IEEE TRANSACTIONS ON ELECTROMAGNETIC COMPATIBILITY, VOL. 30, NO. I , FEBRUARY 1988 31

General Solutions of Maxwell’s Equations for Signals in a Lossy Medium: 11. Electric and Magnetic Field Strengths Due to Magnetic

Exponential Ramp Function Excitation

Abstract-In Part 1 of this series of papers, it was shown that the modification of Maxwell’s equations permits one to derive solutions in lossy media for time-limited signals. Solutions for an electric exponential ramp function excitation were derived and plotted for different locations in a lossy medium. In this paper, solutions for the electric and magnetic field strengths in a lossy medium due to a magnetic exponential ramp function excitation are presented. The solutions are in integral forms, and they are evaluated by numerical integration methods using a digital computer. Computer plots for the electric and magnetic field strengths at different locations in the propagation medium are given. The plots obtained for the transients may be used to represent solutions in lossy media for signals that can be represented in terms of time-series expansion of the transients. This topic is discussed in Part I l l of this series of papers.

Key Words-Maxwell’s equations, lossy medium, nonsinusoidal waves.

Index Code-J3c/d.

I. INTRODUCTION OLUTIONS for waves with general time variation S propagating in a lossy medium have been calculated from

the following modified Maxwell’s equations [ 1 , ch. 21, [2] :

curl H=EdE/at+uE (1)

curl E = p a H / a t + sH (2)

E div E = p div H=O (3)

where E and H a r e the electric and magnetic field strengths, E is the permittivity, p the permeability, u the electric conductiv- ity, and s the magnetic conductivity. The term sH in (2) does not exist with the usual Maxwell’s equations. Its presence in (2) will make it possible to obtain general solutions for waves in a lossy medium with u # 0. The transition s 4 0 can be made after convergent solutions for E and Hare obtained from (1)-(3) without encountering the problem of divergencies [ 11, PI .

For a planar transverse electromagnetic (TEM) wave propagating in the direction of the y axis of a Cartesian coordinate system, (1)-(3) yield two second-order differential

Manuscript received June 20, 1986; revised September 15, 1987. This work was supported by the Research Management Unit of Kuwait University under Project EE027.

The author is with the Department of Electrical and Computer Engineering, Kuwait University, Code 13060, State of Kuwait. Tel. 4834506.

IEEE Log Number 8718469.

equations for E and H :

a w a y 2 - pEa2E/at2- ES)aE/at - W E = o (4)

a2H/ay2-pEa2H/at2-(pu+ ES)aH/at-suH=o. ( 5 )

If the electric field strength E is found from (4), it was shown [l], [2] that (1)-(3) yield the following solutions for the magnetic field strength H :

H ( y , t ) = - 1 ( d E / a t + u E ) dy+H,(t) (6)

[ 1 (aE/ay)eSl/p dt + H,(y) (7) I H ( y , t)=e-sf/p -p -1

where H J t ) and Ht( y ) are integration constants which can be determined from the initial and boundary conditions. Com- puter plots for solutions of E and H a t different locations in a lossy medium have been obtained [2] for an electric exponen- tial ramp function excitation.

If the magnetic field strength H is found from (5 ) , the following solutions for the electric field strength E can be calculated from (1)-(3), [ 11, [2] :

E ( y , t ) = - (paH/at+sH) dy+E,(t) (8)

E ( y , t)=e-mf/f [ - 6 - 1 j (aH/ay)egr/f di+E,(y)] . (9)

The integration constants E,(t) and E,( y ) can be determined from initial and boundary conditions.

In the following section, we shall present the steps for determining the electric and magnetic field strengths E and H in a lossy medium due to magnetic exponential ramp function excitations. The solutions are derived from ( 5 ) , (8), and (9), and they are in integral forms. Computer plots of E and H at different locations in the propagation medium are obtained by numerical integrations of the integral solutions. The special case of wave propagation in a loss-free medium will not be considered here since it is well presented in the literature [ 13, [3], [4].

The procedures for calculating solutions for E and H from (4), (6), and (7) are the same for (3, (8), and (9) since they are of the same form. Note that (4), (6), and (7) yield E and H when the boundary conditions are specified for E while (3,

OO18-9375/88/0200-0037$01 .OO O 1988 IEEE

38 IEEE TRANSACTIONS ON ELECTROMAGNETIC COMPATIBILITY, VOL. 30, NO. I , FEBRUARY 1988

procedure for obtaining a solution for u ( y , t) is presented in [ l , ch. 21 and [2] and will not be repeated here. The resulting solution of the magnetic field strength HH( y, t) is of the form

H H ( y , t )=HoQ[(l -e -20f )e-Y/L+u(y , t ) ] . (17)

The solution of u ( y, t) is given in the following integral form:

I.

Fig. 1 . A magnetic exponential ramp function excitation applied to the boundary of a lossy medium.

(8), and (9) yield E and H when the boundary conditions are specified for H. Here, we shall not be concerned with detailed derivations since they are already given in [l] and [2] . We shall outline the methodology used in arriving at the final solutions and present the computer plots obtained from the integral solutions of E and H, for magnetic exponential ramp function excitation.

11. ELECTRIC AND MAGNETIC FIELDS DUE TO MAGNETIC EXFQNENTIAL RAMP FUNCTION EXCITATION

Consider the exponential ramp function shown in Fig. 1. Let this function represent the magnetic field strength H ( t ) applied to the boundary of a lossy medium at the plane y = 0. Thus, H(t ) can be expressed as follows:

H(0, t)=Hor(t)=O =HoQ(l -e-f/7)=H1(l

for t z 0 , H I = H o Q (10) where Q = (1 - H(0, t) = HO at t = AT. The boundary condition

I , 7 > 0, is a constant which yields

H(m, t) = finite (1 1)

holds for the plane y -+ 03. If E and H are zero for y L 0 at the time t = 0, the following initial conditions hold:

E ( y , O)=H(y, O ) = O , y>O (12)

Assume that a solution of (5) is of the form

H ( Y , ~ ) = H H ( Y , t ) = H o Q [ u ( ~ , t ) +(1 - e - f / T ) F ( y ) ] , t>0 (15)

where F( y ) is a function of y only and not t. By inserting (15) into (5) one obtains a solution

F ( y ) = e-YlL, L = (su)-Il2 (16)

which satisfies the boundary condition (ll), and a second- order differential equation for u( y, t) of the form given in (5). The differential equation can be solved for u ( y , t) by the separation of variables using Bernoulli's product method. The

b2

where sh is the hyperbolic sine function, and the parameters a, b, c, 0, and K are defined as follows:

a = (1 / 2 ) ( p ~ ) - ' ( p u + E S ) = (c /2)( Zu + sZ- ') = - + - (19) 2E 2 p

u s

b= [(2?rk)2+su]1'2 (20) C = (pE) - 112

2 = (p/E)

1 Z C

z p = -

C

E = -

(21)

p = 2 a k , k = wavenumber (22)

K = ( ~ T ) - ' ( ~ ~ / c ~ - s u ) ~ / ~ . (23)

The electric field strength EH( y, t) in the lossy medium due to the magnetic excitation function can be calculated by inserting (17) into (8) and (9), and by applying the initial and boundary conditions in (1 1)-( 14) to determine the integration constants E,(t) and Et( y). The insertion of (1 7) into (9) yields the following solution for the case u > 0 and s > 0:

2rK

HUSSAIN: GENERAL SOLUTIONS OF MAXWELL'S EQUATIONS: I1 39

where ch is the hyperbolic cosine function and El( y)e-"'/' = 0, becomes

(25) EH(y, t ) =HI z [ [. (%) - z a y - z (%) 'I2] e-2ar CY = Zca/2 = U/2€.

Inserting (17) into (8) yields the same solution for E H ( y , t ) given by (24) with the term E,(y)e-"'/' replaced by E,,(t). Hence, the condition

Et ( y ) = =E,( t ) = EHoe-ut /E (26) = H l Z [ -Zaye-2a' +ZAl(y, t ) -ZH2(y, t ) ] .

must be satisfied. The initial condition of (12) requires EHO = (31)

the electric and magnetic field strengths H ~ ( ~ , t ) and EH( y , t ) can be written in a normalized form, suitable

the transition -+ for > can be made in (24). for evaluation by a digital computer, by using the following

substitutions:

0, which takes care of the integration constant in the solution OfEH(Y, 0.

The first term yields

For

lim s-0 Z - l ( ~ / a ) ' / ~ ( l + Z2a~-1e -20t ) exp [ - (sa) 1/2y] 11 = p ~ / ~ ~ = 4 n - k / Z a = ( 4 n - k / a ) ( ~ / ~ ) " ~ (32)

= Z [ ( U / S ) ' / ~ - ~yle-~"'. (27) 6 = a t = Zact/2 = Ut/2€ (33)

The first integral in (24) is denoted ZHI(y, t ) for s -+ 0: = CYy/c = z a y / 2 = (/L/€) %y/2 (34)

With the above substitutions, (17) and (18) yield the following normalized form of solutions for HH( y , t ) and U( y , t ) , for s = Oanda > 0:

+2" e-*' ,:," [ch ((y2-P2~2)1/2t n-

] $+: e-2al H H ( [ , 8)=Hl[l - e - 2 8 + u ( t , e) ] (36) CY sh ( C Y ~ - @ ~ C ~ ) ' / ~ ~

- (& - p2c2) 112

I sh (1 -172)1'26 sin [ y u([, e)=- e-8 d17

-4 n- [ s 0 (1 - ) 7 2 ) ' / 2 17 n-

1: [ch (CY2-p2c2)1/2t

The solution of the electric field strength EH( y , t ) in (3 1) becomes 1 EH( t , 6 ) = HI Z [ - 2 t e -28 + ZAl ( 5 , 8 ) - ZH2 ( E , e) ] (38)

CY sh (CY2-f12C2)1/2t - ( C Y 2 - p Z c 2 ) 112

(28)

The function ZHI(y, t ) without the term Z(a/s)'/2e-2a1 is denoted --Zh1(y, t ) :

The second integral in (24) is denoted Z ~ 2 ( y , t ) for s-+O, which remains finite if the electric conductivity a of the propagation medium is larger than zero:

2 Za ZH2(y, t)=- cat

n-

CY sin (/32c2-a2)1'2t (p2c2-CY2)1/2 - ] dp. (30)

(39)

The electric field strength E H ( y , t ) of (24), with the term 6 = 2 d / Z a e 1. (40)

40 IEEE TRANSACTIONS ON ELECTROMAGNETIC COMPATIBILITY, VOL. 30, NO. I , FEBRUARY 1988

03 II \

I I I J . . . . . . . . . . . . . . . . . . . . . . . 1 . ’ .

0 5 ’0 8 - 15 20 25 30 e- Fig. 2. Electric field strength due to excitation by a magnetic exponential

ramp function for the locations [ = 0, 1, 2, and 3 in the time range 0 5 8 Fig. 4. Magnetic field strength due to excitation by a magnetic exponential

ramp function for the locations [ = 0, 1, 2, and 3 in the time range 0 d 8 I 30. I 30.

Ow 10 20 30 LO 50 60 70 80 90 100 e- Fig. 3. Electric field strength due to excitation by a magnetic exponential

ramp function for the locations [ = 0, 1, 2, and 3 in the time range 0 I 8 5 loo. Computer plots of EH([ , O)/H,Z for the locations [ = 0, 1,

2, and 3 are shown in Fig. 2 for the time range 0 I 8 I 30, and in Fig. 3 for the time range 0 I 8 I 100. According to the plots, the function E&, 8)/H1Z rises instantly to its peak value and then drops to reach a steady-state value as 8 increases. For E 2 3, E&, B)/H,Z increases to reach a steady-state value without any significant drop in amplitude.

Computer plots of “ ( E , @/HI for the locations 4 = 0, 1 , 2, and 3 are shown in Fig. 4 for the time range 0 I 8 I 30, and in Fig. 5 for the time range 0 I 8 I 100. The plots in Figs. 4 and 5 differ by a constant from those given in [2] that are obtained for the electric field strength due to an electric exponential ramp function excitation.

111. CONCLUSIONS Solutions and computer plots for the electric and magnetic

field strengths due to a magnetic exponential ramp function

Fig. 5. Magnetic field strength due to excitation by a magnetic exponential ramp function for the locations 5 = 0, 1, 2, and 3 in the time range 0 d 8 5 loo.

excitation applied at the boundary of a lossy medium are presented. The plots will be used later to obtain plots for the electric and magnetic field strengths due to a magnetic sinusoidal pulse excitation by means of time-series expansion.

REFERENCES [ 11 H. Harmuth, Propagation of Nonsinusoidal Electromagnetic

Waves. New York: Academic, 1986. [2] M. G. M. Hussain, “General solutions of Maxwell’s equations for

waves in a lossy medium: I. Electric and magnetic field strengths due to electric exponential ramp function excitation,” IEEE Trans. Electro- magn. Compat., this issue, pp. 29-36.

[3] J. D. Kraus and K. R. Carver, Electromagnetics, 2nd ed.. New York: McGraw-Hill, 1973.

[4] J. A. Stratton, Electromagnetic Theory. New York: McGraw-Hill, 1941.