12

Difference Equations and Functional Equations in Transmission-line Theory^†

RAYMOND REDHEFFER

PROFESSOR OF MATHEMATICS
UNIVERSITY OF CALIFORNIA, LOS ANGELES

12.1    Introduction

Historically, the subject of this chapter starts in 1862 with a paper by G. G. Stokes, in which difference equations governing the optical behavior of m + n identical glass plates are written down and solved by treating the discrete variable n as continuous. It is probable that Stokes attached no physical significance to the continuous problem but regarded it merely as a mathematical artifice. If one applies Stokes’ method to a dielectric sheet of thickness m + n, however, one gets the same equations as before, except that m and n are really continuous. The analysis can thus be thought of as solving the cascade problem by the construction of a suitable dielectric medium and looking at the behavior for integral values of the thickness.

Not much more appears to have been done along these lines until the advent of microwave technology during and after the Second World War. In this period, it was found that equations similar to Stokes’ can be used to form an algebra of obstacles, which yields an interesting approach to certain aspects of transmission-line theory. Further impetus was given by the simultaneous development of the theory of the scattering matrix, and the subject has by now attained maturity.

The reader unfamiliar with engineering terminology may welcome a brief digression at this point. What we have in mind is a “disturbance” or “wave” that propagates through an obstacle. The scattering matrix specifies the obstacle by its transmission and reflection coefficients, which are defined as follows. The left-hand transmission coefficient t is the complex amplitude of the wave emerging from the right when a wave of amplitude 1 is incident at the left; thus, |t| is the amplitude of the wave, while ∠t is its phase. The left-hand reflection r is the complex amplitude of the reflected wave when a wave of amplitude 1 is incident at the left, and similar definitions hold for the right-hand transmission τ and reflection ρ. The obstacle is supposed to be followed by a matched termination; that is, the terminating reflection r₁ is 0 when r is measured, and similarly for ρ. Also, the propagation takes place in a uniform transmission line. To specify each transmission and reflection by a single number, it is assumed that only one mode is relevant at the given frequency, which remains fixed throughout the discussion.

Since the frequency is fixed, the main interest is in a space variable giving position along the line. Adding obstacles in succession leads to difference equations for the coefficients, whereas adding to the thickness of a continuous medium leads to functional equations. These equations, naturally, must be distinguished from those of network analysis and synthesis, in which the frequency is varied while the network is held constant. Their closest affinity is not to the methods of circuit analysis, but to a general class of methods for which Bellman has coined the descriptive phrase invariant imbedding. The latter have proved useful not only in circuit theory but also in probability, in neutron diffusion, and in other areas of mathematical physics.

The meaning of the term “invariant imbedding” is aptly illustrated by the functional equations for an inhomogeneous dielectric medium. The response of the part of the medium from x to y, with y > x, is determined as if this part were in free space—in other words, without regard to the rest of the medium in which it is imbedded. The resulting equations, naturally, are invariant as far as this imbedding is concerned, and one speaks of the “method of invariant imbedding.”

In the early part of this chapter, we present a résumé of certain topics in transmission-line theory using the combination of networks† with each other as leading idea. The discussion is then generalized, first to continuous variation, then to matrices, and finally to operators on a Hilbert space (see Chap. 4). Occasion is taken by the way to indicate an unexpected analogy between transmission-line theory and the theory of probability. The practical bearing of these methods is illustrated by examples in the form of problems, and the specific physical situation is kept firmly in mind throughout the discussion. There nevertheless emerges a unified mathematical structure, which has applications to several fields of technology.

THE ALGEBRAIC FOUNDATIONS

12.2    An Instructive Special Case

Some interesting conclusions concerning the interaction of networks are suggested by the following example. Suppose a plane panel transmits 1 per cent of the power in a monochromatic plane wave incident on it and reflects 99 per cent, no power being absorbed. Converting from power to amplitude requires taking the square root, so that

represent the absolute value of the amplitude transmission and reflection coefficients, respectively.

Fig. 12.1 Two lossless plane sheets spaced for maximum transmission.

Let this panel be followed by another just like it, as shown in Fig. 12.1, at such a distance as to maximize the over-all transmission. For maximum transmission, the round-trip path from p₁ to p₂ in the figure (with due regard to the phase shift introduced by the two reflections) must be a whole number of wavelengths. In that case all the waves at p₁, p₂, p₃, … add in phase, so that the resulting amplitude of the wave moving from left to right between the panels is

Upon letting this wave traverse the second sheet, we get the value (10)(0.1) = 1.

The result has an air of paradox. It is as if we had a security guard who lets only 1 per cent of the visitors through; to make security doubly sure, that guard is followed by a second one with the same duty; and then we find, to our surprise, that everyone gets past both guards.

The reason for this odd behavior in the optical example is not hard to find. The two panels form a resonant cavity, so that the field between them is much stronger than the incident field. The attenuation introduced by the second panel is just what is needed to cut that stronger field down to the incident value, 1.

Since the over-all transmission is 100 per cent, the over-all reflection must be zero. The reflection depends, however, on the phase shifts associated with the complex reflection coefficient r and transmission coefficient t. Only for certain values of ∠t − ∠r does the over-all reflection in fact reduce to 0. Thus, we see that a relation involving phases can be deduced even though the underlying physical principle, conservation of energy, involves the magnitudes alone.

Fig. 12.2 Resultant amplitudes of waves in Fig. 12.1.

The foregoing results can be obtained without introduction of infinite series. In Fig. 12.2, the quantities z_i represent the absolute value of the amplitudes of the waves traveling in the indicated directions and regions. By superposition, z₃ is equal to the part of the incident wave, 1, that is transmitted, plus the part of z₄ that is reflected:

Similarly, and . These equations imply z₅ = 1, as before.

12.3    The Composition of Networks in General

Let the obstacle with coefficients t, τ, r, and ρ in Fig. 12.3 be placed adjacent to a second obstacle having coefficients t₁, τ₁, r₁, and ρ₁ as shown in Fig. 12.4. If the incident wave has amplitude 1, then the complex amplitude of the wave moving from left to right between the obstacles is

Fig. 12.3 An obstacle with left-hand coefficients (t,r) and right-hand coefficients (τ,ρ).

Fig. 12.4 Two adjacent obstacles.

provided |r₁ρ| < 1, which is assumed without comment henceforth. Upon letting this resultant wave progress through the second obstacle, we get

for the transmitted amplitude. Similarly, letting the wave A be reflected at the second obstacle and then transmitted from right to left through the first gives

as the amplitude of the reflected wave.

Since the incident amplitude was 1, these expressions give the left-hand transmission and reflection coefficients, respectively. In the same way, we can determine the right-hand coefficients. Upon introducing the scattering matrix

we can summarize the four equations as follows:

Here, the star product denotes the scattering matrix for the composite obstacle consisting of the first immediately followed by the second.

Fig. 12.5 Resultant amplitudes of waves in the presence of an obstacle.

This star multiplication imposes a definite algebraic structure on our obstacles—a structure so detailed that it can be used as the definition of the obstacles themselves. From that standpoint a linear four-terminal network is a quadruple of complex numbers that combines with other quadruples in the manner prescribed by Eq. (12.3).

A somewhat different approach to Eq. (12.3) proves useful for the further mathematical development. In Fig. 12.5, the complex numbers z_i represent amplitudes of waves propagating as shown, so that odd subscripts refer to propagation from left to right and even subscripts from right to left. Superposition combines with the definition of t, τ, r, and ρ to yield

as in the discussion accompanying Fig. 12.2. Thus, the scattering matrix is associated with a certain linear transformation.

Fig. 12.6 Amplitudes of waves in the presence of two adjacent obstacles.

Let us determine the transformation associated with the star product. In Fig. 12.6, we have the relation (12.4) and also

Upon solving for z₅ and z₂ in terms of z₁ and z₆, we get

Since the scattering matrix is wholly determined by the associated linear transformation, the matrix in the foregoing equation must be the scattering matrix for the composite obstacle. This yields a new proof of Eq. (12.3). Also, Eq. (12.3) has been shown consistent with Eqs. (12.4) and (12.5), just as the physics of the problem suggests.

EXERCISES

1. Show that the scattering matrix for a length x of transmission line with propagation constant k is (tτrρ) = (e^jkxe^jkx00). The line is lossless if |t| ≡ 1, passive if |t| ≤ 1 for x > 0. What can you say about k in these two cases?

2. An obstacle is in shunt if the electrical fields (E fields) on each side are always equal. Show that (tτrρ) is in shunt if and only if

t = 1 + randτ = 1 + ρ

(In the notation of Fig. 12.5, the desired condition is z₁ + z₂ ≡ z₃ + z₄.)

3. In the notation of Fig. 12.7, show that

where g is the reflection coefficient looking toward the generator at point A and r is the reflection coefficient looking toward the load at point B. If the propagation constant is k, then

Fig. 12.7 Measuring probe, with arbitrary probe reflection, generator mismatch, and load mismatch.

(The probe is symmetric, and the generator power to a matched line is unity. See Exercise 6.)

4. In Exercise 3 the probe is connected to a detector and voltmeter such that the reading V is proportional to |E + E′|². The power standing-wave ratio (SWR)² is defined by (SWR)² = V_max/V_min as the probe is moved along the line. Assuming a lossless line and neglecting the effect of the probe, show that the load reflection is

independently of the generator reflection G. (Take p = 0, t = 1 in the result of Exercise 3 and vary x.)

5. In Exercise 12.4 suppose that the probe reflection is not negligible but that R = 0. Show that the apparent standing-wave ratio is

and thus describe an experimental method of getting G.

6. A generator with complex reflection g (looking toward the generator) would deliver a wave of amplitude A₀ to a matched line at its right. If the line is not matched but is terminated by the complex reflection R₁, as seen from the generator, show that the complex amplitudes of waves moving from left to right and from right to left at the generator are, respectively,

Obtain the field by adding these expressions.

12.4    Matrix Multiplication

The algebra of four-terminal networks is known to be equivalent to the algebra of 2 by 2 complex matrices, but the multiplication in Eq. (12.3) is certainly not matrix multiplication. To reconcile these two observations, note that Eq. (12.4) can be written as a relation between (z₂,z₁) and (z₄,z₃):

When the first obstacle is followed by a second as in Fig. 12.6, we have a similar relation between (z₄,z₃) and (z₆,z₅). The matrix taking (z₂,z₁) into (z₆,z₅) is the ordinary matrix product, and this gives an alternative description of Eq. (12.6).

It follows that the correspondence

where the prime denotes the transpose, is an isomorphism. The star product of two matrices on the left corresponds, under this relationship, to the matrix product of the two matrices on the right. This shows, incidentally, that star multiplication is associative, i.e., that

S₁ * (S₂ * S₃) = (S₁ * S₂) * S₃

for 2 by 2 complex matrices S_i. A physical interpretation is readily given.

For purposes of algebraic description, either the star product or the matrix product is equally appropriate. The star product applies if our four-terminal networks are specified by their reflection and transmission coefficients t, τ, r, ρ, whereas the matrix product must be used if the network is specified by the four complex numbers

Sooner or later it must be stated that τ ≠ 0, if we wish to contemplate the foregoing expressions without uneasiness. A good many of the following results also require τ ≠ 0, or t ≠ 0, or both, and these conditions are hereby assumed once and for all. By Eq. (12.3), tτ ≠ 0 in the star product if tτ ≠ 0 in each factor.

EXERCISES

1. With d = tτ − rρ, let the obstacle be specified by

where and i, j, k are the units of quaternion algebra. Show that the star-product rule for (tτrρ) makes these symbols combine by the rules of quaternion multiplication.

2. Show that the angles ϕ₀ = ∠t + ∠τ − ∠r − ∠ρ and ϕ₁ = ∠t − ∠τ are invariant under addition of lossless line lengths; i.e., they are unchanged if (tτrρ) is replaced by

(aa00) * (tτrρ) * (bb00)   with |a| = |b| = 1

3. Prove that the expressions ϕ₀ and ϕ₁ of Exercise 2 are the only phases invariant under addition of line lengths, in the following sense: If f(∠t, ∠τ, ∠r, ∠ρ) is such an invariant, then

f(∠t, ∠τ, ∠r, ∠ρ) ≡ F(ϕ₀,ϕ₁)

for some function F.

12.5    Lossless Networks and the Reciprocity Theorem

Since power is proportional to |amplitude|², the condition for a network to be lossless is

If we agree that the composition of two lossless networks is lossless, and that a network remains lossless when turned end for end,† we can deduce a reciprocity theorem, and also a relation among the phases.

Let (tτrρ) and (τtρr) be spaced for maximum transmission (Fig. 12.8). The maximum transmission is, in magnitude,

as in the discussion of Eq. (12.1). Since the transmission does not exceed 1, we get

|t| |τ| ≤ 1 − |ρ|² = |τ|²

Fig. 12.8 Obstacle followed by its mirror image at such a distance as to maximize the over all power transmission.

Thus, |t| ≤ |τ|. Turning both networks end for end in Fig. 12.8 and repeating the argument gives |τ| ≤ |t|. These two inequalities together show that |t| = |τ|.

The full reciprocity theorem t = τ cannot be deduced by any argument involving power transfer at the terminals only; the equality ∠t = ∠τ requires information about the interior of the obstacle. If, however, the two obstacles in Fig. 12.8 are separated by an electrical distance x of lossless line, the transmission

is maximum when x = − ∠ρ. (This choice of x leads to the situation analyzed in the foregoing derivation.) Since

|t|² = |τ|² = 1 − |ρ|²

the maximum value is 1, so that the over-all reflection is

The latter equation is equivalent to

where n is an integer. Here is a relationship among the phases that does follow from energy considerations at the terminals, even though the expected equality ∠t = ∠τ does not.

EXERCISES

1. Prove that every lossless obstacle (tτrρ) with r ≠ 0 satisfies the equation

tτ − rρ = − exp [j(∠r + ∠ρ)]

2. If the lossless obstacle (tτrρ) is followed by the complex reflection r₁, show that the over-all left-hand reflection is

(Note that |r| = |ρ| since |t| = |τ|, and use Exercise 1.)

3. A lossless obstacle is followed by a second one at such a distance as to maximize the over-all left-hand reflection. Show that the phase shift associated with the left-hand reflection is the same as it was before the second obstacle was added. What if the spacing is such as to minimize the over-all reflection? (Use Exercise 2 with ϕ variable.)

4. The transformation from normalized impedance to reflection is

If this function is denoted by , show that and

Show also that .

5. The standing-wave ratio associated with a reflection r is ; cf. Sec. 12.3, Exercise 4. If two lossless obstacles are spaced for maximum reflection, show that the corresponding standing-wave ratio is the product of the individual standing-wave ratios. What if they are spaced for minimum reflection? (Use Exercise 2 with ϕ variable, and also Exercise 4.)

6. Show that for a symmetrical lossless obstacle, r and t are related as in Fig. 12.9.

Fig. 12.9 Relationship of complex transmission and complex reflection for symmetrical, lossless obstacles.

12.6    Passive Networks

A dissipative or passive network produces no energy, so that Eqs. (12.9) are replaced by

Now, these inequalities are not preserved by star multiplication; hence they are not the only conditions needed to ensure passivity. The additional conditions can be obtained by noting that

in Fig. 12.5, since the power output does not exceed the power input. By Eq. (12.4) we get

together with the inequalities (12.11) as necessary and sufficient conditions for the inequality (12.12).

It is clear from the derivation that the conditions (12.11) and (12.13) are preserved by matrix multiplication, and we show that they are also preserved by star multiplication. Indeed, if the networks of Fig. 12.6 are dissipative, then we have, besides the inequality (12.12),

|z₄|² + |z₅|² ≤ |z₃|² + |z₆|²

Addition of this and (12.12) yields

|z₂|² + |z₅|² ≤ |z₁|² + |z₆|²

and hence the star product is dissipative, (z₁,z₆) being arbitrary.

To compare these results with those for the lossless case, we introduce the power-absorption coefficients,

a² = 1 − |t|² − |r|²   α² = 1 − |τ|² − |ρ|²

Then the inequality (12.13) yields, after brief calculation,

and, with ϕ₀ = ∠t + ∠τ − ∠r − ∠ρ,

The former reduces to |t| = |τ| and the latter to Eq. (12.10) when aα = 0. Since the inequality (12.14) compares the magnitudes of the coefficients from opposite sides, and the inequality (12.15) compares their phases, the two together constitute a reciprocity theorem.

EXERCISES

1. A shunt obstacle (tτrρ) satisfies t = 1 + r, τ = 1 + ρ (Sec. 12.3, Exercise 2). Show that such an obstacle is passive if and only if t = τ and |r| + cos (∠r) ≤ 0, so that r lies in the shaded circle of Fig. 12.10. Thus show that a passive shunt obstacle is always symmetrical.

2. A figure of merit for the performance of a measuring probe (Fig. 12.7) is M = fa²/(2|p|), where p is the probe reflection coefficient, a² is the probe absorption coefficient, and f is the fraction of the absorbed power that contributes to the voltmeter reading. Show that, for a passive shunt probe,

Fig. 12.10 Possible locus of the complex reflection for a passive shunt obstacle.

3. A passive obstacle (tτrρ) has small power-absorption coefficients a² and α². With d = tτ − rρ and ϕ₀ as in the text, show that

and

apart from higher-order terms in a and α.

4. Show that the over-all power absorption of (ttrr) * (ttrr) is 2a²/|t|², apart from terms in a⁴.

12.7    The Associated Linear Fractional Transformation

In Fig. 12.11, an obstacle is followed at the right by a variable reflection z. Symbolically, the situation is described by

Fig. 12.11 Obstacle followed by a reflection of constant magnitude but variable phase.

where the blank entries are irrelevant; hence the over-all left-hand reflection, given by Eq. (12.3), is

In just the same way, if the variable reflection z is introduced on the left, then the expression

gives the right-hand reflection of the composite obstacle.

The expressions (12.16) and (12.17) are linear fractional transformations of the complex variable z. To get a geometric interpretation, recall that such transformations take circles into circles. The circle |z| = a < |ρ|⁻¹ is mapped by w = R(z) onto a circle with center and radius, respectively,

This gives the response to a terminating reflection of constant magnitude a but variable phase. Since , the center of the circle can be found as the response to the terminating reflection .

The particular form of C_a yields an interpretation of the angle

ϕ₀ = ∠t + ∠τ − ∠r − ∠ρ

Fig. 12.12 Image of the circle |z| = constant after modification by an arbitrary obstacle (tτrρ).

that played a role in our energy considerations. Namely, ϕ₀ is the positive angle from the line Or produced to the line rC_a, no matter what value a may have. See Fig. 12.12.

EXERCISES

1. Prove that the origin lies inside the circle traced out by R(z), for |z| = a, if and only if |r/a| < |tτ − rρ|.

2. Let

z = ae^{i(θ−∠ρ−π)}

where a > 0 and θ is a real variable. Let ϕ be the angle between the radius to w = R(z) and the line joining r and C_a (Fig. 12.12). Show that

independently of the variables t, τ, and r.

3. In Fig. 12.7, let the line be lossless, let G = 0, and let the probe be in shunt, so that t = 1 + p. Show that the measured standing-wave ratio gives

for the apparent value of the load reflection R. Show also that the formula remains valid when the probe is not in shunt provided P, R, P′ are replaced respectively by

(See Sec. 12.3, Exercises 3 and 4. The values V_max and V_min can be found by computing the radius and center of the image of the unit circle under a certain linear fractional transformation.)

4. In Exercise 3, let t = 1 + p and let η be the error in minimum position due to the probe reflection. Show that sin 2kη satisfies a certain quadratic equation, and thus obtain the approximation formulas

for and , respectively. Show also that the result for t ≠ 1 + p can be obtained by the substitution given in Exercise 3.

12.8    Another Characterization of Passive Networks

A discussion of energy can be based on R(z) and P(z). We say that the network is reflectively dissipative from the left,† or left-dissipative, if |R(z)| < 1 whenever |z| < 1. Evidently, the condition implies |r| < 1 and |ρ| < 1. When that is the case, R(z) is analytic for |z| < 1, so that the maximum of |R(z)| occurs when |z| = 1. The desired condition is then equivalent to

|C_a| + R_a ≤ 1   for a = 1

which becomes

when we use Eqs. (12.18)

The inequality (12.19), together with |r| < 1 and |ρ| < 1, is necessary and sufficient to ensure |R(z)| < 1 for |z| < 1. Since the conditions are unaltered by interchanging t with τ, and r with ρ, the same ensures |P(z)| < 1 for |z| < 1. In other words, the network is left-dissipative if and only if it is right-dissipative.

To contrast this notion of passivity with that of Sec. 12.6, observe that the inequality (12.13) can be written in the form

Comparing the inequalities (12.19) and (12.20) shows that dissipative networks are also reflectively dissipative and that the converse is true provided |t| = |τ|; the equivalence of the two notions, however, hinges on this reciprocity theorem.

It is always possible to choose a positive constant h so that

|ht| = |h⁻¹τ|

Since

(ht)(h⁻¹τ) = tτ

we deduce that the network specified by t, τ, r, and ρ is reflectively dissipative if and only if there is a positive constant h for which

is dissipative in the sense of Sec. 12.6.

EXERCISES

1. Prove that if a passive object (tτrρ) has ρ = 0, then |tτ| + |r| ≤ 1. (Let the object be backed by a perfect reflector, |r₁| = 1, with such a phase ∠r₁ as to maximize the over-all reflection.)

2. For ρ = 0 as in Exercise 1, use the inequality (12.13) to get the relationship

|r|² ≤ (1 − |t|²)(1 − |τ|²)

This inequality is stronger than that of Exercise 1 unless |t| = |τ|.

Fig. 12.13 Obstacle matched by a symmetric lossless tuner.

3. A tuner is a lossless network (t₁t₁r₁ − r₁) such that (tτrρ) * (t₁t₁r₁ −r₁) has zero right-hand reflection (Fig. 12.13). The network (tτrρ) with the tuner is then said to be matched from the right. Show that a match is obtained if and only if . In this case, the over-all transmissions of the matched obstacle are, in magnitude,

|t|(1 − |ρ|²)^−½   and   |τ|(1 − |ρ|²)^−½

(See Sec. 12.5, Exercise 2.)

4. Show that the left-hand reflection R of the matched obstacle in Exercise 3 satisfies the inequality

where a² and α² are the power-absorption coefficients. (Assume that the obstacle is passive and use Exercise 2.)

5. A homogeneous, lossy dielectric sheet reflects 10 per cent of the incident power and transmits 80 per cent. Two identical sheets of this type are spaced for maximum transmission. Show that the over-all power reflection coefficient |R|² is at most 0.0124, although the over-all transmission T satisfies 1 − |T|² = 0.210. (Note that R is unchanged if the second sheet is replaced by a lossless one having the same reflection coefficient, and use Exercise 4.)

6. Show that if (tτrρ) is lossless, then the image of |z| = a under R(z) or P(z) is a circle with radius and distance from the origin to its center respectively

and thus that interchanging a and |r| has the effect of interchanging |C_a| and R_a. [Use Eqs. (12.18) and Sec. 12.5.]

7. In Exercise 6 deduce a|C_a|² = (a − R_a)(1 − aR_a); see Fig. 12.14. Show that the result for P(z) remains valid even when the lossless tuner is accompanied by an arbitrary fixed network at its left as in Fig. 12.13.

Fig. 12.14 The Silver crescent and the Rieke banana.

12.9    Fixed Points and Commutativity

The condition ξ = R(ξ) in Eq. (12.16) leads to

The two† roots ξ₁ and ξ₂ are fixed points of the transformation w = R(z), and their reciprocals are fixed points of w = P(z). The values ξ₁ and ξ₂ are the particular terminating reflectances that are not changed by the network.

If two networks both have the same fixed point ξ, then so also does their star product. Now the fixed points are determined by, and determine, the two ratios

Hence if these equations hold for each factor, A and B being constant, then they hold for the star product.

A different interpretation of Eqs. (12.23) is obtained when we consider networks that permute—i.e., that give the same result independently of their order in the line. The equation

is equivalent to r/ρ = r₁/ρ₁ together with

Hence if each object of a series satisfies Eqs. (12.23), where A and B are constant, the objects permute with each other. It follows that each permutes with the whole series, and therefore the coefficients of the series satisfy Eqs. (12.23). This invariance of Eqs. (12.23) under star multiplication is an aid in the study of cascaded networks.

EXERCISES

1. Prove that if a lossless network takes a match into a match, then it takes any reflection into a reflection of the same magnitude. [If R(0) = 0 in Eq. (12.16), then r = 0. Hence |t| = 1, hence |τ| = 1, hence ρ = 0, and the result follows.]

2. Prove that if the linear fractional transformation w = R(z) has the distinct fixed points z₁ and z₂, then it can be written in the form

where H is a complex constant.

3. Suppose two transformations R₁(z) and R₂(z) of the type considered in Exercise 2 have the same fixed points but different constants H₁ and H₂. Show that the iterated transformation w = R₂{R₁(z)} has the same fixed points and the constant H₁H₂. Generalize to n iterations.

12.10    Series of Obstacles; the Cascade Problem

The scattering matrix for a series of obstacles can be obtained by a continued star product, by a continued matrix product, or by compounding the transformations

z_n+1 = P_n(z_n)

(The latter can be made to lead again to a matrix product.) If the obstacles are not adjacent, one regards the line lengths separating them as being themselves obstacles and computes the product accordingly. The scattering matrix for a line of length x is

where k and κ are the propagation constants in the two directions.

Dividing

shows that the quotient t/τ for two obstacles can be got by multiplying the individual quotients; hence, the analogous result is true for any number. [Since the determinant of the right-hand matrix (12.8) is t/τ, this remark also follows from the fact that the determinant of the matrix product is the product of the determinants.]

No other relations of the foregoing kind can be stated for the cascade problem in full generality, though statements can be given in important special cases. For example, since the star product of two lossless networks is lossless, the relations

|t|² + |r|² = |τ|² + |ρ|² = 1   |t| = |τ|   ∠t + ∠τ − ∠r − ∠ρ = π

are valid for the series if they are for each member of the series.

Another example is the commutativity relations (12.23). Let the over-all coefficients for a series of n obstacles be denoted by t_n, τ_n, r_n, and ρ_n. If Eqs. (12.23) hold for each obstacle, A and B being constant, we can express t_n, τ_n, and r_n in terms of ρ_n, namely,

and t_n/τ_n is known, since it is equal to the product Π(t_k/τ_k).

Sometimes it is desirable to find the limit of the transmission and reflection as the number of obstacles increases indefinitely. If there is enough dissipation to make t_nτ_n → 0, then Eqs. (12.25) give

where ξ₁ξ₂ = A and ξ₁ + ξ₂ = −2B. Thus, the ξ_i are the roots of

ξ² + 2Bξ + A = 0

That is, they are the fixed points (12.22) that were discussed previously.

EXERCISES

1. A network is bilateral (or reciprocal) if t = τ. Prove by induction or otherwise: If each object in a series is bilateral, so is the series as a whole. Is the analogous statement for reflection also true?

2. Prove that the reflection from a series of lossless objects is maximal when the objects are so spaced that the reflection from each adjacent pair is maximal. (See Sec. 12.5, Exercise 3. In this and the following problems, the objects are in a lossless line.)

3. Show that the standing-wave ratio for a series of lossless objects cannot exceed the product of the individual standing-wave ratios and that this value is always attained for some spacing. (See Sec. 12.5, Exercise 5.)

4. Show that, if the amplitude reflection for any array of lossless objects is equal to |R| with the objects in any given order, then for any other order there exists a spacing such that the reflection is again equal to |R|. (Consider the interchange of adjacent objects.)

5. Show that a series of lossless obstacles can be spaced for zero reflection if and only if the largest standing-wave ratio does not exceed the product of the others. (By Exercise 4, the largest can be put at the end. For zero reflection, the reflection of all but the last must be equal in magnitude to that of the last. Use Exercise 3, noting that the reflection is a continuous function of the spacing variables.)

12.11    Identical Networks in Cascade

Let the obstacle specified by t, τ, r, and ρ be repeated n times, so that the over-all scattering matrix is

Since identical obstacles permute, Eqs. (12.25) hold with A and B as in Eqs. (12.23), and also t_n/τ_n = (t/τ)ⁿ. All that is lacking is the expression for ρ_n, which can be obtained by the difference equation

or by a matrix product. Solution of Eq. (12.28) is facilitated by introducing the fixed points, and the nth power of a matrix can be obtained by writing it in diagonal form. †

The result of this calculation is

where δ is defined by

Expressions for r_n and t_n are found by interchanging r with ρ and t with τ.

As we shall presently see, a less pedestrian proof of Eqs. (12.29) can be based on the identity

[Equation (12.31) states that m + n obstacles can be regarded as a group of m, adjacent to a group of n.] Written out in full, Eq. (12.31) yields the difference equations

and their analogues for ρ_m+n and τ_m+n. For n = 1, the equations become recurrence formulas, which show that the four coefficients are determined uniquely by Eq. (12.31) together with the initial values

Thus, any method whatever of solving Eq. (12.31) subject to the conditions (12.32) must yield Eqs. (12.29), the desired result.

EXERCISES

1. Show that for lossless media, Eq. (12.30) becomes .

2. Let (t_nt_nr_nρ_n) be a bilateral network for which the limits

lim δ_nn = jp   lim (t_n − 1)n = jq

exist as n → ∞, with δ_n as in Eq. (12.30). This network is repeated n times to give

(T_nT_nR_nP_n) = (t_nt_nr_nρ_n)ⁿ

Show that the over-all transmission T_n satisfies

FUNCTIONAL EQUATIONS

12.12    Homogeneous Anisotropic Media

The dielectric medium shown in cross section in Fig. 12.15 is anisotropic, in the sense that it has different properties for waves traveling from right to left and from left to right. (This anisotropy should be distinguished from that in which the medium properties depend on polarization.) Since a medium of thickness x + y can be thought of as a medium of thickness x adjacent to a medium of thickness y, the transmission and reflection coefficients t(x), τ(x), r(x), and ρ(x) satisfy the equation

Fig. 12.15 Uniform dielectric medium of thickness x + y, in cross section.

Writing out in full gives the functional equations

and similarly for ρ(x + y) and τ(x + y). We assume the initial conditions

t(0) = τ(0) = 1   r(0) = ρ(0) = 0

and also existence of the derivatives

Subtracting r(x) from the first equation (12.34) and dividing by y yields

Upon letting y → 0 we get†

In just the same way the remaining equations lead to

where 2b = b₁ + b₂.

12.13    Solution of the Equations

Since the medium is homogeneous, the two matrices on the right of Eq. (12.33) can be permuted; hence, by Sec. 12.9,

Letting x → 0 shows that the constants are c/a and b/a, respectively, so that

Equations (12.37) and (12.38) give (log t)′ and (log τ)′, and by subtraction,

These three relationships among the four coefficients of our dielectric medium were obtained by quite general principles, which presuppose no knowledge of the solution. Of course one can actually solve Eq. (12.39), since the coefficients are constant; the result is

Equation (12.41) now gives

We get r from Eq. (12.40) and t and τ by combining Eq. (12.44) with Eq. (12.42):

EXERCISE

Show that a homogeneous isotropic dielectric sheet is in shunt (Sec. 12.3, Exercise 2) only if t = 1 and r = 0.

12.14    Application to the Cascade Problem

The expressions obtained in the foregoing discussion satisfy the functional equations (12.33). If m and n are integers and

t_n = t(n)   τ_n = τ(n)   r_n = r(n)   ρ_n = ρ(n)

the choice x = m and y = n makes Eq. (12.33) identical with Eq. (12.31). In other words, the expressions satisfy not only the functional equations for the anisotropic medium but also the difference equations for the iterated network.

The physical basis of the relationship is clear from Fig. 12.16. If a dielectric medium of unit thickness has the prescribed coefficients t, τ, r, and ρ, then the nth iterate t_n, τ_n, r_n, ρ_n corresponds to the same dielectric medium at a thickness n. This remark gives a new method of solving the cascade problem.

Fig. 12.16 Cross section of n identical dielectric sheets, stacked.

The desired initial conditions

t(1) = t   τ(1) = τ   r(1) = r   ρ(1) = ρ

lead to four equations in the four unknowns a, b₁, b₂, and c. If a new unknown X is defined by

Δ cosh Δ = X sinh Δ

then Eqs. (12.40), (12.41), and (12.43) with x = 1 give

where A and B are as in Eqs. (12.23). The condition b² − ac = Δ² reduces to

cosh² Δ = (tτ)⁻¹(1 + ρB)²

The coefficients r(1), ρ(1), and t(1) τ(1) now have the correct values, and t(1) = t, τ(1) = τ determine b_i through

Thus, the cascade problem is solved by Eq. (12.43) with these values of the parameters.

EXERCISES

1. Given any solution a, b₁, b₂, and c, show that all others can be found by replacing Δ, b₁, and b₂, respectively, by

±Δ + n₁πj   b₁ + n₂πj   and   b₂ − n₂πj

where n₁ and n₂ are integers.

2. Show that the particular solution for which Δ = δ in Eq. (12.30) gives

Thus obtain Eqs. (12.29) from Eq. (12.43).

12.15    Interpretation of the Constants

Although Eqs. (12.35) give the mathematical meaning of the four parameters a, b₁, b₂, c, their physical significance requires clarification. Let the dielectric medium of Fig. 12.17 have the propagation constant k for waves traveling from left to right and κ for the opposite direction. The interface reflections at the left are r₁ and ρ₁, and those at the right are r₂ and ρ₂. Continuity of the E field gives the corresponding transmissions,

t_i = 1 + r_i   τ_i = 1 + ρ_i

Fig. 12.17 Interface reflections and propagation constants for homogeneous anisotropic dielectric sheet.

The multiple-reflection argument of Sec. 12.3, or Eq. (12.3) with the matrix (12.24), leads to

with corresponding expressions for τ(x) and ρ(x).

Since t(0) = τ(0) = 1 and r(0) = ρ(0) = 0, the constants r_i and ρ_i are not independent, but satisfy the equations

Thus, as before, our problem is characterized by four complex constants: k, κ, r₁, and ρ₂.

Differentiating and setting x = 0 yields

with similar expressions for b₂ and c. As a check, we note the relationships

which agree with the general principles set forth previously.

The fact that solutions of the functional equations always involve just four complex constants is not surprising. From the point of view of Maxwell’s equations, the behavior is characterized by ⁺, ⁻, μ⁺, μ⁻ the dielectric constants and permeabilities in the two directions. In terms of the free-space wavelength λ, we have

The quantities

determine r₁ = (1 − Z)/(1 + Z) and ρ₂, with the result

Also Eqs. (12.46) hold, as they should. For isotropic media, we have

With these results it is easy to show (if we had not known it before) that the method of functional equations is consistent with the methods of field theory.

It should be mentioned, in conclusion, that t = τ leads to k = κ, but r₁ = ρ₂ only if r = ρ. Thus the representation even of bilateral lossless networks by the process of Sec. 12.14 requires an anisotropic medium.

EXERCISES

1. In books on electromagnetic theory it is shown that the formulas governing behavior of a dielectric sheet at normal incidence can be used at arbitrary incidence θ provided the propagation constant is taken as

and provided the quantity Z determining the interface reflection is replaced by

at perpendicular and parallel polarization, respectively.

Show that the transmission t of an isotropic, homogeneous dielectric sheet at incidence θ and thickness x₀ is given by

2. For any dielectric sheet, the derivatives r′(0), ρ′(0), … depend on the behavior as the thickness approaches zero, while for lossy sheets the interface reflections r₁, ρ₁, … depend on the behavior as the thickness approaches infinity. Prove that these quantities are related by the formulas

[Apply L’Hospital’s rule to Eqs. (12.48).]

3. A homogeneous isotropic dielectric sheet with thickness x and propagation constant k is preceded by a metal plate having right-hand reflection K = −1 (see Fig. 12.22). Show that the over-all right-hand reflection can be written in the form

where ξ is constant. [At normal incidence, coth² ξ = (/₀)/(μ/μ₀).]

4. Show that if ξ and k are real in Exercise 3, then |ρ| = 1 and

tan ∠ρ = 2 tan kx tanh ξ

12.16    Nonuniform Dielectric Media

Let the quantities ⁺, ⁻, μ⁺, μ⁻ of Sec. 12.15 be arbitrary functions of the coordinate s perpendicular to the interfaces. We shall regard these parameters as being given for − ∞ < s < ∞. The dielectric slab is obtained by taking that part of the infinite medium that extends from† x to y > x, as shown in Fig. 12.18. In other words, ⁺(s), μ⁺(s), ⁻(s), and μ⁻(s) are used for x < s < y, but ₀ and μ₀ elsewhere. If the transmission and reflection coefficients are denoted by

t(x,y)   τ(x,y)   r(x,y)   ρ(x,y)

Fig. 12.18 Cross section of dielectric sheet obtained by taking a slice from an infinite inhomogeneous anisotropic medium.

then the same procedure as that employed previously yields

for z > y > x. Writing out in full gives four functional equations, of which

is typical. The initial conditions are

t(x,x) = τ(x,x) = 1r(x,x) = ρ(x,x) = 0

and it is assumed that the partial derivatives exist for y = x:

In terms of ⁺(x), ⁻(x), μ⁺(x), and μ⁻(x), these quantities have the values assigned in Sec. 12.15.

Subtracting r(x,y) from both sides of Eq. (12.52), dividing by z − y, and letting z → y, we obtain

just as in the proof of Eqs. (12.36). Similarly,

where 2b = b₁ + b₂. In these equations, t, τ, r, and ρ are evaluated at (x,y), and a, b₁, b₂, and c at y.

The system has the same structure as Eqs. (12.36) to (12.39) except that the coefficients are variable. It might be expected that t, τ, and r can be expressed in terms of ρ, just as before, and we shall see that this is the case. Equations (12.55) and (12.56) give

and these, together with Eq. (12.54), yield the desired formulas:

EXERCISES

1. If the medium is isotropic, prove that the coefficients satisfy t = τ and

2. Let t_n, r_n, and ρ_n be the transmission and reflection coefficients obtained for an isotropic inhomogeneous dielectric sheet of thickness x₀ when the free-space wavelength λ is replaced by nλ (Fig. 12.19). If a(x) and b(x) are given in terms of (x) and μ(x) by Eqs. (12.50), show that, as n → ∞,

apart from terms of order 1/n³. [If a(x) and b(x) correspond to λ, note that a(x)/n and b(x)/n correspond to nλ. By the method of Sec. 12.15, Exercise 1, the result remains valid for arbitrary incidence θ.]

3. Two lossless, homogeneous, isotropic media of dielectric constants ₁ and ₃ and permeabilities μ₀ are separated by a similar medium of constants ₂ and μ₀. Show that, for zero over-all reflection, ₂² = ₁₃. Show further that the electrical thickness of the central medium must be a quarter wave.

4. In Exercise 3, let the media 1 and 3 have the constant separation d and let the intervening medium consist of many matched quarter-wave steps of the type there described. Show that, as λ → 0, the limiting behavior of the central medium is such that (1/)^½ is a linear function of position x, agreeing with (1/₁)^½ and (1/₃)^½ at x = 0 and x = d, respectively. (Note that the layers have different thicknesses.)

5. Express the reflection of the transition step of Exercise 4 in closed form as a function of λ.

Fig. 12.19 An inhomogeneous, lossless, isotropic sheet in cross section.

12.17    Linearization

The nonlinear equation (12.57) can be linearized by the familiar substitution ρ = −q_y/(cq). But a linearization more tightly knit with the underlying physics is determined by the matrix

of the correspondence (12.8). If

then Eqs. (12.54) to (12.57) are found to be equivalent to

The initial condition is M(x,x) = I, the identity matrix.

If M and C were scalars rather than matrices, we should obtain

by inspection of Eq. (12.58). The latter would give

which, by the isomorphism (12.8), is equivalent to Eq. (12.51).

Although the foregoing considerations can be justified even when C(s) is a matrix, it is preferable to give a direct proof. We have

upon noting that (M⁻¹)_z = −M⁻¹M_zM⁻¹ and using Eq. (12.58). Since the expression has the value M(x,y) when z = y, Eq. (12.60) holds. What this shows is that the functional equations follow from the differential system.

EXERCISES

1. For an isotropic homogeneous dielectric sheet, the interface reflections satisfy

(Sec. 12.15). If χ is defined by r₁ = tan ½χ, show that

2. In Exercise 1, let X = kx be the electrical thickness of the sheet. Show that

where d = t² − r².

3. A medium, lossless but anisotropic with respect to polarization, is specified in the notation of Exercise 2 by X₁ and χ₁ at a certain polarization and by X₂ and χ₂ when the polarization is rotated through 90°. To construct a quarter-wave plate giving circular polarization, it is desired to make

Show that this condition reduces to

cos χ₁ cos χ₂ + tan X₁ tan X₂ = 0

Show also that a measure of the transmitting efficiency is

|t₁|⁻² + |t₂|⁻² = 2 + sin² X₁ tan² χ₁ + sin² X₂ tan² χ₂

12.18    Conditions for a Passive Solution

We inquire next: What conditions on a, b₁, b₂, and c ensure that the dielectric medium represents a passive network for y ≥ x? The answer to this question can be found by physical considerations, as follows. Regard the medium as composed of many thin layers. If each layer is dissipative, Sec. 12.6 shows that the star product is also dissipative, and the latter gives the scattering matrix for the whole medium.

When y − x = Δy is small, the differential system (12.54) to (12.57) together with the initial conditions yields

where o(Δy) denotes an expression such that o(Δy)/Δy → 0 as Δy → 0. For the matrix (12.61) to satisfy the criterion (12.12) as Δy → 0 through positive values, it is necessary that

Since z₁ and z₄ are unrestricted, the condition reduces to

For the lossless case, the three inequalities are replaced by equalities.

When a, b₁, b₂, and c are piecewise continuous on a given interval J, it is not difficult to put the foregoing considerations into proper mathematical form.† The inequalities (12.62) are necessary and sufficient conditions that the scattering matrix be dissipative for x in J, y in J, and y > x. This is an example, par excellence, of a theorem in which the mathematician’s task is facilitated by a modicum of physical insight.

EXERCISES

1. Prove that an isotropic dielectric medium is lossless if and only if (s) and μ(s) are real, −∞ < s < ∞.

2. Let r₁, r₀, t₁, t₀, ρ₁, and ρ₀ be six real functions of the real variable y for y ≥ 0, satisfying the following conditions:

   a. and are piecewise continuous.

   b. 1 − r₁² = t₁² = 1 − ρ₁² ≠ 0, and the common value is continuous.

   c. 2t₀ − r₀ − ρ₀ = (2n + 1)π for an integer n.

       d. .

Show that these functions can be realized as coefficients

of a lossless dielectric medium having piecewise continuous (y) and μ(y). [With

it will be found that the desired differential equations hold. The medium is lossless by Exercise 1.]

3. Prove conversely that conditions a to d of Exercise 2 must be satisfied if r₁, r₀, … can be realized as coefficients of a dielectric medium of the type described. (Use the results of Sec. 12.5. It is necessary to give a suitable interpretation to the phases r₀ and ρ₀ when r₁ or ρ₁ is zero.)

4. A lossless, isotropic, dielectric medium (Fig. 12.19) is contracted by a factor n; then n of these contracted media are stacked together as shown in Fig. 12.20. If T_n is the transmission coefficient in Fig. 12.20, find t = lim T_n as n → ∞. (The limits needed in Sec. 12.11, Exercise 2, are given by the result of Sec. 12.16, Exercise 2. See also Sec. 12.11, Exercise 1.)

5. Let the symbol [ ] denote “mean value of,” so that

Fig. 12.20 Sheet obtained by contracting and stacking the sheets shown in Fig. 12.19.

Show that the limiting medium considered in Exercise 4 is equivalent† to a homogeneous medium with dielectric constant, permeability, and thickness

at perpendicular polarization, while for parallel polarization the respective values are

(Extend the result of Exercise 4 to arbitrary θ as in Sec. 12.15, Exercise 1, and compare your expression for t⁻¹ with the one given there. Equality of the reflections follows from Fig. 12.9.)

6. In Exercise 5 let (x) be continuous and not constant, and let μ ≡ μ₀. Show that the equivalent permeability μ_e for the limit medium satisfies μ_e < μ₀ at parallel polarization. (Use the Schwarz inequality.)

12.19    Probability: A Reinterpretation

A particle moves at random on a line L in such a way that it can be transmitted, reflected, or absorbed when it encounters a given segment J of L. If x and y, with y > x, give the coordinates of the ends of J, the probabilities of transmission and reflection are denoted by t(x,y) and r(x,y) when the particle is incident on J from the left, and by τ(x,y) and ρ(x,y) when the particle is incident from the right. (For purposes of this definition, the segment J is thought to be withdrawn from the rest of the line.) When the line is homogeneous, e.g., has a uniform mass distribution on it, then the probabilities are functions of the length y − x only. Use of two variables allows for inhomogeneity, as in the foregoing discussion of dielectric media.

The probabilities associated with a given segment are assumed independent of all information about adjacent segments. With this understanding, one can get functional equations as follows: Consider the interval (x,y) and an adjacent interval (y,z), z > y > x. The particle can pass through the interval (x,z) in the mutually exclusive ways suggested by Fig. 12.21; namely, it can pass with no internal reflections, with two, with four, and so on. The assumed independence yields

Fig. 12.21 Modes of passage through two adjacent segments.

for the probabilities of these various events, and the sum represents the probability of getting through the segment (x,z). Thus we are led to the functional equation

Analogous considerations give τ(x,z), r(x,z), and ρ(x,z). The resulting functional equations are identical with those of transmission-line theory; in other words, Eq. (12.51) holds.

Because of this equivalence, much of the foregoing analysis applies without change to the probability problem. The differential system (12.54) to (12.57) is valid, as are the results of Sec. 12.13 for uniform media. Interpreting the constants as in Sec. 12.15 enables us even to assign a “dielectric constant” and “permeability.” It is only in conditions for the dissipative or passive character of the process that the difference in the two physical models becomes mathematically significant. Here these conditions result from the fact that our probabilities cannot exceed 1, whereas for the transmission line they result from conservation of energy. Analysis of the inequalities is facilitated by introducing the scattering matrix, which we proceed to do.

12.20    The Scattering Matrix

Let the segment (x,y) = J be withdrawn from the line L, and let the particle be presented to J in a random manner. Specifically, z₁ represents the probability that the particle is tossed at J from the left, and z₄ that it is tossed from the right (see Fig. 12.5). In these circumstances, the molecule can emerge from the right in two mutually exclusive ways: It can be presented from the left and then transmitted, or from the right and then reflected. If z₃ is the corresponding probability, we have

z₃ = tz₁ + ρz₄

upon writing t for t(x,y) and ρ for ρ(x,y). A similar equation gives z₂, the probability of emerging at the left. Together the equations lead to Eq. (12.4), which is repeated for convenience:

The role of the probabilities z₁ and z₄ is now clear: They give a suitable vector for our scattering matrix to scatter.

The conditions of physical realizability can be presented from two viewpoints. On the one hand, we have

because of the probability interpretation. On the other hand, if z₁ ≥ 0 and z₄ ≥ 0 in Eq. (12.63) then we expect

It is readily seen that the inequalities (12.65) and (12.64) are equivalent. When they hold, we say that the matrix is p dissipative (p for probability). The matrix is p lossless if the two inequalities on the right of (12.64), and the one on the right of (12.65), are replaced by equalities. There is then zero probability of absorption.

The criterion (12.65) shows at once that the star product of two matrices is p dissipative if each factor is; we simply add the inequalities

z₂ + z₃ ≤ z₁ + z₄andz₅ + z₄ ≤ z₃ + z₆

referring to Fig. 12.6. This fact makes it possible to pass from relationships in the small to those in the large, just as in the transmission-line case. Upon combining Eq. (12.61) with the inequalities (12.64), we get

as necessary and sufficient conditions on the coefficients of the differential system for the solution matrix to be p dissipative in the region x in L, y in L, y > x. Replacing the last two inequalities in (12.66) by equalities makes the matrix be p lossless.†

It appears, then, that the sole difference between one-dimensional wave propagation and one-dimensional diffusion lies in the constraints that the coefficients must satisfy. In the former case, they are complex and satisfy the inequalities (12.62), whereas in the latter they are real and satisfy the inequalities (12.66).

EXERCISE

Show that, in the lossless isotropic case,

12.21    The Underlying Closure Principle

Let ρ(x,y,K) be the solution of

ρ_y = a + 2bρ + cρ²

that equals K when y = x. Figure 12.22 and Eq. (12.17) suggest that

where ρ, t, τ, and r are given by Eqs. (12.54) to (12.57); the result is verified by direct substitution. Here we have a concrete physical interpretation of the familiar fact that all solutions of Riccati’s equation can be expressed in terms of a single solution by quadrature.

Let ρ(x,y,K) have the value K₁ at (x₀,y₁), so that

ρ(x₀,y,K) = K₁   at y = y₁

Since ρ(x,x,K) ≡ K we have also

as shown in Fig. 12.23; hence, by uniqueness,

ρ(x₀,y,K) ≡ ρ(y₁,y,K₁)

The result becomes

when Eq. (12.68) is used and the variables are renamed.

Fig. 12.22 Dielectric sheet backed by a termination of arbitrary reflection K.

Depending on one’s preference, Eq. (12.69) is a uniqueness theorem, a semigroup property (see Chap. 4), or “the major premise of Huygens’ principle according to Hadamard.” It has to do with the consistency of our differential equation as a description of the underlying process, and it forms the starting point for the Lie-group theory of that equation. On the other hand, the result is also meaningful without reference to differential equations. By the foregoing proof, Eq. (12.69) holds for any function ρ(x,y,K), provided ρ(x,x,K) ≡ K and provided the curves traced out by ρ(x,y,K), as y varies, form a field in a suitable region. (A family of curves forms a field in a given region if there is one and only one curve of the family through each point of the region.)

Fig. 12.23 Trajectory traced out by the complex solution of a first-order differential equation.

If Eq. (12.69) is written in terms of Eq. (12.67), the choice K = 0 yields

while K = 1/r(x,z) gives

The choice K = 1/r(x,y), together with the two preceding equations, leads to

where T(x,y) = [t(x,y)τ(x,y)]^½. Conversely, these three relationships ensure the validity of Eq. (12.69).

If t = τ, the foregoing equations are identical with the functional equations (12.51). Since these have formed the basis of our whole discussion, a substantial part of transmission-line theory appears to be contained in the general closure property (12.69). Here we have a unifying principle behind the diversity of special methods.

To see what becomes of this principle in the nonbilateral case, let a function h = h(x,y) be defined by

T(x,y) = h(x,y)t(x,y)

Then ht = T and h⁻¹τ = T both satisfy Eq. (12.72), and hence

satisfies Eq. (12.51). In other words, there exists a function h such that the desired equations hold for ht, h⁻¹τ, r, and ρ, though they may not hold for t, τ, r, and ρ.

Nothing more can be deduced from the Riccati equation alone. But if, in addition, Eqs. (12.54) to (12.57) hold, a brief calculation shows that h is constant and hence, by the initial conditions, h = 1. Thus the functional equations express the closure property of Eq. (12.57) in the bilateral case, and the closure property of the system (12.54) to (12.57) in the general case.

REVIEW EXERCISES

1. An isotropic inhomogeneous dielectric slab specified by (x)/₀ ≡ and μ(x)/μ₀ ≡ m extends from the plane x = 0 to the plane x = x. If the reflection at incidence θ is ρ = ρ(x), define

w = (1 + ρ)(1 − ρ)⁻¹   Y_⊥ = w sec θ   Y_|| = w cos θ

where the subscript specifies polarization perpendicular or parallel to the plane of incidence, respectively. Prove that

where λ is the free-space wavelength. (See Sec. 12.15, Exercise 1.)

2. The reflection at near-grazing incidence is important in the study of radio-wave propagation over distances that are small compared with the earth’s radius. In Exercise 1, let α and β be those solutions of the Riccati equations

that satisfy α(0) = β(0) = 0. Prove that

3. A microwave absorber consists of a thin layer of material having parameters (x) and μ(x), backed by a metal plate at x = 0 (see Fig. 12.22). Suppose |(x)μ(x)|/(₀μ₀) is so large that the quantity m − sin² θ in Exercise 1 can be replaced, with negligible error, by m − ½ as θ varies. Show that the absorber satisfies

w_⊥ sec θ = constw_|| cos θ = const

as θ varies, the thickness x being fixed. (The result follows at once from the uniqueness theorem for first-order equations.)

4. In Exercise 3, the performance of the absorber, reflection vs. θ, is given by the explicit formula

where T = tan² (θ/2) and where ρ_⊥ = Ae^iB at θ = 0. The result for ρ|| is obtained by adding π to B. Prove these results.

5. Show that if an absorber of the type considered in Exercise 3 has zero reflection at a given angle θ and polarization, then the reflection at the angle θ for the other polarization is

6. An absorber of the type considered in Exercise 3 is designed for use at both polarizations simultaneously. Show that the performance is optimum when r(0) is pure imaginary, in which case the reflection is independent of polarization. Show that the optimum performance possible is

TRANSMISSION AND REFLECTION OPERATORS

12.22    Transmission, Reflection, and Scattering Matrices

Let the obstacle be supplied with 2n terminal pairs, of which n are thought to be at the left and n more at the right. At the high frequencies for which the analysis is especially suited, these terminal pairs are presumably waveguides, but only one mode is considered relevant.† Also, without real loss of generality we can, and do, assume that the wave-guides all have the same characteristic impedance.

Fig. 12.24 Obstacle with 2n terminal pairs.

If a wave of complex amplitude zⁱ is incident on the ith guide at the left, as in Fig. 12.24, let

t_ijzⁱandr_ijzⁱ

represent the amplitudes of the transmitted waves and reflected waves at the jth guide on the right or left, respectively. When the amplitudes z¹, z², …, zⁿ are simultaneously present, the input is represented by a column vector z having components zⁱ. Superposition gives

for the transmitted and reflected vectors, respectively.

In terms of the transposed matrices t = (t_ji) and r = (r_ji), the vectors are tz and rz, respectively. Similarly, τz and ρz represent the vectors transmitted or reflected when the vector z is incident from the right.

If the vector z₁ is incident from the left and simultaneously z₄ from the right, as in Fig. 12.5, then the vector z₃ emerging from the right can be computed by taking the part of z₁ that is transmitted together with the part of z₄ that is reflected. The calculation is done at first for the jth terminal, but the single matrix equation

describes the result for each value of j. Upon computing z₂, the wave reflected at the left, we get

just as in Eq. (12.4).

The n by n matrices t, τ, r, and ρ play the same algebraic role as the complex numbers t, τ, r, and ρ used previously. Hence, t and τ are called the transmission coefficients and r and ρ the reflection coefficients. Here, however, these quantities are matrices, i.e., operators on the space of n-dimensional complex vectors. The 2n by 2n matrix S in Eq. (12.74) is the scattering matrix, as in the case n = 1. In order to use the previous formulas, we denote the n by n identity matrix by 1.

12.23    The Star Product and Closure

Distinguishing n terminal pairs as input and n as output enables us to combine obstacles (and that is the reason for making the distinction). If the obstacle of the foregoing discussion is adjacent to a second one, the over-all coefficients can be computed in principle by multiple reflection. But it is much simpler to use the alternative derivation, the one leading to Eq. (12.6) by way of Fig. 12.6. We get the same formula (12.3) in the matrix case; in fact, the order of factors in Eq. (12.3) was chosen to obviate rewriting the result for n > 1. The isomorphism (12.8) is also valid, and proved in the same way. Hence, star multiplication is associative.

So much for the algebra of star products. In the case n = 1, we found that there was also an associated calculus; namely, the functional equation (12.51) turned out to be the closure or semigroup property for a certain differential system. The analogous result is true for arbitrary n, as will be seen next. For precision of statement, we say that a region R of the xy plane is admissible if it is open, connected, and such that the closed line segment joining (x,x) to (x,y) is in R whenever (x,y) is in R.

THEOREM 12.1.   Let t, τ, r, and ρ be complex n by n matrix functions of the two real variables x and y, with tτ nonsingular, and set

If S(x,x) equals the identity matrix, if S_y(x,x) exists, and if

in an admissible region R, then there are complex n by n matrix functions a(y), b₁(y), b₂(y), and c(y) such that the entries of S satisfy

throughout R. Conversely, if t, τ, r, ρ is any solution of this differential system, with tτ nonsingular, then Eq. (12.75) holds.

It should be noticed that no continuity hypothesis on the coefficients is needed. In Theorems 12.3 and 12.4, below, which assume continuity, the fact that tτ is nonsingular can be deduced from the differential system.

Derivation of the differential system is similar to that in the case n = 1. To proceed in the opposite direction, consider the equivalent linear system M_y = MC, where

The matrix M is nonsingular since

Hence, M(x,z) = M(x,y)M(y,z) follows from

as in Sec. 12.17, and Eq. (12.75) follows from this.

If K is a complex n by n matrix such that the expression ρ(x,y,K) of Eq. (12.67) is well defined, then the latter satisfies the equations

It is possible to discuss functional equations by means of the closure principle (12.69), which is proved for matrices just as for scalars. The result of an elementary but long calculation is that Eq. (12.69) holds if and only if there is a scalar function h such that

satisfies the relationship S_h(x,z) = S_h(x,y) * S_h(y,z). The system (12.76) implies h = 1, giving another proof of Eq. (12.75). At the same time, we see to what extent the closure principle for the whole system is more demanding than that for Eqs. (12.77) alone.

EXERCISES

1. The identity I is a matrix such that S * I = I * S = S for all 2n by 2n scattering matrices S. The inverse S⁻¹ satisfies S⁻¹ * S = S * S⁻¹ = I. Show that the identity and inverse under star multiplication are the same as those under matrix multiplication. [Set z₅ = z₁ and z₆ = z₂ in Eqs. (12.4) to (12.6). In this and the following problems the results are to be established for arbitrary n.]

2. The network (tτrρ) has left-hand reflection R when followed by an open circuit r₁ = 1. The network (tτrρ) * (τtρr) has left-hand reflection R₀ and transmission T₀. Show that

R = R₀ + T₀

and interpret by the method of images (Fig. 12.25). How would the result change if (tτrρ) were backed by a short circuit, r₁ = −1?

3. Show that the solutions of Eqs. (12.76) satisfy the equations

and interpret physically. (Refer the proof to the linear system. The minus sign is needed because the “slab thickness” increases with increasing y but decreases with increasing x.)

Fig. 12.25 Obstacle backed by a perfect reflector, and its mirror image.

4. Prove that in Sec. 12.7 we have R(z₁) = z₂⁻¹ if and only if P(z₂) = z₁⁻¹. (Assume that the needed inverses exist.)

5. Show that the star product of two symmetric matrices is symmetric.

12.24    The Norm and Energy Transfer

If A = (A₁, A₂, … , A_n) is a complex row vector, its length is defined by

By the Schwarz inequality, we have

where z ranges over all complex column vectors (z¹, z², … , zⁿ) of unit length.

The merit of this alternative description is that the latter applies to any matrix A, provided A has n columns. The resulting measure is called the norm of A. Since |A| reduces to the ordinary absolute value when A has a single element, or to the length of A when A is a vector, there is no need to introduce a new symbol.

If A* denotes the adjoint of A, then

Hence |A|² is given by the Rayleigh quotient,

If A and B are self-adjoint matrices, the condition A ≤ B means

z*Az ≤ z*Bzfor all z

It can be shown that this relationship has many of the properties of ordinary inequality. By Eq. (12.80), the condition |A| ≤ 1 is equivalent to A*A ≤ 1. If A* A = 1, that is, if A is unitary, then Eq. (12.79) gives |Az| = |z| for all z.

A network specified by the four matrices t, τ, r, and ρ is dissipative if and only if the scattering matrix S of (12.74) is such that

|z₃|² + |z₂|² ≤ |z₁|² + |z₄|²

for all vectors z₁ and z₄. In terms of the foregoing notation, the condition is simply |S| ≤ 1. When equality holds—i.e., when S is unitary—the network is lossless.

As in Sec. 12.8, the network is reflectively dissipative from the left if the matrix

satisfies |R(K)| < 1 whenever |K| < 1. (Here K is an n by n complex matrix, not a vector.) The corresponding condition from the right involves

If, in addition, R(K) is unitary for all unitary K, the network is reflectively lossless from the left.

EXERCISES

1. Prove that if a unitary scattering matrix (tτrρ) has |r| < 1, then also |ρ| < 1.

2. Show that when |S₁| ≤ 1 and |S₂| ≤ 1, then |S₁ * S₂| ≤ max (|S₁|, |S₂|). (In this and the following exercises assume that the relevant star products exist.)

3. If two matrices are unitary, show that their star product also is unitary.

4. If S is unitary, show that the corresponding bilinear function R(K) is unitary for all unitary K. [Consider (tτrρ) * (00KK) and use the result of Exercise 3.]

12.25    The Matching Problem

As the reader will recall, for n = 1 a lossless network with right-hand reflection gives zero left-hand reflection when followed by a reflection r. The lossless network is called a tuner, and the new network, tuner plus load, is said to be matched from the left.

Construction of a lossless tuner for arbitrary n hinges on the following fact of matrix theory: Given any n by n complex matrix K, with |K| ≤ 1, there is an n by n matrix X such that

If |K| < 1, then X is nonsingular. The proper choice of K for match is K = r*, as is shown by a brief calculation. In other words, if |r| < 1 then

The problem of matching a network simultaneously from both sides is more subtle. We try to choose reflections r₀ and ρ₀ with corresponding unitary scattering matrices such that

By Eq. (12.83), the attempt is successful if simultaneously

with R and P as in Eqs. (12.81) and (12.82). Now, the Brouwer fixed-point theorem guarantees existence of a solution (r₀, ρ₀) of Eq. (12.85), provided P and R map the unit operator sphere into itself, i.e., provided

|R(K)| < 1and|P(K)| < 1

for all |K| < 1. The latter, in turn, is precisely the condition for the network to be reflectively dissipative.†

EXERCISES

1. A scattering matrix S = (tτrρ) with |ρ| < 1 is matched from the right by each of two lossless networks, T₁ and T₂. Show that there are unitary matrices t₀ and τ₀ such that

T₁ = T₂ * (t₀τ₀00)

For n = 1, interpret the result as stating that a tuner is unique, except for a length of anisotropic lossless line. (Let S * T₁ = S₁ and S * T₂ = S₂ both have right-hand reflections ρ = 0. The matrix T₀ = T₂⁻¹ * T₁ is unitary by Exercise 3 of Sec. 12.24, and it satisfies S₂ * T₀ = S₁.)

2. A network (tτrρ) is lossless from the left provided R(K) is unitary for all unitary K. Assuming tτ nonsingular, show that (tτrρ) is lossless from the left if and only if (hth⁻¹τrρ) is lossless for some positive scalar h. [To investigate the converse of Exercise 4, Sec. 12.24, match (tτrρ) from the right. A matrix that commutes with every unitary matrix is necessarily scalar.]

3. Show that when tτ is nonsingular, the network (tτrρ) is lossless from the left if and only if it is from the right. (Use Exercise 2.)

4. Let (tτrρ) be a lossless network with tτ nonsingular. If K runs through the set of all unitary matrices, show that R(K) does also. [Let K₀ be unitary. With h as given by Exercise 2, solve the equation

(hth⁻¹τrρ) * (00KK₁) = (00K₀K₀)

for (00KK₁). The fact that K is unitary follows from Exercise 3, Sec. 12.24.]

5. Let S = (tτrρ) be a scattering matrix with |ρ| < 1. Show that the operators

I₁ = t*(1 − ρρ*)⁻¹t   I₂ = τ(1 − ρ*ρ)⁻¹τ*   I₃ = r + τρ*(1 − ρρ*)⁻¹t

are energy invariants; i.e., they are unchanged if S is replaced by S * U, where U is any lossless network with nonsingular transmissions. (Establish the invariance when U is a tuner T, and also when ρ = 0. The desired result then follows from S * U = S * T * T⁻¹ * U = S₁ * U₁.)

6. The operators I in Exercise 5 are the sole energy invariants in the following sense: If f(t,τ,r,ρ) is such an invariant, then

f(t,τ,r,ρ) = f₁(I₁,I₂,I₃)

for some function f₁. [Let S * T = (t₁τ₁r₁0), where T is a tuner. If the polar decompositions for t₁ and τ₁ are t₁ = u₁h₁, τ₁ = h₂u₂, consider S * T * (u₁*u₂*00).]

7. For n = 1 obtain invariants by inspection of Eqs. (12.18). Show that there is the additional invariant ∠t − ∠τ if the lossless network U is required to be bilateral. (See Sec. 12.10. The geometric interpretation can be extended to arbitrary n by use of Exercise 4.)

12.26    Further Discussion of Passive Networks

If a dissipative obstacle has tτ nonsingular, it is easily shown that |ρ| < 1 and |r| < 1, so that the star product of two such obstacles is well defined. Indeed,

|ρr₁| ≤ |ρ| |r₁| < 1

so that

(1 − ρr₁)z ≠ 0   for z ≠ 0

Therefore, 1 − ρr₁ is nonsingular. Proceeding as in the case n = 1 (Sec. 12.6), we deduce that if two networks are dissipative and have nonsingular transmissions, the same is true of their star product. The analogous statement for reflective dissipativity from the right or left is also true. These considerations lead to the following result:

THEOREM 12.2.   Let t, τ, r, and ρ be complex n by n matrices with tτ nonsingular. Then the following statements a, b, and c are equivalent:

a. |r + τK(1 − ρK)⁻¹t| < 1 whenever |K| < 1

b. |ρ + tK(1 − rK)⁻¹τ| < 1 whenever |K| < 1

c. There is a positive scalar h such that

Also, if the latter matrix is unitary, then the former are unitary for all unitary K, and conversely.

In other words, if the network is reflectively dissipative (or lossless) from the left, then it is also from the right, but it need not be dissipative or lossless in the general sense, unless h in statement c happens to be 1.

A proof of Theorem 12.2 can be based on the theory of quadratic forms.† Here, however, we sketch an argument more closely related to the engineering situation.

To show that statement c implies statements a and b, choose X so that the matrix containing it is unitary and consider

[One gets P(K) by Eq. (12.3), since the scalar h commutes.] Being the product of two dissipative matrices, the matrix containing P(K) is dissipative; hence, |P(K)| < 1 by consideration of

Thus statement c implies b; similarly, c implies a.

We show next that statement b implies a. The proof depends on the fact that a condition like b gives not only the obvious inequality |ρ| < 1 got by setting K = 0, but also (provided tτ is nonsingular) |r| < 1. With this in mind, observe that the product

is reflectively dissipative from the right, since each factor is. Hence, by the foregoing remark, the left-hand reflection for the product cannot have norm as large as 1. This gives the desired condition, |R(K)| < 1.

It remains to show that b implies c. Since a holds when b does, we can use the bilateral matching of Eq. (12.84). The resulting matrix is reflectively dissipative both from the left and from the right, since each factor is. By computing the associated functions P₁(K) and R₁(K), we see that

|t₁Kτ₁| < 1whenever |K| < 1

Because K is arbitrary, the latter is equivalent to |t₁| |τ₁| ≤ 1.

Since t₁ and τ₁ are nonsingular, their norms are not zero. If

we have

h|t₁| = h⁻¹|τ₁|

Each quantity is ≤1, since the product is; i.e., we have

Now, it is easily shown that the inverse under star multiplication is the same as that under matrix multiplication. In particular, the inverse of a unitary matrix is its adjoint. By solving Eq. (12.84) for the (tτrρ) matrix and writing ht for t, and h⁻¹τ for τ, we get

By the inequalities (12.86), the matrix involving t₁ and τ₁ is dissipative; therefore, the whole product is dissipative. This completes the proof.

EXERCISES

1. Suppose we say that (tτrρ) is reflectively dissipative from the left if |R(K)| ≤ 1 whenever |K| ≤ 1 and 1 − ρK is nonsingular. Show that this definition is equivalent to that of the text provided tτ is nonsingular.

2. (In this and the following problems, “reflectively dissipative” has the meaning given in Exercise 1.) Show that (tτrρ) is reflectively dissipative from the left if

|z₂| ≤ |ρz₂ + tz₁|implies|z₁| ≤ |τz₂ + rz₁|

for all complex n vectors z_i. Show further that the converse is true provided |ρ| ≤ 1 and t ≠ 0.

3. Let t ≠ 0, τ ≠ 0, |r| ≤ 1, |ρ| ≤ 1. Show that (tτrρ) is reflectively dissipative from the left if and only if it is from the right. (Use Exercise 2.)

4. If there is a positive scalar h such that

show that (tτrρ) is reflectively dissipative from the left. (Use Exercise 2. Note that tτ is not assumed nonsingular.)

5. In Exercise 4, prove that the converse holds provided t ≠ 0, τ ≠ 0, and |ρ| ≤ 1. (Use Exercise 2 together with the theory of pencils of quadratic forms.)

12.27    Inequalities for the Differential System

If Δy = y − x is small, the system (12.76) gives the estimate (12.61) for the scattering matrix S(x,y). The condition |S| ≤ 1 is equivalent to SS* ≤ 1. Since Δy > 0, and since (Δy)² is negligible compared with Δy, we get

as the desired inequality. Such is the condition for the matrix-propagating medium to be passive in the small.

Since the algebra of the star product for arbitrary n is the same as that for n = 1, passing to relationships in the large causes no more trouble now than it did in Sec. 12.18. Without further ado we state the following result:

THEOREM 12.3.   Let the hypothesis of Theorem 12.1 hold for x in J and y in J, where J is a given interval. Suppose also that the matrix coefficients a(y), b₁(y), b₂(y), and c(y) are piecewise continuous. Then

|S(x,y)| ≤ 1for x in J, y in J, and y > x

if and only if the inequality (12.87) holds for y in J. In case equality holds in (12.87) for all y, then S(x,y) is unitary, and conversely.

The analogous criterion for reflectively dissipative networks is based on part c of Theorem 12.2. It leads to the condition

where H is an arbitrary real scalar matrix. The inequality (12.88) is necessary and sufficient for the function ρ(x,y,K) of Eq. (12.67) to satisfy

|ρ(x,y,K)| < 1

whenever |K| < 1, x is in J, y is in J, and y > x.

REVIEW EXERCISES

1. If a one-to-one correspondence between n by n matrices is denoted by L = f(S), we can define a star product by

S₁ * S₂ = f⁻¹[f(S₁)f(S₂)]

where the product on the right is ordinary matrix multiplication. Prove that all such star products are associative.

2. Show that the particular star product of this chapter is generated by

f(S) = [(1 − S) + (1 + S)J]⁻¹[(S − 1) + (S + 1)J]

where

3. Denoting the n by n zero matrix by 0, we define

and B₁, B₂, C by analogy with A. Show that the entire system (12.76) is equivalent to the single Riccati equation

W_y = A + B₁W + WB₂ + WCW

together with the initial condition W(x,x) = J.

4. A two-sided invariant is a function (tτrρ) that is unchanged when S is replaced by U₁ * S * U₂, where U₁ and U₂ are any lossless networks with nonsingular transmissions. Prove that any two-sided invariant must be of the form

where t₁ and τ₁ are given in Eq. (12.84) and where σ means “the spectrum of.” (Proceed as in Exercise 6 of Sec. 12.25.)

12.28    The Probability Scattering Matrix

The analogy between transmission-line theory and diffusion that was noted for n = 1 is also valid in the general case. Let a particle move at random on n parallel lines in such a way that it can be transmitted, reflected, or absorbed when it encounters a given segment. The particle can also jump from one line to another, but only in a direction perpendicular to the lines. If J is a system of aligned parallel segments (Fig. 12.26), let t_ij denote the probability that the particle emerges from the jth segment at the right when the particle is introduced into the ith segment at the left. Under the same hypothesis, r_ij represents the probability of emerging from the jth segment at the left. Similarly, one can define the right-hand coefficients τ_ij and ρ_ij.

Let the particle be introduced at random from the left in such a way that zⁱ, the ith component of a column vector z, gives the probability that we toss the particle at the ith segment. The laws of total and compound probability lead to

for the probabilities of transmission and reflection, respectively, at the jth segment. In terms of the transposed matrices t = (t_ji) and r = (r_ji), the results become tz and rz, just as in the electromagnetic case.

Fig. 12.26 System of parallel segments, permitting formulation of diffusion problem.

If the column vector z₁ describes the probabilities of presentation at the left, and z₄ at the right, then the vectors z₂ and z₃ giving probabilities of egress at the left and right, respectively, are obtained from Eq. (12.74). Hence the algebraic structure of this problem is identical with that of the transmission-line problem considered previously. The role taken by the complex amplitude vectors in the former case is now taken by the real probability vectors z_i.

So far the set of intervals J was regarded as fixed. If the left-hand ends are at x and the right-hand ends at y, with y > x, then the coefficients t, τ, r, and ρ are functions of x and y, and so is the scattering matrix S. Assuming independence of the phenomena for adjacent sets of intervals, we get the same functional equations as in the electromagnetic case. Theorem 12.1 now shows that the probabilities are characterized by the differential system (12.76). Thus, as for n = 1, here too the sole distinction between the two models lies in the constraints.

The probability interpretation gives

with equality in (12.90) when there is no possibility of absorption. On the other hand, if z₁ and z₄ have nonnegative components then the same must be true of z₃ and z₂ in Eq. (12.74). Also, we have

with equality when there is zero probability of absorption.

It is easily seen that the latter conditions are equivalent to the former. When they hold, the system is called p dissipative or p lossless, just as in the case n = 1. The formulation (12.91) shows that the conditions are preserved by the star product, and hence that the foregoing methods can be used. The result is the following:

THEOREM 12.4.   Let the hypothesis of Theorem 12.1 hold for x in J and y in J, where J is an interval of real values. Suppose also that the matrix coefficients a(y), b₁(y), b₂(y), and c(y) are piecewise continuous. Then the matrix S(x,y) is p dissipative for x in J, y in J, y > x, if and only if the matrix

is such that its off-diagonal elements are nonnegative and its column sums nonpositive for y in J. The condition for S(x,y) to be p lossless is that the column sums be zero.

12.29    A More General Interpretation

It is small comfort to the engineer who asks a mathematician for help when the latter fails to solve the problem and then proceeds to imbed the problem in a broad class of similar ones that he also cannot solve. A penchant for facile generalization must not be used as a cloak for perplexity.

The problems posed in the foregoing discussion were actually solved; therefore, it is only fair to claim a certain latitude in this matter of abstraction. We wish to regard t, τ, r, and ρ as operators on a Hilbert space—i.e., on a vector space of infinitely many dimensions in which the notions of “length” and “angle” are defined. The resulting mathematical structure is related to potential theory, to the theory of heat flow, and to other areas of mathematical physics; see Chap. 4.

The particular Hilbert space to be used consists of functions in L², that is, functions z such that z is Lebesgue integrable and

is finite. To see how this space enters the situation, one can imagine a continuous rather than a discrete variation across the transmission line or across the diffusion medium. In that case, sums such as

are replaced by integrals,

Much of the foregoing analysis goes through without change,† but the n-dimensional vector space is now the Hilbert space of complex L² functions z.

If the cross section of the generalized transmission line is uniform, the result depends on ξ − ξ₁ only; hence it can be written as a convolution,

Assuming that the functions are in L², we introduce the Fourier transform T and inverse transform T⁻¹ defined by

If Λ(u) = Tt₀, then the convolution theorem, applied to Eq. (12.92), gives T(tz) = ΛTz, or

Thus, the operator t admits the representation t = T⁻¹ΛT.

12.30    A Special Case and Examples

Having generalized the transmission and reflection operators, let us specialize the rest of the model. If there is no multiple reflection, the functional equation gives

where, as before, t(x,y) is the operator t that effects the transmission from the cross section at x to the cross section at y, with y > x. When the transmission line is longitudinally uniform, we have

t(x,y) = t(y − x)

On writing x for y − x and y for z − y as in Sec. 12.12, we see that Eq. (12.94) leads to a one-parameter semigroup:

Let the function Λ(u) corresponding to t(x) be denoted by Λ_x(u). Substituting Eq. (12.95) into Eq. (12.93) gives

The advantage of this formulation over that of Eq. (12.95) is that here we have ordinary multiplication, while in Eq. (12.95) there is a more elusive, operator multiplication. Subject to mild continuity restrictions, Eq. (12.96) gives Λ_x = (Λ₁)^x, so that

The propagation process is wholly characterized by Λ(u) ≡ Λ₁(u), a single complex function of the real variable u.

Formula (12.97) represents a generalization of transmission-line theory in that the propagating quantity z is an element of Hilbert space and the transmission coefficient t is an operator on that space. But the formula is a specialization of transmission-line theory in that the transmission line is uniform both longitudinally and transversely, and multiple internal reflection is absent.

The function Λ(u) in Eq. (12.97) is determined by the partial differential equation underlying the propagation process. For instance, the Laplace equation†

z_xx + z_yy = 0

gives

Λ(u) = e^−|u|

when suitable behavior at ∞ is required. The resulting formula

z(x,y) = T⁻¹e^−|u|xTz(0,ξ)

represents a harmonic function for x > 0 that reduces to z(0,ξ) for x = 0⁺, provided the latter is in L². Returning to Eq. (12.92) via the convolution theorem gives the Poisson formula for a half plane:

The reader may experience a certain dismay at seeing the Dirichlet problem treated as a problem concerning a propagation or transmission-line phenomenon. On the other hand, the equation

z_xx + z_yy + k²z = 0

seems eminently suited for such treatment; it describes the radiation field of an antenna when the initial illumination z(0,ξ) is a known function in L². In this case, the differential equation gives

when suitable regularity conditions are postulated, and the resulting formula (12.97) is the plane-wave expansion of the antenna field.† Since we get Dirichlet’s problem upon letting k → 0, it is no wonder that the two topics can both be treated in the same conceptual framework.

Sometimes the propagation occurs in time rather than space. The absence of multiple reflection then means that the future does not influence the past. As an illustration, the equation

z_x = kz_yyx = time

for heat flow yields

and Eq. (12.97) gives the temperature at time x > 0 in terms of that at time 0. The familiar form

is obtained from Eq. (12.97) by the convolution theorem.

The implication to be drawn from these examples is that the development presents many points of contact with other disciplines, as soon as one takes a sufficiently broad viewpoint. Transmission-line theory not only is of interest in its own right but serves to illuminate adjacent areas of mathematical physics.

EXERCISES

1. With u = k sin θ, show that Huygens’ wavelets give F(u) = Tz for the secondary pattern of a cylindrical antenna with primary illumination density equal to . Deduce the formula of the text,

from the fact that z(x,y) must give the same secondary pattern as does z(0,y), except for a phase shift due to the change in origin. (In Fig. 12.27, the change of origin introduces a phase shift

for the plane waves traveling in direction θ. Convergence of the infinite integrals is assumed in this and the following problems.)

Fig. 12.27 Computation of antenna far field.

2. The effect of a thin, plane obstacle at x = x₀ is approximated by applying a complex transmission function T₀(y) to the near field z(x,y) and neglecting all other effects. Show that the secondary pattern with obstacle can be obtained by applying the operator

Λ^−x(u)TT₀(y)T⁻¹Λ^x(u)

to the original unperturbed secondary pattern F(u). [The pattern with obstacle is

F₀(u) = Λ^−x(u)TT₀(y)z(x,y)

Use Exercise 1.]

3. In Exercise 2, let t₀(u) = TT₀(y). Show that the far pattern with obstacle is given in terms of that without obstacle by

(Change the order of integration indiscriminately.)

4. Let the function t₀(u,u₁) characterize an obstacle in the following sense: Given an incident plane wave of amplitude 1 in direction sin⁻¹ (u/k), the amplitude of the transmitted wave in direction sin⁻¹ (u₁/k) is t₀(u,u₁) du₁. Show that the pattern with obstacle in terms of that without obstacle is

if all multiple reflection between obstacle and antenna can be neglected.

5. Prove that the obstacle of Exercise 4 can be specified by a single transmission function T₀(y), after the manner of Exercise 2, if and only if t₀(u,u₁) is a function of the single variable u − u₁. (Prove a converse to Exercise 3 by use of the convolution theorem.)

6. If the propagation process of the text is inhomogeneous, so that Eq. (12.94) holds instead of Eq. (12.95), then Eq. (12.97) becomes

when suitable auxiliary conditions are postulated. Show that this form satisfies Eq. (12.94).

REFERENCES

1.   Ambarzumian, V., Diffuse Reflection of Light by a Foggy Medium, Compt. rendus Acad. Sci. U.R.S.S., vol. 38, pp. 229–232, 1943.

2.   Barrar, R. B., and R. M. Redheffer, On Nonuniform Dielectric Media, I.R.E. Trans. on Antennas and Propagation, vol. AP-3, pp. 101–107, 1955.

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† One new feature is that the fixed-point theorem invoked for the double-matching problem of Sec. 12.25 is no longer valid.

† The y here is measured along the y axis while that in t(x,y) is measured along the x axis.

† The need for infinitely many modes to satisfy the boundary conditions gives the problem an infinite-dimensional character, even though the free-space region outside the antenna is a “single” transmission line.

† The preparation of this chapter was sponsored, in part, by the Office of Naval Research. Reproduction in whole or in part is permitted for any purpose of the United States government.

† The colorless words obstacle and object are often used, since network conjures up a too vivid image of a resistance-inductance-capacitance (RLC) circuit. The obstacles need not be bilateral.

† The proof given here tacitly assumes the further hypothesis: “An arbitrary length of lossless line is lossless.” The reader sufficiently steeped in pure mathematics to be bothered by this assumption will easily show that it is dispensable.

† The overworked term “stable” is sometimes used.

† We refuse to worry about other possibilities.

† The two techniques are not unrelated. The characteristic vectors needed for the diagonalization are the fixed directions associated with the matrix transformation.

† The result could also be found by differentiating formally with respect to y and then putting y = 0. The method of the text has the merit that existence of r′(x) is deduced rather than assumed, and this feature, irrelevant physically, is a great comfort to the mathematician.

† The thickness of the slab is not x as in the previous discussion, but y − x.

† In each interval of continuity, approximate the coefficients by continuously differentiable ones that satisfy the given conditions (12.62) with strict inequality.

† That is, the complex transmission and reflection coefficients for the one medium equal those for the other, as functions of both θ and λ.

† The mathematical details cause no trouble when a, b₁, b₂, and c are piecewise continuous.

† Finitely many modes can be accounted for by imagining extra (fictitious) terminal pairs, one for each mode.

† To ensure that the product (12.84) is well defined, it is necessary that we have |ρ₀| < 1 and |r₀| < 1. These inequalities can be attained by arbitrarily small changes in t, r, r, and ρ; hence, their failure to hold does not invalidate the conclusions we shall draw later (Theorem 12.2).

† See the problems at the end of this section.