2 4 the_smith_chart_package

2/7/2005

2_4 The Smith Chart

1/2

2.4 – The Smith Chart
Reading Assignment: pp. 64-73
The Smith Chart

The Smith Chart provides:
1)

2)
The most important fact about the Smith Chart is:

HO: The Complex Γ plane
Q: But how is the complex Γ plane useful?
A:
HO: Transformations on the Complex Γ Plane
Q: But transformations of Γ are relatively easy—
transformations of line impedance Z is the difficult one.

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005

2_4 The Smith Chart

2/2

A:

HO: Mapping Z to Γ
HO: The Smith Chart

HO: Zin Calculations using the Smith Chart
Example: The Input Impedance of a Shorted Transmission
Line
Example: Determining the Load Impedance of a
Transmission Line
Example: Determining the Length of a Transmission Line

Expressing a load or line impedance in terms of its admittance
is sometimes helpful. Additionally, we can easily map
admittance onto the Smith Chart

HO: Impedance and Admittance
Example: Admittance Calculations with the Smith Chart
Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005

The Complex Gamma Plane

1/6

The Complex Γ Plane
Resistance R is a real value, thus we can indicate specific resistor values
as points on the real line:

R =20 Ω

R =0

R =50 Ω

R

R =5 Ω
Likewise, since impedance Z is a complex value, we can indicate specific
impedance values as point on a two dimensional complex plane:

Im {Z }

Z =30 +j 40 Ω
Re {Z }

Z =60 -j 30 Ω

Note each dimension is defined by a single real line: the horizontal line
(axis) indicating the real component of Z (i.e., Re {Z } ), and the vertical
line (axis) indicating the imaginary component of impedance Z (i.e.,
Im {Z } ). The intersection of these two lines is the point denoting the
impedance Z = 0.
* Note then that a vertical line is formed by the locus of all points
(impedances) whose resistive (i.e., real) component is equal to, say, 75.
* Likewise, a horizontal line is formed by the locus of all points
(impedances) whose reactive (i.e., imaginary) component is equal to -30.

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005


2/6

Im {Z }

R =75
Re {Z }

X =-30
If we assume that the real component of every impedance is positive,
then we find that only the right side of the plane will be useful for
plotting impedance Z—points on the left side indicate impedances with
negative resistances!
Im {Z }

Invalid
Region
(R<0)

Valid
Region
(R>0)

Re {Z }

Moreover, we find that common impedances such as Z = ∞ (an open
circuit!) cannot be plotted, as their points appear an infinite distance
from the origin.
Im {Z }

Z =0

(short)

Z =Z0

Z = ∞ (open)

(matched)

Re {Z }

Jim Stiles

The Univ. of Kansas

Somewhere way the
heck over there !!

Dept. of EECS

2/7/2005


3/6

Q: Yikes! The complex Z plane does not appear to be a very helpful.

Is there some graphical tool that is more useful?

A: Yes! Recall that impedance Z and reflection coefficient Γ are
equivalent complex values—if you know one, you know the other.

We can therefore define a complex Γ plane in the same manner
that we defined a complex impedance plane. We will find that there
are many advantages to plotting on the complex Γ plane, as opposed
to the complex Z plane!
Im {Γ }
Γ =0.3 +j 0.4
Γ =-0.5 +j 0.1

Re {Γ }
Γ =0.6 -j 0.3

Note that the horizontal axis indicates the real component of Γ ( Re {Γ } ),

while the vertical axis indicates the imaginary component of Γ ( Im {Γ } ).
We could plot points and lines on this plane exactly as before:

Im {Γ }

Re {Γ}=0.5
Re {Γ }

Im {Γ} =-0.3
Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005


4/6

However, we will find that the utility of the complex Γ pane as a graphical
tool becomes apparent only when we represent a complex reflection
coefficient in terms of its magnitude ( Γ ) and phase (θ Γ ):

Γ = Γ e jθΓ
In other words, we express Γ using polar coordinates:

Γ = 0.6 e

j 3π 4

Im {Γ }

Γ

Γ

θΓ

Re {Γ }

Γ = 0.7 e j 300
Note then that a circle is formed by the locus of all points whose
magnitude Γ equal to, say, 0.7. Likewise, a radial line is formed by the
locus of all points whose phase

θ Γ is equal to 135 .

θ Γ = 135

Im {Γ }

Γ = 0.7
Re {Γ }

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005


5/6

Perhaps the most important aspect of the complex Γ plane is its validity
region. Recall for the complex Z plane that this validity region was the
right-half plane, where Re {Z } > 0 (i.e., positive resistance).
The problem was that this validity region was unbounded and infinite in
extent, such that many important impedances (e.g., open-circuits) could
not be plotted.
Q: What is the validity region for the complex Γ plane?
A:

Recall that we found that for Re {Z } > 0 (i.e., positive

resistance), the magnitude of the reflection coefficient was
limited:

0 < Γ <1

Therefore, the validity region for the complex Γ plane consists of
all points inside the circle Γ = 1 --a finite and bounded area!

Im {Γ }

Invalid
Region
( Γ > 1)

Valid
Region
( Γ < 1)
Re {Γ }

Γ =1

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005


6/6

Note that we can plot all valid impedances (i.e., R >0) within this finite
region!
Im {Γ }

Γ = e j π = −1.0

Γ=0

(matched)

(short)

Re {Γ }

Γ = e j 0 = 1.0

(open)

Γ =1
(Z = jX → purely reactive)

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005

Transformations on the Complex

1/7

Transformations on the
Complex Γ Plane
The usefulness of the complex Γ plane is apparent when we
consider again the terminated, lossless transmission line:

z =0

z =−
Z0, β

Γ in

Z0, β

Γ

L

Recall that the reflection coefficient function for any location z
along the transmission line can be expressed as (since z L = 0 ):
Γ ( z ) = ΓL e j 2 β z
= ΓL e

j (θ Γ +2 β z )

And thus, as we would expect:
Γ(z = 0) = ΓL

and

Γ(z = − ) = Γ Le - j 2 β = Γin

Recall this result “says” that adding a transmission line of length
to a load results in a phase shift in θ Γ by −2 β radians, while
the magnitude Γ remains unchanged.

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005


2/7

Q: Magnitude Γ and phase θ Γ --aren’t those the

values used when plotting on the complex Γ plane?

A: Precisely! In fact, plotting the
transformation of ΓL to Γin along a transmission
line length has an interesting graphical
interpretation. Let’s parametrically plot Γ ( z )
from z = z L (i.e., z = 0 ) to z = z L −

(i.e., z = − ):

Im {Γ }

Γ (z )

θL

Γ (z = 0 ) = ΓL

ΓL

Γ (z = − ) = Γ

Re {Γ }

θin = θL − 2β
in

Γ =1

Since adding a length of transmission line to a load ΓL modifies
the phase θ Γ but not the magnitude ΓL , we trace a circular arc
as we parametrically plot Γ ( z ) ! This arc has a radius ΓL and an
arc angle 2β radians.

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005


3/7

With this knowledge, we can easily solve many interesting
transmission line problems graphically—using the complex Γ
plane! For example, say we wish to determine Γin for a
transmission line length = λ 8 and terminated with a short
circuit.

Z0, β

z =0

z =−

Γ in

Z0 , β

ΓL = −1

=λ 8

The reflection coefficient of a short circuit is ΓL = −1 = 1 e j π ,
and therefore we begin at that point on the complex Γ plane.
We then move along a circular arc −2 β = −2 (π 4 ) = − π 2
radians (i.e., rotate clockwise 90 ).

Im {Γ }

Γ (z )

Γin = 1 e

+jπ2

Re {Γ }
ΓL = 1 e

+ jπ

Γ =1

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005


4/7

When we stop, we find we are at the point for Γin ; in this case
Γin = 1 e j π 2 (i.e., magnitude is one, phase is 90 o ).

Now, let’s repeat this same problem, only with a new
transmission line length of = λ 4 . Now we rotate clockwise
2β = π radians (180 ).
Im {Γ }
Γ (z )

Γin = 1 e

+j π2

Re {Γ }
ΓL = 1 e

+ jπ

Γ =1

For this case, the input reflection coefficient is Γin = 1 e j 0 = 1 :
the reflection coefficient of an open circuit!

Our short-circuit load has been transformed into an open
circuit with a quarter-wavelength transmission line!
But, you knew this would happen—right?

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005


z =0

z =−

Z0, β

Γin = 1
(open)

5/7

Γ L = −1

Z0, β

(short)

=λ 4

Recall that a quarter-wave transmission line was one of the
special cases we considered earlier. Recall we found that the
input impedance was proportional to the inverse of the load
impedance.
Thus, a quarter-wave transmission line transforms a short into
an open. Conversely, a quarter-wave transmission can also
transform an open into a short:
Im {Γ }
Γ =1

ΓL = 1 e

+jπ2

Re {Γ }
Γin = 1 e

+ jπ

Γ (z )
Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005


6/7

Finally, let’s again consider the problem where ΓL = −1 (i.e.,

short), only this time with a transmission line length

=λ 2 (a

half wavelength!). We rotate clockwise 2β = 2π radians (360 ).

Hey look! We came clear
around to where we started!
Im {Γ }

Γ (z )

ΓL = 1 e

+ jπ

Re {Γ }
Γin = 1 e

+ jπ

Γ =1

Thus, we find that Γin = ΓL if

= λ 2 --but you knew this too!

Recall that the half-wavelength transmission line is likewise a
special case, where we found that Zin = Z L . This result, of

course, likewise means that Γin = ΓL .

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005


7/7

Now, let’s consider the opposite problem. Say we know that the
input impedance at the beginning of a transmission line with
length = λ 8 is:
Γin = 0.5 e j 60

Q: What is the reflection coefficient of the load?
A: In this case, we begin at Γin and rotate COUNTERCLOCKWISE along a circular arc (radius 0.5) 2β = π 2 radians
(i.e., 60 ). Essentially, we are removing the phase shift
associated with the transmission line!
Im {Γ }
Γ (z )
θ
in

θL = θin + 2β

0.5
Γ in
ΓL

= 0.5 e j 60

Re {Γ }

= 0.5 e j 150

Γ =1

The reflection coefficient of the load is therefore:
ΓL = 0.5 e j 150
Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005

Mapping Z to Gamma

1/8

Mapping Z to Γ
Recall that line impedance and reflection coefficient are
equivalent—either one can be expressed in terms of the other:

Γ (z ) =

Z (z ) − Z 0
Z (z ) + Z 0

and

⎛ 1 + Γ (z ) ⎞
⎟
1 − Γ (z ) ⎠
⎝

Z (z ) = Z 0 ⎜

Note this relationship also depends on the characteristic
impedance Z0 of the transmission line. To make this relationship
more direct, we first define a normalized impedance value z ′
(an impedance coefficient!):

z ′ (z ) =

Z (z ) R (z )
X (z )
=
+j
= r (z ) + j x (z )
Z0
Z0
Z0

Using this definition, we find:

Γ (z ) =

Z (z ) − Z 0
Z (z ) + Z 0

=
=

Jim Stiles

Z (z ) Z 0 − 1
Z (z ) Z 0 + 1
z ′ (z ) − 1
z ′ (z ) + 1

The Univ. of Kansas

Dept. of EECS

2/7/2005

Mapping Z to Gamma

2/8

Thus, we can express Γ ( z ) explicitly in terms of normalized
impedance z ′ --and vice versa!

z ′ (z ) − 1
Γ (z ) =
z ′ (z ) + 1

z ′ (z ) =

1 + Γ (z )
1 − Γ (z )

The equations above describe a mapping between coefficients
z ′ and Γ . This means that each and every normalized impedance
value likewise corresponds to one specific point on the complex
Γ plane!
For example, say we wish to mark or somehow indicate the
values of normalized impedance z’ that correspond to the
various points on the complex Γ plane.
Some values we already know specifically:

case

z′

Γ

1

∞

∞

1

2

0

0

-1

3

Z0

1

0

4

j Z0

j

j

5
Jim Stiles

Z

−j Z0

j

−j

The Univ. of Kansas

Dept. of EECS

2/7/2005

Mapping Z to Gamma

3/8

Therefore, we find that these five normalized impedances map
onto five specific points on the complex Γ plane:
Γi
Γ =1
z ′ = −j

z′ = 1

( Γ=− j )
z′ = ∞

( Γ=0 )

( Γ=1)

Γr

z′ = 0

( Γ=−1)
z ′ = −j

( Γ=− j )

Or, the five complex Γ map onto five points on the normalized
impedance plane:

x

z ′ = −j

z′ = 0

z′ = ∞

( Γ=− j )

( Γ=1)

( Γ=−1)

z ′ = −j

( Γ=− j )

Jim Stiles

r
z′ = 1

( Γ=0 )

The Univ. of Kansas

Dept. of EECS

2/7/2005

Mapping Z to Gamma

4/8

Now, the preceding provided examples of the mapping of points
between the complex (normalized) impedance plane, and the
complex Γ plane.
We can likewise map whole contours (i.e., sets of points)
between these two complex planes. We shall first look at two
familiar cases.

Z =R

In other words, the case where impedance is purely real, with no
reactive component (i.e., X = 0 ).
Meaning that normalized impedance is:

z′ =r + j0

(i .e., x

= 0)

where we recall that r = R Z 0 .
Remember, this real-valued impedance results in a real-valued
reflection coefficient:
r −1
Γ=
r +1
I.E.,:

Γr

Jim Stiles

Re {Γ} =

r −1
r +1
The Univ. of Kansas

Γi

Im {Γ} = 0

Dept. of EECS

2/7/2005

Mapping Z to Gamma

5/8

Thus, we can determine a mapping between two contours—one
contour ( x = 0 ) on the normalized impedance plane, the other
( Γi = 0 ) on the complex Γ plane:

x =0

Γi = 0

⇔
Γi

Γ =1
x =0

( Γi =0 )
Γr

x

r
x =0

( Γi =0 )

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005

Mapping Z to Gamma

6/8

Z = jX

In other words, the case where impedance is purely imaginary,
with no resistive component (i.e., R = 0 ).
Meaning that normalized impedance is:

(i .e., r

z ′ = 0 + jx

= 0)

where we recall that x = X Z 0 .
Remember, this imaginary impedance results in a reflection
coefficient with unity magnitude:

Γ =1

Thus, we can determine a mapping between two contours—one
contour ( r = 0 ) on the normalized impedance plane, the other
( Γ = 1 ) on the complex Γ plane:

r =0

Jim Stiles

⇔

The Univ. of Kansas

Γ =1

Dept. of EECS

2/7/2005

Mapping Z to Gamma

7/8

Γi

Γ =1

r =0

( Γ =1)

Γr

x
r =0

( Γ =1)
r

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005

Mapping Z to Gamma

8/8

Q: These two “mappings” may

very well be fascinating in an
academic sense, but they seem
not particularly relevant, since
actual values of impedance
generally have both a real and
imaginary component.
Sure, mappings of more general
impedance contours (e.g., r = 0.5
or x = −1.5 ) onto the complex Γ
would be useful—but it seems
clear that those mappings are
impossible to achieve!?!

A: Actually, not only are mappings of more general impedance
contours (such as r = 0.5 and x = −1.5 ) onto the complex Γ
plane possible, these mappings have already been achieved—
thanks to Dr. Smith and his famous chart!

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/7/2005

The Smith Chart

1/11

The Smith Chart
Say we wish to map a line on the normalized complex impedance
plane onto the complex Γ plane.
For example, we could map the vertical line r =2 ( Re{z ′} = 2 ) or
the horizontal line x =-1 ( Im{z ′} = −1 ).
Im {z ′}

r =2
Re {z ′}

x =-1

Recall we know how to map the vertical line r =0; it simply maps
to the circle Γ = 1 on the complex Γ plane.

Likewise, we know how to map the horizontal line x = 0; it simply
maps to the line Γi = 0 on the complex Γ plane.
But for the examples given above, the mapping is not so straight
forward. The contours will in general be functions of both
Γr and Γi (e.g., Γ2 + Γi2 = 0.5 ), and thus the mapping cannot be
r

stated with simple functions such as Γ = 1 or Γi = 0 .

Jim Stiles

The Univ. of Kansas

Dept. Of EECS

2/7/2005

The Smith Chart

2/11

As a matter of fact, a vertical line on the normalized impedance
plane of the form:
r = cr ,
where cr is some constant (e.g. r = 2 or r = 0.5 ), is mapped onto
the complex Γ plane as:
2

2

⎛
⎛ 1 ⎞
c ⎞
Γr − r ⎟ + Γi2 = ⎜
⎜
⎟
1 + cr ⎠
⎝
⎝ 1 + cr ⎠

Note this equation is of the same form as that of a circle:
2
2
( x − x c ) + ( y − yc ) = a 2

where:

a = the radius of the circle
Pc ( x = xc , y = yc )

⇒ point located at the center of the circle

Thus, the vertical line r = cr maps into a circle on the complex Γ
plane!
By inspection, it is apparent that the center of this circle is
located at this point on the complex Γ plane:
⎛

Pc ⎜ Γr =
⎝

Jim Stiles

⎞
cr
, Γi = 0 ⎟
1 + cr
⎠

The Univ. of Kansas

Dept. Of EECS

2/7/2005

The Smith Chart

3/11

In other words, the center of this circle always lies somewhere
along the Γi = 0 line.
Likewise, by inspection, we find the radius of this circle is:

a=

1
1 + cr

We perform a few of these mappings and see where these
circles lie on the complex Γ plane:
Γi
r = −0.3

Γ =1

r = 0. 0
r = 1.0

r = 0.3

Jim Stiles

The Univ. of Kansas

Γr
r = 3. 0

Dept. Of EECS

2/7/2005

The Smith Chart

4/11

We see that as the constant cr increases, the radius of the
circle decreases, and its center moves to the right.
Note:
1. If cr > 0 then the circle lies entirely within the circle
Γ = 1.
2. If cr < 0 then the circle lies entirely outside the circle
Γ = 1.
3. If cr = 0 (i.e., a reactive impedance), the circle lies on
circle Γ = 1 .
4. If cr = ∞ , then the radius of the circle is zero, and its
center is at the point Γr = 1, Γi = 0 (i.e., Γ = 1 e j 0 ). In

other words, the entire vertical line r = ∞ on the
normalized impedance plane is mapped onto just a single
point on the complex Γ plane!
But of course, this makes sense! If r = ∞ , the impedance is
infinite (an open circuit), regardless of what the value of
the reactive component x is.

Now, let’s turn our attention to the mapping of horizontal lines
in the normalized impedance plane, i.e., lines of the form:

x = ci

Jim Stiles

The Univ. of Kansas

Dept. Of EECS

2/7/2005

The Smith Chart

5/11

where ci is some constant (e.g. x = −2 or x = 0.5 ).
We can show that this horizontal line in the normalized
impedance plane is mapped onto the complex Γ plane as:

( Γr

− 1)

2

2

⎛
1⎞
1
+ ⎜ Γi − ⎟ = 2
ci ⎠ ci
⎝

Note this equation is also that of a circle! Thus, the horizontal
line x = ci maps into a circle on the complex Γ plane!
By inspection, we find that the center of this circle lies at the
point:
⎛

Pc ⎜ Γr = 1, Γi =
⎝

1⎞
⎟
ci ⎠

in other words, the center of this circle always lies somewhere
along the vertical Γr = 1 line.
Likewise, by inspection, the radius of this circle is:

a=

Jim Stiles

1

ci

The Univ. of Kansas

Dept. Of EECS

2/7/2005

The Smith Chart

6/11

We perform a few of these mappings and see where these
circles lie on the complex Γ plane:
Γr = 1

Γi
x = 0 .5

x = 1 .0

x = 2 .0

x = 3. 0

Γ =1

Γr

x = −3 . 0
x = −0 . 5

x = −1.0

x = −2.0

We see that as the magnitude of constant ci increases, the
radius of the circle decreases, and its center moves toward the
point ( Γr = 1, Γi = 0 ) .
Note:
1. If ci > 0 (i.e., reactance is inductive) then the circle lies
entirely in the upper half of the complex Γ plane (i.e.,
where Γi > 0 )—the upper half-plane is known as the
inductive region.

Jim Stiles

The Univ. of Kansas

Dept. Of EECS

2/7/2005

The Smith Chart

7/11

2. If ci < 0 (i.e., reactance is capacitive) then the circle
lies entirely in the lower half of the complex Γ plane (i.e.,
where Γi < 0 )—the lower half-plane is known as the
capacitive region.

3. If ci = 0 (i.e., a purely resistive impedance), the circle
has an infinite radius, such that it lies entirely on the line
Γi = 0 .
4. If ci = ±∞ , then the radius of the circle is zero, and its

center is at the point Γr = 1, Γi = 0 (i.e., Γ = 1 e j 0 ). In other

words, the entire vertical line x = ∞ or x = −∞ on the
normalized impedance plane is mapped onto just a single
point on the complex Γ plane!

But of course, this makes sense! If x = ∞ , the impedance
is infinite (an open circuit), regardless of what the value of
the resistive component r is.
5. Note also that much of the circle formed by mapping
x = ci onto the complex Γ plane lies outside the circle

Γ = 1.

This makes sense! The portions of the circles laying
outside Γ = 1 circle correspond to impedances where the
real (resistive) part is negative (i.e., r < 0).

Thus, we typically can completely ignore the portions of the
circles that lie outside the Γ = 1 circle !
Jim Stiles

The Univ. of Kansas

Dept. Of EECS

2/7/2005

The Smith Chart

8/11

Mapping many lines of the form r = cr and x = ci onto circles on

the complex Γ plane results in tool called the Smith Chart.
Im{ Γ }

Re{ Γ }

Jim Stiles

The Univ. of Kansas

Dept. Of EECS

2/7/2005

The Smith Chart

9/11

Note that around the outside of the Smith Chart there is a
scale indicating the phase angle θ Γ , from −180 < θ Γ < 180 .
However, there is another scale that also directly indicates the
equivalent transmission line distance ∆z associated with phase
shift ∆θ Γ = 2β ∆z , in terms of λ (i.e., the electrical distance).
The two scales are related by the equation:
∆z =

∆θ Γ ⎛ ∆θ Γ ⎞
=⎜
⎟λ
2β ⎝ 4π ⎠

Note the Smith Chart is simply the vertical lines r = cr and

horizontal lines x = ci of the normalized impedance plane,
mapped onto the two types of circles on the complex Γ plane.
Note for the normalized impedance plane, a vertical line r = cr

and a horizontal line x = ci are always perpendicular to each
other when they intersect. We say these lines form a
rectilinear grid.

However, a similar thing is true for the Smith Chart! When a
mapped circle r = cr intersects a mapped circle x = ci , the two
circles are perpendicular at that intersection point. We say
these circles form a curvilinear grid.
In fact, the Smith Chart is formed by distorting the rectilinear
grid of the normalized impedance plane into the curvilinear grid
of the Smith Chart!

Jim Stiles

The Univ. of Kansas

Dept. Of EECS

2/7/2005

The Smith Chart

x

I.E.,:

10/11

x =1

r =0

r

x = −1

x =0

x = −1

Distorting this rectilinear grid:

x

r

Jim Stiles

The Univ. of Kansas

Dept. Of EECS

2/7/2005

The Smith Chart

11/11

And then distorting some more—we have the curvilinear grid of
the Smith Chart!

x
r

Jim Stiles

The Univ. of Kansas

Dept. Of EECS

2/8/2005

Zin Calculations using the Smith Chart.doc

1/7

Zin Calculations using
the Smith Chart
′
z0 = 1

′
zin

z = −

z L′

z = 0

′
The normalized input impedance zin of a transmission line length
, when terminated in normalized load z L′ , can be determined as:
′
zin =

Zin
Z0

⎛ Z L + j Z 0 tan β ⎞
⎟
Z 0 ⎝ Z 0 + j Z L tan β ⎠
Z Z + j tan β
= L 0
1 + j Z L Z 0 tan β
=

=

1

Z0 ⎜

z L′ + j tan β
1 + j z L′ tan β

Q: Evaluating this unattractive expression

looks not the least bit pleasant. Isn’t there a
′
less disagreeable method to determine zin ?

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


2/7

A: Yes there is! Instead, we could determine this normalized
input impedance by following these three steps:
1. Convert z L′ to ΓL , using the equation:

Zin − Z 0
Zin + Z 0
Z Z −1
= in 0
Zin Z 0 + 1
z ′ −1
= in
′
zin + 1

Γin =

2. Convert ΓL to Γin , using the equation:
Γin = Γ L e − j 2 β
′
3. Convert Γin to zin , using the equation:

z L′ =

Z L 1 + ΓL
=
Z 0 1 − ΓL

Q: But performing these three

calculations would be even more
difficult than the single step
you described earlier. What
short of dimwit would ever use
(or recommend) this approach?

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


3/7

A: The benefit in this last approach is that each of the three
steps can be executed using a Smith Chart—no complex
calculations are required!
1. Convert z L′ to ΓL
Find the point z L′ from the impedance mappings on your

Smith Chart. Place you pencil at that point—you have now
located the correct ΓL on your complex Γ plane!
For example, say z L′ = 0.6 − j 1.4 . We find on the Smith

Chart the circle for r =0.6 and the circle for x =-1.4. The
intersection of these two circles is the point on the
complex Γ plane corresponding to normalized impedance
z L′ = 0.6 − j 1.4 .

This point is a distance of 0.685 units from the origin, and
is located at angle of –65 degrees. Thus the value of ΓL is:
ΓL = 0.685 e − j 65

2. Convert ΓL to Γin
Since we have correctly located the point ΓL on the

complex Γ plane, we merely need to rotate that point
clockwise around a circle ( Γ = 0.685 ) by an angle 2β .
When we stop, we are located at the point on the complex
Γ plane where Γ = Γin !
Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


4/7

For example, if the length of the transmission line
terminated in z L′ = 0.6 − j 1.4 is = 0.307 λ , we should

rotate around the Smith Chart a total of 2β = 1.228π
radians, or 221 . We are now at the point on the complex
Γ plane:
Γ = 0.685 e + j 74

This is the value of Γin !
′
3. Convert Γin to zin
When you get finished rotating, and your pencil is located
at the point Γ = Γin , simply lift your pencil and determine

the values r and x to which the point corresponds!

For example, we can determine directly from the Smith
Chart that the point Γin = 0.685 e + j 74 is located at the

intersection of circles r =0.5 and x =1.2. In other words:
′
zin = 0.5 + j 1.2

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


5/7

Step 1

Γ = 0.685

ΓL = 0.685 e − j 65

θ Γ = −65

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


6/7

Step 2

2

= 0.147 λ

Γin = 0.685 e − j 74

Γ = 0.685

ΓL = 0.685 e − j 65

1

= 0.16λ

=

1

+

2

= 0.160λ + 0.147 λ = 0.307 λ

2β = 221
Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


7/7

Step 3

′
zin = 0.5 + j 1.2

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005

Example Shorted Transmission Line.doc

1/3

Example: The Input
Impedance of a Shorted
Transmission Line
Let’s determine the input impedance of a transmission line that
is terminated in a short circuit, and whose length is:
a)

= λ 8 = 0.125λ

b)

= 3λ 8 = 0.375λ

′
zin

2β = 90

⇒

⇒

2β = 270

′
z0 = 1

z = −

Jim Stiles

z L′ = 0

z = 0

The Univ. of Kansas

Dept. of EECS

2/8/2005

a)

= λ 8 = 0.125λ


⇒

2/3

2β = 90

′
Rotate clockwise 90 from Γ = −1.0 = e j 180 and find zin = j .

zin = j

ΓL = −1 = e j 180

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005

b)


= 3λ 8 = 0.375λ

⇒

3/3

2β = 270

′
Rotate clockwise 270 from Γ = −1.0 = e j 180 and find zin = − j .

ΓL = −1 = e j 180

zin = j

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005

Example The Load Impedance.doc

1/2

Example: Determining the
Load Impedance of a
Transmission Line
Say that we know that the input impedance of a transmission
line length = 0.134λ is:
′
zin = 1.0 + j 1.4
Let’s determine the impedance of the load that is terminating
this line.

′
zin =
1 + j 1. 4

′
z0 = 1

z L′ = ??

= 0.134λ

z = −

z = 0

′
Locate zin on the Smith Chart, and then rotate counter-

clockwise (yes, I said counter-clockwise) 2β = 96.5 .
Essentially, you are removing the phase shift associated with
the transmission line. When you stop, lift your pencil and find
z L′ !

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005

Example The Load Impedance.doc

2/2

= 0.134λ
2β = 96.5

′
zin = 1 + j 1.4

z L′ = 0.29 + j 0.24

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005

Example Determining the tl length.doc

1/7

Example: Determining
Transmission Line Length
A load terminating at transmission line has a normalized
impedance z L′ = 2.0 + j 2.0 . What should the length of
transmission line be in order for its input impedance to be:

a) purely real (i.e., xin = 0 )?
b)

have a real (resistive) part equal to one (i.e., rin = 1.0 )?

Solution:
a) Find z L′ = 2.0 + j 2.0 on your Smith Chart, and then rotate

clockwise until you “bump into” the contour x = 0 (recall this is
contour lies on the Γr axis!).
When you reach the x = 0 contour—stop! Lift your pencil and
note that the impedance value of this location is purely real
(after all, x = 0 !).
Now, measure the rotation angle that was required to move
clockwise from z L′ = 2.0 + j 2.0 to an impedance on the x = 0
contour—this angle is equal to 2β !
You can now solve for , or alternatively use the electrical
length scale surrounding the Smith Chart.
Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


2/7

One more important point—there are two possible solutions!

Solution 1:
2β = 30
= 0.042λ

z L′ = 2 + j 2
Γ (z )

′
zin = 4.2 + j 0

x =0

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


3/7

Solution 2:

z L′ = 2 + j 2
′
zin = 0.24 + j 0

x =0

Γ (z )

2β = 210
= 0.292λ

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


4/7

b) Find z L′ = 2.0 + j 2.0 on your Smith Chart, and then rotate

clockwise until you “bump into” the circle r = 1 (recall this circle
intersects the center point or the Smith Chart!).
When you reach the r = 1 circle—stop! Lift your pencil and note
that the impedance value of this location has a real value equal
to one (after all, r = 1 !).
Now, measure the rotation angle that was required to move
clockwise from z L′ = 2.0 + j 2.0 to an impedance on the r = 1
circle—this angle is equal to 2β !
You can now solve for , or alternatively use the electrical
length scale surrounding the Smith Chart.
Again, we find that there are two solutions!

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


5/7

Solution 1:

z L′ = 2 + j 2

Γ (z )

r =1

′
zin = 1.0 − j 1.6

2β = 82
= 0.114λ

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


6/7

Solution 2:

′
zin = 1.0 + j 1.6

Γ (z )

z L′ = 2 + j 2

r =1

2β = 339
= 0.471λ

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/8/2005


7/7

′
Q: Hey! For part b), the solutions resulted in zin = 1 − j 1.6 and
′
zin = 1 + j 1.6 --the imaginary parts are equal but opposite! Is

this just a coincidence?

A: Hardly! Remember, the two impedance solutions must result
in the same magnitude for Γ --for this example we find
Γ ( z ) = 0.625 .
Thus, for impedances where r =1 (i.e., z ′ = 1 + j x ):
Γ=

jx
z ′ − 1 (1 + jx ) − 1
=
=
z ′ + 1 (1 + jx ) + 1 2 + j x

and therefore:
2

Γ =

jx

2

2+ j x

2

x2
=
4 +x2

Meaning:

x =
2

4 Γ

2

1− Γ

2

of which there are two equal by opposite solutions!

x = ±

2 Γ
1− Γ

2

Which for this example gives us our solutions x = ±1.6 .

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/10/2005

Admittance.doc

1/4

Admittance
As an alternative to impedance Z, we can define a complex
parameter called admittance Y:

Y =

I
V

where V and I are complex voltage and current, respectively.
Clearly, admittance and impedance are not independent
parameters, and are in fact simply geometric inverses of each
other:
1
1
Y =
Z =

Z

Y

Thus, all the impedance parameters that we have studied can
be likewise expressed in terms of admittance, e.g.:

Y (z ) =

1

Z (z )

YL =

1

ZL

Yin =

1

Zin

Moreover, we can define the characteristic admittance Y0 of a
transmission line as:

I + (z )
Y0 = +
V (z )

And thus it is similarly evident that characteristic impedance
and characteristic admittance are geometric inverses:
Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/10/2005

Admittance.doc

Y0 =

1

Z0 =

Z0

2/4

1

Y0

As a result, we can define a normalized admittance value y ′ :

y′ =

Y
Y0

An therefore (not surprisingly) we find:

y′ =

Y Z0 1
=
=
Y0 Z
z′

Note that we can express normalized impedance and
admittance more compactly as:

y ′ = Y Z0

and

z ′ = Z Y0

Now since admittance is a complex value, it has both a real and
imaginary component:

Y = G + jB

where:

Re {Y } G = Conductance
Im {Z }

Jim Stiles

B = Susceptance

The Univ. of Kansas

Dept. of EECS

2/10/2005

Admittance.doc

3/4

Now, since Z = R + jX , we can state that:

G + jB =

1
R + jX

Q: Yes yes, I see, and from this
we can conclude:
G =

1

R

and

B=

−1

X

and so forth. Please speed this up
and quit wasting my valuable time
making such obvious statements!

A: NOOOO! We find that G ≠ 1 R and B ≠ 1 X
(generally). Do not make this mistake!

In fact, we find that

G + jB =

R − jX
1
R + jX R − jX

R − jX
R2 + X 2
R
X
= 2
−j 2
R +X2
R +X2
=

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/10/2005

Admittance.doc

4/4

Thus, equating the real and imaginary parts we find:

G =

R

and

R2 + X 2

B=

−X
R2 + X 2

Note then that IF X = 0 (i.e., Z = R ), we get, as expected:

G =

1

R

and

B =0

And that IF R = 0 (i.e., Z = R ), we get, as expected:

G =0

and

B=

−1

X

I wish I had a
nickel for every
time my software
has crashed—oh
wait, I do!

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/17/2005

Example Admittance Calculations with the Smith Chart.doc

1/9

Example: Admittance
Calculations with the
Smith Chart
Say we wish to determine the normalized admittance y1′ of the
network below:

z 2′ =
1 .7 − j 1 . 7

y1′

z L′ =
1.6 + j 2.6

′
z0 = 1
= 0.37 λ

z = 0

z = −

First, we need to determine the normalized input admittance of
the transmission line:

′
yin

z L′ =
1.6 + j 2.6

′
z0 = 1
= 0.37 λ

z = −

Jim Stiles

z = 0

The Univ. of Kansas

Dept. of EECS

2/17/2005


2/9

There are two ways to determine this value!

Method 1
First, we express the load z L = 1.6 + j 2.6 in terms of its
admittance y L′ = 1 z L . We can calculate this complex value—or
we can use a Smith Chart!
z L = 1.6 + j 2.6
y L = 0.17 − j 0.28

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/17/2005


3/9

The Smith Chart above shows both the impedance mapping
(red) and admittance mapping (blue). Thus, we can locate the
impedance z L = 1.6 + j 2.6 on the impedance (red) mapping, and
then determine the value of that same ΓL point using the
admittance (blue) mapping.
From the chart above, we find this admittance value is
approximately y L = 0.17 − j 0.28 .
Now, you may have noticed that the Smith Chart above, with
both impedance and admittance mappings, is very busy and
complicated. Unless the two mappings are printed in different
colors, this Smith Chart can be very confusing to use!
But remember, the two mappings are precisely identical—they’re
just rotated 180 with respect to each other. Thus, we can
alternatively determine y L by again first locating z L = 1.6 + j 2.6

on the impedance mapping :

z L = 1.6 + j 2.6

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/17/2005


4/9

Then, we can rotate the entire Smith Chart 180 --while keeping
the point ΓL location on the complex Γ plane fixed.
y L = 0.17 − j 0.28

Thus, use the admittance mapping at that point to determine
the admittance value of ΓL .
Note that rotating the entire Smith Chart, while keeping the
point ΓL fixed on the complex Γ plane, is a difficult maneuver to
successfully—as well as accurately—execute.
But, realize that rotating the entire Smith Chart 180 with
respect to point ΓL is equivalent to rotating 180 the point ΓL
with respect to the entire Smith Chart!
This maneuver (rotating the point ΓL ) is much simpler, and the

typical method for determining admittance.
Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/17/2005


5/9

z L = 1.6 + j 2.6

y L = 0.17 − j 0.28

′
Now, we can determine the value of yin by simply rotating
clockwise 2β from y L′ , where = 0.37 λ :

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/17/2005


6/9

2β

yin = 0.7 − j 1.7

y L = 0.17 − j 0.28

Transforming the load admittance to the beginning of the
′
transmission line, we have determined that yin = 0.7 − j 1.7 .
Method 2
Alternatively, we could have first transformed impedance z L′ to
′
the end of the line (finding zin ), and then determined the value

′
of yin from the admittance mapping (i.e., rotate 180 around the

Smith Chart).

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/17/2005


7/9

′
zin = 0.2 + j 0.5

z L = 1.6 + j 2.6

2β

The input impedance is determined after rotating clockwise
′
2β , and is zin = 0.2 + j 0.5 .
Now, we can rotate this point 180 to determine the input
′
admittance value yin :

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/17/2005


8/9

2β
′
zin = 0.2 + j 0.5

yin = 0.7 − j 1.7
The result is the same as with the earlier method-′
yin = 0.7 − j 1.7 .
Hopefully it is evident that the two methods are equivalent. In
method 1 we first rotate 180 , and then rotate 2β . In the
second method we first rotate 2β , and then rotate 180 --the
result is thus the same!
Now, the remaining equivalent circuit is:

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2/17/2005


z 2′ =
1 .7 − j 1 . 7

y1′

9/9

′
yin =
0.7 − j 1.7

Determining y1′ is just basic circuit theory. We first express
z 2′ in terms of its admittance y2′ = 1 z 2′ .
Note that we could do this using a calculator, but could likewise
′
use a Smith Chart (locate z 2 and then rotate 180 ) to
accomplish this calculation! Either way, we find that
y2′ = 0.3 + j 0.3 .

y2′ =
0.3 + j 0.3

y1′

′
yin =
0.7 − j 1.7

Thus, y1′ is simply:
′
y1′ = y2′ + yin

= ( 0.3 + j 0.3) + ( 0.7 − j 1.7 )
= 1 .0 − j 1 . 4

Jim Stiles

The Univ. of Kansas

Dept. of EECS

2 4 the_smith_chart_package

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