Question

Question: Rework Problem 13.9 if the source voltage is a pulse of magnitude and duration ; that is,$e_G\left(t\right)=E\left[u_{-1}\left(t\right)-u_{-1}\left(t-t/10\right)\right]$ . and $Z_R=5Z_c$are the $Z_G=z_c/3$same as in Problem 13.9. Also plot versus time t for $0⩽t⩽6\tau$.

Short answer

Answer

The Bewley Lattice diagram as,

Plot the voltage verses .

Step-by-step solution

Write the given data from the question.

The source voltage,$e_G\left(t\right)=E\left[u_{-1}\left(t\right)-u_{-1}\left(t-\frac{\tau }{10}\right)\right]$

The Source impedance,$Z_G=\frac{Z_c}{3}$

The receiving end impedance,$Z_R=5Z_c$

 Step 2: Determine the equation to draw the Bewley lattice diagram.

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

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

Here, is the receiving end impedance and is the characteristics impedance.

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

${\Gamma }_S\left(s\right)=\frac{\frac{Z_G\left(s\right)}{Z_c}-1}{\frac{Z_G\left(s\right)}{Z_c}+1}$ …… (2)

The equation to calculate the first forward travelling wave is given as follows.

$V_1\left(s\right)=E_G\left(s\right)\left[\frac{Z_c}{Z_c+Z_G}\right]$

Draw the Bewley lattice diagram.

The Laplace of the source voltage,

$e_G\left(t\right)=E\left[u_{-1}\left(t\right)-u_{-1}\left(t-\frac{\tau }{10}\right)\right]$

Apply the Laplace transform.

$$\begin{gathered} E_G\left(s\right)=E\left[\frac{1}{s}-\frac{e^{-s\left(\frac{\tau }{10}\right)}}{s}\right] \\ {E}_G\left(s\right)=\frac{E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\text{V} \end{gathered}$$

Calculate the receiving end voltage reflection.

Substitute for into equation (1).

$$\begin{gathered} {\Gamma }_R\left(s\right)=\frac{\frac{5Z_c}{Z_c}-1}{\frac{5Z_c}{Z_c}+1} \\ {\Gamma }_R\left(s\right)=\frac{5-1}{5+1} \\ {\Gamma }_R\left(s\right)=\frac{4}{6} \\ {\Gamma }_R\left(s\right)=\frac{2}{3}\text{pu} \end{gathered}$$

Calculate the sending end voltage reflection.

Substitute for into equation (2).

$$\begin{gathered} {\Gamma }_S\left(s\right)=\frac{\frac{Z_c}{3Z_c}-1}{\frac{Z_c}{3Z_c}+1} \\ {\Gamma }_S\left(s\right)=\frac{\frac{1}{3}-1}{\frac{1}{3}+1} \\ {\Gamma }_S\left(s\right)=\frac{-1}{2}\text{pu} \end{gathered}$$

Calculate the first forward travelling wave.

Substitute $\frac{Z_c}{3}$for $Z_G\left(s\right)$and for into equation (3).

$$\begin{gathered} V_1\left(s\right)=\frac{E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\left[\frac{Z_c}{Z_c+\frac{Z_c}{3}}\right] \\ {V}_1\left(s\right)=\frac{E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\times \frac{3}{4} \end{gathered}$$

…… (4)

Calculate the second backward travelling wave.

$V_2\left(s\right)={\Gamma }_R\left(s\right)V_1\left(s\right)$

Substitute for and for into above equation.

$\frac{0.75E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\text{V}$

…… (5)

Calculate the third forward travelling wave.

$V_2\left(s\right)={\Gamma }_R\left(s\right)V_1\left(s\right)$

Substitute localid="1656161663168" $\frac{0.5E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\text{V}$for V₂ (S) and -1/2 p.u for${\Gamma }_S\left(s\right)$into above equation.

localid="1656161671222" $$\begin{gathered} V_3\left(s\right)=V_2\left(s\right)=\frac{0.5E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\times \frac{-1}{2} \\ {V}_3\left(s\right)=-\frac{0.25E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right] \end{gathered}$$ …… (6)

Calculate the fourth backward travelling wave.

localid="1656161683272" $V_4\left(s\right)={\Gamma }_R\left(s\right)V_3\left(s\right)$

Substitutelocalid="1656161690210" $-\frac{0.25E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]$ for V₃ (S) and 2/3 P.u forlocalid="1656161920064" ${\Gamma }_R\left(s\right)$into above equation.

localid="1656161965861" $$\begin{gathered} V_4\left(s\right)=-\frac{0.25E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\times \frac{2}{3} \\ {V}_4\left(s\right)=-\frac{0.1667E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\text{V} \end{gathered}$$

…… (7)

Calculate the fifth forward travelling wave.

localid="1656161979031" $V_5\left(s\right)={\Gamma }_S\left(s\right)V_4\left(s\right)$

Substitutelocalid="1656161989818" $-\frac{0.1667E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\text{V}$ for V₃ (S) and2/3 pu for${\Gamma }_R\left(s\right)$into above equation.

localid="1656162001756" $$\begin{gathered} V_5\left(s\right)=-\frac{0.1667E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\times \frac{-1}{2} \\ {V}_5\left(s\right)=\frac{0.0834E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\text{V} \end{gathered}$$

…… (8)

Calculate the sixth backward travelling wave.

localid="1656162013012" $V_6\left(s\right)={\Gamma }_R\left(s\right)V_5\left(s\right)$

Substitutelocalid="1656162027931" $\frac{0.0834E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\text{V}$for and for into above equation.

localid="1656162038956" $$\begin{gathered} V_6\left(s\right)=\frac{0.0834E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\times \frac{2}{3} \\ {V}_6\left(s\right)=\frac{0.0556E}{s}\left[1-e^{-s\left(\frac{\tau }{10}\right)}\right]\text{V} \end{gathered}$$

…… (9)

Draw the Bewley Lattice diagram as,

From the equation (4).

$V_1\left(s\right)=0.75E\left[\frac{1}{s}-\frac{e^{-s\left(\frac{\tau }{10}\right)}}{s}\right]$$v_1\left(x,t\right)=0.75E\left[u_{-1}\left(t\right)-u_{-1}\left(t-\frac{\tau }{10}\right)\right]$

Take the Laplace inverse of the above equation,

$v_1\left(x,t\right)=0.75E\left[u_{-1}\left(t\right)-u_{-1}\left(t-\frac{\tau }{10}\right)\right]$

The voltage waveform cuts the dotted line at the time period .

$$\begin{gathered} v_1\left(\frac{l}{3},\frac{\tau }{10}\right)=0.75E\left[u_{-1}\left(\frac{\tau }{10}\right)-u_{-1}\left(\frac{\tau }{10}-\frac{\tau }{10}\right)\right] \\ {v}_1\left(\frac{l}{3},\frac{\tau }{10}\right)=0.75E\left[u_{-1}\left(\frac{\tau }{10}\right)-u_{-1}\left(0\right)\right]\text{V} \end{gathered}$$

Therefore, the pulse waveform of the amplitude having the duration from to .

From the equation (5).

$V_2\left(s\right)=0.5E\left[\frac{1}{s}-\frac{e^{-s\left(\frac{\tau }{10}\right)}}{s}\right]$

Take the Laplace inverse of the above equation,

$v_2\left(x,t\right)=0.5E\left[u_{-1}\left(t\right)-u_{-1}\left(t-\frac{\tau }{10}\right)\right]$

The voltage waveform cuts the dotted line at the time period .

$$\begin{gathered} 2\tau -\frac{\tau }{10}=\frac{19\tau }{10} \\ {v}_2\left(\frac{l}{3},\frac{19\tau }{10}\right)=0.5E\left[u_{-1}\left(\frac{19\tau }{10}\right)-u_{-1}\left(\frac{19\tau }{10}-\frac{\tau }{10}\right)\right] \\ {v}_2\left(\frac{l}{3},\frac{19\tau }{10}\right)=0.5E\left[u_{-1}\left(\frac{19\tau }{10}\right)-u_{-1}\left(\frac{18\tau }{10}\right)\right]\text{V} \end{gathered}$$

Therefore, the pulse waveform of the amplitude having the duration from to .

From the equation (6).

$V_3\left(s\right)=-0.25E\left[\frac{1}{s}-\frac{e^{-s\left(\frac{\tau }{10}\right)}}{s}\right]$

Take the Laplace inverse of the above equation,

$V_1\left(s\right)=0.75E\left[\frac{1}{s}-\frac{e^{-s\left(\frac{\tau }{10}\right)}}{s}\right]$

The voltage waveform cuts the dotted line at the time period .

role="math" localid="1656162676787" $$\begin{gathered} 2\tau +\frac{\tau }{10}=\frac{21\tau }{10} \\ {v}_3\left(\frac{l}{3},\frac{21\tau }{10}\right)=-0.25E\left[u_{-1}\left(\frac{21\tau }{10}\right)-u_{-1}\left(\frac{21\tau }{10}-\frac{\tau }{10}\right)\right] \\ {v}_3\left(\frac{l}{3},\frac{21\tau }{10}\right)=-0.25E\left[u_{-1}\left(\frac{21\tau }{10}\right)-u_{-1}\left(\frac{20\tau }{10}\right)\right]\text{V} \end{gathered}$$

Therefore, the pulse waveform of the amplitude having the duration from 2 hr to 2.1 hrs.

From the equation (7).

$V_4\left(s\right)=-0.1667E\left[\frac{1}{s}-\frac{e^{-s\left(\frac{\tau }{10}\right)}}{s}\right]$

Take the Laplace inverse of the above equation,

$v_4\left(x,t\right)=-0.1667E\left[u_{-1}\left(t\right)-u_{-1}\left(t-\frac{\tau }{10}\right)\right]$

The voltage waveform cuts the dotted line at the time period .

$$\begin{gathered} 4\tau -\frac{\tau }{10}=\frac{39\tau }{10} \\ {v}_4\left(\frac{l}{3},\frac{39\tau }{10}\right)=-0.1667E\left[u_{-1}\left(\frac{39\tau }{10}\right)-u_{-1}\left(\frac{39\tau }{10}-\frac{\tau }{10}\right)\right] \\ {v}_4\left(\frac{l}{3},\frac{39\tau }{10}\right)=-0.1667E\left[u_{-1}\left(\frac{39\tau }{10}\right)-u_{-1}\left(\frac{38\tau }{10}\right)\right]\text{V} \end{gathered}$$

Therefore, the pulse waveform of the amplitude having the duration from to .

From the equation (8).

Take the Laplace inverse of the above equation,

The voltage waveform cuts the dotted line at the time period .

Therefore, the pulse waveform of the amplitude having the duration from to .

From the equation (9).

Take the Laplace inverse of the above equation,

The voltage waveform cuts the dotted line at the time period .

Therefore, the pulse waveform of the amplitude having the duration from to .

Plot the voltage verses .