Impedance on lossless lines
√
• For lossless lines, γ = jω L0 C 0 = jβ; we will focus on lossless
lines almost always from here on out
EE 311 - Lecture 7
• Thus
Ṽ (z) = V0+ e−jβz + V0− ejβz
¡
¢
˜ = 1 V + e−jβz − V − ejβz
I(z)
0
0
Z0
• Impedance on lossless lines
• Reflection coefficient
(1)
(2)
where V0+ and V0− are two phasor constants and Z0 =
for a lossless line
• Impedance equation
• Shorted line example
p
L/C
• The impedance on a lossless line is then
· + −jβz
¸
V e
Ṽ (z)
+ V0− ejβz
= Z0 0+ −jβz
Z(z) =
˜
V0 e
− V0− ejβz
I(z)
Assigned reading: Sec 2.5.1 of Ulaby
which is a function of z!
• Thus, the impedance on a transmission line changes depending
where you measure it on the line
1
2
Alternate form
Finding gamma
• Note that we can factor out
get:
Z(z) =
V + e−jβz
Z0 0+ −jβz
V0 e
= Z0
where we have defined Γ =
coefficient”
V0+ e−jβz
µ



on the top and bottom to
1+
1−
1 + Γe2jβz
1 − Γe2jβz
V0−
V0+
• Consider a Xmission line of length l terminated in an
impedance ZL . A picture is below

V0− 2jβz
e
V0+

V0− 2jβz
e
V0+
¶
• The two parallel lines in this picture represent a transmission
line. It could be any type of line (parallel plate, coax, etc.) but
we are told the Z0 and the µ, ² inside the line which is all we
need to know to talk about voltages, currents, and impedances

• Since we’ve drawn a circuit impedance ZL at z = 0, the
impedance on the transmission line at z = 0 must be ZL
as a “voltage reflection
• Note that if Γ = 0 (we have no V0− wave), Z(z) = Z0 meaning
that the impedance is no longer a function of z
• This is usually the case when we talk about infinitely long
lines, where + propagating waves can propagate forever
without generating a − traveling wave
• Finite length lines terminated in an impedance, however, will
typically have Γ 6= 0 as we will see next...
3
Zg
~
Vg
+
~
Ii
Transmission line
+
+
~
Vi Zin
Z0
~
VL
~
IL
ZL
-
-
Generator
Load
z = -l
⇓
z=0
4
• Using our previous formula for Z(z) and plugging in z = 0 we
get
1+Γ
= ZL
Z(0) = Z0
1−Γ
which can be solved to obtain
ZL − Z 0
Γ=
ZL + Z 0
Impedance equation without Γ
• We can also plug in Γ =
ZL −Z0
ZL +Z0
Z(z) = Z0
µ
so that
• Note Γ in general is complex; Ulaby writes Γ = |Γ| ejθr
Z(−l) = Z0
• Knowing Γ we can find the impedance anywhere else on the
line, for example at z = −l, by using
µ
¶
1 + Γe2jβz
Z(z) = Z0
1 − Γe2jβz
and simplify some to get
¶
ZL − jZ0 tan(βz)
Z0 − jZL tan(βz)
µ
ZL + jZ0 tan(βl)
Z0 + jZL tan(βl)
¶
• This last formula is probably the most useful of all the
equations since it directly gives the impedance at z = −l if we
know Z0 , ZL , β, and l
since all quantities in this equation are now known (note
√
√
β = ω LC = ω µ²)
• Notice the dependence on l occurs only in the tan(βl) terms,
which are periodic in βl with period π. Since β = 2π/λ, the
impedance is periodic in l with period λ/2.
• Another way of writing this equation is
¶
µ
1 + Γ(z)
Z(z) = Z0
1 − Γ(z)
• Thus impedances on lossless transmission lines repeat
themselves every half wavelength
where Γ(z) = Γe2jβz
6
5
Shorted line example
• Consider a transmission line which is short circuited (Z L = 0)
• Plugging ZL = 0 into our formula gives
·
¸
jZ0 tan(βl)
Z(−l) = Z0
= jZ0 tan(βl)
Z0
Shorted line impedance
• Now let’s plot the impedance versus position on the line:
(c)
• This is purely imaginary since both Z0 and tan(βl) are real;
makes sense because a real impedance allows power dissipation
and there is no way to dissipate power in a lossless line
terminated in a short!
sc
Impedance
Zin
jZ0
l
Z0
short
circuit
z
-λ
⇒
7
-3λ
4
-λ
2
-λ
4
(d) 8
• Our plot is just a plot of tan(βl). Note that a shorted line
looks like a positive reactance for 0 < l < λ/4 but looks like an
open circuit when l = λ/4. This is not what you expect from
circuit theory!
• Reasonable because transmission line theory only applies when
lines become a significant fraction of a wavelength long (i.e.
λ/4 as above)
• This rarely happens (except in very long power system
transmission lines) at low frequencies (as in circuits) because
the wavelengths are so long at low frequencies
• Transmission line effects however are extremely important
when dealing with higher frequency systems where wire lengths
can be appreciable fractions of a wavelength
• Note by choosing an appropriate length of our short circuited
line we can get any possible value of reactance we desire, i.e.
the line can look like any inductor or capacitor at a specific
frequency as long as l is chosen correctly. See Example 2-7 in
book.
EE 311 - Lecture 8
• More on shorted line
• Voltage and current on shorted line
• Standing waves: VSWR
• Shorted line example
• Other examples
Assigned reading: Sec 2.5.2 of Ulaby
9
Reduction to circuit theory
• Note our shorted line can be useful because often lumped
element inductors or capacitors are not available at high
frequencies and transmission lines represent the only way of
obtaining the necessary impedances
• If we think a little more about the circuit theory limit (where
the line is much shorter than a wavelength, so l/λ << 1 and βl
is also << 1) we can see that transmission line theory gives us
the right answer since
µ
¶
ZL + jZ0 tan(βl)
lim Z0
= ZL
βl→0
Z0 + jZL tan(βl)
as βl → 0,tan(βl) = 0
• Thus the impedance measured on a wire connected to ZL is
just ZL as long as βl << 1 (the line is much shorter than a
wavelength)
11
10
Improved circuit theory approximation
• We can do one step better than this for our shorted line by
using a better approximation for the tangent:
lim tan(βl) ≈ βl
βl→0
• This gives Z(−l) = jZ0 tan(βl) ≈ jZ0 βl
p
√
• Recalling Z0 = L/C and β = ω LC gives
p
√
Z(−l) ≈ j L/Cω LCl = jωLl, which shows that a short
circuited line that is small compared to the wavelength has the
same reactance as an inductor with inductance Ll, which is
inductance per unit length times length!
• This makes sense because a shorted line looks like a small turn
of wire = an inductor in the circuit theory limit.
• However, when βl is not << 1, the graph of Z(−l) shows that
the shorted line no longer follows our circuit theory
expectations
12
Voltage and current plots
Voltages and currents on a shorted line
• It is also interesting to examine the voltages and currents on a
L −Z0
short circuited line. Note for a short, Γ = Z
ZL +Z0 = −1 since
ZL = 0
• Plots of |V (z)| and |I(z)| on a shorted line look like:
(a) ZL = Z0
λ/2
2|V0+|
• This gives
V (z) = V0+ e−jβz − V0+ ejβz = 2jV0+ sin(βz)
¢ 2V +
1 ¡ + −jβz
I(z) =
+ V0+ ejβz = 0 cos(βz)
V0 e
Z0
Z0
(3)
(4)
-λ
which apply for a shorted line only!
• If we plot |V (z)| and |I(z)| versus z, we observe a “standing
wave” pattern, meaning that |V (z)| and |I(z)| oscillate
periodically on the line as z is changed
-λ
13
(b) ZL = 0
-3λ
4
(b) Z = 0
λ/2
(short circuit)
-λ
-λ
2
4
(short circuit)
-3λ
4
(c) Z =
-λ
-λ
2
4
14 circuit)
(open
0
z
~
|V(z)|
+
2|V0 |
0
z
VSWR
Voltage and current plots
• Note that |V (z)| and |I(z)| also repeat themselves every λ/2 on
the line, which is true for all lossless transmission line voltage
and current amplitudes, and for impedances
• To describe the standing wave voltage pattern on a Xmission
line, a term known as the voltage standing wave ratio (VSWR)
is used
• This is defined to be the ratio of the maximum to minimum
voltage amplitude on the line
15
• The VSWR on a shorted line is infinite since the minimum
voltage is zero
• In general however we can write
¯
¯
Max ¯V0+ e−jβz + V0− ejβz ¯
¯
¯
V SW R =
Min ¯V0+ e−jβz + V0− ejβz ¯
which can be rewritten as
¯¯
¯¤
£¯
Max ¯V0+ e−jβz ¯ ¯1 + Γe2jβz ¯
¯
£¯
¤
V SW R =
Min ¯V0+ e−jβz ¯ |1 + Γe2jβz |
• Thus we are looking for the maximum and minimum of
¯
¯
¯1 + Γe2jβz ¯ since the V + term cancels. This is the amplitude
0
of a sum of two complex numbers (i.e. |a + b|)
16
“Matched” line example
VSWR equation
• Thus, we’ll get the largest amplitude when the two numbers
are exactly in phase, the smallest amplitude when the two are
out of phase.
¯
¯
• The maximum value of ¯1 + Γe2jβz ¯ will therefore occur when
Γe2jβz has phase 0 degrees, giving Γe2jβz = |Γ|.
• The minimum will occur when Γe2jβz = − |Γ|
• Thus on a “matched” line the impedance is not a function of z
but rather is everywhere Z0
• Note this occurs because Γ = 0 since ZL = Z0 so V0− = 0 and
there is no reflected wave
• The VSWR can therefore be written in general as
V SW R =
• A “matched” line is a line terminated in its own characteristic
impedance, i.e. ZL = Z0 . Then,
¶
µ
ZL + jZ0 tan(βl)
= Z0
Z(−l) = Z0
Z0 + jZL tan(βl)
1 + |Γ|
1 − |Γ|
• The same equation for VSWR is obtained if the ratio of the
maximum to minimum current on the line is considered
• People working with transmission lines usually try to make all
loads matched loads to avoid varying impedances on the line.
Unfortunately this is not always possible so we need
transmission line theory for the unmatched cases.
17
18
Quarter wave line
• A Quarter wave line has length l = λ/4, so plugging this in to
π
λ
the impedance equation and noting that βl = 2π
λ 4 = 2 , we get
¶
µ
Z2
ZL + jZ0 tan(π/2)
= 0
Z(−λ/4) = Z0
Z0 + jZL tan(π/2)
ZL
since tan(π/2) =
sin(π/2)
cos(π/2)
=∞
• Thus the impedance on a lossless Xmission line λ/4 away from
a load is proportional to the inverse of the load impedance.
Our short circuited line was an example since a short (ZL = 0)
became an open (Z(−λ/4) = ∞) a quarter wavelength away
• Recall that two quarter wave lines form a λ/2 line which should
result in Z(−λ/2) = ZL
Z02
ZL . Using the equation again to
Z02 ZL
= ZL , which agrees with our
Z02
• Let’s check: ZA =
Z02
ZA
get ZB =
=
expectations!
19
EE 311 - Lecture 9
• A sample problem
• Shorted lossy line
• Power on lines
Assigned reading: Secs. 2.6-2.7 of Ulaby
find ZB we
20
Sample problem continued
A sample problem
• Because the line is lossless, this is the same as the power
dissipated in the load
• For this circuit, how much power is dissipated in the load?
• Answer: Here we’ve got TL and ccts mixed together, so we
need to use TL theory to replace the line with an impedance
Zin , then use circuit theory to find the power
• Thus we find Zin from TL theory, replace the right hand side
of the picture with Zin , and then use circuit theory to find the
power dissipated in Zin
• Goal: replace previous picture with picture below, then circuit
theory tells us the power into the load is
µ
¶¾
½
1
1
1
(1)Zin
∗
Re {VL IL } = Re
∗
2
2
100 + Zin 100 + Zin
=
R G =100 Ω
2 |100 + Zin |
2
R G =100 Ω I
L
Z0 =100 Ω
VG=1 V
f=1 MHz
Zin
Re {Zin }
100Ω
z=0
z=−λ/8
16 µH
VG=1 V
f=1 MHz
Zin V
L
21
22
Finding Zin
Impedance on lossy lines
• To use the transmission line equation to find Zin = Z(−λ/8),
first we need to find the load impedance
• Placing the resistance and inductance in parallel we find
ZL = 100 Ω k jω(16 × 10−6 ) ≈ 50 + j50
• Then
¶
ZL + jZ0 tan(βl)
Z(−λ/8) = Z0
Z0 + jZL tan(βl)
#
"
λ
50 + j50 + j100 tan( 2π
λ 8)
= 100(2 + j)
= 100
λ
100 + j(50 + j50) tan( 2π
λ 8)
µ
• Thus the power dissipated in the load is
Re {200 + 100j}
2 |100 + 200 + 100j|
2
= 0.001 Watts
• Note we could also have found V (z) and I(z) everywhere on
the line and then taken 12 Re {V (0)I(0)∗ }, but this would have
been a lot harder!
23
• So far we’ve been talking about impedance on lossless lines
only. We can generate equations to handle lossy lines just by
realizing that γ = α + jβ for lossy lines instead of just jβ
• Our old equation for impedance
¶
µ
ZL + jZ0 tan(βl)
Z(−l) = Z0
Z0 + jZL tan(βl)
becomes
Z(−l) = Z0
for a lossy line
µ
ZL + Z0 tanh(γl)
Z0 + ZL tanh(γl)
¶
• Here tanh is the hyperbolic tangent function:
¢ ¡
¢
¡ γl
e − e−γl / eγl + e−γl and is taken of a complex argument
γl
24
Short circuited lossy line impedance
Short circuited lossy line
• The plots below compare impedances on lossy and lossless
short circuited lines:
• For a shorted lossy line, ZL = 0, so the general equation
becomes Z(−l) = Z0 tanh(γl) which can be expanded as
Lossy line
sinh(αl) cos(βl) + j cosh(αl) sin(βl)
Z(−l) = Z0
cosh(αl) cos(βl) + j sinh(αl) sin(βl)
40
where sinh and cosh are hyperbolic sine and cosine functions
nπ
• Also if l = nλ
4 for n odd, βl = 2 and cos(βl) = 0. Thus
nλ
Z(− 4 ) = Z0 / tanh(αl) for n odd
Real
Imaginary
40
Real
Imaginary
30
Impedance/Z0
• For example, if l = nλ
2 , βl = nπ and sin(βl) = 0. Thus
Z(− nλ
)
=
Z
tanh(αl)
0
2
50
30
Impedance/Z0
• This is quite complicated compared to the lossless answer
jZ0 tan(βl), but we can still find some points of interest
Lossless line
50
20
10
20
10
0
0
−10
−10
−20
0
0.5
1
Distance along line (λ)
−20
0
0.5
1
Distance along line (λ)
25
26
Power on lines
Analysis of impedance
• Note that the impedance now has both real and imaginary
parts since the line is lossy. Also note that the “resonance”
effects at λ/4, 3λ/4, etc. get smaller as the line gets longer due
to attenuation of the reflected wave
• Eventually as l → ∞ , Z(−l) should approach Z0 (which may
be complex) on a lossy line because all the reflected wave gets
attenuated before returning to z = −l
• This is the extent of our study of lossy lines; we will focus on
lossless lines only from here on...
• We have both voltages and currents on a transmission line, so
the line carries power
• We’re working with sinusoidal steady state quantities, so we
should use our AC circuit power concepts
• The instantaneous power is p(z, t) = v(z, t)i(z, t). However as
with AC circuits this will contain a term that oscillates rapidly
on top of a DC offset.
• The time average power is what we are interested in. It can be
shown that this is given by:
o
1 n
˜ ∗
Pav = Re Ṽ (z)I(z)
2
as in AC circuit theory
27
28
Power on terminated lines
• On a terminated lossless line:
¡
¢
Ṽ (z) = V0+ e−jβz + Γejβz
¡
¢
˜ = V + /Z0 e−jβz − Γejβz
I(z)
0
• Plugging these in gives a result of the form
Pav =
Pav =
1 ¯¯ + ¯¯2
Re {(a + b) (a∗ − b∗ )}
V
2Z0 0
¢ª
©¡
1 ¯¯ + ¯¯2
V
Re |a|2 − |b|2 + ba∗ − ab∗
2Z0 0
´
1 ¯¯ + ¯¯2 ³
2
1 − |Γ|
Pav =
V0
2Z0
EE 311 - Lecture 10
• Gamma plane analysis
• Voltage and current in the Gamma plane
• Impedance in the Gamma plane
Assigned reading: Secs. 2.8-2.9.2 of Ulaby
• The power carried by a lossless line is the same everywhere on
the line, even though the voltages and currents may vary.
• This is because a lossless line cannot absorb any power.
29
30
Complex Gamma plane
Gamma plane analysis
• One of our earlier ways of writing Z(z) was
¶
µ
1 + Γ(z)
Z(z) = Z0
1 − Γ(z)
• Let’s draw a picture of how Γ(z) varies by sketching the
complex Gamma plane
• If we take z = 0, then Γ(z) = Γ, which is just some particular
complex number (a point in the Gamma plane)
where Γ(z) = Γe2jβz
Im{Γ(z)}
• We can also write
¡
¢
V (z) = V0+ e−jβz +V0− ejβz = V0+ e−jβz 1 + Γe2jβz = V0+ e−jβz (1 + Γ(z))
I(z) =
¢ V + e−jβz
1 ¡ + −jβz
− V0− ejβz = 0
(1 − Γ(z))
V0 e
Z0
Z0
Γ(−d)
• Taking a closer look at Γ(z) = Γe2jβz , we can see that |Γ(z)|
never changes on the line as we vary z, only the phase changes
due to the e2jβz term
31
Re{Γ(z)}
32
Moving in the Gamma plane
• Note that we can think either in terms of rectangular,
Γ(z) = ΓR (z) + jΓI (z), or polar coordinates, Γ(z) = |Γ(z)|ejθr
in the Gamma plane
• What happens when z =
6 0? Since Γ(z) = Γe2jβz , only the
phase of Γ(z) changes, not its magnitude
• Thus, we just rotate around a circle of radius |Γ(z)|
• Note that as z becomes more negative (as in z = −l), the angle
of Γ(z) decreases, so we rotate clockwise in the Gamma plane
• As we move down the line, we rotate through an angle
corresponding to e−2jβl which is 2βl radians clockwise= 4πl/λ
radians. Note this repeats every time l increases by λ/2
33
Interesting things about the Gamma
plane:
L −Z0
1. Since Γ = Z
ZL +Z0 , it can be shown that the maximum possible
|Γ| is 1. Thus all our possible Γ’s will fit inside a circle of
radius 1 in the Gamma plane
2. The origin of the Gamma plane means Γ = 0, so there are no
reflections at the load and ZL must be Z0
3. Real axis: we wrote V (z) = V0+ e−jβz (1 + Γ(z)) so
|V (z)| = |V0+ ||1 + Γ(z)|. Just like our VSWR analysis, the
maximum voltage will occur when 1 and Γ(z) are in phase.
This is when Γ(z) is on the positive real axis
4. Minimum voltage amplitude will occur when 1 and Γ(z) are out
of phase, this is when Γ(z) is on the negative real axis. Note
that the reverse is true for the current since it goes as |1 − Γ(z)|
34
Visualizing voltage and currents
• These statements are summarized below.
“Paddle wheel” diagram
• Note we can use our knowledge about the location of maxima
and minima to find Γ if we know |Γ| and the distance l from
the load to a voltage max or min. We just “spin” Γ(z) back
from the max or min point thorugh 4πl/λ radians to get to ΓL
• A graphical interpretation of voltage variations on line which
are proportional to |1 + Γ(z)| is the “paddle wheel” diagram
Im{Γ(z)}
Maximum Current
Minimum Voltage
|1+Γ(z)|
• It is clear from the diagram that we’ll get a maximum length of
the paddle arm (|1 + Γ(z)|) when Γ(z) is on the positive real
axis
• It is also clear we’ll get a minimum length of the paddle arm
when Γ(z) is on the negative real axis
Γ(−d)
Re{Γ(z)}
1
• Note that the length |1 + Γ(z)| will vary as we move on the line
and Γ(z) rotates
• Also we can see that the maximum possible voltage on the line
is 2|V0+ |, minimum possible is 0, both of which can only occur
if |Γ(z)| = 1, i.e. short or open circuit load
Maximum Voltage
Minimum Current
35
36
Im{Γ(z)}
Re{Γ(z)}
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SS O P
S.W LO (C F,
L. AK EF
I
RF . PE CO
.
or
E
M
S.W NS
F,
A
EF
O
TR
.C
SM
N
A
TR
0.28
39
Im{Γ(z)}
Constant Normalized Reactance
Re{Γ(z)}
Constant Normalized Resistance
• It turns out the resulting contours are off-center circles in the
Gamma plane.
R
O
),
Zo
X/
• What are these contours? To find them solve
1+ΓR (z)+jΓI (z)
Zn (z) = 1−Γ
by setting the real or imaginary parts
R (z)−jΓI (z)
of Zn (z) to a constant to find a relationship between ΓR and ΓI
R
,O
o)
120
0.39
100
0.8
0.4
110
E
NC
TA
EP
SC
SU
VE
TI
CI
PA
A
C
0
-65 .5
• The Smith Chart is a Gamma plane with labeled contours of
constant Zn (z)
3
0.4
0
13
7
0.0
2
0.4
0.3
65
0.5
8
0.0
0.6
Smith Chart
0.0 —> WAVELE
0.49
N
G
T
H
S TOW
ARD
0.48
0.0
D <—
0.49
G
E
RD LOA
NER
TOWA
0.48
± 180
AT
THS
0.47
OR
170
-170
ENG
—>
VEL
0.47
WA
0.0
160
6
<—
4
-90
90
-160
0.4
0.4
4
85
-85
6
0.0
0.0
150
5
50
5
IN
0.4
)
80
-1
D
-80
U
/Yo
0.4
CTI
5
(-jB
VE
5
0.0
CE
R
E
AN
AC
PT
TA
0.0
CE
0.1
75
NC
-75
6
14
US
EC
40
0.4
0
ES
-1
O
IV
4
M
CT
PO
DU
N
EN
IN
70
T
(+
jX
/Z
0.2
1
0.4
0.6 60
0.4
50
0.9
9
0.0
0.7
0.5
0.4
0.7
1.2
1.4
55
-55
1.4
2.0
0.12
1.6
1.6
0.11
5.0
0.1
0.8
0.8
0.39
0.11
-100
37
1.0
1.0
0.38
0.9
1.2
0.12
-90
0.13
0.37
-4
5
1.0
1.0
1.6
1.0
0.14
-80
-4
0
0.36
• Thus we can overlay scales of Zn (z) on our Gamma plane to
get impedances graphically if we know Γ(z)
Assigned reading: Secs 2.9.2-2.9.4 of Ulaby
0.35
0.15
-70
0.2
0.1
0.4
1
-110
0
.0
9
0.4
2
0.0
-1
8
2
0
CAP
0.4
AC
ITI
3
VE
0.0
RE
7
AC
-1
TA
30
NC
E
C
OM
PO
N
EN
T
(-j
• The above equation provides a “map” between Γ(z) and Zn (z).
A little math shows that a given Γ(z) yields a unique value of
Zn (z) and vice-versa
• Properties of Smith Chart
• Smith Chart example
3.0
4
6
0.1
0.3
-35
1.4
1.8
-60
-30
1.8
2.0
4.0
7
0.4
0.1
as a “normalized impedance”, i.e. the impedance relative to
the characteristic impedance of the line
• Smith Chart
0.3
0.1
0.6
0.2
Z(z)
1 + Γ(z)
=
Z0
1 − Γ(z)
0.1
9
3
0.3
0.8
0.4
0.2
Zn (z) =
EE 311 - Lecture 11
2
0.3
8
0.1
0
-5
-25
1.0
0.6
0.3
• We’ve seen that looking in the Gamma plane makes it easier to
see the behavior of voltages and currents on lines. What about
impedance?
³
´
1+Γ(z)
• We know Z(z) = Z0 1−Γ(z)
, now define
10
-20
3.0
0
6
0.0
4.0
-4
4
0.4
0.2
-70
1
-30
-60
0.2
0
-5
0.22
1.2
-20
-15
5.0
1
-10
0.3
40
20
10
9
0.1
30
0.28
2.0
9
0.2
20
0.3
0.8
0.4
Impedance in the Gamma plane
0.23
0.25
0.26
0.24
0.27
0.25
0.24
0.26
0.23
0.27
REFLECTION COEFFICIENT IN DEG
REES
LE OF
ANG
SMISSION COEFFICIENT IN
F TRAN
DEGR
O
E
L
EES
ANG
50
0.2
1
20
0.2
50
0.22
Work on example
Smith Chart example
• Circles of constant real part of Zn (z) are centered on the real
axis, while constant imaginary part circles are centered off the
chart (positive above real axis, negative below)
• Thus, the Smith Chart enables us to find impedances on
transmission lines graphically as opposed to using the formulas
• We can work entirely in terms of impedances without worrying
about Γ(z) even though the chart actually is just a labeled Γ(z)
plane
• Repeat our earlier problem but now use the Smith Chart
R G =100 Ω
Z0 =100 Ω
VG=1 V
f=1 MHz
41
0.11
45
50
0.9
55
1.4
0.8
1.8
0.6 60
1.6
2.0
0.0
6
0.4
4
70
14
0
(+
jX
/Z
0.0
5
0.4
3.0
0.088+0.125=0.213
9
15
0.28
0.3
1.0
5.0
20
10
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
0.27
REFLECTION COEFFICIENT IN DEG
REES
LE OF
ANG
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EES
ANG
0.8
0.6
0.2
0.1
0.4
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.2
50
50
20
0.4
-10
1.0
0.6
1.8
2.0
0
-65 .5
0.2
1.4
1.2
1.0
-70
0.14
-80
-4
0
0.15
0.35
5
-4
4
0.9
6
0.1
0.3
0
-5
3
-35
0.11
-100
-90
0.13
0.36
0.12
0.37
0.8
-60
-55
7
0.1
1.6
-30
0.3
0.6
2
0.3
0.7
1
0.4
0.39
0.38
0.7
0.6
1.6
5
4
3
4
0.5
0.4
5
6
0.3
7
8
0.2
9
10
0.1
12
1.4
3
14
0.05
1.2 1.1 1
2
20
1
15
TOWARD LOAD —>
10
7
5
1 1
30 ∞ 0
43
0.1
0.01
0 0
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 1
0.99
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
CENTER
1
1.1
ORIGIN
0.8
1.8
6
4
1.1
1.2
1.3 1.4
0.2
0.4
0.6
1.2
1.3
0.95
1.2
1.4
0.8
1.5
0.9
1.3
1.8
1.5
2
3
2
3
1.6 1.7 1.8 1.9 2
0.8
1.4
<— TOWARD GENERATOR
2
1
3
1.6
1
0.7
1.5
1.6
4
5
4
0.5
1.7
10
5
2.5
0.6
3
0.4
1.8
20
∞
10 15 ∞
6
4
0.3
0.2
1.9
10 ∞
5
0.1
0
2
TR
TR S.W RF S.W A
T
L
A
A
N
N . P . L . L TEN
SM
SM EA O O
.
.C
. C K SS [ SS C [dB
O
O (C dB O ]
EF
EF O ]
EF
F,
F, NS
F
E
P T.
or
P)
I
0.1
0.4
1
-110
0
.0
9
0.4
2
0.0
-1
8
2
0
CAP
0.4
AC
ITI
3
VE
0.0
RE
7
AC
-1
TA
30
NC
E
C
OM
PO
N
EN
T
(-j
0.4
8
0.1
0
-5
-25
0.3
0.9
2
2
8
• Using the chart in this problem we get Zn (−λ/8) = 2 + j so
Z(−l) = 100(2 + j), exactly the same as before! But using the
chart required almost no calculations!
0.1
9
1
2.5
7. Read off new value of Zn (−l) at this point. Multiply by Z0 to
get Z(−l) = Z0 Zn (−l)
0
-60
-20
3.0
4
0.0
0
-15 -80
0.8
5
0.0
4.0
9
-4
R BS B] , P r I
SW d S [d EFF , E o
S
O CO EFF
.L .
N FL C O
RT R FL.
R
0
3
10
6. Since we know |Γ(z)| remains constant, the distance from the
center of the chart to the new point should be the same as the
old point. Measure and label the location of the new point on
the chart.
0.3
0.4
0.2
5
0.4
0.2
0.4
1.0
-15
0.47
5.0
0.22
0.2
1
-30
0.3
0.28
o)
jB/Y
E (NC
TA
EP
4
SC
-75
0.4
SU
40
6
-1
VE
TI
0.0
C
DU
IN
R
-70
O
),
Zo
X/
0.8
-20
-85
10
0.6
0.2
RADIALLY SCALED PARAMETERS
1
0.22
IND
UCT
IVE
1
0.2
30
4.0
0.2
RE
AC
TA
75
NC
EC
OM
PO
N
EN
T
0.4
5
0.2
0.0
4
0.4
0.3
0.4
6
150
0.1
8
1.0
4
5. Draw a line from the center of the chart to this new location on
the “wavelengths toward generator” scale
2
40
80
0.3
50
25
0.8
5
4. Add l/λ (here 0.125) to this number, and find that number of
the “wavelengths toward generator” scale. We rotated
clockwise like we know we are supposed to in the Gamma
plane. Here we are finding 0.088 + 0.125 = 0.213
0.1
7
30
20
15
3. Draw a line from the center of the chart through this point out
to the scales on the outer edge of the chart. Read the number
off the “wavelengths toward generator” scale. Here it is 0.088
0.3
3
60
0.2
0.6
10
2. Plot ZnL on the chart: find circle centered on real axis labeled
real part 1/2, circles centered off chart above real axis labeled
imaginary part 1/2. See next page for plot on the chart
1
85
0.3
4
35
0.2
20
= 1/2 + j1/2 in this problem
0.1
6
70
)
/Yo
(+jB
0.1
6
0.35
40
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
20
50+j50
100
0.3
0.0 —> WAVELE
0.49
NGTH
S TOW
ARD
0.48
0.0
—
0.49
GEN
D LOAD <
ERA
OWAR
0.48
± 180
HS T
TO
170
NGT
R—
-170
ELE
V
0.47
>
WA
160
<—
-90
90
-160
0.36
80
20
∞ 40 30
=
0.15
10
∞ 100 40
ZL
Z0
9
R
,O
o)
1. Calculate ZnL =
0.1
E
NC
TA
EP
SC
SU
VE
TI
CI
PA
A
C
Steps for using chart to find impedance on a terminated line:
0.14
90
65
0.5
3
0.4
0
13
120
0.37
0.7
2
0.4
7
0.0
110
1
0.4
0.13
0.38
0.39
100
0.4
9
0.0
but
42
0.12
1.2
0.1
1.0
0.088
ZL −Z0
ZL +Z0
z=0
z=−λ/8
Zin
8
0.0
50+j50Ω =ZL
To get into the Gamma plane, we could calculate Γ =
we don’t have to because the chart does it for us!
44
Things to notice about the Smith Chart
More things to notice...
1. Notice the scales inside the “wavelengths toward generator”
scale. One is just labeled in the opposite direction so it is
“wavelengths toward load”, while the innermost scale is a label
of degrees in the Gamma plane, i.e. “angle of reflection
coefficient Γ(z)”
4. Thus if we know |Γ(z)| on a line (which you can measure just
by measuring the distance from your impedance point to the
center of the chart) you can just as easily read the VSWR by
reading off the VSWR scale, note it is labeled from the center
outward. In our example problem, we find |Γ(z)| = 0.44 and
VSWR=2.6 just by using these scales
2. Notice the scales in wavelengths only range between 0 and 0.5,
this is because impedances on lines repeat every 0.5
wavelengths. If you add the length of the line to your original
number on the scale, then can’t find the result, remember that
the scale “wrapped around” at 0.5 so you need to keep rotating
by the remainder.
5. Notice that purely real Zn (z) is on the real axis of the Gamma
plane, purely imaginary Zn (z) is on the |Γ(z)| = 1 outermost
circle
3. Notice the “radially scaled parameters” on the side of the chart.
These all relate to |Γ(z)| since that is our radius in the Gamma
plane. The two of most interest to us are the “Refl-vol” scale,
which is a scale for the magnitude of the reflection coefficient
|Γ(z)|, and the “standing wave-vol ratio” which is the VSWR
7. Plotting the reciprocal of an impedance on the chart gives
Zn −1
n −1
Γ0 = 1/Z
1/Zn +1 = − Zn +1 = −Γ. Thus Γ becomes −Γ for the
reciprocal impedance, which means a reversal in Γ, i.e. a flip
through the center of the chart.
6. Plotting the conjugate of an impedance Zn ∗ (z) on the chart
∗
∗
L −Z0
flips a point over the real axis since Γ0 = Z
ZL ∗ +Z0 = Γ
45
46
And more things to notice...
EE 311 - Lecture 12
8. Thus a normalized admittance (= 1/Zn ) plotted on the chart is
just rotated 180◦ from the same normalized impedance
9. This fact makes it such that the chart can be used exactly the
same way if we work in terms of normalized admittance,
Yn = YYL0 = Z0 YL . This is sometimes easier if we working with
circuits or transmission lines connected in parallel, since
admittances add in parallel.
47
• Slotted line example
• Impedance matching networks
Assigned reading: Sec. 2.10 of Ulaby
48
A “slotted line” example
Example
• A “slotted line” is a transmission line with a slot cut into it to
allow measurements of voltage and/or current as a function of
z along the line.
• Such a measurement is made for the following circuit, and the
resulting current amplitude plotted
R G =100 Ω
Current amplitude measured along line
Z0 =50 Ω
Air filled
VG
ZL
z=0
z=−L
Question 2: Find |Γ|.
Question 3: What is ZL ?
• Question 1: We know that voltages and current amplitudes,
and impedances on a lossless transmission line repeat
themselves every half wavelength. By examining the current
plot, the distance between two adjacent minima is 1 m. Thus
λ/2 = 1, so λ = 2 m. Since the line is air filled,
√
β = 2π/λ = ω µ0 ²0 and f = ω/(2π) = 2c = 150 MHz
• Question 2: We know that the VSWR on the line is the ratio of
the maximum to minimum voltage or current amplitude on the
line. Here the maximum is 3 mA while the minimum is 1 mA,
1+|Γ|
giving VSWR=3. Since VSWR= 1−|Γ|
, we can solve to find
4
3.5
3
|I(z)| (mA)
Question 1: What is the frequency of operation, f ?
2.5
|Γ| =
2
VSWR−1
VSWR+1
=
1
2
• We can also find |Γ| using the VSWR=3 distance on the
“Standing wave-vol” scale measured on the “Refl-vol” scale
1.5
1
0.5
0
−2
−1.75
−1.5
−1.25
−1
−0.75
z 49
(m)
−0.5
−0.25
0
Solution for Question 3
• We can use the Smith Chart to find ZL .
• First, we know ZnL must be on a circle in the Gamma plane
with radius |Γ(z)| = 0.5. We can find this distance either using
the “Refl-vol” scale or the VSWR=3 scale on the side of the
chart.
• We can locate one point on the circle by recognizing that there
is a current minimum 25 cm from the load, or 25 cm/ 2 m =
λ/8 from the load.
• Recall that a current minimum is at the same location as a
voltage maximum, which is on the positive real axis in the
Gamma plane
• Thus, 25 cm up the line from the load, we are on the positive
real axis. To get to ZnL we rotate λ/8 toward the load
(counter-clockwise) since we are going back to the load from
the current minimum
51
50
Finish example
• We can then read off the value of ZnL and un-normalize (i.e.
multiply by Z0 ) to get the load impedance ZL . See the chart
on the next page
• Final answer: ZL = 50(0.6 + j0.8) = 30 + j40 Ohms
• Notice here that finding the complex valued ZL requires us to
know the complex valued Γ. Although we can obtain |Γ| from
knowledge of the VSWR on the line, VSWR alone does not
provide the phase of Γ.
• The additional information regarding the location of a current
(or voltage) maximum or minimum is therefore necessary to
determine the phase of Γ which then provides ZL
• The slotted line procedure we have studied is a basic procedure
for determining the impedance of an unknown load.
52
45
1.4
1.2
1.0
50
0.9
55
0.8
1.8
2.0
65
0.5
0.0
6
0.4
4
70
14
0
(+
jX
/Z
0.0
5
0.4
0.4
20
3.0
0.6
1
0.2
9
0.2
30
0.3
0.8
4.0
15
1.0
5.0
20
0.2
IND
UCT
IVE
0.28
1.0
10
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
0.27
REFLECTION COEFFICIENT IN DEG
REES
LE OF
ANG
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EES
ANG
0.8
0.6
10
0.1
0.4
20
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.2
50
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
50
0.2
20
0.4
0.1
-10
1.0
0.1
0.4
1
-110
0.0
9
0
.4
2
0
.0
-12
8
0
CAP
0.4
AC
ITI
3
VE
0.0
RE
7
AC
-1
TA
30
NC
E
CO
M
PO
N
EN
T
(-j
0.4
2.0
1.8
1.6
1.4
1.2
1.0
5
-4
0.14
-80
-4
0
0.15
0.11
-100
-90
0.13
0.36
0.12
0.37
0.4
0.39
0.38
R BS B] , P r I
SW d S [d EFF , E o
S
O CO EFF
.L .
N FL CO
RT R FL.
R
0
10
1
0.9
5
20
0.8
2
0.7
4
3
15
0.6
2.5
2
1.8
1.6
8
6
5
4
10
3
4
0.5
0.4
5
6
0.3
7
8
0.2
9
10
0.1
12
1.4
3
14
0.05
1.2 1.1 1
2
1
20
TOWARD LOAD —>
10
7
5
15
1 1
30 ∞ 0
0.1
53
0.01
0 0
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 1
0.99
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
CENTER
1
1.1
• This is the goal of impedance matching networks
0.35
0.9
-70
4
0
-5
6
0.1
0.3
0.8
-35
-55
-60
RADIALLY SCALED PARAMETERS
20
∞ 40 30
• This would be nice because reflected waves carry power back to
the source, not to the load where we usually want power to go.
However, we can’t always get ZL = Z0 so we have to think of
other ways to get as much power into our loads as we can
4
1.1
1.2
1.3 1.4
0.2
0.4
0.6
1.2
1.3
0.95
1.2
1.4
0.8
1.5
1.8
1.5
2
3
2
3
1.6 1.7 1.8 1.9 2
0.9
1.3
<— TOWARD GENERATOR
2
1
3
1.6
1
0.8
1.4
0.7
1.5
1.6
4
5
4
0.5
1.7
10
5
2.5
0.6
3
0.4
1.8
20
∞
10 15 ∞
6
4
0.3
0.2
10 ∞
5
0.1
1.9
0
2
TR
TR S.W RF S.W A
T
L
A
A
N . P . L . L TEN
N
SM EA O O
SM
.
. C K SS [ SS C [dB
.C
O (C dB O ]
O
EF O ]
EF
EF
F, NS
F,
F
P T.
E
or
P)
I
0.6
0.2
-30
7
0.1
3
0.3
-60
2
0.3
0.7
0.3
1
-65
9
0.5
0.1
8
0.1
0
-5
-25
5
0.0
0.6
4
0.0
0
-15 -80
0.8
-20
3.0
6
0.4
1.0
4.0
9
0.3
-4
0
5
0.4
0.2
0.2
0.4
• As mentioned previously, ideally we’d like to terminate all our
transmission lines in their characteristic impedance so we never
get any reflected waves or impedance variations
• First let’s consider how we can design our systems to provide
maximum power transfer by looking back into circuit theory.
Remember Xmission line problems are usually worked by using
Xmission line theory to replace the lines with circuit
impedances, and then using circuit theory from there.
-15
0.47
5.0
0.22
0.2
1
-30
0.3
0.28
o)
jB/Y
E (NC
TA
EP
4
SC
-75
0.4
SU
40
6
-1
VE
TI
0.0
C
DU
IN
R
-70
O
),
Zo
X/
0.8
-20
-85
10
0.6
0.2
∞ 100 40
1
0.22
RE
AC
TA
75
NC
EC
OM
PO
N
EN
T
0.4
5
Impedance matching
0.3
2
50
25
0.2
0.0
4
0.2
0.3
0.4
6
150
0.1
8
30
40
80
0.1
7
0.3
3
60
1
85
0.3
4
35
0.3
0.0 —> WAVELE
0.49
NGTH
S TOW
ARD
0.48
0.0
—
0.49
GEN
D LOAD <
ERA
OWAR
0.48
± 180
HS T
TO
170
NGT
R—
-170
ELE
V
0.47
>
WA
160
<—
-90
90
-160
Yo)
jB/
E (+
NC
TA
EP
SC
SU
0.1
6
70
40
9
0.1
R
,O
o)
VE
TI
CI
PA
CA
0.35
80
1.6
120
0.6 60
2
0.4
7
0.0
0.15
0.36
90
0.7
8
0.0
3
0.4
0
13
110
1
0.4
0.14
0.37
0.38
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100
0.4
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0.12
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9
0.0
54
ORIGIN
Conjugate matching
Maximum power transfer
• Taking the derivatives gives
• Consider the generalized circuit below
• If ZS and VS are fixed, what value of ZL will result in
maximum power dissipation in ZL ?
(RS + RL )2 + (XL + XS )2 − 2(RL + RS )RL
((RL + RS )2 + (XL + XS )2 )
• Maximize this by setting
∂
∂RL
= 0 and
∂
∂XL
= 0.
RL + RS − 2RL = 0 , XL + XS = 0
• Thus, we find RL = RS and XL = −XS will provide maximum
power dissipation. In other words, ZL = ZS ∗
• This is called a “conjugate match” between a source and a load.
• We will usually try to design our impedance matching networks
to create a conjugate match
ZS
M
Zg
VS
ZL
~
Vg
+
Z0
M’
Transmission line
56
A
Matching
network
Zin
Generator
55
=0
which can be solved to find
• Find using calculus. For ZL = RL + jXL ,
1 2
Dissipated power = |I| RL
2
¸
·
RL
|VS |2
|VS |2
RL
=
=
2 |ZS + ZL |2
2
(RS + RL )2 + j(XS + XL )2
2
ZL
A’
Load
Impedance matching networks
• We know Xmission lines are very useful for turning one
impedance into another, so we can design impedance matching
networks using Xmission lines. Note we don’t necessarily have
to have ZS = Z0 but this will often be the case
• In general, we’d like to be able to change an arbitrary ZL into
some other impedance ZS ∗
• You can see that we’re going to need at least 2 degrees of
freedom in our network since we have to match both the real
and imaginary parts of the two impedances
• We also want to use a lossless network because we don’t want
to dissipate power in the matching network, only in the load.
• While it is possible to make impedance matching networks out
of lossless circuit elements (i.e. inductors and capacitors),
circuit elements are often hard to find at high frequencies
making Xmission line networks essential
57