Question

For the case of a lossless line, how would the results of Problem 5.33 change? In terms of ${\mathit{Z}}_{\mathbf{c}}$, which is a real quantity for this case, express ${\mathit{P}}_{\mathbf{12}}$in terms$\left|I_1\right|$ and $\left|V_1\right|$.

Short answer

(a) The driving point impedance$\frac{V_1}{I_1}$is $Z_c$.

(b) The voltage gain$\frac{\left|V_2\right|}{\left|V_2\right|}$ is.1

(c) The current gain$\frac{\left|I_2\right|}{\left|I_1\right|}$ is $1$.

(d) The complex power gain$\frac{\left|I_2\right|}{\left|I_1\right|}$ is.1

(e) The real power gain $\eta$is $1$.

Step-by-step solution

Determine the general equation of sending end voltage and current.

The equation for sending end voltage is given as follows.

${\mathit{V}}_{\mathbf{1}}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\left(\gamma l\right){\mathit{V}}_{\mathbf{2}}\mathbf{+}{\mathit{Z}}_{\mathbf{c}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\left(\gamma l\right){\mathit{I}}_{\mathbf{2}}$ …… (1)

The equation for sending end current is given as follows.

${\mathit{I}}_{\mathbf{1}}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\left(\gamma l\right)\left(\frac{1}{Z_c}\right){\mathit{V}}_{\mathbf{2}}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\left(\gamma l\right){\mathit{I}}_{\mathbf{2}}$ ……. (2)

Here ${\mathit{V}}_{\mathbf{1}}$and ${\mathit{I}}_{\mathbf{1}}$are the sending end voltage and current respectively, ${\mathit{V}}_{\mathbf{2}}$and${\mathit{I}}_{\mathbf{2}}$are the receiving end voltage and current respectively$Z_c$and is the characteristics impedance.

Derive the equation for driving point impedance.

(a)

Write the receiving end voltage is given as follows.

$\begin{array}{l}{V}_2=I_2{Z}_c\\ I_2=\frac{V_2}{Z_c}\end{array}$ …… (3)

Substitute the expression of $I_2$into equation (1).

$\begin{array}{l}{V}_1=\cosh \left(\gamma l\right)V_2+Z_c\sinh \left(\gamma l\right)\frac{V_2}{Z_c}\\ V_1=\cosh \left(\gamma l\right)V_2+\sinh \left(\gamma l\right)V_2\end{array}$

$V_1=V_2\left[\cosh \left(\gamma l\right)+\sinh \left(\gamma l\right)\right]$…… (4)

$V_1=V_2\left[\frac{e^{\gamma l}+e^{-\gamma l}}{2}+\frac{e^{\gamma l}-e^{-\gamma l}}{2}\right]$

Solve further as,

$\begin{array}{l}{V}_1=V_2\left[\frac{e^{\gamma l}+e^{-\gamma l}+e^{\gamma l}-e^{-\gamma l}}{2}\right]\\ V_1=V_2\left[\frac{2e^{\gamma l}}{2}\right]\end{array}$

$V_1=V_2{e}^{\gamma l}$

Substitute $\alpha +j\beta$ for$\gamma$ into above equation.

$\begin{array}{l}{V}_1=V_2{e}^{\left(\alpha +j\beta \right)l}\\ V_1=V_2{e}^{\alpha l}{e}^{j\beta l}\end{array}$ …… (5)

Substitute the equation (3) into equation (2).

$I_1=\sinh \left(\gamma l\right)I_2+\cosh \left(\gamma l\right)I_2$

$I_1=\left[\sinh \left(\gamma l\right)+\cosh \left(\gamma l\right)\right]I_2$ …… (6)

$\begin{array}{l}{I}_1=\left[\frac{e^{\gamma l}+e^{-\gamma l}}{2}+\frac{e^{\gamma l}-e^{-\gamma l}}{2}\right]I_2\\ I_1=I_2{e}^{\gamma l}\end{array}$

Substitute $\alpha +j\beta$ for$\gamma$ into above equation.

$I_1=I_2{e}^{\alpha l}{e}^{j\beta l}$ …… (7)

Divide the equation (5) by equation (7).

$\begin{array}{l}\frac{V_1}{I_1}=\frac{V_2{e}^{\alpha l}{e}^{j\beta l}}{I_2{e}^{\alpha l}{e}^{j\beta l}}\\ \frac{V_1}{I_1}=\frac{V_2}{I_2}\end{array}$

From equation (3),

$\frac{V_1}{I_1}=Z_c$

Hence, the driving point impedance$\frac{V_1}{I_1}$is$Z_c$ .

Derive the equation for voltage gain.

(b)

Calculate the magnitude of the sending end voltage from equation (4).

$\begin{array}{l}\left|V_1\right|=\left|V_2\right|\sqrt{{\sin }^2\left(\gamma l\right)+{\cos }^2\left(\gamma l\right)}\\ \left|V_1\right|=\left|V_2\right|\end{array}$

Calculate the voltage gain from the above equation,

$\frac{\left|V_2\right|}{\left|V_1\right|}=1$

Hence, the voltage gain $\frac{\left|V_2\right|}{\left|V_2\right|}$is $1$.

 Step 4: Derive the equation for current gain.

(c)

Calculate the magnitude of the sending end current from equation (6).

$\begin{array}{l}\left|I_1\right|=\left|I_2\right|\sqrt{{\sin }^2\left(\gamma l\right)+{\cos }^2\left(\gamma l\right)}\\ \left|I_1\right|=\left|I_2\right|\end{array}$

Calculate the current gain from the above equation,

$\frac{\left|I_2\right|}{\left|I_1\right|}=1$ …… (8)

Hence, the voltage gain $\frac{\left|I_2\right|}{\left|I_1\right|}$is$1$ .

Calculate the complex power gain.

(d)

The expression for the complex is given as follows,

$-S_{21}=V_2{I}_2^{*}$

Substitute$V_1{e}^{-j\beta l}$ for $V_2$ and $I_1{e}^{-j\beta l}$for $I_2$ above equation.

$\begin{array}{l}-S_{21}=\left(V_1{e}^{-j\beta l}\right)\left(I_1^{*}{e}^{j\beta l}\right)\\ -S_{21}=\left(V_1{I}_1^{*}\right)\\ -S_{21}=S_{12}\end{array}$

Calculate the complex power gain

$\frac{-S_{21}}{S_{12}}=1$

Hence, the complex power gain$\frac{-S_{21}}{S_{12}}$ is.$1$

Calculate the real power gain.

The real power gain is given as,

$\begin{array}{l}\eta =-\frac{P_{21}}{P_{12}}\\ \eta =\frac{\text{real\hspace{0.33em}}\left(Z_c{I}_1^2\right)}{\text{real\hspace{0.33em}}\left(Z_c{I}_2^2\right)}\\ \eta =\frac{I_1^2}{I_2^2}\\ \eta ={\left(\frac{I_1}{I_2}\right)}^2\end{array}$

From equation (8).

$\begin{array}{l}\eta =-\frac{P_{21}}{P_{12}}\\ \eta =1\end{array}$

Hence, the real power gain$\eta$ is $1$.