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Equivalent-π model of a long line

Short answer

An equivalent-π model is a lumped π circuit that exactly matches the long-line distributed-parameter ABCD constants. Series Z' = Z (sinh γl)/(γl) and each shunt Y'/2 = (Y/2)(tanh(γl/2))/(γl/2), where Z is total series impedance and Y total shunt admittance.

What is an equivalent π-model of a long transmission line? Derive expressions for parameters of this circuit in terms of line parameters.

An equivalent-Π model of a long line is a lumped Π circuit (series Z′Z', shunt Y′/2Y'/2 at each end) whose ABCD constants are exactly equal to those of the long line. It gives exact terminal results while being as simple as a nominal-Π circuit.

 IS          Z'                 IR
 o-->--+---[ Z' ]----+-->--o
       |             |
 VS  [Y'/2]       [Y'/2]      VR
       |             |
 o-----+-------------+-----o

ABCD of the Π circuit and of the long line

For any Π circuit:

A=D=1+Z′Y′2,B=Z′,C=Y′(1+Z′Y′4)A = D = 1 + \frac{Z'Y'}{2}, \quad B = Z', \quad C = Y'\left(1 + \frac{Z'Y'}{4}\right)

For the long line:

A=D=cosh⁡γl,B=Zcsinh⁡γl,C=sinh⁡γlZcA = D = \cosh\gamma l, \quad B = Z_c\sinh\gamma l, \quad C = \frac{\sinh\gamma l}{Z_c}

Series arm Z′Z'

Equating BB:

Z′=Zcsinh⁡γl=zysinh⁡γlZ' = Z_c\sinh\gamma l = \sqrt{\frac{z}{y}}\sinh\gamma l

Multiply and divide by lzyl\sqrt{zy} (note Z=zlZ = zl, γl=lzy\gamma l = l\sqrt{zy}):

Z′=zl sinh⁡γllzy=Z sinh⁡γlγlZ' = zl\,\frac{\sinh\gamma l}{l\sqrt{zy}} = Z\,\frac{\sinh\gamma l}{\gamma l}

Shunt arm Y′/2Y'/2

Equating AA:

1+Z′Y′2=cosh⁡γlY′2=cosh⁡γl−1Zcsinh⁡γl=1Zctanh⁡γl2\begin{aligned} 1 + \frac{Z'Y'}{2} &= \cosh\gamma l \\ \frac{Y'}{2} &= \frac{\cosh\gamma l - 1}{Z_c\sinh\gamma l} = \frac{1}{Z_c}\tanh\frac{\gamma l}{2} \end{aligned}

using cosh⁡γl−1=2sinh⁡2γl2\cosh\gamma l - 1 = 2\sinh^2\frac{\gamma l}{2} and sinh⁡γl=2sinh⁡γl2cosh⁡γl2\sinh\gamma l = 2\sinh\frac{\gamma l}{2}\cosh\frac{\gamma l}{2}. Writing 1Zc=y/z\frac{1}{Z_c} = \sqrt{y/z} and multiplying and dividing by γl2\frac{\gamma l}{2}:

Y′2=Y2 tanh⁡(γl/2)γl/2\frac{Y'}{2} = \frac{Y}{2}\,\frac{\tanh(\gamma l/2)}{\gamma l/2}

(The CC equation is then automatically satisfied, because AD−BC=1AD - BC = 1.)

Summary

ElementNominal-ΠEquivalent-Π
SeriesZZZsinh⁡γlγlZ\frac{\sinh\gamma l}{\gamma l}
Each shuntY2\frac{Y}{2}Y2tanh⁡(γl/2)γl/2\frac{Y}{2}\frac{\tanh(\gamma l/2)}{\gamma l/2}

The correction factors sinh⁡γlγl\frac{\sinh\gamma l}{\gamma l} and tanh⁡(γl/2)γl/2\frac{\tanh(\gamma l/2)}{\gamma l/2} tend to 1 as γl→0\gamma l \to 0, so for shorter lines the equivalent Π reduces to the nominal Π. For long lines, Z′Z' is slightly less than ZZ and Y′Y' slightly more than YY.

✓ Answer checked by Claude Opus 5.5 (high effort)

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