Question

(a) The series expansions of the hyperbolic functions are given by

$$\begin{gathered} \mathbf{cosh}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{4}}}{\mathbf{24}}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{6}}}{\mathbf{720}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.} \\ \mathbf{sinh}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{2}}}{\mathbf{6}}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{4}}}{\mathbf{120}}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{6}}}{\mathbf{5040}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.} \end{gathered}$$

For the ABCD parameters of a long transmission line represented by an equivalent ${}^{\mathbf{\pi }}$circuit, apply the above expansion considering only the first two terms, and express the result in terms of Y and Z.

(b) For the nominal ${}^{\mathbf{\pi }}$and equivalent ${}^{\mathbf{\pi }}$circuits shown in Figures 5.3 and 5.7 of the text, show that

role="math" localid="1655374470951" $\frac{\mathbf{A}\mathbf{‐}\mathbf{1}}{\mathbf{B}}\mathbf{=}\frac{\mathbf{Y}}{\mathbf{2}}\mathbf{}$and $\mathbf{}\frac{\mathbf{A}\mathbf{‐}\mathbf{1}}{\mathbf{B}}\mathbf{=}\frac{\mathbf{Y}\mathbf{\text{'}}}{\mathbf{2}}$

Short answer

(a)Therefore, the ${}^{\mathbf{A}\mathbf{B}\mathbf{C}\mathbf{D}}$parameter in terms of Y and Z is:

$$\begin{gathered} \mathit{A}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{Z}\mathbf{Y}}{\mathbf{2}} \\ \mathit{B}\mathbf{=}{\mathit{Z}}_{\mathbf{c}}\left(1+\frac{ZY}{6}\right) \\ \mathit{C}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Z}}_{\mathbf{c}}}\left(1+\frac{ZY}{6}\right) \\ \mathit{D}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{Z}\mathbf{Y}}{\mathbf{2}} \end{gathered}$$

(b) It is proved that ${}^{\frac{\mathbf{A}\mathbf{-}\mathbf{1}}{\mathbf{B}}\mathbf{=}\frac{\mathbf{Y}}{\mathbf{2}}}$for nominal ${}^{\mathbf{\pi }}$circuit.

It is proved that ${}^{\frac{\mathbf{A}\mathbf{-}\mathbf{1}}{\mathbf{B}}\mathbf{=}\frac{\mathbf{Y}\mathbf{\text{'}}}{\mathbf{2}}}$for equivalent ${}^{\mathbf{\pi }}$circuit.

Step-by-step solution

Define the given information

The series expansions of the hyperbolic functions is,

$$\begin{gathered} \mathbf{cosh}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{4}}}{\mathbf{24}}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{6}}}{\mathbf{720}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.} \\ \mathbf{sinh}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{2}}}{\mathbf{6}}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{4}}}{\mathbf{120}}\mathbf{+}\frac{{\mathbf{\theta }}^{\mathbf{6}}}{\mathbf{5040}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.} \end{gathered}$$ …… (1)

Define the ABCD Parameter in Long Transmission Line

The ABCD parameter for long transmission line

$$\begin{gathered} \mathit{A}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{}\gamma \mathit{l} \\ \mathit{B}\mathbf{=}{\mathit{Z}}_{\mathbf{c}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{}\mathbf{}\gamma \mathit{l} \\ \mathit{C}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Z}}_{\mathbf{c}}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{}\gamma \mathit{l} \\ \mathit{D}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{}\gamma \mathit{l} \end{gathered}$$

Here, $\mathit{l}$is a length of transmission line

The expression for the propagation constant is,

$$\begin{gathered} \mathit{\gamma }\mathbf{=}\sqrt{\mathbf{Z}\mathbf{Y}} \\ \mathit{\gamma }\mathbf{=}\mathit{\alpha }\mathbf{+}\mathit{J}\mathit{\beta } \end{gathered}$$

Here, ${}^{\gamma }$is a Propagation constant and ${}^{\mathit{\alpha }}$is the attenuation constant in Neper per sec. and ${}^{\mathit{\beta }}$is a Phase constant in rad per km

The expression for the phase constant is,

$\mathit{\beta }\mathbf{=}\mathit{\omega }\sqrt{\mathbf{L}\mathbf{C}}$

Obtain the ‘ ’ Parameter in Terms of Y and Z

(a).

The expression of impedance angle of line,

$$\begin{gathered} \mathit{\theta }\mathbf{=}\mathit{\gamma }\mathit{l} \\ \mathbf{}\mathbf{}\mathbf{=}\sqrt{\mathbf{Z}\mathbf{Y}} \end{gathered}$$

Here, ${}^{\mathbf{\theta }}$is an impedance angle of line and ${}^{\mathbf{Z}}$is total impedance and is a total admittance.

The ABCD parameter of long transmission line

$$\begin{gathered} \mathit{A}\mathbf{=}\mathbf{cosh}\mathbf{}\mathit{\gamma }\mathit{l} \\ \mathit{B}\mathbf{=}{\mathit{Z}}_{\mathbf{c}}\mathbf{}\mathbf{sinh}\mathbf{}\mathit{\gamma }\mathit{l} \\ \mathit{C}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{}\mathbf{sinh}\mathbf{}\mathit{\gamma }\mathit{l} \\ \mathit{D}\mathbf{=}\mathbf{cos}\mathbf{}\mathit{\gamma }\mathit{l} \end{gathered}$$

Substitute ${}^{\sqrt{\mathbf{Z}\mathbf{Y}}}$for ${}^{\mathbf{\theta }}$and A for ${}^{\mathbf{\gamma }\mathbf{l}}$in the equation (1)

$$\begin{gathered} \mathit{A}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{Z}\mathbf{Y}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{Z}}^{\mathbf{2}}{\mathbf{Y}}^{\mathbf{2}}}{\mathbf{24}}\mathbf{+}\frac{{\mathbf{Z}}^{\mathbf{3}}{\mathbf{Y}}^{\mathbf{3}}}{\mathbf{720}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.} \\ \mathit{B}\mathbf{=}{\mathit{Z}}_{\mathbf{c}}\left[1+\frac{ZY}{6}+\frac{Z^2{Y}^2}{120}+\frac{Z^3{Y}^3}{5040}+.......\right] \\ \mathit{C}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Z}}_{\mathbf{c}}}\left[1+\frac{ZY}{6}+\frac{Z^2{Y}^2}{120}+\frac{Z^3{Y}^3}{5040}+.......\right] \\ \mathit{D}\mathbf{=}\mathit{A} \end{gathered}$$

Observe the first two terms in the above equation:

$$\begin{gathered} \mathit{A}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{Z}\mathbf{Y}}{\mathbf{2}} \\ \mathit{B}\mathbf{=}{\mathit{Z}}_{\mathbf{c}}\left(1+\frac{ZY}{6}\right) \\ \mathit{C}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Z}}_{\mathbf{c}}}\left(1+\frac{ZY}{6}\right) \end{gathered}$$

Therefore, the ${}^{\mathbf{A}\mathbf{B}\mathbf{C}\mathbf{D}}$parameter in terms of Y and Z are,

$$\begin{gathered} \mathit{A}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{ZY}}{\mathbf{2}} \\ \mathit{B}\mathbf{=}{\mathit{Z}}_{\mathbf{c}}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{ZY}}{\mathbf{6}}\mathbf{)} \\ \mathit{C}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{ZY}}{\mathbf{6}}\mathbf{)} \\ \mathit{D}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{ZY}}{\mathbf{2}} \end{gathered}$$

Obtain the value of Equivalent and nominal Circuit

(b).

The expression of nominal ${}^{\mathbf{\pi }}$circuit in ABCD parameter,

$$\begin{gathered} \mathit{A}\mathbf{=}\mathit{D} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{Z}\mathbf{Y}}{\mathbf{2}} \\ \mathit{B}\mathbf{=}\mathit{Z} \\ \mathit{C}\mathbf{=}\mathit{Y}\left(1+\frac{ZY}{6}\right) \end{gathered}$$

Substitute ${}^{\mathbf{1}\mathbf{+}\frac{\mathbf{ZY}}{\mathbf{2}}}$for A and ${}^{\mathbf{Z}}$for B in ${}^{\frac{\mathbf{A}\mathbf{-}\mathbf{1}}{\mathbf{B}}}$ ,

$$\begin{gathered} \frac{\mathbf{A}\mathbf{-}\mathbf{1}}{\mathbf{B}}\mathbf{=}\frac{\mathbf{1}\mathbf{+}{\displaystyle \frac{\mathbf{Z}\mathbf{Y}}{\mathbf{2}}}\mathbf{-}\mathbf{1}}{\mathbf{Z}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\displaystyle \frac{\mathbf{Z}\mathbf{Y}}{\mathbf{2}}}}{\mathbf{Z}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{Z}\mathbf{Y}}{\mathbf{2}\mathbf{Z}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{Y}}{\mathbf{2}} \end{gathered}$$

Therefore, it is proved that ${}^{\frac{\mathbf{A}\mathbf{-}\mathbf{1}}{\mathbf{B}}\mathbf{=}\frac{\mathbf{Y}}{\mathbf{2}}}$for nominal ${}^{\mathbf{\pi }}$circuit.

The expression of nominal ${}^{\mathbf{\pi }}$circuit in ABCD parameter,

$$\begin{gathered} \mathit{A}\mathbf{=}\mathit{D} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{Z}\mathbf{\text{'}}\mathbf{Y}\mathbf{\text{'}}}{\mathbf{2}} \\ \mathit{B}\mathbf{=}\mathit{Z}\mathbf{\text{'}} \\ \mathit{C}\mathbf{=}\mathit{Y}\mathbf{\text{'}}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{Z}\mathbf{\text{'}}\mathbf{Y}\mathbf{\text{'}}}{\mathbf{6}}\mathbf{)} \end{gathered}$$

Substitute ${}^{\mathbf{1}\mathbf{+}\frac{\mathbf{Z}\mathbf{\text{'}}\mathbf{Y}\mathbf{\text{'}}}{\mathbf{2}}}$for A and ${}^{\mathbf{Z}\mathbf{\text{'}}}$for B in ${}^{\frac{\mathbf{A}\mathbf{-}\mathbf{1}}{\mathbf{B}}}$ ,

$$\begin{gathered} \frac{\mathbf{A}\mathbf{-}\mathbf{1}}{\mathbf{B}}\mathbf{=}\frac{\mathbf{1}\mathbf{+}{\displaystyle \frac{\mathbf{Z}\mathbf{\text{'}}\mathbf{Y}\mathbf{\text{'}}}{\mathbf{2}}}\mathbf{-}\mathbf{1}}{\mathbf{Z}\mathbf{\text{'}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\displaystyle \frac{\mathbf{Z}\mathbf{\text{'}}\mathbf{Y}\mathbf{\text{'}}}{\mathbf{2}}}}{\mathbf{Z}\mathbf{\text{'}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{Z}\mathbf{\text{'}}\mathbf{Y}\mathbf{\text{'}}}{\mathbf{2}\mathbf{Z}\mathbf{\text{'}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{Y}\mathbf{\text{'}}}{\mathbf{2}} \end{gathered}$$

Therefore, it is proved that ${}^{\frac{\mathbf{A}\mathbf{-}\mathbf{1}}{\mathbf{B}}\mathbf{=}\frac{\mathbf{Y}\mathbf{\text{'}}}{\mathbf{2}}}$for equivalent ${}^{\mathbf{\pi }}$circuit.