Question

The single-phase, two-wire lossless line in Figure 13.3 has a series inductance $L=0.999\times {10}^{-6}H/m$, a shunt capacitance $C=1.112\times {10}^{-11}F/m$, and a $60Km$line length. The source voltage at the sending end is a ramp $e_G\left(t\right)=Et{u}_{-1}\left(t\right)=E{u}_{-2}\left(t\right)kV$with a source impedance equal to the characteristic impedance of the line. The receiving-end load consists of a $150\Omega$resistor in parallel with a $1\mu F$capacitor. The line and load are initially unenergized. Determine (a) the characteristic impedance in $\Omega$, the wave velocity in , and the transit time in for this line; (b) the sending- and receiving-end voltage reflection coefficients in per-unit; (c) the Laplace transform of the sending-end voltage, localid="1656144662132" $V_S\left(s\right)$; and (d) the sending-end voltage localid="1656144667884" $v_S\left(t\right)$as a function of time.

Short answer

(a) The value of characteristics impedance, velocity and transit time are$299.73\Omega$, $3\times {10}^8\frac{\text{m}}{\text{s}}$and $0.2\text{\hspace{0.33em}ms}$respectively.

(b) The sending-and receiving-end voltage reflection coefficients are $0\text{\hspace{0.33em}pu}$and $\frac{-s-3330}{s+10003}\text{\hspace{0.33em}pu}$respectively.

(c) The sending end voltage $v_S\left(t\right)$is .$V_S\left(s\right)=\frac{E}{2}\left[\frac{1}{s^2}+\left(-\frac{0.333}{s^2}-\frac{6.67\times {10}^{-5}}{s}+\frac{6.67\times {10}^{-5}}{s+10003}\right)e^{-2s\tau }\right]$

(d) The sending end voltage $v_S\left(t\right)$is

$v_S\left(t\right)=\frac{E}{2}\left\{t{u}_{-2}\left(t\right)-\left[0.333\left(t-0.4\right)+6.67\times {10}^{-5}-6.67\times {10}^{-5}{e}^{-10003\left(t-0.4\right)}\right]\right\}{u}_{-1}\left(t-0.4\right)$

Step-by-step solution

Write the given data from the question.

The series inductance $L=0.999\times {10}^{-6}\frac{\text{H}}{\text{m}}$,

The shunt capacitance$C=1.112\times {10}^{-11}\frac{\text{F}}{\text{m}}$,

The length of the line,$l=60\text{\hspace{0.33em}km}$

The source voltage,

role="math" $\begin{array}{l}{e}_G\left(t\right)=Et{u}_{-1}\left(t\right)\text{\hspace{0.33em}kV}\\ e_G\left(t\right)=E{u}_{-2}\left(t\right)\text{\hspace{0.33em}kV}\end{array}$

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

The receiving end Capacitor,$C=1\mu \text{F}$

The receiving end resistance,$R=150\Omega$

Determine the equations to calculate the characteristic impedance, the wave velocity, and the transit time, the sending- and receiving-end voltage reflection coefficients, the Laplace transform of the sending-end voltage and the sending-end voltage  vSt 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.

$v=\frac{1}{\sqrt{LC}}$ …… (2)

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

$v=\frac{1}{\sqrt{LC}}$ …… (3)

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

${\Gamma }_R\left(s\right)=\frac{\frac{Z_R\left(s\right)}{Z_c}-1}{\frac{Z_R\left(s\right)}{Z_c}+1}$ …… (4)

Here, $Z_R\left(s\right)$is the receiving end impedance and $Z_c$is the characteristics impedance.

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

${\Gamma }_S\left(s\right)=\frac{\frac{Z_G\left(s\right)}{Z_c}-1}{\frac{Z_G\left(s\right)}{Z_c}+1}$ …… (5)

The equation to calculate the voltage at distance$x$ is given as follows.

$V\left(x,s\right)=E_G\left(s\right)\left[\frac{Z_c}{Z_c+Z_G\left(s\right)}\right]\left[\frac{e^{-\frac{sx}{v}}+{\Gamma }_R\left(s\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1-{\Gamma }_R\left(s\right){\Gamma }_S\left(s\right)e^{-2s\tau }}\right]$ …… (6)

 Step 3: Calculate the characteristic impedance, the wave velocity, and the transit time.

(a)

Calculate the characteristics impedance.

Substitute $0.999\times {10}^{-6}\frac{\text{H}}{\text{m}}$for $L$, and $1.112\times {10}^{-11}\frac{\text{F}}{\text{m}}$for$C$into equation (1).

$\begin{array}{l}{Z}_c=\sqrt{\frac{0.999\times {10}^{-6}}{1.112\times {10}^{-11}}}\\ Z_c=\sqrt{8.983\times {10}^4}\\ Z_c=299.73\Omega \end{array}$

Calculate the value of the velocity.

Substitute $0.999\times {10}^{-6}\frac{\text{H}}{\text{m}}$for $L$, and $1.112\times {10}^{-11}\frac{\text{F}}{\text{m}}$for $C$into equation (2).

$\begin{array}{l}v=\frac{1}{\sqrt{0.999\times {10}^{-6}\times 1.112\times {10}^{-11}}}\\ v=\sqrt{9\times {10}^{16}}\\ v=3\times {10}^8\frac{\text{m}}{\text{s}}\end{array}$

Calculate the value of the transit time.

Substitute $60\text{\hspace{0.33em}km}$for $l$and $3\times {10}^8\frac{\text{m}}{\text{s}}$for$v$ into equation (3).

$\begin{array}{l}\tau =\frac{60\times {10}^3}{3\times {10}^8}\\ \tau =0.2\text{\hspace{0.33em}ms}\end{array}$

Hence the value of characteristics impedance, velocity and transit time are $299.73\Omega$, $3\times {10}^8\frac{\text{m}}{\text{s}}$and $0.2\text{\hspace{0.33em}ms}$respectively

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

(b)

The value of the impedance at the receiving end.

$\begin{array}{l}{Z}_R\left(s\right)=150\parallel \frac{1}{1\times {10}^{-6}s}\\ Z_R\left(s\right)=\frac{150\times \left(\frac{1}{1\times {10}^{-6}s}\right)}{150+\left(\frac{1}{1\times {10}^{-6}s}\right)}\\ Z_R\left(s\right)=\frac{150}{1+150\times {10}^{-6}s}\Omega \end{array}$

Calculate the receiving end voltage reflection.

Substitute $\frac{150}{1+150\times {10}^{-6}s}\Omega$for $Z_R\left(s\right)$and $299.73\Omega$for $Z_c$into equation (4).

$\begin{array}{l}{\Gamma }_R\left(s\right)=\frac{\left(\frac{\frac{150}{1+150\times {10}^{-6}s}}{299.73}\right)-1}{\left(\frac{\frac{150}{1+150\times {10}^{-6}s}}{299.73}\right)+1}\\ {\Gamma }_R\left(s\right)=\frac{\frac{150}{299.73\left(1+150\times {10}^{-6}s\right)}-1}{\frac{150}{299.73\left(1+150\times {10}^{-6}s\right)}+1}\\ {\Gamma }_R\left(s\right)=\frac{150-299.73-0.04495s}{150+299.73+0.04495s}\\ {\Gamma }_R\left(s\right)=\frac{-s-3330}{s+10003}\text{\hspace{0.33em}pu}\end{array}$

Calculate the sending end voltage reflection.

Substitute $299.73\Omega$for$Z_G\left(s\right)$ and$Z_c$ into equation (5).

$\begin{array}{l}{\Gamma }_S\left(s\right)=\frac{\frac{299.73}{299.73}-1}{\frac{299.73}{299.73}+1}\\ {\Gamma }_S\left(s\right)=\frac{1-1}{1+1}\\ {\Gamma }_S\left(s\right)=0\text{\hspace{0.33em}pu}\end{array}$

Hence the sending-and receiving-end voltage reflection coefficients are $0\text{\hspace{0.33em}pu}$and $\frac{-s-3330}{s+10003}\text{\hspace{0.33em}pu}$respectively.

Calculate the Laplace transform of the sending end current, .

(c)

Laplace of source voltage,$E_G\left(s\right)=\frac{E}{s^2}$

Calculate the Laplace of voltage at distance .$x$

Substitute $299.73\Omega$for $Z_c$and, $Z_G\left(s\right)$,$\frac{E}{s^2}$for $\frac{-s-3330}{s+10003}\text{\hspace{0.33em}pu}$for ${\Gamma }_R$and $0$for ${\Gamma }_S$into equation (6).

$\begin{array}{l}{V}_S\left(s\right)=\frac{E}{s^2}\left(\frac{299.73}{299.73+299.73}\right)\left[\frac{e^{-\frac{sx}{v}}-\left(\frac{-s-3330}{s+10003}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1-\left(0\right)\left(\frac{-s-3330}{s+10003}\right)e^{-2s\tau }}\right]\\ V_S\left(s\right)=\frac{E}{2s^2}\left[\frac{e^{-\frac{sx}{v}}+\left(\frac{-s-3330}{s+10003}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1}\right]\\ V_S\left(s\right)=\frac{E}{2}\left[\frac{e^{-\frac{sx}{v}}}{s^2}+\frac{1}{s^2}\left(\frac{-s-3330}{s+10003}\right)e^{s\left[\frac{x}{v}-2\tau \right]}\right]\end{array}$

At the sending end the value of $x$is equal to $0$.

Substitute $0$for $x$into above equation.

$\begin{array}{l}{V}_S\left(s\right)=\frac{E}{2}\left[\frac{e^{-\frac{s\left(0\right)}{v}}}{s^2}+\frac{1}{s^2}\left(\frac{-s-3330}{s+10003}\right)e^{s\left[\frac{\left(0\right)}{v}-2\tau \right]}\right]\\ V_S\left(s\right)=\frac{E}{2}\left[\frac{1}{s^2}+\frac{1}{s^2}\left(\frac{-s-3330}{s+10003}\right)e^{-2s\tau }\right]\end{array}$

Use the partial fraction,

$V_S\left(s\right)=\frac{E}{2}\left[\frac{1}{s^2}+\left(-\frac{0.333}{s^2}-\frac{6.67\times {10}^{-5}}{s}+\frac{6.67\times {10}^{-5}}{s+10003}\right)e^{-2s\tau }\right]$

Hence the sending end voltage is

$V_S\left(s\right)=\frac{E}{2}\left[\frac{1}{s^2}+\left(-\frac{0.333}{s^2}-\frac{6.67\times {10}^{-5}}{s}+\frac{6.67\times {10}^{-5}}{s+10003}\right)e^{-2s\tau }\right]$

Calculate the sending-end voltage   as a function of time.

The sending end voltage expression.

$V_S\left(s\right)=\frac{E}{2}\left[\frac{1}{s^2}+\left(-\frac{0.333}{s^2}-\frac{6.67\times {10}^{-5}}{s}+\frac{6.67\times {10}^{-5}}{s+10003}\right)e^{-2s\tau }\right]$

Take the Laplace inverse of the above equation.

$v_S\left(t\right)=\frac{E}{2}\left\{t{u}_{-2}\left(t\right)-\left[0.333\left(t-0.4\right)+6.67\times {10}^{-5}-6.67\times {10}^{-5}{e}^{-10003\left(t-0.4\right)}\right]\right\}{u}_{-1}\left(t-0.4\right)$

Hence the sending end voltage$v_S\left(t\right)$is .

$v_S\left(t\right)=\frac{E}{2}\left\{t{u}_{-2}\left(t\right)-\left[0.333\left(t-0.4\right)+6.67\times {10}^{-5}-6.67\times {10}^{-5}{e}^{-10003\left(t-0.4\right)}\right]\right\}{u}_{-1}\left(t-0.4\right)$

Hence, the receiving end current $v_R\left(s\right)$is

$v_R\left(t\right)=50\left\{2-3.265e^{-2000\left(t-\tau \right)}\sin \left[2449.9\left(t-\tau \right)\right]\right\}{u}_{-1}\left(t-0.5\right)\text{\hspace{0.33em}kV}$