T Parameters: What Are They? (Examples Problems And How to Convert T Parameters to other Parameters)

what are t parameters

What are T Parameters?

T parameters are defined as transmission line parameters or ABCD parameters. In a two-port network, port-1 is considered as sending end and port-2 is considered as receiving end. In the network diagram below, port-1 terminals represent the input (sending) port. Similarly, port-2 terminals represent the output (receiving) port.

two port network t parameter

T-parameter in a Two-port Network

For the above two-port network, equations of T-parameters are;

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

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

Where;

VS = Sending end voltage
IS = Sending end current
VR = Receiving end voltage
IR = Receiving end current

These parameters are used to make mathematical modeling of a transmission line. Parameter A and D are unitless. The unit of parameter B and C is ohm and mho, respectively.

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

To find the value of T-parameters, we need to open and short circuit the receiving end. When the receiving end is open-circuited, receiving end current IR is zero. Put this value in the equations and we get the value of A and C parameters.

$I_R=0$

open circuit condition

From equation-1;

$V_S=AV_R + B(0)$

$V_S=AV_R$

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

From equation-2;

$I_S = CV_R + D(0)$

$I_S = CV_R$

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

When the receiving end is short-circuited, the voltage across the receiving terminals VR is zero. By putting this value in the equation, we can get the values of B and D parameters.

$V_R = 0$

short circuit condition

From equation-1;

$V_S=A(0) + BI_R$

$V_S = BI_R$

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

From equation-2;

$I_S=C (0) + DI_R$

$I_S = DI_R$

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

T Parameters Solved Example Problem

Consider an impedance is connected between the sending end and receiving end terminals as shown in the below figure. Find T-parameters of given network.

t parameter example

T-parameter Example

Here, the sending end current is the same as the receiving end current.

$I_S = I_R$

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

Now, we apply KVL to the network,

$V_S = V_R + I_S Z_1$

$V_S = V_R + I_R Z_1$

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

Compare equation-1 and 4;

$A = 1, \, B = Z_1$

Compare equation-2 and 3;

$C = 0, \, D = 1$

T Parameters of a Transmission Line

According to the length of line, transmission lines are classified as;

Short transmission line
Medium transmission line
Long transmission line

Now, we find T-parameters for all types of transmission lines.

Short Transmission Line

The transmission line having a length of less than 80km and voltage level less than 20kV is considered a short transmission line. Due to the small length and lower voltage level, the capacitance of the line is neglected.

Therefore, we are considering only resistance and inductance while modeling a short transmission line. The graphical representation of the short transmission line is as shown below figure.

t parameter of short transmission line

T-parameter of Short Transmission Line

Where,
IR = Receiving end current
VR = Receiving end voltage
Z = Load Impedance
IS = Sending end current
VS = Sending end voltage
R = Line resistance
L = Line inductance

When current flows through the transmission line, IR drop occurs at line resistance and IXL drop occurs at inductive reactance.

From the above network, the sending end current is the same as receiving end current.

$I_S = I_R$

$V_S = V_R + I_R Z$

Now, compare these equations with the equations of the T-parameters (equation 1 & 2). And we get values of A, B, C, and D parameters for a short transmission line.

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

Medium Transmission Line

The transmission line having a length of 80km to 240km and voltage level is 20kV to 100kV is considered as a medium transmission line.

In the case of a medium transmission line, we cannot neglect the capacitance. We must consider the capacitance while modeling a medium transmission line.

According to the placement of capacitance, the medium transmission lines are classified into three methods;

End Condenser Method
Nominal T method
Nominal π method

End Condenser Method

In this method, the capacitance of the line is assumed to be lumped at end of a transmission line. The graphical representation of End condenser method is shown below figure.

t parameter of end condenser method

T-parameter of End Condenser Method

Where;
IC = Capacitor current = YVR

From the above figure,

$I_S = I_C + I_R$

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

By KVL, we can write;

$V_S = V_R + Z I_S$

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

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

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

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

Now, compare equations-5 and 6 with equations of T parameters;

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

Nominal T Method

In this method, the capacitance of the line is placed at the mid-point of the transmission line. The graphical representation of the Nominal T method is as shown below figure.

t parameter of nominal t method

T-parameter of Nominal T Method

Where,
IC = Capacitor current = YVC
VC = Capacitor voltage

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

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

From KCL;

$I_S = I_R + I_C$

$I_S = I_R + Y V_C$

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

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

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

Now,

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

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

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

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

Now, compare equations-7 and 8 with equations of T parameter and we get,

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

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

$C = Y$

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

Nominal π Method

In this method, the capacitance of the transmission line is divided into halves. One half is placed at sending end and the second half is placed at receiving end. Graphical representation of nominal π method is as shown below figure.

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 above figure, we can write;

$V_S = V_R + I_1 Z$

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

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

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

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

Now,

$I_S = I_1 + I_{C2}$

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

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

Put the value of VS in this 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) $\begin{equation*} I_S = I_R \left[ 1 + \frac{YZ}{2} \right] + Y V_R \left[ 1 + \frac{YZ}{4} \right] \end{equation*}$

By comparing equations-9 and 10 with equations of T parameters, we get;

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

$B = Z$

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

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

Long Transmission Line

The long transmission line is modeled as a distributed network. It cannot be assumed as a lumped network. The distributed model of a long transmission line is as shown below figure.

t parameter of long transmission line

T-parameter of Long Transmission Line

The length of a line is X km. To analyze the transmission line, we consider a small part (dx) of the line. And it is as shown below figure.

long transmission line t parameter

Zdx = series impedance
Ydx = shunt impedance

The voltage increases over the length increases. So, the rise of voltage is;

$dV = IZdx$

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

Similarly, current drawn by element is;

$dI = VYdx$

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

Differentiating above equations;

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

The general solution of above equation is;

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

Now, differentiate this equation with respect to X,

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

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

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

Now, we need to find constants K1 and K2;

For that assume;

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

Putting these values in above equations;

$V_R = K_1 cosh 0 + K_2 sinh 0$

$V_R = K_1 + 0$

$K_1 = V_R$

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

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

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

Therefore,

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

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

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

Where,

ZC = Characteristic Impedance
ɣ = Propagation Constant

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

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

Compare these equations with the equations of T-parameters;

$A=cosh \gamma x$

$B=Z_C sinh \gamma x$

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

$D=\cos \gamma x$

Conversion of T parameters to other Parameters

We can find other parameters from the equations of T parameters. For that, we need to find a set of equations of other parameters in terms of T parameters.

Consider the generalized two-port network as shown below figure.

conversion of t parameters to other parameters

In this figure, the direction of receiving end current is changed. Therefore, we consider few changes in equations of T parameters.

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

Equations of T parameters is;

(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

The following set of equations represents Z parameters.

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

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

Now, we will find the equations of Z parameters in terms of T parameters.

$CV_2 = I_1 + DI_2$

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

Now compare equation-14 with equation-15

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

Now,

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

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

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

Compare equation-13 with equation-16;

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

T parameter to Y parameters

The set of equations of Y parameters is;

(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 equation-12;

$DI_2 = CV_2 - I_1$

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

Put this value in equation-11;

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

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

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

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

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

Compare this equation with equation-17;

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

From equation-11;

$BI_2 = AV_2 - V_1$

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

Compare this equation with equation-18;

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

T parameter to H parameters

The set of equations of H parameters is;

(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 equation-12;

$DI_2 = CV_2 - I_1$

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

Compare this equation with equation-22;

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

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

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

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

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

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