Download Even/Odd Mode Analysis of the Wilkinson Divider

4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
1/14
Even/Odd Mode Analysis
of the Wilkinson Divider
Consider a matched Wilkinson power divider, with a source at
port 2:
Port 2 Z0
λ
+
Vs
-
4
2 Z0
Port 1
2 Z0
Z0
2 Z0
λ
Port 3
Z0
4
Too simplify this schematic, we remove the ground plane, which
includes the bottom conductor of the transmission lines:
Port 2
λ
Z0
4
2 Z0
Port 1
Z0
+Vs-
2 Z0
2 Z0
λ
4
Port 3
Z0
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
2/14
Q: How do we analyze this circuit ?
A: Use Even-Odd mode analysis!
Remember, even-odd mode analysis uses two important
principles:
a) superposition
b) circuit symmetry
To see how we apply these principles, let’s first rewrite the
circuit with four voltage sources:
Vs
Vs
V2
λ
Z0
2 Z0
2 Z0
λ
2
Z0
4
2 Z0
V1
2
− Vs 2
V3
4
Vs
2
Z0
Turning off one positive source at each port, we are left with
Vs
o
2
an odd mode circuit:
V2
λ
V1 o
2 Z0
Z0
Jim Stiles
2 Z0
2 Z0
λ
Odd Mode Circuit
Z0
4
4
− Vs 2
V3o
Z0
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
3/14
Note the circuit has odd symmetry, and thus the plane of
symmetry becomes a virtual short, and in this case, a virtual
Vs
ground!
o
2
V2
λ
o
2Z0 V1
Z0
4
Z0
2 Z0
V=0
2Z0 V1
o
Z0
2 Z0
λ
− Vs 2
V3o
4
Z0
Dividing the circuit into two half-circuits, we get:
λ
Z0
4
+
2Z 0
V1 o
+
2 Z0
Z0
−
2
Z0
4
+
V1 o
Vs
−
λ
2Z 0
+
-
V2o
+
2 Z0
−
Z0
V3o
+
-
− Vs 2
−
Note we have again drawn the bottom conductor of the
transmission line (a ground plane) to enhance clarity (I hope!).
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
4/14
Analyzing the top circuit, we find that the transmission line is
terminated in a short circuit in parallel with a resistor of value
2Z0. Thus, the transmission line is terminated in a short circuit!
λ
Z0
4
+
+
V1 o = 0
Z0
2 Z0
V2o
−
+
-
Vs
2
−
This of course makes determining V1 o trivial (hint: V1 o = 0 ).
Now, since the transmission line is a quarter wavelength, this
short circuit at the end of the transmission line transforms to
an open circuit at the beginning!
Z0
+
Z0
V2o
+
-
Vs
2
As a result, determining voltage V2o
is nearly as trivial as determining
voltage V1 o . Hint:
−
V2o =
Vs
Z0
V
= s
2 Z0 + Z0 4
And from the odd symmetry of the circuit, we likewise know:
V
V3o = −V2o = − s
4
Now, let’s turn off the odd mode sources, and turn back on the
even mode sources.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
5/14
Vs
V2e
Even Mode Circuit
λ
V1 e
Z0
4
2 Z0
Z0
2 Z0
2 Z0
λ
2
Vs
V3e
4
2
Z0
Note the circuit has even symmetry, and thus the plane of
symmetry becomes a virtual open.
Vs
V2e
λ
2Z0 V1
e
2
Z0
4
2 Z0
Z0
I=0
2Z0 V1
e
2 Z0
λ
Z0
Vs
V3e
4
2
Z0
Dividing the circuit into two half-circuits, we get:
λ
+
2Z 0
V1 e
2 Z0
−
Jim Stiles
Z0
4
Z0
+
V2e
+
-
Vs
2
−
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
λ
Z0
4
+
2Z 0
V1 e
6/14
Z0
2 Z0
−
+
+
-
V3e
Vs
2
−
Note we have again drawn the bottom conductor of the
transmission line (a ground plane).
Analyzing the top circuit, we find that the transmission line is
terminated in a open circuit in parallel with a resistor of value
2Z0. Thus, the transmission line is terminated in a resistor
valued 2Z0.
λ
Z0
4
+
+
2Z 0
V1 e
V2e
2 Z0
+
-
Vs
2
−
−
Now, since the transmission line is a quarter wavelength, the
2Z0 resistor at the end of the transmission line transforms to
this value at the beginning:
Zin =
Z0
+
e
Z0
V2
−
+
-
Vs
2
2Z 0
)
2
= Z0
Voltage V2 e can again be determined by
voltage division:
V2e =
Jim Stiles
(
2Z 0
The Univ. of Kansas
Vs
Z0
V
= s
2 Z0 + Z0 4
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
7/14
And then due to the even symmetry of the circuit, we know:
V3e =V2e =
Vs
4
Q: What about voltage V1 e ? What is its value?
A: Well, there’s no direct or easy way to find this value. We
must apply our transmission line theory (i.e., the solution to the
telegrapher’s equations + boundary conditions) to find this
value. This means applying the knowledge and skills acquired
during our scholarly examination of Chapter 2!
λ
+
+
2Z 0
V1 e
Z0
4
2 Z0
V2e
+
-
Vs
2
−
−
If we carefully and patiently analyze the above transmission
line circuit, we find that (see if you can verify this!):
V1 e =
− jVs
2 2
And thus, completing our superposition analysis, the voltages
and currents within the circuit is simply found from the sum of
the solutions of each mode:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
V1 = V1 o +V1 o = 0 −
V2 =V2o +V2o =
jVs
2 2
Vs
4
+
=−
Vs
4
=
8/14
jVs
2 2
Vs
2
V V
V3 =V3o +V3o = − s + s = 0
4
λ
V1 =
− jVs
Z0
V2 = Vs 2
Z0
4
2 Z0
2 2
4
+Vs-
2 Z0
2 Z0
λ
4
V3 = 0
Z0
Note that the voltages we calculated are total voltages—the
sum of the incident and exiting waves at each port:
V1 V1 ( z1 = z1P ) =V1 + ( z1 = z1P ) +V1 − ( z1 = z1P )
V2 V2 ( z2 = z2P ) =V2+ ( z2 = z2P ) +V2− ( z2 = z2P )
V3 V3 ( z3 = z3P ) =V3+ ( z3 = z3P ) +V3− ( z3 = z3P )
Since ports 1 and 3 are terminated in matched loads, we know
that the incident wave on those ports are zero. As a result, the
total voltage is equal to the value of the exiting waves at those
ports:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
9/14
− jVs
V1 + ( z1 = z1P ) = 0
V1 − ( z1 = z1P ) =
V3+ ( z 3 = z 3P ) = 0
V3− ( z3 = z 3P ) = 0
2 2
The problem now is to determine the values of the incident and
exiting waves at port 2 (i.e., V2 + ( z 2 = z 2P ) and V2 − ( z 2 = z 2P ) ).
Recall however, the specific case where the source impedance
is matched to transmission line characteristic impedance (i.e.,
Z s = Z 0 ). We found for this specific case, the incident wave
“launched” by the source always has the value Vs 2 at the
source:
Z0
+
Vs
+
-
V
+
Z0
( z = z s ) = Vs 2
−
z
z=zs
Now, if the length of the transmission line connecting a source
to a port (or load) is electrically very small (i.e., β A 1 ), then
the source is effectively connected directly to the source (i.e,
βz s = βz P ):
Z0
And thus the total voltage is:
+
Vs
+
-
V
−
z=zs=zP
Jim Stiles
Zin
V = V + ( z = z P ) +V − ( z = z P )
=V
=
Vs
+
(z = zS ) +V − (z = z P )
2
+V
The Univ. of Kansas
−
(z = zP )
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
10/14
For the case where a matched source (i.e. Z s = Z 0 ) is connected
directly to a port, we can thus conclude:
V + ( z = z P ) = Vs 2
V − ( z = z P ) = V −Vs 2
Thus, for port 2 we find:
V2+ ( z 2 = z2P ) =Vs 2
V2− ( z2 = z2P ) =V2 −Vs 2 = Vs 2 −Vs 2 = 0
Now, we can finally determine the scattering parameters
S12 , S22 , S32 :
V1 − ( z1 = z1P ) ⎛ − jVs ⎞ 2 − j
S12 = +
=
=
V2 ( z2 = z2P ) ⎜⎝ 2 2 ⎟⎠Vs
2
V2− ( z2 = z2P )
2
S22 = +
= (0) = 0
V2 ( z2 = z2P )
Vs
V3− ( z 3 = z3P )
2
S32 = +
= (0) = 0
V2 ( z2 = z2P )
Vs
Q: Wow! That seemed like a lot of hard work, and we’re only
1
3
of the way done. Do we have to move the source to port 1 and
then port 3 and perform similar analyses?
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
11/14
A: Nope! Using the bilateral symmetry of the circuit
( 1 → 1, 2 → 3, 3 → 2 ), we can conclude:
S13 = S12 =
−j
2
S33 = S22 = 0
S23 = S32 = 0
and from reciprocity:
S21 = S12 =
−j
S31 = S13 =
2
−j
2
We thus have determined 8 of the 9 scattering parameters
needed to characterize this 3-port device. The remaining
holdout is the scattering parameter S11. To find this value, we
must move the source to port 1 and analyze.
Port 2
λ
-Vs+
Z0
4
2 Z0
Port 1
Z0
2 Z0
2 Z0
λ
4
Port 3
Z0
Note this source does not alter the bilateral symmetry of the
circuit. We can thus use this symmetry to help analyze the
circuit, without having to specifically define odd and even mode
sources.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
12/14
Since the circuit has (even) bilateral symmetry, we know that
the symmetry plane forms a virtual open.
V2
λ
−Vs +
V1
Z0
4
Z0
2 Z0
2Z0
I=0
2Z0
Z0
2 Z0
V1
−Vs +
λ
V3
4
Z0
Note the value of the voltage sources. They have a value of Vs
(as opposed to, say, 2Vs or Vs/2) because two equal voltage
sources in parallel is equivalent to one voltage source of the
same value. E.G.:
+
+
5V
+
-
+
-
5V
5V
−
+
-
5V
−
Now splitting the circuit into two half-circuits, we find the top
half-circuit to be:
λ
Vs
4
2Z 0 +
+
-
V1
2 Z0
Z0
Z0
−
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
13/14
Which simplifies to:
λ
4
2Z 0 +
Vs
Z0
2 Z0
V1
+
-
−
And transforming the load resistor at the end of the
line back to its beginning:
Vs
2Z 0
+
-
+
2Z 0
V1
λ
4
wave
Finally, we use voltage division to
determine that:
−
⎞ Vs
2Z 0
⎟=
Z
+
Z
2
2
2
0 ⎠
⎝ 0
⎛
V1 = Vs ⎜
Port 2
Thus,
λ
-Vs+
V1 = V
s
Z0
4
2 Z0
2
Z0
2 Z0
2 Z0
λ
4
Port 3
And since the source is matched:
Z0
V1 + ( z1 = z1P ) =Vs 2
V1 − ( z1 = z1P ) =V1 −Vs 2 =Vs 2 −Vs 2 = 0
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/14/2009
Wilkinson Divider Even and Odd Mode Analysis.doc
14/14
So our final scattering element is revealed!
V1 − ( z1 = z1P )
2
S11 = +
= (0) = 0
V1 ( z1 = z1P )
Vs
So the scattering matrix of a Wilkinson power divider has been
confirmed:
⎡ 0
⎢
S = ⎢− j 2
⎢⎣ − j 2
−j
0
0
2
−j
⎤
⎥
0 ⎥
0 ⎥⎦
2
His worst
handout ever!
Oh no, I’ve seen
much worse.
So, what’d
ya think?
Jim Stiles
The Univ. of Kansas
Dept. of EECS