《电磁学》参考书籍英文电子书（Edward J. Rothwell Michael J. Cloud）Appendix-A

Re(x) Figure A 3: Complex plane technique for evaluating a principal value integral. plane. In many problems of interest the integral of f around the large semicircle tends to zero as r-o and the integrals around the small semicircles are well-behaved E-0. It may then be shown that f(x)dx=兀j rk+ 2TJ UHP where rk is the residue at the kth simple pole. The first sum on the right accounts for the contributions of those poles that lie on the real axis; note that it is associated with a factor j instead of 2 j, since these terms arose from integrals over semicircles rather than over full circles. The second sum, of course, is extended only over those poles that reside in the upper half-plane Fourier transform solution of the 1-D wave equation Successive applications of the Fourier transform can reduce a partial differential equa- tion to an ordinary differential equation, and finally to an algebraic equation. After the algebraic equation is solved by standard techniques, Fourier inversion can yield a solution to the original partial differential equation. We illustrate this by solving the one-dimensional inhomogeneous wave equation a- y(x, y, z, t)=S(x, y, z, t) A.1 where the field y is the desired unknown and S is the known source term. For uniqueness of solution we must specify y and ay/az over some z=constant plane. Assume that f(r, y, t) a(,y,2,D)=0=8(x,y,D A.18 We begin by positing inverse temporal Fourier transform relationships for y and S v(x,y,z,1)= 0 2001 by CRC Press LLC

Figure A.3: Complex plane technique for evaluating a principal value integral. plane. In many problems of interest the integral of f around the large semicircle tends to zero as R → ∞ and the integrals around the small semicircles are well-behaved as ε → 0. It may then be shown that P.V. ∞ −∞ f (x) dx = π j n k=1 rk + 2π j UHP rk where rk is the residue at the kth simple pole. The first sum on the right accounts for the contributions of those poles that lie on the real axis; note that it is associated with a factor π j instead of 2π j, since these terms arose from integrals over semicircles rather than over full circles. The second sum, of course, is extended only over those poles that reside in the upper half-plane. Fourier transform solution of the 1-D wave equation Successive applications of the Fourier transform can reduce a partial differential equa￾tion to an ordinary differential equation, and finally to an algebraic equation. After the algebraic equation is solved by standard techniques, Fourier inversion can yield a solution to the original partial differential equation. We illustrate this by solving the one-dimensional inhomogeneous wave equation  ∂2 ∂z2 − 1 c2 ∂2 ∂t 2  ψ(x, y,z, t) = S(x, y,z, t), (A.16) where the field ψ is the desired unknown and S is the known source term. For uniqueness of solution we must specify ψ and ∂ψ/∂z over some z = constant plane. Assume that ψ(x, y,z, t)    z=0 = f (x, y, t), (A.17) ∂ ∂z ψ(x, y,z, t)    z=0 = g(x, y, t). (A.18) We begin by positing inverse temporal Fourier transform relationships for ψ and S: ψ(x, y,z, t) = 1 2π ∞ −∞ ψ( ˜ x, y,z,ω)e jωt dω,

s(x, y, z,t)= .1,Z, 一D Substituting into(A 16), passing the derivatives through the integral, calculating the derivatives, and combining the inverse transforms. we obtain k2)v(x 3 o)eJor do=0 here k=o/c. By the Fourier integral theorem +k2)ψ(x,y,z,O)-S( )=0. 19) We have thus converted a partial differential equation into an ordinary differential equa- tion. A spatial transform on z will now convert the ordinary differential equation into an algebraic equation. We write ψ(x,y,z,o) y2(r, y, kz, o)e /kz2 dkz S( S(x, y, k in(A 19), pass the derivatives through the integral sign, compute the derivatives, and set the integrand to zero to get k2)y(x -S(x,y,k2,o)=0; (x,y,k2,)= (k2-k)(k2+k) (A.20) The price we pay for such an easy solution is that we must now perform a two- dimensional Fourier inversion to obtain y(x, y, z, t) from y(x, y, kz, o). It turns out to be easiest to perform the spatial inverse transform first, so let us examine Wy0)=广(xyk By(A 20)we have ψ(x,y,z,o)= [S-(x, y, kz, o)1 k)(2 +kejkdkx where the integrand involves a product of two functions. With 8(kz, o) (k2-k)( z +k) the convolution theorem gives s(x, y, S, og(a-s, o)ds (A.21) 0 2001 by CRC Press LLC

S(x, y,z, t) = 1 2π ∞ −∞ S˜(x, y,z,ω)e jωt dω. Substituting into (A.16), passing the derivatives through the integral, calculating the derivatives, and combining the inverse transforms, we obtain 1 2π ∞ −∞  ∂2 ∂z2 + k2  ψ( ˜ x, y,z,ω) − S˜(x, y,z,ω) e jωt dω = 0 where k = ω/c. By the Fourier integral theorem  ∂2 ∂z2 + k2  ψ( ˜ x, y,z,ω) − S˜(x, y,z,ω) = 0. (A.19) We have thus converted a partial differential equation into an ordinary differential equa￾tion. A spatial transform on z will now convert the ordinary differential equation into an algebraic equation. We write ψ( ˜ x, y,z,ω) = 1 2π ∞ −∞ ψ˜ z (x, y, kz,ω)e jkzz dkz, S˜(x, y,z,ω) = 1 2π ∞ −∞ S˜ z (x, y, kz,ω)e jkzz dkz, in (A.19), pass the derivatives through the integral sign, compute the derivatives, and set the integrand to zero to get (k2 − k2 z )ψ˜ z (x, y, kz,ω) − S˜ z (x, y, kz,ω) = 0; hence ψ˜ z (x, y, kz,ω) = − S˜ z (x, y, kz,ω) (kz − k)(kz + k) . (A.20) The price we pay for such an easy solution is that we must now perform a two￾dimensional Fourier inversion to obtain ψ(x, y,z, t) from ψ˜ z (x, y, kz,ω). It turns out to be easiest to perform the spatial inverse transform first, so let us examine ψ( ˜ x, y,z,ω) = 1 2π ∞ −∞ ψ˜ z (x, y, kz,ω)e jkzz dkz. By (A.20) we have ψ( ˜ x, y,z,ω) = 1 2π ∞ −∞ [S˜ z (x, y, kz,ω)]  −1 (kz − k)(kz + k)  e jkzz dkz, where the integrand involves a product of two functions. With g˜z (kz,ω) = −1 (kz − k)(kz + k) , the convolution theorem gives ψ( ˜ x, y,z,ω) = ∞ −∞ S˜(x, y,ζ,ω)g˜(z − ζ,ω) dζ (A.21)

Re(kal Figure A 4: Contour used to compute inverse transform in solution of the 1-D wave where 8(z,) 8(kz,) 2J(k2-k)(k2+k) To compute this integral we use complex plane techniques. The domain of integration extends along the real k z-axis in the complex kz-plane; because of the poles at kz=tk, we must treat the integral as a principal value integral. Denoting I(k2) 2T(k, -k)(k, +k) I(k2)dk,= lim I(k2)dk 一k-8 = lim I(k2)dkz lin 1(k2)dk;+lim/1(k2)dk2 where the limits take8→0and△→∞. Our kx-plane contour takes detours around the poles using semicircles of radius 8, and is closed using a semicircle of radius A(Figure A. 4). Note that if z>0, we must close the contour in the upper half-plane By Cauchy's integral theorem I(k,)dk,+I(k,)dk+I(k2)dk,+I(k2 Thus I(k )dk. I(k2)dk, I(k2)dk2-lim I(k2)dk2 The contribution from the semicircle of radius A can be computed by writing k, in polar coordinates as k-=△e 1 lim I(k,)dk,= (△e-k)(△e+k) JAeje de 0 2001 by CRC Press LLC

Figure A.4: Contour used to compute inverse transform in solution of the 1-D wave equation. where g˜(z,ω) = 1 2π ∞ −∞ g˜z (kz,ω)e jkzz dkz = 1 2π ∞ −∞ −1 (kz − k)(kz + k) e jkzz dkz. To compute this integral we use complex plane techniques. The domain of integration extends along the real kz-axis in the complex kz-plane; because of the poles at kz = ±k, we must treat the integral as a principal value integral. Denoting I(kz) = −e jkzz 2π(kz − k)(kz + k) , we have ∞ −∞ I(kz) dkz = lim r I(kz) dkz = lim −k−δ − I(kz) dkz + lim k−δ −k+δ I(kz) dkz + lim  k+δ I(kz) dkz where the limits take δ → 0 and  → ∞. Our kz-plane contour takes detours around the poles using semicircles of radius δ, and is closed using a semicircle of radius  (Figure A.4). Note that if z > 0, we must close the contour in the upper half-plane. By Cauchy’s integral theorem r I(kz) dkz + 1 I(kz) dkz + 2 I(kz) dkz +  I(kz) dkz = 0. Thus ∞ −∞ I(kz) dkz = − lim δ→0 1 I(kz) dkz − lim δ→0 2 I(kz) dkz − lim →∞  I(kz) dkz. The contribution from the semicircle of radius  can be computed by writing kz in polar coordinates as kz = e jθ : lim →∞  I(kz) dkz = 1 2π lim →∞ π 0 −e jze jθ (e jθ − k)(e jθ + k) je jθ dθ.

Using Eulers identity we can write m I(k2)dkz 2丌△→∞ Thus, as long as z>0 the integrand will decay exponentially as△→∞,and (kz)dk2→0. Similarly,r I(2)dk2-0 when z <0 if we close the semicircle in the lower half-plane Thus, I(k2)dkz =-lim I(k2)dkz -lim I(k2)dk2 (A.22) The integrals around the poles can also be computed by writing k, in polar coordinates Writing k2=-k+Sej6 we find I(k2)dk,=o (一k+8e1°-k)(-k+8el6+k) Similarly, using k2=k+8eJ8, we obtain 1)k= Substituting these into(A. 22)we have 8(,0j、jp1snkx alid for z>0. For z <0, we close in the lower half-plane instead and get 8(z,)=--,sink (A.24) Substituting(A 23)and(A 24)into(A 21)we obtain k ψ(x,y,z,ω)=)s(x,y,,) S(x,y,5,O) d where we have been careful to separate the two cases considered above. To make things a bit easier when we apply the boundary conditions, let us rewrite the above expression Splitting the domain of integration we write (:0)=广5gm①x=+o 0 2001 by CRC Press LLC

Using Euler’s identity we can write lim →∞  I(kz) dkz = 1 2π lim →∞ π 0 −e−z sin θ e jz cos θ 2e2 jθ je jθ dθ. Thus, as long as z > 0 the integrand will decay exponentially as  → ∞, and lim →∞  I(kz) dkz → 0. Similarly,  I(kz) dkz → 0 when z < 0 if we close the semicircle in the lower half-plane. Thus, ∞ −∞ I(kz) dkz = − lim δ→0 1 I(kz) dkz − lim δ→0 2 I(kz) dkz. (A.22) The integrals around the poles can also be computed by writing kz in polar coordinates. Writing kz = −k + δe jθ we find lim δ→0 1 I(kz) dkz = 1 2π lim δ→0 0 π −e jz(−k+δe jθ )jδe jθ (−k + δe jθ − k)(−k + δe jθ + k) dθ = 1 2π π 0 e− jkz −2k j dθ = − j 4k e− jkz. Similarly, using kz = k + δe jθ , we obtain lim δ→0 2 I(kz) dkz = j 4k e jkz. Substituting these into (A.22) we have g˜(z,ω) = j 4k e− jkz − j 4k e jkz = 1 2k sin kz, (A.23) valid for z > 0. For z < 0, we close in the lower half-plane instead and get g˜(z,ω) = − 1 2k sin kz. (A.24) Substituting (A.23) and (A.24) into (A.21) we obtain ψ( ˜ x, y,z,ω) = z −∞ S˜(x, y,ζ,ω) sin k(z − ζ) 2k dζ − 1 2k ∞ z S˜(x, y,ζ,ω) sin k(z − ζ) 2k dζ where we have been careful to separate the two cases considered above. To make things a bit easier when we apply the boundary conditions, let us rewrite the above expression. Splitting the domain of integration we write ψ( ˜ x, y,z,ω) = 0 −∞ S˜(x, y,ζ,ω) sin k(z − ζ) 2k dζ + z 0 S˜(x, y,ζ,ω) sin k(z − ζ) k dζ − − ∞ 0 S˜(x, y,ζ,ω) sin k(z − ζ) 2k dζ.

Expansion of the trigonometric functions then gives v(x, y, z,)=S(x, y, s,o) sin k(z-5) ds COS K S(x, y, S, o)cos ks ds S(r, y, s, a)sin ks ds SIn Kz S(x, y, s, o)cos ks da cos kz S(x, y, s, o)sin ks ds The last four integrals are independent of z, so we can represent them with functions constant in z. Finally, rewriting the trigonometric functions as exponentials we have +a +b(x, y, o) This formula for y was found as a solution to the inhomogeneous ordinary differential solutions of the homogeneous differential equation. Since these are exponentia .g equation(A 19). Hence, to obtain the complete solution we should add any possible in fact represents the complete solution, where A and B are considered unknown and can be found using the boundary conditions If we are interested in the frequency-domain solution to the wave equation, then we are done. However, since our boundary conditions(A. 17)and(A 18) pertain to the time domain, we must temporally inverse transform before we can apply them. Writing the sine function in(A. 25)in terms of exponentials, we can express the time-domain solution V(x,y,z,t)=/F S(x, y, 5, o) S(x, y, 5, o) ds J-I (A.26) A combination of the Fourier integration and time-shifting theorems gives the general (x,y,,a) y (A.27) where we have assumed that S(x, y, 5, 0)=0. Using this in(A 26) along with the time shifting the y(x,y,z,1) S(r, y, s, t)d S(, y,s, t)dr ds (xy1-2)+b(y+ p5=E/xd+a(=x1-)+b(x+) A.28 a(x,y, t)=F [A(, y, o). 0 2001 by CRC Press LLC

Expansion of the trigonometric functions then gives ψ( ˜ x, y,z,ω) = z 0 S˜(x, y,ζ,ω) sin k(z − ζ) k dζ + + sin kz 2k 0 −∞ S˜(x, y,ζ,ω) cos kζ dζ − cos kz 2k 0 −∞ S˜(x, y,ζ,ω)sin kζ dζ − − sin kz 2k ∞ 0 S˜(x, y,ζ,ω) cos kζ dζ + cos kz 2k ∞ 0 S˜(x, y,ζ,ω)sin kζ dζ. The last four integrals are independent of z, so we can represent them with functions constant in z. Finally, rewriting the trigonometric functions as exponentials we have ψ( ˜ x, y,z,ω) = z 0 S˜(x, y,ζ,ω) sin k(z − ζ) k dζ + A˜ (x, y,ω)e− jkz + B˜ (x, y,ω)e jkz. (A.25) This formula for ψ˜ was found as a solution to the inhomogeneous ordinary differential equation (A.19). Hence, to obtain the complete solution we should add any possible solutions of the homogeneous differential equation. Since these are exponentials, (A.25) in fact represents the complete solution, where A˜ and B˜ are considered unknown and can be found using the boundary conditions. If we are interested in the frequency-domain solution to the wave equation, then we are done. However, since our boundary conditions (A.17) and (A.18) pertain to the time domain, we must temporally inverse transform before we can apply them. Writing the sine function in (A.25) in terms of exponentials, we can express the time-domain solution as ψ( ˜ x, y,z, t) = z 0 F−1 c 2 S˜(x, y,ζ,ω) jω e j ω c (z−ζ) − c 2 S˜(x, y,ζ,ω) jω e− j ω c (z−ζ) dζ + + F−1  A˜ (x, y,ω)e− j ω c z  + F−1  B˜ (x, y,ω)e j ω c z  . (A.26) A combination of the Fourier integration and time-shifting theorems gives the general identity F−1 S˜(x, y,ζ,ω) jω e− jωt0  = t−t0 −∞ S(x, y,ζ,τ) dτ, (A.27) where we have assumed that S˜(x, y,ζ, 0) = 0. Using this in (A.26) along with the time￾shifting theorem we obtain ψ(x, y,z, t) = c 2 z 0  t− ζ−z c −∞ S(x, y,ζ,τ) dτ − t− z−ζ c −∞ S(x, y,ζ,τ) dτ  dζ + + a  x, y, t − z c  + b  x, y, t + z c  , or ψ(x, y,z, t) = c 2 z 0 t+ z−ζ c t− z−ζ c S(x, y,ζ,τ) dτ dζ + a  x, y, t − z c  + b  x, y, t + z c  (A.28) where a(x, y, t) = F−1 [A˜ (x, y,ω)], b(x, y, t) = F−1 [B˜ (x, y,ω)].