The primary constants for a coaxial cable at 1 GHz are $$ \mathrm{L}=250 \mathrm{nh} / \mathrm{m}, \mathrm{C}=90 $$ $$ \mathrm{pf} / \mathrm{m} $$. $$ \mathrm{R}=0.06 \mathrm{ohm} / \mathrm{m} $$, and $$ \mathrm{G}=0.005 \mathrm{mho} / \mathrm{m} $$ Determine (a) the attenuation coefficient a, (b) the phase constant b , (c) the phase velocity vp, (d) relative permittivity हr for a length of 10 m , when the innut power is 500 wate

Question

UpStudy AI · Accepted Answer

We are given the following per‐unit‐length parameters at $$ f=1\,\mathrm{GHz} $$:

- $$ L=250\,\mathrm{nH/m}=250\times10^{-9}\,\mathrm{H/m} $$
- $$ C=90\,\mathrm{pF/m}=90\times10^{-12}\,\mathrm{F/m} $$
- $$ R=0.06\,\Omega/\mathrm{m} $$
- $$ G=0.005\,\mathrm{mho/m} $$

Also, the frequency is 
$$
f=1\times10^9\,\mathrm{Hz}\quad\Longrightarrow\quad \omega=2\pi f=2\pi\times10^9\,\mathrm{rad/s}\,.
$$

The general propagation constant for a transmission line is 
$$
\gamma=\sqrt{(R+j\omega L)(G+j\omega C)}\,.
$$
For a low‐loss line (i.e. when $$R$$ and $$G$$ are small compared with $$\omega L$$ and $$\omega C$$), we may use the approximate formulas
$$
\alpha\approx\frac{1}{2}\left(\frac{R}{Z_0}+G\,Z_0\right)\quad,\quad
\beta\approx\omega\sqrt{LC}\,,
$$
where the characteristic impedance is
$$
Z_0=\sqrt{\frac{L}{C}}\,.
$$

Below we work through each part.

---

**(a) Attenuation Coefficient $$\alpha$$**

First, calculate $$Z_0$$:
$$
Z_0=\sqrt{\frac{L}{C}}=\sqrt{\frac{250\times10^{-9}}{90\times10^{-12}}}=\sqrt{\frac{250}{90}\times10^3}\,.
$$
Compute the ratio:
$$
\frac{250}{90}\approx2.7778,\quad\text{so}\quad Z_0\approx\sqrt{2.7778\times10^3}\,.
$$
Since 
$$
\sqrt{2777.8}\approx52.7\,\Omega\,,
$$
we have $$Z_0\approx52.7\,\Omega$$.

Now, the approximate attenuation is
$$
\alpha\approx\frac{1}{2}\left(\frac{R}{Z_0}+G\,Z_0\right)=\frac{1}{2}\left(\frac{0.06}{52.7}+0.005\times52.7\right)\,.
$$

Calculate each term:
$$
\frac{0.06}{52.7}\approx0.00114\quad\text{and}\quad0.005\times52.7\approx0.2635\,.
$$
Then,
$$
\alpha\approx\frac{1}{2}(0.00114+0.2635)=\frac{1}{2}(0.26464)\approx0.1323\,\mathrm{np/m}\,.
$$

For a $$10\,\mathrm{m}$$ line the total attenuation in nepers is
$$
\alpha_{10}=\alpha\times10\approx1.323\,\mathrm{np}\,.
$$
(Recall that $$1\,\mathrm{np}\approx8.686\,\mathrm{dB}$$, so this corresponds to roughly $$11.5\,\mathrm{dB}$$ of loss.)

---

**(b) Phase Constant $$\beta$$**

For a low‐loss line the phase constant is approximately
$$
\beta\approx\omega\sqrt{LC}\,.
$$
First compute $$\sqrt{LC}$$:
$$
LC=250\times10^{-9}\times90\times10^{-12}=(250\times90)\times10^{-21}=22500\times10^{-21}=2.25\times10^{-17}\,.
$$
Then,
$$
\sqrt{LC}=\sqrt{2.25\times10^{-17}}=1.5\times10^{-8.5}\,.
$$
Recall that 
$$
10^{-8.5}=10^{-8}\times10^{-0.5}\quad \text{and} \quad 10^{-0.5}\approx0.316\,,
$$
so
$$
\sqrt{LC}\approx1.5\times0.316\times10^{-8}\approx0.474\times10^{-8}\,,
$$
or
$$
\sqrt{LC}\approx4.74\times10^{-9}\,.
$$

Now,
$$
\beta=\omega\sqrt{LC}=(2\pi\times10^9)(4.74\times10^{-9})\,.
$$
Since
$$
2\pi\times4.74\approx29.8\,,
$$
we obtain
$$
\beta\approx29.8\,\mathrm{rad/m}\,.
$$

---

**(c) Phase Velocity $$v_p$$**

The phase velocity is given by
$$
v_p=\frac{\omega}{\beta}=\frac{1}{\sqrt{LC}}\,.
$$
Since we already found $$\sqrt{LC}\approx4.74\times10^{-9}$$, then
$$
v_p\approx\frac{1}{4.74\times10^{-9}} \approx2.11\times10^8\,\mathrm{m/s}\,.
$$

---

**(d) Relative Permittivity $$\varepsilon_r$$**

For a TEM wave in a dielectric, the phase velocity in the medium is related to the speed of light $$c$$ and the relative permittivity $$\varepsilon_r$$ by
$$
v_p=\frac{c}{\sqrt{\varepsilon_r}}\,.
$$
Thus,
$$
\varepsilon_r=\left(\frac{c}{v_p}\right)^2\,.
$$
Using $$ c\approx3.00\times10^8\,\mathrm{m/s} $$ and $$ v_p\approx2.11\times10^8\,\mathrm{m/s} $$,
$$
\varepsilon_r=\left(\frac{3.00\times10^8}{2.11\times10^8}\right)^2\approx\left(1.421\right)^2\approx2.02\,.
$$

---

For completeness, note that if the loss in the transmission line is taken into account and the input power is $$500\,\mathrm{W}$$ over a $$10\,\mathrm{m}$$ line, then the power at the output (assuming no reflections) would be reduced by the factor
$$
P_{\text{out}}=P_{\text{in}}\;e^{-2\alpha l}\,,
$$
with $$ l=10\,\mathrm{m} $$ and $$\alpha\approx0.1323\,\mathrm{np/m}$$. Since
$$
2\alpha l\approx2\times0.1323\times10\approx2.646\,,
$$
we have
$$
P_{\text{out}}\approx500\,e^{-2.646}\,.
$$
Numerically,
$$
e^{-2.646}\approx0.071\quad\Longrightarrow\quad P_{\text{out}}\approx35.5\,\mathrm{W}\,.
$$
Even though this part was not explicitly asked, it uses the given length and input power.

---

Final answers:

$$ \displaystyle \alpha\approx0.1323\,\mathrm{np/m} $$ (or about $$1.323\,\mathrm{np}$$ over $$10\,\mathrm{m}$$),  
$$ \displaystyle \beta\approx29.8\,\mathrm{rad/m}$$,  
$$ \displaystyle v_p\approx2.11\times10^8\,\mathrm{m/s}$$,  
$$ \displaystyle \varepsilon_r\approx2.02$$.
Attenuation coefficient $$ \alpha \approx 0.1323\,\mathrm{np/m} $$ (or about $$1.323\,\mathrm{np}$$ over $$10\,\mathrm{m}$$), Phase constant $$ \beta \approx 29.8\,\mathrm{rad/m} $$, Phase velocity $$ v_p \approx 2.11\times10^8\,\mathrm{m/s} $$, Relative permittivity $$ \varepsilon_r \approx 2.02 $$.

The primary constants for a coaxial cable at 1 GHz are
. , and Determine (a) the attenuation
coefficient a, (b) the phase constant b , © the phase velocity vp, (d) relative
permittivity हr for a length of 10 m , when the innut power is 500 wate

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Engineering Questions

The primary constants for a coaxial cable at 1 GHz are . , and Determine (a) the attenuation coefficient a, (b) the phase constant b , © the phase velocity vp, (d) relative permittivity हr for a length of 10 m , when the innut power is 500 wate