Question

Referring to Figure 13.3, the source voltage at the sending end is a step ${\mathbf{e}}_G\left(\mathbf{t}\right)={\mathbf{Eu}}_{-\mathbf{1}}\left(\mathbf{t}\right)$with an inductive source impedance ${\mathbf{Z}}_G\left(\mathbf{s}\right)={\mathbf{sL}}_G$, where ${\mathbf{L}}_G/{\mathbf{Z}}_c=\mathrm{\tau }/\mathbf{3}$. At the receiving end, ${\mathbf{Z}}_R={\mathbf{Z}}_c/\mathbf{4}$. The line and source inductance are initially unenergized.

(a) Draw the Bewley lattice diagram for $\mathbf{0}\le \mathbf{t}\le \mathbf{5}\tau$.

(b) Plot$\upsilon (\mathbf{l}\mathbf{,}\mathbf{t})$ versus time$\mathit{t}$ for role="math" localid="1656411981539" $\mathbf{0}\le \mathbf{t}\le \mathbf{5}\tau$.

Short answer

(a) Draw the Bewley lattice diagram for $\text{0}\le \text{t}\le \text{5}\tau$.

(b) Plot$\upsilon \left(\text{l,t}\right)$ versus timerole="math" localid="1656412142724" $\mathbf{t}$ for $\text{0}\le \text{t}\le \text{5}\tau$.

Step-by-step solution

Write the given data from the question.

Refer the figure 13.3 from the test book.

Consider the following data.

The source voltage at sending end is,${\mathrm{e}}_{\mathrm{G}}\left(\mathrm{t}\right)={\mathrm{Eu}}_{-1}\left(\mathrm{t}\right)$ .

The inductive source end impedance,${\mathrm{Z}}_{\text{R}}=\frac{{\mathrm{Z}}_{\mathrm{c}}}{4}$ .

Determine the formula of sending end coefficient (Γs) and receiving end coefficient (ΓR).

Write the formula of sending end coefficient $\left({\mathrm{\Gamma }}_{\text{s}}\right)$.

${\mathrm{\Gamma }}_s=\frac{\frac{{\mathbf{Z}}_G\left(\mathrm{s}\right)}{{\mathbf{Z}}_c}-\mathbf{1}}{\frac{{\mathbf{Z}}_G\left(\mathbf{s}\right)}{{\mathbf{Z}}_c}+\mathbf{1}}$ …… (1)

Here, ${\mathbf{Z}}_G$is source impedance and ${\mathbf{Z}}_c$is characteristics impedance.

Write the formula ofreceiving end reflection coefficient ${\Gamma }_{\text{R}}$.

${\mathrm{\Gamma }}_R\left(\mathbf{s}\right)=\frac{\frac{{\mathbf{Z}}_c}{\mathbf{4}{\mathbf{Z}}_c}-\mathbf{1}}{\frac{{\mathbf{Z}}_c}{\mathbf{4}{\mathbf{Z}}_c}+\mathbf{1}}$ …… (2)

Here,${\mathbf{Z}}_R$ is receiving end impedance and ${\mathbf{Z}}_c$is characteristics impedance.

(a) Draw the Bewley lattice diagram for 0≤t≤5τ.

Determine the sending end coefficient $\left({\mathrm{\Gamma }}_{\text{s}}\right)$.

Substitute$s{L}_{\text{G}}$ for$Z_{\text{G}}\left(s\right)$ and$Z_{\text{c}}$ for$Z_{\text{c}}$ into equation (1).

$\begin{array}{c}{\Gamma }_{\text{s}}=\frac{\frac{s{L}_{\text{G}}}{Z_c}-1}{\frac{s{L}_{\text{G}}}{Z_c}+1}\\ =\frac{s-\frac{Z_{\text{c}}}{L_{\text{G}}}}{s+\frac{Z_{\text{c}}}{L_{\text{G}}}}\end{array}$ …… (3)

Determine the receiving end reflection coefficient $\left({\mathrm{\Gamma }}_{\text{R}}\right)$.

Substitute$1$ for$\frac{{\mathrm{Z}}_{\mathrm{c}}}{4{\mathrm{Z}}_{\mathrm{c}}}$ into equation (2).

$\begin{array}{c}{\mathrm{\Gamma }}_{\text{R}}\left(\mathrm{s}\right)=\frac{1-4}{1+4}\\ =\frac{-3}{5}\end{array}$

Apply voltage domain method and determine the voltage wave.

${\text{V}}_{\text{1}}=\left[{\text{E}}_{\text{G}}\right]\left[\frac{{\mathrm{Z}}_{\mathrm{c}}}{{\mathrm{Z}}_{\mathrm{c}}+{\mathrm{Z}}_{\mathrm{g}}}\right]$

Apply Laplace transform of above equation.

${\mathrm{V}}_1\left(\mathrm{s}\right)=\frac{{\mathrm{E}}_{\text{G}}\left(\mathrm{s}\right)}{\mathrm{s}}\left(\frac{{\mathrm{Z}}_{\text{c}}}{{\mathrm{Z}}_{\text{c}}+\mathrm{s}{\text{L}}_{\text{G}}}\right)$

Divide both numerator and denominator by ${\text{L}}_{\text{G}}$.

${\text{V}}_{\text{1}}\left(\text{s}\right)=\left[\frac{{\text{E}}_{\text{G}}\left(\mathrm{s}\right)}{\mathrm{s}}\right]\left(\frac{{\mathrm{Z}}_{\mathrm{c}}}{{\mathrm{Z}}_{\mathrm{c}}+\mathrm{s}{\text{L}}_{\text{G}}}\right)$

Take partial fraction for equation (3).

${\mathrm{V}}_1\left(\mathrm{s}\right)={\text{E}}_{\text{G}}\left(\text{s}\right)\left(\frac{1}{\mathrm{s}}-\frac{1}{\mathrm{s}+\frac{{\text{Z}}_{\text{c}}}{{\text{L}}_{\text{G}}}}\right)$ ...... (4)

Draw the Bewley lattice diagram for $\text{0}\le \text{t}\le \text{5}\tau$.

Figure 1

(b) Plot υ(l,t) versus time t for 0≤t≤5τ.

Determine the expression for voltage equation.

…… (5)

Substitute the values of $V_1\left(s\right)$and ${\Gamma }_{\text{s}}$from equation (3) and equation (4) into equation (5).

$\mathrm{V}\left(\text{1,s}\right)=\frac{2\text{E}}{5}\left(\frac{1}{\mathrm{s}}-\frac{1}{\mathrm{s}+\frac{{\text{Z}}_{\text{c}}}{{\text{L}}_{\text{g}}}}\right){\mathrm{e}}^{-\mathrm{s\tau }}-\frac{6\text{E}}{25}\left(\frac{1}{\mathrm{s}}\right)\left(\frac{\mathrm{s}-\frac{{\text{Z}}_{\text{c}}}{{\text{L}}_{\text{G}}}}{\mathrm{s}+\frac{{\text{Z}}_{\text{c}}}{{\text{L}}_{\text{G}}}}\right)\left(\frac{\frac{{\text{Z}}_{\text{c}}}{{\text{L}}_{\text{G}}}}{\mathrm{s}+\frac{{\text{Z}}_{\text{c}}}{{\text{L}}_{\text{G}}}}\right){\mathrm{e}}^{-\mathrm{s}\left(3\mathrm{\tau }\right)}$

Take partial fractions for second part of the equation.

$\mathrm{V}(1,\mathrm{s})=\frac{2\text{E}}{5}\left(\frac{1}{\mathrm{s}}-\frac{1}{\mathrm{s}+\frac{{\text{Z}}_{\text{c}}}{{\text{L}}_{\text{G}}}}\right){\mathrm{e}}^{-\mathrm{s\tau }}-\frac{6\text{E}}{25}\left(\frac{1}{\mathrm{s}}-\frac{1}{\mathrm{s}+\frac{{\text{Z}}_{\text{c}}}{{\text{L}}_{\text{G}}}}-2\frac{\frac{{\mathrm{Z}}_{\mathrm{c}}}{{\mathrm{L}}_{\mathrm{G}}}}{{(\mathrm{s}+\frac{{\text{Z}}_{\text{c}}}{{\text{L}}_{\text{G}}})}^2}\right)$

Apply inverse Laplace transform.

$V(1,t)=\frac{2\text{E}}{5}\left(1-e^{\frac{(t-3\tau )}{{\text{L}}_{\text{G}}/{\text{Z}}_{\text{c}}}}\right){\text{U}}_{-1}(t-\tau )+\frac{6\text{E}}{25}\left(1-e^{\frac{(t-3\tau )}{{\text{L}}_{\text{G}}/{\text{Z}}_{\text{c}}}}-\frac{2(t-3\tau )}{{\text{L}}_{\text{G}}/{\text{Z}}_{\text{c}}}{e}^{\frac{(t-3\tau )}{{\text{L}}_{\text{G}}/{\text{Z}}_{\text{c}}}}\right){\text{U}}_{-1}(t-3\tau )$

Plot $\nu (l,t)$versus time $t$for $0\le t\le 5\tau$.

Figure 2

Figure 2 shows the wave's nature, roughly equal to one step delayed by $\tau$. As seen in the first section, it has a steady-state value of $\frac{\text{2E}}{\text{5}}$. The second component will be seen after a $3\mathrm{\tau }$time delay, and the steady-state value is $\frac{\text{16E}}{\text{25}}$.

Therefore, the plot is just similar to unit step function.