circuit characterization
TRANSCRIPT
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Microwave circuit
characterization
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Microwave component
1
2
4
5
Portesports
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3 6
Elments de circuit
Circuit elements
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Reference planes
On each access line i of a
z
n
znz1
1
component, a coordinate axis zi
is defined. The origin of this axis
lies in the reference plane of the
port i.
Assumptions :
3LEMA-EPFL
i
They support only the dominant mode
The reference plane is distant enough
From discontinuities to ensure that the
Higher order modes are attenuated
Available models
Currents and voltages ill defined
Well known from low frequency circuit analysis
Not adapted to system approach
Scattering matrix
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Well suited to microwaves
Not known from low frequency circuit analysis
Adapted to system analysis
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Kirchhoffs model
i1 i4
1
2
3
4
5
6
i2
i3
i5
i6
u1
u2
u3
u4
u5
u6
5LEMA-EPFL
[Z]
Current, voltages and impedances
Current voltages and impedances
Current and voltage difficult to measure
Current and voltage not always uniquelydefined
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Example : TEM line
V= E
+EH
V= E dl+
I= H dlC+
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-
Zc =V
I=
L
C
Example : TEM line
Fields identical to static fields
Zero cutoff frequency
Needs at least two separate conductors
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Non-TEM line : the rectangular
waveguide
V j= E A
Ey x,y,z( )= E0ja
sin
x
ae
jz = E0ey x,y( )ejz
=ja x jz = jz
Example : mode TE10 of a rectangular
waveguide
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x , ,
ay ,
V=E0j a
sin
x
ae
jzdy
y
Thusa
b
Non TEM voltage and current definition
Onl defined for one mode
The voltage is proportional to the transverseelectric field
The current is proportional to the transversemagnetic field
The characteristic impedance is equal to U/I. Wechoose the characteristic im edance e ual to the
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wave impedance
The power flux is given by the product of thevoltage and the current
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Arbitrary line
Et x,y,z( ) = et x,y( ) Eo+e
jz + Eoe
jz( )=
et x,y( )C1
V+ejz + Vejz(
Ht x,y,z( ) = ht x,y( ) Eo+ejz Eo
ejz(
V z( ) = V+ejz + Vejz
I z( ) = I+ejz Iejz
Zc =V
+
I+=
V
I=
C1Eo+
C2Eo+ =
C1
C2
11LEMA-EPFL
= t x,yC2
I+ejz Iejz( )C1
C2= Zmod
Different impedances
Wave impedance :
Line characteristic impedance :
Impedance matrix of a circuit
o
Zmod =Et
HtZc = V
+
I+ =V
I =LC
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Impedance matrix
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Single port element
V
Circuit une porteaccs
plan de rfrence
Zin
I
V
n
S
Ref. plane
Single port
Zin =I
Power delivered to the component
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port
P =1
2E H* ds= Pr+ 2j Wm We( )
s
(Poynting)
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Single port element
Et x,y,z( ) = V z( )
et x,y( )C1
ejz
Ht x,y,z( ) = I z( )ht x,y( )
C2e
jz
with1
et ht ds=1 thus P = 1 VI*et ht ds =1VI
*
15LEMA-EPFL
1 2 s 1 2 s
and
Zin = R + jX=V
I=
VI*
I2=
2P
I2
=2 Pr+ 2j Wm We( )( )
I2
Single port element
dissipated in the system (Losses)
X is proportional to the mean reactivestored energy in the system
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Impedance and admittance matrix
v1+,i1+
v1-,-i1-
v2+,i2+
v2-,-i2-
v3+,i3+
v4+,i4+
v4-,-i4-
v5+,i5+
v5-,-i5-
v6+,i6+
t2
t3
et
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[Z]
v3-,-i3- v6-,-i6-
Vn = Vn+ + Vn
In = In+ In
At each port ti
Impedance and admittance matrix
obtained in open
circuit conditions
Admittance matrix is
obtained in short
circuit conditions
=
I[ ] = Y[ ] V[ ]
Zij =ViI
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k
Yij =Ii
Vj Vk=0 pour kj
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Properties of the impedance and
admittance matrix
Yi = Yi
Lossless circuit :
Zij = Zji
Re Zmn{ }= 0
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Example : transmission line
= =dI(0) I(d)
U(0)U(d
( ) ( )2
, ,
, 2U U d I I d = =
( )
( )
e e
e e
z z
z z
U z U U
I z I I
++
++
= +
=
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1 2
U1 U2( ) ( )
( ) ( )
0 e e
0 e e
d d
d d
U U U U d U U
I I I I d I I
++ +
++ +
= + = +
= =
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Example : transmission line
U Z Z I dI(0) I(d)
U(0) U(d ( ) ( )
( )( )
2 21 22 2
1
2
1coth
sinh
1coth
sinh
c
U Z Z I
dd I
ZI
dd
=
=
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1 2
U1 U2( )
( )
( )( )
2 21 22 2
1
2
1coth
sinh
1coth
sinh
c
I Y Y U
dd U
YU
dd
=
=
Equivalent T circuit of a reciprocal
two-port
11 12Z12 Z22
I1 I2
U1 U2
Za Zb
Zc
Za = Z11 Z12Zb = Z22 Z12
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c 12
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Equivalent T circuit of a transmisson
line
I1 I2
U1 U2
Za Za
Zc
Za = Zcaract coth d( )1
sinh d( )
= Zcaractch d( ) 1sinh d( )
= Zcaracttanh
d
2
23LEMA-EPFL
Zc = caractsinh d( )
Equivalent circuit of a reciprocaltwo-port
11 12Y12 Y22
I1 I2Yc Ya = Y11 + Y12
=
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1 2a b
Yc = Y12
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Equivalent circuit of atransmisson line
I1 I2
U1 U2Ya Ya
Yc
Ya = Ycaract coth d( )1
sinh d( )
= Ycaractch d( )1sinh d( )
= Ycaracttanh
d
2
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Yc = Ycaractsinh z( )
Scattering parameters
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Normalized wave amplitudes :
definition
ai =vi + Zciii2 Zci
, bi =vi Zciii2 Zci
[W1/2]
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vi = Zci ai + bi( ) , ii =ai bi( )
Zci
Normalized wave amplitudes :
properties
Transmission lines
vi = vi+
e
jz
+ vi
e
+ jz
ii = ii+ejz + ii
e+ jz
ai
=vi
+
Zcie
jz
vi
+ z
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i =Zci
ai : progressive wave
bi : retrograde wave
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Power at one component port
Pi = Re viii*[ ]= Re ai + bi( ) ai* bi*( )[ ]= ai 2 bi 2
/a /2 : ower enterin ort i
29LEMA-EPFL
/bi/2 : power leaving port i
Signal model
Plans de rfrenceReference planes
a1
b1a2
b2
a4
b4a5
b
t1
t2
t4
t5
b[ ]= S[ ] a[ ]
=bi
30LEMA-EPFL
[S]
a3
b3
a6
b6
t3 t6 aj ak=0 , kj
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Scattering matrix : definition
ai =vi + Zciii2 Zci
, bi =vi Zciii2 Zci
orma ze wave amp u e
vi = Zci ai + bi( ) , ii =ai bi( )
Zci
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b[ ]= S[ ] a[ ] sij =bi
aj ak=0 , kj
Signal model
Signals are easy to measure
Well suited to a system approach (transferfunctions)
Drawbacks :
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Not known at low frequencies
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Scattering matrix : properties
Depends on the component and its
properties
To change the connecting lines implies to
chna e the scatterin matrix
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The scattering matrix is obtained termiating
the ports by matched loads
Reciprocity
ass ve, near, so rop c c rcu
The circuit is reciprocal
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zij = zji
sij = sji
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Lossless circuit
* tai = bi a[ ] a[ ] b b[ ] = 0 a[ ] = a
Moreoverb[ ] = S[ ] a[ ]
b[ ]= a[ ] S[ ]a[ ] a[ ] a[ ] S[ ]S[ ] a[ ] = 0a[ ] 1[ ] S[ ]S[ ]{ }a[ ] = 0S S[ ] = 1[ ]
35LEMA-EPFL
sij*sik = jk
i=1
N
jk =1 si j = k
0 si j k
Matched circuit
Sii = 0
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Change of reference planeti
aib
[S].
.
.
tn
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[S'].
.
.
t i
t'n
z
b'i
a'i
Change of reference plane
a' i = aie i
b' i = bieji
i = izi
thus s' ii = siiej2i
And in general
a[ ] = diag ej[ ]a'[ ] j
38LEMA-EPFL
a'[ ] = diag e j[ ]a[ ]b[ ] = diag ej[ ]b'[ ]b'[ ] = diag ej[ ]b[ ]
diag ej[ ]=
...
0 ej2 :
:
0 ... ejn
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Change of reference plane
[ ] [ ] [ ]
[ ] [ ]
' e e '
' e e
j j
j j
b diag S diag a
S diag S diag
=
=
39LEMA-EPFL
' e i jjij ijs s +=
Passive circuits : questions
.
2. Reciprocity
3. Losslessness
4. Matched
40LEMA-EPFL
. ymme ry
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[S]->[Z][b] = [S] [a] ai= (Ui+ZciIi)/2Zci Ui= Zci(ai + bi)
[U] = [Z] [I] bi= (Ui ZciIi)/2Zci Ii= (ai bi)/Zci
We define two auxiliary matrices
[G] = [diag(Zci)] [F] = [diag(1/2Zci)]
And we can write
41LEMA-EPFL
[a] = [F] {[U] + [G] [I] } ([a] + [b]) = 2 [F] [U]
[b] = [F] {[U] [G] [I] } ([a] [b]) = 2 [F] [G] [I]
[S]->[Z]
[a] = [F] {[U] + [G] [I] } -> [a] = [F] {[Z] + [G] } [I]
-> = a
[b] = [F] {[U] [G] [I] } -> [b] = [F] {[Z] [G] } [I]
-> [b] = [F] {[Z] [G] } {[Z] + [G] }1 [F]1 [a] = [S] [a]
= 1 1
42LEMA-EPFL
In one dimension, we have = S11 = (Zt Zc)/(Zt + Zc)
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[Z]->[S]([a] + [b]) = 2 [F] [U] -> [U] = (2 [F])1([a] + [b]) = (2 [F])1([1] + [S]) [a]
a = = a-> a =
regrouping, we obtain:
[U] = (2 [F])1([1] + [S]) ([1] [S])1 2 [F] [G] [I] = [Z] [I]
1 1
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For a single acess component, we have
Zt = Zc(1 + r)/(1 r)
Flow charts
are represented as arrows
a sij b
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Flow charts
Two port
a1
b1
b2
a2s11
s21
s22s12
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Flow charts
a1 b2s21
b
1
a
2
s11 s22
s
12
s13s32
s23s31
Three port
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a3b3
s33
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Flow chartsa1 b2s21
b1 a2
11 22
a3 b4
s13 s42
4-port
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b3 a4
s33
s34
s44
Flow chart reduction rules
1 multi lication
ab
s1 s2
a s1s2 b
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Flow chart reduction rules
2 addition
a b
s1
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s2
a b
s1+s2
Flow chart reduction rules
3 retroaction
ab
s1
s2
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a b
1-s1s2
s1
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Example
a1 b'2s21 s'21b2=a'1
b1 a'2
s11 s22s12
s'11 s'22s'12
a2=b'1
51LEMA-EPFL
Example
a1 s21 b2='a1
b1
s11 s22
s12
s'11
52LEMA-EPFL
a2=b'1
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Examplea1 s21
b1
s11
s12
a1
s'11
1-s'11s22
53LEMA-EPFL
s'11
1-s'11s22b1
s11+s21s12
Example
a1 b'2s21 s'21b2=a'1
s22 s'11
a2=b'1
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Example1
a1 b'2s21 s'21
1-s'11s22
a
s21
1-s'11s22
s'21
b'
55LEMA-EPFL
Example
b2=a'1
b1 a'2
s22
s12
s'11
s'12
a2=b'1
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Example
''
1
a'2
b12s12 s 121-s22s'11
s12s'12
1-s s'
57LEMA-EPFL
Example
b'2s'21b2=a'1
a'2
s22 s'11 s'22s'12
58LEMA-EPFL
a2=b'1
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Exampleb'2s'21
a'2
s'22s'12
22
1-s'11
s22
59LEMA-EPFL
b'2
a'2
s'22+s'12s'21
s22
1-s'11
s22
Single port components
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Single port components
single portcircuitacces
Zin
I
V Single portcircuitacces
a
s11b
61LEMA-EPFL
[ ] [ ] [ ], , have only one term. They are scalars Z Y S
Single port components
s11 =
Zin ZcZin + Zc
, Zin =1+ s111 s11
Equivalences
a1
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Flow chart b1s11, ,
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Examples
short circuit : s11 = -1
open circuit : s11 = 1
total reflection : s11 = ej
matched sin le ort com onents :
63LEMA-EPFL
matched load
Others : antennas, loads, etc.
Short circuits
spring contact
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g/4
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Matched loads
65LEMA-EPFL
Matched load
1
b1a
2
b2a3
4
b4
a5b5
a6
=0
=0
The measurement ofs21 has to be done
by terminating all non
concerned orts b
66LEMA-EPFL
[S]
3
s21 =b2
a1 ai =0, i1
matched loads
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Two port circuits
67LEMA-EPFL
Two port circuits
[S]
a1
b1
b2
a2
a1
b1
b2
a2s11
s21
s22s12
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[ ] 11 1221 22
s sS
s s
=
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Examples
, ,
Lossy : attenuators, isolators
69LEMA-EPFL
Phase shifters
matched
lossless
=0 e
j
70LEMA-EPFL
ej
o
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Phase shifters
Matched transmission line
L
71LEMA-EPFL
S[ ] = 0 e jL
e jL 0
Phase shifters
Fox' phase shifter
Non reciprocal phase shifters
S[ ] =0 e
j1
ej2 0
1 2
72LEMA-EPFL
Gyrators
S[ ] =0 1
1 0
1 2 =
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Fox' phase shifters
/2
/4
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/4
Fox' phase shifters
mode 1mode 2pol. lin.pol. cir.
(1- 2)l= /2
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quarter wave line
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Fox' phase shifters
mode 1mode 2
pol. cir. (1- 2)l=
pol. cir inverse
75LEMA-EPFL
rotation of + = phase shift of + for both components
Fox' phase shifters
mode 1mode 2
(1- 2)l= /2
pol. lin.pol. cir.
76LEMA-EPFL
rotation of - = phase shift of + for both components
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Attenuators
Reciporcal
Lossy
Attenuation level given by:
S[ ] =0 s12
s12 0
s12
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Attenuators using absorbers
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Rotative attenuator
lame mobile
lame fixefixed slab
rotatin slab
80LEMA-EPFL
lame fixefixed slab
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Resistive attenuators (T
attenuators)
and R2 in order to be
matched
Then we can obtain
the attenuation
R2
83LEMA-EPFL
Resistive attenuators
R1 R1I
I2
R2
port 1 port 2
ZoZin
u1 u2
84LEMA-EPFL
1
2 1
1 1in o
o
Z R Z
R R Z
= + =+
+
12
12
oZ RRR
=
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Resistive attenuatorsR1 R1I
II2a1 b2
R2
port 1 port 2
ZoZin
u1 u2
b1 a2
85LEMA-EPFL
12 221
1 1 1
o
o
Z Rb us
a u Z R
= = =
+
Other solution
Impedance matrix of the 2 port
[ ] 1 2 2
2 1 2
R R RZ
R R R
+ =
+
from which we obtain the scatterin arameter matrix
86LEMA-EPFL
[ ]( )
( )
( )
2 2 21 2 2 2
2 22 2 21 2 2 2 1 2 2
21
2
o o
o o o
R R Z R R Z S
R R Z R R Z R R Z R
+ = + + +
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Other solution
= =
2 21
212
oZ RRR
=
,
and1
21oZ Rs
Z R
=
+
87LEMA-EPFL
so finally, the attenuation level is given by
( ) 1 1211 1
[ ] 20log 20log 20logo o
o o
Z R Z R LA dB s
Z R Z R
+= = =
+
Isolators
non reciprocal
lossy
0 0
88LEMA-EPFL
=1 0