Tellegen’s Theorem
Circuit:
φ0
i2
–
+
v1
v2
–
+
i0
i1
+
v0
–
i3
φ1
+ v3 –
φ2
i4
–
+
v5
–
i5
φ3
+
v4
v0 = φ0 – φ3, v1 = φ1 – φ0, v2 = φ0 – φ2,
v3 = φ3 – φ2, v4 = φ1 – φ3, v5 = φ3 – φ2.
Incidence Matrix:
node: 0
element: 0
1
2
3
4
5
1
+1 0
–1 +1
2
3
element: 0
0 –1
0 0
+1 0 –1 0
0 +1 –1 0
0 +1 0 –1
0 0 –1 +1
node: 0
2
3
4
5
+1 –1 +1
0
0
0
0 +1
0 +1 +1 0
= AT
0 0 –1 –1 0 –1
–1 0 0
0 –1 +1
1
= A
1
2
3
Potential, Voltage, and Current Vectors
φ0
φ =
φ1
φ2
φ3
v =
v0
i0
v1
i1
v2
i =
i2
v3
i3
v4
i4
v5
i5
Then,
5
v = Aφ, i·Aφ = i·v =
∑i v
n n
n =0
Also, by KCL
⎡0 ⎤
⎢0 ⎥
⎢ ⎥
ATi = ⎢0⎥ , φ·ATi = 0.
⎢ ⎥
⎣0 ⎦
.
But
i·Aφ ≡ φ·ATi,
leading to Tellegen’s Theorem
5
∑i v
n n
= 0.
n =0
The only requirement is that all the in be for one set a of elements in the circuit so that KCL
holds, and all the vn be for another set b of elements in the circuit so that KVL holds (a set of
potentials φ can be assigned). When set a is the same as set b, the result is simply the
conservation of power. But Tellegen’s Theorem is more general and leads to many other results
such as reciprocity theorems. See Tellegen's Theorem and Electrical Networks (MIT research
monograph no. 58) by Paul Penfield, Robert Spencer, S. Duinker.
Example of Using Tellegen’s Theorem
Consider two networks with the same topology and, inside their respective two-port boxes, the
same set of elements—passive complex impedances zn(s). The outside elements differ—an open
circuit at port 0 and a source at port 1 in one case, and a source at port 0 and a short circuit at
port 1 in the other case.
ib1
ia1
ia0
ib0
zn(s)
+
va0
+
+ v
a1
–
–
n = 2 ... N
–
–ib0
zn(s)
+
vb0 +
–
–
+
vb1
n = 2 ... N
–
Choosing the currents from network a and the voltages from network b for Tellegen’s Theorem,
N
0=
∑
N
ian vbn = ia 0 vb 0 + ia1vb1 +
n =0
N
=
∑
N
ian vbn = 0 ⋅ vb 0 + ia1 ⋅ 0 +
n=2
∑
n=2
N
ian vbn =
∑i
an vbn
n=2
∑i
an ibn zn ( s ).
n=2
Choosing the voltages from network a and the currents from network b for Tellegen’s Theorem,
N
0=
∑
N
van ibn = va 0ib 0 + va1ib1 +
n =0
∑
n=2
N
van ibn = va 0 ib 0 + va1ib1 +
∑
n=2
N
ian zn ( s)ibn = va 0 ib 0 + va1ib1 +
∑i
an ibn zn ( s )
n=2
= va 0ib 0 + va1ib1.
Therefore we have a reciprocity of the reverse open-circuit voltage transfer equaling the forward
short-circuit current transfer:
va 0
i
= b1 .
va1 −ib 0