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
Input impedance for finite length line
› Quarter wavelength line
› Half wavelength line

Smith chart
› A graphical tool to solve transmission line
problems
› Use for measuring reflection coefficient,
VSWR, input impedance, load impedance,
the locations of Vmax and Vmin
Ex1 A 0.269- long lossless line with Z0 = 50  is terminated
in a load ZL = 60+j40 . Use the Smith chart to find
a) L
b) VSWR
c) Zin
d) the distance from the load to the first voltage maximum




To minimize power reflection from load
Zin = Z0
Matching techniques
1. Quarter - wave transformers Z S  Z 0 RL for real
load
2. single - stub tuners
3. lumped – element tuners
The capability of tuning is desired by having variable
reactive elements or stub length.
EX2, a load 10-j25  is terminated in a 50  line. In
order for 100% of power to reach a load, Zin must
match with Z0, that means Zin = Z0 = 50 .
 Distance d WTG = (0.5-0.424) +0.189 
= 0.265 
to point 1+ j2.3. Therefore cut TL and insert a
reactive element that has a normalized reactance
of -j2.3.
 The normalized input impedance becomes
1+ j2.3 - j2.3 = 1
which corresponds to the center or the Smith chart.
 The value of capacitance can be evaluated by
known frequency, for example, 1 GHz is given.
1
XC 
  j 2.3  50   j115 
jC
1
C 
 1.38 pF
 j115
 Working with admittance (Y) since it is more convenient
to add shunt elements than series elements
 Stub tuning is the method to add purely reactive
elements
 Where is the location of y on Smith chart?
Ex3 let z = 2+j2, what is the admittance?
1
y
 0.25  j 0.25
2  j2
We can easily find the admittance on the Smith chart by
moving 180 from the location of z.

There are two types of stub tuners
1. Shorted end, y =  (the rightmost of the Y chart)
2. opened end, y = 0 (the leftmost of the Y chart)
Short-circuited shunt stub
Open-circuited shunt stub

Procedure
1. Locate zL and then yL. From yL, move clockwise to
1  jb circle, at which point the admittance yd = 1  jb.
On the WTG scale, this represents length d.
2. For a short-circuited shunt stub, locate the short end
at 0.250 then move to 0 jb, the length of stub is
then l and then yl = jb.
3. For an open-circuit shunt stub, locate the open end
at 0,then move to 0 jb.
4. Total normalized admittance ytot = yd+yl = 1.


The most popular transmission line since it can be
fabricated using printed circuit techniques and it is
convenient to connect lumped elements and
transistor devices.
By definition, it is a transmission line that consists of
a strip conductor and a grounded plane
separated by a dielectric medium


The EM field is not contained entirely in dielectric so
it is not pure TEM mode but a quasi-TEM mode that
is valid at lower microwave frequency.
The effective relative dielectric constant of the
microstrip is related to the relative dielectric
constant r of the dielectric and also takes into
account the effect of the external EM field.
Typical electric field lines
Field lines where the air and
dielectric have been replaced
by a medium of effective
relative permittivity, eff
Therefore in this case u p 
c
 eff
2 f

up
and
m/s
rad / m
up
0
g  
f
 eff
m.

Consider t/h < 0.005 and assume no dependence of frequency,
the ratio of w/h and r are known, we can calculate Z0 as
 eff 
r 1
2

 r 1
h
2 1  12
w
8h w
60
ln(  ) 
for w / h  1, Z 0 
w 4h
 eff
for w / h  1, Z 0 
120
1
 eff w  1.393  0.667 ln( w  1.444)
h
h

Assume t is negligible, if Z0 and r are known, the
ratio w/h can be calculated as
w
8e A
for w / h  2,
 2A
h e 2
 r 1 
w 2
0.61  
for w / h  2,
  B  1  ln(2 B  1) 
ln( B  1)  0.39 

h 
2 r 
 r  

where
and
Z0  r  1  r  1
0.11
A

(0.23 
)
60
2
r 1
r
377
B
2Z 0  r
The value of r and the dielectric thickness (h) determines
the width (w) of the microstrip for a given Z0.
Assume t/h  0.005,
1/ 2


 
r

for w / h  0.6,   0 
 r 1  0.6(  1)( w )0.0297 
r
h


1/ 2


 
r

for w / h  0.6,   0 
 r 1  0.63(  1)( w )0.1255 
r
h


 conductor loss
 dielectric loss
 radiation loss
tot  c   d
where c = conductor attenuation (Np/m)
d = dielectric attenuation (Np/m
Rskin
c 
Zo w
( Np / m)
Rskin
 c  8.686
Zo w
Rskin 
1

(dB / m)

If the conductor is thin, then the more accurate
skin resistance can be shown as
Rskin
1

 (1  et /  )
.
2 f  r ( eff  1)
d 
tan 
c 2  eff ( r  1)
Np / m