Quarter wave transformer
• The quarter wave transformer is a useful matching network for
matching a purely real ZL to a purely real ZS or a complex ZL
to a purely real ZS
EE 311 - Lecture 13
• The topology when both ZS = RS and ZL = RL are purely
real is shown below
• Since we desire Zin = RS , we can just use the equation for
impedance on a λ/4 line, which is Zin = Z02 /RL .
√
• Thus, if we set RS = Z02 /RL , we can solve for Z0 = RS RL .
• Quarter wave transformer
• Example
• Thus choosing a λ/4 length line with this characteristic
impedance will provide the desired match
• Single stub tuner
• Example
RS
Assigned reading: Sec. 2.10 of Ulaby
VS
RL
Z0
Zin
z=0
z=−λ/4
2
1
Quarter wave transformer: complex ZL
• The topology when ZL is complex and ZS is real is below
Quarter wave transformer example
• In this case, we use an extra length of line l to convert ZL into
a purely real impedance ZB . Then we match ZB to ZS using
our previous quarter wave design
0
√
0
Z 2
• Since Zin = Z0B , we find Z0 = RS ZB
Design a quarter wave transformer to match a load impedance
100 + j100 Ohms to a source with a 25 Ohm resistance. Assume all
Xmission lines are air filled and that Z0 of the non-quarter wave
line is 100 Ohms. Specify all line lengths in meters, not
wavelengths, for frequency 2 GHz.
• We find ZB and l using the Smith Chart. Plot ZL /Z0 on the
chart, then rotate clockwise until the real axis is reached where
Zin is purely real
• Read off value of impedance here and unnormalize to find ZB .
The distance rotated to get the real axis is l
RS
Zin=RS
λ /4
Z0
ZB
All lines are air filled
Z0 =100 Ω
Z0’
VS
Z0’
VS
25 Ω
ZL
Zin=RS
λ /4
ZB
l
4
3
l
Z L=100+j100Ω
45
50
1.2
1.0
0.9
55
0.8
1.6
1.8
2.0
0.0
6
0.4
4
70
14
0
(+
jX
/Z
0.0
5
0.4
20
3.0
0.6
0.3
30
0.8
4.0
15
1.0
20
0.2
IND
UCT
IVE
0.28
5.0
10
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
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EES
ANG
0.8
0.23
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
10
0.8
-10
1.0
2.0
1.8
1.6
1.4
1.2
0.9
1.0
5
-4
0.12
0.37
0.41
0.1
0.11
-100
-90
0.13
0.36
6
0.0
0.6
0
-5
0.14
-80
-4
0
0.15
0.35
0.4
0.39
0.38
RADIALLY SCALED PARAMETERS
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
5
20
∞ 40 30
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
12
0.1
1.4
3
14
0.05
1.2 1.1 1
2
20
1
15
TOWARD LOAD —>
10
7
5
1 1
1.1
30 ∞ 0
0.01
0 0
0.1
6
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
CENTER
1
1.1
0.2
1.2
0.99
1.1
1.3
0.95
1.2
4
1.2
1.3 1.4
0.4
0.6
1.4
1
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
0.8
1.5
0.8
1.4
0.7
1.5
4
5
4
0.5
1.6
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
N . P . L . L TEN
SM
SM EA O O
.
.C
. C K SS [ SS C [dB
O (C dB O ]
O
EF O ]
EF
EF
F, NS
F,
F
P T.
E
or
P)
I
R
O
),
Zo
X/
0.7
-70
0.8
0.34
0.16
-55
-35
-60
-60
-75
-70
0.2
-30
7
0.1
3
0.3
4
0
-65 .5
1
0.4
0.3
2
0.3
40
-1
-110
0.09
0.4
2
0.0
-12
8
0
CAP
0.4
AC
ITI
3
VE
0.0
RE
7
AC
-1
T
A
30
NC
EC
OM
PO
N
EN
T
(-j
0.4
0.1
9
8
0.1
0
-5
-25
5
0.6
4
0.0
0
-15 -80
0.8
-20
3.0
0.0
4.0
6
0.4
1.0
-15
0.3
-4
0
5
0.4
5.0
0.2
9
0.2
0.4
0.2
1
-30
0.3
0.28
0.22
0.47
-20
o)
jB/Y
E (NC
TA
EP
SC
SU
VE
TI
C
DU
IN
0.6
0.2
∞ 100 40
1
0.22
1.0
1
0.2
9
0.2
RE
AC
TA
75
NC
EC
OM
PO
N
EN
T
0.4
5
25
0.4
0.3
150
0.1
8
2
0.2
80
0.3
50
40
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
0.47
>
AV
W
0.0
160
<—
4
-90
90
-160
0.4
85
-85
6
30
0.2
0.1
• Finally, since λ in an air filled line at 2 GHz is
c/(2 × 109 ) = 0.15 m, the quarter wave line is 3.75 cm long
while the 0.088 wavelength line is 1.32 cm
0.3
3
60
1
• Distance (using scales on chart) is 0.088 wavelengths,
ZB = 100(2.6) = 260 Ohms
p
0
• Next find Z0 = (260)(25) = 80.62 Ohms
0.1
7
35
0.3
• First design the line to convert ZL into ZB using the Smith
Chart. Plot ZnL = 100+j100
= 1 + j on the chart and rotate to
100
reach the real axis.
R
,O
o)
0.16
0.34
70
40
9
0.1
Work on design
)
/Yo
(+jB
CE
AN
PT
CE
US
ES
IV
T
CI
PA
CA
65
0.5
3
0.4
0
13
0.6 60
7
0.0
0
12
0.15
0.35
80
1.4
2
0.4
0.14
0.36
90
0.7
8
0.0
110
0.37
0.38
0.39
100
0.4
0.13
0.12
0.11
0.1
0.09
0.41
ORIGIN
Final network:
Single stub matching network
• Resulting design is below. Note that a quarter wave
transformer can require us to have some funny values for line
characteristic impedances
• A matching network that avoids funny Z0 values is the single
stub tuner, used with a real source impedance Z0 and an
arbitrary load impedance
• Xmission line manufacturers typically only produce a limited
set of Z0 values, for example 50, 75, or 100 Ohms, so using
quarter wave transformers with manufactured lines can be
problematic
• The topology is the network is below. A piece of transmission
line of length d is short circuited and connected in parallel with
another line. The short circuited line is called the “stub”
• In other situations, for example with microstrip lines, the
circuit designer can fabricate any particular desired Z0 , so
quarter wave transformers are not problematic
25 Ω
VS
• Since we have things connected in parallel here it is easier to
work in terms of admittance than impedance. We can use the
Smith Chart in exactly the same way for admittance.
All lines are air filled
d
Z0 =100 Ω
Z0’ =80.62Ω
Z0
100+j100 Ω
Z0
3.75 cm
1.32 cm
VS
Z0
8
7
Zin
l
ZL
Single stub design continued
Single stub network design
• For a short circuit line of length d, Z(−d) = jZ0 tan(βd). Thus
the input admittance looking into the short is
Y (−d) = 1/Z(−d) = −j
Z0 cot(βd), which is purely imaginary
• We want our matching network to transform admittance
YL = 1/ZL into admittance Y0 = Z10 . Since Y0 normalized on
the Smith Chart is Y0 /Y0 = 1, we want to move YL to the
center of the chart
• Thus, at z = −l, without the stub, our unnormalized
admittance would be Y0 (1 + jBn ) = Y0 + jY0 Bn
• We now design the stub to cancel the reactive part of this
admittance, i.e. we set
Y0 Bn = Y0 cot(βd)
and solve for the length d
• However, since |Γ| 6= 0, we can never reach the center of the
chart just by moving down a single transmission line.
• Notice there are multiple possible values for the lengths l and
d. Typically we try to choose the shortest ones to minimize the
size of our networks
• We can however choose the line length l so that we match the
real part of the normalized admittance. In other words, choose
the first line length l to reach the point Yn (z) = 1 + jBn
• Single stub tuners are usually constructed as part of a
particular high frequency circuit
• This translates into rotating on the Smith Chart until we
intersect the Gn = 1 circle on the chart. The length l is the
length we had to rotate to reach this circle.
• However, it is possible to create a sliding stub with a variable
length device that can be inserted into any circuit then
adjusted to match to any given load
10
9
Work on design
• To design the network we need to specify l and d
Single stub design example
• Using admittance, YL =
Design a single stub network to match a load impedance 250 + j250
Ohms to a source impedance Z0 = 100 Ohms.
YnL =
YL
Y0
1
250+j250
= 100(YL ) = 0.2 − j0.2
= 2 − j2 mMhos, so
• Plot this on the chart and rotate until intersecting the G n = 1
circle. Distance rotated is 0.216 wavelengths from scales on
chart = l
• Admittance here is ≈ 1 + j1.8, so unnormalized = Y0 + j1.8Y0
d
• Design stub to cancel imaginary part: −j1.8Y0 = −jY0 cot(βd),
solve to find cot(βd) = 1.8, βd = 0.5071, so
d = 0.5071(λ/(2π)) = 0.081 wavelengths
100Ω
100 Ω
VS
100Ω
Zin
250 + j250Ω
See chart on next page for details.
l
12
11
0.2
0.0
0.2
0.6
0.1
2
3
0.3
0.7
0.5
5
4
0.4
0.4
0.6
4
0.5
0.5
6
3
7
0.2
8
2.5
8
0.6
0.4
10
0.7
0.3
0.1
5
1.8
14
0.8
0.2
0.05
12
4
1.6
0.6
0.8
0.8
3
0.9
0.1
0.01
1
1.1
1.1
0.99
0.1
1.2
1.2
0.95
0.2
0 1
0 0
13
30 ∞ 0
CENTER
1
1.3
1.4
1.3
0.9
0.4
1.2
TOWARD LOAD —>
10
7
5
1.1
15
1 1
1.2 1.1 1
20
2
RADIALLY SCALED PARAMETERS
1.4
1.0
35
70
4
0.8
1.4
1.5
0.6
1.3 1.4
0.16
0.34
1.6
3
1.5
1.8
0.1
7
0.3
3
2
0.8
1.5
0.7
2
1.6
1.6 1.7 1.8 1.9 2
1
30
60
25
0.1
0.3
2
50
8
20
3.0
15
4.0
5.0
15
0.6
1.7
0.5
2.5
3
3
3
0.3
5
5
1.8
0.4
4
4
6
1.9
0.2
4
10
<— TOWARD GENERATOR
2
1
Assigned reading: Sec. 2.11-2.11.1 of Ulaby
• Example #2
• Example #1
9
6
2
0.4
0.6
1.8
EE 311 - Lecture 14
0.3
5
10
• Transients on lines
ORIGIN
0.7
0.8
1
0.9
0.8
0.9
0
1
1
15
0.2
0.4
0.15
0.35
1
10
0.4
0.2
40
10
50
20
10
5
∞
0.1
2
0
10 ∞
10 15 ∞
20
0.3
20
0.6
0.14
9
20
0.7
0.36
80
0.2
9
∞ 40 30
50
0.9
0.8
0.13
0.37
0.2
∞ 100 40
0.3
Yo)
jB/
E (+
NC
TA
EP
SC
SU
110
0.9
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
R
,O
o)
120
VE
TI
CI
PA
CA
2
0.4
0.3
3
0.4
0
13
0.2
8
0.0
0.4
45
1.2
1.4
65
0.5
0
-65 .5
1.4
2.0
90
] F
dB F
. [ OE
EN S C ] . P)
TT S B T
A . LO [d NS
SS O P
S.W LO (C F,
K
.
L A EF
I
RF . PE CO
.
or
E
M
S.W NS
F,
A
EF
O
TR
.C
SM
N
A
TR
0.28
SW
RT
N
R
.
d
R L
RF FL OSS BS
L. . C [d
C OE B]
O
EF FF
F, , P
E
or
I
0.1
4
0.0
0
-15
0.38
1.6
1.6
0.12
5.0
7
0.0
0.0 —> W
A
V
E
L
E
0.49
N
GTHS
TOW
A
0.48
R
0.0
—
<
D
D
0.49
GEN
RD LOA
ERA
TOWA
0.48
± 180
TO
THS
0.47
170
R—
-170
ENG
VEL
0.47
>
WA
0.0
160
<—
4
-90
90
-160
0.4
85
-85
6
0.0
150
5
IND
80
UCT
0.4
IVE
5
R
E
AC
TA
0.0
0.1
75
N
6
14
C
EC
0.4
0
OM
4
PO
N
EN
70
T
(+
jX
/Z
o)
-80
jB/Y
E (NC
TA
EP
SC
SU
E
IV
CT
DU
IN
R
O
),
Zo
X/
-70
40
-1
6
20
55
-55
0.1
6
0.4
0.6 60
0.6
0.41
5
0.0
0.7
0.5
0.7
0.4
0.1
-110
0.09
0.4
2
0
.0
-12
8
0
CAP
0.4
AC
ITI
3
VE
0.0
RE
7
AC
-1
T
A
30
NC
EC
OM
PO
N
EN
T
(-j
0.2
5
0.4
0.8
0.8
0.39
0.11
-100
0.3
-75
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
0.2
0.39
100
0.35
0.15
-70
0.4
0.11
3.0
0.34
0.16
-35
1.4
1.8
-60
-30
1.8
2.0
4.0
7
0.1
0.6
0.2
0.4
0.3
3
0.3
0.8
0.4
0.1
2
0.3
8
0.1
0
-5
-25
1.0
0.6
0.4
0.0
-20
3.0
0.1
4
0.4
4.0
-4
0
-60
0.2
1
-30
0
-5
0.22
1.2
-20
-15
5.0
40
10
0.3
1
-10
9
0.8
0.09
0.41
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
20
0.1
9
0.2
30
20
50
0.3
10
0.2
0.28
50
0.22
2.0
1
0.2
14
100Ω
0.216 λ
100Ω
250 + j250Ω
16
• We know the basic effects already: time delays between end of
line, reflections of an incoming wave from the load.
• For the next two lectures, we will work only in the time
domain; no more sinusoidal signals, phase, j, or complex
values; lossless lines only
• Pulses on transmission lines model wires in digital circuits;
pulses are the signal type in digital circuits
• We will consider a simple unit step function excitation of a line
and see what happens; this can be extended to pulses on lines
by using two step functions to model a pulse
• For non-dispersive transmission lines, all frequencies propagate
with the same phase velocity. This means we can talk about
time domain signals as well, which are the sum of many
frequency components
Transients on transmission lines
VS
100 Ω
0.081 λ
Final design
Time domain telegrapher’s equations
• The time domain Telegrapher’s equations are:
∂
∂
v(z, t) = −L0 i(z, t)
∂z
∂t
• Take
∂
∂z
Time domain waves
∂
∂
i(z, t) = −C 0 v(z, t)
∂z
∂t
of the first equation, use
∂
∂t
of the second equation:
2
∂2
0 0 ∂
v(z,
t)
=
L
C
v(z, t)
∂z 2
∂t2
which is a time domain wave equation
• The result for time domain voltages and currents in similar to
that in the sinusoidal steady state
• The voltage is still the sum of waves propagating in the plus z
(V0+ ) and minus z directions (V0− ), only now they don’t have
to be sinusoidal; both propagate at the phase velocity u p
• Solutions for v(z, t) and i(z, t) are
v(z, t) = V0+ (t − z/up ) + V0− (t + z/up )
¤
1 £ +
V0 (t − z/up ) − V0− (t + z/up )
Z0
where V0+ (t − z/up ) and V0− (t + z/up ) are arbitrary functions
√
√
of their arguments and up = 1/ L0 C 0 = 1/ µ²
i(z, t) =
• The current i(z, t) is the difference of the plus and minus going
waves divided by Z0
p
• Z0 is still the characteristic impedance of the line L0 /C 0
• Note above that the parenthesis indicate functional
dependencies, not multiplications!
18
17
Solution of first example
First example:
• Clearly in this diagram we aren’t considering the sinusoidal
steady state because there is no frequency or wavelength
mentioned and because things happen at a particular point in
time (i.e. t = 0)
• If the line is initially discharged, i.e. v(z, t) = 0 for t < 0, we
can see that the boundary condition of continuous voltage at
z = 0 will cause 1 volt to appear across the line at t = 0:
v(0, t) = u(t)
where u(t) is the unit step function (1 for t > 0, 0 otherwise)
t=0
.....
VG=1 V
+
Z0 =100 Ω
air filled
−
.....
z=0
• Since a source at the left end of the line cannot generate a
negative traveling wave, the voltage source initially creates only
a positive traveling wave (V0− = 0):
v(0, t) = V0+ (t) = u(t)
• For other values of z, we know v(z, t) = V0+ (t − z/up ), so this
step function will propagate to the right with velocity u p = c
since the line is air filled
• In a time ∆t the edge of the pulse will propagate a distance
c∆t, and we can make sketches of the voltage on the line as a
function of space at different times to illustrate this
Voltage on line (Volts)
Consider an infinitely long line excited by a voltage step
2
2
2
1
1
0
0
1
0
ct2
ct
−1
Distance along line
−1
1
Distance along line
20
19
t=t2
t=t1
t=0
−1
Distance along line
First example continued
• We can see the leading edge of the voltage step traveling to the
right on the line with velocity c
• We also know
i(z, t) =
¤
1 £ +
V (t − z/up ) − V0− (t + z/up )
Z0 0
which here since V0− = 0 is just
1 +
V (t − z/up )
i(z, t) =
Z0 0
Thus plots of current on the line in this example are exactly
the same as plot of the voltage except divided by Z0
• This example is simple because the line is infinitely long, so the
plus going wave propagates forever and never generates a
minus going wave
Second example:
Consider again an infinitely long line but in this case excited by a
unit step source with source resistance RG
• In this case, the boundary condition at z = 0 becomes:
1 − RG i(0, t) = v(0, t)
since the current i(0, t) must be continuous between the
Xmission line and the circuit elements
RG
VG=1 V
t=0
.....
+
Z0 =100 Ω
air filled
−
z=0
• However if we have a terminated line, we should expect that we
can have reflections off the termination
22
21
Second example continued
• Again since there will be no V0− in this problem, we can find
µ + ¶
V0 (t)
= V0+ (t)
1 − RG
Z0
EE 311 - Lecture 15
which can be solved to find
V0+ (t) = (1)
Z0
Z0 + RG
• This looks like a voltage divider betwen Z0 and RG
• Plots of voltage and current on the line in this case are the
0
same as in example 1 but reduced by Z0Z
+RG
• Transients on terminated lines
• Circuit theory limit
• An example
Assigned reading: Sec. 2.11.2 of Ulaby
• Thus a transmission line initially looks like a resistance Z 0
• This is always true for an infinitely long line or for a matched
load ZL = Z0 . It is only true for finite times on a terminated
line as well shall see in the next example...
24
23
.....
Terminated lines in transients problems
Consider a terminated line excited by a step function
• If RL 6= Z0 we will have to generate a
v(0, t)/i(0, t) = RL
V0−
at the load to satisfy
• We’ve already seen that initially the transmission line looks like
a resistance Z0 . This is only true until enough time has passed
for a signal to propagate from the source to the load
• Thus for times between t = 0 and t = l/c (the amount of time
it takes to reach the load) the voltage on the line is just the
0
same as in Example 2, i.e. a 1 V voltage step times RGZ+Z
0
propagating to the right with velocity c
Terminated line continued
• However, when we reach the load, our v(z, t) and i(z, t) will not
have the correct ratio to match RL (unless RL = Z0 ) so we
have to generate a reflected wave
• At z = 0 (now the location of the load) we have
v(0, t)/i(0, t) = Z0
where Γ =
V0+ (t) + V0− (t)
V + (t)
= Z0 0+
+
−
V0 (t) − V0 (t)
V0 (t)
¶
• Thus we get the same equation as in the steady state case:
t=0
1+Γ
RL − Z 0
, Γ=
1−Γ
RL + Z 0
when solved
+
Z0 =100 Ω
air filled
−
RL
• The minus traveling wave generated at the load will propagate
back toward the source end of the line.
z=0
z=−l
26
25
Terminated line continued
Terminated line at longer times
• For l/c < t < 2l/c , voltages and currents on the line are below
• The arrows in the plots above indicate a minus traveling wave
with velocity c, i.e. the leading edge of the − wave will be at
z = 0 at time l/c and at z = −c∆t at time t = l/c + ∆t
¡
¢
• Note that the current involves the difference Z10 V0+ − V0−
+
−
V0
2
V0
2
t=l/c+∆
t=l/c+∆
c∆
Volts
1
Volts
1
Γ
0
0
−1
−1
Distance along line
Total Current
20
t=l/c+∆
1
c∆
0
−1
Current on line (mA)
1+Γ
27
• It can be shown that the reflection coefficient at the source is
G −Z0
given by ΓS = R
RG +Z0 , where RG is the Thevanin resistance
seen looking into the source circuit
t=l/c+∆
10
(1−Γ)/Z
0
0
−10
Distance along line
• The minus traveling wave generated at the load now travels to
the left until it encounters the source at t = 2l/c. If the
impedance of the source RG is not equal to Z0 , yet another
plus going wave will be generated which adds to our original
0
.
plus going wave RGZ+Z
0
• Note that a voltage or current source in the source circuit is
handled the same way as usual: voltage sources become shorts
and current sources become opens.
Distance along line
Total voltage
2
Volts
VG=1 V
1+Γ
1−Γ
V0− (t)
V0+ (t)
RL = Z0
RG
µ
c∆
Distance along line
28
Circuit theory limit
New plots of voltages and currents
• Voltages and currents after source end reflection are below
• The new plus traveling wave generated will propagate to the
load where it will reflect and generate a new minus traveling
wave which adds to the original minus traveling wave, and so
on ad infinitum
V+
−
V0
0
2
2
1+(1) Γ
S
Volts
Volts
0
Γ
0
t=2 l/c+∆
t=2 l/c+∆
−1
Distance along line
Distance along line
Total voltage
Total Current
20
Volts
1+ (1) Γ + (1) Γ
S
Γ
0
t=2 l/c+∆
Current on line (mA)
2
1+Γ
−1
• If we think about these pulse problems, we know that as time
gets longer and longer we should see longer and longer time
scale phenomena (i.e. lower and lower frequencies) until we
reach DC at time infinity
• Since wavelengths become very long at lower frequencies, the
transmission line will look short compared to λ no matter how
long it is. A short transmission line compared to λ acts like a
wire!
1
c∆
1
• We know that transmission lines only act like transmission
lines when their length is comparable to or larger than a
wavelength in the sinusoidal steady state
Γ
1
−1
• We can ask ourselves if this process ever reaches a steady state.
10
(1+(1) ΓS Γ − (1) Γ)/Z0
0
(1−(1) Γ)/Z0
t=2 l/c+∆
−10
Distance along line
• Thus for very long times in transient problems, we can replace
our transmission lines with wires and use circuit theory to find
the limits as time approaches infinity
Distance along line
30
29
A transient example problem
For the circuit shown below, sketch and dimension v(z, t) and i(z, t)
at
1. t = 2.4 × 10
−8
sec
2. t = 7.2 × 10
−8
sec
• The first thing to notice is that the phase velocity on the line is
√
up = 1/ µ0 2²0 here = 2.12 × 108 m/s
• Thus for the first part, the initial pulse generated by the step
function current at the source will have propagated
up ∆t = 5.088 m down the line (not reaching the load yet.)
3. t = 12 × 10−8 sec
4. What is vL as t → ∞?
• In the second part, the pulse will have propagated 15.26 m
which is 10 m down to the load and 5.26 m back up
t=0
Z0=50 Ω
ε=2ε0 µ=µ0
1A
Initial work on solution
vL
100Ω
• In the third part, the pulse will have propagated 25.44 m which
is 10 m down, 10 m back, and 5.44 m back down the line
z=0
z=−10 m
32
31
Solution for t = 7.2 × 10−8 sec
Solution for t = 2.4 × 10−8 sec
• Initially there is only a forward going wave that has propagated
5.088 m. The current source says that the current amplitude of
the pulse is 1 Amp, so the voltage is 50 volts since
I+ = V0+ /Z0 . Sketching these:
V− at t=24 nsec
60
5.088 m
20
20
0
−5
0
z (m)
Total voltage at t=24 nsec
−10
100
2
80
1.5
Amps
60
40
20
80
1
5.26 m
40
20
50/3
0
−5
0
z (m)
Total voltage at t=72 nsec
−10
100
0.5
−0.5
−10
0
60
40
−10
−5
0
z (m)
Total current at t=72 nsec
2
1.5
50(4/3)
60
40
1
0.5
20
0
−5
z (m)
80
60
0
−5
0
z (m)
Total current at t=24 nsec
0
−10
80
20
Volts
0
−10
Volts
40
0
100
Volts
60
V− at t=72 nsec
100
Amps
80
= 13 , so the voltages and
0
Volts
80
100−50
100+50
V+ at t=72 nsec
0
100
Volts
Volts
0
100
40
• The reflection coefficient is Γ =
currents look like:
2/3
0
0
−5
z (m)
0
−10
−5
z (m)
33
V+ at t=120 nsec
V− at t=120 nsec
0
0
100
Volts
80
100
80
50(4/3)
60
40
20
5.44 m
−10
40
50/3
0
−5
0
z (m)
Total voltage at t=120 nsec
−10
100
80
60
20
0
Volts
• Thus a new V0+ wave is generated which adds to the initial V0+
wave we had. The amplitude of the new V0+ wave is
ΓS Γ(50) = 50
3 volts
0
• Sketches of the voltages and currents are:
• In this case, we have propagated 25.44 m = 10 m down, 10 m
back, and 5.44 m back down.
• Thus, our V0− wave is going to have encountered the source,
where yet another reflection occurs. Note the Thevanin
equivalent of a current source is an open circuit, so RG = ∞
G −Z0
and ΓS = R
RG +Z0 = 1.
−5
z (m)
Plots for t = 12 × 10−8 sec
Volts
Solution for t = 12 × 10−8 sec
−0.5
−10
0 34
−5
0
z (m)
Total current at t=120 nsec
2
1.5
50(5/3)
60
Amps
V+ at t=24 nsec
• In this case, we have propagated 15.26 m = 10 m down and
5.26 m back up. When the initial V0+ wave encountered the
load, a reflected wave was generated because RL = 100 6= Z0 .
50(4/3)
40
20
1
2/3
0.5
0
0
−10
−5
z (m)
−0.5
−10
0
36
35
−5
z (m)
0
Solution as t → ∞
• This bouncing around of waves repeats itself forever, i.e. the
new 50/3 V0+ wave goes to the load where it generates a new
V0− wave of 50/3(1/3) = 50/9 volts that adds to the previous
V0− wave
• However, we know that as t → ∞, the transmission line will act
like a wire and we can use circuit theory
• Clearly we should get 100 volts across the load as t → ∞
• Our last plot showed a voltage of 50(5/3), this will approach
100 volts in ever smaller steps as t → ∞
1A
vL
EE 311 - Lecture 16
1. Scalars and vectors
2. Vector algebra
3. Scalar product
4. Cross product
5. Orthogonal coordinate systems
100Ω
38
37
Scalars and vectors
Introduction
• We’ve now finished our study of transmission lines, both in
time and frequency domains
• Hopefully we now have a better appreciation of the “wire”
element in circuit theory
• We now want to work on a better understanding of the
capacitor, inductor, and resistor elements
• To do this, we need the theories of electro- and magnetostatics; these involve electric and magnetic fields
• These “fields” are 3-D vector functions of both space and time;
need to review basic vector operations to get ready to work
with fields
A scalar is a quantity described by a number only, which may be a
function of space
Example: Temperature in a room T (x, y, z) in degrees Celsius,
described only by a number, but a function of space
Our Xmission line voltages and currents were scalar also, functions
of one space dimension and time
A vector is a quantity that has both magnitude and direction
Example: The force exerted by the wind F (x, y, z) has both a
magnitude (in Newtons) and direction, both of which can be
functions of space
Notation: Here an overbar denotes a vector (F ), while scalars have
no overbar. Ulaby uses bold face type for vectors
40
39
Specifying a vector
• To specify a vector we need to specify both its magnitude and
direction
• Since we live in a 3d world, in general we need 3 numbers to
specify a vector
• These numbers depend on our choice of a “coordinate system”.
The same vector could have different sets of 3 numbers
depending on how the numbers are referenced
• However, the laws of physics should not depend on how these
numbers are chosen (i.e. on the choice of a particular
coordinate system). Like calling the same person by two
different names: still the same person!
Magnitude and direction of a vector
We can separate a vector into its magnitude and direction as
A = âA
where â is a “unit vector” in the direction of A, which has
magnitude one and no units by definition
Here A is the magnitude of vector A, also written as |A| which is a
scalar having the same units as A but no direction
Note to get a unit vector in the direction of a vector take
â =
• We should get the same physical answer for a problem
regardless of the choice of a coordinate system; sometimes one
choice may be more convenient than others though
A
|A|
42
41
Vector algebra
• Vector algebra involves the fundamental operations we can
perform on vectors. Possibilities are addition, subtraction,
“multiplication”, and “division”
• We can figure all of these out graphically without resorting to
any specific coordinate system.
• To do this, represent vector A as a straight line segment of
length |A| and in direction â as below
• Two vectors are called “equal” only if they have both the same
magnitude and the same direction
Adding or subtracting two vectors
• Vectors can be added graphically using the “head-to-tail” rule:
1. Draw vector A
2. Draw vector B placing the “tail” of B at the “head” of A
3. The vector from the tail of A to the head of B is the sum
C =A+B
• It is easy to see that A + B = B + A
• Also A + (B + C) = (A + B) + C
• To subtract two vectors, C = A − B, we just add
C = A + (−B), where −B is a vector with the same magnitude
as B but the opposite direction
A
C =A + B
B
A
A
44
43
Scalar product basics
Scalar product
Now that we’ve learned how to add and subtract vectors
graphically, what about multiplication? There a three types of
multiplication operations involving vectors:
[1] Multiplication of a vector by a scalar
[2] Scalar product of two vectors
[3] Cross product of two vectors
• [2], the scalar product of two vectors, is a little more involved
• The scalar product of two vectors results in a scalar number
• The resulting number is the product of three factors: the
magnitudes of each of the two vectors and the cosine of the
angle between the directions of the vectors
• This can be written mathematically as
[1] is simple: a vector multiplied by a scalar just produces a vector
with the same direction and its magnitude multiplied by the scalar.
We can write this as
¡
¢
kA = kâ|A| = â k|A|
where k is a scalar and note that the final answer has the same
direction as vector A but a scaled magnitude
A · B = |A||B| cos θAB
where θAB is the angle between the directions of vectors A and
B
• Note the notation for the scalar product (also called “dot”
product): A · B
46
45
Properties of scalar products
• Note that the dot product of two vectors can be positive or
negative depending on the angle between the two vectors
Properties of scalar products
We can make a table of information about the scalar product:
θAB
• Note also that the absolute value of the dot product is always
less than or equal to the product of the magnitudes of the two
vectors
• The dot product of two vectors can be interpreted as the
product of the magnitude of one vector times the projection of
the second vector onto the first one, as shown:
• Note the projection of B onto A has length |B| cos θAB
0
◦
0◦ < θAB < 90◦
90
B
cos θAB
|A||B|
vectors in same
than |A||B|
0
90 < θAB < 180
◦
vectors are
negative, greater
than −|A||B|
−|A||B|
48
47
direction (parallel)
perpendicular
◦
180◦
A
Comment
positive, less
◦
B
θAB
A·B
vectors are in
opposite directions
Cross products
• Another “multiplication” operation is [3], the cross product of
two vectors
Operations with scalar products
• Note if we dot a vector with itself we get the magnitude of the
vector squared:
A · A = |A|2
since cos θAA = 1
• We can show graphically that A · B = B · A
• The cross product of two vectors results in a third vector,
whose magnitude is |A||B| sin θAB
• The direction of the cross product is perpendicular to the plane
containing A and B in the sense determined by the
“right-hand” rule
• Notation: C = A × B indicates that vector C is the cross
product (also called “vector product”) of vectors A and B
• We can also show graphically that A · (B + C) = A · B + A · C
B
• Thus the scalar product is both commutative and distributive
C =A X B
A
50
49
Cross product explanation
• The magnitude of the cross product is similar to the scalar
product so that is clear
• For the direction, we need to realize:
1. Any two non-parallel vectors determine a unique plane
which contains both vectors
2. This plane then has a single direction perpendicular to it,
although it could point up or down
3. The up or down choice is made by the “right-hand rule”
• To use the “right-hand” rule, point your right hand fingers in
the direction of A, then curl your fingers toward B from A.
Your right hand thumb then points in the direction of the cross
product.
Properties of cross product
• Because of the right hand rule, the cross product is NOT
commutative: A × B = −B × A
• It is distributive: A × (B + C) = A × B + A × C but not
associative: A × (B × C) 6= (A × B) × C
• A few useful identities:
¡
¢
A· B×C
=
¡
¢
A× B×C
=
¡
¢
¡
¢
B· C ×A =C · A×B
¡
¢
¡
¢
B A·C −C A·B
The latter is the “back-cab” rule
• Note the only division operation possible with vectors is
dividing a vector by a scalar, identical to multiplying a vector
by the inverse of the scalar
52
51
Coordinate systems
Orthogonal coordinate systems
• So far we’ve just been doing vector operations graphically, but
this gets tedious after a while. Try to use numbers instead.
• We know in 3d space we need three numbers to describe a
vector; also we need three numbers to describe the location of a
point in space
• A coordinate system tells us how to choose three numbers to
describe the location of a point by locating a point at the
intersection of 3 surfaces
• The three surfaces chosen differ with the coordinate system
used
• If the normal vectors to the three surfaces used are orthogonal
to each other at every point, we have an “orthogonal
coordinate system” (OCS)
• We are most familiar with the Cartesian coordinate system,
where the three surfaces are planes parallel to the xy, yz, and
xz planes respectively. However we can choose any three
surfaces as long as their normals are perpendicular
• To specify the directions of vectors, we use three unit vectors,
one for each of the coordinates. Call these âu1 , âu2 , and âu3
• In an OCS, these 3 unit vectors are perpendicular to each other
(when defined at the same point) so that
âu1 · âu2 = âu1 · âu3 = âu2 · âu3 = 0
and are arranged so that
âu1 × âu2 = âu3
âu2 × âu3 = âu1
• We can represent any vector at a point in an OCS as the sum
of its components along the three OCS unit vectors:
A = âu1 Au1 + âu2 Au2 + âu3 Au3
54
53
• From our previous equations and the definitions of the scalar
and cross products we can show that
q
|A| =
A2u1 + A2u2 + A2u3
Au1
= âu1 · A
Au2 = âu2 · A
Au3 = âu3 · A
• Also if we have two vectors at the same point:
A = âu1 Au1 + âu2 Au2 + âu3 Au3 and
B = âu1 Bu1 + âu2 Bu2 + âu3 Bu3 we can show that
A·B
A×B
= Au1 Bu1 + Au2 Bu2 + Au3 Bu3
= âu1 (Au2 Bu3 − Au3 Bu2 ) + âu2 (Au3 Bu1 − Au1 Bu3 ) +
âu3 (Au1 Bu2 − Au2 Bu1 )
• The cross product can be written more compactly as a 3 × 3
determinant:
¯
¯
¯ â
¯
¯ u1 âu2 âu3 ¯
¯
¯
¯ Au1 Au2 Au3 ¯
¯
¯
¯
¯
¯ Bu1 Bu2 Bu3 ¯
55
âu3 × âu1 = âu2