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 γ\gamma using the given parameters of the transmission line.

γ2=(R+jωL)(G+jωC)\gamma^{2} = (R+j\omega L)(G+j\omega C) γ2=RG+jωRC+jωGL+j2ω2LC\gamma^{2} = RG + j\omega RC + j\omega GL + j^{2}\omega^{2}LC γ2=(RG−ω2LC)+j(ωRC+ωGL)\gamma^{2} = (RG -\omega^{2}LC) + j(\omega RC + \omega GL)

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

γ2=(−ω2LC)+j(ωRC)\gamma^{2} = (-\omega^{2}LC) + j(\omega RC) γ2=(−(8π×106)2×2×10−6×5.56×10−12)+j(8π×106×25×10−3×5.56×10−12)\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}) γ2=−7.02×10−3+j3.49×10−6\gamma^{2}= -7.02 \times 10^{-3} + j 3.49\times 10^{-6} γ2=(−7.02×10−3)2+(3.49×10−6)2∠tan⁡−1(3.49×10−6−7.02×10−3)\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}}) γ2=7.02×10−3∠179.9715034∘\gamma^{2}= 7.02 \times 10^{-3} \angle 179.9715034^{\circ} γ=7.02×10−3∠179.9715034∘2\gamma= \sqrt{ 7.02 \times 10^{-3}} \angle \frac{179.9715034^{\circ}}{2} γ=83.8×10−3∠89.98575170∘\gamma= 83.8 \times 10^{-3} \angle 89.98575170^{\circ}

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

γ=α+jβ\gamma = \alpha + j \beta

The real part α\alpha is the attenuation constant.

α=Re[γ]\alpha= Re[\gamma] α=∣γ∣cos⁡(γϕ)\alpha = |\gamma| \cos(\gamma_{\phi}) α=83.8×10−3×cos⁡(89.98575170∘)\alpha= 83.8 \times 10^{-3} \times \cos (89.98575170^{\circ}) α=20.8μNp/m\alpha = 20.8 \mu Np/m

The imaginary part β\beta is the phase constant.

β=Im[γ]\beta= Im[\gamma] β=∣γ∣sin⁡(γϕ)\beta=|\gamma| \sin(\gamma_{\phi}) β=2.65×10−3×sin⁡(89.98575170∘)\beta= 2.65 \times 10^{-3} \times \sin (89.98575170^{\circ}) β=83.8mrad/m\beta= 83.8 mrad/m

Question 2

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

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

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

a) f=2.00f= 2.00 kHz

b) f=20.0f=20.0 MHz

Zo=R+jωLG+jωCZ_{o}= \sqrt{\frac{R+j\omega L}{G+j\omega C}}

Zo=R2+(ωL)2∠tan⁡−1(ωLR)G2+(ωC)2∠tan⁡−1(ωCG)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})}}

Zo=R2+(ωL)2G2+(ωC)2∠tan⁡−1(ωLR)−tan⁡−1(ωCG)2Z_{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=2.00f= 2.00 kHz:

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

∣Zo∣=R2+(ωL)2G2+(ωC)2|Z_{o}|= \sqrt{\frac{\sqrt{R^{2}+(\omega L)^{2}}}{\sqrt{G^{2}+(\omega C)^{2}}}}

∣Zo∣=(2.5×10−3)2+(4×π×103×2×10−6)2(4×π×103×5.56×10−12)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}}}}

∣Zo∣=712Ω|Z_{o}|= 712 \Omega

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

Zo∠=tan⁡−1(ωLR)−tan⁡−1(ωCG)2Z_{o} \angle = \frac{\tan^{-1}(\frac{\omega L}{R})-\tan^{-1}(\frac{\omega C}{G})}{2}

Zo∠=tan⁡−1(4×π×103×2×10−62.5×10−3)−tan⁡−1(4×π×103×5.56×10−120)2Z_{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}

Zo∠=−22.4∘Z_{o} \angle =-22.4^{\circ}

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

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

∣Zo∣=R2+(ωL)2G2+(ωC)2|Z_{o}|= \sqrt{\frac{\sqrt{R^{2}+(\omega L)^{2}}}{\sqrt{G^{2}+(\omega C)^{2}}}}

∣Zo∣=(2.5×10−3)2+(40×π×106×2×10−6)2(40×π×106×5.56×10−12)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}}}}

∣Zo∣=600Ω|Z_{o}|= 600 \Omega

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

Zo∠=tan⁡−1(ωLR)−tan⁡−1(ωCG)2Z_{o} \angle = \frac{\tan^{-1}(\frac{\omega L}{R})-\tan^{-1}(\frac{\omega C}{G})}{2}

Zo∠=tan⁡−1(40×π×106×2×10−625×10−3)−tan⁡−1(40×π×106×5.56×10−120)2Z_{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}

Zo∠=(−2.85×10−3)∘Z_{o} \angle = (-2.85 \times 10^{-3})^{\circ}

Question 3

A transmission line has the following parameters:

f=1.00f= 1.00 kHz, R=4.00mΩ/m R= 4.00 m\Omega/m, L=3.00μH/m L=3.00 \mu H/m, C=15.0pF/mC=15.0 pF/m and G=1.00nS/mG=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 γ\gamma using the given parameters of the transmission line.

∣γ∣=(RG−ω2LC)2+(ωRC+ωGL)2|\gamma|= \sqrt{\sqrt{(RG -\omega^{2}LC)^{2} + (\omega RC + \omega GL)^{2}}} ∣γ∣=(4×10−3×10−9−4×106×π2×3×10−6×15×10−12)2+(2×103×π[4×10−3×15×10−12+10−9×3×10−6])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}}} ∣γ∣=42.6×10−6|\gamma|= 42.6 \times 10^{-6}

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

γϕ=tan−1(ωRC+ωGLRG−ω2LC)2\gamma_{\phi} =\frac{tan^{-1}(\frac{\omega RC + \omega GL}{RG-\omega^{2} LC})}{2} γϕ=tan−1(2×103×π×4×10−3×15×10−12+2×103×π×10−9×3×10−64×10−3×10−9−4×106×π2×3×10−6×15×10−12)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} γϕ=83.7∘\gamma_{\phi}= 83.7^{\circ}

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

γ=α+jβ\gamma = \alpha + j \beta

The real part α\alpha is the attenuation constant.

α=Re[γ]\alpha= Re[\gamma] α=∣γ∣cos⁡(γϕ)\alpha = |\gamma| \cos(\gamma_{\phi}) α=42.6×10−6×cos⁡(83.7∘)\alpha= 42.6 \times 10^{-6} \times \cos (83.7^{\circ}) α=4.67μNp/m\alpha = 4.67 \mu Np/m

The imaginary part β\beta is the phase constant.

β=Im[γ]\beta= Im[\gamma] β=∣γ∣sin⁡(γϕ)\beta=|\gamma| \sin(\gamma_{\phi}) β=42.6×10−6×sin⁡(83.7∘)\beta= 42.6 \times 10^{-6} \times \sin (83.7^{\circ}) β=42.3μrad/m\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−αz0.75 = e^{-\alpha z} ln(0.75)=−αzln(0.75) = -\alpha z ln(0.75)=−αzln(0.75) = -\alpha z z=−ln(0.75)αz=- \frac{ln(0.75)}{\alpha} z=−ln(0.75)4.67×10−6z=- \frac{ln(0.75)}{4.67 \times 10^{-6}} z=61.6kmz= 61.6 km

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

μp=ωβ \mu_{p} = \frac{\omega}{\beta} μp=2×103×π42.3×10−3 \mu_{p} = \frac{2 \times 10^{3}\times \pi}{42.3 \times 10^{-3}} μp=2×103×π42.3×10−3 \mu_{p} = \frac{2 \times 10^{3}\times \pi}{42.3 \times 10^{-3}} μp=148×106m/s \mu_{p} = 148 \times 10^{6} m/s μp=1.48×108m/s \mu_{p} = 1.48 \times 10^{8} m/s

#ElectricityAndMagnetism

Note by Brody Acquilano
1 year, 7 months ago

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