8/27/2007
The Terminated Transmission Line notes
1/2
B. The Terminated, Lossless Transmission Line
We now know that a lossless transmission line is completely
characterized by real constants Z 0 and β .
Likewise, the 2 waves propagating on a transmission line are
completely characterized by complex constants V0+ and V0− .
Q: Z 0 and β are determined from L, C, and ω . How do we
find V0+ and V0− ?
A: Apply Boundary Conditions!
Every transmission line has 2 “boundaries”
1)
2)
At one end of the transmission line.
At the other end of the transmission line!
Typically, there is a source at one end of the line, and a load
at the other.
Æ The purpose of the transmission line is to get power from
the source, to the load!
Let’s apply the load boundary condition!
HO: The Terminated, Lossless Transmission Line
HO: Special Values of Load Impedance
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
The Terminated Transmission Line notes
2/2
Q: So the line impedance at the end of a line must be load
impedance ZL (i.e., Z ( z = z L ) = Z L ); what is the line
impedance at the beginning of the line (i.e.,
Z ( z = z L − A ) = ? )?
A: The input impedance !
HO: Transmission Line Input Impedance
Q: You said the purpose of the transmission line is to
transfer E.M. energy from the source to the load. Exactly
how much power is flowing in the transmission line, and how
much is delivered to the load?
A: HO: Power Flow and Return Loss
Note that we can specify a load with:
1) its impedance ZL
2) its reflection coefficient Γ L
3)
return loss
A fourth alternative is VSWR.
HO: VSWR
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
The Terminated Lossless Transmission
1/8
The Terminated, Lossless
Transmission Line
Now let’s attach something to our transmission line. Consider a
lossless line, length A , terminated with a load ZL.
I(z)
+
V (z)
IL
Z0, β
-
A
z = zL − A
+
VL
ZL
-
z
z = zL
Q: What is the current and voltage at each and every point on
the transmission line (i.e., what is I ( z ) and V ( z ) for all points
z where z L − A ≤ z ≤ z L ?)?
A: To find out, we must apply boundary conditions!
In other words, at the end of the transmission line ( z = z L )—
where the load is attached—we have many requirements that all
must be satisfied!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
The Terminated Lossless Transmission
2/8
1. To begin with, the voltage and current ( I ( z = z L ) and
V ( z = z L ) ) must be consistent with a valid transmission line
solution:
V (z = z L ) =V + (z = z L ) + V − (z = z L )
= V0+ e − j β zL + V0− e + j β zL
V0+ ( z = z L ) V0− ( z = z L )
I (z = zL ) =
−
Z0
Z0
V0+ − j β zL V0− + j β zL
=
e
−
e
Z0
Z0
2. Likewise, the load voltage and current must be related by
Ohm’s law:
VL = Z L I L
3. Most importantly, we recognize that the values I ( z = z L ) ,
V ( z = z L ) and IL, VL are not independent, but in fact are
strictly related by Kirchoff’s Laws!
I(z=zL)
+
Z0 , β
Jim Stiles
+
V (z=zL) VL
-
z = zL − A
IL
ZL
-
z = zL
The Univ. of Kansas
Dept. of EECS
8/27/2007
The Terminated Lossless Transmission
3/8
From KVL and KCL we find these requirements:
V ( z = z L ) =VL
I (z = z L ) = IL
Combining these equations and boundary conditions, we find
that:
VL = Z L I L
V (z = z L ) = Z L I (z = z L )
V + (z = zL ) + V − (z = zL ) =
ZL
V + (z = zL ) − V − (z = zL )
(
Z0
)
Rearranging, we can conclude:
V − (z = z L ) Z L − Z 0
=
+
V (z = z L ) Z L + Z 0
Q: Hey wait as second! We earlier defined V − ( z ) V + ( z ) as
reflection coefficient Γ ( z ) . How does this relate to the
expression above?
A: Recall that Γ ( z ) is a function of transmission line position z.
The value V
−
(z
= zL ) V
+
(z
= z L ) is simply the value of function
Γ ( z ) evaluated at z = z L (i.e., evaluated at the end of the line):
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
The Terminated Lossless Transmission
4/8
V − (z = zL )
ZL − Z0
=
Γ
z
=
z
=
(
)
L
V + (z = zL )
ZL + Z0
This value is of fundamental importance for the terminated
transmission line problem, so we provide it with its own special
symbol ( ΓL ) !
ΓL Γ (z = z L ) =
ZL − Z0
ZL + Z0
Q: Wait! We earlier determined that:
Γ (z ) =
Z (z ) − Z 0
Z (z ) + Z 0
so it would seem that:
ΓL = Γ ( z = z L ) =
Z (z = z L ) − Z 0
Z (z = z L ) + Z 0
Which expression is correct??
A: They both are! It is evident that the two expressions:
ΓL =
are equal if:
Jim Stiles
ZL − Z0
ZL + Z0
and
ΓL =
Z (z = z L ) − Z 0
Z (z = z L ) + Z 0
Z (z = z L ) = Z L
The Univ. of Kansas
Dept. of EECS
8/27/2007
The Terminated Lossless Transmission
5/8
And since we know that from Ohm’s Law:
ZL =
and from Kirchoff’s Laws:
VL
IL
VL V ( z = z L )
=
IL I (z = z L )
and that line impedance is:
V (z = zL )
= Z (z = zL )
I (z = zL )
we find it apparent that the line impedance at the end of the
transmission line is equal to the load impedance:
Z (z = zL ) = Z L
The above expression is essentially another expression of the
boundary condition applied at the end of the transmission line.
Q: I’m confused! Just what are were we
trying to accomplish in this handout?
A: We are trying to find V(z) and I(z) when a
lossless transmission line is terminated by a
load ZL!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
The Terminated Lossless Transmission
6/8
We can now determine the value of V0− in terms of V0+ . Since:
V − ( z = z L ) V0−e + j β zL
ΓL = +
=
V ( z = z L ) V0+e − j β zL
We find:
V0− = e −2 j β zL Γ L V0+
And therefore we find:
V − ( z ) = (e −2 j β zL ΓL V0+ ) e + j β z
V ( z ) = V0+ ⎡⎣e − j β z + (e −2 j β zL ΓL ) e + j β z ⎤⎦
V0+ ⎡ − j β z
−2 j β z L
+ j βz
⎤
I (z ) =
e
e
e
−
Γ
(
)
L
⎣
⎦
Z0
where:
ΓL =
Jim Stiles
ZL − Z0
ZL + Z0
The Univ. of Kansas
Dept. of EECS
8/27/2007
The Terminated Lossless Transmission
7/8
zL = 0
Now, we can further simplify our analysis by arbitrarily
assigning the end point zL a zero value (i.e., z L = 0 ):
I(z)
+
V (z)
IL
+
Z0 , β
VL
-
-
A
z = −A
ZL
z
z = 0
If the load is located at z =0 (i.e., if z L = 0 ), we find that:
V (z = 0) =V + (z = 0) + V − (z = 0)
= V0+ e − j β ( 0) + V0− e + j β ( 0)
= V0+ + V0−
V0+ ( z = 0 ) V0− ( z = 0 )
I (z = 0) =
−
Z0
Z0
V0+ − j β ( 0) V0− + j β ( 0)
e
e
=
−
Z0
Z0
V0+ −V0−
=
Z0
⎛V0+ + V0− ⎞
Z (z = 0) = Z 0 ⎜ +
− ⎟
⎝ V0 −V0 ⎠
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
The Terminated Lossless Transmission
8/8
Likewise, it is apparent that if z L = 0 , ΓL and Γ 0 are the same:
V − ( z = 0 ) V0 −
ΓL = Γ ( z = z L ) = +
=
= Γ0
V ( z = 0 ) V0 +
Therefore:
ΓL =
ZL − Z0
= Γ0
ZL + Z0
Thus, we can write the line current and voltage simply as:
V ( z ) = V0+ ⎡⎣e − j β z + Γ 0 e + j β z ⎤⎦
⎡⎣for z L = 0 ⎤⎦
V0+ − j β z
⎡⎣e
− Γ 0 e + j β z ⎤⎦
I (z ) =
Z0
Q:
But, how do we determine V0+ ??
A:
We require a second boundary condition to determine V0+ .
The only boundary left is at the other end of the transmission
line. Typically, a source of some sort is located there. This
makes physical sense, as something must generate the incident
wave !
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Special Values of Load Impedance
1/10
Special Values of
Load Impedance
It’s interesting to note that the load ZL enforces a boundary
condition that explicitly determines neither V(z) nor I(z)—but
completely specifies line impedance Z(z)!
e − j β z + ΓL e + j β z
Z L cos β z − jZ 0 sin β z
Z (z ) = Z0 − j β z
=
Z
0
e
− ΓL e + j β z
Z 0 cos β z − jZ L sin β z
Γ ( z ) = ΓL e + j 2β z =
Z L − Z 0 + j 2β z
e
ZL + Z0
Likewise, the load boundary condition leaves V + ( z ) and V −( z )
undetermined, but completely determines reflection
coefficient function Γ ( z ) !
Let’s look at some specific values of load impedance
Z L = RL + jX L and see what functions Z(z) and Γ ( z ) result!
1. Z L = Z 0
In this case, the load impedance is numerically equal to the
characteristic impedance of the transmission line. Assuming
the line is lossless, then Z0 is real, and thus:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Special Values of Load Impedance
RL = Z 0
and
2/10
XL = 0
It is evident that the resulting load reflection coefficient is
zero:
Z − Z0 Z0 − Z0
ΓL = L
=
=0
ZL + Z0 Z0 + Z0
This result is very interesting, as it means that there is no
reflected wave V − ( z ) !
Thus, the total voltage and current along the transmission line
is simply voltage and current of the incident wave:
V ( z ) =V + ( z ) =V0+e − j β z
V0+ − j β z
I (z ) = I (z ) =
e
Z0
+
Meaning that the line impedance is likewise numerically equal
to the characteristic impedance of the transmission line for
all line position z:
V (z )
V0+e − j β z
Z (z ) =
= Z 0 + − j βz = Z 0
I (z )
V0 e
And likewise, the reflection coefficient is zero at all points
along the line:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Special Values of Load Impedance
3/10
V − (z )
0
Γ (z ) = +
= +
=0
V (z ) V (z )
We call this condition (when Z L = Z 0 ) the matched condition,
and the load Z L = Z 0 a matched load.
2. Z L = jXL
For this case, the load impedance is purely reactive (e.g. a
capacitor of inductor), the real (resistive) portion of the load
is zero:
RL = 0
The resulting load reflection coefficient is:
ΓL =
Z L − Z 0 jXL − Z 0
=
Z L + Z 0 jX L + Z 0
Given that Z0is real (i.e., the line is lossless), we find that this
load reflection coefficient is generally some complex number.
We can rewrite this value explicitly in terms of its real and
imaginary part as:
⎛ 2 Z 0 XL ⎞
jX L − Z 0 ⎛ XL2 − Z 02 ⎞
ΓL =
=⎜ 2
+
j
⎟
⎜ 2
2 ⎟
jX L + Z 0 ⎝ X L + Z 02 ⎠
⎝ XL + Z 0 ⎠
Yuck! This isn’t much help!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Special Values of Load Impedance
4/10
et’s instead write this complex value ΓL in terms of its
magnitude and phase. For magnitude we find a much more
straightforward result!
ΓL
2
=
jX L − Z 0
jX L + Z 0
2
2
X L2 + Z 02
= 2
=1
X L + Z 02
Its magnitude is one! Thus, we find that for reactive loads,
the reflection coefficient can be simply expressed as:
ΓL = e j θΓ
where
⎡ 2 Z 0 XL ⎤
2
2⎥
⎣ XL − Z 0 ⎦
θ Γ = tan −1 ⎢
We can therefore conclude that for a reactive load:
V0− = e j θ V0+
Γ
As a result, the total voltage and current along the
transmission line is simply (assuming z L = 0 ):
V ( z ) =V0+ (e − j β z + e + j θL e + j β z )
(
=V0+ e + j θΓ 2 e
− j ( β z +θ Γ 2 )
+e
+ j ( β z +θ Γ 2 )
)
= 2V0+ e + j θΓ 2 cos ( β z + θ Γ 2 )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Special Values of Load Impedance
5/10
V0+ − j β z
− e + j βz )
I (z ) =
e
(
Z0
V0+ + j θL 2 − j ( β z +θL 2)
+ j β z +θ L 2 )
=
−e (
e
e
Z0
(
= −j
2V0+
Z0
)
e + j θL 2 sin ( β z + θ L 2 )
Meaning that the line impedance can be written in terms of a
trigonometric function:
Z (z ) =
V (z )
= j Z 0 cot ( β z + θ Γ 2 )
I (z )
Note that this impedance is purely reactive—V(z) and I(z) are
90D out of phase!
Finally, the reflection coefficient function is:
V − ( z ) V0+e + j θ e + j β z
+ j 2( β z +θ
Γ (z ) = +
=
=
e
j
z
−
β
V (z )
V0+e
Γ
Γ
2)
Meaning that for purely reactive loads:
Γ (z ) = e
+ j 2( β z +θ Γ 2 )
=1
In other words, the magnitude reflection coefficient function
is equal to one—at each and every point on the transmission
line.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Special Values of Load Impedance
6/10
3. Z L = RL
For this case, the load impedance is purely real (e.g. a
resistor), and thus there is no reactive component:
XL = 0
The resulting load reflection coefficient is:
ΓL =
ZL − Z0 R − Z0
=
ZL + Z0 R + Z0
Given that Z0 is real (i.e., the line is lossless), we find that
this load reflection coefficient must be a purely real value!
In other words:
Re {ΓL } =
R − Z0
R + Z0
Im {ΓL } = 0
So a real-valued load ZL results in a real valued load reflection
coefficient GL .
Now let’s consider the line impedance Z ( z ) and reflection
coefficient function Γ ( z ) .
Q: I bet I know the answer to this one! We know that a
purely imaginary (i.e., reactive) load results in a purely
reactive line impedance.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Special Values of Load Impedance
7/10
Thus, a purely real (i.e., resistive) load will result in a purely
resistive line impedance, right??
A: NOPE! The line impedance resulting from a real load is
complex—it has both real and imaginary components!
Thus the line impedance, as well as reflection coefficient
function, cannot be further simplified for the case where
Z L = RL .
Q: Why is that?
A: Remember, a lossless transmission line has series
inductance and shunt capacitance only. In other words, a
length of lossless transmission line is a purely reactive device
(it absorbs no energy!).
* If we attach a purely reactive load at the end of the
transmission line, we still have a completely reactive system
(load and transmission line). Because this system has no
resistive (i.e., real) component, the general expressions for
line impedance, line voltage, etc. can be significantly
simplified.
* However, if we attach a purely real load to our reactive
transmission line, we now have a complex system, with both
real and imaginary (i.e., resistive and reactive) components.
This complex case is exactly what our general expressions
already describes—no further simplification is possible!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Special Values of Load Impedance
8/10
4. Z L = RL + jX L
Now, let’s look at the general case, where the load has both a
real (resitive) and imaginary (reactive) component.
Q: Haven’t we already determined all the general
expressions (e.g., ΓL ,V ( z ) , I ( z ) , Z ( z ) , Γ ( z ) ) for this general
case? Is there anything else left to be determined?
A: There is one last thing we need to discuss. It seems
trivial, but its ramifications are very important!
For you see, the “general” case is not, in reality, quite so
general. Although the reactive component of the load can be
either positive or negative ( −∞ < X L < ∞ ), the resistive
component of a passive load must be positive ( RL > 0 )—there’s
no such thing as negative resistor!
This leads to one very important and useful result. Consider
the load reflection coefficient:
ΓL =
ZL − Z0
ZL + Z0
(RL + jXL ) − Z 0
(RL + jXL ) + Z 0
(R − Z 0 ) + jXL
= L
(RL + Z 0 ) + jXL
=
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Special Values of Load Impedance
9/10
Now let’s look at the magnitude of this value:
ΓL
2
(R − Z 0 ) + jXL
= L
(RL + Z 0 ) + jXL
2
RL − Z 0 ) + X L2
(
=
2
(RL + Z 0 ) + XL2
R
(
=
(R
(R
=
(R
2
L
2
− 2RL Z 0 + Z 02 ) + X L2
2
+ 2RL Z 0 + Z 02 ) + X L2
2
+ Z 02 + X L2 ) − 2RL Z 0
L
L
2
L
+ Z 02 + X L2 ) + 2RL Z 0
It is apparent that since both RL and Z 0 are positive, the
numerator of the above expression must be less than (or equal
to) the denominator of the above expression.
Æ In other words, the magnitude of the load reflection
coefficient is always less than or equal to one!
ΓL ≤ 1
(for RL ≥ 0 )
Moreover, we find that this means the reflection coefficient
function likewise always has a magnitude less than or equal to
one, for all values of position z.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Special Values of Load Impedance
Γ (z ) ≤ 1
10/10
(for all z)
Which means, of course, that the reflected wave will always
have a magnitude less than that of the incident wave
magnitude:
V − (z ) ≤ V + (z )
(for all z)
We will find out later that this result is consistent with
conservation of energy—the reflected wave from a passive
load cannot be larger than the wave incident on it.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Transmission Line Input Impedance
1/9
Transmission Line
Input Impedance
Consider a lossless line, length A , terminated with a load ZL.
I(z)
IL
+
+
Z0 , β
V (z)
-
VL
ZL
-
A
z = −A
z = 0
Let’s determine the input impedance of this line!
Q: Just what do you mean by input impedance?
A: The input impedance is simply the line impedance seen
at the beginning ( z = −A ) of the transmission line, i.e.:
Zin = Z ( z = −A ) =
V ( z = −A )
I ( z = −A )
Note Zin equal to neither the load impedance ZL nor the
characteristic impedance Z0 !
Zin ≠ Z L
Jim Stiles
and
The Univ. of Kansas
Zin ≠ Z 0
Dept. of EECS
8/27/2007
Transmission Line Input Impedance
2/9
To determine exactly what Zin is, we first must determine the
voltage and current at the beginning of the transmission line
( z = −A ).
V ( z = − A ) = V0+ ⎡⎣e + j β A + Γ L e − j β A ⎤⎦
V0+ + j β A
⎡⎣e
− Γ L e − j β A ⎤⎦
I ( z = −A ) =
Z0
Therefore:
⎛ e + j β A + ΓL e − j β A ⎞
V ( z = −A )
= Z0 ⎜ + j βA
Zin =
⎟
I ( z = −A )
− ΓL e − j β A ⎠
⎝e
We can explicitly write Zin in terms of load ZL using the
previously determined relationship:
ΓL =
ZL − Z0
ZL + Z0
Combining these two expressions, we get:
Zin =
ZL + Z 0 ) e + j βA + (ZL − Z 0 ) e − j βA
(
Z0
(ZL + Z 0 ) e + j βA − (ZL − Z 0 ) e − j βA
⎛ Z L (e + j β A + e − j β A ) + Z 0 (e + j β A − e − j β A ) ⎞
= Z0 ⎜
⎟
⎜ Z L (e + j β A + e − j β A ) − Z 0 (e + j β A − e − j β A ) ⎟
⎝
⎠
Now, recall Euler’s equations:
e + j β A = cos β A + j sin β A
e − j β A = cos β A − j sin β A
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Transmission Line Input Impedance
3/9
Using Euler’s relationships, we can likewise write the input
impedance without the complex exponentials:
⎛ Z L cos β A + j Z 0 sin β A ⎞
⎟
A
A
+
β
β
cos
sin
Z
j
Z
0
L
⎝
⎠
Zin = Z 0 ⎜
⎛ Z + j Z 0 tan β A ⎞
= Z0 ⎜ L
⎟
A
+
β
tan
Z
j
Z
L
⎝ 0
⎠
Note that depending on the values of β , Z 0 and A , the input
impedance can be radically different from the load impedance
ZL !
Special Cases
Now let’s look at the Zin for some important load impedances
and line lengths.
Æ You should commit these results to memory!
1. A = λ
2
If the length of the transmission line is exactly one-half
wavelength ( A = λ 2 ), we find that:
βA =
meaning that:
cos β A = cos π = −1
Jim Stiles
2π λ
=π
λ 2
and
The Univ. of Kansas
sin β A = sin π = 0
Dept. of EECS
8/27/2007
Transmission Line Input Impedance
and therefore:
4/9
⎛ Z L cos β A + j Z 0 sin β A ⎞
⎟
⎝ Z 0 cos β A + j Z L sin β A ⎠
Zin = Z 0 ⎜
⎛ Z ( − 1) + j Z L (0) ⎞
= Z0 ⎜ L
⎟
(
1)
(0)
−
+
Z
j
Z
L
⎝ 0
⎠
= ZL
In other words, if the transmission line is precisely one-half
wavelength long, the input impedance is equal to the load
impedance, regardless of Z0 or β.
Zin = Z L
Z0, β
A = λ
2. A = λ
ZL
2
4
If the length of the transmission line is exactly one-quarter
wavelength ( A = λ 4 ), we find that:
βA =
meaning that:
cos β A = cos π 2 = 0
Jim Stiles
2π λ π
=
λ 4 2
and
The Univ. of Kansas
sin β A = sin π 2 = 1
Dept. of EECS
8/27/2007
Transmission Line Input Impedance
5/9
and therefore:
⎛ Z L cos β A + j Z 0 sin β A ⎞
Zin = Z 0 ⎜
⎟
+
cos
sin
A
A
β
β
Z
j
Z
L
⎝ 0
⎠
⎛ Z (0) + j Z 0 (1) ⎞
= Z0 ⎜ L
⎟
⎝ Z 0 (0) + j Z L (1) ⎠
(Z )
=
2
0
ZL
In other words, if the transmission line is precisely one-quarter
wavelength long, the input impedance is inversely proportional to
the load impedance.
Think about what this means! Say the load impedance is a short
circuit, such that Z L = 0 . The input impedance at beginning of
the λ 4 transmission line is therefore:
Z )
(
=
2
Zin
0
ZL
Z )
(
=
2
0
0
=∞
Zin = ∞ ! This is an open circuit! The quarter-wave transmission
line transforms a short-circuit into an open-circuit—and vice
versa!
Zin = ∞
Z0, β
A = λ
Jim Stiles
ZL=0
4
The Univ. of Kansas
Dept. of EECS
8/27/2007
Transmission Line Input Impedance
6/9
3. Z L = Z 0
If the load is numerically equal to the characteristic impedance
of the transmission line (a real value), we find that the input
impedance becomes:
⎛ Z L cos β A + j Z 0 sin β A ⎞
⎟
Z
j
Z
+
A
A
β
β
cos
sin
L
⎝ 0
⎠
Zin = Z 0 ⎜
⎛ Z cos β A + j Z 0 sin β A ⎞
= Z0 ⎜ 0
⎟
+
Z
j
Z
A
A
β
β
cos
sin
0
⎝ 0
⎠
= Z0
In other words, if the load impedance is equal to the
transmission line characteristic impedance, the input impedance
will be likewise be equal to Z0 regardless of the transmission
line length A .
Zin = Z 0
Z0 , β
ZL=Z0
A
4. Z L = j X L
If the load is purely reactive (i.e., the resistive component is
zero), the input impedance is:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Transmission Line Input Impedance
7/9
⎛ Z L cos β A + j Z 0 sin β A ⎞
⎟
A
A
cos
sin
+
β
β
Z
j
Z
L
⎝ 0
⎠
Zin = Z 0 ⎜
⎛ j X L cos β A + j Z 0 sin β A ⎞
= Z0 ⎜
⎟
2
⎝ Z 0 cos β A + j X L sin β A ⎠
⎛ X cos β A + Z 0 sin β A ⎞
= j Z0 ⎜ L
⎟
A
A
−
cos
sin
β
β
Z
X
L
⎝ 0
⎠
In other words, if the load is purely reactive, then the input
impedance will likewise be purely reactive, regardless of the
line length A .
Z in = j X in
Z0 , β
ZL=jXL
A
Note that the opposite is not true: even if the load is purely
resistive (ZL = R), the input impedance will be complex (both
resistive and reactive components).
Q: Why is this?
A:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Transmission Line Input Impedance
8/9
5. A λ
If the transmission line is electrically small—its length A is
small with respect to signal wavelength λ --we find that:
βA =
and thus:
cos β A = cos 0 = 1
and
2π
λ
A = 2π
A
λ
≈0
sin β A = sin 0 = 0
so that the input impedance is:
⎛ Z L cos β A + j Z 0 sin β A ⎞
⎟
Z
j
Z
β
β
A
A
cos
sin
+
L
⎝ 0
⎠
Zin = Z 0 ⎜
⎛ Z (1) + j Z L (0) ⎞
= Z0 ⎜ L
⎟
Z
j
Z
+
(1)
(0)
L
⎝ 0
⎠
= ZL
In other words, if the transmission line length is much smaller
than a wavelength, the input impedance Zin will always be equal
to the load impedance Z L .
This is the assumption we used in all previous circuits courses
(e.g., EECS 211, 212, 312, 412)! In those courses, we assumed
that the signal frequency ω is relatively low, such that the
signal wavelength λ is very large ( λ A ).
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Transmission Line Input Impedance
9/9
Note also for this case ( the electrically short transmission
line), the voltage and current at each end of the transmission
line are approximately the same!
V (z = −A) ≈ V (z = 0)
and
I(z = −A) ≈ I (z = 0) if A λ
If A λ , our “wire” behaves exactly as it did in EECS 211 !
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Power Flow and Return Loss
1/5
Power Flow and
Return Loss
We have discovered that two waves propagate along a
transmission line, one in each direction (V + ( z ) and V − ( z ) ).
V0+ − j β z
⎡⎣e
− Γ 0 e + j β z ⎤⎦
I (z ) =
Z0
+
V (z ) =V ⎡⎣e
+
0
− j βz
+ Γ0 e
-
z = −A
+ j βz
⎤⎦
A
IL
+
VL
ZL
-
z = 0
The result is that electromagnetic energy flows along the
transmission line at a given rate (i.e., power).
Q: How much power flows along a transmission line, and
where does that power go?
A:
Jim Stiles
We can answer that question by determining the
power absorbed by the load!
The Univ. of Kansas
Dept. of EECS
8/27/2007
Power Flow and Return Loss
2/5
The time average power absorbed by an impedance ZL is:
1
2
1
= Re {V (z = 0) I (z = 0)∗ }
2
1
=
Re V0+ ⎡⎣e − j β 0 + Γ 0 e + j β 0 ⎤⎦
2 Z0
Pabs = Re {VL I L ∗ }
{(
=
=
V0+
2
2 Z0
V0+
2
2 Z0
{
Re 1 − ( Γ 0∗ − Γ 0 ) − Γ 0
2
) (V
0
+
⎡⎣e
−jβ 0
− Γ0 e
+jβ0
⎤⎦
)}
∗
}
(1 − Γ )
2
0
The significance of this result can be seen by rewriting the
expression as:
Pabs =
=
=
V0+
2
2
2 Z0
V0+
0
2
2 Z0
V0+
(1 − Γ )
−
2
2 Z0
−
V0+ Γ 0
2
2 Z0
V0−
2
2 Z0
The two terms in above expression have a very definite physical
meaning. The first term is the time-averaged power of the
wave propagating along the transmission line toward the load.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Power Flow and Return Loss
3/5
We say that this wave is incident on the load:
Pinc = P+ =
V0+
2
2Z 0
Likewise, the second term of the Pabs equation describes the
power of the wave moving in the other direction (away from
the load). We refer to this as the wave reflected from the
load:
Pref = P− =
V0−
2
2Z0
2
=
Γ 0 V0+
2Z0
2
2
= Γ L Pinc
Thus, the power absorbed by the load (i.e., the power delivered
to the load) is simply:
Pabs = Pinc − Pref
or, rearranging, we find:
Pinc = Pabs + Pref
This equation is simply an expression of the conservation of
energy !
It says that power flowing toward the load (Pinc) is either
absorbed by the load (Pabs) or reflected back from the load
(Pref).
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Power Flow and Return Loss
4/5
Pabs
Pinc
Pref
ZL
2
Note that if Γ L = 1 , then Pinc = Pref, and therefore no power is
absorbed by the load.
This of course makes sense !
The magnitude of the reflection coefficient ( ΓL ) is equal to
one only when the load impedance is purely reactive (i.e., purely
imaginary).
Of course, a purely reactive element (e.g., capacitor or inductor)
cannot absorb any power—all the power must be reflected!
RREETTUURRN
NL
S
SS
LO
OS
The ratio of the reflected power to the incident power is known
as return loss. Typically, return loss is expressed in dB:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
Power Flow and Return Loss
5/5
⎡P ⎤
2
R .L. = −10 log10 ⎢ ref ⎥ = −10 log10 ΓL
⎣ Pinc ⎦
For example, if the return loss is 10dB, then 10% of the
incident power is reflected at the load, with the remaining 90%
being absorbed by the load—we “lose” 10% of the incident
power
Likewise, if the return loss is 30dB, then 0.1 % of the incident
power is reflected at the load, with the remaining 99.9% being
absorbed by the load—we “lose” 0.1% of the incident power.
Thus, a larger numeric value for return loss actually indicates
less lost power! An ideal return loss would be ∞ dB, whereas a
return loss of 0 dB indicates that Γ L = 1 --the load is reactive!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
VSWR
1/3
VSWR
Consider again the voltage along a terminated transmission line,
as a function of position z :
V ( z ) = V0+ ⎡⎣e − j β z + Γ L e + j β z ⎤⎦
Recall this is a complex function, the magnitude of which
expresses the magnitude of the sinusoidal signal at position z,
while the phase of the complex value represents the relative
phase of the sinusoidal signal.
Let’s look at the magnitude only:
|V ( z )| = |V0+| |e − j β z + Γ L e + j β z |
= |V0+||e − j β z ||1 + Γ L e + j 2 β z |
= |V0+||1 + Γ L e + j 2 β z |
ICBST the largest value of |V (z)| occurs at the location z
where:
ΓL e + j 2β z = ΓL + j 0
while the smallest value of |V (z)|occurs at the location z
where:
ΓL e + j 2β z = − ΓL + j 0
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
VSWR
2/3
As a result we can conclude that:
V ( z ) max = V0+ (1 + ΓL
)
V ( z ) min = V0+ (1 − ΓL
)
The ratio of V ( z ) max to V ( z ) min is known as the Voltage
Standing Wave Ratio (VSWR):
VSWR V ( z ) max
1 + ΓL
=
V ( z ) min
1 − ΓL
∴
1 ≤ VSWR ≤ ∞
Note if Γ L = 0 (i.e., Z L = Z 0 ), then VSWR = 1. We find for this
case:
V (z ) max = V (z ) min = V0+
In other words, the voltage magnitude is a constant with
respect to position z.
Conversely, if Γ L = 1 (i.e., Z L = jX ), then VSWR = ∞ . We find
for this case:
V (z ) min = 0
and
V (z ) max = 2V0+
In other words, the voltage magnitude varies greatly with
respect to position z.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8/27/2007
VSWR
3/3
As with return loss, VSWR is dependent on the magnitude of ΓL
(i.e, |ΓL|) only !
|V(z)|
∆z = λ
2
|V(z)|max
|V(z)|min
z
Jim Stiles
The Univ. of Kansas
Dept. of EECS