Load variation on transmission line, Smith Chart and load matching methods.pdf
SVGhaisas
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Sep 26, 2025
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About This Presentation
Load as seen from different points away from the termination on a transmission line varies periodically. Smith chart is a handy tool in designing the transmission line with load. For efficient power delivery different methods for load matching are discussed.
Slide Content
Load variation along transmission line, λ/4 transformer , Smith Chart ,
Single and double stub impedance matching.
By
Dr.S.V.GHAISAS
Single and double stub impedance matching.
By
Dr.S.V.GHAISAS
As we have seen, the device connectivity at Rf must include
the effects of wave propagation in the direction of
propagation. In particular, the reflection of waves and
formation of standing wave patterns need to be avoided while
delivering the power to the load. Various ways are developed
to deliver the power efficiently.
By studying, variation of load along the TL, relations
between impedances and reflection coefficients at different
ports, strategies to deliver power efficiently can be devised.
the effects of wave propagation in the direction of
propagation. In particular, the reflection of waves and
formation of standing wave patterns need to be avoided while
delivering the power to the load. Various ways are developed
to deliver the power efficiently.
By studying, variation of load along the TL, relations
between impedances and reflection coefficients at different
ports, strategies to deliver power efficiently can be devised.
Load variation along the Transmission Line (TL) -
Considera load �
??????at z=0 on a transmission line.
The load as seen at a distance z from the load is
��=
??????
0
+
??????
−??????????????????
+??????
0
−
??????
??????β??????
??????
0
+
??????
−??????β??????
+??????
0
−
??????
??????β??????
for a lossless line. Since, From the definition
of �
0 (1)
�
0=
??????
0
+
??????
0
+=−
??????
0
+
??????
0
−, and reflection coefficient Γ=
??????
0
−
??????
0
+, substituting in (1)
the Z(z) becomes, ��=�
0
??????
−??????β??????
+Γ??????
??????β??????
??????
−??????β??????
−Γ??????
??????β??????
(2)
substituting Γ=
�
??????−�
0
�
??????+�
0
Considera load �
??????at z=0 on a transmission line.
The load as seen at a distance z from the load is
��=
??????
0
+
??????
−??????????????????
+??????
0
−
??????
??????β??????
??????
0
+
??????
−??????β??????
+??????
0
−
??????
??????β??????
for a lossless line. Since, From the definition
of �
0 (1)
�
0=
??????
0
+
??????
0
+=−
??????
0
+
??????
0
−, and reflection coefficient Γ=
??????
0
−
??????
0
+, substituting in (1)
the Z(z) becomes, ��=�
0
??????
−??????β??????
+Γ??????
??????β??????
??????
−??????β??????
−Γ??????
??????β??????
(2)
substituting Γ=
�
??????−�
0
�
??????+�
0
we get by dividing by Z
0, the NORMALIZED impedance
Z
??????�=
�??????−??????�0??????�??????β�
�0−??????�????????????�??????β�
. (3)
When, load is at z=0 and we are measuring input impedance at -l ,
The distance measured from the load, toward source,
Z
n−�=
�
??????+??????�
0??????�??????β??????
�0+??????�????????????�??????β??????
. (4)
This is the input normalized impedance on TL at -l distance.
Fig. 1-Termination of a transmission line at
z=0. The load is measured along the TL at –l.
The choice of z direction is consistent with
the �
−????????????�
dependence of the incident wave.
0, the NORMALIZED impedance
Z
??????�=
�??????−??????�0??????�??????β�
�0−??????�????????????�??????β�
. (3)
When, load is at z=0 and we are measuring input impedance at -l ,
The distance measured from the load, toward source,
Z
n−�=
�
??????+??????�
0??????�??????β??????
�0+??????�????????????�??????β??????
. (4)
This is the input normalized impedance on TL at -l distance.
Fig. 1-Termination of a transmission line at
z=0. The load is measured along the TL at –l.
The choice of z direction is consistent with
the �
−????????????�
dependence of the incident wave.
Impedance matching –quarter wave transformer.
TL with characteristic impedance Z0 is to be
Terminated on load ZL. We need to insert
another TL With appropriate impedance so that Zin (Z1 in the Figure) at the
termination of TL will be Z0.
�
??????−�=
�??????+??????�0??????�??????β??????
�
0+??????�
????????????�??????β??????
. (6)
for l= λ/4, away from the load ZL, a piece of TL of length λ/4 and
characteristic impedance Z1 is inserted. The expected Zin should be Z0 at λ/4
away from the load.
tan(βλ/4)=tan(π/2)=∞. As a result,
Fig. 2 (a)-A TL is terminated in the
impedance ZL. To avoid reflection
another piece of TL is introduced.
TL with characteristic impedance Z0 is to be
Terminated on load ZL. We need to insert
another TL With appropriate impedance so that Zin (Z1 in the Figure) at the
termination of TL will be Z0.
�
??????−�=
�??????+??????�0??????�??????β??????
�
0+??????�
????????????�??????β??????
. (6)
for l= λ/4, away from the load ZL, a piece of TL of length λ/4 and
characteristic impedance Z1 is inserted. The expected Zin should be Z0 at λ/4
away from the load.
tan(βλ/4)=tan(π/2)=∞. As a result,
Fig. 2 (a)-A TL is terminated in the
impedance ZL. To avoid reflection
another piece of TL is introduced.
�
0��������??????��������
=�
1�ℎ�������??????��??????�??????��������������ℎ??????���������.
�
1
�
??????
=
�
1
2
�
??????
or
�
1
2
=�
0�
?????? (7)
By choosing matching element of length λ/4 and characteristic impedance ??????
�=
??????
�??????
??????Ωthe �
????????????at the input terminals (red dots) is �
0resulting into Γ=0 condition.
The reflections between �
??????and input terminal impedance as seen from the load do
not allow wave in the backward direction on the TL.
Note that, the reflection coefficient at the entrance of TL with characteristic
impedance �
1is zero for the wavelength used for making ??????/4transformer. For
wavelengths close by, the condition is deviated and Γ≠0. Thus, the bandwidth over
which Γ<0.3is small.
0��������??????��������
=�
1�ℎ�������??????��??????�??????��������������ℎ??????���������.
�
1
�
??????
=
�
1
2
�
??????
or
�
1
2
=�
0�
?????? (7)
By choosing matching element of length λ/4 and characteristic impedance ??????
�=
??????
�??????
??????Ωthe �
????????????at the input terminals (red dots) is �
0resulting into Γ=0 condition.
The reflections between �
??????and input terminal impedance as seen from the load do
not allow wave in the backward direction on the TL.
Note that, the reflection coefficient at the entrance of TL with characteristic
impedance �
1is zero for the wavelength used for making ??????/4transformer. For
wavelengths close by, the condition is deviated and Γ≠0. Thus, the bandwidth over
which Γ<0.3is small.
It is illustrative to obtain the no-reflection condition from the wave reflections
from different interfaces.
Fig. 2 (b) –The reflecting and
transmitting interfaces for the
situation depicted in Fig. 2(a). The
reflection coefficients
Γ
1�
0���
1,Γ
1
′
�
1���
0,Γ
2(����)
, reflections.
The S-matrix for a TL of characteristic impedance �
1connected to TL with �
0is
given by,
�1−�0
�
1+�
0
2�1�0
�
1+�
0
2�
1�
0
�1+�0
�0−�1
�1+�0
=
Γ�
�−Γ
. Thus, Γ
1=−Γ
1′and T from both sides is
same.
from different interfaces.
Fig. 2 (b) –The reflecting and
transmitting interfaces for the
situation depicted in Fig. 2(a). The
reflection coefficients
Γ
1�
0���
1,Γ
1
′
�
1���
0,Γ
2(����)
, reflections.
The S-matrix for a TL of characteristic impedance �
1connected to TL with �
0is
given by,
�1−�0
�
1+�
0
2�1�0
�
1+�
0
2�
1�
0
�1+�0
�0−�1
�1+�0
=
Γ�
�−Γ
. Thus, Γ
1=−Γ
1′and T from both sides is
same.
Note that the square root terms are the result of the power amplitudes like
�
1=
??????
0
+
�
0
that define the s-matrix elements.
For summing all the reflection contributions, we note that a wave crossing
over from the first interface , reflecting from the load and crossing the first
interface again like number 2 ray shown in Fig 2(b) has a phase change of π.
This results in to a negative sign. A wave reflecting twice will have a change of
phase 2π, and so on. We denote the resultant reflection coefficientΓ
????????????.
The contribution from
ray 1 is Γ
1,
Ray 2 is −�
2
Γ
2
Ray 3 is −Γ
1Γ
2
2
�
2
………
Γ
????????????=Γ
1−Γ
2�
2
(1+Γ
1Γ
2+Γ
1Γ
2
2
+⋯)
=Γ
1−
Γ
2??????
2
1−Γ1Γ2
(8)
�
1=
??????
0
+
�
0
that define the s-matrix elements.
For summing all the reflection contributions, we note that a wave crossing
over from the first interface , reflecting from the load and crossing the first
interface again like number 2 ray shown in Fig 2(b) has a phase change of π.
This results in to a negative sign. A wave reflecting twice will have a change of
phase 2π, and so on. We denote the resultant reflection coefficientΓ
????????????.
The contribution from
ray 1 is Γ
1,
Ray 2 is −�
2
Γ
2
Ray 3 is −Γ
1Γ
2
2
�
2
………
Γ
????????????=Γ
1−Γ
2�
2
(1+Γ
1Γ
2+Γ
1Γ
2
2
+⋯)
=Γ
1−
Γ
2??????
2
1−Γ1Γ2
(8)
Using Eq. (7) , Γ
1=
�
??????−�
0
�??????+�0
=Γ
2.
Substituting in (8)
Γ
????????????=0
This exercise shows that under the condition obtained (geometric mean),
the phases of the reflected and transmitted waves render a destructive
interference in the region of source to load region and reflection is nulled.
1=
�
??????−�
0
�??????+�0
=Γ
2.
Substituting in (8)
Γ
????????????=0
This exercise shows that under the condition obtained (geometric mean),
the phases of the reflected and transmitted waves render a destructive
interference in the region of source to load region and reflection is nulled.
The λ/4 section matching is only for a particular wavelength. The bandwidth
(BW) over which matching occurs is small. To increase the BW, more sections
of quarter transformers can be added.
In the figure, two sections of λ/4 sections
are added.
Effectively, we cover wavelengths between λ
and2λ increasing the BW for matching. By
applying Geometric means,
we obtain �
1
3
=�
??????
2
�
0and �
2
3
=�
0
2
�
??????. (9)
There are various methods to match the load over wide BW using multiple
sections and also continuous variation in dimensions for antennas.
What is the condition on Z
Land Z
0for applying geometric mean condition?
Back
Fig. 3 –double quarter wave transformer.
(BW) over which matching occurs is small. To increase the BW, more sections
of quarter transformers can be added.
In the figure, two sections of λ/4 sections
are added.
Effectively, we cover wavelengths between λ
and2λ increasing the BW for matching. By
applying Geometric means,
we obtain �
1
3
=�
??????
2
�
0and �
2
3
=�
0
2
�
??????. (9)
There are various methods to match the load over wide BW using multiple
sections and also continuous variation in dimensions for antennas.
What is the condition on Z
Land Z
0for applying geometric mean condition?
Back
Fig. 3 –double quarter wave transformer.
The method of geometrical mean is exact for single transformer. For two
or higher ??????/4elements it is approximate. Although with increasing
elements BW increases.
Evaluate Γ for two element λ/4 transformer -
In this case the apparent impedance at EF must be �
0
for no reflection condition. The TLs with impedances
Z1 and Z2 are of the length ??????/2. So normalized
impedance at AB is
�
??????
�1
, at CDbecomes
�
1
�??????
. So absolute
impedance is
�
1
2
�??????
.
The normalized impedance at CD is
�
1
2
�2�??????
.So, the normalized impedance at EF
is
�
2�
??????
�
1
2
. The absolute impedance is
�
2
2
�
??????
�
1
2. (10)
This is not �
0.So, Γ≠0. Note that if�
1=�
2=�
0, then at EF �=�
??????as
expected at −�=??????/2.
Fig. 4 –Double quarter wave transformer reflecting
interfaces.
or higher ??????/4elements it is approximate. Although with increasing
elements BW increases.
Evaluate Γ for two element λ/4 transformer -
In this case the apparent impedance at EF must be �
0
for no reflection condition. The TLs with impedances
Z1 and Z2 are of the length ??????/2. So normalized
impedance at AB is
�
??????
�1
, at CDbecomes
�
1
�??????
. So absolute
impedance is
�
1
2
�??????
.
The normalized impedance at CD is
�
1
2
�2�??????
.So, the normalized impedance at EF
is
�
2�
??????
�
1
2
. The absolute impedance is
�
2
2
�
??????
�
1
2. (10)
This is not �
0.So, Γ≠0. Note that if�
1=�
2=�
0, then at EF �=�
??????as
expected at −�=??????/2.
Fig. 4 –Double quarter wave transformer reflecting
interfaces.
Continued ……
Let, �
0=50Ω,�
??????=100Ω.
Substituting from (8) in Eq. (9) we get the impedance at EF from geometrical
mean method is,
�=�
0
2
3
�
??????
1
3
(11)
Hence, Γ=
�
0
2
3
�
??????
1
3
−�0
�
0
2
3
�
??????
1
3
+�0
=
12.93
112.93
=0.114 (12)
Thus, the reflection is low but not zero. However, the BW increases as measured
below Γ=0.3.
Let, �
0=50Ω,�
??????=100Ω.
Substituting from (8) in Eq. (9) we get the impedance at EF from geometrical
mean method is,
�=�
0
2
3
�
??????
1
3
(11)
Hence, Γ=
�
0
2
3
�
??????
1
3
−�0
�
0
2
3
�
??????
1
3
+�0
=
12.93
112.93
=0.114 (12)
Thus, the reflection is low but not zero. However, the BW increases as measured
below Γ=0.3.
Consider a shorted load, Z
L=0, Z
n(0)=0. At -βl=π/4, �
??????=
0+??????�0
�0+??????0
=??????so, the shorted
load appears at distance -l=π/4βto be purely inductive with unit magnitude. At
-l=π/2β, Z
n tends to real infinity, at -l=3π/4β Z
n=-i, and at -l=π/β Z
n=0. This shows
that the load changes as one moves away from the load on a TL of characteristic
impedance Z
0. The load pattern repeats after phase π or after the distance of λ/2. The
pattern is periodic with a period of λ/2.
Similarly, for an open load, Z
L= ∞. At -l=π/4β, Z
n=-i, at -l=π/2β Z
n=0 , at -l=3π/4β
Z
n=i, -l=π/β Z
n=∞.
Smith Chart -It is a representation of Eq. (4) on a complex Γplane rendering
various essential parameters like Γ, VSWR, admittance, etc. immediately readable
from the plot.
Recall Eq. (4), depicting variation of the load toward source along TL,
Z
n−�=
�
??????+??????�
0??????�??????β??????
�
0+??????�
????????????�??????β??????
. (4)
L=0, Z
n(0)=0. At -βl=π/4, �
??????=
0+??????�0
�0+??????0
=??????so, the shorted
load appears at distance -l=π/4βto be purely inductive with unit magnitude. At
-l=π/2β, Z
n tends to real infinity, at -l=3π/4β Z
n=-i, and at -l=π/β Z
n=0. This shows
that the load changes as one moves away from the load on a TL of characteristic
impedance Z
0. The load pattern repeats after phase π or after the distance of λ/2. The
pattern is periodic with a period of λ/2.
Similarly, for an open load, Z
L= ∞. At -l=π/4β, Z
n=-i, at -l=π/2β Z
n=0 , at -l=3π/4β
Z
n=i, -l=π/β Z
n=∞.
Smith Chart -It is a representation of Eq. (4) on a complex Γplane rendering
various essential parameters like Γ, VSWR, admittance, etc. immediately readable
from the plot.
Recall Eq. (4), depicting variation of the load toward source along TL,
Z
n−�=
�
??????+??????�
0??????�??????β??????
�
0+??????�
????????????�??????β??????
. (4)
It may be noted that at -l=π/2β a shorted TL impedance is ∞and for an open TL
at -l=π/2β it is 0. Since admittance �=
1
�
, it indicates that at
??????
2??????
=
??????
4
distance,
the impedance of load appears as admittance. This can be seen from the
expression,
Z
??????
??????
2??????
=
�??????+??????�0tan
??????
2
�0+??????�??????tan
??????
2
=
��+????????????�??????
??????
2
1+??????��tan
??????
2
=
1
��
=�
??????, (13)
since tan
??????
2
=∞.
Also,
Γ�=
??????
0
−
??????
??????????????????
??????
0
+
??????
−??????????????????
=Γ0�
2????????????�
(14)
Note that at -l=π/2β Z
n=0 which is ADMITTANCE.
So, at -l=π/2β we can obtain Admittance of the load.
at -l=π/2β it is 0. Since admittance �=
1
�
, it indicates that at
??????
2??????
=
??????
4
distance,
the impedance of load appears as admittance. This can be seen from the
expression,
Z
??????
??????
2??????
=
�??????+??????�0tan
??????
2
�0+??????�??????tan
??????
2
=
��+????????????�??????
??????
2
1+??????��tan
??????
2
=
1
��
=�
??????, (13)
since tan
??????
2
=∞.
Also,
Γ�=
??????
0
−
??????
??????????????????
??????
0
+
??????
−??????????????????
=Γ0�
2????????????�
(14)
Note that at -l=π/2β Z
n=0 which is ADMITTANCE.
So, at -l=π/2β we can obtain Admittance of the load.
Constructing Smith Chart -
Γ=(ZL–Z0)/(ZL+Z0)invertingtheequationforΓ,
�??????
�
0
=
1+Γ
1−Γ
=Z
n.
Γisacomplexquantity.WhenplottedonacomplexplaneforagivenZ
LandZ
0,
itformsacirclewhoseradiusis|Γ|.Thelargestvalueis1,soallthechanges
thatarepossibleonaTLwillliewithinthisunitcircle.Since,Γ(z)=Γ(0)e
2iβz
from
(16),where,β=2π/λwhentheobservationpointmovesonthiscircleitsphase
changesandphaseisperiodicwithperiodofλ/2.Anyothervalueofreflection
willbe<1,soallsuchobservationswillfallonconcentriccircleswithradii<1.
Thecenterpointistheno-reflectionpointwithcircleofzeroradius.
SmithChartisZ
nplottedonthecomplexΓplane.
Let,Γ=x+iy,Z
n=R
n+iX
n.FromtherelationbetweenZ
nandΓ,
Γ=(ZL–Z0)/(ZL+Z0)invertingtheequationforΓ,
�??????
�
0
=
1+Γ
1−Γ
=Z
n.
Γisacomplexquantity.WhenplottedonacomplexplaneforagivenZ
LandZ
0,
itformsacirclewhoseradiusis|Γ|.Thelargestvalueis1,soallthechanges
thatarepossibleonaTLwillliewithinthisunitcircle.Since,Γ(z)=Γ(0)e
2iβz
from
(16),where,β=2π/λwhentheobservationpointmovesonthiscircleitsphase
changesandphaseisperiodicwithperiodofλ/2.Anyothervalueofreflection
willbe<1,soallsuchobservationswillfallonconcentriccircleswithradii<1.
Thecenterpointistheno-reflectionpointwithcircleofzeroradius.
SmithChartisZ
nplottedonthecomplexΓplane.
Let,Γ=x+iy,Z
n=R
n+iX
n.FromtherelationbetweenZ
nandΓ,
R
n+iX
n=
1+�+??????�
1−�−??????�
.By,multiplyinganddividingby(1-x)+iy,theresultant
expressioncanbeseparatedinrealandimaginarypartsexpressingR
nandX
n
asfunctionsof(x,y).Theseare,
??????
�
??????�+1
−�
2
+�
2
=
1
??????�+1
2
(15)
ForgivenR
nitisacirclewithradius(1/(R
n+1))andcentreat(R
n/R
n+1,0)on
theΓplane.Thesearecircleswithcentresalongtherealaxisbetweenx=1to
x=0.TheimportantcircleisforR
n=1.Itisacirclewithradius½andcentreat
(1/2,0)PASSINGTHROUGH(0,0)thepointofzeroreflection.
1−�
2
+�−
1
??????�
2
=
1
??????
�
2 (16)
Thesearecircleswithradii1/X
nandcentersat(1,1/X
n).
WhenthesecirclesareplottedonΓ=x+iyplane,itformsSmithChart.
n+iX
n=
1+�+??????�
1−�−??????�
.By,multiplyinganddividingby(1-x)+iy,theresultant
expressioncanbeseparatedinrealandimaginarypartsexpressingR
nandX
n
asfunctionsof(x,y).Theseare,
??????
�
??????�+1
−�
2
+�
2
=
1
??????�+1
2
(15)
ForgivenR
nitisacirclewithradius(1/(R
n+1))andcentreat(R
n/R
n+1,0)on
theΓplane.Thesearecircleswithcentresalongtherealaxisbetweenx=1to
x=0.TheimportantcircleisforR
n=1.Itisacirclewithradius½andcentreat
(1/2,0)PASSINGTHROUGH(0,0)thepointofzeroreflection.
1−�
2
+�−
1
??????�
2
=
1
??????
�
2 (16)
Thesearecircleswithradii1/X
nandcentersat(1,1/X
n).
WhenthesecirclesareplottedonΓ=x+iyplane,itformsSmithChart.
Finding |Γ| -
Since, Γ=�+??????�,Γ=��Γ=�and y=0 on the real axis.
Note that for a given �
??????, Γ= constant, is a circle on the x+iyplane as we
move away from the load.
Since, 0≤Γ≤1the value of x on the real axis between 0 –1 range is the Γ
for the given �
??????.
Finding VSWR -
����=
??????
max
??????
�??????�
=
1+|Γ|
1−|Γ|
. (17)
On the real axis x in the complex Γ plane, Γ=
��−1
��+1
becomes Γ=
??????�−1
??????�+1
since,
�
??????=0on real axis. Substituting for Γin (17)
����=�
??????. (18)
Thus, the intersection of the Γ= constant circle on the real axis of the
(x,iy) complex plane and the �
??????=��������curve is VSWR from the load �
??????.
Since, Γ=�+??????�,Γ=��Γ=�and y=0 on the real axis.
Note that for a given �
??????, Γ= constant, is a circle on the x+iyplane as we
move away from the load.
Since, 0≤Γ≤1the value of x on the real axis between 0 –1 range is the Γ
for the given �
??????.
Finding VSWR -
����=
??????
max
??????
�??????�
=
1+|Γ|
1−|Γ|
. (17)
On the real axis x in the complex Γ plane, Γ=
��−1
��+1
becomes Γ=
??????�−1
??????�+1
since,
�
??????=0on real axis. Substituting for Γin (17)
����=�
??????. (18)
Thus, the intersection of the Γ= constant circle on the real axis of the
(x,iy) complex plane and the �
??????=��������curve is VSWR from the load �
??????.
fig. 5 –Smith Chart
In the following slide Smith chart is shown. It consists of the unit circle Γ=1.
Equations (15) and (16) are plotted within the circle. �
??????=��������form full
circles with the radius Γ=1, �
??????=��������forms partial circles within Γ=
1circle. �
??????=1is the ONLY circle that passes through the origin Γ=1, i.e. the
condition of NO REFLECTION. There fore, while addressing the impedance
matching the constructions are performed on the TL such that on the Smith Chart
one lands upon the �
??????=1circle.
Further, the discussion following the equation (4) shows that, if a normalized
load is represented on the Smith chart as a point say (R’, iX’) at the intersection of
the curves �
??????=�
′
����
??????=�′then when one measures the load along the TL
, away from the load, the point (R’, iX’) moves CLOCKWISE on the circle passing
through the (R’, iX’) point and with the centerat 0. In the figure, point (0.6, i0.4)
is shown that moves clockwise on the green circle as one moves away from the
load.
In the following slide Smith chart is shown. It consists of the unit circle Γ=1.
Equations (15) and (16) are plotted within the circle. �
??????=��������form full
circles with the radius Γ=1, �
??????=��������forms partial circles within Γ=
1circle. �
??????=1is the ONLY circle that passes through the origin Γ=1, i.e. the
condition of NO REFLECTION. There fore, while addressing the impedance
matching the constructions are performed on the TL such that on the Smith Chart
one lands upon the �
??????=1circle.
Further, the discussion following the equation (4) shows that, if a normalized
load is represented on the Smith chart as a point say (R’, iX’) at the intersection of
the curves �
??????=�
′
����
??????=�′then when one measures the load along the TL
, away from the load, the point (R’, iX’) moves CLOCKWISE on the circle passing
through the (R’, iX’) point and with the centerat 0. In the figure, point (0.6, i0.4)
is shown that moves clockwise on the green circle as one moves away from the
load.
Circle of �
??????=1
(1,0) X
+iY??????/8
??????/4
Circle of �
??????=−3
-i3??????/8
(-1,0) ??????/2
Γ=1
P(0.6, i0.4)
??????=1
(1,0) X
+iY??????/8
??????/4
Circle of �
??????=−3
-i3??????/8
(-1,0) ??????/2
Γ=1
P(0.6, i0.4)
Fig. 6 For clarity only part of the Smith
chart is displayed in the figure.
A NORMALZED load (0.6, i0.4) is
represented by a red dot. A green circle
drawn with the radius connecting the
origin and the point is touching the
circle Rn=2 on the right side of the
origin, therefore the VSWR is 2 from
(17) and (18). Along the green circle
VSWR and |Γ| is Constant. A linear
scale from 0 to -1 is shown in the figure.
It helps reading |Γ|directly from the
chart. In the present case, the green
circle cuts the real axis at (-0.343, 0).
Thus, magnitude Γ=0.343, taking
absolute value on the real axis.
Note that we are plotting on the Γ(x, iy) plane Zn.
So magnitude of Γis the real axis value of the
circle and real axis intersection.
chart is displayed in the figure.
A NORMALZED load (0.6, i0.4) is
represented by a red dot. A green circle
drawn with the radius connecting the
origin and the point is touching the
circle Rn=2 on the right side of the
origin, therefore the VSWR is 2 from
(17) and (18). Along the green circle
VSWR and |Γ| is Constant. A linear
scale from 0 to -1 is shown in the figure.
It helps reading |Γ|directly from the
chart. In the present case, the green
circle cuts the real axis at (-0.343, 0).
Thus, magnitude Γ=0.343, taking
absolute value on the real axis.
Note that we are plotting on the Γ(x, iy) plane Zn.
So magnitude of Γis the real axis value of the
circle and real axis intersection.
Fig. 7 The point (0.6, i0.4) is the complex load AT the l=0 distance.
At distances away from the load, the complex value changes keeping
VSWR and |Γ| constant.
The radius of the green circle is |Γ|
(0.343), the angle ϕ(tan
−1
0.4
0.6
)is
the phase of complex number Γ.
This is the value of Γat z=0 denoted
by red dot.
Back
By inverting the expression
��=�
0
??????
−??????????????????
+Γ??????
??????????????????
??????
−??????????????????
−Γ??????
??????????????????
we get
Γ�
−??????2????????????
=
��−??????−1
�
�−??????+1
at z=-l.
At distances away from the load, the complex value changes keeping
VSWR and |Γ| constant.
The radius of the green circle is |Γ|
(0.343), the angle ϕ(tan
−1
0.4
0.6
)is
the phase of complex number Γ.
This is the value of Γat z=0 denoted
by red dot.
Back
By inverting the expression
��=�
0
??????
−??????????????????
+Γ??????
??????????????????
??????
−??????????????????
−Γ??????
??????????????????
we get
Γ�
−??????2????????????
=
��−??????−1
�
�−??????+1
at z=-l.
As one moves away from the
load, above expression shows ,
how Γchanges with z. In
particular at –l , which is λ/11 in
terms of operation wavelength of
the signal. The effect of moving
away from the load on Γis only
change in phase. The change is
−2??????�.Since the convention of
phase measurement on a
complex plane is anti-clockwise
from the real axis. The negative
sign indicates a clockwise
movement on the green circle up
to the blue dot, changing the
phase by 2??????�as shown in the
figure.
Back
Fig. 7 -(repeated for the sake of clarity)
load, above expression shows ,
how Γchanges with z. In
particular at –l , which is λ/11 in
terms of operation wavelength of
the signal. The effect of moving
away from the load on Γis only
change in phase. The change is
−2??????�.Since the convention of
phase measurement on a
complex plane is anti-clockwise
from the real axis. The negative
sign indicates a clockwise
movement on the green circle up
to the blue dot, changing the
phase by 2??????�as shown in the
figure.
Back
Fig. 7 -(repeated for the sake of clarity)
Impedancematchingwithstubs-
AstubisaTLwitheitheropenendorshortedendononesideandconnected
tosignalcarryingTLinseriesorparallel.Normally,thecharacteristicimpedance
ofstubissameassignalcarryingTL,Z
0.Ifnotpropernormalizationisneeded
duringcalculations.ForanyshortedloadΓ=1,andasonemovesawayfromthe
short,ontheSmithchart,theloadpointmovesfrom(-1,0)intheclockwise
direction.e.g.forthelengthofλ/8,itrepresentsapureinductivenormalized
loadofi.Atλ/4itispurelyresistiveand∞.
Theadmittanceoftheshortedstubappearsatdiagonallyoppositepointon
theSmithchartat(1,0).Aswemoveawayfromtheshort,thepointonthe
chartmovesclockwisefrom(1,0).Anormalizedloadwith�
??????≠�
0willappear
asapointonthechart.Asonemoveawayfromtheload,thepointwilltracea
circleclockwise,showningreeninthefigure,centredat(0,0).Aconstant
radiusisaresultofconstantVSWR.
AstubisaTLwitheitheropenendorshortedendononesideandconnected
tosignalcarryingTLinseriesorparallel.Normally,thecharacteristicimpedance
ofstubissameassignalcarryingTL,Z
0.Ifnotpropernormalizationisneeded
duringcalculations.ForanyshortedloadΓ=1,andasonemovesawayfromthe
short,ontheSmithchart,theloadpointmovesfrom(-1,0)intheclockwise
direction.e.g.forthelengthofλ/8,itrepresentsapureinductivenormalized
loadofi.Atλ/4itispurelyresistiveand∞.
Theadmittanceoftheshortedstubappearsatdiagonallyoppositepointon
theSmithchartat(1,0).Aswemoveawayfromtheshort,thepointonthe
chartmovesclockwisefrom(1,0).Anormalizedloadwith�
??????≠�
0willappear
asapointonthechart.Asonemoveawayfromtheload,thepointwilltracea
circleclockwise,showningreeninthefigure,centredat(0,0).Aconstant
radiusisaresultofconstantVSWR.
At 3λ/8 it is purely capacitive -i, and at λ/2, it is back to (-1,0). So, a
shorted stub can be a purely inductive or capacitive depending on its
length. Similarly for open stub. It is therefore used for compensating
imaginary part of the load. When stub is connected parallel to
the TL, admittances add up, while in series impedances add up. As one
goes away to a distance -I on the TL, admittance on Smith chart also
moves clockwise.
Fig. 8 –Open and
shorted stubs on a
transmission line.
shorted stub can be a purely inductive or capacitive depending on its
length. Similarly for open stub. It is therefore used for compensating
imaginary part of the load. When stub is connected parallel to
the TL, admittances add up, while in series impedances add up. As one
goes away to a distance -I on the TL, admittance on Smith chart also
moves clockwise.
Fig. 8 –Open and
shorted stubs on a
transmission line.
Single Stub matching -Given 50 Ohm TL, Z
L= 25 –i50 Ohm. Using single
shorted stub, match the load and TL. Find the length of the stub l, and
connecting distance -d from the load.
Fig. 9 -Single shorted stub
use for impedance
matching of the load �
??????.
Notethatthedistance
measuredfromtheloadis
negative to be consistent
with the convention of sign
used in the phase i(??????�−
??????.??????)
L= 25 –i50 Ohm. Using single
shorted stub, match the load and TL. Find the length of the stub l, and
connecting distance -d from the load.
Fig. 9 -Single shorted stub
use for impedance
matching of the load �
??????.
Notethatthedistance
measuredfromtheloadis
negative to be consistent
with the convention of sign
used in the phase i(??????�−
??????.??????)
0.115??????
0.178??????
0.25??????
0.34??????
P
Q
R
S
Fig. 10 –Smith chart
corresponding to the
situation depicted in Fig. 9.
Blue circle provides VSWR
for the given load. Brown
circle is Rn=1 that passes
through (0,0). As one moves
away from the load, the
impedance is given by the
common intersection of
blue circle, Rn and Xn
circles.
0.178??????
0.25??????
0.34??????
P
Q
R
S
Fig. 10 –Smith chart
corresponding to the
situation depicted in Fig. 9.
Blue circle provides VSWR
for the given load. Brown
circle is Rn=1 that passes
through (0,0). As one moves
away from the load, the
impedance is given by the
common intersection of
blue circle, Rn and Xn
circles.
Normalized load Z
n= 0.5 –iPoint P. A circle is drawn through P, on this
circle Γ and SWR are constant. At 180
0
on the circle opposite to P, point Q
is located that gives normalized admittance
Y
n= 0.4 + i0.8. On the circumference of Smith Chart, the distance from
load in terms of λ is given (shown in figure only at relevant points for
clarity). Point Q is located at 0.115λ. The constant Γ circle passes through
R
n=1 circle at R and S. For admittance, moving away from load is moving
clockwise on the constant Γ circle. At point R, Y
n= 1 + i1.6 and its location
is at 0.178λ. So, the stub needs to be placed at a distance d = (0.178 -
0.115)λ= 0.063λ from the load.
n= 0.5 –iPoint P. A circle is drawn through P, on this
circle Γ and SWR are constant. At 180
0
on the circle opposite to P, point Q
is located that gives normalized admittance
Y
n= 0.4 + i0.8. On the circumference of Smith Chart, the distance from
load in terms of λ is given (shown in figure only at relevant points for
clarity). Point Q is located at 0.115λ. The constant Γ circle passes through
R
n=1 circle at R and S. For admittance, moving away from load is moving
clockwise on the constant Γ circle. At point R, Y
n= 1 + i1.6 and its location
is at 0.178λ. So, the stub needs to be placed at a distance d = (0.178 -
0.115)λ= 0.063λ from the load.
To obtain no-reflection condition, admittance should be unity. The length
of the stub be chosen such that imaginary part becomes zero in the
admittance at 0.063λ from the load. We find that a shorted load has
admittance of –i1.6 at point T on the chart and its position is 0.34λ. A
shortedloadwillberepresentedbyapointat(0,0) on Γ=1circle. The
admittance will be a point at (1,0). On the Smith chart it is 0.25λ. Hence,
the stub length is l = (0.34 -0.25) λ= 0.09λ.
By similar analysis, for point S, d = 0.321λ and l = 0.41λ.
Back
of the stub be chosen such that imaginary part becomes zero in the
admittance at 0.063λ from the load. We find that a shorted load has
admittance of –i1.6 at point T on the chart and its position is 0.34λ. A
shortedloadwillberepresentedbyapointat(0,0) on Γ=1circle. The
admittance will be a point at (1,0). On the Smith chart it is 0.25λ. Hence,
the stub length is l = (0.34 -0.25) λ= 0.09λ.
By similar analysis, for point S, d = 0.321λ and l = 0.41λ.
Back
DOUBLE STUB MATCHING -
In single stub matching it may not be easy to locate stub at any point on
TL. There are specific positions where TL may be available for insertion of a
stub. Hence, double stub method becomes handy.
We will illustrate this for the same example above for single stub. Z
L=25-
i50, Z
0=50. Further, stubs can be attached parallelly at z=0 and z=λ/8. The
situation is depicted in the figure (next slide). Corresponding Smit chart is
also shown,
In single stub matching it may not be easy to locate stub at any point on
TL. There are specific positions where TL may be available for insertion of a
stub. Hence, double stub method becomes handy.
We will illustrate this for the same example above for single stub. Z
L=25-
i50, Z
0=50. Further, stubs can be attached parallelly at z=0 and z=λ/8. The
situation is depicted in the figure (next slide). Corresponding Smit chart is
also shown,
0,??????/2
??????/8
??????/4
3??????/4
P
Q
G
H
�
??????=1
�
??????=0.4
����=4.3
circle
Two stub compensation for load
termination.
“Note that VSWR can be
Read from the intersection
Of blue circle on the right
side of the origin as the
�
??????value. “
Fig. 11 –The double stub compensation is
depicted above. Corresponding Smith chart
is on the left.
i3.2
-i0.313 G’
??????/8
??????/4
3??????/4
P
Q
G
H
�
??????=1
�
??????=0.4
����=4.3
circle
Two stub compensation for load
termination.
“Note that VSWR can be
Read from the intersection
Of blue circle on the right
side of the origin as the
�
??????value. “
Fig. 11 –The double stub compensation is
depicted above. Corresponding Smith chart
is on the left.
i3.2
-i0.313 G’
As before at Q the admittance is Y
n= 0.4+i0.8 and is located at 0.115λ. We have to
connect a stub at z=0. We have pulled the R
n=1 circle to λ/8 position in anticipation
that the next stub will be connected at λ/8 away from the load. When stub at z=0 is
connected, the imaginary part of the admittance changes but real part is unchanged.
Hence, the admittance point will move on the circle valued 0.4, shown in yellow
color. It cuts at G and H. At G admittance is 0.4+i1.8. So a stub with admittance +i1.0
is needed to make the admittance point move to the point G ,0.4+i1.8, after
connecting at z=0 in parallel. Impedance of a shorted stub of length
3??????
??????
is –i. Its
admittance is +i. So, the length l1 = �??????/??????.
n= 0.4+i0.8 and is located at 0.115λ. We have to
connect a stub at z=0. We have pulled the R
n=1 circle to λ/8 position in anticipation
that the next stub will be connected at λ/8 away from the load. When stub at z=0 is
connected, the imaginary part of the admittance changes but real part is unchanged.
Hence, the admittance point will move on the circle valued 0.4, shown in yellow
color. It cuts at G and H. At G admittance is 0.4+i1.8. So a stub with admittance +i1.0
is needed to make the admittance point move to the point G ,0.4+i1.8, after
connecting at z=0 in parallel. Impedance of a shorted stub of length
3??????
??????
is –i. Its
admittance is +i. So, the length l1 = �??????/??????.
Now, the point G is located at 0.172λ. After rotating through 90
0
it arrives at 1-i3.2
Ohm at point G’ on Rn=1 circle. A shorted stub with +i3.2 admittance is needed to
attain the no-reflection condition. The point corresponding to +i3.2 occurs on the
Γ=1 circle at 0.202λ.Asbefore,ashortstub with impedance
1
??????3.2
=−??????0.313must
be connected at λ/8 distance. Point corresponding to –i0.313 occurs diagonally
opposite to +i3.2 at 0.202λ. Thus, the length of the second stub l2 is (�.�??????+
�.���)λ=0.452λ.
Similarly, one can compute for point H . The first stub length is 0.165λ, the second,
λ/8.
back
0
it arrives at 1-i3.2
Ohm at point G’ on Rn=1 circle. A shorted stub with +i3.2 admittance is needed to
attain the no-reflection condition. The point corresponding to +i3.2 occurs on the
Γ=1 circle at 0.202λ.Asbefore,ashortstub with impedance
1
??????3.2
=−??????0.313must
be connected at λ/8 distance. Point corresponding to –i0.313 occurs diagonally
opposite to +i3.2 at 0.202λ. Thus, the length of the second stub l2 is (�.�??????+
�.���)λ=0.452λ.
Similarly, one can compute for point H . The first stub length is 0.165λ, the second,
λ/8.
back
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