneous Maxwell equations. The total fieldsgenerated by the source can then be expressed as an infinite sumof convolutions of expansion coefficients and the time-domain vector

mode functions. The approach based on wave splitting seems rigorousbut very complicated. In addition it is based on the field expansionsin terms of the longitudinal field components. Consequently thetheory is only suitable for a hollow waveguide and cannot be appliedto transverse electromagnetic (TEM) transmission lines whose crosssection is multiple connected.

However a hollow waveguide is not suitable for carrying awideband signal since it blocks all the low frequency componentsbelow the cut-off frequency of dominant mode, which may cause severedistortion. For this reason, a hollow waveguide is only appropriatefor narrow band applications where the frequency spectrum of thetransmitted signal falls in between the cut-off frequency of thedominant mode and the cut-off frequency of the next higher ordermode. On the other hand the cut-off frequency of the dominant modeof a TEM transmission line is zero. Therefore it is often used forwideband applications.

Hence it is necessary to develop a more general time-domaintheory for the waveguide, which is suitable for hollow waveguides aswell as TEM transmission lines. It is the purpose of present paper toachieve this goal. A more straightforward and concise time-domainapproach for the analysis of waveguides has been presented in thispaper. Instead of starting from eigenvalue theory of the longitudinal


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field components, we use the eigenvalue theory of the transverseelectric field. The eigenfunctions of this transverse eigenvalue problemconstitute a complete set and can be used to expand all field

components in an arbitrary waveguide (hollow or multiple connected).A set of inhomogeneous Klein-Gordon equations for the field expansioncoefficients can be obtained when these field expansions are substitutedinto Maxwell equations. As a result, the excitation problems maybe reduced to the solution of these inhomogeneous scalar equations.On the contrary, all the previous publications deal with homogeneousKlein-Gordon equations and leave the excitation problem in thewaveguide very complicated [21]. To solve the inhomogeneous Klein-Gordon equation, the Greens function method has been used. The new

time-domain theory has been applied to study the transient process inboth hollow waveguide and TEM transmission line with the effects ofhigher order mode being taken into account. It should be mentionedthat, as far as the author can determine, there is no a single numericalexample in previous publications that has really considered the higherorder mode effects when dealing with the transients in a waveguide.

For the purpose of validating the new time-domain theory for thewaveguide, a typical excitation problem by a sinusoidal wave, which isturned on at t = 0, has been investigated. Our theoretical prediction

shows that the steady state response of the field distribution tends tothe well-known result derived from time-harmonic theory as t .Other excitation waveforms are also discussed, which indicates thatthe higher order modes play a significant role in the field distributionsexcited by a wideband signal. To further validate the theory threeappendices have been attached. Appendix A gives a simple proof of thecompleteness of the transverse eigenvalue problem for the waveguide.Appendices B and C investigate the numerical examples studied inSection 4 by directly solving Maxwell equations subject to the initialand boundary conditions in the waveguide. It is shown that bothapproaches give the exactly same results.

2. FIELD EXPANSIONS BY EIGENFUNCTIONS

Consider a perfect conducting waveguide, which is uniform along z-axisand filled with homogeneous and isotropic medium. The cross-sectionof the waveguide is denoted by and the boundary of by . Notethat the cross-section can be multiple connected. The transient

electrical field in a source free region of the waveguide will satisfy the


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following equations

2E(r, t) 2E(r, t)/c2t2 = 0, r

E(r, t) = 0, r un E(r, t) = 0, r (2)

where un is the outward normal to the boundary . The solutioncan then be expressed as the sum of a transverse component and alongitudinal component, both of which are the separable functions oftransverse coordinates and the longitudinal coordinate with time, i.e.,

E(r, t) = (et + uzez)u(z, t)

where et and ez are functions of transverse coordinates only. Fora hollow waveguide (i.e., the cross-section is simple connected),the transverse fields may be expressed in terms of the longitudinalfields. Therefore the analysis for the hollow waveguide can be basedon the longitudinal fields and this is the usual way to study thetransient process in the hollow waveguide [2124]. If the cross-sectionof the waveguide is multiple connected, such as multiple conductortransmission line for which the dominant mode has no longitudinalfield components at all, the above analysis is no longer valid. To ensure

the theory to be applicable to the general situations we may use thetransverse fields. Inserting the above equation into (2) and taking theboundary conditions into account, we obtain

et et k2cet = 0, r un et = et = 0, r

where k2c is the separation constant. The above system of equationsconstitutes an eigenvalue problem and has been studied by Kurokawa[25]. It has been be shown that the system of eigenfunctions {etn|n =1, 2, . . .} or vector mode functions is complete in the product spaceL2() L2(), where L2() stands for the Hilbert space consistingof square-integrable real functions. Kurokawas approach is rigorousenough for engineering purposes although he does not provide the proofof the existence of eigenfunctions of the above eigenvalue problem.In Appendix A of present paper, a more rigorous proof on thecompleteness of the eigenfunctions of the above eigenvalue problemhas been presented.

The corresponding eigenvalues {k2cn 0, kcn+1 kcn|n =1, 2, . . .} are the cut-off wavenumbers. These mode functions canbe arranged to fall into three different categories (1) TEM mode, etn = 0, etn = 0; (2) TE mode, etn = 0, etn = 0;


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and (3) TM mode, etn = 0, etn = 0. It should be noted thatonly TEM modes correspond to a zero cut-off wavenumber. The cut-off wavenumbers for TE and TM modes are always great than zero.

From the complete set {etn} other three complete systems may beconstructed{uz etn|un uz etn = 0, r }{ etn/kcn| etn/kcn = 0, r }

{uz ( etn/kcn), c|un [uz ( etn/kcn)] = 0, r }where c is a constant. Hereafter we assume that the vector modefunctions are orthonormal, i.e.,

etm etnd = mn. If the set

{etn} is orthonormal, then all the above three complete sets are alsoorthonormal. According to the boundary conditions that the vectormode functions must satisfy, {etn} are electric field-like and mostappropriate for the expansion of the transverse electric field; {uz etn}are magnetic field-like and most appropriate for the expansion ofthe transverse magnetic field; { etn/kcn} are electric field-likeand most appropriate for the expansion of longitudinal electric field;{ etn/kcn} are magnetic field-like and most appropriate for theexpansion of longitudinal magnetic field. Notice that E is magneticfield-like while H is electric field-like. It should be notified thatall the modes and cut-off wavenumbers are independent of time andfrequency and they only depend on the geometry of the waveguide.Therefore these modes can be used to expand the fields in bothfrequency and time domain. Following a similar procedure describedin [25] the transient electromagnetic fields in the waveguide can berepresented by

E =

n=1

vnetn + uz

n=1

etn

kcn ezn (3)H =

n=1

inuz etn + uz

uz H

d +

n=1

etnkcn

hzn (4)

E =

n=1

vnz

+ kcnezn

uz etn +

n=1

kcnvn

etnkcn

(5)

H =

n=1in

z+ kcnhzn

etn + uz

n=1kcnin

etnkcn

(6)

where

vn(z, t) =

E etnd, in(z, t) =

H uz etnd


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hzn(z, t) =

H etn

kcn

d, ezn(z, t) =

uz E etn

kcn

d

We will call vn and in the time-domain modal voltage and time-domainmodal current respectively. Substituting the above expansions into thetime-domain Maxwell equations

E(r, t) = H(r, t)

t Jm(r, t)

H(r, t) = E(r, t)t

+ J(r, t)

and comparing the transverse and longitudinal components, we obtain

inz

+ kcnhzn = vnt

+

J etnd (7)

kcnin = ezn

t+

uz J etn

kcn

d, for TM modes (8)

vnz

+ eznkcn = int

Jm uz etnd (9)

kcnvn = hznt

(uz Jm)uz etn

kcn

d, for TE modes

(10)

t

H uz

d =

uz Jm

d, for TE modes (11)

It is easy to show that the modal voltage and modal current for TEM

mode satisfy the one-dimensional wave equation

2vT EMnz2

1c2

2vT EMnt2

=

t

J etnd z

Jm uz etnd

2iT EMnz2

1c2

2iT EMnt2

= z

J etnd + t

Jm uz etnd

(12)from (7) and (8). For TE modes the modal voltage vT En satisfies the

following one-dimensional Klein-Gordon equation, i.e.,

2vT En2z

1c2

2vT Ent2

k2cnvT En =

t

J etnd z

Jm uz etnd


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+kcn

(uz Jm)uz etn

kcn

d (13)

from (7), (9) and (10). The modal current iT En can be determined bya time integration of vT En /z. For TM modes, it can be shown that

the modal current iT Mn also satisfies the Klein-Gordon equation

2iT Mn2z

1c2

2iT Mnt2

k2cnvT Mn =

z

J etnd+ t

Jm uz

etndkcn

uz Jetn

kcn

d (14)

from (7), (8) and (9). The modal voltage vT Mn can be determinedby a time integration of iT Mn /z. Now we can see that theexcitation problem in a waveguide is reduced to the solution of aseries of inhomogeneous Klein-Gordon equations. Compared to thetreatment in previous publications, where only homogeneous Klein-Gordon equations are involved but very complicated time-domainvector mode functions have to be introduced to solve an excitationproblem [e.g., 21], our approach is much simpler, and at the same time

more general since it can be applied to a hollow waveguide as well asa multiple conductor transmission line.

3. SOLUTIONS OF INHOMOGENEOUSKLEIN-GORDON EQUATION

To get the complete solution of the transient fields in the waveguide,we need to solve the inhomogeneous Klein-Gordon equation. This canbe done by means of the retarded Greens function.

3.1. Retarded Greens Function of Klein-Gordon Equation

The retarded Greens function for Klein-Gordon equation is defined by

2

z2 1

c22

t2 k2cn

Gn(z, t; z

, t) = (z z)(t t)

Gn(z, t; z, t)t


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The inverse Fourier transform may then be written as

Gn(z, t; z, t) =

c2

(2)2

ejp(zz)dp

ej(tt)

(2

p2

c2

k2

cnc2

)

d

To calculate the second integral we may extend to the complexplane and use the residue theorem in complex variable analysis. Thereare two simple poles 1,2 =

p2c2 + k2cnc

2 in the integrand. Tosatisfy the causality condition, we only need to consider the integralofej(tt)/(2 p2c2 k2cnc2) along a closed contour consisting of thereal axis from to and an infinite semicircle in the upper halfplane. Note that the poles on the real axis have been pushed up a little

bit by changing them into 1,2 = p2c2 + k2cnc2 + j to satisfy thecausality condition, where will be eventually made to approach zero.The contour integral along the large semicircle will be zero for t > t.Using the residue theorem and [27]

0

sin q

x2 + a2

x2 + a2

cos bxdx =

2J0

a

q2 b2

H(q b),

a > 0, q > 0, b > 0

we obtain

Gn(z, t; z, t) =

c

0

sin

c(t t)p2 + k2cnp2 + k2cn

cosp(z z)dp

=c

2J0

kcnc

(t t)2 |z z|2/c2

H(t t |z z|/c) (16)

where J0(x) is the Bessel function of first kind and H(x) the unit stepfunction. Note that when kcn = 0 this reduces to Greens function of

wave equation.

3.2. Solution of Inhomogeneous Klein-Gordon Equation

Consider the inhomogeneous Klein-Gordon equation2

z2 1

c22

t2 k2cn

un(z, t) = f(z, t) (17)

and (15). Both equations can be transformed into the frequencydomain by Fourier transform

2

z2+ 2cn

un(z, ) = f(z, )


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left-traveling wave

J

right-traveling wave

z=a z=b

Figure 1. Left-traveling wave and right-traveling wave in waveguide.

2

z2+ 2cn

Gn(z, ; z

, t) = (z z)ejt

where 2

cn

= 2/c2

k2

cn

. Multiplying the first equation and the second

equation by Gn and u respectively and then subtracting the resultantequations yield

un(z, )2Gn(z, ; z

, t)z2

Gn(z, ; z, t) 2un(z, )

z2

= (z z)un (z, )ejt f(z, )Gn(z, ; z, t) (18)We assume that the source distribution f(z, t) is limited in a finiteinterval (a, b) as shown in Figure 1. Taking the integration of (18) overthe interval [a, b] and then taking the inverse Fourier transform withrespect to time, we obtain

un(z, t) =

Gn(z, t; z, t)

un(z, t)

zdt

un(z

, t)Gn(z, t; z

, t)z

dt

b

z=a

b

a

dz

f(z, t)Gn(z, t; z, t)dt, z (a, b) (19)

where we have used the symmetry of Greens function about z and z.Note that if a , b , the above expression becomes

un(z, t) =

dz

f(z, t)Gn(z, t; z, t)dt, z (, ) (20)

Next we will show that the solution in the region (z+, +)(z+ b)and (, z)(z a) may be expressed in terms of its boundary


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values at b and a. Without loss of generality, we assume that a < 0and b > 0. Taking the integration of (18) over [z+, +] with z+ band using integration by parts, we obtain

Gn(z+, ; z, t)

un(z+, )

z un(z+, ) Gn(z+, ; z

, t)z

= un(z, )ejt

where we have used the radiation condition at z = +. Taking theinverse Fourier transform and letting z = z+ lead to

un(z, t) + cJ0(kcnct)H(t)

un(z, t)

z

= 0, z

b > 0 (21)

where we have replaced z+ by z and t t by t since z+ and t arearbitrary, and denotes the convolution with respect to time. Thisequation is called right-traveling condition of the wave [21]. Similarlytaking the integration of (18) over (, z] with z a, we mayobtain left-traveling condition

un(z, t) cJ0(kcnct)H(t) un(z, t)z

= 0, z a < 0 (22)

Both (21) and (22) are integral-differential equations. If the source isturned on at t = 0, all the fields must be zero when t < 0. In thiscase, (21) and (22) can be solved by single-sided Laplace transform.Denoting the Laplace transform of un by un we have

un(z, s) +

(s/c)2 + k2cn

1/2 un(z, s)z

= 0, z b > 0

un(z, t) (s/c)2 + k2cn1/2 un(z, s)

z = 0, z a < 0The solutions of the above equations can be easily found. Making useof the inverse Laplace transforms listed in [26], the solutions of (21)and (22) can be expressed as

un(z, t) = un

b, t z b

c

ckcn(z b)

t zb

c0

J1

kcnc

(t )2

(z b)2

/c2

(t )2 (z b)2/c2 un(b, )d

(z b > 0)


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un(z, t) = un

a, t +

z ac

+ ckcn(z a)

t+ za

c0

J1 kcnc(t )2

(z

a)2/c2

(t )2 (z a)2/c2 un(a, )d

(z a < 0)The above two equations have also been derived by the methodof impulse response function [16] and wave splitting technique [21].Once the input signal is known the output signal after traveling acertain distance in the waveguide can be determined by a convolutionoperation. Some numerical results can be found in references [16, 21].

However there is a common mistake in studying the propagation oftransients in a waveguide, which assumes a wideband input signal (suchas a step function or a rectangular pulse) and a single-mode operationat the same time. The electric field at the output is then determinedby the product of the response determined by one of the above twoequations and the transverse mode function. This process totallyignores the contributions from higher order modes and is not accurateenough to describe the actual response in the waveguide. Indeed if theexcitation pulse is wideband, the waveguide will be overmoded and the

effects of higher order modes cannot be neglected in most situations.This will be demonstrated below.

4. TRANSIENT PROCESS IN WAVEGUIDES

Now the above general theory will be applied to study the propagationcharacteristics of electromagnetic pulse in waveguides. In general awideband pulse in the waveguide will generate a number of higherorder modes, and the field distributions inside the waveguide shouldbe determined by (3) and (4), which will be approximated by a finitesummation of m terms.

4.1. Transient Process in a Rectangular Waveguide

Let us first consider a rectangular waveguide of width a and height bas depicted in Figure 2. If the waveguide is excited by a line currentextending across the waveguide located at x0 = a/2, then currentdensity is given by

J(r, t) = uy(x x0)(z z0)f(t)Since the line current density is uniform in y direction, the field excitedby the current will be independent of y. As a consequence, only TEn0


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o x

y

0 a

b

J

x

Figure 2. Cross-section of a rectangular waveguide.

mode will be excited. In this case we have

kcn =n

a , etn(x, y) = eT En0 (x, y) = uy 2

ab1/2

sinnx

a , n = 1, 2, 3 . . .

From (13), (20) and the above equation, we obtain

vT En (z, t) =b

2

2

ab

1/2sin

n

ax0

t|zz0|/c

df(t)

dtJ0 kcnc(t

t)2

|z

z0

|2/c2 dt (23)

where =

/. Thus the time-domain voltages vT En for TEn0(n =2, 4, 6, . . .) vanish. From (3) the total electric field in the waveguidemay be approximated by a finite summation of m terms

E = uyEy = uy

2

ab

1/2 mn=1

vT En sinnx

a

= uym

n=1

a

sin na

x0 sin nxa

t|zz0|/c

df(t)dt

J0

kcnc

(t t)2 |z z0|2/c2

dt (24)

The above expression has been validated in Appendix B by usinga totally different approach. The theory can also be validated by

considering the time-domain response to a continuous sinusoidal waveturned on at t = 0. We should expect that the time-domain responsetends to the well-known steady state response as time goes to infinity.Hereafter we assume that x0 = a/2, and z0 = 0 for all numerical


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examples. To validate the theory let f(t) = H(t)sin t. Then (23)may be expressed as

vT En (z, t) = vT En (z, t)

steady

+ vT En (z, t)transient

, t >|z|/c

where k = /c and vT En (z, t)steady

represents the steady-state part of

the response and vT En (z, t)transient

the transient part of the response

vT En (z, t)steady

=b

2

2

ab

1/2ka sin

n

2

|z|/acos ka(ct/a u)J0 kcnau2 |z|2/a2 du

vT En (z, t)transient

= b2

2

ab

1/2ka sin

n

2

ct/a

cos ka(ct/a u)J0

kcna

u2 |z|2/a2

du

Note that the transient part of the response approaches to zero ast . To investigate the steady-state part of the response thefollowing calculations are needed [27]

a

J0

b

x2a2

sin cxdx =

0, 0 < c < b

cos

a

c2 b2

/

c2 b2, 0 < b < c

a

J0 bx2a2 cos cxdx =

expab2c2

/

b2c2, 0 < c < b

sin ac2b2 /c2b2, 0 < b < cThus we have

vT En (z, t)steady

=b

2

2

ab

1/2 ka sin n2

(ka)2 (kcna)2

sin

ka

ct

a |z|

a

(ka)2 (kcna)2

, k > kcn

cos

ka

ct

a

exp|z/a|(kcna)2(ka)2 , k < kcn

Therefore only those modes satisfying k > kcn will propagate in the

steady state. When vT En

steady

are inserted into (24) it will be found


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0 2 4 6 8 104

2

0

2

4

vTE1

ct/a{

z/a = 2, ka = 1.2}

vTE3

Figure 3. Time-domain modal voltages excited by a sinusoidal wave.

0 5 10 15 20

0

3

4.5

ct/a{x = a/2, z = 2a, ka = 1.2}

Ey{m = 1} Ey{m = 3}

Figure 4. Electric fields excited by a sinusoidal wave.

that the steady state response of the electric field agrees with thetraditional time harmonic theory of waveguides (see Eqn. (74), Chapter5 of [20]). The variation of vT En =

2vT En /

b/a and Ey = aEy/

with the time at z = 2a for different modes have been shown in Figure 3and Figure 4 respectively, where m stands for the number of terms

chosen in (24). The operating frequency has been chosen in betweenthe first cut-off frequency and the second one. It can be seen that thecontribution from the higher order modes are negligible. However thisis not true for other wideband excitation pulse.


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282 Geyi

0 2 4 6 8 101

0

1

ct/a

{z = 2a

}

vTE1

vTE5

Figure 5. Time-domain modal voltages excited by a unit stepwaveform.

0 2 4 6 8 10

2

0

2

{ }/ /2 , 2ct a x a z a= =

{ }1yE m = { }39yE m =

Figure 6. Electric fields excited by a unit step waveform.

Let us consider a unit step pulse, i.e., f(t) = H(t). The time-domain voltages vT En for the first and fifth mode are shown in Figure 5,which clearly indicates that the voltage for the higher order modescannot be ignored in this case. As indicated by Figure 6, the electricfields at z = 2a, obtained by assuming m = 1 (only the dominant modeis used) and m = 39 (the first 39 modes are used), are quite different

due to the significant contributions from the higher order modes. Itis seen that time response of the field is totally different from theoriginal excitation pulse (i.e., a unit step function) due to the fact thata hollow waveguide is essentially a high pass filter, which blocks all


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the low frequency components below the first cut-off frequency. Inaddition the waveguide exhibits severe dispersion. It should be notedthat the singularities in the electric field distributions come from the

time derivatives of the excitation waveform in (23), and the periodic-like performance of the field distribution results from the behavior ofBessel functions.

Thus it is clear that a hollow metal waveguide is not an idealmedium to carry a wideband signal that contains significant lowfrequency components below the first cut-off frequency. In this caseone should use a multi-conductor transmission line supporting a TEMmode whose cut-off frequency is zero.

0

a

y

x

mJ

b

Figure 7. Cross-section of a coaxial waveguide.

4.2. Transient Process in a Coaxial Cable

To see how a pulse propagates in a TEM transmission line as well asthe effects of the higher order modes we may consider a coaxial lineconsisting of an inner conductor of radius a and an outer conductorof radius b, as shown in Figure 7. Let the coaxial line be excited by amagnetic ring current located at z = 0, i.e.,

Jm(r, t) = uf(t)(z)( 0), a < 0 < bwhere (,,z) are the polar coordinates and u is the unit vectorin direction. According to the symmetry, only TEM mode andthose TM0q modes that are independent of will be excited. Theorthonormal vector mode functions for these modes are given by [28]

kc1 = 0, et1(, ) = ur 1

2 ln c1

kcn =na

,


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284 Geyi

etn(, ) = u

2

na

J1(n/a)N0(n) N1(n/a)J0(n)J20 (n)/J

20 (c1n) 1

, n 2

where c1 = b/a, u is the unit vector in direction, and nis the nth nonvanishing root of the eqauation J0(nc1)N0(n) N0(nc1)J0(n) = 0. It follows from (12), (14) and (20) that

iT EM1 =

2 ln c1f(t |z z0|/c)

iT Mn = n0

2a

J1(n0/a)N0(n) N1(n0/a)J0(n)

J20 (n)/J

20 (c1n) 1

t|zz0|/c

df(t)dt

J0

kcnc

(t t)2 |z z0|2/c2

dt

Therefore if the highest frequency component of the excitationwaveform is below the cut-off frequency of the first higher mode, thecoaxial line will be an ideal medium for a distortion-free transmissionof signals. From (4) the magnetic field in the coaxial cable is given by

H = uH = uiT EMn 1

2 ln c1

+u

n=2

iT Mn

n2a

J1(n/a)N0(n) N1(n/a)J0(n)J20 (n)/J

20 (c1n) 1

=u

2 ln c1f(t |z z0|/c)

u

n=2

22n0

4a2

J1(n0/a)N0(n) N1(n0/a)J0(n)J20 (n)/J20 (c1n) 1

J1(n/a)N0(n) N1(n/a)J0(n)J20 (n)/J

20 (c1n) 1

t|zz0|/c

df(t)dt

J0

kcnc

(t t)2 |z z0|2/c2

dt (25)

In the following all numerical examples are based on the assumptionsthat 0 = (a + b)/2, and z0 = 0. The time-domain currents for the firstthree modes excited by a unit step waveform f(t) = H(t) are depicted

in Figure 8, where iT EM1 = iT EM1 and iT Mn = iT Mn . Figure 9 gives


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0 2 4 6 8 10

0

1

2

iTEM1

ct/a{z = 2a}

iTM3 i

TM2

Figure 8. Time-domain modal currents excited by unit stepwaveform.

0 2 4 6 8 100

0.5

1

1.5

ct/a

{r = 1.1a, z = 2a

}

H{m = 5}

H{m = 1}

Figure 9. Magnetic fields excited by unit step waveform.

the magnetic fields H = H at r = 1.1a, z = 2a, where m denotesthe number of terms chosen in (25). If we neglect the contribution ofthe higher modes by letting m = 1, the magnetic field is a perfect stepfunction. When higher modes are included (m = 5, the first five modesare used), some ripples will occur in the response of the magnetic field,

which stands for the contribution of higher order modes excited by thehigh frequency components of the unit step waveform. In this situationthe signal has been distorted when it propagates in the coaxial line.

Fig. 10 and Fig. 11 give the time-domain currents and the


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286 Geyi

0 2 4 6 8 10

0

1

2

i TEM1

i TM2

iTM3

ct/a{z = 2a, cT/a = 3}

Figure 10. Time-domain modal currents excited by a rectangularpulse.

0 2 4 6 8 100.5

0

0.5

1

H{m = 5}

H{m = 1}

ct/a

{r = 1.1a, z = 2a, cT/a = 3

}Figure 11. Magnetic fields excited by rectangular pulse.

magnetic fields in the coaxial line excited by a rectangular pulse f(t) =H(t) H(t T) respectively. It can be seen that the ripples occurnot only inside the pulse but also outside the pulse. Therefore when arectangular pulse train passes through a transmission line the pulse willspread in time, which causes the pulse to smear into the time intervals

of succeeding pulses. This phenomenon is very similar to the situationof a pulse train passing through a bandlimited channel, where thepulses will also spread in time, introducing intersymbol interference.Note that the time spread in the transmission line is caused by the


21/31


higher order modes while the time spread in a bandlimited channel isdue to the shortage of bandwidth. To reduce the effects of time spreadin both cases, the pulse shaping techniques can be used to restrain the

high frequency components.

5. CONCLUSIONS

In this paper a new and concise approach for the time-domain theoryof waveguides has been presented. The theory has been applied tothe time-domain analysis of excitation problems in typical waveguides.Numerical results for various typical excitation pulses have beenexpounded, which give a real physical picture of the transient process in

a waveguide. In a special case where the excitation pulse is a sinusoidalwave turned on at t = 0, the steady state response is shown to approachto the well-known result from time-harmonic theory, which validatesour theory. More validations can be found in the Appendices.

Our analysis results have also indicated that the contributionsfrom the higher order modes excited by a wideband waveform aresignificant. Therefore the input signal at a given reference plane cannotsimply be written as a single term of a separable function of spaceand time when studying the propagation of wideband signals in a

waveguide or a feeding line of an antenna. Instead the expressionof the input signal should consist of a number of such terms givenby (3) and (4), each term having a different time variation than theothers. The number of terms to be selected depends on the accuracyrequired as well as the bandwidth of excitation waveform. The widerthe bandwidth of the excitation waveform, the more terms of the higherorder modes must be included.

APPENDIX A. PROOF OF THE COMPLETENESS OF

THE TRANSVERSE EIGENFUNCTIONS

To prove the completeness of the transverse eigenfunctions we need thefollowing theorem [29, pp. 284].

Theorem: Let H be a real separable Hilbert space with dim H = .Let B : D(B) H H be a linear, symmetric operator. We furtherassume that B is strongly monotone, i.e., there exists a constant c1such that (Bu, u) > c1

u

2 for all u

D(B). Let BF be the Friedrichs

extension ofB and HB be the energy space of operator B. We furtherassume that the embedding HB H is compact. Then the followingeigenvalue problem

BFu = u, u D(BF)


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288 Geyi

has a countable eigenfunctions {un}, which form a completeorthonormal system in the Hilbert space H, with un HB. Eacheigenvalue corresponding to un has finite multiplicity. Furthermore we

have 1 2 and n n .Now consider the following eigenvalue problem for the transverseelectric field in the waveguide

B(e) = e ( e) = k2ce, r un e = e = 0, r (A1)

where k2c = 2 2, and B = . The domain of

definition of operator B is defined as follows

D(B) =ee (C())2,un e = e = 0 on

where C() stands for the set of functions that have continuouspartial derivatives of any order. Let L2() stand for the spaceof square-integrable functions defined in and H = (L2())2 =L2() L2(). For two transverse vector fields e1 and e2 in (L2())2,the usual inner product is defined by

(e1

, e2

) =

e1 e2

d.

and the corresponding norm is denoted by = (, )1/2. In the abovea bar over a letter is used to represent the complex conjugate. Now wemodify (A1) as equivalent form by adding a term e on both sides

A(e) = e ( e) + e = k2c + e, r = un e = e = 0, r

where is an arbitrary positive constant. In order to apply the previoustheorem we need to prove that the operator A is symmetric, stronglymonotone and the embedding HA H is compact. The proof will becarried out in three steps.

Step 1: We first show that A is symmetric, strongly monotone. For alle1, e2 D(A) = D(B), we have

(e1

, e2

)A = (A(e1

), e2

) =

[

e1

(

e1

) + e1

]e2

d

=

[ e1 e2 + ( e1)( e2) + e1 e2]d (A2)


23/31


Therefore the new operator A is symmetric. Thus we can assume thate is real. A is also strongly monotone since

(A

(e

),e

) =

[ e

e

+ ( e

)( e

) + e

e

]d e2

Step 2: We then demonstrate some important properties of energyspace HA. The energy space HA is the completion of D(A) with

respect to the norm A = (, )1/2A . Now let e HA, and bydefinition, there exists admissible sequence {en D(A)} for e suchthat

en e n 0

and {en} is a Cauchy sequence in HA. From (A2) we obtainen em2A = en em2 + en em2 + en em2Consequently { en} and { en} are Cauchy sequences in H. Asa result, there exist h H, and L2() such that

en n h, en n From integration by parts

en d =

en d, (C0 ())2

( en)d =

en d, C0 ()

we obtain

h d =

e d, (C0 ())2

d =

e d, C0 ().

In the above C0 () is the set of all functions in C() that vanishoutside a compact set of . Therefore e = h and e = in the generalized sense. For arbitrary e1, e2 HA, there are twoadmissible functions {e1n} and {e2n} such that e1n e1 n 0 ande2n e2 n 0. We define

(e1,e2)A = limn(e1n,e2n)A

=

[ e1 e2 + ( e1)( e2) + e1 e2] d


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290 Geyi

where the derivatives must be understood in the generalized sense.

Step 3: Finally we prove that the embedding HA H is compact. LetJ(e) = e, e HA

Then the linear operator J : HA H is continuous sinceJ(e)2 = e2 1

e2 + e2 + e2

= 1eHA.

A bounded sequence {en} HA impliesen2HA = en2 + en2 + en2

=

(enx)2 + (eny)2 + (enx)2 + (eny)2

d c

where c is a constant. So the compactness of the operator J followsfrom the above inequality and the following Rellichs theorem.

Rellichs theorem [25]: Any sequence {fn}, which satisfies fn2 =

f2nd c1 and fn2 =

(fn)2d c2, where c1 and c2

are constants, has a subsequence still denoted by {fn} such thatlim

m,n

w(fm fn)2d = 0, where w is the weight function.From the previous theorem, the set of eigenfunctions {en}

constitutes a complete system in (L2())2. The proof is completed.

APPENDIX B. A DIFFERENT APPROACH TO THETRANSIENT RESPONSE OF RECTANGULARWAVEGUIDE

To validate (24), let us take a totally different approach to theexcitation problem considered in Section 4.1. The electromagneticfields in the waveguide satisfy the wave equations

E+ 2E

t2= J

t, H+

2H

t2= J

From the wave equations and the property of exciting source only they component of the electric field E is excited. The y component of the

electric field Ey satisfies2

x2+

2

z2

2

t2

Ey = f

(t)(x x0)(z z0)


25/31


and the boundary condition Ey(x,z,t)

x=0,a= 0. Making use of the

Fourier transform pair with respect to z and t,

Ey(x,p,) = F(Ey) =

Ey(x,z,t)ejpzjtdzdt

Ey(x,z,t) = F1(Ey) =

1

(2)2

Ey(x,p,)ejpz+jtdpd

we obtain

2

x2p2 + 2

c2

Ey = jf()e

jpz0(x x0)

Ey(x,p,)

x=0,a= 0

(B1)

where c = 1/

and f() is the Fourier transform off(t). Accordingto the boundary condition we may use the well-known complete

orthonormal set {2a sin

na x|n = 1, 2, . . .} to expand the electric field

[30]. So we have

Ey =

n=1

en(z, t)

2

asin

n

ax (B2)

Inserting this into (B1) we get

n=1

en(z, t)2

asin

n

ax

n

a 2

p2 + 2

c2 = jf()ejpz0(xx0)

Multiplying both sides by

2/a sin(nx/a) and taking the integrationover [0, a] gives

en(z, t) =jc2f()

2/a sin(nx0/a)e

jpz0

2 c2(n/a)2 p2c2

The electric field can be obtained by taking the inverse Fourier

transformEy =

n=1

F1 [en(z, t)]

2

asin

n

ax


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292 Geyi

where

F1 [en(z, t)] = c22

asin

nx0a

F1 jf()ejpz0

2

c2(n/a)2

p2c2

=c2

(2)2

2

asin

nx0a

f(t)

ejpzjpz0dp

ejt

2 p2c2 c2(n/a)2d

The integral with respect to can be evaluated by residue theoremand result is

ejtd

2

p2

c2

c2

(n/a)2

=

2 sin t

p2c2 + c2(n/a)2

p2

c2

+ c2

(n/a)2

Thus we have

F1[en(z, t)] = c2

2

asin

nx0a

f(t)

0

cosp(z z0) sin tc

p2 + (n/a)2

p2 + (n/a)2

dp

= 2

2a

sin nx0

af(t)

J0

nc

t2 |z z0|/c2

H(t |z z0|/c)

Finally the electric field is given by

E = uyEy

= uy

n=1

a sin

nx0a sin

n

a x

t|zz0|/c

df(t)dt

J0

kcnc

(t t)2 |z z0|2/c2

dt (B3)

which is exactly the same as (24).

APPENDIX C. A DIFFERENT APPROACH TO THETRANSIENT RESPONSE OF COAXIAL WAVEGUIDE

To validate (25), the excitation problem discussed in Section 4.2 willbe studied by directly solving Maxwell equations. The electromagnetic


27/31


fields in the coaxial waveguide satisfy

H+ 2H

t2= Jm

t, E+

2E

t2= Jm

(C1)From the above equations and the symmetry of the source only thefield components E, Ez, and H are excited. Thus we have

H= Hz

u +1

(H)uz

The boundary condition un E = 0 on the conductor requires

(H)=a,b = 0. Equation (C1) reduces to2H

2+

1

H

H2

+2H

z2

2Ht2

= f(t)(z z0)( 0)

Making use of the Fourier transform pair with respect to z and t,

H(,,p,) = F(H) =

H(,,z,t)ejpzjtdzdt

H(,,z,t) = F1(H) = 1

(2)2

H(,,p,)ejpz+jtdpd

we get

2H2

+1

H

+

2 1

2

H = jf()e

jpz0( 0)(H)

=a,b

= H

+ H=a,b

= 0

(C2)where 2 = 2p2. To solve this equation, we may use the method ofeigenfunction expansion [30]. We first consider the following eigenvalueproblem

2h2

+1

h

+

2 1

2

h = 0

h

+ h=a,b = 0(C3)

This is a typical eigenvalue problem of Sturm-Liouville type. Thegeneral solution of the above equation is

h = a1J1() + a2N1()


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294 Geyi

Applying the boundary conditions yields

a1/a2 = N0(a)/J0(a) = N0(b)/J0(b)

The above equation determines the eigenvalues n, n = 1, 2, . . . , whichsatisfyJ0(c

n)N0(n) J0(n)N0(cn) = 0where n = na is the nth root of the above equation, and c

= b/a.The eigenfunctions corresponding to n are given by

hn() = bn[J1(n/a)N0(n) N1(n/a)J0(n)]The constants bn can be determined by using the normalized conditionb

a h2nd = 1, which gives bn =n2a

J20 (n)J20(cn) 11/2. The normalized

eigenfunctions may be expressed as

hn() =n

2a

[J1(n/a)N0(n) N1(n/a)J0(n)]J20 (n)/J

20 (c

n) 11/2 , n = 1, 2, . . .

In solving (C3) we have assumed that = 0. If = 0, (C3) reduces to

2h

2+

1

h

h

2= 0

h

+ h

=a,b

= 0

The general solution of the above equation is h = c1 + c2/, wherec1 and c2 are two constants. The boundary condition requires thatc1 = 0. Therefore h = c2/. Applying the normalized condition weobtain eigenfunction corresponding to = 0

h0 =1

ln cThus the set of eigenfunctions {hn, n = 0, 1, 2, . . .} is complete bySturm-Liouville theory [30], and can be used to expand the solution of(C2)

H =

n=0

gnhn (C4)

where gn are the expansion coefficients to be determined. Substituting

this into (C2) we obtain

n=0

gn

2hn

2+

1

hn

+

2 1

2

hn

= jf()ejpz0( 0)


29/31


i.e.,

n=0gn

2 2n

hn = jf()e

jpz0( 0)

Multiplying both sides by hn and taking the integration over [a, b],we get

gn =jc20f()e

jpz0hn(0)2 p2c2 2nc2

The magnetic field can then be obtained by taking the inverse Fouriertransform of (C4)

H =

n=0

F1(gn)hn

where

F1(gn) = c20hn(0)F1

jf()ejpz0

2 p2c2 2nc2

Similar to the discussion in Appendix B, we may obtain

F1(gn) = 0hn(0)2

t|zz0|/c

df(t)

dtJ0 kcnc(t t

)2 |z z0|2/c2 dt

The magnetic field is given by

H = uH =u

2 ln cf(t |z z0|/c)

u

n=1

22n04a2

[J1(n0/a)N0(n) N1(n0/a)J0(n)]J20 (n)/J

20 (c

n) 11/2

[J1(n/a)N0(n)

N1(n/a)J0(n)]

J20 (n)/J20 (c

n) 11/2

t|zz0|/c

df(t)dt

J0

kcnc

(t t)2 |z z0|2/c2

dt

which is exactly the same as (25).

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296 Geyi

2. Namiki, M. and K. Horiuchi, On the transient phenomena in thewave guide, J. Phys. Soc. Japan, Vol. 7, 190193, 1952.

3. Ito, M., Dispersion of very short microwave pulses in waveguide,

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Documents

W. Geyi- A Time-Domain Theory of Waveguide