Question

The single-phase, two-wire lossless line in Figure 13.3 has a series inductance$\mathit{L}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{3}\mathbf{)}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{H}\mathbf{/}\mathit{m}$, a shunt capacitance, and a$\mathbf{50}\mathit{K}\mathit{m}$line length. The source voltage at the sending end is a step${\mathit{e}}_{\mathbf{G}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{100}{\mathit{u}}_{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathit{k}\mathit{V}$with${\mathit{Z}}_{\mathbf{G}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathit{\Omega }$.The receiving-end load consists of a $\mathbf{100}\mathit{\Omega }$resistor in parallel with $\mathbf{2}\mathit{m}\mathit{H}$a inductor. The line and load are initially unenergized. Determine (a) the characteristic impedance in ohms, the wave velocity in$\mathit{m}\mathbf{/}\mathit{s}$, and the transit time in$\mathit{m}\mathit{s}$for this line; (b) the sending-and receiving-end voltage reflection coefficients in per-unit; (c) the Laplace transform of the receiving-end current${\mathit{I}}_{\mathbf{R}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$, ; and (d) the receiving-end current ${\mathit{i}}_{\mathbf{R}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$as a function of time.

Short answer

(a) The value of characteristics impedance, velocity and transit time are$100\Omega$ , $3\times {10}^8\frac{\text{m}}{\text{s}}$and $0.1667\text{\hspace{0.33em}ms}$respectively.

(b) The sending-and receiving-end voltage reflection coefficients are $0\text{\hspace{0.33em}pu}$and $\frac{-25000}{s+25000}\text{\hspace{0.33em}pu}$respectively.

(c) The receiving end current$I_R\left(s\right)$ is$\frac{500}{s}\left[2e^{-s\tau }-\frac{s{e}^{-s\tau }}{s+25000}\right]\text{\hspace{0.33em}A}$ .

(d) The value of$i_R\left(t\right)$ is $500\left(2-e^{-25000\left(t-\tau \right)}\right)u_{-1}\left(t-\tau \right)\text{\hspace{0.33em}A}$.

Step-by-step solution

Write the given data from the question.

The series inductance$L=\frac{1}{3}\times {10}^{-6}\frac{\text{H}}{\text{m}}$,

The shunt capacitance$C=\frac{1}{3}\times {10}^{-10}\frac{\text{F}}{\text{m}}$,

The length of the line$l=50\text{\hspace{0.33em}Km}$,

The source voltage$e_G\left(t\right)=100u_{-1}\left(t\right)\text{\hspace{0.33em}kV}$,

The source impedance,$Z_G\left(s\right)=100\Omega$

The receiving end resistor$R=100\Omega$,

The receiving end inductor$L=2\text{\hspace{0.33em}mH}$,

Determine the equations to calculate the characteristic impedance, the wave velocity, the transit time, the sending-and receiving-end voltage reflection coefficients, Laplace transform of the receiving-end current, IRsand the receiving-end current  iRt as a function of time.

The equation to calculate the characteristic impedance is given as follows.

$Z_c=\sqrt{\frac{L}{C}}$ …… (1)

The equation to calculate the value of velocity is given as follows.

$\mathit{v}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{L}\mathbf{C}}}$ …… (2)

The equation to calculate the transit time is given as follows.

$\mathit{\tau }\mathbf{=}\frac{\mathbf{l}}{\mathbf{v}}$ …… (3)

The equation to calculate the receiving end voltage reflection is given as follows.

${\mathit{\Gamma }}_{\mathbf{R}}\left(s\right)\mathbf{=}\frac{\frac{{\mathbf{Z}}_{\mathbf{R}}\left(s\right)}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{-}\mathbf{1}}{\frac{{\mathbf{Z}}_{\mathbf{R}}\left(s\right)}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{+}\mathbf{1}}$ …… (4)

Here, ${\mathit{Z}}_{\mathbf{R}}\left(s\right)$is the receiving end impedance and ${\mathit{Z}}_{\mathbf{c}}$is the characteristics impedance.

The equation to calculate the sending end voltage reflection is given as follows.

${\mathit{\Gamma }}_{\mathbf{S}}\left(s\right)\mathbf{=}\frac{\frac{{\mathbf{Z}}_{\mathbf{G}}\left(s\right)}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{-}\mathbf{1}}{\frac{{\mathbf{Z}}_{\mathbf{G}}\left(s\right)}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{+}\mathbf{1}}$ …… (5)

The equation to calculate the current at distance is given as follows.

…… (6)

${\mathit{I}}_{\mathbf{R}}\left(s\right)\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{G}}\left(s\right)}{{\mathbf{Z}}_{\mathbf{c}}\mathbf{+}{\mathbf{Z}}_{\mathbf{G}}\left(s\right)}\left[\frac{e^{-\frac{sx}{v}}-{\Gamma }_R\left(s\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1-{\Gamma }_R\left(s\right){\Gamma }_S\left(s\right)e^{-2s\tau }}\right]$

Calculate the characteristic impedance, the wave velocity, and the transit time.

(a)

Calculate the characteristics impedance.

Substitute $\frac{1}{3}\times {10}^{-6}\frac{\text{H}}{\text{m}}$for $L$, and $\frac{1}{3}\times {10}^{-10}\frac{\text{F}}{\text{m}}$for$C$into equation (1).

$\begin{array}{l}v=\frac{1}{\sqrt{\frac{1}{3}\times {10}^{-6}\times \frac{1}{3}\times {10}^{-10}}}\\ v=\sqrt{\frac{3^2}{{10}^{-16}}}\\ v=\sqrt{9\times {10}^{-16}}\\ v=3\times {10}^8\frac{\text{m}}{\text{s}}\end{array}$

Calculate the value of the transit time.

Substitute$50\text{\hspace{0.33em}km}$for$l$ and$3\times {10}^8\frac{\text{m}}{\text{s}}$ for$v$ into equation (3).

$\begin{array}{l}\tau =\frac{50\times {10}^3}{3\times {10}^8}\\ \tau =0.1667\text{\hspace{0.33em}ms}\end{array}$

Hence the value of characteristics impedance, velocity and transit time are$100\Omega$ , $3\times {10}^8\frac{\text{m}}{\text{s}}$and $0.1667\text{\hspace{0.33em}ms}$respectively.

Calculate the sending-and receiving-end voltage reflection coefficients.

(b)

Calculate the value of the impedance at the receiving end.

$\begin{array}{l}{Z}_R\left(s\right)=100\parallel 2\times {10}^{-3}s\\ Z_R\left(s\right)=\frac{100\times 2\times {10}^{-3}}{100+s2\times {10}^{-3}}\\ Z_R\left(s\right)=\frac{100s}{\frac{100}{2\times {10}^{-3}}+s}\\ Z_R\left(s\right)=\frac{100s}{s+50000}\Omega \end{array}$

Calculate the receiving end voltage reflection.

Substitute$\frac{100s}{s+50000}\Omega$ for $Z_R\left(s\right)$and$100\Omega$for $Z_c$into equation (4).

$\begin{array}{l}{\Gamma }_R\left(s\right)=\frac{\left(\frac{\frac{100s}{s+50000}}{100}\right)-1}{\left(\frac{\frac{100s}{s+50000}}{100}\right)+1}\\ {\Gamma }_R\left(s\right)=\frac{\frac{s}{s+50000}-1}{\frac{s}{s+50000}+1}\\ {\Gamma }_R\left(s\right)=\frac{s-s-50000}{s+s+50000}\\ {\Gamma }_R\left(s\right)=\frac{-25000}{s+25000}\text{\hspace{0.33em}pu}\end{array}$

Calculate the sending end voltage reflection.

Substitute$100\Omega$ for $Z_G\left(s\right)$and $100\Omega$for $Z_c$into equation (5).

$\begin{array}{l}{\Gamma }_S\left(s\right)=\frac{\frac{100}{100}-1}{\frac{100}{100}+1}\\ {\Gamma }_S\left(s\right)=\frac{1-1}{1+1}\\ {\Gamma }_S\left(s\right)=0\text{\hspace{0.33em}pu}\end{array}$

Hence the sending-and receiving-end voltage reflection coefficients are $0\text{\hspace{0.33em}pu}$and$\frac{-25000}{s+25000}\text{\hspace{0.33em}pu}$ respectively.

Calculate the Laplace transform of the receiving-end current, IR(s).

(c)

Laplace of source voltage$E_G\left(s\right)=\frac{100}{s}\text{\hspace{0.33em}kV}$,

Calculate the Laplace of current at distance$x$ .

Substitute$100\Omega$ for$Z_c$ , $100\Omega$for$Z_G\left(s\right)$ , $\frac{100}{s}\text{\hspace{0.33em}kV}$for$E_G$ , $\frac{-25000}{s+25000}\text{\hspace{0.33em}pu}$for${\Gamma }_R$ and $0$for${\Gamma }_S$ into equation (6).

$\begin{array}{l}{I}_R\left(s\right)=\frac{100\times {10}^3}{s}\left(\frac{1}{100+100}\right)\left[\frac{e^{-\frac{sx}{v}}-\left(\frac{-25000}{s+25000}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1-\left(0\right)\left(\frac{-25000}{s+25000}\text{\hspace{0.33em}pu}\right)e^{-2s\tau }}\right]\\ I_R\left(s\right)=\frac{100\times {10}^3}{200s}\left[\frac{e^{-\frac{sx}{v}}+\frac{25000}{s+25000}{e}^{s\left[\frac{x}{v}-2\tau \right]}}{1}\right]\end{array}$

At the receiving end the value of$x$ is equal to$l$ .

Substitute$l$ for $x$into above equation.

$I_R\left(s\right)=\frac{500}{s}\left[e^{-\frac{sl}{v}}+\left(\frac{s+25000}{s+25000}-\frac{s}{s+25000}\right)e^{s\left[\frac{l}{v}-2\tau \right]}\right]$

Substitute$\frac{l}{v}$ for$\tau$ into above equation.

$\begin{array}{l}{I}_R\left(s\right)=\frac{500}{s}\left[e^{-s\tau }+\left(\frac{s+25000}{s+25000}-\frac{s}{s+25000}\right)e^{s\left[\tau -2\tau \right]}\right]\\ I_R\left(s\right)=\frac{500}{s}\left[e^{-s\tau }+\left(\frac{s+25000}{s+25000}-\frac{s}{s+25000}\right)e^{-s\tau }\right]\\ I_R\left(s\right)=\frac{500}{s}\left[e^{-s\tau }+e^{-s\tau }-\frac{s{e}^{-s\tau }}{s+25000}\right]\\ I_R\left(s\right)=\frac{500}{s}\left[2e^{-s\tau }-\frac{s{e}^{-s\tau }}{s+25000}\right]\text{\hspace{0.33em}A}\end{array}$

Hence the receiving end current$I_R\left(s\right)$ is $\frac{500}{s}\left[2e^{-s\tau }-\frac{s{e}^{-s\tau }}{s+25000}\right]\text{\hspace{0.33em}A}$.

Calculate the receiving-end current   as a function of time.

(d)

The receiving end currents.

$I_R\left(s\right)=500\left[2\frac{e^{-s\tau }}{s}-\frac{e^{-s\tau }}{s+25000}\right]$

Take Laplace inverse of the above equation.

$I_R\left(s\right)=500\left[2\frac{e^{-s\tau }}{s}-\frac{e^{-s\tau }}{s+25000}\right]$

Hence, the value of$i_R\left(t\right)$ is$500\left(2-e^{-25000\left(t-\tau \right)}\right)u_{-1}\left(t-\tau \right)\text{\hspace{0.33em}A}$ .