Chapter 11, Problem 11P

(a)

To determine

Calculate the values of Z and $\gamma$.

Answer to Problem 11P

The values are $Z=644.3-j97\Omega$ and $\gamma =\left(5.415+j33.97\right)\times {10}^{-3}\frac{1}{\text{mi}}$.

Explanation of Solution

Calculation:

Consider the general expression for characteristic impedance of the line.

$Z_{\text{o}}=\sqrt{\frac{R+j\omega L}{G+j\omega C}}$        (1)

Here,

R, L, G, and C are line parameters.

Consider the general expression for angular frequency.

$\omega =2\pi f$

Here,

f is the frequency.

Substitute 1 kHz for f in above equation.

$\begin{array}{c}\omega =2\pi \left(1\times {10}^3\text{Hz}\right)\\ =6283.18\text{Hz}\end{array}$

Simplify the value of L.

$\begin{array}{c}L=3.4\left(\frac{{10}^{-3}\text{H}}{\text{mi}}\right)\\ =3.4\times {10}^{-3}\frac{\text{H}}{\text{mi}}\end{array}$

Simplify the value of C.

$\begin{array}{c}C=8.4\left(\frac{{10}^{-9}\text{F}}{\text{mi}}\right)\\ =8.4\times {10}^{-9}\frac{\text{F}}{\text{mi}}\end{array}$

Simplify the value of G.

$\begin{array}{c}G=0.42\left(\frac{{10}^{-6}\text{S}}{\text{mi}}\right)\\ =4.2\times {10}^{-7}\frac{\text{S}}{\text{mi}}\end{array}$

Substitute $6283.18$ for $\omega$, $3.4\times {10}^{-3}$ for L, $8.4\times {10}^{-9}$ for C, 6.8 for R, and $4.2\times {10}^{-7}$ for G in Equation (1).

$\begin{array}{c}{Z}_{\text{o}}=\sqrt{\frac{\left(6.8\right)+j\left(6283.18\right)\left(3.4\times {10}^{-3}\right)}{\left(4.2\times {10}^{-7}\right)+j\left(6283.18\right)\left(8.4\times {10}^{-9}\right)}}\\ =\sqrt{\frac{6.8+j21.36}{4.2\times {10}^{-7}+j5.278\times {10}^{-5}}}\\ =\sqrt{424698.18\angle -17.20°}\\ =651.68\angle -8.6°\Omega \end{array}$

Convert polar form into rectangular form in above equation.

$\begin{array}{l}{Z}_{\text{o}}=644.3-j97\Omega \\ Z=644.3-j97\Omega \end{array}$

Consider the general expression for propagation constant.

$\gamma =\sqrt{\left(R+j\omega L\right)\left(G+j\omega C\right)}$        (2)

Substitute $6283.18$ for $\omega$, $3.4\times {10}^{-3}$ for L, $8.4\times {10}^{-9}$ for C, 6.8 for R, and $4.2\times {10}^{-7}$ for G in Equation (2).

$\begin{array}{c}\gamma =\sqrt{\left[\left(6.8\right)+j\left(6283.18\right)\left(3.4\times {10}^{-3}\right)\right]\left[\left(4.2\times {10}^{-7}\right)+j\left(6283.18\right)\left(8.4\times {10}^{-9}\right)\right]}\\ =\sqrt{\left(6.8+j21.36\right)\left(4.2\times {10}^{-7}+j5.278\times {10}^{-5}\right)}\\ =\sqrt{1.1831\times {10}^{-3}\angle 161.885°}\\ =0.0344\angle 80.9425°\end{array}$

Convert polar form into rectangular form in above equation.

$\gamma =\left(5.415+j33.97\right)\times {10}^{-3}\frac{1}{\text{mi}}$        (3)

Conclusion:

Thus, the values are $Z=644.3-j97\Omega$ and $\gamma =\left(5.415+j33.97\right)\times {10}^{-3}\frac{1}{\text{mi}}$.

(b)

To determine

Calculate the phase velocity.

Answer to Problem 11P

The phase velocity is $1.85\times {10}^5\frac{\text{mi}}{\text{s}}$.

Explanation of Solution

Calculation:

Consider the general expression for propagation constant.

$\gamma =\alpha +j\beta$        (4)

Refer to Part (a).

On comparing imaginary part of Equation (3) and (4),

$\beta =33.97\times {10}^{-3}$

Consider the general expression for wave velocity.

$u=\frac{\omega }{\beta }$        (5)

Here,

$\beta$ is linear function of frequency.

Substitute $6283.18$ for $\omega$ and $33.97\times {10}^{-3}$ for $\beta$ in Equation (5).

$\begin{array}{c}u=\frac{6283.18}{33.97\times {10}^{-3}}\\ =184962\frac{\text{mi}}{\text{s}}\\ =1.84962\times {10}^5\frac{\text{mi}}{\text{s}}\\ \cong 1.85\times {10}^5\frac{\text{mi}}{\text{s}}\end{array}$

Conclusion:

Thus, the phase velocity is $1.85\times {10}^5\frac{\text{mi}}{\text{s}}$.

(c)

To determine

Calculate the wavelength.

Answer to Problem 11P

The wavelength is $185.02\text{mi}$.

Explanation of Solution

Calculation:

Consider the general expression for wavelength.

$\lambda =\frac{2\pi }{\beta }$        (6)

Substitute $33.97\times {10}^{-3}$ for $\beta$ in Equation (6).

$\begin{array}{c}\lambda =\frac{2\pi }{33.97\times {10}^{-3}}\\ =185.02\text{mi}\end{array}$

Conclusion:

Thus, the wavelength is $185.02\text{mi}$.