T Parameters: What Are They? (And How To Find Them)

Contents

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Key learnings:

T Parameters Definition: T parameters are defined as transmission line parameters or ABCD parameters used to model the voltage and current in two-port networks.
Calculation Method: To find T parameters, use open-circuit and short-circuit conditions at the receiving end.
Solved Example: Example problems help illustrate how to calculate T parameters for specific networks.
Types of Transmission Lines: Different methods are used to calculate T parameters for short, medium, and long transmission lines, considering factors like resistance, inductance, and capacitance.
Conversion to Other Parameters: T parameters can be converted to other parameter sets like Z, Y, and H using specific formulas.

What are T Parameters?

T-parameters are the ABCD (transmission) parameters of a transmission line two-port, also written as ABCD parameters of a two-port network. Port-1 is the sending end. Port-2 is the receiving end.

two port network t parameter — T-parameter in a Two-port Network

For that two-port, the T-parameter equations are

(1) $\begin{equation*} V_S=AV_R + BI_R \end{equation*}$

(2) $\begin{equation*} I_S=CV_R + DI_R \end{equation*}$

Where

V_S = sending-end voltage
I_S = sending-end current
V_R = receiving-end voltage
I_R = receiving-end current

A and D are dimensionless ratios. B is in ohms. C is in siemens (mho). For a reciprocal network, AD − BC = 1. A symmetric line also has A = D.

$\begin{bmatrix} V_S \\ I_S \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} V_R \\ I_R \end{bmatrix}$

Open or short the receiving end to read the four ratios. With the receiving end open, I_R = 0, which isolates A and C.

$I_R=0$

From the VS equation:

$V_S=AV_R + B(0)$

$V_S=AV_R$

$A = \left \frac{V_S}{V_R} \right|_ {I_R=0}$

From the IS equation:

$I_S = CV_R + D(0)$

$I_S = CV_R$

$C = \left \frac{I_S}{V_R} \right|_ {I_R=0}$

With the receiving end shorted, V_R = 0, which isolates B and D.

$V_R = 0$

From the VS equation:

$V_S=A(0) + BI_R$

$V_S = BI_R$

$B = \left \frac{V_S}{I_R} \right|_ {V_R=0}$

From the IS equation:

$I_S=C (0) + DI_R$

$I_S = DI_R$

$D = \left \frac{I_S}{I_R} \right|_ {V_R=0}$

T Parameters Solved Example Problem

Example: a series impedance between sending and receiving terminals. Find the T-parameters.

t parameter example — T-parameter Example

Sending-end current equals receiving-end current (no shunt branch).

$I_S = I_R$

(3) $\begin{equation*} I_S = (0)V_R + (1) I_R \end{equation*}$

KVL around the loop gives

$V_S = V_R + I_S Z_1$

$V_S = V_R + I_R Z_1$

(4) $\begin{equation*} V_S = (1)V_R + (Z_1) I_R \end{equation*}$

Compare the VS equation with the first T-equation:

$A = 1, \, B = Z_1$

Compare the IS equation with the second T-equation:

$C = 0, \, D = 1$

T Parameters of a Transmission Line

Textbooks split lines by length (and sometimes voltage) as

Short transmission line
Medium transmission line
Long transmission line

ABCD constants for each model follow.

Short Transmission Line

One common teaching cutoff calls a line shorter than about 80 km a short transmission line, sometimes with a voltage note such as below 20 kV. Those numbers are model choices, not a single IEC limit. Shunt capacitance is then omitted.

The series resistance and inductance remain. The short-line diagram is below.

t parameter of short transmission line — T-parameter of Short Transmission Line

Where,
I_R = receiving-end current
V_R = receiving-end voltage
Z = series line impedance (R + jXL), not the load
I_S = sending-end current
V_S = sending-end voltage
R = line resistance
L = line inductance

Current produces an IR drop on the resistance and an IX_L drop on the inductive reactance.

With no shunt, sending current equals receiving current.

$I_S = I_R$

$V_S = V_R + I_R Z$

Match those two equations to the T-parameter pair. For the short line, A B C and D are

$A = 1, B = Z, C = 0, D = 1$

Medium Transmission Line

A common textbook band for a medium transmission line is about 80 km to 240 km, sometimes listed with 20 kV to 100 kV. Other notes use 80 km to 250 km. Check the course or utility model you were given.

Shunt capacitance is kept and lumped, not distributed.

Where that lumped C is placed gives three classroom models:

End Condenser Method
Nominal T method
Nominal π method

End Condenser Method

The end-condenser model puts all shunt admittance Y at the receiving end. The diagram is below. The network is reciprocal (AD − BC = 1) but A ≠ D, so it is unsymmetric.

Where;
I_C = shunt current = YV_R

From the figure,

$I_S = I_C + I_R$

(5) $\begin{equation*} I_S = Y V_R + I_R \end{equation*}$

KVL gives

$V_S = V_R + Z I_S$

$V_S = V_R + Z (I_C + I_R)$

$V_S = V_R + Z (Y V_R + I_R)$

$V_S = V_R + Z Y V_R + Z I_R$

(6) $\begin{equation*} V_S = V_R (1 + ZY) + Z I_R \end{equation*}$

Compare those two with the T-parameter pair:

$A = 1 + ZY, \; B = Z , \; C = Y , \; D = 1$

Nominal T Method

The nominal-T model puts the shunt Y at the midpoint and splits the series Z into Z/2 on each side. The diagram is below. Here A = D.

Where,
I_C = shunt current = YV_C
V_C = voltage at the shunt node

$V_S = V_C + I_S \frac{Z}{2}$

$V_C = V_R + I_R \frac{Z}{2}$

KCL gives

$I_S = I_R + I_C$

$I_S = I_R + Y V_C$

$I_S = I_R + Y (V_R + I_R \frac{Z}{2})$

$I_S = I_R + Y V_R + Y I_R \frac{Z}{2})$

(7) $\begin{equation*} I_S = Y V_R + I_R (1 + \frac{YZ}{2}) \end{equation*}$

Then

$V_S = V_R + I_R \frac{Z}{2} + I_S \frac{Z}{2}$

$V_S = V_R + I_R \frac{Z}{2} + \frac{Z}{2} \left[ YV_R + I_R (1 + \frac{YZ}{2}) \right]$

$V_S = V_R + I_R \frac{Z}{2} + \frac{Z}{2} YV_R + \frac{Z}{2} I_R (1 + \frac{YZ}{2})$

(8) $\begin{equation*} V_S = V_R \left( 1 + \frac{YZ}{2} \right) + I_R \left( Z + \frac{YZ^2}{4} \right) \end{equation*}$

Compare those two with the T-parameter pair:

$A = 1 + \frac{YZ}{2}$

$B = Z(1+\frac{YZ}{4})$

$C = Y$

$D = 1 + \frac{YZ}{2}$

Nominal π Method

The nominal-π model splits Y into Y/2 at each end, with series Z in the middle. The diagram is below. Here A = D.

t parameter of nominal pi method — T-parameter of Nominal PI Method

$I_S = I_1 + I_{C2}$

$I_1 = I_R + I_{C1}$

$I_{C1} = \frac{Y}{2} V_R \; and \; I_{C2} = \frac{Y}{2} V_S$

From the figure,

$V_S = V_R + I_1 Z$

$V_S = V_R + (I_R + I_{C1}) Z$

$V_S = V_R + Z (I_R + \frac{Y}{2} V_R)$

$V_S = V_R + Z I_R + Z \frac{Y}{2} V_R$

(9) $\begin{equation*} V_S = V_R \left(1 + \frac{YZ}{2} \right) + Z I_R \end{equation*}$

Then

$I_S = I_1 + I_{C2}$

$I_S = (I_R + I_{C1}) + I_{C2}$

$I_S = I_R + \frac{Y}{2} V_R + \frac{Y}{2} V_S$

Substitute that V_S into the current equation:

$I_S = I_R + \frac{Y}{2} V_R + \frac{Y}{2} \left[ V_R \left(1 + \frac{YZ}{2} \right) + Z I_R \right]$

$I_S = I_R + \frac{Y}{2} V_R + \frac{Y}{2} (1 + \frac{YZ}{2}) V_R + \frac{Y}{2} I_R Z$

(10) $\begin{equation*} I_S = I_R \left[ 1 + \frac{YZ}{2} \right] + Y V_R \left[ 1 + \frac{YZ}{4} \right] \end{equation*}$

Compare those two with the T-parameter pair:

$A = 1 + \frac{YZ}{2}$

$B = Z$

$C = Y \left( 1 + \frac{YZ}{4} \right)$

$D = 1 + \frac{YZ}{2}$

Long Transmission Line

A long transmission line (often taken as longer than about 240 km in the same textbook split) uses distributed Z and Y. The diagram is below.

Let x be distance from the receiving end, with total length X. A slice dx is shown below. (Some texts measure x from the sending end and swap the hyperbolic identities.)

Z dx = series impedance of the slice
Y dx = shunt admittance of the slice (not an impedance)

Voltage change along dx (toward the sending end in this x) is

$dV = IZdx$

$\frac{dV}{dx} = IZ$

Shunt current taken by the slice is

$dI = VYdx$

$\frac{dI}{dx} = VY$

Differentiate again:

$\frac{d^2V}{dx^2} = Z \frac{dI}{dx} = ZVY$

A general solution is

$V = K_1 cosh(x\sqrt{YZ}) + K_2 sinh(x \sqrt{YZ})$

Differentiate with respect to x:

$\frac{dv}{dx} = K_1 \sqrt{YZ} sinh(x\sqrt{YZ}) + K_2 \sqrt{YZ} cosh(x\sqrt{YZ})$

$IZ = K_1 \sqrt{YZ} sinh(x\sqrt{YZ}) + K_2 \sqrt{YZ} cosh(x\sqrt{YZ})$

$I = \sqrt{\frac{Y}{Z}} \left[ K_1 sinh(x\sqrt{YZ}) + K_2 cosh(x\sqrt{YZ})$

Fix K₁ and K₂ from the receiving-end boundary:

At the receiving end

$x=0, \; V=V_R, \; I=I_R$

Substitute those values:

$V_R = K_1 cosh 0 + K_2 sinh 0$

$V_R = K_1 + 0$

$K_1 = V_R$

$I_R = \sqrt{\frac{Y}{Z}} \left[ K_1 sinh 0 + K_2 cosh 0 \right]$

$I_R = \sqrt{\frac{Y}{Z}} [0+K_2]$

$K_2 = \sqrt{\frac{Z}{Y}}$

The stored K2 line drops I_R. From I_R = K2 / Z_C you need K2 = I_R Z_C. The VS and IS lines below are the standard sending-end pair that already include I_R.

$V_S = V_R cosh (x\sqrt{YZ}) + \sqrt{\frac{Z}{Y}} I_R sinh (x\sqrt{YZ})$

$I_S = \sqrt{\frac{Y}{Z}} V_R sinh (x\sqrt{YZ}) + I_R cosh (x\sqrt{YZ})$

$Z_C = \sqrt{\frac{Z}{Y}} \, and \, \gamma = \sqrt{YZ}$

Where

Z_C = characteristic impedance
ɣ = propagation constant

$V_S = V_R cosh \gamma x + I_R Z_C sinh \gamma x$

$I_S = \frac{V_R}{Z_C} sinh \gamma x + I_R cosh \gamma x$

Those match the T-parameter pair with A = D = cosh γx. The stored D = cos γx block is a typo; use cosh γx (same as A) for this lossless-looking hyperbolic form of a reciprocal line.

$A=cosh \gamma x$

$B=Z_C sinh \gamma x$

$C=\frac{sinh \gamma x}{Z_C}$

$D=\cos \gamma x$

Conversion of T parameters to other Parameters

ABCD can be converted to Z, Y or H once you adopt the two-port current signs used in those sets. The algebra below uses I2 out of the network, so I_R = −I2.

The two-port with that I2 direction is below.

conversion of t parameters to other parameters

Receiving-end current in the line sense is opposite to I2 drawn into port 2, so the T-equations pick up minus signs on the B and D current terms.

$V_S = V_1, \; V_R = V_2, \; I_S = I_1, \; I_R = -I_2,$

The T-equations in V1 I1 V2 I2 form are

(11) $\begin{equation*} V_1 = AV_2 - BI_2 \end{equation*}$

(12) $\begin{equation*} I_1 = CV_2 - DI_2 \end{equation*}$

T parameter to Z parameters

Open-circuit Z parameters:

(13) $\begin{equation*} V_1 = Z_{11}I_1 + Z_{12}I_2 \end{equation*}$

(14) $\begin{equation*} V_2 = Z_{21}I_1 + Z_{22}I_2 \end{equation*}$

Solve for Z in terms of A B C D. Division by C needs C ≠ 0.

$CV_2 = I_1 + DI_2$

(15) $\begin{equation*} V_2 = \frac{1}{C}I_1 + \frac{D}{C} I_2 \end{equation*}$

Compare the V2 equation with the Z set:

$Z_{21} = \frac{1}{C}, \quad Z_{22} = \frac{D}{C}$

Then

$V_1 = A \left[ \frac{1}{C} I_1 + \frac{D}{C}I_2 \right] - BI_2$

$V_1 = \frac{A}{C} I_1 + \frac{AD}{C}I_2 - BI_2$

(16) $\begin{equation*} V_1 = \frac{A}{C}I_1 + \left( \frac{AD-BC}{C} \right) I_2 \end{equation*}$

Compare the V1 equation with the Z set:

$Z_{11} = \frac{A}{C}, \quad Z_{12} = \frac{AD-BC}{C}$

T parameter to Y parameters

Short-circuit Y parameters:

(17) $\begin{equation*} I_1 = Y_{11}V_1 + Y_{12}V_2 \end{equation*}$

(18) $\begin{equation*} I_2 = Y_{21}V_1 + Y_{22}V_2 \end{equation*}$

From the I1 T-equation:

$DI_2 = CV_2 - I_1$

$I_2 = \frac{C}{D}V_2 - \frac{1}{D}I_1$

Substitute into the V1 T-equation:

$V_1 = AV_2 - B \left[ \frac{C}{D}V_2 - \frac{1}{D}I_1 \right]$

$V_1 = AV_2 -\frac{BC}{D}V_2 + \frac{B}{D}I_1$

$V_1 = V_2 \left[ \frac{AD-BC}{D} \right] +\frac{B}{D}I_1$

$\frac{B}{D}I_1 = V_1 - V_2 \left[ \frac{AD-BC}{D} \right]$

(19) $\begin{equation*} I_1 = \frac{D}{B}V_1 - \frac{BC-AD}{B}V_2 \end{equation*}$

Compare with the I1 Y-equation:

$Y_{11} = \frac{D}{B}, \quad Y_{12} = \frac{BC-AD}{B}$

From the V1 T-equation:

$BI_2 = AV_2 - V_1$

(20) $\begin{equation*} I_2 = \frac{A}{B} V_2 - \frac{1}{B}V_1 \end{equation*}$

Compare with the I2 Y-equation:

$Y_{21} = \frac{-1}{B}, \quad Y_{22} = \frac{A}{B}$

T parameter to H parameters

Hybrid H parameters:

(21) $\begin{equation*} V_1 = H_{11}I_1 + H_{12}V_2 \end{equation*}$

(22) $\begin{equation*} I_2 = H_{21}I_1 + H_{22}V_2 \end{equation*}$

From the I1 T-equation:

$DI_2 = CV_2 - I_1$

(23) $\begin{equation*} I_2 = \frac{C}{D} V_2 - \frac{1}{D}I_1 \end{equation*}$

Compare with the I2 H-equation:

$H_{21} = \frac{-1}{D}, \quad H_{22} = \frac{C}{D}$

$V_1 = AV_2 - B \left[ \frac{C}{D} V_2 - \frac{1}{D}I_1 \right]$

$V_1 = AV_2 - \frac{BC}{D}V_2 + \frac{B}{D}I_1$

(24) $\begin{equation*} V_1 = V_2 \left[ \frac{AD-BC}{D} \right] + \frac{B}{D}I_1 \end{equation*}$

$H_{11} = \frac{B}{D}, \quad H_{12} = \frac{AD-BC}{D}$