Propagation velocity and loss of a lossy line: 22-Elec-A7 Electromagnetics, May 2016, Q7

Question 7: Propagation velocity and loss of a lossy line (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. All four distributed parameters of a uniform line are supplied, together with the operating frequency, so the complex propagation constant can be formed directly.

Given data
Quantity	Symbol	Value
Series inductance per metre	$L'$	$25\ \mu\text{H/m}$
Shunt capacitance per metre	$C'$	$160\ \text{pF/m}$
Series resistance per metre	$R'$	$0.01\ \Omega/\text{m}$
Shunt conductance per metre	$G'$	$10^{-7}\ \text{S/m}$
Frequency	$f$	$1\ \text{MHz}$

Find. The phase (propagation) velocity and the attenuation of the line at 1 MHz, expressed in nepers and decibels per unit length.

Figure 7.1 — the distributed equivalent circuit of one metre of line, and the split of the attenuation between conductor and dielectric loss.

Approach. Confirm that the line is in the low-loss regime, in which the reactive elements alone fix $Z_0$ and $v_p$; then evaluate the attenuation as the sum of a series-loss term and a shunt-loss term, and check both against the exact complex propagation constant.

Test the low-loss condition. With $\omega = 2\pi f = 6.2832\times10^{6}\ \text{rad/s}$, $$\frac{R'}{\omega L'} = \frac{0.01}{157.1} = 6.37\times10^{-5}, \qquad \frac{G'}{\omega C'} = \frac{10^{-7}}{1.005\times10^{-3}} = 9.95\times10^{-5}$$ Both are far below unity, so the low-loss approximations are excellent here.
Characteristic impedance. $$Z_0 \simeq \sqrt{\frac{L'}{C'}} = \sqrt{\frac{25\times10^{-6}}{160\times10^{-12}}} = 395.3\ \Omega$$
Propagation velocity. $$v_p \simeq \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(25\times10^{-6})(160\times10^{-12})}} = \boxed{1.581\times10^{7}\ \text{m/s}}$$
Attenuation. The series and shunt losses contribute independently: $$\alpha \simeq \frac{R'}{2Z_0} + \frac{G'Z_0}{2} = 1.265\times10^{-5} + 1.976\times10^{-5} = \boxed{3.241\times10^{-5}\ \text{Np/m}}$$
Convert to decibels. Multiplying by $8.686\ \text{dB/Np}$, $$\alpha = 2.815\times10^{-4}\ \text{dB/m} = \boxed{0.2815\ \text{dB/km}}$$

The exact calculation confirms the approximations to five significant figures. Forming $\gamma = \sqrt{(R' + j\omega L')(G' + j\omega C')}$ gives $\alpha = 3.2413\times10^{-5}\ \text{Np/m}$ and $\beta = 0.3974\ \text{rad/m}$, whence $v_p = \omega/\beta = 1.5811\times10^{7}\ \text{m/s}$ and a wavelength on the line of $15.81\ \text{m}$. The exact characteristic impedance is $395.3 + j0.007\ \Omega$ — real for all practical purposes, which is itself the signature of a low-loss line.

Two remarks on the numbers. The velocity is only 5.3 % of the speed of light, which tells you this is not a simple two-conductor line in a homogeneous dielectric — a heavily loaded or coiled cable of this kind trades velocity for a high impedance. And the loss divides roughly 39 % to the conductors and 61 % to the dielectric, so on this line it is the shunt leakage, not the copper, that dominates; improving the insulation would buy more than thickening the conductors.

Final results
Quantity	Value
Angular frequency	$\omega = 6.283\times10^{6}\ \text{rad/s}$
Loss tangents $R'/\omega L'$, $G'/\omega C'$	$6.37\times10^{-5}$, $9.95\times10^{-5}$
Characteristic impedance	$Z_0 = 395.3\ \Omega$
Propagation velocity	$\mathbf{v_p = 1.581\times10^{7}\ \text{m/s}}$ $(0.053c)$
Attenuation	$\mathbf{\alpha = 3.241\times10^{-5}\ \text{Np/m}}$
Attenuation in decibels	$\mathbf{0.2815\ \text{dB/km}}$
Phase constant and wavelength	$\beta = 0.3974\ \text{rad/m}$, $\lambda = 15.81\ \text{m}$
Split of the loss (conductor : dielectric)	$39\% : 61\%$

Topic. The telegrapher's equations for a lossy line, and the low-loss expansion that separates attenuation into conductor and dielectric contributions.

Key relations. $\gamma = \alpha + j\beta = \sqrt{(R'+j\omega L')(G'+j\omega C')}$; $Z_0 = \sqrt{(R'+j\omega L')/(G'+j\omega C')}$; low-loss limits $Z_0 \simeq \sqrt{L'/C'}$, $v_p \simeq 1/\sqrt{L'C'}$, $\alpha \simeq R'/2Z_0 + G'Z_0/2$; $1\ \text{Np} = 8.686\ \text{dB}$.

Why this works. When the loss elements are small compared with the reactances they sit beside, the square root can be expanded to first order and the real and imaginary parts separate cleanly: the reactive elements alone set the velocity and the impedance, and the loss elements contribute only attenuation. The two loss terms enter with opposite dependence on $Z_0$ — series loss divided by it, shunt loss multiplied by it — which is why a high-impedance line is relatively tolerant of series resistance but very sensitive to leakage.

Common pitfall. Quoting the attenuation in nepers when decibels are wanted, or applying the voltage factor 8.686 as if it were the power factor 4.343. Also resist the temptation to skip the low-loss test: the same formulas applied to, say, seawater or a poorly insulated cable at low frequency can be wrong by tens of per cent, and the exact complex root is then the only safe route.

Source. D. M. Pozar, Microwave Engineering, 4th ed., §2.1–2.7 (transmission-line theory, standing waves, impedance transformation). M. N. O. Sadiku, Elements of Electromagnetics, 7th ed., ch. 11 (transmission lines) and ch. 12 (waveguides).

Question 7 of 8: Propagation velocity and loss of a lossy line

Question 7: Propagation velocity and loss of a lossy line (20 marks)