ECE 5324/6324 Spring 2017 Assignment 1

2.1-1 We start with equation (2-1)

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Then use the relationship between the transient variables on the phasor (complex) variables

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Substitute in

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One can always interchange linear operators, such as the real part operation and differential operators.

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Taking the derivatives we obtain

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Now we note that this equation is valid for any value of the time variable, t. Letting ! we find

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Letting ! we find

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where we have used the fact that the imaginary part of j times a complex number is equal to the real part. Since the real and imaginary parts of the equation are valid, the complex equation is valid

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∇×E = -∂B∂t

E = Re Ee jωt( )B = Re Be jωt( )

∇×Re Ee jωt( )= -∂∂tRe Be jωt( )

Re ∇× Ee jωt( )( )= Re − ∂∂tBe jωt( )⎛

⎝⎜⎞⎠⎟

Re e jωt∇×E( )= Re −e jωt jωB( )ωt = 0

Re ∇×E( )= Re − jωB( )

ωt = −π / 2

Re − j∇×E( )= Re −ωB( )Im − jj∇×E( )= Im − jωB( )Im ∇×E( )= Im − jωB( )

∇×E= − jωB

2.2-X1 (a) We can solve the continuity equations for the charges

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Plugging in the given current response functions we find the charge response functions for the medium

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Now, assuming homogenous media, the divergence operator need only apply to the field

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(b) Just substituting the given current response functions and the charge response functions from part (a) into the provided Maxwell’s equations, we obtain

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Simplifying and segregating the fields to the left side and the impressed currents to the right we obtain

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This is a set of equations without any unknown charge dynamics. We can solve these. One way is to find wave equations for each field separately, which is part (c).

!qev = − 1jω

∇⋅ !J

!qmv = − 1jω

∇⋅ !Jm

!qev E,H( )= − 1jω

∇⋅Ji −∇⋅ !ε − ε0( )E⎡⎣ ⎤⎦

!qmv E,H( )= − 1jω

∇⋅Jmi −∇⋅ !µ − µ0( )H⎡⎣ ⎤⎦

!qev E,H( )= − 1jω

∇⋅Ji − !ε − ε0( )∇⋅E

!qmv E,H( )= − 1jω

∇⋅Jmi − !µ − µ0( )∇⋅H

ε0∇⋅E = − 1jω

∇⋅Ji − !ε − ε0( )∇⋅E

µ0∇⋅H = − 1jω

∇⋅Jmi − !µ − µ0( )∇⋅H

∇×H − jωε0E = Ji + jω !ε − ε0( )E−∇×E− jωµ0H = Jmi + jω !µ − µ0( )H

− jω !ε∇⋅E = ∇⋅Ji− jω !µ∇⋅H = ∇⋅Jmi

∇×H − jω !εE = Ji−∇×E− jω !µH = Jmi

(c) Taking the curl of the curl equations, we obtain

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where again we have used the assumption of homogenous media, so that the curl operates only on the field variable. Using the usual vector identity and substituting

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Using the divergence equations above, and rearranging we find the driven wave equations

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One is free to excite the electric and magnetic currents separately and combine the resulting fields. Also, one can solve either the electric field equation, or the magnetic field equation, and obtain the other field using a Maxwell curl equation, typically this will be done in a source free region.

−∇×∇×E− jω !µ∇×H = ∇× Jmi ∇×∇×H − jω !ε∇×E = ∇× Ji

∇2E−∇ ∇⋅E( )− jω !µ Ji + jω !εE( ) = ∇× Jmi −∇2H +∇ ∇⋅H( )− jω !ε −Jmi − jω !µH( ) = ∇× Ji

∇2E+ω 2 !µ !εE = ∇× Jmi −1jω !ε

∇ ∇⋅Ji( )+ jω !µJi

∇2H +ω 2 !ε !µH = −∇× Ji −1jω !µ

∇ ∇⋅Jmi( )+ jω !εJmi

2.3-2 (a) The magnetic field of an ideal dipole is given by equation (2-73a)

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To use equation (2-63) we need to take the curl. When the vector field has only a 𝜙-component, and depends only on r and θ, equation (C-35), for the curl in spherical coordinates, reduces to

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Using this and our expression for H in equation (2-63)

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Simplifying a bit

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Here we have used the first of the following two results. These results are worth remembering

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Taking the derivatives

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Simplifying and rearranging

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H = IΔz4π

jβ 1+ 1jβr

⎛⎝⎜

⎞⎠⎟e− jβr

rsinθϕ

∇×H = r 1r sinθ

∂∂θ

Hφ sinθ( )− θ 1r∂∂r

rHφ( )

E = 1jωε

∇×H = 1jωε

IΔz4π

jβ r 1r sinθ

∂∂θ

1+ 1jβr

⎛⎝⎜

⎞⎠⎟e− jβr

rsinθ sinθ

⎛⎝⎜

⎞⎠⎟− θ 1

r∂∂r

r 1+ 1jβr

⎛⎝⎜

⎞⎠⎟e− jβr

rsinθ

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

E = IηΔz4π

r 1r sinθ

1+ 1jβr

⎛⎝⎜

⎞⎠⎟e− jβr

r∂∂θsin2θ − θ 1

rsinθ ∂

∂r1+ 1

jβr⎛⎝⎜

⎞⎠⎟e− jβr

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

βη=ω εµ

µε

=ωε and βη =ω εµ µε=ωµ

E = IηΔz4π

−r 1r sinθ

1+ 1jβr

⎛⎝⎜

⎞⎠⎟e− jβr

r2sinθ cosθ − θ 1

rsinθ − jβ 1+ 1

jβr⎛⎝⎜

⎞⎠⎟e− jβr − 1

jβr2e− jβr

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

E = IηΔz4π

e− jβr

r1r− j 1

βr2⎛⎝⎜

⎞⎠⎟2cosθ r + jβ + 1

r− j 1

βr2⎛⎝⎜

⎞⎠⎟sinθθ

⎡

⎣⎢

⎤

⎦⎥

Finally, using dimensionless groups, except in the pre-factor (which now has the obviously correct SI units for electric field, volts per meter) we arrive at my preferred form of equation (2-73b)

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(b) Now we start from the vector potential, equation (2-65)

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This vector field is decomposed into cartesian components but uses spherical coordinates. We need to have it all one or the other to use standard expressions for the divergence and gradient. I think it is easier to go all spherical. Use equation (C-3)

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We then have

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Since we do not have a 𝜙-component, we only need the first two terms of equation (C-34) for the divergence in spherical coordinates

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Substituting in our vector potential

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Simplifying

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Perform the differentiation

E = j Iηβ4π

βΔz( ) e− jβr

βr1jβr

+ 1jβr( )2

⎛

⎝⎜⎞

⎠⎟2cosθ r + 1+ 1

jβr+ 1

jβr( )2⎛

⎝⎜⎞

⎠⎟sinθθ

⎡

⎣⎢⎢

⎤

⎦⎥⎥

A = µIe− jβr

4πrΔzz

z = rcosθ − θsinθ

A = µIe− jβr

4πrΔz rcosθ − θsinθ( )

∇⋅A = 1r2

∂∂r

r2Ar( )+ 1r sinθ

∂∂θ

Aθ sinθ( )

∇⋅A = 1r2

∂∂r

r2 µIe− jβr

4πrΔzcosθ

⎛⎝⎜

⎞⎠⎟ +

1r sinθ

∂∂θ

µIe− jβr

4πrΔz −sinθ( )sinθ⎛

⎝⎜⎞⎠⎟

∇⋅A = µIΔz4π

cosθr2

∂∂r

re− jβr( )− 1r sinθ

e− jβr

r∂∂θsin2θ⎡

⎣⎢⎤⎦⎥

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Simplify

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The gradient, equations (C-33), with no 𝜙 dependence is

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Then we have

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Simplify

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Differentiate

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Simplify

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Now substitute into equation (2-46), and begin the algebra circus

∇⋅A = µIΔz4π

cosθr2

r − jβ( )e− jβr + e− jβr( )− 1r sinθ

e− jβr

r2sinθ −cosθ( )⎡

⎣⎢⎤⎦⎥

∇⋅A = − µIΔz4π

e− jβr

r21+ jβr( )cosθ

∇g = r ∂g∂r

+ θ 1r∂g∂θ

∇ ∇⋅A( ) = r ∂∂r

− µIΔz4π

e− jβr

r21+ jβr( )cosθ⎡

⎣⎢⎤⎦⎥+ θ 1

r∂∂θ

− µIΔz4π

e− jβr

r21+ jβr( )cosθ⎡

⎣⎢⎤⎦⎥

∇ ∇⋅A( ) = − µIΔz4π

rcosθ ∂∂r

e− jβr 1r2

+ jβr

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥+ θ e

− jβr

r1r2

+ jβr

⎛⎝⎜

⎞⎠⎟∂∂θcosθ

⎡⎣⎢

⎤⎦⎥

∇ ∇⋅A( ) = − µIΔz4π

rcosθ e− jβr − 2r3

− jβr2

⎛⎝⎜

⎞⎠⎟ − jβe

− jβr 1r2

+ jβr

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥− θ e

− jβr

r1r2

+ jβr

⎛⎝⎜

⎞⎠⎟ sinθ

⎡⎣⎢

⎤⎦⎥

∇ ∇⋅A( ) = µIΔz4π

e− jβr

rr 2r2

+ 2 jβr

− β 2⎛⎝⎜

⎞⎠⎟ cosθ + θ 1

r2+ jβr

⎛⎝⎜

⎞⎠⎟ sinθ

⎡⎣⎢

⎤⎦⎥

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Finally obtain the same result. I think the method of part (a) is somewhat easier.

E = − jωA − j∇ ∇⋅A( )ωεµ

E = − jω µIe− jβr

4πrΔz rcosθ − θsinθ( )⎡

⎣⎢⎤⎦⎥− j 1

ωεµµIΔz4π

e− jβr

rr 2r2

+ 2 jβr

− β 2⎛⎝⎜

⎞⎠⎟ cosθ + θ 1

r2+ jβr

⎛⎝⎜

⎞⎠⎟ sinθ

⎡⎣⎢

⎤⎦⎥

E = − j IηβΔz4π

e− jβr

rrcosθ − θsinθ( )+ r 2

β 2r2+ 2 jβr

−1⎛⎝⎜

⎞⎠⎟cosθ + θ 1

β 2r2+ jβr

⎛⎝⎜

⎞⎠⎟sinθ⎡

⎣⎢

⎤

⎦⎥

⎡

⎣⎢

⎤

⎦⎥

E = − j IηβΔz4π

e− jβr

rr 1

β 2r2+ jβr

⎛⎝⎜

⎞⎠⎟2cosθ + θ 1

β 2r2+ jβr

−1⎛⎝⎜

⎞⎠⎟sinθ⎡

⎣⎢

⎤

⎦⎥

E = j Iηβ4π

βΔz( ) e− jβr

βr1jβr

+ 1jβr( )2

⎛⎝⎜

⎞⎠⎟2cosθ r + 1+ 1

jβr+ 1

jβr( )2⎛⎝⎜

⎞⎠⎟sinθθ⎡

⎣⎢

⎤

⎦⎥

2.4-2 The radial distance from the source point to the field observation point, R, is given by

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Expanding the dot product, and using unit vectors, we obtain

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Factoring out the radius from origin to field point, r, we can then express in terms of a small parameter, x

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To eliminate the square root, we approximate with a Taylor series expansion. We find the following Taylor series

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Since we are only interested in the first term of the Taylor series, we can get same result, by dropping the x squared term in the square root and using the more familiar Taylor series

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Either way leads to

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Note that if you wanted a second order term (say for estimating the error of using just the first order approximation), you need to use the first Taylor series above.

R = r − ′r = r − ′r( ) ⋅ r − ′r( )[ ]1/2

R = r ⋅r − 2r ⋅ ′r + ′r ⋅ ′r[ ]1/2 = r2 − 2r ′r r ⋅ ˆ′r + ′r 2[ ]1/2

R = r 1− 2r ⋅ ˆ′r x + x2[ ]1/2 where x ≡ ′r

r

1− 2ax + x2[ ]1/2 = 1− ax + 1− a2

2x2 +O x3( )

1+ bx[ ]1/2 = 1+ b2x − b

2

8x2 +O x3( )

R ! r 1− r ⋅ ˆ′r ′rr

⎛⎝⎜

⎞⎠⎟ = 1− r ⋅ ′r

2.5-6 We have the normalized antenna patter given by

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First, calculate the beam solid angle, defined by

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where the second integral uses standard spherical coordinates for the integration variables. Plug in our antenna pattern, and perform the integration

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We have used the fact that the integrand does not depend on 𝜙 (i.e. the antenna is omni directional or azimuthally invariant) to immediately perform that integration to obtain 2π. The θ integration is also straightforward. The directivity is inversely related to the beam solid angle.

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For Example 2-3, the value of α is π/6

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The same result is obtained as in the example.

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F θ( ) = 1 π2−α <θ < π

2+α

0 elsewhere

⎧⎨⎪

⎩⎪

ΩA = F θ ,φ( ) 2 dΩsphere∫∫ = F θ ,φ( ) 2 sinθ dθ dφ

0

π

∫−π

π

∫

ΩA = sinθ dθ dφπ /2−α

π /2+α

∫−π

π

∫ = 2π sinθ dθπ /2−α

π /2+α

∫ = 2π −cosθ[ ]π /2−απ /2+α

= 2π −cos π / 2 +α( )+ cos π / 2 −α( )[ ]= 2π sinα + sinα[ ]= 4π sinα

D = 4πΩA

= 4π4π sinα

= cscα

π2− π

6= π

3 and π

2+ π

6= 2π

3

D = csc π6

⎛⎝⎜

⎞⎠⎟ = 2