Transmission Linespol/pdf/transmissionLines.pdftransformer, Slotted line, Single stub VI....

1

Transmission Lines

Dr. Sandra Cruz-Pol ECE Dept. UPRM

Introduction to Transmission Lines (T.L.) n Hi frequency or hi power requires T.L. n TEM waves propagate thru T.L. n We will develop T.L. theory to see how

waves propagate thru them

Exercise 11.3 n  A 40-m long TL has Vg=15 Vrms, Zo=30+j60 Ω, and VL=5e-j48o Vrms. If the

line is matched to the load and the generator, find: the input impedance Zin, the sending-end current Iin and Voltage Vin, the propagation constant γ.

n  Answers:

ZL Zg

Vg

+ Vin

-

Iin

+ VL

-

Zo=30+j60 γ=α +j β

40 m

Zin = 30+ j60Ω, Iin = 0.112∠− 63.4o A,

Vin = 7.5∠0oVrms, γ = 0.0101+ j0.02094 m−1

7.45∠−111oe−γ 40 = 0.112∠− 63o

Seem to be in parallel, but these V are not equal!

Don’t worry about the details, I’ll teach you

about solving this type of problems pretty soon.

Transmission Lines I.  TL parameters (R’,L’,C’, G’) II.  TL Equations (for V and I) III.  3 Concepts:

I.  Input Impedance, II.  Reflection Coefficient and III.  Characteristic Impedance

IV.  SWR, Power V.  Smith Chart VI.  Applications - Quarter-wave

transformer, Slotted line, Single stub VI.  Microstrips

Transmission Lines (TL) n  TL have 2

conductors in parallel with a dielectric separating them

n  They transmit TEM waves inside the lines for insulation

outer conductor

Common Transmission Lines Two-wire (ribbon)

Coaxial

Microstrip

Stripline (Triplate)

twinlead

2

Other TL (higher order) [Chapter 12] Fields inside the TL

n V proportional to E, n  I proportional to H

∫∫⋅=

⋅−=

dlHI

dlEV

Distributed parameters The parameters that characterize the TL are

given in terms of per length. n R = ohms/meter n  L = Henries/ m n C = Farads/m n G = mhos/m

cmckmc

GHz

Hz

15000,000,2000/000,560/

2

60

==

==

λ

λ

At high frequencies we’re dealing with wavelengths comparable to the size of

the circuits.

Common Transmission Lines R’, L’, G’, and C’ depend on the particular transmission line structure and the material properties. R, L, G, and C can be calculated using fundamental EMAG techniques.

Parameter Two-Wire Line Coaxial Line Parallel-Plate Line

Unit

R’ L’ G’ C’

€

1πaσ condδ

€

12πσ condδ

1a

+1b

⎛

⎝ ⎜

⎞

⎠ ⎟

€

2wσ condδ

€

Ω /m

€

µπacosh D

2a⎛

⎝ ⎜

⎞

⎠ ⎟

€

µ2πln b

a⎛

⎝ ⎜ ⎞

⎠ ⎟

€

µdw

€

H /m

€

πσ diel

acosh D / 2a( )( )

€

2πσ diel

ln b /a( )

€

σ dielwd

€

S /m

€

πεacosh D / 2a( )( )

€

2πεln b /a( )

€

εwd

€

F /m

TL representation Distributed line parameters Using KVL:

3

Distributed parameters n  Taking the limit as Δz tends to 0 leads to

n  Similarly, applying KCL to the main node gives

ttzILtzRI

ztzV

∂

∂+=

∂

∂−

),(),(),(

ttzVCtzGV

ztzI

∂

∂+=

∂

∂−

),(),(),(

−∂V∂z

= RI + L ∂I∂t

−∂I∂z=GV +C ∂V

∂t0 2

2

2

=−∂

∂s

s VzV

γ

Wave equation

n  Using phasors

n  The two expressions reduce to

Wave Equation for voltage

−∂Vs∂z

= R+ jωL( ) Is

−∂Is∂z

= G + jωC( )Vs

])(Re[),(

])(Re[),(tj

s

tjs

ezItzIezVtzVω

ω

=

=

( )( )CjGLjR ωωγ ++=2





TL Equations n Note that these are the wave eq. for

voltage and current inside the lines.

n The propagation constant is γ and the wavelength and velocity are

d2Vsdz2

−γ 2Vs = 0d 2Isdz2

−γ 2Is = 0

γ =α +β j = (R+ jωL)(G + jωC)

λ = 2π β u =ω β = f λ

Waves move through line n  The general solution is

n  In time domain is

Vs =V+e−γz + V −eγz

z

V (z, t) = Re[Vs (z)ejω t ]

=V +e−αz cos(ωt −βz)+V −e+αz cos(ωt +βz)

Waves move through line n  For Current

n  Similarly for time-domain, I

Is = I+e−γz + I −eγz

z

I(z, t) = I +e−αz cos(ωt −βz)+ I −eαz cos(ωt +βz)

4





We will define 3 concepts:

n Characteristic impedance, Zo

n Reflection Coefficient, ΓL n  Input Impedance, Zin




Characteristic Impedance of a Line, Zo n  Is the ratio of positively traveling voltage wave to

current wave at any point on the line

−dV (z)dz

= (R+ jωL)I(z)

z

Zo =V +

I +=R+ jωL

γ=

R+ jωLG + jωC

= Ro + jXo = −V −

I −

substitutingV (z) =V +e−γz

I(z) = I +e−γz

−(−γV +e−γz ) = (R+ jωL)I +e−γz

Characteristic Impedance, Zo

Zo =V +

I += −

V −

I −=

R+ jωLG + jωC

Is = I+e−γz + I −eγz

Is =V +

Zoe−γz +

V −

−Zoeγz

Different cases of TL

n  Lossless

n Distortionless

n  Lossy

Transmission line

Transmission line

Transmission line

Find the characteristic impedance and propagation constant for each:

5

Lossless Lines (R=0=G) Have perfect conductors and a perfect dielectric

medium between them.

n  Propagation:

n  Wavelength & Velocity:

n  Impedance

α = 0, γ = jβ, β =ω LC

u = ωβ

= 1LC

= f λ, λ =2πβ

Zo = Ro =LC

Xo = 0

Distortionless line (R/L = G/C) Is one in which the attenuation is independent on

frequency.

n  Propagation:

n  Velocity:

n  Impedance

γ =α + jβ

α = RG β =ω LC

u = ωβ

= 1LC

= f λ

Xo = 0 Zo = Ro =LC=

RG

Summary γ = α + jβ

Propagation ConstantZo

Characteristic Impedance General (Lossy)

Lossless

Distortionless RC = GL LCjRG ωγ +=

LCjωγ += 0

CjGLjRZo ω

ω+

+=

GR

CLRZ oo ===

))(( CjGLjR ωωγ ++=

CLRZ oo ==

u = ωβ= 707 km / s

P.E. 11.2 A telephone line has R=30 Ω/km, L=100 mH/km, G=0, and C= 20µF/km. At 1kHz, FIND: the characteristic impedance of the line, the propagation constant, the phase velocity. n  Is it distortionless? Solution: Zo =

R+ jωLG + jωC

=30+ j2π (1k)(100m)0+ j2π (1k)20 ⋅10−6

= 70.75∠−1.37oΩ

γ = R+ jωL( ) G + jωC( )

= 30+ j2π (100)( ) 0+ j2π20 ⋅10−3( )= 0.21+ j8.88 / km

No







6

Reflection coefficient at the load, ΓL

Vs (z) =V+e−γz +V −e+γz

Load is usually taken at z=0

and generator at z= -l

ΓL =V −

V +

( )zLz

s eeVzV γγ +−+ Γ+=)(

z=0 z= -l

For a Lossless TL terminated with a load

( )zLz

s eeVzV γγ +−+ Γ+=)(

( )zLz

os ee

ZV

zI γγ +−+

Γ−=)(

( )( )zL

z

zL

z

os

s

eeeeZ

zIzVzZ γγ

γγ

+−

+−

Γ−

Γ+==)()()(

Then,

Similarly,

The impedance anywhere along the line is given by

The impedance at the load end, ZL, is given by

Z(l = 0) = ZL = Zo1+ΓL( )1−ΓL( )

Terminated, Lossless TL

LCjj ωβγ =+= 0

Solving for ΓL

Conclusion: The reflection coefficient is a function of the load impedance and the characteristic impedance.

Recall for the lossless case,

Then ( )zjLzj

s eeVzV ββ +−+ Γ+=)(

( )zjLzj

os ee

ZV

zI ββ +−+

Γ−=)(

oL

oLL ZZ

ZZ+

−=Γ

Definition: Matched line

n Means that Zo=ZL

n Therefore there are no reflection!

ΓL=0 oL

oLL ZZ

ZZ+

−=Γ

What happens when you connect the wrong TL to a speaker?

Terminated, Lossless TL

( )

( )djLdjdj

o

djL

djdj

eeeZV

dI

eeeVdV

βββ

βββ

−+++

−++

Γ−=

Γ+=

)(

)(

Using convention the coordinate system, z = - l , at input.

Rewriting the expressions for voltage and current, we have

Rearranging,

( )

( )ljLlj

o

ljL

lj

eeZV

lI

eeVlV

ββ

ββ

−++

−+

Γ−=−

Γ+=−

)(

)(

( )

( )ljLlj

o

ljL

lj

eeZV

lI

eeVlV

ββ

ββ

2

2

1)(

1)(

−++

−++

Γ−=−

Γ+=−

-z

z = - l

7

Voltage anywhere on the Line

( )

( )djLdjdj

o

djL

djdj

eeeZV

dI

eeeVdV

βββ

βββ

−+++

−++

Γ−=

Γ+=

)(

)(

Recall,-z

z = - l

Vin =V (−l)VL =V (0)

V (l = λ / 4) =V + e jπ /2 + (0)e− jπ /2( )

Input voltage= sending end,

Load Voltage = receiving end

V (l) =V + e jβ l +ΓLe− jβ l( )

Voltage quaterwave from matchedload

λ / 4





Impedance (Lossless line) ( )( )ljL

ljL

o eeZ

lIlVlZ β

β

2

2

11

)()()(

−

−

Γ−

Γ+==

( )( ))(1

)(1)(llZlZ o Γ−

Γ+=

( )( )ljZZ

ljZZZZlZLo

oLoin β

βtantan)(

+

+==

ljjL

ljL eeel βθβ 22)( −− ΓΓ=Γ=Γ

The reflection coefficient at any point along the line:

Conclusion: The load impedance is “transformed” as we move away from the load.

After some algebra, an alternative expression for the impedance is given by

Then, the impedance can be written as.


The reflection coefficient can be modified as follows

Then, the impedance can be written as

After some algebra, an alternative expression for the impedance is given by

Conclusion: in lossy lines, we end up with the hyperbolic tangent.

Impedance (Lossy line) ( )( )lL

lL

o eeZ

lIlVlZ γ

γ

2

2

11

)()()(

−

−

Γ−

Γ+==

( )( ))(1

)(1)(llZlZ o Γ−

Γ+=

( )( )lZZ

lZZZZlZLo

oLoin γ

γtanhtanh)(

+

+==

)()( 222 ljlL

lL eeel βαγ −−− Γ=Γ=Γ

Example: Matched Case A TL has Vg=10 Vrms, Zo=50Ω If the line is matched to the load and the generator, find: the input impedance Zin, the sending-end Voltage Vin

n  Answers:

ZL Zg

Vg

+ Vin

-

Iin

+ VL

-

Zo γ=j β

l

Zg = ZL = Z0 ⇒ Zin = 50Ω

ΓL =ZL − ZoZL + Zo

= 0

Zin = ZoZL + jZo tanβl( )Zo + jZL tanβl( )

= Zo

Γ = 0VoltageDivider Vin = 5∠0

o

Example: Matched Case A λ/8 long TL has VL=5 < 30o, Zo=50Ω If the line is matched to the load, find: the input impedance Zin, the sending-end Voltage Vin, the propagation constant γ.

n  Answers:

ZL Zg

Vg

+ Vin

-

Iin

+ VL

-

Zo γ=j β

l

ZL = Z0 ⇒ Zin = 50Ω




Γ = 0

VL = 5∠30o =V

l=0=V + e jβ 0 + (0)e− jβ 0( ),

Solve for : V + = 5∠30o

V (l = λ / 8) = 5∠30o e jπ /4 + (0)e− jπ /4( )

Vin = 5∠30o e jπ /4( ) = 5∠75o

8

Exercise 1: Not Matched A 2cm lossless TL has VL=10 ej30o, Zg=60 Ω, ZL=50 Ω and Zo=100Ω, λ=10cm . Find: the input impedance Zin, the sending-end Voltage Vin,

n  Use this equation at load and at input, find V+ n  Find Vin n  Find Zin (at input)

ZL Zg

Vg

+ Vin

-

Iin

+ VL

-

Zo γ=j β

l

V (l) =V + e jβ l +ΓLe− jβ l( ) ΓL =

ZL − ZoZL + Zo


Exercise 2: using formulas n  A 2cm lossless TL has Vg=10 Vrms, Zg=60 Ω, ZL=100+j80 Ω and Zo=40Ω,

λ=10cm . find: the input impedance Zin, the sending-end Voltage Vin,

ZL Zg

Vg

+ Vin

-

Iin

+ VL

-

Zo γ=j β

l

( )52tan)80100(4052tan40)80100(

40)2(π

π

jj

jjcmZZin

++

⎟⎠

⎞⎜⎝

⎛ ++==

Ω−= 17.212.12 jZinVoltage Divider:

radZZZV

Vgin

ingin 766.030.3 −∠=

+=

Example 3: Not matched to load A generator with 10Vrms and Rg=50, is connected to a 75Ω load thru a 0.8λ, 50Ω-lossless line. n  Find VL

PowerDividerVin =∠

V + = 5.006∠1.26radVL =V (l = 0)



Zo = 50Ω ZL = 75, l = .8λ


ZL Zg

Vg

+ Vin

-

Iin + VL

-

Zo γ=j β

l

ΓL = 0.2Zin = 35.+ j8.75Ω

Exercise 11.3: Matched Case A 40-m long TL has Vg=15 Vrms, Zo=30+j60 Ω, and VL=5e-j48o Vrms. If the line is matched to the load and the generator, find: the input impedance Zin, the sending-end current Iin and Voltage Vin, the propagation constant γ.

n  Answers:

ZL Zg

Vg

+ Vin

-

Iin

+ VL

-

Zo=30+j60 γ=α +j β

40 m

oorms

oin

oinin

emjVV

AIjZ

63112.011145.7

02094.00101.0,05.7

,4.63112.0,6030

40

1

−∠=−∠

+=∠=

−∠=Ω+=

−

−

γ

γ





Power n  The average input power at a distance l from the load is given

by

n  which can be reduced to

n  The first term is the incident power and the second is the reflected power. Maximum power is delivered to load if Γ=0

[ ])()(Re21 * lIlVPave =

( )22

12

Γ−=+

o

oave Z

VP

9

SWR or VSWR or s Whenever there is a reflected wave, a standing wave will form out of the combination of incident and reflected waves.

The (Voltage) Standing Wave Ratio - SWR (or VSWR) is defined as

L

Ls

II

VV

SWRs

Γ−

Γ+=

===

11

min

max

min

max

Summary

n  Input Impedance

n Reflection Coef

n SWR


s =1+ ΓL1− ΓL


Three (3) common Cases of line-load combinations: n  Shorted Line (ZL=0)

n  Open-circuited Line (ZL=∞)

n  Matched Line (ZL = Zo)

ljZZ oin βcot−=

oin ZZ =

Zin = 0+ jZo tanβl = jb

∞==Γ sL ,1

1,0 ==Γ sL

ΓL = −1, s =∞

Standing Waves –Short (ZL=0)

n  So substituting in V(z)

∞=−=Γ= sljZZ Loin ,1 , tanβ

-z -λ/4 -λ/2 -λ

|V(z)|

])1([)( ljlj eeVzV ββ −+ −+=

)sin2()( ljVzV β+=

( )lVzV βsin2)( +=

⎟⎠

⎞⎜⎝

⎛= + lVzVλπ2sin2)(

Voltage maxima

*Voltage minima occurs at same place that impedance has a

minimum on the line

Standing Waves –Open(ZL=∞)


|V(z)|

Zin = − jZo cotβ l , ΓL = +1, s =∞

-z -λ/4 -λ/2 -λ

Voltage minima

V (z) =V +[e jβl + (+1)e− jβl ]V (z) =V +(2cosβl)

V (z) = V + 2cos βl( )

V (z) = V + 2cos 2πλl

⎛

⎝⎜

⎞

⎠⎟

Standing Waves –Matched (ZL = Zo)


|V(z)| V (z) =V +[e jβl + (0)e− jβl ]V (z) =V +e jβl

V (z) = V + e jβl

V (z) = V +-z -λ/4 -λ/2 -λ

Zin = Zo , ΓL = 0, s =1

10

Java applets

n  http://www.amanogawa.com/transmission.html

n  http://physics.usask.ca/~hirose/ep225/ n  http://www.educatorscorner.com/

index.cgi?CONTENT_ID=2483





The Smith Chart

Smith Chart

n Commonly used as graphical representation of a TL.

n Used in hi-tech equipment for design and testing of microwave circuits

n One turn (360o) around the SC = to λ/2

What can be seen on the screen?

Network Analyzer Smith Chart n  Use the reflection coefficient real and imaginary parts .

and define the normalized ZL:

( )( )

ir

irL

L

L

L

L

o

LL

jjjxrz

zz

jxrZZz

Γ−Γ+

Γ+Γ+=+=

+

−=Γ

+=Γ−

Γ+==

11

1111

( )( )oL

oLir ZZ

ZZj+

−=Γ+Γ=∠Γ=Γ Γθ

Γi

Γr

|Γ|

11

Now relating to z =r +jx

n  After some algebra, we obtain two eqs.

n  Similar to general equation of a circle of radius a, center at (h,k)

Γr −r1+ r

⎛

⎝⎜

⎞

⎠⎟2

+Γi2 =

11+ r⎡

⎣⎢⎤

⎦⎥

2

Γr −1[ ]2 + Γi −1x

⎡

⎣⎢⎤

⎦⎥

2

=1x⎡

⎣⎢⎤

⎦⎥

2

Circles of r

Circles of x

(x − h)2 + (y− k)2 = a2

Examples of circles of r and x

2

2

1Radius 1,1 Center

11Radius 0,

1 Center

⎥⎦

⎤⎢⎣

⎡=⎟⎠

⎞⎜⎝

⎛=

⎥⎦

⎤⎢⎣

⎡+

=⎟⎠

⎞⎜⎝

⎛+

=

xx

rrr

⎥⎦

⎤⎢⎣

⎡+

=⎟⎠

⎞⎜⎝

⎛+

=rr

r1

1Radius 0,1

Center

Circles of r Circles of x

⎥⎦

⎤⎢⎣

⎡=⎟⎠

⎞⎜⎝

⎛=xx1Radius 1,1 Center

Examples

Find z= r+jxgiven Γ

Also Find Γ

Given Z

Examples of circles of r and x

Circles of r Circles of x

rrr

+=⎟

⎠

⎞⎜⎝

⎛+

=1

1Radius 0,1

Center

Find SWR on the SC

n Numerically s = r on the +axis of Γr in the SC Proof:

1s1-s0 but then

11)when (

11

+≡+Γ=Γ

+

−===

+

−=Γ

j

rrrz

zz

r

L

L

Moving on the TL on the SmithC n A lossless TL is represented as

a circle of constant radius, |Γ|, or constant s

n Moving along the line from the load toward the generator, the phase decrease, therefore, in the SC equals to moves clockwisely.

ljjL

ljL eeel βθβ 22)( −− ΓΓ=Γ=Γ To

generator

12

One-turn on the Smith Chart

n  One turn (360o) around the SC = to λ/2 because in the formula below, if you substitute length for half-wavelength, the phase changes by 2π, which is one turn.

n  Find the point in the SC where Γ=+1,-1, j, -j, 0, 0.5 n  What is r and x for each case?

ljLel β2)( −Γ=Γ

Fun facts : Admittance in the SC n  The admittance, y=YL/Yo where Yo=1/Zo, can be

found by moving ½ turn (λ/4) on the TL circle

( )( )

( )( ) Γ+

Γ−=

Γ+

Γ−=

Γ+

Γ−⎟⎠⎞

⎜⎝⎛==

Γ+

Γ−=

−Γ−

−Γ+=

Γ−

Γ+==

Γ−

Γ+=

Γ−

Γ+=

Γ−

Γ+=

Γ−

Γ+===

=⎟⎠

⎞⎜⎝

⎛=

+

+

+

+

11

11

1

11)0(

11

)1(1)1(1

11)4/(

11

11

11

1

11)0(

4

222

0

0

2

2

02

02

2

2

2

2

j

j

lj

lj

o

o

j

j

j

j

lj

lj

lj

o

lj

oo

L

ee

eV

eZV

Yly

eelz

ee

ee

eZV

eVZZ

Zlz

l

β

β

π

π

β

β

β

β

λ

πλ

λπ

β

Vmax and Vmin on the SmithC n  The Γr +axis, where r > 0 corresponds to Vmax

n  The Γr -axis, where r < 0 corresponds to Vmin

Vmax (Maximum impedance)

Vmin

Exercise: using S.C. n  A 2cm lossless TL has Vg=10 Vrms, Zg=60 Ω, ZL=100+j80 Ω and Zo=40Ω,


n  Load is at .2179λ @ S.C. n  Move .2λ and arrive to .4179λn  Read

ZL Zg

Vg

+ Vin

-

Iin

+ VL

-

Zo γ=j β

2 cm

55.3. jzin −=

Voltage Divider:

radZZZV

Vgin

ingin 775.032.3 −∠=

+=

25.24080100 jjzL +=

+= λλ 2.=l

Ω−= 2212 jZin

oL 5.23622.0 ∠=Γ

ocm 120622.0)2( −∠=Γ

Ω−= 4.176. jyin

zL

zin 0.2λ

Exercise: cont….using S.C. n  A 2cm lossless TL has Vg=10 Vrms, Zg=60 Ω, ZL=100+j80 Ω and Zo=40Ω,


n  Distance from the load (.2179λ) to the nearest minimum & max n  Move to horizontal axis toward the generator and arrive to .5λ

(Vmax) and to .25λ for the Vmin.

n  Distance to min=.5-.2179=.282λ n  Distance to 2st voltage maximum is .282λ +.25λ=.482 See drawing

ZL Zg

Vg

+ Vin

-

Iin

+ VL

-

Zo γ=j β

2 cm

25.2 jzL += λλ 2.=l oL 5.23622.0 ∠=Γ

13

Exercise : using formulas n  A 2cm lossless TL has Vg=10 Vrms, Zg=60 Ω, ZL=100+j80 Ω and Zo=40Ω,


ZL Zg

Vg

+ Vin

-

Iin

+ VL

-

Zo γ=j β

2 cm

( )52tan)80100(4052tan40)80100(

40)2(π

π

jj

jjcmZZin

++

⎟⎠

⎞⎜⎝

⎛ ++==

Ω−= 17.212.12 jZinVoltage Divider:

radZZZV

Vgin

ingin 766.030.3 −∠=

+=

o

oL

oLL ZZ

ZZ 4.2362.0 ∠=+

−=Γ

ooL

ljLecm

6.12062.0144

)2( 2

−∠=−∠Γ=

Γ=Γ − β

Another example: n  A 26cm lossless TL is connected to load ZL=36-j44 Ω and Zo=100Ω,

λ=10cm . find: the input impedance Zin

n  Load is at .427λ @ S.C. n  Move .1λ and arrive to .527λ (=.027λ)

n  Read

ZL Zg

Vg

+ Vin

-

Iin

+ VL

-

Zo γ=j β

26cm

16.31. jzin +=

Distance to first Vmax:

44.36. jzL −= λλλλ 1.)5(.56.2 +==l

Ω+= 1631 jZin

ΓL = 0.54∠−127o

λλλ 028.0427.5.0min =−=lλλλ 278.25.028.0max =+=l

Exercise 11.4

n  A 70 Ω lossless line has s =1.6 and at the load θΓ =300o. If the line is 0.6λ long, obtain Γ, ZL, Zin and the distance of the first minimum voltage from the load.

n  Answer

n  The load is located at: n  Move to .4338 λ and draw line from n  center to this place, then read where it crosses you TL circle. n  Distance to Vmin in this case, lmin =.5λ-.3338λ= ~

oL 30023.0 ∠=Γ

1s1-s+

=Γ

Ω−==

−=

6.335.8048.15.1

jzZZjz

LoL

L

6/

5.176.4725.68.0

λ

Ω−=

−=

jZjz

in

inλ3338.

Java Applet : Smith Chart n  http://education.tm.agilent.com/index.cgi?CONTENT_ID=5

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