Example 2.2The load impedance of 40+j70 Ω terminates at 100Ω tx

line that is 0.1λ long. Find Γ at the load, Γ at input terminals, zin, SWR and return loss.

Solution:Normalized input impedance is zL=ZL/Zo=0.4+0.7j

Mark the point on the chartDraw a radial line form center to plot edge through

marked point, and read the angle 104o

Open compass from center to that point (radius)Measure SWR and |Γ| from the scale below the chart

1

Example 2.2

2

104o

ZL connected to λ/4 Line If an arbitrary ZL is connected to λ/4 line

Under normalized condition

Length of λ/4 corresponds to 180o rotation in Smith chart

3

2tan( )

tan( )L o o

in oo L L

Z jZ l ZZ Z

Z jZ l Z

1in

L

zz

(Admittance)

Zin when ZL at λ/4

4

ZL

Zin=1/ZL

The Admittance Smith chart

5

The Slotted lineIt is a device to measure arbitrary load attached to lineIt’s a tx line configuration that is used to find E field

amplitude of standing wave on tx lineDevice is replaced by VNAsSWR and 1st null from load can be measured

6

1 | |

1 | |SWR

1

1

SWR

SWR

min2 l

| | je where

Quarter Wave TransformerCase when Zo of tx line and ZL are irreplaceable

Impedance needs to be matched to avoid reflections

(1) λ/4 transformation (2) Stub matchingInitially Zo of tx line and ZL are given

Insert a section of lossless tx line between tx line ZL

Length of segment is λ/4 and impedance Z1

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In order for Γ = 0, we put Zin = Zo

In this way SWR=1 on tx lineThe impedance will be matched only for one frequencyThe method is limited to real loads onlyComplex loads can be made real by adding proper

lengths of tx lines8

1 o LZ Z R

21

inL

ZZ

R

where βl = π/2tan( )

tan( )L o

in oo L

Z jZ lZ Z

Z jZ l

Example 2.5Match a load of 100 Ω with a 50 Ω line using quarter

wave transformer at frequency 3GHzSolution:

Now λ for 3GHz is 0.1m, λ/4 = 0.025m Frequency response of Γ(quarter wave transformation)

9

1 (50)(100) 70.71Z

Quarter wave transformer: multiple reflection view point

10

11

1

o

o

Z Z

Z Z

0 12

0 1

Z Z

Z Z

13

1

L

L

R Z

R Z

11

1

21

2

2o

o

o

ZT

Z Z

ZT

Z Z

Generator and Load MismatchWe discussed terminated tx lines irrespective of

generatorInvestigation of maximum power transfer to load

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( 2 )

( 2 )

tan( ) 1

tan( ) 1

j lL o l

in o o j lo L l

Z jZ l eZ Z Z

Z jZ l e

* 21 1 1Re{ } | | Re

2 2in in inin

P V I VZ

| | in

in gin g

ZV V

Z Z

Let Zin = Rin+jXin and Zg=Rg+jXg then

Consider Zl = Zo, matched load with tx line

Max power is delivered under conjugate matchingRin = Rg, Xin = -Xg or Zin=Zg*

It means only half of Pin is delivered to the load

12

21| |

2 ( ) ( )in

gin g in g

RP V

R R X X

21 1| |

2 4gg

P VR

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The Terminated Lossy LineIn this case γ = α + jβHere

( )

( )

z zo

z zo

o

V z V e e

VI z e e

Z

2( ) | | ll e

tan( )

tan( )L o

in oo L

Z Z lZ Z

Z Z l