Solution 1 Antenna

Solutions 1 – Fundamentals Antennas and Propagation, Frühjahrssemester 2010

Problem 1.1 : If a transmitter produces 50 W of power, express the transmit power in units of a) dBm, b) dBW. Note: For the unit dBmW, the abbreviation dBm is commonly used. Transmitter power is 50WtP = .

a) [ ]

[ ]3,dBmmW

10 log 10 log 50 10 47.0dBm1mWt

tP

P⎡ ⎤⎢ ⎥= ⋅ = ⋅ ⋅ =⎢ ⎥⎢ ⎥⎣ ⎦

b) [ ]

[ ],dBWW

10 log 10 log 50 17.0dBW1Wt

tP

P⎡ ⎤⎢ ⎥= ⋅ = ⋅ =⎢ ⎥⎢ ⎥⎣ ⎦


Problem 1.2 : If 50 watts are transmitted by a unity gain antenna at 900 MHz carrier frequency, find the received power (assuming unity gain receiver antenna) in [dBm] and [W] at a free space distance of: a) 100 m from antenna, b) 10 km from antenna.

Power density as a function of distance: 24tPpowerS

area Rπ= =

Unity-gain / isotropic receive antenna described by its effective area: 2

4isotropicA λπ

=

Received power: 2

4rec tP S A PR

λπ

⎛ ⎞= ⋅ = ⎜ ⎟⎝ ⎠

a) received power = 3.5 x 10–6 W ≅ –24.5 dBm b) received power = 3.5 x 10–10 W ≅ –64.5 dBm


Problem 1.3 : Assume an air-filled metallic rectangular waveguide of cross-section 22.86 mm x 10.16 mm. Calculate a) the monomode frequency range, b) the guided wavelength at 10 GHz, c) group velocity and phase velocity at 10 GHz. a)

From [FuK II, 4.39], with a = 22.86 mm : , 101 6557

2c TEf MHza με

= =

Similarly, fcTE20 = 13114 MHz and fcTE01 = 14753 MHz. Therefore, the monomode frequency range is limited by the cutoff frequencies of modes TE10 and TE20, respectively. b)

From [FuK II, 4.26], considering TE10 and β=2π/λg : 2 22

g c aπ ω π

λ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

, λg = 39.7 mm

c) From [FuK II, 4.7, 4.9] :

803.97 10 1.32m

phase g sv f x cλ= = = ⋅ 2

802.27 10 0.76m

group sphase

cv x cv

= = = ⋅


Problem 1.4 : A hypothetical isotropic antenna is radiating in free space. At a distance of 100 m from the antenna, the total electric field is measured to be 5 V/m. a) Find the power density at this location. b) Determine the total power radiated by the antenna. a) power density (vectorial!)

2 2* 2

rad1 5 0.03315 W/m2 2 2(120 )r r r

EW E H a a aη π

⎡ ⎤= × = = =⎢ ⎥⎣ ⎦ b) radiating power

22

rad rad0 0

22 2

rad0 0

rad

0.03315 sin

0.03315 (100) sin 2 0.03315 100 2

4166.67 W

SP W dS r d d

P d d

P

π π

φ θ

π π

θ θ φ

θ θ φ π

= =

= ⋅ = ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

=

∫∫ ∫ ∫

∫ ∫


Problem 1.5 : A dipole of length 3λ/2 is resonant at f = 150 MHz. Calculate its mechanical length a) in air, b) in water (εrel = 81). a)

0

r r

c fλε μ

= with λ=2m, the dipole length in air is 3 m.

b) with λ=2/9 m, the dipole length in air is 1/3 m = 0.33 m.


Problem 1.6 : A charge Q is placed at distance h over an extended, perfectly conducting plate. a) Name the boundary conditions of this problem. b) How can the electric field in the halfspace of the charge be calculated? c) How can charge distribution and total charge on the plate be determined? a) boundary at infinity field decaying to zero with r–1

boundary on metal plate Dirichlet boundary for tangential E-field (Etan = 0) b) use of –Q mirror charge at 2h distance. Superposition of E-fields.

2 2 20 1 2

1 1( )4 4

ir i r i r i

i i

Q Q QE r e e er r r r r rπε πε

⎛ ⎞= = ⎜ − ⎟

⎜ ⎟− − −⎝ ⎠∑

c) first calculate E-field on metal plate. Use cylindrical coordinates.

20

14 r

QE erπε

= , 1 12 20

14 r

QE ehπε ρ

=+

, 1 2 1( ) 2 cos zE E E E eρ α= + = − , 2 2

cos ah

αρ

=+

2 2 3/ 20

( )2 ( ) z

hQE eh

ρπε ρ

−=+

charges influenced on the plate according to normal electric field: 0 normal FE qε =

surface charge density: 2 2 3/ 2( )2 ( )F

hQqh

ρπ ρ

−=+

total influenced charge obtained by integration: 2

0 0

( )Influenz FQ q d d Qπ

ρ ρ ρ ϕ∞

= = = −∫ ∫ …

r

1r2r

Q

1r r−2r r−

Q

ρ

–Q +Q

h

1E

2E

α

z

1re

h


Problem 1.7 : Calculate for the fields A, A (in Cartesian coordinates) a) div curl A b) curl grad A

x y zA A Agrad A e e ex y z

∂ ∂ ∂= + +∂ ∂ ∂

yx z

AA Adiv Ax y z

∂∂ ∂= + +∂ ∂ ∂

y yx xz z

x y z

A AA AA Acurl A e e ey z z x x y

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

a)

y yx xz zA AA AA Adiv curl A

x y z y z x z x y∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂∂ ∂∂ ∂ ∂⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

2 22 22 2

0y yx xz zA AA AA Adiv curl A

x y x z y z x y x z y z∂ ∂∂ ∂∂ ∂= − + − + − =

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

b)

x y z

A AA A A Ay yz x z xcurl grad A e e e

y z z x x y

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

2 2 2 2 2 2

0x y zA A A A A Acurl grad A e e e

y z y z x z x z x y x y⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂= − + − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠


Problem 1.8 : Calculate the surface resistance and skin depth of tin (σ = 8.7 x 106

S/m) and silver (σ = 61 x 106

S/m) at frequencies 100 MHz and 10 GHz.

Skin depth δ : 1f

δπ μσ

= . surface resistance RS : 1S

fR π μσδ σ

= =

Tin at 100 MHz : δ = 17.1 μm RS = 6.7 mΩ Tin at 10 GHz : δ = 1.71 μm RS = 67 mΩ Silver at 100 MHz : δ = 6.5 μm RS = 2.5 mΩ Silver at 10 GHz : δ = 0.65 μm RS = 25 mΩ

Solution 1 Antenna

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