IntroductionA microwave circuit is an interconnection of components whose size is comparable with the wavelength at the operation frequency
Type of Components:• Interconnection: it is not an ideal connection (zero resistance) as
in the case of lumped-element networks• Pseudo-lumped components (they present, in a first
approximation, the same frequency behavior of the lumped components, having however a finite size)
• Distributed components (transmission line sections or stubs arbitrarily terminated)
In microwave circuits the junction between two (or more) component do not correspond to an ideal node (as in lumped circuits). The junction must be suitably represented (typically by means of its scattering parameters) in order to take into account the phenomena which always arise at a discontinuity in a transmission line.
Discontinuity between two transmission lines
Zc Zc
Discontinuity
Zc ZcS Matrix
Coax 1: R1=5 cm, r1=2.17 cm
Coax 2: R2=3 cm, r1=1.3 cm
1 2
1 2
60 ln 60 ln 50 cR RZr r
= ⋅ = ⋅ = Ω
No good!
Correct model. Why?
L1=10 cm
L2=10 cm
Higher modes excitation
TEM wave
Transverse E, H fields
At the discontinuity the E field cannot be transverse
Higher modes are then generated, allowing the fulfillment of the continuity conditions. These modes however cannot propagate (they are below cutoff)
Incident TEM wave
Transmitted TEM wave
Amplitude of higher modes excited
The energy of the em field associated to the higher modes is confined closely to the discontinuity, and it affects the incident wave like a lumped reactance connected at the junction of the two coaxial lines
Scattering Matrix of the discontinuityThe scattering parameters of the following configuration can be evaluated by means of an em simulator (numerical solution of the em fields). It has (f0=1 GHz, Zc=50 Ω):
11 22 12 210.1792 20 , 0.9838 109.61 S S S S= = ∠ ° = = ∠ °
Discontinuity50 Ω 50 Ω
10 cm10 cm
Microwave network
Port 1 Port 2
To get the S parameters of discontinuity we need to move the reference sections (ports 1 and 2) by 10 cm inwards:
( )210 10 2.094 120fcm cm radπφ βν
= − ⋅ = ⋅ = − − °
( ) ( )11 22 11 12 21 12exp 2 0.1792 100 , exp 2 0.9838 10.39 S S S j S S S jφ φ′ ′ ′ ′= = − = ∠− ° = = − = ∠− °
L1=10 cm
L2=10 cm
Port 1
Port 2
Frequency dependance
500 600 700 800 900 1000 1100 1200 1300 1400 1500Frequency (MHz)
Return Loss vs. Frequency
-22
-20
-18
-16
-14
-12
-10
DB(|S(1,1)|)CoaxStep
The effects of the discontinuities are generally frequency-dependent
List of some discontinuities and components in the library of MWOffice
Example: microstrip implementation of a double stub matching network
Ideal scheme
Zc, L1
ZS1, LS1 ZS2, LS2
1 2Zc, L0Zc, L0
T junction
Open
Discontinuities inclusion
MLINID=TL3W=2.2 mmL=10 mm
MLINID=TL6W=0.622 mmL=79.68 mm
MOPEN$ID=TL9
1 2
3
MTEE$ID=TL5
MLINID=TL7W=1.33 mmL=32.92 mm
PORTP=1Z=50 Ohm
MOPEN$ID=TL2
1 2
3
MTEE$ID=TL4
MLINID=TL1W=2.2 mmL=5 mm
MLINID=TL8W=1.22 mmL=8 mm
PORTP=2Z=50 Ohm
MSUB
Tand=Rho=
T=H=Er=
0 1 .035 mm0.8 mm2.55
Comparison
0.9 0.95 1 1.05 1.1Frequency (GHz)
-50
-40
-30
-20
-10
0
1 2
Z
IMPEDID=Z1R=17 OhmX=66 Ohm
DoppioStub
Microtrip
Ideal
Computing the models of the discontinuities
• Equivalent circuits (lumped and/or distributed elements, numerically evaluated)
• Analytical formulas (simplest cases, low accuracy)
• Electromagnetic analysis (by means of em simulators, very time consuming)
• Different models for the same discontinuity are often available (high accuracy=long computation time)
Microstrip discontinuities: junctions
Step (2-port)Tee (3-port) Cross
(4-port)
S1-S2
S2S1
S3S3S1
S4
S2
MLINID=TL1W=1 mmL=4 mm
1
2
3
4
MCROSS$ID=TL2
MLINID=TL4W=2 mmL=5 mm
MLINID=TL5W=2.5 mmL=5 mm
MLINID=TL3W=2 mmL=4 mm
MLINID=TL1W=1 mmL=4 mm
MLINID=TL4W=2 mmL=5 mm
1 2
3
MTEE$ID=TL2
MLINID=TL3W=2 mmL=4 mm
MSTEPX$ID=MS1Offset=0 mm
MLINID=TL1W=1 mmL=4 mm
MLINID=TL3W=2 mmL=4 mm
1 2
Bend (2-port)
MLINID=TL1W=2 mmL=4 mm
MLINID=TL3W=2 mmL=4 mm
MBENDAID=TL2W=2 mmANG=90 Deg
MLINID=TL1W=2 mmL=4 mm
MLINID=TL3W=2 mmL=4 mm
MCURVEID=TL2W=2 mmANG=45 DegR=2 mm
MLINID=TL1W=2 mmL=4 mm
MLINID=TL3W=2 mmL=4 mm
MUBEND$ID=TL4S=2 mmM=0.5
S2
S1
S2
S1
S2
S1
Terminations (1-port)
MLINID=TL1W=2 mmL=4 mm
MOPENX$ID=MO1
Open end Via hole
MLINID=TL1W=2 mmL=4 mm
VIA1PID=V1D=1.5 mmH=1 mmT=0.05 mmRHO=1
MLINID=TL1W=2 mmL=4 mm
MRSTUB2WID=TL2W=2 mmRo=7 mmTheta=50 Deg
Radial Stub
S1
S1
S1
Pseudo-lumped componentsSeries inductor
Ls
1 1
1 1
s sL
s s
j L j LY
j L j L
ω ω
ω ω
−
=−
( ) ( )
( ) ( )
1 1tan sin
1 1
sin tan
c s c sS
c s c s
jZ l jZ lY
jZ l jZ l
β β
β β
−
=−
( ) ( ) 0 tan sin
L S s s s s
s c s
c ss
Y Y l l l lL Z l
Z lLv
β β β βω β
≅ ⇒ ≅ ⇒
≅
≅
Zc
ls
Cp
1 1
1 1
p pC
p p
j C j CZ
j C j C
ω ω
ω ω
= ( ) ( )
( ) ( )
1 1tan sin
1 1
sin tan
c s c sS
c s c s
jY l jY lZ
jY l jY l
β β
β β
=
( ) ( ) 0 tan sin
L S s s s s
p c s
c sp
Z Z l l l lC Y l
Y lCv
β β β βω β
≅ ⇒ ≅ ⇒
≅
≅
Zc
ls
Shunt capacitor:
Other componentsInterdigital capacitor
MLINID=TL1W=2 mmL=4 mm
MICAP$ID=MI1W=1 mmS=1 mmG=1 mmL=10 mmN=4WP=1 mm
MLINID=TL2W=2 mmL=4 mm
S2S1
Spiral Inductors
EPSB=1TDB=0TB=0.001 mmRhoB=1
MLINID=TL1W=1 mmL=2 mm
MLINID=TL2W=1 mmL=2 mm
Two-port circuits
• S matrix for reciprocal, lossless networks:
11 22
2 212 11
11 22 21
12
S S
S Sφ φ φ π
=
= −
+ − = ±⋅ =*S S U
Only three real parameters are needed for defining S (for instance |S11|, φ11, φ22)
How to get a transformer at microwaves?
• The transformer is a 2-port widely used in many circuital applications
• The ideal component is reciprocal and lossless.• At microwave frequencies it is however very difficult to
be realized. In those application where the goal is the impedance (admittance) scaling, the component employed is the impedance (admittance) inverter:
n:1
ZL
2LZ inZ n=
Ideal Transformer
K ZL
2
L
ZinKZ =
Ideal Impedance inverter
S parameters of the impedance inverter
( )( )
2 2 20 0 0
11 22 2 2200 0
Z Z=
Z Z
K K ZS SK ZK
− −= =
++
Z0 K Z0
S12 has been obtained by imposing the lossless conditions:
22 2 0
12 11 2 20
11 22 12 12 11
1 2
2 2 2
K ZS SK Z
π πφ φ φ π φ φ
⋅= − = +
+ − = ± ⇒ = ± = ±
012 21 2 2
0
2 K ZS S jK Z
⋅= = ±
+
( )11 22 0 φ φ π= = 12 21 2πφ φ= = ±
Practical implementation of invertersNote: the parameter K of the ideal inverter is independent on frequency. The physical implementation however can only approximate this condition in a limited frequency band
Zc=K
λ0/4
-jK -jK
jK
jJ
-jJ-jJ
J=1/K
φ φ
XZc Zc
φ φ
Yc Yc
jB
( )1
21 2tan , 2 1c c
X KXZ K Z
φ − = − =
− ( )1
21 2tan , 2 1c c
B JBY J Y
φ − = − =
−
Note: for X (B) positive, φ is <0 and K/Zc (J/Yc) is <1.
Equivalent model for a 2-port lossless circuit
S Kφ1 φ2
Zc Zc
11
11
1
1c
SK Z
S−
=+
11 221 2,
2 2S Sπ πφ φ−∠ −∠
= =
11
11
1
1c
SK Z
S+
=−
11 221 2,
2 2S Sφ φ∠ ∠
= − = −
Choice 1 (|K|<Zc):
Choice 2 (|K|>Zc):
12 1 22K S π φ φ∠ = ∠ − + + (must be 0 or π)
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