Question

As shown in Figure 13.32, a single-phase two-wire lossless line with ${\mathbf{Z}}_c=\mathbf{400}\mathbf{\Omega }$,$\nu =\mathbf{3}\times {\mathbf{10}}^8\mathbf{m}/\mathbf{s}$ and$\mathbf{l}=\mathbf{100}\mathbf{km}$ has a$\mathbf{400}-\mathbf{\Omega }$ resistor, denoted${\mathbf{R}}_J$ , installed across the center of the line, thereby dividing the line into two sections, A and B. The source voltage at the sending end is a pulse of magnitude$\mathbf{100}\mathbf{V}$ and duration$\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{ms}$ . The source impedance is${\mathbf{Z}}_G={\mathbf{Z}}_c=\mathbf{400}\mathbf{\Omega }$ , and the receiving end of the line is short-circuited, (a) Show that

Figure 13.32

For an incident voltage wave arriving at the center of the line from either line section, the voltage reflection and refraction coefficients are given by

${\mathrm{\Gamma }}_{\mathrm{BB}}={\mathrm{\Gamma }}_{\mathrm{AA}}=\frac{\left(\frac{Z_{\mathrm{eq}}}{Z_c}\right)-\mathbf{1}}{\left(\frac{Z_{\mathrm{eq}}}{Z_c}\right)+\mathbf{1}}$ ${\mathbf{\Gamma }}_{\mathrm{AB}}\mathbf{=}{\mathbf{\Gamma }}_{\mathrm{BA}}\mathbf{=}\frac{\mathbf{2}\left(\frac{Z_{\mathrm{eq}}}{Z_c}\right)}{\left(\frac{Z_{\mathrm{eq}}}{Z_c}\right)\mathbf{+}\mathbf{1}}$

Where

${\mathbf{Z}}_{\mathrm{eq}}=\frac{{\mathbf{R}}_J{\mathbf{Z}}_c}{{\mathbf{R}}_J+{\mathbf{Z}}_c}$

(b) Draw the Bewley lattice diagram for $\mathbf{0}\le \mathbf{t}\le \mathbf{6}\mathrm{{\rm T}}$.

(c) Plot $\upsilon (\mathbf{1}/\mathbf{2}\mathbf{,}\mathbf{t})$versus time$\mathbf{t}$ for$\mathbf{0}\le \mathbf{t}\le \mathbf{6}\mathrm{{\rm T}}$ and plot$\upsilon (\mathbf{x},\mathbf{6}\mathbf{t})$ versus$\mathbf{x}$ for $\mathbf{0}\mathbf{\le }\mathbf{x}\mathbf{\le }\mathbf{l}$.

Short answer

(a)

The value of voltage reflection coefficient ${\text{Γ}}_{\text{AA}}$is $\frac{\left(\frac{Z_{\text{eq}}}{Z_{\text{c}}}-1\right)}{\left(\frac{Z_{\text{eq}}}{Z_{\text{c}}}+1\right)}$.

The value of voltage refraction coefficient ${\Gamma }_{\text{BA}}$is $\frac{2\left(\frac{Z_{\text{eq}}}{Z_{\text{c}}}\right)}{\frac{Z_{\text{eq}}}{Z_{\text{c}}}+1}$.

(b) Draw the Bewley lattice diagram for $\text{0}\le \text{t}\le 6{\rm T}$.

(c)

Plot the voltage graph of$\nu \left(\frac{1}{2},t\right)$ versus time$t$ .

Plot the voltage graph of $(x,6\tau )$versus time$t$ .

Step-by-step solution

Write the given data from the question.

Refer figure 13.32 from the text book.

Consider the following data:

Characteristics impedance of the line,$Z_{\text{c}}=400\Omega$ .

Length of the cable line,$l=100\text{\hspace{0.17em}km}$ .

Resistor installed across the centre of the line, ${\text{R}}_{\text{J}}=400\Omega$.

Source impedance, ${\text{Z}}_{\text{G}}=400\Omega$.

Determine the formula of voltage reflection coefficients.

Write the formula of voltage reflection coefficient.

${\mathrm{\Gamma }}_{\mathrm{AA}}=\frac{{\mathbf{V}}_A^{-}}{{\mathbf{V}}_A^{+}}$ …… (1)

Here, ${\mathbf{V}}_A^{-}$is reflected component of voltage wave and ${\mathbf{V}}_A^{+}$is voltage wave.

Write the formula of voltage refraction coefficient.

${\mathrm{\Gamma }}_{\mathrm{BA}}=\frac{{\mathbf{V}}_B^{+}}{{\mathbf{V}}_A^{-}}$ …… (2)

Here,${\mathbf{V}}_B^{+}$ is reflected component of voltage wave and${\mathbf{V}}_A^{-}$ is reflected component of voltage wave.

(a) Determine the value of voltage reflection coefficient ΓAA and voltage refraction coefficient  ΓBA.

The resistance connected across line is parallel to impedance of line.

$\begin{array}{c}{Z}_{\text{eq}}={\text{R}}_{\text{J}}\parallel Z_c\\ =\frac{\left({\text{R}}_{\text{J}}\right)\left({\text{Z}}_{\text{c}}\right)}{{\text{R}}_{\text{J}}+{\text{Z}}_{\text{c}}}\end{array}$ …… (3)

If the voltage wave ${\mathrm{V}}_{\mathrm{A}}^{+}$is assumed to be incident at the junction, then ${\mathrm{V}}_{\mathrm{A}}^{-}$ and ${\mathrm{V}}_{\mathrm{B}}^{+}$will represent the incident wave's reflected and refracted components, respectively.

Apply Kirchhoff’s voltage law to the voltage wave.

role="math" localid="1656409698189" ${\mathrm{V}}_{\mathrm{A}}^{+}+{\mathrm{V}}_{\mathrm{A}}^{-}={\mathrm{V}}_{\mathrm{B}}^{+}$

Apply Kirchhoff’s current law to the voltage wave.

$\begin{array}{c}{\text{I}}_{\text{A}}^{+}+{\text{I}}_{\text{A}}^{-}=\frac{{\text{V}}_{\text{B}}^{+}}{{\text{R}}_{\text{J}}}+{\text{I}}_{\text{B}}^{+}\\ \frac{{\text{V}}_{\text{A}}^{+}}{Z_c}+\left(\frac{-{\text{V}}_{\text{A}}^{-}}{Z_{\text{eq}}}\right)=\frac{{\text{V}}_{\text{B}}^{+}}{{\text{R}}_{\text{J}}}+\frac{{\text{V}}_{\text{B}}^{+}}{{\text{Z}}_c}\\ ={\text{V}}_{\text{B}}^{+}\left(\frac{1}{{\text{R}}_{\text{J}}}+\frac{1}{Z_{\text{c}}}\right)\\ ={\text{V}}_{\text{B}}^{+}\left(\frac{{\text{R}}_{\text{J}}+Z_{\text{c}}}{\left({\text{R}}_{\text{J}}\right)\left(Z_{\text{c}}\right)}\right)\end{array}$

Solve further as

${\text{I}}_{\text{A}}^{\text{+}}+{\text{I}}_{\text{A}}^{-}=V_{\text{B}}^{+}\left(\frac{1}{\frac{\left({\text{R}}_{\text{J}}\right)\left(Z_{\text{c}}\right)}{{\text{R}}_{\text{J}}+Z_{\text{c}}}}\right)$ …… (4)

Substitute equation (3) into equation (4).

$\frac{{\text{V}}_{\text{A}}^{+}}{{\mathrm{Z}}_{\mathrm{c}}}-\frac{{\mathrm{V}}_{\text{A}}^{-}}{{\mathrm{Z}}_{\mathrm{c}}}=\frac{{\mathrm{V}}_{\text{B}}^{+}}{{\mathrm{Z}}_{\text{eq}}}$

Substitute ${\text{V}}_{\text{A}}^{\text{+}}+{\text{V}}_{\text{A}}^{-}$ for ${\text{V}}_{\text{B}}^{\text{+}}$into above equation.

$\begin{array}{c}\frac{{\text{V}}_{\text{A}}^{\text{+}}}{{\text{Z}}_{\text{c}}}-\frac{{\text{V}}_{\text{A}}^{-}}{{\text{Z}}_{\text{c}}}=\frac{V_{\text{A}}^{+}+{\text{V}}_{\text{A}}^{-}}{Z_{\text{eq}}}\\ \frac{{\text{V}}_{\text{A}}^{\text{+}}}{{\text{Z}}_{\text{c}}}-\frac{{\text{V}}_{\text{A}}^{-}}{{\text{Z}}_{\text{c}}}=\frac{V_{\text{A}}^{+}}{Z_{\text{eq}}}+\frac{V_{\text{A}}^{-}}{Z_{\text{eq}}}\\ V_{\text{A}}^{+}\left(\frac{1}{Z_c}-\frac{1}{Z_{\text{eq}}}\right)=V_{\text{A}}^{-}\left(\frac{1}{Z_c}+\frac{1}{Z_{\text{eq}}}\right)\\ \frac{V_{\text{A}}^{-}}{V_{\text{A}}^{+}}=\frac{\left(\frac{1}{Z_c}-\frac{1}{Z_{\text{eq}}}\right)}{\left(\frac{1}{Z_c}+\frac{1}{Z_{\text{eq}}}\right)}\end{array}$ …… (5)

Determine the reflection coefficient.

Substitute$\left(\frac{Z_{\text{eq}}}{Z_c}-1\right)$for${\text{V}}_{\text{A}}^{-}$ and$\left(\frac{Z_{\text{eq}}}{Z_{\text{c}}}+1\right)$ for${\text{V}}_{\text{A}}^{+}$into equation (1).

${\mathrm{\Gamma }}_{\text{AA}}=\frac{\left(\frac{{\mathrm{Z}}_{\text{eq}}}{{\mathrm{Z}}_{\mathrm{c}}}-1\right)}{\left(\frac{{\mathrm{Z}}_{\text{eq}}}{{\mathrm{Z}}_{\text{c}}}+1\right)}$

Therefore, the value of voltage reflection coefficient ${\text{Γ}}_{\text{AA}}$is $\frac{\left(\frac{Z_{\text{eq}}}{Z_{\text{c}}}-1\right)}{\left(\frac{Z_{\text{eq}}}{Z_{\text{c}}}+1\right)}$.

Determine the refraction coefficient.

Substitute ${\mathrm{V}}_{\text{A}}^{+}+{\mathrm{V}}_{\text{A}}^{-}$for $V_{\text{B}}^{+}$and $V_{\text{A}}^{+}$ for ${\mathrm{V}}_{\text{A}}^{+}$into equation (2).

${\Gamma }_{\text{BA}}=\frac{V_{\text{A}}^{+}+V_{\text{A}}^{-}}{V_{\text{A}}^{+}}$ …… (6)

Write the expression for$V_{\text{A}}^{-}$from equation (5).

${\mathrm{V}}_{\text{A}}^{-}=\frac{\left(\frac{1}{{\mathrm{Z}}_{\mathrm{c}}}-\frac{1}{{\mathrm{Z}}_{\text{eq}}}\right)}{\left(\frac{1}{{\mathrm{Z}}_{\mathrm{c}}}+\frac{1}{{\mathrm{Z}}_{\mathrm{eq}}}\right)}{\mathrm{V}}_{\text{A}}^{-}$ …… (7)

Substitute equation (7) into equation (6).

role="math" localid="1656410712653" …… (8)

Multiply equation (8) numerator and denominator by$Z_{\text{eq}}$ .

$\begin{array}{c}{\Gamma }_{\text{BA}}=\frac{2\left(\frac{Z_{\text{eq}}}{Z_{\text{c}}}\right)}{\frac{Z_{\text{eq}}}{Z_{\text{c}}}+1}\\ =\frac{2\left(\frac{Z_{\text{eq}}}{Z_{\text{c}}}\right)}{\frac{Z_{\text{eq}}}{Z_{\text{c}}}+1}\end{array}$

Therefore, the value of voltage refraction coefficient ${\mathrm{\Gamma }}_{\text{BA}}$is$\frac{2\left(\frac{Z_{\text{eq}}}{Z_{\text{c}}}\right)}{\frac{Z_{\text{eq}}}{Z_{\text{c}}}+1}$ .

(b) Draw the Bewley lattice diagram for  0≤t≤6Τ.

Write the expression for transit time.

Draw the Bewley lattice diagram for $\text{0}\le \text{t}\le 6\mathrm{{\rm T}}$ .

Figure 1

(c) Plot the voltage graph of  ν12,t  versus time t .

Plot the voltage graph of $\nu \left(\frac{1}{2},t\right)$versus time$t$ .

Figure 2

Plot the voltage graph of $(x,6\tau )$versus time$t$ .

Figure 3