Chapter 11, Problem 4P

(a)

To determine

Find the value of line parameters R, L, G, and C.

Answer to Problem 4P

The value of line parameters are $R=0.25\text{Ω}/\text{m}$, $L=83.8\text{nH}/\text{m}$, $G=15\text{mS}/\text{m}$, and $C=530.5\text{pF}/\text{m}$.

Explanation of Solution

Calculation:

Consider the general expression for line parameter R of a planar line.

$R=\frac{2}{w\delta {\sigma }_c}$        (1)

Here,

${\sigma }_c$ is the conductivity of copper,

$\delta$ is the skin depth, and

w is the width.

Consider the general expression for skin depth.

$\delta =\frac{1}{\sqrt{\pi f{\mu }_c{\sigma }_c}}$

Here,

f is the frequency, and

${\mu }_c$ is the permeability.

Substitute $200\times {10}^6$ for f, $4\pi \times {10}^{-7}$ for ${\mu }_c$, and $5.8\times {10}^7$ for ${\sigma }_c$ in above equation.

$\begin{array}{c}\delta =\frac{1}{\sqrt{\pi \left(200\times {10}^6\right)\left(4\pi \times {10}^{-7}\right)\left(5.8\times {10}^7\right)}}\\ =4.6729\times {10}^{-6}\text{m}\end{array}$

Substitute $4.6729\times {10}^{-6}$ for $\delta$, $30\times {10}^{-3}$ for w, and $5.8\times {10}^7$ for ${\sigma }_c$ in Equation (1).

$\begin{array}{c}R=\frac{2}{\left(30\times {10}^{-3}\right)\left(4.6729\times {10}^{-6}\right)\left(5.8\times {10}^7\right)}\\ =0.25\text{Ω}/\text{m}\end{array}$

Consider the general expression for line parameter L of a planar line.

$L=\frac{{\mu }_{c}d}{w}$

Here,

d is the thickness.

Substitute $4\pi \times {10}^{-7}$ for ${\mu }_c$, $30\times {10}^{-3}$ for w, and $2\times {10}^{-3}$ for d in above equation.

$\begin{array}{c}L=\frac{\left(4\pi \times {10}^{-7}\right)\left(2\times {10}^{-3}\right)}{\left(30\times {10}^{-3}\right)}\\ =83.8\times {10}^{-9}\\ =83.8\text{nH}/\text{m}\end{array}$

Consider the general expression for line parameter G of a planar line.

$G=\frac{\sigma w}{d}$

Substitute ${10}^{-3}$ for $\sigma$, $30\times {10}^{-3}$ for w, and $2\times {10}^{-3}$ for d in above equation.

$\begin{array}{c}G=\frac{\left({10}^{-3}\right)\left(30\times {10}^{-3}\right)}{\left(2\times {10}^{-3}\right)}\\ =15\times {10}^{-3}\\ =15\text{mS}/\text{m}\end{array}$

Consider the general expression for line parameter C of a planar line.

$C=\frac{\epsilon w}{d}$

Substitute $4{\epsilon }_{\text{o}}$ for $\epsilon$ in above equation.

$C=\frac{4{\epsilon }_{\text{o}}w}{d}$

Substitute $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$, $30\times {10}^{-3}$ for w, and $2\times {10}^{-3}$ for d in above equation.

$\begin{array}{c}C=\frac{4\left(\frac{{10}^{-9}}{36\pi }\right)\left(30\times {10}^{-3}\right)}{\left(2\times {10}^{-3}\right)}\\ =530.5\times {10}^{-12}\\ =530.5\text{pF}/\text{m}\end{array}$

Conclusion:

Thus, the value of line parameters are $R=0.25\text{Ω}/\text{m}$, $L=83.8\text{nH}/\text{m}$, $G=15\text{mS}/\text{m}$, and $C=530.5\text{pF}/\text{m}$.

(b)

To determine

Find the value of propagation constant $\gamma$ and characteristic impedance $Z_{\text{o}}$.

Answer to Problem 4P

The value of propagation constant is $\gamma =0.104+j8.378\text{1}/\text{m}$ and characteristic impedance is $Z_{\text{o}}=12.56+j0.13\Omega$.

Explanation of Solution

Calculation:

Find the value of propagation constant.

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

Consider the general expression for angular frequency.

$\omega =2\pi f$

Substitute $200\times {10}^6$ for f in above equation.

$\begin{array}{c}\omega =2\pi \left(200\times {10}^6\right)\\ =1256.64\times {10}^6\text{rad}/\text{s}\end{array}$

Substitute 0.25 for R, $83.8\times {10}^{-9}$ for L, $1256.64\times {10}^6$ for $\omega$, $15\times {10}^{-3}$ for G, and $530.5\times {10}^{-12}$ for C in Equation (2).

$\begin{array}{c}\gamma =\sqrt{\left[0.25+j\left(1256.64\times {10}^6\right)\left(83.8\times {10}^{-9}\right)\right]\left[\left(15\times {10}^{-3}\right)+j\left(1256.64\times {10}^6\right)\left(530.5\times {10}^{-12}\right)\right]}\\ =\sqrt{\left[0.25+j\left(105.3\right)\right]\left[\left(15\times {10}^{-3}\right)+j\left(0.6666\right)\right]}\\ =\sqrt{70.21\angle 178.57°}\\ =8.3791\angle 89.285\end{array}$

The above equation becomes,

$\gamma =0.104+j8.378\text{1}/\text{m}$

Consider the general expression for characteristic impedance.

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

Substitute 0.25 for R, $83.8\times {10}^{-9}$ for L, $1256.64\times {10}^6$ for $\omega$, $15\times {10}^{-3}$ for G, and $530.5\times {10}^{-12}$ for C in above equation.

$\begin{array}{c}{Z}_{\text{o}}=\sqrt{\frac{\left[0.25+j\left(1256.64\times {10}^6\right)\left(83.8\times {10}^{-9}\right)\right]}{\left[\left(15\times {10}^{-3}\right)+j\left(1256.64\times {10}^6\right)\left(530.5\times {10}^{-12}\right)\right]}}\\ =\sqrt{\frac{\left[0.25+j\left(105.3\right)\right]}{\left[\left(15\times {10}^{-3}\right)+j\left(0.6666\right)\right]}}\\ =\sqrt{157.9\angle 1.153°}\\ =12.5658\angle 0.5765\end{array}$

The above equation becomes,

$Z_{\text{o}}=12.56+j0.13\Omega$

Conclusion:

Thus, the value of propagation constant is $\gamma =0.104+j8.378\text{1}/\text{m}$ and characteristic impedance is $Z_{\text{o}}=12.56+j0.13\Omega$.