Question

Consider the parameter of the long line given by $\mathbf{cosh}\mathit{\varphi }$, where$\mathit{\theta }\mathbf{=}\sqrt{\mathrm{ZY}}$ . With $\mathbf{x}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{=}{\mathbf{x}}_1\mathbf{+}{\mathbf{jx}}_2$ and $\mathbf{A}\mathbf{=}{\mathbf{A}}_1\mathbf{+}{\mathbf{jA}}_2$, show that ${\mathbf{x}}_1$ and ${\mathbf{x}}_2$ satisfy the following:

role="math" localid="1655900314891" ${\mathbf{x}}_1^2-{\mathbf{x}}_2^2-\mathbf{2}\left({\mathbf{A}}_1{\mathbf{x}}_1-{\mathbf{A}}_2{\mathbf{x}}_2\right)\mathbf{+}\mathbf{1}=\mathbf{0}$

and role="math" localid="1655900331992" ${\mathbf{x}}_1{\mathbf{x}}_2-\left({\mathbf{A}}_2{\mathbf{x}}_1\mathbf{+}{\mathbf{A}}_1{\mathbf{x}}_2\right)=\mathbf{0}$.

Short answer

The value of $x_1$ satisfy the relation.$x_1^2-x_2^2-2\left({\text{A}}_1{x}_1-{\text{A}}_2{x}_2\right)+1=0$

The value of $x_2$ satisfy the relation .$x_1{x}_2-\left({\text{A}}_2{x}_1+A_1{x}_2\right)=0$

Step-by-step solution

Write the given data from the question.

Consider the following data for a long transmission line$A=\cosh \theta$.

Consider the following data $x=e^{-\theta }=x_1+j{x}_2$ and$A=A_{\text{1}}+j{A}_{\text{2}}$.

Determine the formula to find  x1 and  x2of long transmission line.

Write the formula ofgiven expression ${\mathbf{x}}_1$ and${\mathbf{x}}_2$.

localid="1655900669095" $\mathbf{A}=\mathbf{cosh}\theta$ …… (1)

Here, $\theta =\sqrt{\mathrm{ZY}}$

Here, localid="1655900748673" $\mathbf{Z}$ is total impedance and $\mathbf{Y}$is total admittance.

Step 3:Determine the expression of  x1 and x2 of long transmission line.

Determine the given expression to find $x_1$ and $x_2$of long transmission line.

Substitute $\frac{e^{\theta }+e^{-\theta }}{2}$for $\cosh \theta$into equation (1).

$A=\frac{e^{\theta }+e^{-\theta }}{2}$ …… (2)

Consider the following data

$x=e^{-\theta }$

$x=\frac{1}{e^{\theta }}$

$\frac{1}{x}=e^{\theta }$

Now, substitute $\frac{1}{x}$for $e^{\theta }$and $x$ for $e^{-\theta }$into equation (2).

$\begin{array}{l}A=\frac{\frac{1}{x}+x}{2}\\ 2A=\frac{1}{x}+x\\ 2Ax=1+x^2\\ x^2-2Ax+1=0\end{array}$ …… (3)

Consider the following data.

$\begin{array}{l}x=x_1+j{x}_2\\ A=A_1+j{A}_2\end{array}$

Substitute$x_1+j{x}_2$for $x$ and $A_1+j{A}_2$for $A$into equation (3).

$\begin{array}{l}\left(x_1+j{x}_2\right)-2\left(A_1+j{A}_2\right)\left(x_1+j{x}_2\right)+1=0\\ x_1^2-x_2^2+j2x_1{x}_2-2A_1{x}_1-j2A_1{x}_2-j2A_2{x}_1+2A_2{x}_2+1=0\\ x_1^2-x_2^2+j2x_1{x}_2-2\left[A_1{x}_1-A_2{x}_2+j\left(A_1{x}_2+A_2{x}_1\right)\right]+1=0\end{array}$

Compare the real and imaginary parts of the equation.

$\begin{array}{l}{x}_1^2-x_2^2-2\left(A_1{x}_1+A_2{x}_2\right)+1=0\\ x_1{x}_2-\left(A_2{x}_1+A_1{x}_2\right)=0\end{array}$

Therefore, the value of $x_1$and $x_2$satisfy the relations

$x_1^2-x_2^2-2\left({\text{A}}_1{x}_1-{\text{A}}_2{x}_2\right)+1=0$and $x_1{x}_2-\left({\text{A}}_2{x}_1+A_1{x}_2\right)=0$.