of 27 /27
WELLENAUSBREITUNG Formelsammlung INSTITUT F ¨ UR NACHRICHTENTECHNIK UND HOCHFREQUENZTECHNIK 5. Auﬂage, Sept. 2014 Gregor Lasser
• Author

others
• Category

## Documents

• view

0

0

Embed Size (px)

### Transcript of WELLENAUSBREITUNG Formelsammlung...Reﬂexion an glatten Grenzﬂ¨achen, die Parallelplattenleitung...

• WELLENAUSBREITUNG

Formelsammlung

INSTITUT FÜR NACHRICHTENTECHNIK

UND HOCHFREQUENZTECHNIK

5. Auflage, Sept. 2014

Gregor Lasser

• Maxwellsche Theorie

~∇ · ~S = limV→0

1

V

Σ

~S · d~F = − ∂∂tρ

~∇ · (~∇× ~H) ≡ 0

~∇× ~H = ~S + ∂∂t~D

~∇ · ~S = − ∂∂t

(~∇ · ~D) = − ∂∂tρ

~∇ · ~S + ∂∂tρ = 0

~∇ · ~D= ρ~∇ · ~B=0~∇× ~E =− ∂

∂t~B

~∇× ~H = ~S + ∂∂t~D

Σ

~D · d~F =∫

τ

ρ dV∫

Σ

~B · d~F =0∮

~E · d~l=− ∂∂t

Σ

~B · d~F∮

~H · d~l=∫

Σ

~S · d~F + ∂∂t

Σ

~D · d~F

~F = q(

~E + ~v × ~B)

~S = σ ~E

~S = ρ~v

~D = ε ~E = εrε0 ~E

~B = µ ~H = µrµ0 ~H

∂tρ(~r, t) +

σ

ερ(~r, t) = 0

ρ(~r, t) = ρ0(~r)e−σ

εt

τD =ε

σ

1

• ~∇ · ~E = 0~E(x, y, z, t) = ~E(~r, t) = Re{ ~E(~r) ejωt} = 1

2

(

~E(~r) ejωt + ~E∗(~r) e−jωt)

~∇× ~H = ~S + jω ~D = σ ~E + jωε ~E = jωδ ~E

δ = ε+σ

jω= ε− j σ

ω

tan θ =ε′′

ε′=

σ

ωε

~∇ · ~E = 0~∇ · ~H = 0~∇× ~E = −jωµ ~H~∇× ~H = jωδ ~E

~P (t) = ~E(t)× ~H(t)

~∇ · ~P (t) = −σ ~E(t) · ~E(t)︸ ︷︷ ︸

~E2(t)

− ∂∂t

2~E(t) · ~E(t)︸ ︷︷ ︸

~E2(t)

2~H(t) · ~H(t)︸ ︷︷ ︸

~H2(t)

)

we(t) =ε

2~E2(t)

wm(t) =µ

2~H2(t)

pv(t) = σ ~E2(t)

V

~∇ · ~P (t)dV =∮

Σ

~P (t) · d~F

− ∂∂t

V

(

we(t) + wm(t))

dV =

Σ

~P (t) · d~F +∫

V

pv(t)dV

~E(~r, t) =1

2

(

~E(~r) ejωt + ~E∗(~r) e−jωt)

~E(t) · ~E(t) = 14

(

~E(~r) · ~E(~r) e2jωt + 2 ~E(~r) · ~E∗(~r) + ~E∗(~r) · ~E∗(~r) e−2jωt)

~E(t) · ~E(t) = 12| ~E(t)|2

we(t) = we=ε

4| ~E(t)|2

wm(t) = wm=µ

4| ~H(t)|2

pv(t) = pv =σ

2| ~E(t)|2

~T =1

2~E × ~H∗ = ~Tw + j ~Tb

2

• ~E · d~l = Et1∆l + En1∆x+ En2∆x− Et2∆l −En2∆x−En1∆x

= (Et1 − Et2)∆l − 0En1 + 0En2 = −∂

∂t

F

~B · d~F = 0

Et1 = Et2

Ht1 = Ht2∫

~D · d~F = (Dn1 −Dn2)∆F = ρS∆F

Dn1 −Dn2 = ρS → ε1En1 − ε2En2 = ρSBn1 = Bn2 → µ1Hn1 = µ2Hn2

~n · ~D1 = ρS~n · ~B1 = 0~n× ~E1 = ~0~n× ~H1 = ~K

~n · ( ~D1 − ~D2) = ρS~n · ( ~B1 − ~B2) = 0~n× ( ~E1 − ~E2) = ~0~n× ( ~H1 − ~H2) = ~K

∇2 ~E − µε ∂2

∂t2~E − µσ ∂

∂t~E = 0

∇2 ~E + (ω2µε− jωµσ) ~E = 0∇2 ~E + ω2µδ ~E = 0∇2 ~H + ω2µδ ~H = 0

∇2 = ∂2

∂x2+

∂2

∂y2+

∂2

∂z2

k = ω√

µδ

Ψ(x, y, z) = X(x) Y (y)Z(z)

1

X(x)

∂2

∂x2X(x) +

1

Y (y)

∂2

∂y2Y (y) +

1

Z(z)

∂2

∂z2Z(z) + k2

︸︷︷︸

const.

= 0

1

X(x)

∂2

∂x2X(x) = −k2x

k2x + k2y + k

2z = k

2 = ω2µδ

∂2

∂x2X(x) + k2xX(x) = 0

3

• ∂yHz −

∂zHy = jωδEx

∂zHx −

∂xHz = jωδEy

∂xHy −

∂yHx = jωδEz

∂yEz −

∂zEy =−jωµHx

∂zEx −

∂xEz =−jωµHy

∂xEy −

∂yEx =−jωµHz

Ex =−jκ2

(

kz∂

∂xEz + ωµ

∂yHz

)

Ey =−jκ2

(

kz∂

∂yEz − ωµ

∂xHz

)

Hx =−jκ2

(

kz∂

∂xHz − ωδ

∂yEz

)

Hy =−jκ2

(

kz∂

∂yHz + ωδ

∂xEz

)

4

• Die homogene ebene Welle (HEW)

+∂

∂zey =µ

∂thx

− ∂∂zex =µ

∂thy

0=µ∂

∂thz

− ∂∂zhy = ε

∂tex

+∂

∂zhx = ε

∂tey

0= ε∂

∂tez

∂2

∂z2ex − µε

∂2

∂t2ex = 0

ex(z, t) = c1 f1(z − v t)︸ ︷︷ ︸

=e+x (z,t)

+ c2 f2(z + v t)︸ ︷︷ ︸

=e−x (z,t)

v =1√εµ

k

η =

√µ

ε

η = η0 =

√µ0ε0

≈ 120πΩ ≈ 377Ω

h+x = −e+yη

e+xh+y

= −e+yh+x

= η

~E ⊥ ~H ⊥~ize−xh−y

= −e−yh−x

= −η

we(t) =ε

2(e2x + e

2y)

wm(t) =µ

2(h2x + h

2y)

5

• wm(t) =µ

2

1

η2(e2x + e

2y) = we(t)

p+x ≡ 0p+y ≡ 0p+z = e

+x h

+y − e+y h+x

p+z =1

η(e+x

2+ e+y

2)

~P (t) = ~E × ~H = 1η

(

e+x2 + e+y

2)

~iz =1

(

E2x0 + E2y0

)

~iz

ex(z, t) = Re{Ex(z) ejωt} = E0 cos (k(vt− z)) = E0 cos (ωt− kz)

k =ω

v= ω

√µε

λ =2π

k=

ω√εµ

~E1 = ~Ex =(E1~ix + 0~iy) e−jkz

~E2 = ~Ey =(0~ix + E2~iy) e−jkz

ex(z, t) =E1 cos (ωt− kz)ey(z, t) =E2 cos (ωt− kz + ψ)

ex(0, t)=E1 cos (ωt)

ey(0, t)=E2 cos (ωt+ ψ)

ex(0, t) =E cos (ωt),

ey(0, t) =E cos (ωt∓π

2) = ±E sin (ωt)

ex(z, 0) = E cos (−kz) = E cos (kz)ey(z, 0) = E cos (−kz ±

π

2) = ±E sin (kz)

~E=Ey0 e−jkz~iy

η ~H =−Ey0 e−jkz~ix

P =

~P · d~F = 12Re{

( ~E × ~H∗) · d~F}

=1

2Re{

(ExH∗y −EyH∗x)dF}

=wd

2Re{−Ey0 e−jkz(−

Ey0η

e+jkz)} =E2y02η

wd

P =|U |22ZPV

6

• U =

∫ 0

−d

Eydy = Ey0 d

P =|Ey0|2d22ZPV

~∇× ~E =−jωµ ~H~∇× ~H =σ ~E + jωε ~E

δ = ε− j σω

= ε(1− js)

s =1

Q= tan θ =

σ

εω

η2 =µ

δ=µ

ε

1

1− js

η = R+ jX = ηE1√

1− jsjkz = jω

µδ = jω√µε

1− js = γ = α + jβjkz = jkE

1− js

R = ηE

√√1 + s2 + 1

2(1 + s2)X = ηE

√√1 + s2 − 12(1 + s2)

α = kE

√√1 + s2 − 1

2β = kE

√√1 + s2 + 1

2

η ≈ ηE(1 + js

2) jkz ≈ kE(

s

2+ j)

η ≈ ηE1 + j√

2sjkz ≈ kE

√s

2(1 + j)

d =1

α≈ 1kE

2

s=

√2

ωµσ

7

• Reflexion an glatten Grenzflächen,die Parallelplattenleitung

sinΘ1sinΘ2

=

√ε2ε1

=n2n1

ΓTM =

√ε2 cosΘ1 −

√ε1 cosΘ2√

ε2 cosΘ1 +√ε1 cosΘ2

=n2 cosΘ1 −

n2 − sin2Θ1n2 cosΘ1 +

n2 − sin2Θ1TTM =

2√ε1 cosΘ1√

ε2 cosΘ1 +√ε1 cosΘ2

=2n cosΘ1

n2 cosΘ1 +√

n2 − sin2Θ1

ΓTE =

√ε1 cosΘ1 −

√ε2 cosΘ2√

ε1 cosΘ1 +√ε2 cosΘ2

=cosΘ1 −

n2 − sin2Θ1cosΘ1 +

n2 − sin2Θ1TTE =

2√ε1 cosΘ1√

ε1 cosΘ1 +√ε2 cosΘ2

=2 cosΘ1

cosΘ1 +√

n2 − sin2Θ1

ΓTM = 0 ⇐ tanΘB =√ε2ε1

=n2n1

= n

λ1z0m cosΘm = mπ ⇒ d = −z0m =

λ1m

2 cosΘm=

m

2 f√µ1ε1 cosΘm

λx =λ1

sin Θm

λH,m =λ0

sinΘm

d = mλ02

fG,m =mc

2d

λG,m =c

fG,m=

2d

m

λH,m =λ0

1− (mλ02d

)2

λH,m =λ0

1− ( λ0λG,m

)2

sinΘm =λ0λH,m

=

1− (mλ02d

)2

8

• Die Oberflächenwelle

s1 =σ1ωε1

≫ 1

s2 =σ2ωε2

≪ 1

kE2 = ω√ε2µ0

kz ≈ kE2(

1− j 12s1

ε2ε1

)

α ≈ kE21

2s1

ε2ε1

= kE2

(R1

R2

)2

2

1

s1

ε2ε1

β ≈ kE2 = ω√ε2µ0

kx1≈√ωµ0σ1

2(−1 + j)

kx2≈ωε2√ωµ02σ1

(1− j) = ωε2η1

kx2kx1

≈ −ωε2σ1

Ex1 = kE2 d11− jε2

2s1ε1

−1 + j A1 ejkx1x e−jkzz

Ex1 = −Ez1√ωε22σ1

(1 + j)

Ex2 = −1

1 − jEz2(√

2σ1ωε2

− j√ωε22σ1

)

ZW =ExHy

ZW2 =kzωδ2

=kE2

(

1− j 12s1

ε2ε1

)

ωε2(1− js2)≈ ηE2

(

1 + js22

)

ZW1 =kzωδ1

9

• Iz =

Σ

~S1 · d~F = σ1∞∫

x=0

b∫

y=0

Ez1dxdy

= σ1A1b e−jkzz

∞∫

x=0

ejkx1 xdx

= jσ1A1b

kx1e−jkzz

dUz = IzdZ

dP =1

2|Iz|2dZ

~T =1

2~E × ~H∗ = 1

2

−EzH∗y0

ExH∗y

Tx0 =1

2

(

− Ez0H∗y0)

= −12

ωδ∗1k∗x1

|A1|2

dP = Tx0 b dz

Tx0 b dz =1

2|Iz|2dZ ⇒ dZ = −

ωδ∗1k2x1

σ21

dz

b

dZ ≈ η1dz

b

dZ = dR + jdX, dR =dz

bR1, dX =

dz

bX1

dPW = TW b dz =1

2|Iz|2dR

Iz = −bHy1(0)

dPW =1

2|Hy1(0)|2bR1dz bzw.

dPWdz

=1

2|Hy1(0)|2bR1

p =1

b

dPWdz

=1

2|Htang(0)|2R1

R1 =1

σ1 d1= R� (lies: R square)

R =l

σA

R =

dR =

l∫

0

R�

bdz =

1

σ1d1

l

b∝

√ω

R =l

2π a

√ωµ

2σ1, X =

l

2π a

√ωµ

2σ1

R

R0=

X

X0=a

2

√ωµσ12

=a

2d1∝

√ω ≫ 1

10

• Resonatoren

λG,m,n =1

√(m2a

)2+(

n2b

)2

λH=λ

1−(

λλG

)2

vP =c

1−(

λλG

)2

vG = c

1−( λ

λG

)2

κ2 =(mπ

a

)2

+(nπ

b

)2

≡ ω2εµ− k2z

P =

Re{~T} · d~F =∫

Tzdxdy = −1

2

a∫

0

b∫

0

EyH∗xdxdy =

ωkzµ

2

(aA

π

)2

b

a∫

0

sin2 (π

ax)dx

=ωkzµ

4ab(aA

π

)2

−dP = 12|Htang|2RMdF

− ∂∂zP (z) =

1

2RM

(

2

a∫

0

[

|Hx|2 + |Hz|2]

y=0dx+ 2

b∫

0

[

| Hy︸︷︷︸

0

|2 + |Hz|2]

x=0dy

)

= RMA2(

(kza

π)2

a∫

0

sin2 (π

ax)dx+

a∫

0

cos2 (π

ax)dx+

b∫

0

dy)

= A2RM

(a

2

(1 + (

2a

λH)2)+ b

)

α =π

ωµRM

λHa3b

(a

2

(1 + (

2a

λH)2)+ b

)

c = pλH2, bzw. kz =

λH=pπ

c(mπ

a

)2

+(nπ

b

)2

+(pπ

c

)2

= ω2mnpεµ

11

• ωmnp = πv

√(m

a

)2

+(n

b

)2

+(p

c

)2

Q0,mnp =ωmnpW

P

Q0 =2πW

PTmit T =

1

fmnp

W =1

4

τ

(

ε ~E · ~E∗ + µ ~H · ~H∗)

P =1

2RM

Σ

~Htang · ~H∗tangdF

λ101 =2ac√a2 + c2

ω101 =π√εµ

√a2 + c2

ac

Ey =−2ωma

πA sin (

π

ax) sin (

π

cz)

Hx =2jkza

πA sin (

π

ax) cos (

π

cz)

Hz =−2j A cos (π

ax) sin (

π

cz)

We = Wm=A2µa2 + c2

4c2abc

P =A2RMac(a2 + c2) + 2b(a3 + c3)

c2

Q0 =πη

2RM

b√

(a2 + c2)3

ac(a2 + c2) + 2b(a3 + c3)

Q0 =πη

√2

6RM

12

• Koaxialleitungen

~E = Er ~er~H = Hϕ ~eϕ

∂zU(z) + Z ′I(z) = 0 ,

∂zI(z) + Y ′U(z) = 0

mit Z ′ = R′ + jωL′ , Y ′ = G′ + jωC ′

∂zU(z)− Y ′Z ′U(z) = 0 , ∂

∂zI(z)− Y ′Z ′I(z) = 0

U(z) = Uve−jkzz + Ure

+jkzz , I(z) = Ive−jkzz + Ire

+jkzz

Uv = ZLIv , Ur = −ZLIr

ZL =

Z ′

Y ′bzw. ZL =

R′ + jωL′

G′ + jωC ′

jkz =√Y ′Z ′ =

(G′ + jωC ′)(R′ + jωL′)

R′

L′=G′

C ′

vP =ω

k=

ω

Re{kz}≈ 1√

L′C ′

L =1

I

A

~B · ~nA dA

L′ =1

I

∫ ra

ri

~B · ~nA dr =1

I

∫ ra

ri

Bϕ dr =µ

I

∫ ra

ri

Hϕ dr

Hϕ =I

2πr

L′ =µ

I

∫ ra

ri

I

2πrdr

L′ =µ

2πlnrari

C =Q

C~E · ~er dr

~E =τ

2πε

~err

C ′ =τ

C~E · ~er dr

∫ rari

τ2πε

~err· ~er dr

=2πε

∫ rari

1rdr

13

• C ′ =2πε

ln rari

ZL,verlustlos =η

2πlnrari

mit η =

√µ

ε

R� =

√ωµL2σ

R′ =Rinnen +Raussen

l≈

R�l2πri

+ R�l2πra

l=

R�

2π(1

ri+

1

ra)

R′ =

√ωµL2σ

1

2π(1

ri+

1

ra)

G′ = ωC ′ tan δε = ω2πε

ln rari

tan δε

jkz = γ = α+ jβ =√

(G′ + jωC ′)(R′ + jωL′)

α = αR + αG = (R′

2√

L′

C′︸ ︷︷ ︸

(1)

+G′

√L′

C′

2︸ ︷︷ ︸

(2)

)1

cosh δR−δG2

︸ ︷︷ ︸

(3)

mit sinh δR =R′

ωL′, sinh δG =

G′

ωC ′

αR ≈R′

2√

L′

C′

=R�

2ηra

1 + rari

ln rari

ZL,min. Dämpfung =η0

2π√εr

lnrari

=77 Ω√εr

Umax = Emax ri lnrari

= Emax raln ra

rirari

ZL,max.Spannungsfest =60 Ω√εr

Pmax =U2max2ZL

=πE2maxr

2i

ηlnrari

=πE2maxr

2a

η

ln rari

( rari)2

ZL,max. Leistung =30 Ω√εr

pv(z) = −dP

dz= − d

dzP0e

−2αz = 2αP0e−2αz = 2αP (z)

mit α = αR + αG

14

• Dielektrische Wellenleiter

ξ= kx1d

η= kx2d

−ξ cot ξ = η

ξ2 + η2 = ω2µ0d2(ε1 − ε2) = V 2 ⇒ V =

2πd

λ0

n21 − n22

kx1,m =(2m− 1)π

2d, m = 1, 2, . . .

ωc,m =(2m− 1)π

2d√

ε0µ0(εr1 − εr2)

15

• Streifenleitungen

ZL ≈√µ0ε0

1

2πln(8h

w+w

4h

)

ZW =ZL√εeff

λ =λ0√εeff

εeff = 1 + q(εr − 1)fc,TEM =

c

4h√εr − 1

hmax =λ0

4√εr − 1

fc,QTEM =c

(2w + 0, 8h)√εr

α = αL + αD

− ∂∂zP (z) = |Hx0|2Rw,

R =

√ωµ

2σ,

αL =1

2P (z)

√ωµ

2σw |Hx0|2 =

√ωµ

1

ηh

P =

~P · d~F =∫

Tw,zdxdy =1

2

EyH∗xdxdy =

1

2η|Ey0|2hw

ZW = ηh

w.

αL =

√ωµ

1

ZWw.

α′L = αL

(

1 +2

πarctan (1, 4

d1))

αD = kEs

2,

ε′

ε′′= tanΘ = s

αD =π

λtanΘ

αD =π

λtanΘ

( εrεeff

εeff − 1εr − 1

)

16

• Wellen und Hindernisse

Γrauh = Γglatt exp[−2 (kσ cosΘe)2

]

E/E0 = 1/2− exp (−jπ/4) [C (v) + jS (v)] /√2

C (v) =

v∫

0

cos(πt2/2

)dt S (v) =

v∫

0

sin(πt2/2

)dt

v = h√

2/λ (1/ds + 1/de)

17

• Antennen

~A (~r) = µ

V ′

~Se (~r′) e−jk|~r−~r

′|

4π |~r − ~r ′| dV′

~A (~r) = µe−jkr

4πr

V ′

~Se (~r′) e+jkr

′ cos ϑdV ′ = µe−jkr

4πr~N (ϑ)

|~r − ~r ′| =√

r2 + r ′2 − 2rr ′ cosϑ

=

(r − r ′ cosϑ)2 + r ′2 sin2 ϑ

= (r − r ′ cosϑ)[

1 +1

2

r ′2 sin2 ϑ

(r − r ′ cosϑ)2+ . . .

]

∆α = k∆r =2π

λ

r′ 2 sin2 ϑ

2 (r − r ′ cosϑ)

∆αmax =π

λ

r ′2

r

π

2=π

λ

D2

rR

rR =2D2

λ(+λ)

Eϑ (ϑ, ϕ)

Eϑ (ϑmax, ϕmax)=

Hϕ (ϑ, ϕ)

Hϕ (ϑmax, ϕmax)= f (ϑ, ϕ)

ϑmax = π/2 und

ϕmax = beliebig

EϑEϑ (π/2)

=Hϕ

Hϕ (π/2)= f (ϑ, ϕ) = sin ϑ

φ = r2Re{

~T}

· ~er

~T =1

2~E× ~H⋆

1

2

f

(

~E× ~H⋆)

· ~er df

18

• Pr =1

2Re

∫∫

f

(

~E × ~H⋆)

· ~er df

df = r2 sinϑdϑdϕ = r2dΩ

Pr =

Re{

~T}

r2 · ~er dΩ =∫

φ dΩ = φmax

φ

φmaxdΩ

f (ϑ, ϕ) =E (ϑ, ϕ)

E (ϑmax, ϕmax)

φ

φmax= |f (ϑ, ϕ)|2

Pr = φmax

|f (ϑ, ϕ)|2 dΩ = φmaxΩä

Ωä =

2π∫

0

π∫

0

|f (ϑ, ϕ)|2 sinϑdϑdϕ

D =4π

Ωä=

4π2π∫

0

dϕ∫ π

0|f(ϑ, ϕ)|2 sin ϑdϑ

PLHDPLDUT

=PrHDPrDUT

eDUT = · · ·

|EϑHD | =η |I| s2λr

sinϑ.

PrHD

=πη

3

(s2

λ2

)

|I|2

|EϑHD | =√

3 η

4 π

Pr,HDsin ϑ

r

PrHD

=4πr2

3η|EϑHD |

2 1

sin2 ϑ

|EϑHD ||EϑHD|max

= fHD (ϑ) = sin ϑ

PrHD =4πr2

3η|Eϑ,HD|2max

Eϑ =jηI

2πre−jkr F (ϑ, ϕ)

H⋆ϕ = −jI⋆

2πre+jkrF⋆ (ϑ, ϕ)

PrDUT =1

2Re

f

(

~E× ~H⋆)

· ~er df

=

1

2

2π∫

0

π∫

0

η|I|24π2r2

|F (ϑ, ϕ)|2 r2 sinϑdϑdϕ =

19

• = η|I|28π2

2π∫

0

π∫

0

|F (ϑ, ϕ)|2 sinϑdϑdϕ

|Eϑ|max =η|I|2πr

|F (ϑmax, ϕmax)|

|Eϑ||Eϑ|max

=|F (ϑ, ϕ)|

|F (ϑmax, ϕmax)|= |f (ϑ, ϕ)|

|F (ϑ, ϕ)| = |f (ϑ, ϕ)| |F (ϑmax, ϕmax)| = |f (ϑ, ϕ)|2πr

η|I| |Eϑ|max

PrDUT =r2

2η|Eϑ|2max

2π∫

0

π∫

0

|f (ϑ, ϕ)|2 sinϑdϑdϕ

PrHDPrDUT

=

4πr2

r2

|EϑHD |2max

|Eϑ|2max2π∫

0

π∫

0

|f (ϑ, ϕ)|2 sinϑdϑdϕ

GREF =PLREFPLDUT

· |EϑDUT |2max

|EϑREF |2max

GHD = eDUT8π/3

2π∫

0

π∫

0

|f (ϑ, ϕ)|2 sinϑdϑdϕ

GHD =PLHDPLDUT

· |EϑDUT|2max

|EϑHD |2max= eDUT

PrHDPrDUT

· |EϑDUT|2max

|EϑHD |2max

GHD =8π/3

2π∫

0

π∫

0

|f(ϑ, ϕ)|2 sinϑdϑdϕ

GHD =8π/3

Ωä=

2

3GISO

TE (r) = GPS4πr2

|EE| =A

λrE0

TE (r) =|EE|22η

=

(A

λr

)2E202η

PS =E202η

A

G = 4πr2TE (r)

PS= 4π

A

λ2

GISO =4π

λ2Aw

PE = ATE

20

• GDUT/ISO =EIRP

PLDUT· |EϑDUT|

2max

|EϑISO |2maxEIRP = PLGISO

L = ns

l =πD

cosψ

s = l sinψ = πD tanψ

kwendell − k0s = 2πν ν = 1, 2, 3, . . .

kwendel =ω

v

ω

(l

v− sc0

)

= 2πν ν = 1, 2, 3, . . .

ω = 2πc0λ0

und l ≈ πD ≈ λ0

l = (λ0 + s)v

c03

4λ0 < λ <

4

3λ0

P =1

2|I|2 ZA

~T =1

2~E× ~H⋆

Tr =1

2EϑH

⋆ϕ

Tϑ = −1

2ErHϕ

⋆ ≈ 0

Tϕ ≡ 0|Hϕ| = |Eϑ| /η

Pr =1

2

2π∫

ϕ=0

π∫

ϑ=0

|Eϑ|2η

r2 sinϑdϑdϕ

Eϑ = jηIs

e−jkr

rsin ϑ

Pr = η|I|2 s28λ2

π∫

0

sin3 ϑdϑ

Pr =1

3πη

(s2

λ2

)

|I|2

RA =2

3πη

(s2

λ2

)

m =1 + |ρ|1− |ρ| =

|Umax||Umin|

21

• Z(z) =U(z)

I(z)

Q =ω

2RA

(∂XA∂ω

)∣∣∣∣ω=ω0

mit ZA = RA + jXA

∆ω =ω0Q

Q =ω

2GA

(∂BA∂ω

)∣∣∣∣ω=ω0

mit YA = GA + jBA

r = r0 exp (aψ)

PS1PE2

=PS2PE1

G (ϑ, ϕ) =PLREFPLDUT

· |EϑDUT (ϑ, ϕ)|2

|EϑREF (ϑ, ϕ)|2=

=PLREFPLDUT

|EϑDUT |2max

|EϑREF |2max|fDUT (ϑ, ϕ)|2

|fREF (ϑ, ϕ)|2=

=GREF ·|fDUT (ϑ, ϕ)|2

|fREF (ϑ, ϕ)|2

MEG =

G (ϑ, ϕ) P (ϑ, ϕ) dΩ

22

• Wellen im freien Raum

r.=

4

Te,ISO =Ps

4 π d2

Te =PsGs4 π d2

Pe = TeAe

Pe = TeAe =PsGs4 π d2

Ae

A =λ2

4 πGiso

Pe =PsGs4 π d2

λ2

4 πGe = Ps

4 π d

)2

GsGe

Pe = Ps

(1

λ d

)2

AsAe

L∣∣dB

= 10 logPsPe

Pe∣∣dBW

= Ps∣∣dBW

+Gs∣∣dB

− LISO∣∣dB

+Ge∣∣dB

LISO = −20 log(

λ

4π d

)

Ls = 10 logPsPn

= 10 · log PsPe,min

Pe,minPn

= L∣∣dB

+ SNRmin∣∣dB

Ti =PsGs4πd2

Pe = TeAe =Ti σ

4πd2Ae =

PsGs σ

(4πd2)2λ2

4πGe

PePs

= σ G2s

)21

4πd4

σ = AG = A4π

λ2A = 4 π

A2

λ2

23

• Mehrwegeausbreitung

τ1 = d1/c, und τ2 = d2/c

h(τ) = A1 δ (τ − τ1) +A2 δ (τ − τ2)

H (jω) =

∞∫

0

h(τ) e−jωτ dτ = A1 e−jωτ1 +A2 e

−jωτ2

|H (jω)| =√

A21 + A22 + 2A1A2 cos (ω ·∆τ) mit ∆τ = τ2 − τ1

∆ fNotch =1

∆ τ

H (jω) = |H (jω)| ejφH (jω)

τGr = −dφHdω

~E (~r) = ~E1 e−j~k1~r + ~E2 e

−j~k2~r

~E(t) = ~E0 · cos (ωt− kd)

~E (t) = ~E0 · cos (ωt− k [d0 + vt])= ~E0 · cos (t [ω − kv]− kd0)

= ~E0 · cos(

t

[

2πf − 2πλv

]

− kd0)

= ~E0 · cos(

2πt[

f − vλ

]

− kd0)

∆fD = −v

λ= −f · v

c

∆fD = −v

λcos (γ) = −f · v

ccos (γ)

p(E) =1

σ√2 · π

· e− E2

2·σ2

Varianz: = E2 −(E)2

Varianz = E2 =

∞∫

−∞

E2 · p(E)dE = σ2

σ2 = Re(E)2 = Pm

p(a) =a

σ2· exp

[

− a2

2σ2

]

24

• Mittelwert a = σ√

π2

Varianz a2 − (a)2 = 2 σ2 − σ2 π2= 0.429σ2

Medianwert a50 = σ√2 · ln2 = 1.18 σ

p(a) =a

σ2· exp

[

−a2 + A2

2σ2

]

· I0(aA

σ2

)

quadrat. Mittelwert a2 = 2σ2 + A2

PePr

= Gs ·Ge(

λ

4πd0

)2 (d0d

)n

p(F ) =1

σF√2 · π

· exp[

−(F −M)2

2 · σ2F

]

25