SOLUTIONS MANUAL FOR

by

Numerical Techniques inElectromagneticswith MATLAB,Third Edition

Matthew Sadiku

SOLUTIONS MANUAL FOR

by

Numerical Techniques inElectromagneticswith MATLAB,Third Edition

Matthew Sadiku

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Table of Contents

Chapter 1 1

Chapter 2 10

Chapter 3 56

Chapter 4 83

Chapter 5 99

Chapter 6 119

Chapter 7 142

Chapter 8 151

Chapter 9 164

1

CHAPTER 1 Prob. 1.1

(a)

2 2 2 2 2 2

0x y z

x y z

x y z

y z z y x z z x x y y x

a a a

(b)

2 22 22 2

, ,

0

x y y

y yx x xz

y yx xz z

x y zx y z F F F

F FF F FFx y x y x z z x y

F FF FF Fx y x z y x y z z x z y

� F

(c )

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

( y2 x x x xzx

y y y yx z

yx z z z z

y

FF F F FF) - x x y z x y z

F F F FF Fy x y z x y z

FF F F F Fz x y z x y z

Fy x

�

y

z

F F a

a

a

x x z

y yx x

yx z z

F F Fy z z x

F FF Fz y z x x y

FF F Fx z x y y z

x

y

z

a

a

a

F

Prob. 1.2 Let U V A and apply Stokes’ theorem.

( )

( ) ( )L S

S S

U V d U V d

U V d U V d

� �

� �

� l S

S S

2

Since 0,V ( )

L S

U V d U V d � �� l S

Similarly, we can show that ( ) ( )

L S S

V U d V U d U V d � � �� l S = - S

Thus,

L L

U V d V U d � �� �l l

as required. Prob. 1.3 Using divergence theorem, ( ) ( )

S v

U V d U V dv � �S

But ( ) , where = VU U U � � �A A A A

2

( ) ( )

( )S v

v

U V d U V V U dv

U V U V dv

� � �

�

S

Prob. 1.4 If 0 ,vJ = then Maxwell’s equations become

0 (1)0 (2)

(3)

(4)

t

t

�

�

BD

BE

DH

Since 0 � A for any vector field A,

0

0

t

t

��

��

BE

DH

Showing that (1) and (2) are incorporated in (3) and (4). Thus Maxwell’s equations can be reduced to curl equations (3) and (4). Prob. 1.5 If 0 ,v J

3

0v

t

t

�

�

EH

HE

EH J

2

2t t t

J EE H

22

2 2

1( )t c t

�J EE E

22

2 2

1 ( / )vc t t

E JE

Similarly,

t

H J E

2

22( ) HH H J

t

�

or

2

22 2

1c t

HH J

It is assumed that the medium is free space so that the medium is homogeneous and 1 .u c

Prob. 1.6 (a)

t

t

DH J

HE

( ) ( ) ( )t t t t

HH J D J E J

2

2t

HH J

(b)2

2( ) ( )t t t t t

D J EE H J

2

2t t

E JE

4

Prob. 1.7

t

HE (1)

t

DH J (2)

Dotting both sides of (2) with E gives

(t

� � �

DE H) E J E (3)

But for any arbitrary vectors A and B, ( ) ( ) ( ) � � �A B B A A B Applying this to the left-hand side of (3) by letting A H and B E , we get

1( ) ( ) ( )2 t

� � � �H E H E E J D E (4)

From (1), 1( ) ( ) ( )2t t

� � �

BH E H B H

Substituting this into (4) gives

1 1( ) ( ) ( )2 2t t

� � � �B H E H J E D E

Rearranging terms and then taking the volume integral of both sides,

1( )2v v v

dv dv dvt

� � � �E H E D H B J E

Using the divergence theorem, we get

( )v

Wd dvt

� �� E H S J E

Or

( )S v

W d dvt

� �� E H S E J

as required. Prob. 1.8

0, 0 � �E H

0

10 sin( ) 20 cos( )

y xx y

x y

x y

E Ex y zdz dzE E

k t kz k t kz

E a a

a a

5

1

10cos( ) 20sin( )

o

x yo

t

k t kz t kz a

H E

a

which is the given H. Since all of Maxwell’s equations are satisfied by the fields, they are genuine EM fields. Prob. 1.9

1o

o ot t

H HE E

1 1

( , ) 0 01 sin( )

o o o ox

xy o y

o o o o

x y zt D z t

D D t zz

H D

a a

cos( )oy

D t z

H a

Prob. 1.10

1/ 2 1/ 2

2

1 ( )

1

1

s s z

j joz

joz

dj Hd

H e j e

H j e

H E a

a

a

2

1 1 js os z

H j ej j

HE a

Prob. 1.11 It is evident that 0 and 0 � �E H = are satisfied.

s sjt

BE E H

0 0

j zs o y

j zo

x y z j E eE e

E a

i.e. j z j zoy o y

Ej e j E e

a a

6

(1)

( ) sjt

sDH J H E

0 0

j zos x

j zo

x y z j E eE e

H a

( ) j z j zoo x x

j Ej E e e

a a

( )j j (2) From (1) and (2),

2( ) jj jj

Thus,

,jj

Prob. 1.12 The surface current density is ( ,0, )n n x zH H K a H a At x = 0, , (0, , , )n x z yH y z t a a K a i.e. cos( )o yH t z K a At x = a, n x a a ( , , , ) cos( )cos( )z y o yH a y z t H t z a K a i.e. cos( )o yH t z K a At y =0, n ya a

( ,0, , ) ( ,0, , )

cos( / ) cos( ) sin( / ) sin( )

z x x z

o x o z

H x z t H x z taH x a t z H x a t z

K a a

a a

At y =b, n y a a ( ,0, ) ( , , , ) ( , , , )

cos( / )cos( ) sin( / )sin( )

y x z x z x x

o x o z

H H H x b z t H x b z taH x a t z H x a t z

K a a a

a a

7

Prob. 1.13

(a) 2

22o o t

EE

For E, 2

2 22 cos( ) [ sin( ) cos( )x x xt z t z t z

z z

E a a a

2

2 cos( )o o o o xt zt

E a

Thus, 2 2

o o o o (b)

o dtt

HE H E

sin( )cos( ) 0 0

yx y z t zt z

E a

sin( ) cos( )oo y yt z dt t z

H a a

Prob. 1.14

s sjt t

D EH H E

2

1 1(2sin cos )sin 4

sin 1 14

-j rs r

-j r -j r

IL j er r rIL j j e e

r r r

H a

a

22 2 2

1 12cos sin4

-j rs r

IL j jerj r r r r

E a a

Prob. 1.15

( ) ( ) ( ) ( )

20( )2 2

5

y yx x

x y x y x y x y

jk y jk yjk x jk x

s

j k x k y j k x k y j k x k y j k x k y

e ee eEj j

j e e e e

which consists of four plane waves.

s o sj E H

8

Or

0 0 ( , )

20 sin( )sin( ) cos( )cos( )

s so o

sz

sz szx y

o

y x y x x x y yo

j j x y zE x y

E Ejy x

k k x k y k k x k y

H E

a a

a a

Prob. 1.16 (a) Re( ) sin cos cos( )j t

sI I e x y t z

(b) 2 90 4Re(20 10 )oj x j j t j x j tV e e e e e

2 90 4 2 420 10 20 10oj x j j x j x j x

sV e e e j e e Prob. 1.17 (a) Re( ) cos( 2 ) sin( 2 )j t

s x ye t z t z A A a a (b) Re( ) 10sin sin 5cos( 12 45 )j t o

s x ze x t t z B B a a (c) 3Re( ) 2cos 2 sin( 3 ) cos( 4 )j t x

sC C e x t x e t x Prob. 1.18 Assuming the time factor j te , equation

2

22t t

E JE

becomes 2 2

s s sj E E E or

2 ( ) 0s sj j E E For conducting medium, � so that

2 0s sj E E Prob. 1.19 Let Re( ), Re( ),j t j t

s se V V e A A etc

9

2

22

s

s

jt

t

A A

A A

Similarly,

22

2 sV Vt

Substituting these into eqs. (1.42) and (1.43), we obtain

2 2

2 2

vss s

s s s

V V

A A J

But 2 2k . Thus, 2 2

2 2

vss s

s s s

V k V

k

A A J

Prob. 1.20

(a) a = 1, b = 2, c = 0, b2 – 4ac = -16

Hence, it is elliptic. (b) 2 21, 0, 1,a y b c x 2 2 24 4( 1)( 1) 0b ac x y Hence it is elliptic. (c ) 21, 2cos , (3 sin )a b x c x 2 2 24 4cos 12 4sin 16 0b ac x x Hence it is hyperbolic. (d) 2 2, 2 , ,a x b xy c y 2 2 2 2 24 4 4 0b ac x y x y Hence it is parabolic. Prob. 1.21 (a) 2, 0, 0, 4 0; i.e. it is parabolic.a b c b ac (b) 21, 0, 0, 4 4; i.e. it is elliptic.a b c b ac

(c) 22 2

2 2 0x y

21, 0, 1, 4 4; i.e. it is elliptic.a b c b ac

10

CHAPTER 2 Prob. 2.1 Let ( , ) ( ) ( ).x y X x Y yΦ = '' ' ' '' ' ' 0aX Y bX Y cXY dX Y eXY fXY+ + + + + = Dividing through by XY,

1 1 ' '( '' ') ( '' ') 0bX YaX dX cY eY fX Y XY

+ + + + + =

The PDE is separable if and only if a and d are functions of x only; c and e are functions of y only; b = 0; and f is a sum of a function of x only and a function of y only, i.e. if

1 2( ) ( ) ( ) ( ) [ ( ) ( )] 0xx yy x ya x c y d x c y f x f yΦ + Φ + Φ + Φ + + Φ = Prob. 2.2 (a) Consider the problem as shown below.

This is similar to Example 2.1 with 1 3 2 40, ,o oV V V V V V= = = = − . Hence

odd odd

4 4sinh ( / ) sin( / ) sinh [ ( ) / ]sin( / )( , )sinh( / ) sinh( / )

o o

n n

V Vn x b n y b n a x b n y bV x yn n a b n n a bπ π π π

π π π π

∞ ∞

= =

−= −∑ ∑

But sinh sinh 2cosh sinh2 2

A B A BA B + −− =

odd

4 sin( / )( , ) 2cosh sinh [ ( / 2) / ]sinh( / ) 2

o

n

V n y b n aV x y n x a bn n a b b

π π ππ π

∞

=

= −∑

We now transform coordinates: / 2, / 2x X a y Y b= + = +

x

Vo -Vo

0 a

b y 0

11

odd

sin ( / 2)4 n X( , ) 2cosh sinh

2 b2 sinh cosh2 2

o

n

n Y bV n abV x y

n a n a bnb b

ππ π

π ππ

∞

=

⎛ ⎞+⎜ ⎟⎝ ⎠=⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑

But

sin sin cos( / 2) sin( / 2)cos

2

( 1) cos , oddn

n Y n n Y n Yn nb b b

n Y nb

π π π ππ π

π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞= − =⎜ ⎟⎝ ⎠

odd

( 1) sinh cos( / )4( , )

sinh2

n

o

n

n x n y bV bV x y

n anb

π π

ππ

∞

=

⎛ ⎞− ⎜ ⎟⎝ ⎠=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

(b) Let V(x,y) = X(x)Y(y), / /

1 2( ) sin , ( ) n x b n x bn n

n yY x X x A e A eb

π ππ −= = +

A2 = 0 since X(x) = 0 as x →∞ /( , ) sin( / ) n x b

nV x y a n y b e ππ −=∑

(0, ) sin( / )o nV y V a n y bπ= =∑

o0

0, even2 sin( / ) 4V , odd

b

n o

na V n y b dx

b nn

ππ

=⎧⎪= = ⎨

=⎪⎩∫

/

odd

4 1( , ) sin( / ) n x bo

n

VV x y n y b en

πππ

∞−

=

= ∑

y

x

Y

X

12

Prob. 2.3 (a) This is similar to Example 2.1 with 2 3 4 10, /oV V V V V x a= = = = . Hence,

odd( , ) sin( / )sinh[ ( ) / ]n

nV x y A n x a n b y aπ π

∞

=

= −∑

When y = 0, we obtain

odd

/ sin( / )sinh( / )o nn

V x a A n x a n b aπ π∞

=

= ∑

Multiplying both sides by sin (mπx/a) and integrating over 0 < x < a leads to

0

2

2 2 20

1

2 sin( / )sinh( / )

2 sin( / ) cos( / )sinh( / )

2 ( 1)sinh( / )

ao

n

a

o

no

V xA n x a dxa n b a a

V a axn x a n x aa n b a n n

Vn n b a

ππ

π ππ π π

π π

+

=

⎡ ⎤= −⎢ ⎥

⎣ ⎦−

=

∫

(b) Let V(x,y) = X(x)Y(y). 1 2( ) cos sinX x A x A xβ β= + Since X(-a/2) = 0 = X(a/2) A2 = 0 and

1 1 2( ) cos , ( ) cosh sinhX x A x Y y B y B yβ β β= = + ( )( , ) cos cosh sinhn nV x y x C y D yβ β β= +∑

Substituting ( , / 2) cos( / ) and ( , / 2) cos( / )o oV x b V x a V x b V x aπ π= − = gives n = 1, Dn = 0, and

cosh( / )

on

VCb aπ

=

Thus,

cos( / ) cosh( / )( , )cosh( / )

oV x a y aV x yb a

π ππ

=

Prob. 2.4 (a) Let U(x,y) = X(x) Y(y). 1 1( ) cos sinX x A x B xβ β= +

1

1

'(0) 0 0'( ) 0 0 sin

X BX Aπ β βπ

= → == → = −

i.e. β = n, n = 1,2,3, … 1 2 2( ) cos , cosh sinhX x A nx Y A y B yβ β= = + 2(0) 0 0Y A= → = Hence

13

01 0

( , ) cos sinh cos sinhn nn n

U x y C C nx ny C nx ny∞ ∞

= =

= + =∑ ∑

0( , ) cos sinhn

nU x x C nx nπ π

∞

=

= =∑

Multiplying both sides by cos mx and integrating over 0< x < π yields / 2oC π= and

2

0, even4 , odd

sinhn

nC

nn nπ π

=⎧⎪= ⎨

=⎪⎩

2odd

4 cos sinh( , )2 sinhn

nx nyU x yn n

ππ π

∞

=

= − ∑

(b) First, transform the equation by letting U(x,t) = x + u(x,t)

where t xxu ku=

Subject to u(0,t) = 0 = u(1,t), u(x,0) = -x If u(x,t) = X(x)T(t),

1 2( ) cos sinX x A x A xβ β= +

1(0) 0 0(1) 0 0 sin

X AX β

= → == → =

i.e. β = nπ.

2 2

2( ) sin , kn tX x A n x T Be ππ −= =

2 2

1( , ) sin kn t

nn

u x t C n xe ππ∞

−

=

=∑

1( ,0) sinn

nu x x C n xπ

∞

=

= − =∑

1

0

2( 1)2 sinn

nC x n xdxn

ππ−

= − =∫

2 2

1

2 ( 1)( , ) ( , ) sinn

kn t

nU x t x u x t x n xe

nππ

π

∞−

=

−= + = + ∑

(c ) Let u(x,t) = X(x)T(t). From the boundary conditions, ( ) cos sinX x A x B xβ β= +

(0) 0 0(1) 0 0 sin

X AX β

= → == → =

or β = nπ. 3 4sin , sin cosX A n x T A n at A n atπ π π= = + 4(0) 0 0T A= → =

14

1

( , ) sin sinnn

u x t C n x n atπ π∞

=

=∑

1( ,0) sint n

nu x x n aC n xπ π

∞

=

= =∑

Multiplying both sides by sin mπx and integrating over 0 < x < 1 yields

2

2cos( )n

n aCn a

ππ

= −

2 2 21

2 cos( , ) sin sinn

n au x t n x n ata n

π π ππ

∞

=

= − ∑

Prob. 2.5 (a) Let ( , ) ( ) ( )R Fρ φ ρ φΦ = . From the text, eq. (2.72) onward, 1 2( ) cos( ) sin( )F C Cφ λφ λφ= +

1 0 since ( ,0) ( , )C ρ ρ π= Φ = Φ . Also, n = 1,2,3,…

( ) sin

( )n n

n nn n n

F C nR a b

φ φ

φ ρ ρ−

=

= +

As 0, ( , )ρ ρ φ→ Φ must be finite, i.e. an = 0.

1

sin( , ) n nn

nA φρ φρ

∞

=

Φ =∑

1

(1, ) sin sinnn

A nφ φ φ∞

=

Φ = =∑

which implies that 1 1, 0 if 1nA A n= = ≠ . Thus sin( , ) φρ φρ

Φ =

(b) From the boundary conditions, Φ does not depend on φ, i.e. m= 0 in F’’ + m2F = 0. From eq. (2.89) in the text,

1 2( ) sin cosZ z C z C zμ μ= + If Z(0) = 0 = Z(L), then 2 0 and or /C L n n Lμ π μ π= = = . Also, 2 2 2'' ' 0R R Rρ ρ μ ρ+ − = 2 2 2 2'' ' ( 0) 0R R j Rρ ρ μ ρ+ + − = The solution to this is 0 0( ) ( ) ( )n nR A I B Kρ ρμ ρμ= + Bn = 0 since K0 is infinite at ρ = 0. Thus

01

( , ) ( / ) sin( / )nn

z A I n L n z Lρ πρ π∞

=

Φ =∑

To obtain An, we apply the boundary condition Φ(a,z) = 1, multiply both sides by sin mz/L, and integrate over 0 < x < L. We get

15

0, even4 , odd

( / )n

o

nA

nn I n a Lπ π

=⎧⎪= ⎨ =⎪⎩

0

1,3,5 0

( / )4 sin( / )( , )( / )n

I n L n z LI n a L n

πρ πρ φπ π

∞

=

Φ = ∑

(c ) Let ( , , ) ( ) ( ) ( )t R F T tρ φ ρ φΦ = . Separation of variables leads to

22' 0 ( ) k tT kT T t e μμ −+ = → =

From eq. (2.88) in the text, ( ) cos sinn n nF a n b nφ φ φ= + Since ( , ,0) cos , 0nbρ φ ρ φΦ = = and ( ) cosn nF a nφ φ= Finally, 2 2 2 2'' ' ( ) 0R R n Rρ ρ μ ρ+ + − = with solution 1 2( ) ( ) ( )n nR c J c Yρ ρμ ρμ= + c2 = 0 since R must be finite at ρ = 0. Hence

2

( , , ) ( ) cos k tn n

nt A J n e μ

μμ

ρ φ ρμ φ −Φ =∑∑

But Φ(a,φ,t) = 0 implies that ( ) 0 ( )n n mJ a J Xμ = = where Xm are the roots of Jn and µ = Xm/a . Also, 2( , ,0) cos 2 ( )cosn n

nA J nμ

μ

ρ φ ρ φ ρμ φΦ = =∑∑

It is evident that n = 2 and that 2

2 2 ( )A Jμμ

ρ ρμ=∑

which is the Fourier-Bessel expansion of ρ2. From Table 2.1, property (h),

22 22 2

3 30

2 2( )[ ( )] ( )

a

m m

A J da J a X J Xμ ρ ρμ ρ

μ= =∫

Thus

2 22

1 3

( / )( , , ) 2 cos 2 exp( / )( )

mm

m m m

J X at X kt aX J Xρρ φ φ

∞

=

Φ = −∑

where 2 ( ) 0mJ X = . Prob. 2.6

2

2

U Ux t

∂ ∂=

∂ ∂