Chapter 11, Problem 68P

(a)

To determine

Calculate the value of $Z_{\text{o}}$ and ${\epsilon }_{\text{eff}}$.

Answer to Problem 68P

The value are $Z_{\text{o}}=77.77\Omega$ and ${\epsilon }_{\text{eff}}=1.8$.

Explanation of Solution

Calculation:

Given, width is $w=1.5\text{cm}$ and $h=1\text{cm}$.

The ratio of line width to the substrate thickness is,

$\begin{array}{l}\frac{w}{h}=\frac{1.5}{1}\\ \frac{w}{h}=1.5\end{array}$

Consider the general expression for effective relative permittivity.

${\epsilon }_{\text{eff}}=\frac{\left({\epsilon }_r+1\right)}{2}+\frac{\left({\epsilon }_r-1\right)}{2\sqrt{1+12\frac{h}{w}}}$        (1)

Here,

${\epsilon }_r$ is relative permittivity,

h is substrate thickness, and

w is line width.

Substitute 2.2 for ${\epsilon }_r$ and 1.5 for $\frac{w}{h}$ in Equation (1).

$\begin{array}{c}{\epsilon }_{\text{eff}}=\frac{\left(2.2+1\right)}{2}+\frac{\left(2.2-1\right)}{2\sqrt{1+12\left(\frac{1}{1.5}\right)}}\\ =1.6+0.2\\ =1.8\end{array}$

Consider the general expression for characteristic impedance.

$Z_{\text{o}}=\frac{1}{\sqrt{{\epsilon }_{e\text{ff}}}}\frac{120\pi }{\left[\frac{w}{h}+1.393+0.667\ln \left(\frac{w}{h}+1.444\right)\right]}\frac{w}{h}\ge 1$        (2)

Substitute 1.8 for ${\epsilon }_{e\text{ff}}$ and 1.5 for $\frac{w}{h}$ in Equation (2).

$\begin{array}{c}{Z}_{\text{o}}=\frac{1}{\sqrt{1.8}}\frac{120\pi }{\left[1.5+1.393+0.667\ln \left(1.5+1.444\right)\right]}\\ =\frac{281}{1.5+1.393+0.667\ln \left(2.944\right)}\\ =77.77\Omega \end{array}$

Conclusion:

Thus, the value are $Z_{\text{o}}=77.77\Omega$ and ${\epsilon }_{\text{eff}}=1.8$.

(b)

To determine

Calculate the values of ${\alpha }_c$ and ${\alpha }_d$.

Answer to Problem 68P

The values are ${\alpha }_c=0.223\text{dB}/\text{m}$ and ${\alpha }_d=6.6735\text{dB}/\text{m}$.

Explanation of Solution

Calculation:

Consider the general expression for attenuation due to conductor loss.

${\alpha }_c=8.686\frac{R_s}{w{Z}_{\text{o}}}$        (3)

Consider the general expression for skin resistance of the conductor.

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

Substitute $4\pi \times {10}^{-7}$ for $\mu$, $1.1\times {10}^{-7}$ for ${\sigma }_c$, and $2.5\times {10}^9$ for f in above equation.

$\begin{array}{c}{R}_s=\sqrt{\frac{\left(4\pi \times {10}^{-7}\right)\left(\pi \right)\left(2.5\times {10}^9\right)}{1.1\times {10}^{-7}}}\\ =2.995\times {10}^{-2}\text{Ω}/{\text{m}}^{\text{2}}\end{array}$

Substitute $2.995\times {10}^{-2}$ for $R_s$, $1.5\times {10}^{-2}$ for w, and 77.77 for $Z_{\text{o}}$ in Equation (3).

$\begin{array}{c}{\alpha }_c=\left(8.686\right)\frac{\left(2.995\times {10}^{-2}\right)}{\left(1.5\times {10}^{-2}\right)\left(77.77\right)}\\ =0.223\text{dB}/\text{m}\end{array}$

Consider the general expression for phase velocity.

$u=\frac{c}{\sqrt{{\epsilon }_{e\text{ff}}}}$        (4)

Consider the general expression for line wavelength.

$\begin{array}{l}\lambda =\frac{u}{f}\\ u=f\lambda \end{array}$

Substitute $f\lambda$ for u in Equation (4).

$\begin{array}{l}f\lambda =\frac{c}{\sqrt{{\epsilon }_{e\text{ff}}}}\\ \lambda =\frac{c}{f\sqrt{{\epsilon }_{e\text{ff}}}}\end{array}$

Substitute $1.8$ for ${\epsilon }_{\text{eff}}$, $2.5\times {10}^9$ for f, and $3\times {10}^8$ for c in above equation.

$\begin{array}{c}\lambda =\frac{3\times {10}^8}{2.5\times {10}^9\sqrt{1.8}}\\ =8.944\times {10}^{-2}\text{m}\end{array}$

Consider the general expression for attenuation due to dielectric loss.

${\alpha }_d=27.3\frac{\left({\epsilon }_{\text{eff}}-1\right){\epsilon }_r}{\left({\epsilon }_r-1\right)\sqrt{{\epsilon }_{\text{eff}}}}\frac{\tan \theta }{\lambda }$

Substitute $8.944\times {10}^{-2}$ for $\lambda$, 2.2 for ${\epsilon }_r$, 0.02 for $\tan \theta$, and 1.8 for ${\epsilon }_{e\text{ff}}$ in above equation.

$\begin{array}{c}{\alpha }_d=\left(27.3\right)\frac{\left(1.8-1\right)\left(2.2\right)}{\left(2.2-1\right)\sqrt{1.8}}\frac{\left(0.02\right)}{\left(8.944\times {10}^{-2}\right)}\\ =6.6735\text{dB}/\text{m}\end{array}$

Conclusion:

Thus, the values are ${\alpha }_c=0.223\text{dB}/\text{m}$ and ${\alpha }_d=6.6735\text{dB}/\text{m}$.

(c)

To determine

Calculate the distance down the line before the wave drops by 20 dB.

Answer to Problem 68P

The distance down the line is $\mathcal{l}=2.9\text{m}$.

Explanation of Solution

Calculation:

The total attenuation constant is the sum of the ohmic attenuation constant ${\alpha }_c$ and the dielectric attenuation constant ${\alpha }_d$. That is,

$\alpha ={\alpha }_c+{\alpha }_d$

Substitute 0.223 for ${\alpha }_c$ and 6.6735 for ${\alpha }_d$ in above equation.

$\begin{array}{c}\alpha =0.223+6.6735\\ =6.8965\text{dB}/\text{m}\end{array}$

Given that, the wave drops by 20 dB. That is,

$\alpha \mathcal{l}=20\text{dB}$

Rearrange the above equation.

$\mathcal{l}=\frac{20}{\alpha }$

Substitute 6.8965 for $\alpha$ in above equation.

$\begin{array}{c}\mathcal{l}=\frac{20}{6.8965}\\ =2.9\text{m}\end{array}$

Conclusion:

Thus, the distance down the line is $\mathcal{l}=2.9\text{m}$.