1. a) B ( ρ ,ϕ , z )=ρ (sinϕ ) z3 aρ+ρ3 (cosϕ ) z aϕ+ρ2 (sinϕ ) z2 z

Divergence in cylindrical Co-ordinates

¿ F=1ρ∂(ρ Fρ)

∂ ρ+ 1ρ∂Fϕ

∂ϕ+∂ F z

∂ z

∇ .B= 1ρ∂( ρ ρ ( sinϕ ) z3)

∂ ρ+ 1ρ∂(ρ¿¿3 (cosϕ ) z)

∂ϕ+∂

(ρ¿¿2 (sinϕ ) z2)∂z

¿¿

¿2 sinϕ z3− ρ2 zsinϕ+2ρ2 zsinϕ¿2 z3 sinϕ+ρ2 zsinϕ

Curl in cylindrical Co-ordinates

Curl F= aρ [ 1ρ ∂ F z

∂ϕ−

∂Fϕ

∂ z ]+ aϕ [ ∂ Fρ

∂ z−

∂F z

∂ ρ ]+ az [ 1ρ ∂∂ ρ

( ρFϕ )−1ρ

∂∂ϕ

(Fρ )]

∇ XB= aρ [ 1ρ ∂(ρ2 (sinϕ ) z2)∂ϕ

−∂ (ρ3 (cosϕ ) z )

∂ z ]+ aϕ [ ∂( ρ (sinϕ ) z3)∂z

−∂ (ρ2 (sinϕ ) z2 )

∂ ρ ]+az[ 1ρ ∂∂ ρ

( ρ ρ3 (cosϕ ) z )− 1ρ

∂∂ϕ

( ρ ( sinϕ ) z3 )]

∇ XB= aρ [ ρ z2cosϕ−ρ3 cosϕ ]+aϕ [ z2 ρsinϕ ]+az [4 z ρ2 cosϕ−z3 cosϕ ]

b) At ρ=3 , ϕ=900 , z=2B=24 aρ+0 aϕ+36 z

∇ .B=2 (2 )3 sin 900+(3 )22sin 900

¿16+18=34

∇ XB= aρ [ ρ z2cosϕ−ρ3 cosϕ ]+aϕ [ z2 ρsinϕ ]+az [4 z ρ2 cosϕ−z3 cosϕ ]Ascos900=0∇ XB= aϕ [22 (3 )sin 900 ]=12 aϕ

c) For this to be an acceptable magnetic flux density in a source free region of space it should satisfy the Maxwell’s equations.But we see that ∇ .B≠0Hence it does not satisfy the Maxwell’s equation and is not an acceptable magnetic flux density.

d) At the prescribed point ∇ XB is in the ϕ direction. In general curl of a vector F need not be orthogonal to F.

But in this case we see that the component of B in ϕ direction is zero at the prescribed point and hence it lies in

the ρ z plane and the curl is in ϕ direction and hence orthogonal to the ρ z plane.By definition curl describes the infinitesimal rotation of a 3D vector field. At every point in the field the curl of that field is represented by a vector that characterizes the rotation at that point. Hence at the prescribed point the given vector field has a rotation of magnitude 12 and the axis of rotation is ϕ direction. Also we can note

that the component of B itself in ϕ direction is zero but it is twice the magnitude of curl in ρ direction and thrice the magnitude of curl in z direction.

2. a) Solution in spherical Co-ordinatesf 1 (r )=A1 jn (βr )+B1 yn (βr )Or

f 2 (r )=C1hn(1) (βr )+D1hn

(2) (βr )

Wave function for radial waves in spherical co-ordinates

hn(1) (βr )= jn (βr )+ j yn(βr ) for +r travel (outgoing wave)

Where hn(1) (βr )−Spherical Henkel functionof first kind

jn (βr )∧ yn (βr )−Spherical Bessel functionof first∧second kind respectively

b) Relation to their counterpart in cylindrical co-ordinates

jn (βr )=√ π2 βr

Jn+12

(βr )where Jn+12

(βr ) is the regular Bessel functionof first kind

yn (βr )=√ π2βr

Yn+12

(βr )whereYn+12

(βr ) is theregular Bessel f unctionof second kind

hn(1) (βr )=√ π

2 βrH

n+12

(1) (βr )where Hn+ 12

(1 ) (βr ) isthe Regular Henkel functionof first kind

Matlab code to generate special functions

syms xfor nu = [0, 2, 4, 8] ezplot((pi./(2.*x)).^(1/2).*besselj(nu+0.5, x), [0, 50]) hold onendaxis([0, 25, -0.5, 1.5])grid onylabel('j_p(x)')legend('j_0','j_2','j_4','j_8', 'Location','Best')title('Spherical Bessel Functions')hold off

syms xfor nu = [0, 2, 4, 8] ezplot((pi./(2.*x)).^(1/2).*bessely(nu+0.5, x), [0, 50]) hold onendaxis([0, 25, -0.5, 1.5])grid onylabel('y_n(x)')legend('y_0','y_2','y_4','y_8', 'Location','Best')title('Spherical Bessel Functions of second kind')hold off

syms xfor nu = [0, 2, 4, 8] ezplot(abs((pi./(2.*x)).^(1/2).*(besselj(nu+0.5,x) + 1i*bessely(nu+0.5,x))), [0, 50]) hold onendaxis([0, 25, -0.5, 1.5])grid onylabel('h_p(x)^(^1^)')legend('h_0^(^1^)','h_2^(^1^)','h_4^(^1^)','h_8^(^1^)', 'Location','Best')title('Spherical Hankel functions of the first kind (Absolute Value)')hold off

Behaviour of Spherical Bessel functions of first kind:We can see from the graph that j0 starts at one. The first spherical Bessel function j0 ( x )is also known as the (unnormalized) sinc function. All other functions start at zero. The behaviour is similar to regular Bessel function of first

kind but is scaled by the √ π2 βr

factor. The functions decay as a sine or cosine with the rate of decay proportional to 1

√x

Behaviour of Spherical Bessel functions of second kind:It can be seen from the graph that the functions have a singularity at the origin. They are multivalued. The behaviour is

similar to Regular Bessel functions of second kind but is scaled by the √ π2 βr

factor.

Behaviour of Spherical Hankel functions:From the figure we can see that spherical Hankel functions unlike spherical Bessel functions diverge towards the origin. Furthermore for small arguments, higher orders increase towards the origin with a larger slope. On the other hand for large arguments they decay similarly for all orders

d) Small and large argument behaviour:

As x→∞

We have

jn ( x )=1xsin(x−nπ

2 )yn ( x )=−1

xcos( x−nπ

2 )hn1 ( x )=−i

xexp [i {x−( nπ2 )}]

As x→0

We have

jn ( x )= xn

(2n+1 )‼ (n integral)

yn ( x )=− (2n−1 )‼xn+1 (n integral)

hn1 ( x )=−i

(2n−1 )‼xn+1

3. a) i) Antennas are sometimes coated with dielectric materials to change various properties. Ex-Parabolic Reflector Antennas are coated with different materialsii) An optical coating is one or more thin layers of material deposited on an optical component such as a lens or mirror, which alters the way in which the optic reflects and transmits light. Ex-Sun Glasses which can filter UV Rays.

b) E=E0 ( x+ j y )e− jkz

H=E0η

(− j x+ y )e− jkz

c) Steps to solve multi-layered problems using wave impedance method

i) Start from region n: Γ ( z )=0=¿Z ( z )=ηn

ii) At interface of n and n-1 regions: Z ¿iii) Construct Γ ¿ in n-1 region: Γ ¿iv) Transport Γ todn−1

+¿ ¿ in n-1 region: Γ ¿v) Compute Z ¿vi) Construct Z ¿vii) Repeat the above steps to finally determine

Z ¿

viii) Compute Em−¿¿from : Γ ( z )=

Em−¿

Em+¿e2 γz ¿

¿ in region 1

ix) Apply continuity of E and H fields to every interface to obtain Em−¿¿ and Em

+¿ ¿ in each region

Γ ( z )=−1=¿Z ( z )=0

Z ¿

Γ ¿

Γ ¿

Γ ¿

Z ¿

Z ¿

Γ ¿

Γ ¿

Γ ¿

Γ ¿

Γ ¿

Γ ¿

Z ¿

Z ¿

Z ¿

Z ¿

Γ ¿ Reflection coefficient in region 1

¿

(η2η3 (1−e−2 γ 3d3 )+η22 (1+e−2 γ 3d3 ))+(η2η3 (e−2 γ 2d2−e−2 (γ2d 2+γ 3d3))−η2

2 (e−2 γ 2d2+e−2(γ 2d2+ γ3d 3)))(η3 (1−e−2 γ3d 3 )+η2 (1+e−2 γ3d 3 ))−(η3 (e−2 γ2d2−e−2(γ 2d2+ γ3d3 ))−η2 (e

−2 γ2d2+e−2 (γ2d 2+γ 3d3)))−η1

(η2η3 (1−e−2 γ 3d3 )+η22 (1+e−2 γ 3d3 ))+(η2η3 (e−2 γ 2d2−e−2(γ2d 2+γ 3d3 ))−η2

2 (e−2 γ 2d2+e−2(γ 2d2+γ3d 3)))(η3 (1−e−2 γ 3d3 )+η2 (1+e−2 γ 3d3 ))−(η3 (e−2 γ2d 2−e−2(γ 2d2+ γ3d 3))−η2 (e

−2 γ2d 2+e−2(γ2d 2+γ 3d3 )))+η1

Whereη1=√ μ0ϵ 1

η2=√ μ0ϵ 2

η3=√ μ0ϵ 3

Beacusewe assume lossless regionγ= jβ Hence γ2= j β2=ω√μ0ϵ 2

γ 3= j β3=ω√μ0ϵ 3

d) The polarization of the reflected wave is Right hand Elliptical polarization as the regions are terminated by an infinite PEC which flips the polarization from Left hand to Right hand and because of multiple reflections between interfaces the

circular polarization becomes elliptical polarization as the phase between the components is not an integral multiple of π2

anymore.

e) Total Electric field in region 1

Transmissionco−efficient T12=1+Γ 12T 23=1+Γ 23

Etotal=T12T 23E i

Similarlywe can writeH total=T12T23H i

Js=n (H i+H r )=2 n X H i

2(− z )(T 12T23)E0η

(− j x+ y ) e− jkz

Js=2T 12T23 E0

η1( j y+ x ) e− jkz

T 12=1+Γ ¿

f) When the permittivities are the same:

η1=η2=η3=η

γ 1=γ2=γ3=γ Lossless∧hence γ= jβ

(η2 (1−e−2 γ d3 )+η2 (1+e−2 γ d3 ))+(η2 (e−2 γ d2−e−2 γ (d2+d3 ))−η2 (e−2 γ d2+e−2 γ (d 2+d3 ) ))(η (1−e−2 γd 3 )+η (1+e−2 γ d3 ))−(η (e−2 γd 2−e

−2 γ (d2+d3 ) )−η (e−2 γd 2+e−2 γ (d2+d3 )) )

−η

(η2 (1−e−2 γ d3 )+η2 (1+e−2 γ d3 ))+(η2 (e−2 γd 2−e−2 γ (d2+d3 ) )−η2 (e−2 γd2+e

−2 γ (d2+d 3) ))(η (1−e

−2 γ d3 )+η (1+e−2 γ d3 ))−(η (e−2 γ d2−e−2 γ (d 2+d 3) )−η (e−2 γ d2+e−2 γ (d2+d3 ) ))

+η

¿ −4 η2e−2 γ (d2+d3 )

4 η2

Γ ¿ Reflection co-efficient It is as expected because when the permittivities are the same we have a single region

terminated by a PEC. The reflection co-efficient of PEC is -1 and we can write the reflection coefficient at z '=d2+d3

using Γ ( z ' )=Γ ( z ) e2 γ ( z− z' ) which is same as what we got above.

The polarization of the reflected wave will be RHCP. This is expected because we have a PEC boundary

Js=2(1−e

−2 jβ (d2+d3))E0η1

( j y+ x ) e− jkz

We can write this as we now have a single region of length of d2+d3 and hence reflection coefficient of region 2 doesn’t contribute. We only haveΓ ¿.

4) For the PEC circular plate

Ei=axE0 ejβz

H i=− ay

E0η

e jβz

Js=2 n X H i∨( z=0 )=ax

2E0η

J x=2 E0η

J y=0 J z=0

Nθ=∬ (J xcosθ s cosϕs )ejβ r 'cosψds '

r ' cosψ=x ' sinθs cosϕ s+ y ' sin θs sinϕs

d s '=ρ' d ρ 'dϕ '

Nθ=∬ (J xcosθ s cosϕs )ejβ (x 'sin θscosϕ s+ y 'sin θssin ϕs) ρ'd ρ' dϕ '

¿2E0ηcosθ scos ϕs∫

0

2π

∫0

a

e jβ( x' sinθs cosϕs+ y ' sinθs sinϕ s)ρ' d ρ'dϕ '

x '=ρ' cosϕ'

y '=ρ' sinϕ '

x ' sin θ scos ϕs+ y ' sin θ s sinϕs=ρ' (cos ϕ ' sinθ scos ϕs+sinϕ' sin θ ssin ϕs )

¿ ρ' (Acos ϕ'+Bsinϕ ' )where A=sin θ s cosϕsB=sin θ ssinϕ s

¿ ρ' √A2+B2( A

√A2+B2cosϕ '+

B

√A2+B2sin ϕ' )

Let cos ϕ0=A

√A2+B2sinϕ0=

B

√A2+B2

¿>x ' sinθ s cosϕs+ y' sin θs sinϕs=ρ' √A2+B2 cos (ϕ '−ϕ0)

Nθ=2 E0ηcos θs cos ϕs∫

0

a

[ I ] ρ' dρ'

where I=∫0

2π

e jβρ ' √ A2+B2cos (ϕ '−ϕ 0)dϕ '

e jβρ ' √ A2+B2cos (ϕ '−ϕ0) is a periodic function of ϕ 'with period 2π and the integration of this function yields a bessel

function.

Thuswe canwrite the solution for Nθ=4 π a2E0

ηcosθ s cosϕs ¿

where J1is the bessel functionof first kind

Nϕ=∬ (−J x sinϕs+J y cosϕs )e+ jβ r 'cosψ ds '

Nϕ=2E0ηsinϕs∬ e+ jβ r 'cosψ ds '

Solving it similarly we get

Nϕ=4 π a2 E0

ηsin ϕs¿

Eθ=− jβ e− jβr

4 πr(Lϕ+ηN θ )

Eθ=− jβ e− jβr

4 πr¿

Eθ=− jβ a2 E0 e

− jβr

rcosθ s cosϕs ¿

Eϕ=jβ e− jβr

4 πr(Lθ−η Nϕ )

Eϕ=jβ e− jβr

4 πr¿

Eϕ=− jβ a2E0e

− jβr

rsinϕs ¿

We can finally replace A and B and write

Eθ=− jβ a2 E0 e

− jβr

rcosθ s cosϕs [ J 1 (βasin θ s )

βasinθ s]

Eϕ=jβ a2E0 e

− jβr

rsinϕs[ J1 (βasinθ s )

βasin θs]

|E|=√|Eθ|2+|Eϕ|

2

|E|= βa2E0r √(cosθ scos ϕs )

2+sinϕs2[ J1 (βasinθ s )

βasin θs]

For the square Plate

Ei=axE0 ejβz

H i=− ay

E0η

e jβz

M s=−2n X E i∨( z=0 )=a y2 E0M y=−2 E0M x=0M z=0θ s=0

Lθ=∬(M xcosθ s cosϕs+M y cosθ ssin ϕs−M z sinθ s)ejβ r' cosψds '

Lϕ=∬ (−M xsinϕ s+M ycos ϕs )e+ jβ r' cosψds '

Lθ=−2 E0 cosθ s sinϕs∬ejβ (x 'sin θscos ϕs+ y 'sin θssin ϕ s)dx ' dy '

Lθ=−2 E0b2cosθ s sinϕs[ sin ( X )

X ] [ sin (Y )Y ]

X=βb2sin θ scos ϕs

Y= βb2sinθs sinϕs

In the same manner we can write

Lϕ=−2E0b2 cosϕs[ sin (X )

X ][ sin (Y )Y ]

Eθ=− jβ e− jβr

4 πr(Lϕ+ηN θ )

Eθ=jβ e− jβr

2 πr (E0b2 cosϕs [ sin (X )X ][ sin (Y )

Y ])

Eϕ=jβ e− jβr

4 πr(Lθ−η Nϕ )

Eϕ=− jβ e− jβr

2πr (E0b2cosθ s sinϕs[ sin (X )X ] [ sin (Y )

Y ])|E|=√|Eθ|

2+|Eϕ|2

|E s|=β2πr

b2 E0[ sin (X )X ] [ sin (Y )

Y ]√(cosθ ssinϕ s)2+cos2ϕs

Let’s write the total scattered field for the entire system

Eθ1=jβ e− jβ (r+dcosα )

2πr (E0b2 cosϕs[ sin (X )X ][ sin (Y )

Y ])→Square plate1

Eθ2=− jβ a2E0 e

− jβr

rcosθ s cosϕs[ J 1 (βasin θs )

βasinθ s]→Circular Disc

Eθ3=jβ e− jβ (r−dcosα )

2πr (E0b2cos ϕs[ sin (X )X ] [ sin (Y )

Y ])→Square plate2

Note cosα=sinθsinϕTotal Eθ=Eθ1+Eθ2+Eθ3

Eθ=jβ E0 cosϕs

re− jβr (( b2π [ sin (X )

X ][ sin (Y )Y ] (cos ( βdsinθ ssin ϕs )))−(a2 cosθ s[ J1 (βasinθ s )

βasin θs]))

Eϕ1=− jβ e− jβ (r+dcosα)

2πr (E0b2 cosθ ssin ϕs[ sin (X )X ] [ sin (Y )

Y ])→Square plate1

Eϕ2=jβ a2 E0 e

− jβr

rsin ϕs[ J1 (βasinθ s )

βasinθ s]→Circular disc

Eϕ3=− jβ e− jβ(r−dcosα )

2 πr (E0b2 cosθs sinϕs [ sin (X )X ][ sin (Y )

Y ])→Square plate2

Eϕ=Eϕ1+Eϕ2+Eϕ3

Eϕ=jβ E0 sinϕs

re− jβr (−( b2π [ sin (X )

X ][ sin (Y )Y ]cosθ s (cos (βdsin θs sinϕs ) ))+(a2[ J1 (βasinθ s )

βasinθ s]))

|E s|=√|Eθ|2+|Eϕ|

2

¿β E0r √(cos ϕs(( b2π [ sin (X )

X ][ sin (Y )Y ] (cos ( βdsinθ ssin ϕs )))−(a2 cosθ s[ J1 (βasinθ s )

βasin θs])))

2

+(sin ϕs(−b2

π [ sin (X )X ][ sin (Y )

Y ]cosθ s (cos ( βdsinθ ssinϕ s )))+(a2[ J1 (βasinθ s )βasin θs

]))2

RCS

σ 3D=limr→∞ [4 π r2

|Es|2

|E i|2 ]

σ 3D=4 π β2(cosϕs(( b2π [ sin ( X )X ] [ sin (Y )

Y ](cos (βdsin θs sinϕs ) ))−(a2 cosθ s[ J 1 ( βasinθ s )βasinθ s

])))2

+(sinϕs(−( b2π [ sin ( X )X ] [ sin (Y )

Y ]cosθs (cos (βdsinθ s sinϕs )))+(a2[ J 1 (βasin θs )βasin θ s

])))2

Monostatic θ s=00

σ 3D=4 π β2¿¿

σ 3D=4 π β2(( b2π )2

+( a22 )2

−( a2b2π ))Bi-static θ s=π

σ 3D=4 π β2((cos ϕs(( b2π )+(a2 [0.5 ])))2

+(sin ϕs(−( b2

π )+(a2 [0.5 ] )))2

)σ 3D=4 π β2(( b2π )

2

+( a22 )2

+( a2b2π ))Note: The “0.5” comes from the small argument approximation for Bessel function.

Matlab code for numerical computation of RCS

The values obtained from Matlab have been confirmed with values obtained from the RCS formulas derived above and found to be matching exactly.

c=3*10^8;f=6*10^9;lambda=c/f;beta=2.*pi./lambda;a=0.15; % 0 or 0.15 based on case1 case2 or case3b=0; % 0 or 0.2 based on case1 case2 or case3d=0.35;k=1;theta=0; %Monostatic theta=0 bistatic theta=piphi=pi/2;X=(beta.*b./2).*sin(theta).*cos(phi);Y=(beta.*b./2).*sin(theta).*sin(phi);Etheta=((cos(phi)).*((k).*((b.^2/pi).*(sinc(X/pi)).*(sinc(Y/pi)).*cos(beta*d.*sin(theta).*sin(phi)))-(a.^2.*cos(theta).*(besselj(1,(beta.*a.*sin(theta)))./(beta.*a.*sin(theta))))));Ephi=((sin(phi)).*((k).*(-(b.^2/pi).*(sinc(X/pi)).*(sinc(Y/pi)).*cos(beta*d.*sin(theta).*sin(phi)))+(a.^2.*cos(theta).*(besselj(1,(beta.*a.*sin(theta)))./(beta.*a.*sin(theta)))))); sigma=4*pi*(beta^2).*((Etheta.^2)+(Ephi.^2));

Matlab code for generation scattered fields and polar plots

a=0.15;b=0;d=0.35;

k=1; %(1i*beta).*e^(-1i*beta*r)/rtheta=-pi/2:pi/500:pi/2;phi=pi/2;X=(beta.*b./2).*sin(theta).*cos(phi);Y=(beta.*b./2).*sin(theta).*sin(phi);Etheta=((cos(phi)).*((k).*((b.^2/pi).*(sinc(X/pi)).*(sinc(Y/pi)).*cos(beta*d.*sin(theta).*sin(phi)))-(a.^2.*cos(theta).*(besselj(1,(beta.*a.*sin(theta)))./(beta.*a.*sin(theta))))));Ephi=((cos(phi)).*((k).*(-(b.^2/pi).*(sinc(X/pi)).*(sinc(Y/pi)).*(cos(theta)).*cos(beta*d.*sin(theta).*sin(phi)))+(a.^2.*cos(theta).*(besselj(1,(beta.*a.*sin(theta)))./(beta.*a.*sin(theta))))));Etot=sqrt((Etheta).^2+(Ephi).^2);[E_max,theta_max]=max(Etot);figure(1)Enorm=abs((Etot)./E_max);

plot(theta,Enorm)grid on%view(90,-90)

c) Case1: a = 15 cm, b = 0 (no solar panels)

Monostatic caseθ s=θi=00

σ 3D=25.1151m2

Bistatic case ( forward−z direction )θ s=π

σ 3D=25.1151m2

Case2: a = 0, b = 20 cm (no satellite)

Monostatic caseθ s=θi=00

σ 3D=32.1699m2

Bistatic case ( forward−z direction )θ s=π

σ 3D=32.1699m2

Case3: a = 15 cm, b = 20 cm (both the satellite and two solar panels)

Monostatic caseθ s=θi=00

σ 3D=0.4361m2

Bistatic case ( forward−z direction )θ s=π

σ 3D=114.13m2

5. a) The dual of the problem is to find the scattered field, induced electric current and RCS of a PEC circular cylinder impinged by TMz plane wave. The reason being by using duality theorem we can replace M by J, PMC by a PEC and instead of a TEz wave we have a TMZ wave. This is possible from the boundary conditions associated with a PEC on the surface of the cylinder and from duality we infer that Magnetic currents arising from magnetic charges can be replaced by electric current arising from electric charges in the dual problem without affecting the solution.

b) Scattered fields

H zs=−E0 ∑

n=−∞

n=∞

j−n J n (βa )H n

(2) (βa )H n

(2) (βρ )e jnϕ

Eρs=

−E0jωϵ

1ρ∑n=−∞

n=∞

n j−n+1 J n (βa )H n

(2) (βa )H n

(2 ) (βρ ) e jnϕ

Eϕs=

β E0jωϵ

∑n=−∞

n=∞

j−n J n(βa)

H n(2)(βa)

H n2' (βρ ) e jnϕ

c) far zone scattered field

H zs=−H0√ 2 jπβ

e− jβρ

√ρ ∑n=−∞

n=∞ J n (βa )Hn

(2 ) (βa )e jnϕ

Eρs=

−E0ωϵ

1ρ √ 2 j

πβe− jβρ

√ ρ∑n=−∞

n=∞

nJn (βa )H n

(2) (βa )e jnϕ

Eϕ=β E0jωϵ

∑n=−∞

n=∞

j−n J n (βa )Hn

(2) (βa )12

¿¿

Eϕ=−β E0ωϵ √ 2 j

πβe− jβρ

√ ρ∑

n=−∞

n=∞ J n (βa )H n

(2) (βa )e jnϕ

In the far field Eϕcomponent goes to zero as it’s a localized plane wave.

d) M s= az

2H0

πaωϵ∑n=−∞

n=∞

j−n e jnϕ

H n(2)(βa)

The current is in z direction

e) σ 2D=2 λπ |∑n=0

+∞

ϵ n

Jn (βa )H n

(2) (βa )cos (nϕ )|

2

ϵ n={1n=02n≠0

f)

lambda=1;beta=2*pi/lambda;m=1.25;a=m*lambda;phi=0; %rho=1:1:100;n=1:1:100;sumterm=(1i.^(-n)).*(besselh(n,2,beta.*rho).*besselj(n,beta*a)./besselh(n,2,beta*a));plot(n,sumterm)grid title('Convergence evaluation a=1.25\lambda')axis([0 50 -0.5 0.5])

We see that that the convergence is happening at around a value of 15. So we need -15 to 15 i.e. we need 31 terms for convergence.

m=1.25;k=15;a=m*lambda;phi=pi; %negative x direction

rho=-20*a:a/360:-a; Hz=0;for n=-k:1:k Hz=Hz + (-1).*(1i.^(-n)).*exp(1i*n*phi).*besselh(n,2,beta.*rho).*besselj(n,beta*a)./besselh(n,2,beta*a);end[Hz_max,phi_max]=max(Hz);Hz_norm=Hz./Hz_max;plot(rho,Hz_norm)hold onphi=0; %positive x directionrho=a:a/360:20*a; Hz=0;for n=-k:1:k Hz=Hz + (-1).*(1i.^(-n)).*exp(1i*n*phi).*besselh(n,2,beta.*rho).*besselj(n,beta*a)./besselh(n,2,beta*a);end[Hz_max,phi_max]=max(Hz);Hz_norm=Hz./Hz_max;plot(rho,Hz_norm)grid onaxis([-30 30 -1.5 1.5])title('Scattered Magnetic Field (Normalized)')

Observations:The magnetic field decays almost exponentially in both positive and negative x directions.There exists a discontinuity near the origin as we approach the surface of the cylinder which is as expected as there is no field inside the PMC cylinder.

lambda=1;beta=2*pi/lambda;m=1.25;k=15;a=m*lambda;phi=pi; %negative x directionrho=-20*a:a/360:-a; Erho_temp=0;for n=-k:1:k Erho_temp=Erho_temp + (1./rho).*(-1).*((n).*(1i.^(-n+1))).*exp(1i*n*phi).*besselh(n,2,beta.*rho).*besselj(n,beta*a)./besselh(n,2,beta*a);end Etotal=sqrt(((Erho_temp).^2));

[Etotal_max,phi_max]=max(Etotal);E_norm=Etotal./Etotal_max;plot(rho,E_norm);hold on phi=0; %positive x directionrho=a:a/360:20*a; Erho_temp=0;for n=-k:1:k Erho_temp=Erho_temp + (1./rho).*(-1).*((n).*(1i.^(-n+1))).*exp(1i*n*phi).*besselh(n,2,beta.*rho).*besselj(n,beta*a)./besselh(n,2,beta*a);end Etotal=sqrt(((Erho_temp).^2));[Etotal_max,phi_max]=max(Etotal);E_norm=Etotal./Etotal_max;plot(rho,E_norm);grid ontitle('E field');hold off

Observations:

We see that the field is more uniformly distributed in the negative x direction (forward scattering) and densely distributed in the positive x direction (back scattering). The field decays exponentially in the –x direction but in the +x direction we can see some spikes and the decay is non uniform.

lambda=1;beta=2*pi/lambda;m=1.25;k=15;a=m*lambda;phi=pi; %negative x directionrho=-100*a:a/360:-a; Hz=0;for n=-k:1:k Hz=Hz + (-1).*(1i.^(-n)).*exp(1i*n*phi).*besselh(n,2,beta.*rho).*besselj(n,beta*a)./besselh(n,2,beta*a);end[Hz_max,phi_max]=max(Hz);Hz_norm=Hz./Hz_max;plot(rho,Hz_norm)hold onphi=0; %positive x direction

rho=a:a/360:100*a; Hz=0;for n=-k:1:k Hz=Hz + (-1).*(1i.^(-n)).*exp(1i*n*phi).*besselh(n,2,beta.*rho).*besselj(n,beta*a)./besselh(n,2,beta*a);end[Hz_max,phi_max]=max(Hz);Hz_norm=Hz./Hz_max;plot(rho,Hz_norm)grid onhold on lambda=1;beta=2*pi/lambda;m=1.25;k=20;a=m*lambda;phi=pi; %negative x directionrho=-100*a:a/360:-a; Hz=0;for n=-k:1:k Hz=Hz + (exp(-1i*beta.*rho)./sqrt(rho)).*(-1).*(1i.^(-n)).*exp(1i*n*phi).*besselj(n,beta*a)./besselh(n,2,beta*a);end[Hz_max,phi_max]=max(Hz);Hz_norm=Hz./Hz_max;plot(rho,Hz)hold onphi=0; %positive x directionrho=a:a/360:100*a; Hz=0;for n=-k:1:k Hz=Hz + (exp(-1i*beta.*rho)./sqrt(rho)).*(-1).*(1i.^(-n)).*exp(1i*n*phi).*besselj(n,beta*a)./besselh(n,2,beta*a);end[Hz_max,phi_max]=max(Hz);Hz_norm=Hz./Hz_max;plot(rho,Hz)grid on

Here I have plotted the exact H field and extended the plot to a far distance. In the same figure I have also plotted the far field approximation of the H field. We can observe that the difference between them decreases as the distance increases. Hence we confirm that the given approximation works well at large distances.

g) lambda=1;beta=2*pi/lambda;m=1.25;k=350;a=m*lambda;phi=0:pi/3600:2*pi;Hz=0;for n=-k:1:k Hz=Hz + exp(1i*n*phi).*besselj(n,beta*a)./besselh(n,2,beta*a);end[Hmax,phimax]=max(Hz);Hnorm=abs(Hz)./Hmax;plot(phi,20*log(Hnorm))gridtitle('Scattered H field (in dB)')

Discussions:

We can observe from the plot that most of the field is concentrated in the direction opposite to incident field. This happens because there is a normal incidence and hence reflected field will comprise most part of the scattered field with significantly less amount of field in other directions.

h) Matlab code to find convergence lambda=1; a=1.25*lambda; beta=2*pi/lambda; n=1:1:100; sumterm=(besselj(n,beta*a)./besselh(n,2,beta*a)); plot(n,sumterm) grid axis([1 20 -0.5 1.5]) title('Convergence evaluation a=1.25\lambda')

Matlab code to calculate numerical values of Monostatic and bistatic RCS

lambda=1;beta=2*pi/lambda;a=1.25*lambda;phi=0; %phi=pi for bistatic case and phi=0 for monostatic casesumterm=0;for n=-14:1:14sumterm=sumterm+((besselj(n,beta*a)./besselh(n,2,beta*a)).*exp(1i*n*phi));endsigma=4./beta.*((abs(sumterm)).^2);

Monostatic RCS σ2 D=51.6068m

Bistatic RCS σ2D=3.9619m

Discussions:

We observe from the convergence plot that the series converges at around n=13. So we choose n=14 for the summation limits and find the RCS. We can note that the Monostatic RCS is numerically greater than bi-static RCS. This is expected as the back scattered field is less compared to the forward scattering.

i)

lambda=1;a=1.25*lambda;beta=2*pi/lambda;n=1:1:100;sumterm=(1i.^(-n))./(besselh(n,2,beta*a));plot(n,sumterm)gridtitle('Convergence evaluation a=1.25\lambda')

lambda=1;a=1.25*lambda;beta=2*pi/lambda;phi=0:pi/360:2*pi;sumterm=0; for n=-20:1:20 sumterm=sumterm+((1i.^(-n)).*((exp(1i*n.*phi)./(besselh(n,2,beta*a)))));endplot(phi*180/pi,sumterm);gridtitle('Induced magnetic current')

Physical optics approximationHere we assume that the surface of a cylinder is locally infinite for the wave. Hence we have

M s=cosϕ e− jβρcosϕ

Where ϕ=π2

¿ 3π2

Observations:

We can see that the physical optics approximation agrees with the Exact current on the main lobe (between

¿ π2

¿ 3 π2

) but we loose some information on the side lobes when we do the physical optics approximation

and hence its not in complete agreement.