- 2068 Shrawan · 8 marks
Z and g parameters, neither reciprocal nor symmetrical
Short answer
The 1 Ω shunt and dependent source 3V1 give V1 = −I1, so z11 = −1 Ω, z12 = 0. At port 2, V2 = (2/3)I2 − 2I1 giving z21 = −2 Ω, z22 = 2/3 Ω. g: g11 = −1 S, g12 = 0, g21 = 2, g22 = 2/3 Ω. Not reciprocal and not symmetrical.
For the network shown below, find the Z and g parameters and show that the network is neither reciprocal nor symmetrical. [Figure: port 1 (V1, I1) with a 1 Ω shunt resistor, then a 1 Ω series resistor, then a dependent voltage source 3V1 (+ at top) as a middle shunt branch, then a 1 Ω series resistor and a 2 Ω shunt resistor at port 2 (V2, I2)]
Z-parameters: V1 = z11 I1 + z12 I2, V2 = z21 I1 + z22 I2. g-parameters (hybrid): I1 = g11 V1 + g12 I2, V2 = g21 V1 + g22 I2.
Circuit (from the figure). 1 Ω shunt at port 1; 1 Ω series resistor; a dependent voltage source 3V1 (+ at top) as the middle shunt branch, so the middle node is VM = 3V1; 1 Ω series resistor; 2 Ω shunt at port 2.
Port 1
I1 = V1/1 + (V1 − VM)/1 = V1 + V1 − 3V1 = −V1 → V1 = −I1
Port 2
The current from M into the port 2 node is (VM − V2)/1; KCL at the port 2 node gives I2 + (3V1 − V2) = V2/2:
1.5 V2 = I2 + 3V1
V2 = (2/3) I2 + 2 V1 = (2/3) I2 − 2 I1
Z-parameters
z11 = −1 Ω z12 = 0
z21 = −2 Ω z22 = 2/3 Ω
g-parameters
Use I1 = −V1 and V2 = 2V1 + (2/3) I2:
g11 = I1/V1 |I2=0 = −1 S g12 = I1/I2 |V1=0 = 0
g21 = V2/V1 |I2=0 = 2 g22 = V2/I2 |V1=0 = 2/3 Ω
(Also g11 = 1/z11, g12 = −z12/z11, g21 = z21/z11, g22 = Δz/z11, which give the same values.)
Reciprocity and symmetry
- Reciprocal needs z12 = z21: here z12 = 0 ≠ z21 = −2 (equivalently g12 = −g21 fails: 0 ≠ −2). Not reciprocal.
- Symmetrical needs z11 = z22 (together with reciprocity): −1 ≠ 2/3. Not symmetrical.
Answer: Z = [−1 0; −2 2/3] Ω, g = [−1 0; 2 2/3]; the network is neither reciprocal nor symmetrical.
✓ Answer checked by Claude Opus 5.5 (high effort)
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