Post on 02-Apr-2018
transcript
7/27/2019 Smith Chart - Impedance Matching
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Impedance Matching
& Smith Chart
Microwave EngineeringEE 172
Dr. Ray Kwok
7/27/2019 Smith Chart - Impedance Matching
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Impedance Matching - Dr. Ray Kwok
Technician tuningeven for 50Ω designs
output 50 Ω
50 ΩΩΩΩ
50 ΩΩΩΩ
50 ΩΩΩΩ
input 50 Ω
50 ΩΩΩΩ
Think of a block of cracked glass…
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Impedance Matching - Dr. Ray Kwok
Stub tuning ?
0385.0t7288.0t0
04.0t0123.0t0385.0t7692.0t
t
t2.004.0
t1923.01
7692.0tB
1923.0t)t1923.01(Bt9615.0
t
t2.004.0
t9615.0
t1923.00385.0B
9615.0Bt9615.0)t1923.01(
)1923.0t( j9615.0Bt9615.0)t1923.01(B jt9615.0 j)t1923.01(
t9615.0 j)t1923.01(
)1923.0t( j9615.0
t)1923.0 j9615.0( j1
jt1923.0 j9615.0B j1
tanY j1
tan jYY
1923.0 j9615.0Z1Y
2.0 j1Z
2
22
L
L1
L
L
L
−−=
+−−=−
−=
+
−=−
+=+−
−=
−=−
=+−
++=++−+−
+−
++=
++
++=−
β+
β+=
+==
−=
l
l
ZLShort piece of open stub → capacitor
e.g. ZL = 50 − j 10 Ω
Zin = 50Ω
50ΩC
Length = ?
Y1want Y1 = Yo - jB
ω=
===
−=−=−
λ=π
λ=
=β
β==
<
−=−==
=−=−
λ=π
λ=
=β
β==
0028.0C
0028.050 / 140.0YBB
140.02.0t / 04.0B
094.02
591.0
591.0
tan671.0t
0B
01.050 / 497.0YBB
497.02.0t / 04.0B
009.02
0573.0
0573.0
tan0574.0t
o
o
l
l
l
l
l
l
t = 0.0574 or 0.671
real
imaginary
need inductor !!!!
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Impedance Matching - Dr. Ray Kwok
Matching methods
Stub tuning doesn’t always work
Many other ways to match Lumped elements
Transmission lines
Stubs (open, short, series, shunt) Single Stub
Double Stubs
Quarter-wave transformer
Many combinations Can’t just blindly optimize
Need Smith Chart
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Impedance Matching - Dr. Ray Kwok
SmithChart
A chart of Γ
Γ = u + j v
u
v
relate to Z or Y
ρ = 1
max reflection circle
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Impedance Matching - Dr. Ray Kwok
Constant Resistance Circles
2
2
2
2
22
22
22
2222
22
22
1R
1v
1R
Ru
0v1R
1R
1R
R
1R
Ru
0v1R
1Ru
1R
R2u
0)1R(v)1R(Ru2)1R(u
vu1)vuu21(R
v)u1(
v2 j)vu1( jXR
jv)u1( jv)u1(
jvu1 jvu1
11Z
1Z
1Z
+
=+
+
−
=++
−+
+−
+−
=++
−+
+−
=++−+−+
−−=++−
+−
+−−=+
+−+−⋅
−−++=
Γ −Γ +=
+
−=Γ
real
R & X are normalized
circle for each R
If R = 0, u2 + v2 =1
u
v
R=0
If R = 1, (u - ½)2 + v2 = (½)2
R=1
Family of Constant
Resistance Circles
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Impedance Matching - Dr. Ray Kwok
Constant Reactance Circles
[ ]
222
22
22
22
22
X
1
X
1v)1u(
X
v2v)1u(
v2v)u1(X
v)u1(
v2 j)vu1( jXR
=
−+−
=+−
=+−
+−
+−−=+
imaginary
circle foreach X
If X = 1, (u-1)2
+ (v-1)2
=1
u
vX=1
If X = 2, (u - 1)2 + (v – ½)2 = (½)2
X=2
X=-1
X=-2ρ = 1 circle
Family of Constant Reactance Circles
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Impedance Matching - Dr. Ray Kwok
Constant VSWRCircles
u
v
ρ = 1
max reflection circle
Not shown in Smith Chart
0 < ρ < 1
ρ
θ
θ ρ je=Γ
Any point on Smith Chart
has a definite Z and Γ
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Impedance Matching - Dr. Ray Kwok
Locate Z
e.g. ZL = 50 + j 25 Ω
ALWAYSNORMALIZE
FIRST
50Ω ZL
5.0 j1ZL +=
R = 1circle
X = 0.5circle
ZL
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Impedance Matching - Dr. Ray Kwok
Read off ρρρρand VSWRρ
VSWR =1.6
ρρρρ = |Γ ΓΓ Γ | =0.24
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Impedance Matching - Dr. Ray Kwok
Phase of Γ ΓΓ Γ
76o
Γ = 0.24 (76o)
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Impedance Matching - Dr. Ray Kwok
Move alongTransmissionLinee.g. ZL = 50 + j 25 Ω
50Ω ZL
ZL
0.1λ
Zin
All 50 Ω, constant ρ
Starting phase
0.145λ
ending phase0.245λ
0.1λ along
the Constant
VSWR Circle
Zin(1.65, 0.1)
Zin = 50(1.65 + j 0.1) Ω
4o
Γ in = 0.24 (4o)
0.145 +0.1
= 0.245λ
clockwise
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Impedance Matching - Dr. Ray Kwok
Quarter-Wave
e.g. ZL = 50 + j 25 Ω
50Ω ZL
ZL
λλλλ/4
Zin
Starting phase
0.145λ
ending phase0.395λ
0.25λ along
the Constant
VSWR Circle
Zin(0.8,-0.4)
Zin = 50(0.8 - j 0.4) Ω
0.145 +0.25
= 0.395λ
clockwise
L
L
in
Lin
Lino
YZ1Z
ZZ1
ZZZ
==
=
=
YL = 0.02(0.8 – j 0.4) Ω-1
Convert Z↔ Y easily !!
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Impedance Matching - Dr. Ray Kwok
SeriesConnections
Z
Use Z
Add series L (clockwise)
Add series C (counter-clockwise)
ser-L
ser-C
Add series R (inward)
ser-R
xline
Add transmission line
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Impedance Matching - Dr. Ray Kwok
ParallelConnections
Y
Use Y
Add shunt C (clockwise)
Add shunt L (counter-clockwise)
shunt C
shunt L
Add shunt R (inward)
shunt R
xline
Add transmission line
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Impedance Matching - Dr. Ray Kwok
Move alongthe 3 circlese.g. ZL = 50 + j 25 Ω
ALWAYSNORMALIZE
FIRST
50Ω ZL
5.0 j1ZL +=
R = 1circle
X = 0.5circle
ZL
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Impedance Matching - Dr. Ray Kwok
Exercisee.g. ZL = 20 - j 25 Ω
f = 159 MHz (ω = 109) Find Zin, Γ in, VSWR.
5.0 j4.0ZL −=
ZL
50Ω, λ/8 ZL20Ω
10nH
20pF
YL
B = ωC/Yo = 0.02Zo= 1
Y1
0.192λ
0.317λ
+0.125λ
Y2
Z2
X = ωL/Zo = 10/Zo= 0.2
Z3
Y3
G = G/Yo = Zo /R= 50/20 = 2.5
Yin
Zin
(0.28,0.13)
Zin = 50(0.28+j0.13) = 14 + j7 ΩΩΩΩ
ρ=0.57
164o
Γ ΓΓ Γ in = 0.57 (164o)SWR=7
VSWR = 7 at Y1 junction !!
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Impedance Matching - Dr. Ray Kwok
5.0 j4.0ZL −=
Exercisee.g. ZL = 20 - j25 Ω
f = 159 MHz (ω = 109) Find Zin, Γ in, VSWR.
ZL
50Ω, λ/8 ZL20Ω
10nH
20pF
B = ωC/Yo = 0.02Zo= 1
0.443λ
0.068λ
+0.125λ
Z2
X = ωL/Zo = 10/Zo= 0.2
Z3
G = G/Yo = Zo /R= 50/20 = 2.5
Zin
(0.28,0.13)
Zin = 50(0.28+j0.13) = 14 + j7 ΩΩΩΩ
ρ=0.57
164o
Γ ΓΓ Γ in = 0.57 (164o)SWR=7
VSWR = 7 at Y1 junction !!
Z1
7/27/2019 Smith Chart - Impedance Matching
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Impedance Matching - Dr. Ray Kwok
LOAD
Z=ID=
20-i*25 OhmZ1
CAP
C=ID=
20 pFC1
TLIN
F0=EL=Z0=ID=
0.159 GHz45 Deg50 OhmTL1
IND
L=ID=
10 nHL1
RES
R=ID=
20 OhmR1
PORT
Z=P=
50 Ohm1
Microwave Office (AWR)
0 1 .
0
1 .
0
- 1
.
0
1 0
. 0
10.0
- 1 0 . 0
5 .
0
5. 0
- 5 . 0
2 .
0
2. 0
- 2 .
0
3 .
0
3. 0
- 3 .
0
4 .
0
4. 0
- 4 . 0
0 .
2
0 . 2
- 0. 2
0 .
4
0 .
4
- 0
. 4
0 .
6
0 .
6
- 0
. 6
0 .
8
0
. 8
- 0
. 8
Graph 1Swp Max
0.159GHz
Swp Min0.159GHz
Z[1,1]
Smith Chart Example
S[1,1]Smith Chart Example
Freq ReZ11 ImZ110.159 0.27375 0.13908
One-port impedance
Freq MagS11 AngS11
0.159 0.5771 162.93
Reflection coefficient
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Impedance Matching - Dr. Ray Kwok
Vmax & Vmin
ZL
100Ω, 1.5λ
80-j40ΩZg=100Ω
Vg=10V
θθθθ = 0
Vmax
θθθθ = ππππ
Vmin
0.394λλλλ
0.106λλλλ to Vmin
0.356λλλλ to Vmax
7/27/2019 Smith Chart - Impedance Matching
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Impedance Matching - Dr. Ray Kwok
To what we match?
( ) ( ) ( )22
g
2
g
2
g
2
g
g
2
g
ininlossLoad
gg
2
XR4
RV
2
1
X2R2
RV
2
1RI
2
1PP
+=
+=== −
( ) 2
g
2
go
o
2
g
o
2
total
g
oLineLoadXRZ
ZV
2
1Z
Z
V
2
1ZI
2
1PP
2
++====
Case 1: match ZL = Zo = real → Γ L = 0, VSWR = 1 on the line
Zin lZo ZL
Zg=Rg + jXgVg
Zin
Zg
Vg
Case 2: match Zin = Zg → Γ in = 0, VSWR > 1
g
2
g
2
g
g
2
g
LoadR
V
8
1
R4
RV
2
1P ===
Case 3: match Zin = Z*g → Xin = -Xg
= max power available
conjugate matching
Ideally, match all Zo = Zg = ZL = real, then all 3 PLoad are the same = Pmax.
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Impedance Matching - Dr. Ray Kwok
2-elementmatching
ZLmatchingnetwork
Zin = 50Ω
2 equations, 2 unknowns (Re, Im)
Single frequency matching
Bring ZL to center of chart→
Zo
Many choices
First element → to the “1” circlesR=1G=1
Choices:• lumped elements• transmission line (single, multiple)
• stubs (single, double)
• λ /4• multiple sections
ZL
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Impedance Matching - Dr. Ray Kwok
LumpedElements
R=1G=1
If ZL is inside the G=1 circle,first element cannot be shunt
ZL (0.4, 1)YL (0.34, -0.87)
Z (0.4, 0.48)Y (1, -1.24)
=0.4+j1ZL
jB = +j1.24 = jZoωC
jX = -j0.52 = -j/ ωCZo
Z (0.4, -0.48)Y (1, 1.24)
=0.4+j1ZL
jB = -j1.24 = -jZo / ωL
jX = -j1.48 = -j/ ωCZo
Z (1, 1.4)Y (0.34, -0.48)
jB = +j0.39 = jZoωC
=0.4+j1ZL
jX = -j1.4 = -j/ ωCZo
Z (1, -1.4)Y (0.34, 0.48)
jB = +j1.35 = jZoωC
=0.4+j1ZL
jX = j1.4 = jωL/Zo
If ZL is inside the R=1 circle,first element cannot be in series
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Impedance Matching - Dr. Ray Kwok
Using Stubs
ZL (0.4, 1)YL (0.34, -0.87)
Z (0.4, -0.48)Y (1, 1.24)
=0.4+j1ZL
short shunt stub jB = -j1.24 = -jcotβl
open series stub jX = -j1.48 = -jcotβl
Z (1, 1.4)Y (0.34, -0.48)
open shunt stub jB = +j0.39 = jtanβl
=0.4+j1ZL
open series stub
jX = -j1.4 = -jcotβl
l
l
l
l
β=β−=
β−=
β=
tan jYYcot jYY
cot jZZ
tan jZZ
oop
osh
oop
osh
One way, simply replace lumped elements with stubs
Zo can be
anything here
previous example
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Impedance Matching - Dr. Ray Kwok
ZL (0.4, 1)YL (0.34, -0.87)
=0.4+j1ZL
TransmissionLine Matching
usually requires 1 more element
previous example
0.130λλλλ
0.185λλλλ
0.055λ
50 Ω
jX = -j1.85 = -j/ ωCZo
=0.4+j1ZL
0.435λλλλ
0.305λ
50 Ω
Y (1, 1.9)
short shunt stub jB = -j1.9 = -jcotβl
Z (1, 1.85)
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Impedance Matching - Dr. Ray Kwok
Stub tuning ?
200.0B
03846.0B9615.0
00148.0B0074.0B9615.0B0074.003698.0
B9615.01923.0
0385.0
0385.0B1923.0
B1923.0t
1923.0t)t1923.01(Bt9615.0
B9615.01923.0
0385.0t
9615.0Bt9615.0)t1923.01(
)1923.0t( j9615.0Bt9615.0)t1923.01(B jt9615.0 j)t1923.01(
t9615.0 j)t1923.01(
)1923.0t( j9615.0
t)1923.0 j9615.0( j1
jt1923.0 j9615.0B j1
tanY j1
tan jYY
1923.0 j9615.0Z
1
Y
2.0 j1Z
2
2
L
L1
L
L
L
=
=
−=−+
−=
−
+=
+=−−
−=
=+−
++=+−−+−
+−
++=
++
++=−
β+
β+=
+==
−=
l
l
ZLShort piece of open stub→ shunt C
e.g. ZL = 50 − j 10 Ω
Zin = 50Ω
50ΩC
Length = ?
Y1want Y1 = Yo - jB
ω=
===
λ=
π≈β
=β=−
=
004.0C
004.050 / 200.0YBB4
2
6845tant
)200.0(9615.01923.0
0385.0t
o
l
l
l
real
imaginary
Quadratic equationspossible 2 solutions
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Impedance Matching - Dr. Ray Kwok
Single transmission line matching
( )
( )??R50Z
ZR50ZX
X50)R50(Zt
)ZtX(ZRt50
ZRXt50Z50
) jZt jXR(Z)Xt jRtZ(50
t jXR jZ
jZt jXR
Z50
tan jZZ
tan jZZZZ
2
Lc
cLcin
≠
−=−=
+=
=−
++=−+
++
++
=
β+
β+=
l
l
( )( )
)R50(X50R50
)R50(
X50R50Z
R50
X50ZR50
2
22
22
−>
−−=
−=−
X50
)R50(Z
tan
)R50(
X50R50Z
c
22
c
−
=β
−−=
l
match with just 1 lossless transmission line?
In previous exampleZL = 50(0.4+j1) = 20 + j50
ZLZCZin = 50 Ω
2
X)R50(R
50R
>−
<
R(50 − R) = 600 < X2
Cannot tune match with
just one line !!!
conditions(real)
(imaginary)
If ZL = 25 + j20 Ω,
ZL = 0.5 + j0.4 normalized to 50 Ω
λ=π
λ=
=−
=−
=β
=
−−=−−=
>−
<
0776.02
)488.0(
53.04.0
)5.01(424.0
X
)R1(Ztan
424.0Z
)5.01(4.05.0
)R1(XRZ
X)R1(R
1R
c
c
22
2c
2
l
l
0.5 < 10.25 > (0.4)2
conditions ok
Z = Zc
t = tanβl
ok…. that means Zc is real, and
transmission line is a λ /4 transformer.
solve for Z
Yes, change Zo & length
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Impedance Matching - Dr. Ray Kwok
In SmithChart ?
Last example
ZL = 25 + j20 Ω
Zo = 21.2 ΩLength = 0.078λ
Zin = 50 Ω
ZL = ZL /Zo = 1.19 + j0.94
Zin = 50/21.2 = 2.356
ZL (1.19, 0.94)
21.2 ΩΩΩΩ ZL
0.078λλλλ
Zin = 50Ω
0.172λ
0.250λZin(2.35, 0)
0.172 +0.078
= 0.250λ
clockwise
Matching doesn’t necessary
means center of the chart !!
Depends on Z of the line & system.
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Impedance Matching - Dr. Ray Kwok
Single Stub Tuningrefers to sliding a stub (any kind)
along a transmission line.
previous example
0.130λλλλ
0.064λλλλ
=0.4+j1ZL
0.435λλλλ
0.305λ
50 Ω
Y (1, 1.9)
short shunt stub
jB = -j1.9 = -jcotβl
In practice, usually shunt stubs,short stub for waveguides,
open stub for microstrip.
=0.4+j1ZL
0.434λ
50 Ω
open shunt stub
jB = j1.9 = jtanβl
ZL (0.4, 1)YL (0.34, -0.87)
Y (1, -1.9)
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Impedance Matching - Dr. Ray Kwok
Any Stubtanβl can be “+” or “-”
previous example
0.130λλλλ
0.064λλλλ
=0.4+j1ZL
0.435λλλλ
0.305λ
50 Ω
Y (1, 1.9)
open shunt stub jB = - j1.9 = jtanβlβl = -1.086 + π = 2.055length = 0.327λ > λ/4
can use any type depends on realization.e.g. use shunt open stub only…
=0.4+j1ZL
0.434λ
50 Ω
open shunt stub
jB = j1.9 = jtanβlβl = 1.086
length = 0.173λ
ZL (0.4, 1)YL (0.34, -0.87)
Y (1, -1.9)
I d M t hi D R K k
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Impedance Matching - Dr. Ray Kwok
Double-Stub Tuning• Preferable because it’s not sensitive to the initial line length.• Useful in tuning (especially in waveguide) with variable load.• 2 adjustable stubs connected to a fixed-length transmission line.
• In principle, can use any series or shunt stubs.• In practice, mostly shunt short stubs for waveguide.
• Impedance of the stubs are arbitrary, and they don’t have to be the same.• Impedance of the connecting line doesn’t have to be Zo of the system.• 2 parameters to tune: d1 & d2
ZL
fixed L
Zo
Z2 Z1a d j u s t a b l e d 1
a d j u s t a b l e d 2
Impedance Matching Dr Ray Kwok
7/27/2019 Smith Chart - Impedance Matching
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Impedance Matching - Dr. Ray Kwok
Double Stub Tuninge.g. 2 shunt short stubs (50Ω)
separated by a 50Ω line of 0.2λ
0.3λλλλ
0.5λλλλ
– 0.2λλλλ
• rotate the 1-circle by line length
• adjust d1 along constant-G circle• stop at the rotated blue-circle• xline will bring it to the green circle• adjust d2 along the green circle to Zo
• not for all ZL !! Forbidden zone.
d1d2
ZL
0.2λ
50 Ω =0.4+j1
ZL(0.4,1)YL(0.34,-0.87)
Y (0.34,-0.2)
Y (1,1.22)
-cotβd1 = +0.67
βd1 = -0.98 +π = 2.16d1 = 0.344λ
-cotβd2 = −1.22βd
2= 0.687
d2 = 0.109λ
5 0 Ω
5 0 Ω
Impedance Matching Dr Ray Kwok
7/27/2019 Smith Chart - Impedance Matching
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Impedance Matching - Dr. Ray Kwok
Quarter-Wave Transformer• Zc
2 = ZoZ1
• Transformer real-to-real, complex-to-complex impedance.
• Usually needs one more element to match (before λ /4)(lumped elements, stubs or transmission line).
• normalized to Zo,
ZL
λ /4
Zo
matching
elementZ1Zc
1c ZZ =
Impedance Matching - Dr Ray Kwok
7/27/2019 Smith Chart - Impedance Matching
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Impedance Matching Dr. Ray Kwok
e.g. Quarter-Wavee.g. shunt stub (100Ω) then λ /4
ZL = 20 + j 50 Ω
• move Z to the real axis
• normalized Zc = √Z = √3.1; Zc = 1.76
• Zc = 50(1.76) = 88 Ω
ZL(0.4,1)YL(0.34, -0.87)
Z (3.1,0)Y(0.34,0)
(1/100) tanβd= (0.87)(1/ 50)
βd = 1.05d = 0.167λ
d
ZL
λ /4
Zc =0.4+j150 Ω1 0 0 Ω
Impedance Matching - Dr. Ray Kwok
7/27/2019 Smith Chart - Impedance Matching
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Impedance Matching Dr. Ray Kwok
Advanced Impedance Matching
So far….• single frequency• 1-port network
Need:• wideband matching – multiple sections• multi-ports (simultaneously tuned)
require knowledge of multi-port network analysis
Impedance Matching - Dr. Ray Kwok
7/27/2019 Smith Chart - Impedance Matching
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p g y
Homework
Smith Chart Exercise
Matching ExerciseDouble Stub Exercise