射頻電子 - [第二章] 傳輸線理論

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ଯᓎηၡ Βക ᒡጕፕ ᄪ շ௲ Department of Electronic Engineering National Taipei University of Technology

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Page 1: 射頻電子 - [第二章] 傳輸線理論

Department of Electronic EngineeringNational Taipei University of Technology

Page 2: 射頻電子 - [第二章] 傳輸線理論

••••

Department of Electronic Engineering, NTUT2/47

Page 3: 射頻電子 - [第二章] 傳輸線理論

Two-wire line

Coaxial ( )

Microstrip( )

Department of Electronic Engineering, NTUT3/47

Page 4: 射頻電子 - [第二章] 傳輸線理論

( )

Department of Electronic Engineering, NTUT4/47

Page 5: 射頻電子 - [第二章] 傳輸線理論

( )

Department of Electronic Engineering, NTUT5/47

Page 6: 射頻電子 - [第二章] 傳輸線理論

•( ) :

LL s

s L

Zv v

Z Z=

+

Source

Source impedance Load

impedance

(Transmission line)

sv

sZLZ

l

•? ?

Lv

Department of Electronic Engineering, NTUT6/47

Page 7: 射頻電子 - [第二章] 傳輸線理論

(Distributed Circuit Model)

• l

∆x

• R: Ω/m, L: H/m, C: F/m, G: S/m

G x∆

L x∆R x∆

C x∆sZ

sv LZ

llll

dx dx dx

Department of Electronic Engineering, NTUT7/47

Page 8: 射頻電子 - [第二章] 傳輸線理論

•• :

• :

• R: G:

( ) ( ), , ,v x t i x t

( ) ( ), , ,v x x t i x x t+ ∆ + ∆

( ),v x x t+ ∆

( ),i x x t+ ∆R x∆ L x∆

G x∆ C x∆( ),v x t

( ) ,i x t

Department of Electronic Engineering, NTUT8/47

Page 9: 射頻電子 - [第二章] 傳輸線理論

(I)

• (KVL):

• (KCL):

( ) ( ) ( ) ( ) ( ) ( ),, , ,

i x tv x t v x x t R x i x t L x

t

∂− + ∆ = ∆ + ∆

( ) ( ) ( ) ( ) ( ) ( ),, , ,

v x x ti x t i x x t G x v x x t C x

t

∂ + ∆− + ∆ = ∆ + ∆ + ∆

( ),v x x t+ ∆

( ),i x x t+ ∆R x∆ L x∆

G x∆ C x∆( ),v x t

( ) ,i x t

Department of Electronic Engineering, NTUT9/47

Page 10: 射頻電子 - [第二章] 傳輸線理論

(II)

( ) ( ) ( ), ,,

v x t i x tRi x t L

x t

∂ ∂= − −

∂ ∂

( ) ( ) ( ), ,,

i x t v x tGv x t C

x t

∂ ∂= − −

∂ ∂

( ) ( ) ( ) ( ), , ,,

v x t v x x t i x tRi x t L

x t

− + ∆ ∂= +

∆ ∂

( ) ( ) ( ) ( ), , ,,

i x t i x x t v x x tGv x x t C

x t

− + ∆ ∂ + ∆= + ∆ +

∆ ∂

• ∆x

(transmission-line equation)(telegrapher equation)

Department of Electronic Engineering, NTUT10/47

Page 11: 射頻電子 - [第二章] 傳輸線理論

(III) −

• (KVL):

• (KCL):

( ) ( ) ( ) ( ) ( ) ( )V x V x x R x I x j L x I xω− + ∆ = ∆ + ∆

( ) ( ) ( ) ( ) ( ) ( )I x I x x G x V x x j C x V x xω− + ∆ = ∆ + ∆ + ∆ + ∆

( )V x x+ ∆

( )I x x+ ∆R x∆ j L xω ∆

G x∆1

j C xω ∆( )V x

( )I x

( ) ( ), cosv x t V x tω=

Department of Electronic Engineering, NTUT11/47

Page 12: 射頻電子 - [第二章] 傳輸線理論

(IV) −

( ) ( ) ( )dV xRI x j LI x

dxω= − −

( ) ( ) ( )dI xGV x j CV x

dxω= − −

( ) ( ) ( ) ( )V x V x xRI x j LI x

− + ∆= +

( ) ( ) ( ) ( )I x I x xGV x x j CV x x

− + ∆= + ∆ + + ∆

• ∆x

(transmission-line equation)(telegrapher equation)

Department of Electronic Engineering, NTUT12/47

Page 13: 射頻電子 - [第二章] 傳輸線理論

( ) ( ), ,v x t i x tL

x t

∂ ∂= −

∂ ∂

( ) ( ), ,i x t v x tC

x t

∂ ∂= −

∂ ∂

• R = G = 0

( ),v x x t+ ∆

( ),i x x t+ ∆L x∆

C x∆( ),v x t

( ) ,i x t

( ) ( ) ( ), ,,

v x t i x tRi x t L

x t

∂ ∂= − −

∂ ∂

( ) ( ) ( ), ,,

i x t v x tGv x t C

x t

∂ ∂= − −

∂ ∂

Department of Electronic Engineering, NTUT13/47

Page 14: 射頻電子 - [第二章] 傳輸線理論

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) , cos Re Rej t x j x j tv x t f x t x f x e f x e eω ϕ ϕ ωω ϕ += + = =

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) , cos Re Rej t x j x j ti x t g x t x g x e g x e eω η η ωω η += + = =

( ) ( ) ( )j xI x g x e η=

( ) ( ) ( )j xV x f x e ϕ= ( ) ( ) , Re j tv x t V x e ω=

( ) ( ) , Re j ti x t I x e ω=

• :

f(x) g(x) ( )xϕ ( )xη

time-domain

Department of Electronic Engineering, NTUT14/47

Page 15: 射頻電子 - [第二章] 傳輸線理論

( ) ( )V xj LI x

∂= −

( ) ( )I xj CV x

∂= −

( ) ( )dV xj LI x

dxω= −

( ) ( )dI xj CV x

dxω= −

( ) ( )V x dV x

x dx

∂=

( ) ( ) ( )2

22

d V x dI xj L LCV x

dx dxω ω= − = −

( ) ( ) ( ) ( )2 2

2 22 2

0d V x d V x

LCV x V xdx dx

ω β+ = + =

LCβ ω=•

( ) j x j xV x Ae Beβ β−= + A B (complex constant)

• (Phasor)

(Think that if approaches zero?)ω

( )β

Phasor

Department of Electronic Engineering, NTUT15/47

Page 16: 射頻電子 - [第二章] 傳輸線理論

( ) ( ) ( ) ( )1 1 j x j xdV xI x A j e B j e

j L dx j Lβ ββ β

ω ω− = = − + − −

LCβ ω=

• 0

L L LZ

CLC

ω ωβ ω

= = =

j x j xA e B eL L

β ββ βω ω

− = −

( ) j x j xV x Ae Beβ β−= + ( ) ( )dV xj LI x

dxω= −

Z0

Z0

( )0 0

j x j xA BI x e e

Z Zβ β−= −

Department of Electronic Engineering, NTUT16/47

Page 17: 射頻電子 - [第二章] 傳輸線理論

− (I)

1Z

2Z 3Z 3 1 2Z Z Z= +

1Z

2Z 2Z

1Z

2Z

1Z

3Z

1Z

4Z 5Z

4 2 3||Z Z Z=

Department of Electronic Engineering, NTUT17/47

Page 18: 射頻電子 - [第二章] 傳輸線理論

− (II)

2 01 1 0

2 0

2 0

11 1ab

Z ZZ Z Z Z

Z ZZ Z

= + = + =++

21 1

0 1 22 4

Z ZZ Z Z

= + +

Z0

a-bZ0 c-d

Z0( )

1Z

2Z 2Z

1Z

2Z

1Z 1Zc

d

a

b

a

b

c

d

2Z

1Z

0Z 0Z

a

b

Department of Electronic Engineering, NTUT18/47

Page 19: 射頻電子 - [第二章] 傳輸線理論

− (III)

•2

1 10 1 22 4

Z ZZ Z Z

= + +

L

2 2

0 0lim

4L

L L LZ

C C

ω→

= − =

2 2

0 4

L LZ

C

ω= −4 LCω <

4 LCω >

0, 0lim 4

L CLCω

→ →= = ∞

1Z j Lω=

2

1Z

j Cω=

L

C C

La

b

CCC

2L 2L 2L2La

b

a′

Department of Electronic Engineering, NTUT19/47

Page 20: 射頻電子 - [第二章] 傳輸線理論

( ) ( ) ( ) ( ) , Re Re j x t j x tj tv x t V x e Ae Beβ ω β ωω − − += = +

( ) ( ) ( ) ( )

0 0

, Re Re j x t j x tj t A Bi x t I x e e e

Z Zβ ω β ωω − − +

= = −

( ) ( )cos cosA x t B x tβ ω β ω= − + +

( ) ( )0 0

cos cosA B

x t x tZ Z

β ω β ω= − − +

( ) j x j xV x Ae Beβ β−= +

( )0 0

j x j xA BI x e e

Z Zβ β−= −

• Phasor

Department of Electronic Engineering, NTUT20/47

Page 21: 射頻電子 - [第二章] 傳輸線理論

(Wavelength)

• +x

x (rads)x

j xe β− xβ

2x

x λπ β βλ=

= = 2πλβ

=

( ) ( )1 , Re cosj x j tv x t Ae e A t xβ ω ω β−= = −

x λ=

( ) j xV x Ae β−=

0x = x λ=

0t = t T=

distance

time

phase 0xxβ

=2

xx λβ π

==

For simplification, assume thewave starts from x=0 and t=0.

Department of Electronic Engineering, NTUT21/47

Page 22: 射頻電子 - [第二章] 傳輸線理論

(Wave Velocity)

( )70 4 10 Wb/A-mL µ π −= ⋅≃

( )120 8.85419 10 F/mC ε −⋅≃ ≃

( ) ( )8,

0 0

1 light speed 3 10 /p vacuumv c m s

µ ε= = = ⋅

( )00

0

377 L

ZC

µε

= = = Ω

,0

p vaccumv

fλ =

2 1

2 2pv fT LC

λ ω π ω ωλ λπ β π β

= = ⋅ = = = =

• : ( T )

( )

Department of Electronic Engineering, NTUT22/47

Page 23: 射頻電子 - [第二章] 傳輸線理論

0µ µ≃0,rε ε ε≃

( )8

0 0

1 3 10/p

r r r

cv m s

µ ε ε ε ε⋅= = =

(non-magnetic material)

0p rg

r

cv

f f

ε λλε

= = =

ε r

81.5rε =

( )8

7,

3 103.32 10 /

81.5p waterv m s

⋅= ⋅≃

0, 00.11

81.5g water

λλ λ= ≅

Department of Electronic Engineering, NTUT23/47

Page 24: 射頻電子 - [第二章] 傳輸線理論

@1 GHz * @10 GHz

Air ( ) 1 30 cm − 3 cm −

Alumina( ,96%) 9.6 9.68 cm 0.48 cm 0.968 cm 0.48 mm

Sapphire ( ) 9.4 9.78 cm 0.49 cm 0.978 cm 0.49 mm

Glass ( ) 5 13.42 cm 0.67 cm 1.342 cm 0.67 mm

Polyimide ( ) 3.2 16.77 cm 0.84 cm 1.677 cm 0.84 mm

Quartz ( ) 3.8 15.39 cm 0.77 cm 1.539 cm 0.77 mm

FR4 ( ) 4.5 14.14 cm 0.71 cm 1.414 cm 0.71 mm

RT-duroid 5880 ( ) 2.16 20.41 cm 1.02 cm 2.041 cm 1.02 mm

RT-duroid 6010 ( ) 10.2 9.39 cm 0.47 cm 0.939 cm 0.47 mm

Si ( ) 11.9 8.70 cm 0.44 cm 0.870 cm 0.44 mm

GaAs ( ) 12.85 8.37 cm 0.42 cm 0.837 cm 0.42 mm

LTCC ( ) 7.8 10.74 cm 0.54 cm 1.074 cm 0.54 mm

* : 1/20

Department of Electronic Engineering, NTUT24/47

Page 25: 射頻電子 - [第二章] 傳輸線理論

2t T

tω π=

= 2T

πω

=

( ) ( )1 , cosv x t A x tβ ω= −

0x =

( )1 0, cosv t A tω=

π 2πtω

00

ttω

==

A

A−

( )1 0, cosv t A tω=

0x = x = l

0x =

x = l•

We only pay attention to this point

Department of Electronic Engineering, NTUT25/47

Page 26: 射頻電子 - [第二章] 傳輸線理論

2x

x λβ π=

= 2πλβ

=

( ) ( )1 , cosv x t A x tβ ω= −

0t =

( )1 ,0 cosv x A xβ=

π 2πxβ

00

xxβ

==

A

A−

( )1 ,0 cosv x A xβ=

0x = x = l

0x =

x = l•

We now pay attention to the whole line at any time instant (here, t=0)

Department of Electronic Engineering, NTUT26/47

Page 27: 射頻電子 - [第二章] 傳輸線理論

( )

2λ λx

A

A−

t

t T=

2t T=

x

x

t

Department of Electronic Engineering, NTUT27/47

Page 28: 射頻電子 - [第二章] 傳輸線理論

x =

2λ λx

A

A−

t

t T=

2t T=

t

λ

0x = x = l

We only pay attention to this point

x λ=

Department of Electronic Engineering, NTUT28/47

Page 29: 射頻電子 - [第二章] 傳輸線理論

(Terminated Transmission Line)

LZ LZ0Z 0Z

j xAe β−

j xBe β j xBe β

j xAe β−

0x = x = l 0d =d = l

( ) j x j xV x Ae Beβ β−= +

( )0 0

j x j xA BI x e e

Z Zβ β−= −

( )IN dΓ

( ) 1 1j d j dV d A e B eβ β−= +

( ) 1 1

0 0

j d j dA BI d e e

Z Zβ β−= −

1jA Ae β−= l

1jB Be β= l

incident wave

reflected wave

d x= −l

Department of Electronic Engineering, NTUT29/47

Page 30: 射頻電子 - [第二章] 傳輸線理論

(Reflected Coefficient)

( ) 1 1j d j dV d A e B eβ β−= +

( ) 2 21 10

1 1

j dj d j d

IN j d

B e Bd e e

A e A

ββ β

β

−− −Γ = = = Γ

( ) 10

1

0IN

B

AΓ = Γ =

• (at d = l ) (at d = 0)

d=0 :0Γ ( )0INΓ

Department of Electronic Engineering, NTUT30/47

Page 31: 射頻電子 - [第二章] 傳輸線理論

(I)

( ) ( ) ( )21 0 1 01j d j d j d j dV d A e e A e eβ β β β− −= + Γ = + Γ

( ) ( ) ( )21 10 0

0 0

1j d j d j d j dA AI d e e e e

Z Zβ β β β− −= − Γ = − Γ

( ) ( )( )

00

0

j d j d

IN j d j d

V d e eZ d Z

I d e e

β β

β β

+ Γ= =− Γ

( ) 00

0

10

1IN LZ Z Z+ Γ= =− Γ

00

0

L

L

Z Z

Z Z

−Γ =+

( ) 1 1j d j dV d A e B eβ β−= +

( ) 1 1

0 0

j d j dA BI d e e

Z Zβ β−= −

d

(d = 0)

0 0Γ =0LZ Z=

LZ( )V d

0d =d = l

( )IN dΓ

( )INZ d

( )I d

+

10

1

B

AΓ =

Department of Electronic Engineering, NTUT

0Z

31/47

Page 32: 射頻電子 - [第二章] 傳輸線理論

(II)

( ) ( ) ( )( ) ( )

0 00

0 0

j d j dL L

IN j d j dL L

Z Z e Z Z eZ d Z

Z Z e Z Z e

β β

β β

+ + −=

+ − −

0 00 0

0 0

cos sin tan

cos sin tanL L

L L

Z d jZ d Z jZ dZ Z

Z d jZ d Z jZ d

β β ββ β β

+ += =+ +

00

0

L

L

Z Z

Z Z

−Γ =+

( ) ( )( )

00

0

j d j d

IN j d j d

V d e eZ d Z

I d e e

β β

β β

+ Γ= =− Γ

d = 0 ( )0IN LZ Z=

d = l ( ) 00

0

tan

tanL

INL

Z jZZ Z

Z jZ

ββ

+=+

ll

l

( )

Department of Electronic Engineering, NTUT32/47

Page 33: 射頻電子 - [第二章] 傳輸線理論

(VSWR)

( ) 21 01 j dV d A e β−= + Γ

( )( )

0max

0min

1

1

V dVSWR

V d

+ Γ= =

− Γ

( ) ( ) ( )21 0 1 01j d j d j d j dV d A e e A e eβ β β β− −= + Γ = + Γ

( voltage standing-wave ratio, VSWR):

1/4

( ) ( )1 0max1V d A= + Γ ( ) ( )1 0min

1V d A= − Γ

Department of Electronic Engineering, NTUT33/47

Page 34: 射頻電子 - [第二章] 傳輸線理論

(Matched Line)

• (Matched line):

( )IN LZ d Z=

dVSWR 1

0 0Γ = 1VSWR =

0Z

0LZ Z=

0d =d = l

( ) 0INZ Z=l

( ) 0INZ d Z=

0Z

( ) 00

0

tan

tanL

INL

Z jZ dZ d Z

Z jZ d

ββ

+=+

0LZ Z=

Department of Electronic Engineering, NTUT34/47

Page 35: 射頻電子 - [第二章] 傳輸線理論

• d

( )VSWR

0LZ =

0 1Γ = −

4

λ=l ( )scZ = ∞l

2

λ=l ( ) 0scZ =l

0LZ =

0d =d = l

( ) 0 tanINZ jZ β=l l

( ) 0 tanscZ d jZ dβ=

0Z

( ) 0 tanscZ d jZ dβ=( ) 00

0

tan

tanL

INL

Z jZ dZ d Z

Z jZ d

ββ

+=+

0LZ =

Department of Electronic Engineering, NTUT35/47

Page 36: 射頻電子 - [第二章] 傳輸線理論

4

λ=l ( ) 0ocZ =l

2

λ=l ( )ocZ = ∞l

( )VSWR

0 1Γ =

• dLZ = ∞

( ) 0 cotocZ d jZ dβ= −( ) 00

0

tan

tanL

INL

Z jZ dZ d Z

Z jZ d

ββ

+=+

LZ → ∞

LZ = ∞

0d =d = l

( ) 0 cotINZ jZ β= −l l

( ) 0 cotocZ d jZ dβ= −

0Z

Department of Electronic Engineering, NTUT36/47

Page 37: 射頻電子 - [第二章] 傳輸線理論

¼

• ¼ :

( )204INL

ZZ

Zλ =

0 x LZ Z Z= ⋅

1/4

4d λ= =l

LZ xZ

: 1/475 LZ = Ω 50 Ω

You can simply use a transmission line with 61.2 Ohm characteristic impedance!

( )0 4 50 75 61.2 IN LZ Z Zλ= ⋅ = ⋅ = Ω

( ) 00

0

tan

tanL

INL

Z jZ dZ d Z

Z jZ d

ββ

+=+

LZ

0d =4d λ=

20

4INL

ZZ

Z

λ =

0Z

Department of Electronic Engineering, NTUT37/47

Page 38: 射頻電子 - [第二章] 傳輸線理論

1/2(

)

( )2IN LZ Zλ =2d λ= =l( ) 00

0

tan

tanL

INL

Z jZ dZ d Z

Z jZ d

ββ

+=+

LZ

0d =2d λ=

2IN LZ Zλ =

0Z

Department of Electronic Engineering, NTUT38/47

Page 39: 射頻電子 - [第二章] 傳輸線理論

( ) ( )1 12 sinj d j dV d A e e j A dβ β β−= − =

( ) ( ) 21, Re Re 2 sin

j tj tv d t V d e A d e

πωω β

+

= = ⋅

( ) ( )1 14 2 sin 2 2V j A j Aλ π= =

( ) ( )12 2 sin 0V j Aλ π= =

4d

λ=

2

dλ=

Department of Electronic Engineering, NTUT39/47

Page 40: 射頻電子 - [第二章] 傳輸線理論

( ) 1, 2 sin cos2

v d t A d tπβ ω = ⋅ +

π2

π3

2

π2π

( )( )

max

min

V dVSWR

V d= = ∞

( )V d

( )1 max2A V d=

( )min

0 V d=dβ

d

2

λ4

λ3

4

λλ

2

π3

2

π π2π

( ),v d t

12A

( )min

0 V d=dβ

d

2

λ4

λ3

4

λλ

12A

32t πω =5

4t πω =

3,4 4t π πω = 2t πω =

0,tω π=

Department of Electronic Engineering, NTUT40/47

Page 41: 射頻電子 - [第二章] 傳輸線理論

(I)

100 50sZ j= + Ω50 50LZ j= + Ω

10 0sv °= ∠

( )( ) ( )

50 5010 0 3.92 11.31 (V)

50 50 100 50L

L sL s

jZV V

Z Z j j

+= = ∠ = ∠

+ + + +

Department of Electronic Engineering, NTUT41/47

Page 42: 射頻電子 - [第二章] 傳輸線理論

(II)

• 50 Ohm(load reflection coefficient)

VSWR

10 0sv °= ∠

100 50sZ j= + Ω50 50LZ j= + Ω

( )( )

00

0

50 50 500.447 63.44

50 50 50L

L

jZ Z

Z Z j

+ −−Γ = = = ∠+ + +

( ) ( )( ) ( )50 50 50 tan 45

8 50 100 50 50 50 50 tan 45IN

j jZ j

j jλ

°

°

+ += = − Ω + +

0

0

1 1 0.4472.62

1 1 0.447VSWR

+ Γ += = =− Γ −

0 50Z = Ω

l 8λ=

( )8INZ λ

LV

Department of Electronic Engineering, NTUT42/47

Page 43: 射頻電子 - [第二章] 傳輸線理論

(II) –

( ) ( )( )

( )( ) ( )

8 100 508 10 0 5.59 26.57

8 100 50 100 50IN

sIN s

Z jV V

Z Z j j

λλ

λ−

= = ∠ = ∠ −+ − + +

( ) ( )21 01j d j dV d A e eβ β−= + Γ

( ) 4 218 5.59 26.57 1 0.447 63.44

j jV A e e

π π

λ−°

= ∠ − = + ∠ ⋅

1 3.95 63.44A °= ∠ −

( )0 3.95 63.44 1.77 5 45 (V)LV V ° °= = ∠ − + = ∠

100 50sZ j= + Ω

10 0sv °= ∠( )8 100 50INZ jλ = − Ω

Equivalent circuit

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LZ

+

−sE

sZV

I

+

sourceimpedance

loadimpedance

Phasor

•( ) ( )

2 22 2

2 2

1 1 1

2 2 2s s L

L rms L L Ls L s L s L

E E RP I R I R R

Z Z R R X X

= = = = + + + +

•s LX X= −s LR R= s LZ Z ∗=

• ( ) ( ) ?

( )

2

,max

1

8s

Ls

EP

R=

s

s L

EI

Z Z=

+

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Page 45: 射頻電子 - [第二章] 傳輸線理論

(Power Waves)

0L o

L o

Z Z

Z Z

−Γ =+

( )

s sV E Z I= −

LZ

+

−sE

sZV

I

+

sourceimpedance

loadimpedance

oZ

d = l

LZ

0d =

( )IN dΓ

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Page 46: 射頻電子 - [第二章] 傳輸線理論

LZ

+

−sE

sZV

I

+

LR+

−sE

sR

V

+

sjX LjX

( )( )

( )( )

L s s LL s L ss L s

s L s L s L s L s s L

R R j X XR j X X RZ R Z Z

Z R R j X X R Z Z R R j X X

∗ − − + + + −− − Γ = = = =+ + + + + + + +

L s

L s

Z Z

Z Z

∗−Γ =+

• ( )*L sZ Z= Γ = 0

• (Z0)Rs

LZ+

−sE

sR

V

+

sjX

L LR jX+

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( )

• 1

• 20λ

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