# Transmission Lines

Question 1

Calculate the attenuation constant and the phase constant for an incident voltage wave in a two wire transmission line with the following parameters:

$$f= 4.00$$ MHz, $$R= 25.0 m\Omega/m$$, $$L=2.00 \mu H/m$$, $$C=5.56 pF/m$$ and $$G=1.00fS/m$$.

Find the magnitude of the propagation constant $\gamma$ using the given parameters of the transmission line.

$\gamma^{2} = (R+j\omega L)(G+j\omega C)$ $\gamma^{2} = RG + j\omega RC + j\omega GL + j^{2}\omega^{2}LC$ $\gamma^{2} = (RG -\omega^{2}LC) + j(\omega RC + \omega GL)$

Since $G$ is so small we can say in this case that $G=0$ and get rid of any terms that are multiplied by $G$.

$\gamma^{2} = (-\omega^{2}LC) + j(\omega RC)$ $\gamma^{2} = ( -(8\pi \times 10^{6})^{2}\times 2 \times 10^{-6} \times 5.56 \times 10^{-12}) + j(8 \pi \times 10^6 \times 25 \times 10^{-3} \times 5.56 \times 10^{-12})$ $\gamma^{2}= -7.02 \times 10^{-3} + j 3.49\times 10^{-6}$ $\gamma^{2}= \sqrt{(-7.02 \times 10^{-3})^{2} + (3.49 \times 10^{-6})^{2}} \angle \tan^{-1} (\frac{3.49 \times 10^{-6}}{-7.02 \times 10^{-3}})$ $\gamma^{2}= 7.02 \times 10^{-3} \angle 179.9715034^{\circ}$ $\gamma= \sqrt{ 7.02 \times 10^{-3}} \angle \frac{179.9715034^{\circ}}{2}$ $\gamma= 83.8 \times 10^{-3} \angle 89.98575170^{\circ}$

Separate the propagation constant $\gamma$ into its real and imaginary parts.

$\gamma = \alpha + j \beta$

The real part $\alpha$ is the attenuation constant.

$\alpha= Re[\gamma]$ $\alpha = |\gamma| \cos(\gamma_{\phi})$ $\alpha= 83.8 \times 10^{-3} \times \cos (89.98575170^{\circ})$ $\alpha = 20.8 \mu Np/m$

The imaginary part $\beta$ is the phase constant.

$\beta= Im[\gamma]$ $\beta=|\gamma| \sin(\gamma_{\phi})$ $\beta= 2.65 \times 10^{-3} \times \sin (89.98575170^{\circ})$ $\beta= 83.8 mrad/m$

Question 2

Calculate the characteristic impedance of the transmission line in polar form given the following parameters:

$R= 25.0 m\Omega/m$, $L=2.00 \mu H/m$, $C=5.56 pF/m$ and $G=1.00fS/m$.

Since $G$ is so small we can say in this case that $G=0$ and get rid of any terms that include $G$. If we encounter something like $\tan^{-1}(\frac{\omega C}{G})$ We can say that this tan inverse will be equal to 90 degrees.

a) $f= 2.00$ kHz

b) $f=20.0$ MHz

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

$Z_{o}= \sqrt{\frac{\sqrt{R^{2}+(\omega L)^{2}} \angle \tan^{-1}(\frac{\omega L}{R})}{\sqrt{G^{2}+(\omega C)^{2}} \angle \tan^{-1}(\frac{\omega C}{G})}}$

$Z_{o}= \sqrt{\frac{\sqrt{R^{2}+(\omega L)^{2}}}{\sqrt{G^{2}+(\omega C)^{2}}}} \angle \frac{\tan^{-1}(\frac{\omega L}{R})-\tan^{-1}(\frac{\omega C}{G})}{2}$

a) For the case when $f= 2.00$ kHz:

Find the magnitude of the characteristic impedance $Z_{o}$ using the given parameters for the transmission line.

$|Z_{o}|= \sqrt{\frac{\sqrt{R^{2}+(\omega L)^{2}}}{\sqrt{G^{2}+(\omega C)^{2}}}}$

$|Z_{o}|= \sqrt{\frac{\sqrt{(2.5 \times 10^{-3})^{2}+(4 \times \pi \times 10^{3} \times 2 \times 10^{-6})^{2}}}{\sqrt{(4 \times \pi \times 10^{3} \times 5.56 \times 10^{-12})^{2}}}}$

$|Z_{o}|= 712 \Omega$

Find the phase angle of the characteristic impedance $Z_{o}$ using the given parameters for the transmission line.

$Z_{o} \angle = \frac{\tan^{-1}(\frac{\omega L}{R})-\tan^{-1}(\frac{\omega C}{G})}{2}$

$Z_{o} \angle = \frac{\tan^{-1}(\frac{4 \times \pi \times 10^{3}\times 2 \times 10^{-6}}{2.5 \times 10^{-3}})-\tan^{-1}(\frac{4 \times \pi \times 10^{3} \times 5.56 \times 10^{-12}}{0})}{2}$

$Z_{o} \angle =-22.4^{\circ}$

b) For the case when $f= 20.00$ MHz:

Find the magnitude of the characteristic impedance $Z_{o}$ using the given parameters for the transmission line.

$|Z_{o}|= \sqrt{\frac{\sqrt{R^{2}+(\omega L)^{2}}}{\sqrt{G^{2}+(\omega C)^{2}}}}$

$|Z_{o}|= \sqrt{\frac{\sqrt{(2.5 \times 10^{-3})^{2}+(40 \times \pi \times 10^{6} \times 2 \times 10^{-6})^{2}}}{\sqrt{(40 \times \pi \times 10^{6} \times 5.56 \times 10^{-12})^{2}}}}$

$|Z_{o}|= 600 \Omega$

Find the phase angle of the characteristic impedance $Z_{o}$ using the given parameters for the transmission line.

$Z_{o} \angle = \frac{\tan^{-1}(\frac{\omega L}{R})-\tan^{-1}(\frac{\omega C}{G})}{2}$

$Z_{o} \angle = \frac{\tan^{-1}(\frac{40 \times \pi \times 10^{6}\times 2 \times 10^{-6}}{25 \times 10^{-3}})-\tan^{-1}(\frac{40 \times \pi \times 10^{6} \times 5.56 \times 10^{-12}}{0})}{2}$

$Z_{o} \angle = (-2.85 \times 10^{-3})^{\circ}$

Question 3

A transmission line has the following parameters:

$f= 1.00$ kHz, $R= 4.00 m\Omega/m$, $L=3.00 \mu H/m$, $C=15.0 pF/m$ and $G=1.00 nS/m$.

a) Calculate the distance from the source at which the signal drops to 75% of its amplitude at the sending end.

b) Calculate the velocity at which the signal travels along the line.

Find the magnitude of the propagation constant $\gamma$ using the given parameters of the transmission line.

$|\gamma|= \sqrt{\sqrt{(RG -\omega^{2}LC)^{2} + (\omega RC + \omega GL)^{2}}}$ $|\gamma|= \sqrt{\sqrt{(4 \times 10^{-3} \times 10^{-9} - 4 \times 10^{6} \times \pi^{2}\times 3 \times 10^{-6} \times 15 \times 10^{-12})^{2} + (2 \times 10^{3} \times\pi [4 \times 10^{-3} \times 15 \times 10^{-12} + 10^{-9} \times 3 \times 10^{-6}])^{2}}}$ $|\gamma|= 42.6 \times 10^{-6}$

Find the phase angle of the propagation constant $\gamma$ using the given parameters for the transmission line.

$\gamma_{\phi} =\frac{tan^{-1}(\frac{\omega RC + \omega GL}{RG-\omega^{2} LC})}{2}$ $\gamma_{\phi} =\frac{tan^{-1}(\frac{2\times 10^{3}\times \pi \times 4 \times 10^{-3} \times 15 \times 10^{-12} + 2\times 10^{3}\times \pi \times 10^{-9}\times 3\times 10^{-6}}{4\times 10^{-3}\times10^{-9}-4 \times 10^{6} \times \pi^{2}\times 3 \times 10^{-6} \times 15 \times 10^{-12}})}{2}$ $\gamma_{\phi}= 83.7^{\circ}$

Separate the propagation constant $\gamma$ into its real and imaginary parts.

$\gamma = \alpha + j \beta$

The real part $\alpha$ is the attenuation constant.

$\alpha= Re[\gamma]$ $\alpha = |\gamma| \cos(\gamma_{\phi})$ $\alpha= 42.6 \times 10^{-6} \times \cos (83.7^{\circ})$ $\alpha = 4.67 \mu Np/m$

The imaginary part $\beta$ is the phase constant.

$\beta= Im[\gamma]$ $\beta=|\gamma| \sin(\gamma_{\phi})$ $\beta= 42.6 \times 10^{-6} \times \sin (83.7^{\circ})$ $\beta= 42.3 \mu rad/m$

a) Calculate the distance from the source at which the signal drops to 75% of its amplitude at the sending end.

$0.75 = e^{-\alpha z}$ $ln(0.75) = -\alpha z$ $ln(0.75) = -\alpha z$ $z=- \frac{ln(0.75)}{\alpha}$ $z=- \frac{ln(0.75)}{4.67 \times 10^{-6}}$ $z= 61.6 km$

b) Calculate the velocity at which the signal travels along the line.

$\mu_{p} = \frac{\omega}{\beta}$ $\mu_{p} = \frac{2 \times 10^{3}\times \pi}{42.3 \times 10^{-3}}$ $\mu_{p} = \frac{2 \times 10^{3}\times \pi}{42.3 \times 10^{-3}}$ $\mu_{p} = 148 \times 10^{6} m/s$ $\mu_{p} = 1.48 \times 10^{8} m/s$ Note by Brody Acquilano
12 months ago

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