Transmission Lines and Antennas The Smith Chart - Justification

Introduction to Smith Chart and Applications For lossless lines, as we move away from the
load, reflection coefficient (ratio between
backward and forward wave voltages)
describes a CW circle in the complex plane.

T.L.
l ZL
Javier Leonardo Araque Quijano
Of: 453 – 204
Ext. 14083
jlaraqueq@unal.edu.co

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Impedance Transformation The Smith Chart

●
A general transformation is obtained
if the normalized impedance is used:

|G|
=1

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 Impedance Vs. Admittance charts
●
With load given as normalized admittance y, G is the
negative of that obtained from normalized impedance z, → Working with Impedance Chart Only
original chart can be used by a 180° rotation of the G plane.
● The property above means that we pass from Impedance chart
to admittance and viceversa by the transformation G → -G

Impedance chart: Admittance chart:

z= 0.5 + j0.5 y = 1/z = 1- j
Complex plane is G Complex plane is G
grid gives complex Impedance (normalized) grid gives complex admittance (normalized)

x x

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Matching with Lumped Elements

Effect of Lumped Elements on z (y)
(c'ed)
●Series inductor: increases
Lp
●
Four distinct regions:
Lp Ls reactance only CW in r = const Ls
z
●
Series capacitor: decreases z
r>1:
reactance only, CCW in r = const Lp – Cs or Cp – Ls
Cs Cp ●Parallel inductor: decreases Cs Cp
susceptance only, CCW in g=cnst
g>1:
●
Parallel capacitor: increases Ls-Cp or Cs-Lp
susceptance only, CW in g=cnst
●Around z, solid lines are
x>0, r<1, g<1:
constant-resistance circles, Cp-Cs or Cs-Cp
z
dashed are constant- Cp-Ls or Cs-Lp
conductance. Around y, lines are
constant conductance.
x<0, r<1, g<1:
●Similar behavior for the Lp-Ls or Ls-Lp
y addition of z Lp-Cs or Ls-Cp
resistance/conductance

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 Equations for Exact Lumped
Exercise
Matching
Problem: match a load impedance Z L = RL+j XL = 1/(GL+jBL) to a
generator/line with impedance Z0=1/Y0 using lumped lossless ●
Compute all the single-stage L-section
components (i.e. transform ZL into Z0 for Z0 real using only
capacitor/inductances): matching networks of lumped elements that
GL/Y0<1 RL/Z0<1
allow a power source with Rg = 75 Ohm to
jX jX
deliver maximum power to a load with
Z0 jB ZL Z0 jB ZL
ZL = (33-j50) Ohm. The operation frequency is
90MHz.
●
Compute the -20dB and the -10dB bandwidth
for the reflection coefficient.
●
Select one of the above and support your choice

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Matching with T.L. Length of Shunt element

●
Series TL moves impedance
Cp point to the g=1 circle.
Cp Transmission lines are the Moving to admittance chart
●
●

most convenient way to we see the need to cancel

match loads: e.g. Printed lines Lp
Lp the remaining susceptance.
are inexpensive, able to z
achieve arbitrary component ●
Stub length is obtained by
z values. locating the negative value
● In general T.L. cannot of susceptance (to cancel
simulate series impedances, y out the initial value), discard
y only transformation and shunt g and move ”towards the
reactances (open/shorted load” till a short or open is
stubs) found
●
With fixed caracteristic ● Stopping at y=0 (left) gives
impedance, equal to the length of open stub, at y->
reference impedance, only the
infty length of shorted stub
series-shunt configuration is
possible
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 Example: matching T.L.

Reading Scales in Smith Chart with TL l1 ZL

l2
●
SWR Match the load Zl=(80 + j20) Ohm to a 50
Ohm line using minimum-length lines.
●
RL 1. Locate z = 1.6 + 0.4j
2. Move into admittance chart by reflecting
●
Reflection coefficient (P and E/I) this, results y = 0.59 – j0.15

3. Insert a series line long enough to arrive to

● Transmission coefficient (P and E/I) the g=1 circle (length l1= 0.18l). Susceptance
value therein is b = 0.57.
● Attenuation 4. Set g=0, keep b to reach the susceptance
value that must be cancelled by the stub.

5. Reflect w.r.t. real axis to obtain the

susceptance value that must be achieved by
stub.

6. Move towards load till the first among open

circuit and short circuit. In this case we stop
at y = ∞, meaning a shorted stub, length
l2=0.17l
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Validating the Previous Solution Analytical Solution of the Line-Stub

●
Simulation is performed in
the 0 – 2*f0 band to assess
Matching Problem
performance of the
matching network. 1. Compute the load reflection coefficient w.r.t. Z 0:
●
Impedance is assumed
constant throughout the T.L.
band; it is defined via the ls ZL
equation-defined RF

lp
2. Compute the length of the series line resulting in
component, 1-port and Gin = 1/Z 0:
setting Z11 to the
impedance value given.
●
Return loss is very high at Choose n to obtain the smallest non-negative value
4. Obtain the stub element and termination type as
the design frequency, line follows:
for each sign, then choose the sign resulting in the
lengths have to be slightly smallest length.
tuned.
●
20-dB bandwidth of design 3. Compute the resulting input susceptance Where atan2 is the four-quadrant tangent function.
(use the sign opposite to the one chosen in the
is about log2(1.1/0.92) =
previous step): Choose m to obtain the smallest non-negative
0.26 octaves.
length:
m even → termination is short-circuit
m odd → termination is open-circuit
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 Exercise References
●
Design a line-stub matching network to match a [1] D. M. Pozar, Microwave Engineering. John Wiley
load with Zl = (10-j100) Ohm to a line with Zc & sons, 1998.
= 75 Ohm using both the graphical and the [2] P. H. Smith, “Transmission Line Calculator,”
algebraic methods. Electronics, vol. 12, No. 1, pp. 29–31, January 1939.
●
Compute the 20dB bandwidth of your design.

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