FALLSEM2013-14_CP1630_24-Jul-2013_RM01_L5-L5_FS-2013-14-review-of-TL

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R vi w f Tr n mi i n Lin  SENSE

Transcript of FALLSEM2013-14_CP1630_24-Jul-2013_RM01_L5-L5_FS-2013-14-review-of-TL

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R vi w f Tr n mi i n Lin 

SENSE

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Coaxial cable is an exam le of TL Transmission lines usually consist of 2 parallel

Lumped Element Model for a Transmission Line

conductors.

A short segment  ∆ z  of transmission line can be

modeled as a lumped-element circuit.

Conduction current incentre conductor 

R, and L

z, t

V (z, t)

Displacement current

(conductance, capitance)

 I 

 E

+ + + +

 I + I   I  H 

 I + I 

 I - - - -

V + V 

 I + I 

 I  H 

V + V 

 I + I 

SENSE

The voltage and current vary along the structure in time t and distance z.

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e can

consider on a

small section of 

TL  ∆

z. It is sosmall that

are occurring

instantaneouslyan we can

simply use

lumped circuit

theory.

SENSE

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 I 

+ + + +

 I + I   I  H 

 I + I 

na ys s L j R   ω +),(   t  z I 

dI  I    δ +

 I - - - -

V + V 

 I + I 

 I  H 

V + V 

 I + I 

dz

),(   t  zV  C  jG   ω +dz

dV  zV    δ +

From Kirchoff Volta e Law Kirchoff current law 

 zdt 

dV C  zGV  z

dz

dI  I  I    δ δ δ    +=⎟

 ⎠

 ⎞⎜⎝ 

⎛  +− zdt 

dI  L z RI  z

dz

dV V V    δ δ δ    +=⎟

 ⎠

 ⎞⎜⎝ 

⎛  +−

 zdt 

dI  L z RI  z

dz

dV δ δ δ    +=−   z

dt 

dV C  zGV  z

dz

dI δ δ δ    +=−

dt 

dI  L RI 

dz

dV +=−

dt 

dV C GV 

dz

dI +=−(a) (b)

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dI 

Let’s V=Voe jωt , I = Ioe

 jωt

Therefore

V  j

dt 

ω =dt 

=en

( ) I  L j Rdz

dV ω +=−   ( )V C  jG

dz

dI ω +=−

a

b

Differentiate with respect to z

( )dI 

 L j RV d 

ω +=−2

2   ( )dz

dV C  jG

dz

 I d ω +=−

2

2

 z z( )( )V C  jG L j R

dz

V d ω ω    ++=

2

2

( )( ) I C  jG L j Rdz

 I d ω ω    ++=

2

2

dz

V d    2

2  γ =  I 

dz

 I d    2

2

2

γ =

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The solution of V and I can be written in the form of 

c d z z  Be AeV    γ γ    +− +=o

 z z

 Z  Be Ae I 

γ γ    +−

−=

where

⎥⎦

⎢⎣

+

+=

C  jG

 L j R Z 

o ω 

ω 

( )( )C  jG L j R j   ω ω  β α γ    ++=+=and

Let say at z=0 , V=VL , I=IL and Z=ZL

ef 

 B AV  L   +=o

 L Z 

 B A I    −=

and   L

 L

 L  Z 

 I 

=

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o L L   Z  I V  A

  +=

  o L L   Z  I V  B

  −=

Inserting in equations ( c) and (d) we have

⎥⎥⎤

⎢⎢⎡   −

−⎥⎥⎤

⎢⎢⎡   +

=−+−+

22)(

 z z

o L

 z z

 L

ee Z  I 

eeV  zV 

γ γ γ γ 

⎥⎤

⎢⎡   −

−⎥⎤

⎢⎡   +

=−+−+

)( z z

 L z z

 L

eeV ee I  z I 

γ γ γ γ 

o

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⎤⎡   +   −   z zee

  γ γ    ⎤⎡   −   −   z zee

  γ γ 

⎥⎦⎢⎣

=

2

cos   zγ u

⎥⎦⎢⎣  2

Then we have

)sinh()cosh()(   z Z  I  zV  zV o L L

  γ γ    −= *

)sinh()cosh()(   z Z 

V  z I  z I 

o

 L L   γ γ    −= **

⎟⎟

⎜⎜

−==)sinh()cosh(

)sinh()cosh(

)(

)()(

 zV 

 z I 

 z Z  I  zV 

 z I 

 zV  z Z 

 L L

o L L

γ γ 

γ γ and

o

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⎟⎟ ⎠⎜⎜⎝    −−

= )sinh()cosh(

)sinh()cosh(

)(  z Z  z Z 

 z Z  z Z 

 Z  z Z  Lo

o Lo γ γ 

γ γ or 

⎟⎟

 ⎠

⎜⎜

⎝ 

⎛ 

−=

)tanh(

)tanh()(

 z Z  Z 

 z Z  Z  Z  z Z 

 Lo

o Lo

γ 

γ Or further reduce

For lossless transmission line , γ= jβ since  α=0

⎟⎟ ⎞

⎜⎜⎛    +

=  )tan(

)(  z jZ  Z 

 Z  z Z    o Lo

 β )cos()cosh(   z z j   β  β    =

sinsinh   −=o

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Standin Wave Ratio SWRantinode

node z

 Beγ 

Reflection coefficient

 Ae-γz

Beγz

 z Ae

  γ −=

Γ+=+=   −− 1 z z ze Be AeV    γ γ γ 

Voltage and current in term of reflection coefficient

V    ⎛    Γ+1

−   −−   z z z Ae Be Ae

  γ γ γ 

o L

 L  I    ⎠⎝    Γ−== 1

⎛    Γ+1 Z −==oo

 L Z  Z    ⎠⎝    Γ−

=1o Z 

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For loss-less transmission line   γ = jβ

By substituting in * and ** ,voltage and current amplitude are

2/12 gcos   +++=   z zV 

2/1

2 )2cos(21)(   θ  β    +Γ−Γ+=   z A z I  ho

Voltage at maximum and minimum points are

max  =   min   −=Γ+

==  1Therefore  L Z 

Γ−1 

o Z 

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Other related equations

oo   sZ  Z  I V  Z    =⎟⎟

 ⎠⎜⎜⎝ 

⎛ 

Γ− Γ+== 11

min

maxmax

s Z  Z 

 I V  Z    o

o   =⎟⎟ ⎠

⎜⎜⎝ 

⎛ 

Γ+ Γ−== 11

max

minmin

⎟⎟

 ⎞

⎜⎜

⎛    −

=Γ  o L   Z  Z 

o

From equations (g) and (h), we can find the max and min points

K,4,2,02   π π θ  β    ±±=+ zMaximum

K,,   π π = z

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 Zo ZLZin

lγ tanho L   jZ  Z    +lγ tanh Lo

on jZ  Z   +

o L

o L

 Z  Z  Z  Z 

+−=Γ

Γ−Γ+=

1

1SWR

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Various forms of  Transmission Lines

Two wire

cable   Coaxialcable

Microstripe

line

Rectangular 

waveguideCircular 

waveguideStripline

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Comparison of  Waveguide and Transmission Line Characteristics

Transmission line

•   -

wire, coaxial, microstrip, etc.).• Normal operating mode is the TEM or quasi-TEM mode (can support

.

• No cutoff frequency for the TEM mode. Transmission lines can

transmit signals from DC up to high frequency.• gn can s gna a enua on a g requenc es ue o con uc or

and dielectric losses.

• Small cross-section transmission lines (like coaxial cables) can only

ransm ow power eve s ue o e re a ve y g e s concen ra eat specific locations within the device (field levels are limited by

dielectric breakdown).

arge cross-sec on ransm ss on nes e power ransm ss on nes

can transmit high power levels.

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Comparison of  Waveguide and Transmission Line Characteristics

Waveguide

• Metal waveguides are typically one enclosed conductor filled with an

,

consists of multiple dielectrics.

pera ng mo es are or mo es canno suppor a mo e .

• Must operate the waveguide at a frequency above the respective TE orTM mode cutoff frequency for that mode to propagate.

• Lower signal attenuation at high frequencies than transmission lines.

• Metal waveguides can transmit high power levels. The fields of thepropagating wave are spread more uniformly over a larger cross-

sectional area than the small cross-section transmission line.

• Large cross-section (low frequency) waveguides are impractical due to

large size and high cost.

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SENSE

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SENSE

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 ( )ab L   /ln2π 

µ 

=a

ab

/ln

2   ε π =   c

c

ck 

 f   =( )ab Z o   /ln

2

1

ε 

µ 

π =

k c =  2

Where a = radius of inner conductor 

b = radius of outer conductor