Two-port references, currents and power
Define real electrical ports, keep both currents entering positive voltage terminals, derive absorbed complex power and a reciprocal π-network Y matrix with signed loading and independent power checks.
- Explain the equal/opposite terminal-current condition that defines an electrical port.
- Choose voltage polarities consistently at both terminal pairs.
- Define both port currents entering the positive terminals and distinguish the output-current alternative.
- Derive absorbed time-average power for AoN peak phasors.
- State when a linear two-port supports Y and/or Z matrix representations.
- Derive a reciprocal π-network admittance matrix using signed KCL.
- Distinguish reciprocity from physical port-interchange symmetry.
- Prove passivity through a nonnegative sum of dissipated branch powers.
- Solve an external load using I₂=−Y_LV₂.
- Audit the resulting voltages, currents, matrices and power numerically.
Every microwave network model begins with correctly signed port voltages and currents. This first RF-03 lesson derives the two-port convention from circuit topology and energy flow before introducing matrix coefficients. It then checks every sign with a simple reciprocal resistive pi network. The voltage polarity and current-arrow locations must eventually be drawn with canonical MAJAL primitives, never with an arbitrary SVG.
0103.01.1 — What counts as a port?
A port consists of two externally accessible terminals such that the current entering one terminal equals the current leaving the other, so one scalar port current describes the pair. This is a physical port-current constraint, not a statement that every arbitrary pair of nearby wires forms a valid port.
A two-port has two such terminal pairs; the internal structure may contain many components. At a given frequency, its terminal relation may be represented by a linear map if its underlying device obeys the specified linearity and source conditions. Voltages and currents at one port must obey the same geometrical reference convention in all subsequent derivations.
0203.01.2 — Choose voltage polarity and entering-current references
Let V₁ be the electric potential of port 1's upper positive terminal minus its lower reference terminal. Define V₂ analogously. Define I₁ and I₂ as entering the network at their respective positive terminals. Consequently each current arrow points into the two-port even though the arrows may point toward one another from opposite sides of a drawing.
These reference directions are arbitrary but must be held fixed. A negative I₂ is a physical current leaving the network at port 2, not a sign mistake. The more familiar left-to-right output current is I_out=−I₂ under this convention; this reversal must be made explicit before writing chain/ABCD relations.
Figure 1.1 — Signed electrical references of a two-port
Both I₁ and I₂ are defined ENTERING the network at their positive terminals; each voltage arrow points from minus to plus.
I₁ enters the left port pointing right, while I₂ enters the right port pointing left. V₁ and V₂ are positive on the upper terminals. The conventional output current toward an attached load equals −I₂. These arrows define references, not actual solved current flow.
0303.01.3 — Absorbed power and peak phasors
The passive sign convention associates positive absorbed instantaneous power with current entering the positive-voltage terminal. For the two-port p_abs=v₁i₁+v₂i₂. Under sinusoidal peak phasors at the same angular frequency, the average absorbed power is one half the real part of V₁I₁*+V₂I₂*.
Complex power is S_abs=(1/2)(V₁I₁*+V₂I₂*); its imaginary part indicates net reactive exchange under this phasor convention and is not dissipated average power. Using RMS phasors instead would remove the factor 1/2. A sign reversal at port 2 without updating the power formula can falsely label a passive circuit active.
0403.01.4 — From linearity to two-port matrices
A linear source-free two-port at one angular frequency obeys a linear relation between its terminal phasors. For an admittance description, the port voltages are the independent variables and the currents result. For an impedance description, the currents are independent and voltages result. Neither description is guaranteed to exist for every ideal network or test configuration.
Write I=YV only after defining the sign of both currents. Its entries have siemens units; the corresponding V=ZI representation, when it exists, has ohm units. Matrix inversion Y=Z⁻¹ is valid only when the relevant matrix is invertible. The next chapter will derive individual matrix entries from open/short measurement conditions.
0503.01.5 — Derive a pi network directly by KCL
Take a passive pi network with admittance Y₁ from port 1's positive node to the common reference, Y₂ analogously at port 2, and a bridge admittance Y_b between the positive nodes. The current entering port 1 is the sum of its shunt current Y₁V₁ and bridge current Y_b(V₁−V₂).
The current entering port 2 is Y₂V₂ plus Y_b(V₂−V₁). Writing both expressions in matrix form gives off-diagonal entries −Y_b. That negative sign comes from the entering-current convention and voltage difference across the bridge; it is not optional notation.
0603.01.6 — Reciprocity, geometric symmetry and passivity
The pi network with ordinary reciprocal passive elements has Y₁₂=Y₂₁=−Y_b. This remains true even when Y₁≠Y₂, so reciprocity is not the same as invariance when the two ports are swapped. Geometric port symmetry additionally requires the corresponding self-entries to match, and the pi example has that property only when Y₁=Y₂.
For nonnegative real conductances G₁,G₂,G_b, the average absorbed power is (1/2)[G₁|V₁|²+G₂|V₂|²+G_b|V₁−V₂|²]≥0. This direct sum-of-squares proves passivity at the specified frequency. For an arbitrary linear admittance matrix, the Hermitian part (Y+Y†)/2 must be positive semidefinite for passive sinusoidal power; this single-frequency condition is not the whole positive-real analytic criterion.
0703.01.7 — Output loading and the meaning of negative I₂
Connect a load admittance Y_L across port 2. Let its current I_L=Y_LV₂ leave the network into the external load. Kirchhoff at the interface demands I₂+I_L=0, hence I₂=−Y_LV₂. No extra sign may be inserted on the left port.
Substitute the pi network expression I₂=−Y_bV₁+(Y₂+Y_b)V₂ into this boundary equation. The loaded voltage gain follows immediately: V₂/V₁=Y_b/(Y₂+Y_b+Y_L). The input current is I₁=(Y₁+Y_b)V₁−Y_bV₂, which yields the loaded input admittance.
0803.01.8 — Numerical audit before measurement
Choose a purely resistive network: R₁=100Ω, R₂=200Ω and R_b=50Ω. Thus Y₁=0.010S, Y₂=0.005S and Y_b=0.020S. The two-port admittance matrix is [[0.030,−0.020],[−0.020,0.025]]S. The off-diagonal entries are equal, while the two diagonal entries differ.
With real peak voltages V₁=1V and V₂=0.4V, the entering currents are I₁=22mA and I₂=−10mA. The absorbed average power is 9mW, exactly equal to the sum of the three individual resistor losses. Negative I₂ is expected: the network delivers power through port 2 while still dissipating net positive power. A calibrated 50Ω VNA does not directly measure these terminal Z/Y entries without converting the measured wave variables.
Practice — signed two-port variables
- Explain the terminal-current condition that makes two conductor terminals a physical port.
- Define both port voltages and currents with the passive sign convention, and express the output current leaving the network.
- Using peak phasors, derive the formula for average absorbed two-port power and distinguish it from the RMS convention.
- Write the I=YV and V=ZI relations and state precisely when the matrices can be inverted.
- Derive the admittance matrix of a general pi network by Kirchhoff's current law with both currents entering.
- Show that the pi network can be reciprocal without being geometrically symmetric.
- Prove nonnegative dissipated power for real G₁,G₂,G_b≥0 and arbitrary complex port voltages.
- Connect an external load Y_L to port 2 and derive V₂/V₁, identifying why I₂ is negative.
- For R₁=100Ω, R₂=200Ω, R_b=50Ω, V₁=1V and V₂=0.4V, compute currents and power by two independent methods.
- Find the Z matrix of the given pi network and its loaded input resistance for R_L=100Ω.
Worked solutions
1. A valid port satisfies the port-current condition: if I_{k,+} enters the positive terminal and I_{k,−} enters the negative terminal, their sum is zero. Therefore one current describes that pair. KCL for the whole two-port does not by itself guarantee an arbitrary pair of internal wires is a valid externally defined port.
2. Define V₁=φ₁+−φ₁− and V₂=φ₂+−φ₂−. Both I₁ and I₂ enter the network at the positive terminals. If a diagram instead labels output current exiting toward the load, it is I_out=−I₂. Changing this arrow without changing the algebra reverses power signs and ABCD entries.
3. Instantaneous absorbed power is p=v₁i₁+v₂i₂. For peak phasors, averaging Re(Ve^{jωt})Re(Ie^{jωt}) over a cycle gives (1/2)Re(VI*), so P=(1/2)Re(V₁I₁*+V₂I₂*). RMS phasors absorb the factor 1/√2 in both voltage and current, giving P=Re(V_rms I_rms*).
4. A linear admittance description is [I₁,I₂]^T=Y[V₁,V₂]^T, with Y in siemens; the impedance description is [V₁,V₂]^T=Z[I₁,I₂]^T, with Z in ohms. Only if det(Y)≠0 can Z=Y⁻¹ be used. Singular ideal networks may not admit both representations.
5. At node 1, bridge current is Y_b(V₁−V₂), so I₁=(Y₁+Y_b)V₁−Y_bV₂. At node 2, bridge current is Y_b(V₂−V₁), so I₂=−Y_bV₁+(Y₂+Y_b)V₂. Therefore Y₁₂=Y₂₁=−Y_b by direct KCL.
6. Y₁₂=Y₂₁ follows from the one reciprocal bridge, independently of Y₁ and Y₂. But swapping ports exchanges the diagonals Y₁+Y_b and Y₂+Y_b, which are unequal when Y₁≠Y₂. Reciprocity therefore does not imply port-exchange symmetry.
7. Expand P=(1/2)Re(V₁I₁*+V₂I₂*) with the pi-network currents and real nonnegative G values. Cross terms combine into G_b|V₁−V₂|². Thus P=(1/2)[G₁|V₁|²+G₂|V₂|²+G_b|V₁−V₂|²]≥0 for arbitrary complex voltages.
8. The external load draws I_L=Y_LV₂ out of the network. Since the two currents flow into opposite sides of the same connection, I₂=−I_L=−Y_LV₂. Substitute −Y_bV₁+(Y₂+Y_b)V₂=−Y_LV₂ to obtain V₂/V₁=Y_b/(Y₂+Y_b+Y_L).
9. Conductances are 0.010, 0.005 and 0.020 S. Then I₁=0.030(1)−0.020(0.4)=0.022 A, and I₂=−0.020(1)+0.025(0.4)=−0.010 A. Port power is P=(1/2)(0.022−0.004)=0.009 W. Resistor power is (1/2)[1²/100+0.4²/200+(1−0.4)²/50]=(1/2)(0.010+0.0008+0.0072)=0.009 W.
10. With Y=[[0.030,−0.020],[−0.020,0.025]] S, detY=0.030·0.025−0.020²=0.00035 S². Thus Z=Y⁻¹=[[71.429,57.143],[57.143,85.714]]Ω. For R_L=100Ω, Y_L=0.010 S and V₂/V₁=0.020/(0.005+0.020+0.010)=4/7. Then Y_in=0.030−0.020(4/7)=0.0185714 S, giving R_in≈53.846Ω.
The following RF-03 chapter will derive Z and Y parameters one by one from ideal open- and short-circuit constraints, then study singular cases and reciprocity.