ABCD transmission matrices and cascading

Choose V₁ and V₂ from positive to negative terminals, while I₁ and I₂ enter the positive terminals. The physical current traveling to the right at the output equals −I₂. Define [V₁,I₁]ᵀ=T[V₂,−I₂]ᵀ with T=[A B;C D]. A and D are dimensionless, B has units Ω, and C has units S. This definition must be shown before any conversion formula.

The two-port convention is a definition of measurement coordinates, not a statement that actual power always enters at both ends. On an energized passive transmission chain, I₂ is often negative because the device supplies the load. The minus sign belongs to the state vector even in that case.

Assume Z₂₁≠0, and start from V₁=Z₁₁I₁+Z₁₂I₂ and V₂=Z₂₁I₁+Z₂₂I₂. Solve I₁=(V₂−Z₂₂I₂)/Z₂₁ and substitute into V₁. Comparing coefficients of V₂ and −I₂ gives A=Z₁₁/Z₂₁, B=(Z₁₁Z₂₂−Z₁₂Z₂₁)/Z₂₁, C=1/Z₂₁, and D=Z₂₂/Z₂₁. The condition Z₂₁≠0 is substantive: the conversion can fail even when the physical network exists.

From the second impedance equation, solve for the input current without reversing either port reference. Inserting that expression into the first equation yields the ABCD coefficients and the identity AD−BC=Z₁₂/Z₂₁; note that the pivot condition Z₂₁≠0 must hold.

Across a series impedance Z_s, the through-current is the same and the input voltage differs from output voltage by the series drop: V₁=V₂+Z_s I_through and I₁=I_through=−I₂. Therefore T_series=[1 Z_s;0 1]. An ideal pure resistor uses Z_s=R; an inductor uses Z_s=jωL with AoN's +jωt phasor convention.

Derive the series block from Kirchhoff's voltage law using the current flowing in the series path. The output current reference is entering the box from the right, so that current equals −I₂. All four matrix entries follow directly.

For a shunt admittance Y_p the voltage stays V₁=V₂. Kirchhoff's current law says I₁=I_through+Y_pV₂, with I_through=−I₂, so T_shunt=[1 0;Y_p 1]. A capacitor uses Y_p=jωC. The shunt element is referenced between signal and return conductors, and its current arrow must match the passive sign convention.

For a shunt element connected across the two conductors, input and output voltages coincide, but the input current contains both the continuing load current and shunt branch current. Derive the lower-left ABCD term from Kirchhoff's current law.

Connect two blocks in order, with block 1 followed by block 2. The output vector of block 1 is exactly the input vector of block 2 after reversing their local entering-current convention at the connection. It follows that T_total=T₁T₂. Matrix multiplication is generally noncommutative: a series element followed by a shunt has T=[1+Z_sY_p, Z_s;Y_p,1], but the reversed order has T=[1,Z_s;Y_p,1+Y_pZ_s].

Multiplication order preserves topology. Apply the second block to its own output state first, then the first block to the resulting intermediate state. The diagonal entries differ when the physical order of a series and shunt element is reversed.

The determinant from the Z conversion is AD−BC=Z₁₂/Z₂₁. For a reciprocal two-port with Z₁₂=Z₂₁, AD−BC=1. Thus determinant one is a reciprocity signature under these conventions and regular conversion assumptions; it does not by itself prove passivity or that the network is lossless. Both series and shunt blocks have unit determinant, so their cascade also does.

Substituting the Z-to-ABCD formulas into AD−BC cancels the products Z₁₁Z₂₂ and leaves Z₁₂/Z₂₁. Reciprocal networks therefore have unit determinant when this representation exists, but a positive resistor remains lossy despite the same determinant.

A uniform transmission-line segment of length l has ABCD entries A=D=cosh(γl), B=Z₀sinh(γl), C=sinh(γl)/Z₀. In the lossless limit γ=jβ, so A=D=cos(βl), B=jZ₀sin(βl), and C=j sin(βl)/Z₀. The line field or circuit impedance Z₀ must correspond to its actual mode; it is not automatically the reference impedance of a VNA.

The transmission-line chain form follows from the two traveling-wave solutions and the telegrapher equations. For a line with γ=α+jβ, finite losses lead to hyperbolic factors. Only in the lossless case does γ=jβ and the cosine/sine expression follow.

Take Z_s=10 Ω and Y_p=0.02 S at a fixed frequency. The series-then-shunt matrix is [1.2,10 Ω;0.02 S,1], with determinant 1. For an output V₂=5 V and through-current −I₂=0.1 A, the input has V₁=1.2×5+10×0.1=7 V and I₁=0.02×5+0.1=0.2 A. Verify both KVL and KCL directly instead of relying on the matrix alone. Near singular network representations, matrix-to-matrix conversions need numerical care.

Interpret the numerical example in the physical sequence: 0.2 A traverses the input series resistor, the shunt branch takes Y_pV₂=0.1 A, and the remaining through-current is 0.1 A. The series drop is 2 V, so V₁=7 V. This independent KCL/KVL check guards against multiplying matrices in the wrong order.

1. Derive the ABCD convention from both port currents entering.
2. Derive A, B, C and D from the Z equations.
3. Write the chain matrix for a series resistor R=25 Ω.
4. Write the chain matrix for a shunt capacitor C with +jωt convention.
5. Multiply a series Z followed by a shunt Y and reverse the order.
6. For Z=[4 1;1 3] Ω, calculate ABCD and its determinant.
7. Show that the series and shunt matrices have determinant one.
8. Calculate the ABCD of a lossless line for βl=π/2.
9. For Z_s=10 Ω and Y_p=0.02 S, compute V₁,I₁ given V₂=5 V, −I₂=0.1 A.
10. Does AD−BC=1 prove the two-port is lossless? Explain.

Next we will distinguish reciprocity from passivity, derive the power-based matrix constraints and connect terminal-network models to measured scattering parameters with explicit real reference impedances.