Download Example: Odd-Even Mode Circuit Analysis

3/6/2009
Example Odd Even Mode Circuit Analysis
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Example: Odd-Even Mode
Circuit Analysis
Carefully (very carefully) consider the symmetric circuit
below.
+
Z 0 = 50 Ω
4.0 V
λ
50 Ω
50 Ω
50 Ω
λ
50 Ω
2
Z 0 = 50 Ω
+
50 Ω
v1
-
50 Ω
The two transmission lines each have a characteristic
impedance of .
Use odd-even mode analysis to determine the value of
voltage v1 .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3/6/2009
Example Odd Even Mode Circuit Analysis
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Solution
To simplify the circuit schematic, we first remove the bottom
(i.e., ground) conductor of each transmission line:
50 Ω
Z 0 = 50 Ω
4.0 V
+
-
λ
50 Ω
50 Ω
50 Ω
λ
50 Ω
+
v1
-
2
50 Ω
Z 0 = 50 Ω
Note that the circuit has one plane of bilateral symmetry:
λ
50 Ω
50 Ω
50 Ω
50 Ω
+
v1
50 Ω
λ
2
50 Ω
Thus, we can analyze the circuit using even/odd mode analysis
(Yeah!).
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3/6/2009
Example Odd Even Mode Circuit Analysis
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The even mode circuit is:
λ
50 Ω
2.0 V
+
-
50 Ω
50 Ω
50 Ω
+
-
+
2.0 V
v1e
-
λ
50 Ω
50 Ω
2
I=0
Whereas the odd mode circuit is:
λ
50 Ω
2.0 V
+
-
50 Ω
50 Ω
50 Ω
+ -2.0 V
+
-
v1o
50 Ω
λ
2
50 Ω
V =0
We split the modes into half-circuits from which we can
determine voltages v1e and v1o :
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3/6/2009
λ
Example Odd Even Mode Circuit Analysis
50 Ω
2
50 Ω
v
-
λ
+
2.0 V
e
1
Recall that a A = λ 2 transmission
line terminated in an open
circuit has an input impedance
of Z in = ∞ —an open circuit!
Likewise, a transmission line A = λ 4
terminated in an open circuit has an
input impedance of Z in = 0 —a short
50 Ω
4
+
-
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circuit!
Therefore, this half-circuit simplifies to:
And therefore the voltage v1e is easily
50 Ω
determined via voltage division:
50 Ω
+
-
+
2.0 V
v1e
-
⎛ 50 ⎞
⎟ = 1.0 V
+
50
50
⎝
⎠
v1e = 2 ⎜
50 Ω
Now, examine the right half-circuit of the odd mode:
λ
50 Ω
2
50 Ω
+
v1o
λ
4
50 Ω
-2.0 V
+
-
Recall that a A = λ 2 transmission
line terminated in a short
circuit has an input impedance
of Z in = 0 — a short circuit!
Likewise, a transmission line A = λ 4
terminated in an short circuit has an
input impedance of Z in = ∞ — an open
circuit!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3/6/2009
Example Odd Even Mode Circuit Analysis
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50 Ω
This half-circuit simplifies to Æ
It is apparent from the circuit
that the voltage v1o = 0 !
50 Ω
+
+
-
-2.0 V
v1o
-
50 Ω
Thus, the superposition of the odd and even modes leads to
the result:
v1 = v1e + v1o = 1.0 + 0 = 1.0V
Jim Stiles
The Univ. of Kansas
Dept. of EECS