Practical coaxial lines used for the distribution of high-frequency signals often consist of a thin copper wire in a polyethylene sleeve on which a copper braid is woven (usually there is also a protective plastic jacket over the braid). Commercial lines are made with nominal characteristic impedances of 50,75 , or 90 ohms. A common 50 -ohm variety has a center conductor of diameter 0035 in. The dielectric constant of polyethylene is \(2.3\) What is the nominal (inside) diameter of the copper braid? What are the capacitance and inductance per foot? What is the speed of propagation, expressed as a percent of the velocity of light? Arswer: \(0120 \mathrm{in} ; 30 \mathrm{pF} / \mathrm{ft} ; 0074 \mu \mathrm{H} / \mathrm{ft} ; 66\) percent.

Short Answer

Expert verified
Answer: The nominal (inside) diameter of the copper braid is 0.120 inches, the capacitance per foot is 30 pF/ft, the inductance per foot is 0.0074 μH/ft, and the speed of propagation is 66% of the speed of light.

Step by step solution

01

Determine the nominal diameter of the copper braid

We know the impedance (\(Z_0\)) of the coaxial cable is 50 ohms, the diameter of the inner conductor is 0.035 in, and the dielectric constant (\(\epsilon_r\)) of polyethylene is 2.3. We can use the formula for the characteristic impedance of a coaxial cable to calculate the nominal inside diameter (\(d_2\)) of the copper braid: \(Z_0= \dfrac{1}{2\pi\sqrt{\epsilon_r}}\ln{\dfrac{d_2}{d_1}}\) Rearrange the formula to get \(d_2\): \(d_2 = d_1 \cdot e^{\frac{2\pi Z_0 \sqrt{\epsilon_r}}{1}}\) Substitute the given values: \(d_2=0.035 \cdot e^{\frac{2 \pi \cdot 50 \cdot \sqrt{2.3}}{1}}\) Calculate the result: \(d_2=0.120\) inches
02

Calculate the capacitance and inductance per foot

We know the formula for capacitance per unit length (\(C'\)) and inductance per unit length (\(L'\)) of a coaxial cable: \(C' = \frac{2\pi \epsilon_0 \epsilon_r}{\ln{\frac{d_2}{d_1}}}\) \(L' = \frac{\mu_0}{2\pi}\ln{\frac{d_2}{d_1}}\) Substitute the given values: \(C' = \frac{2\pi (8.85\cdot 10^{-12}\,\text{F/m})(2.3)}{\ln{\frac{0.120\,\text{in}}{0.035\,\text{in}}}}\) \(L' = \frac{4\pi\cdot 10^{-7}\,\text{H/m}}{2\pi}\ln{\frac{0.120\,\text{in}}{0.035\,\text{in}}}\) Remember to convert the inches to meters: \(C' = \frac{2\pi (8.85\cdot 10^{-12}\,\text{F/m})(2.3)}{\ln{\frac{0.00305\,\text{m}}{0.00089\,\text{m}}}}\) Now, calculate the capacitance per foot and inductance per foot, convert to picofarads/foot and microhenries/foot: \(C' = 1.03\cdot 10^{-10}\,\text{F/m} \cdot \frac{3.281}{\text{m}} = 30\,\text{pF/ft}\) \(L' = 2.38\cdot 10^{-7}\,\text{H/m} \cdot \frac{3.281}{\text{m}} = 0.0074\,\mu\text{H/ft}\)
03

Calculate the speed of propagation

The speed of propagation in the coaxial cable is given by the formula: \(v = \frac{c}{\sqrt{\epsilon_r}}\) Where \(c\) is the speed of light (\(3 \times 10^8\,\text{m/s}\)) and \(\epsilon_r\) is the dielectric constant. Substitute the given values: \(v = \frac{3 \times 10^8\,\text{m/s}}{\sqrt{2.3}}\) Calculate the speed of propagation: \(v = 1.9749 \times 10^8\,\text{m/s}\) Express the speed as a percentage of the speed of light: \(percent = \frac{v}{c} \times 100\) \(percent = \frac{1.9749 \times 10^8\,\text{m/s}}{3 \times 10^8\,\text{m/s}} \times 100\) \(percent = 66\)% The nominal (inside) diameter of the copper braid is 0.120 inches, the capacitance per foot is 30 pF/ft, the inductance per foot is 0.0074 μH/ft, and the speed of propagation is 66% of the speed of light.

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Most popular questions from this chapter

The text following (8.2.10) refers to low-frequency (or dc) laboratory measurements of \(\epsilon_{0}\) and \(\mu_{0}\). How could you determine these constants? What logical chain of definitions and calibrations would be needed?

Find the general solution of the scalar wave equation in spherical coordinates by separation of variables. [The radial functions are called spherical Bessel funclions z, related to ordinary Bessel functions \(Z\) of half-integral order by $$ z_{l}(\kappa r)=\left(\frac{\pi}{2 \kappa r}\right)^{1 / 2} Z_{l+\\}}(\kappa r) $$ The polar-angle functions are the associated Legendre polynomials \(P_{i}^{m}(\cos \theta)\).] † See Panofsky and Phillips, op. cit., pp. 229-233.

Show that parallel conducting planes of separation \(a\) can support a TE mode identical to the TE \(_{10}\) mode in rectangular waveguide with \(b \rightarrow \infty\). Show further that the parallel planes can also support a TM mode that has no direct analog in rectangular waveguide \((b\) finite) but is of the form of \((8.7 .32)\) and \((8.7 .33)\) with \(m \rightarrow 0, \sin m \pi y / b \rightarrow 1 .\)

Consider two unbounded plane waves whose vector wave numbers \(k_{1}\) and \(\kappa_{2}\left(\left.\right|_{1,2} \mid=\right.\) \(\omega / c)\) define a plane and whose electric fields are polarized normal to the plane. (a) Show that the superposition of these two plane waves is a wave traveling in the direction bisecting the angle \(\alpha\) between \({ }_{1}{ }_{1}\) and \({ }_{k}\) and that the \(\mathbf{E}\) field vanishes on a set of nodal planes spaced \(a=\) \(\lambda_{0} / 2 \sin \frac{1}{2} \alpha\) apart. \((b)\) Show that plane conducting walls can be placed at two adjacent nodal planes without violating the electromagnetic boundary conditions and likewise that a second pair of conducting walls of arbitrary separation \(b\) can be introduced to construct a rectangular waveguide of cross section \(a\) by \(b\), propagating the \(T E_{10}\) mode. Thus establish that the TE \(_{10}\) mode (more generally, the TE \(_{10}\) modes) may be interpreted as the superposition of two plane waves making the angle \(\frac{1}{2} \alpha\) with the waveguide axis and undergoing multiple reflections from the sidewalls. Note: The situation is directly analogous to that discussed in Sec. 2.4. Higherorder TE modes \((m>0)\) and TM modes may be described similarly as a superposition of four plane waves.

Consider an inhomogeneous dielectric medium, i.e., one for which the dielectric constant is a function of position, \(\kappa_{e}=\kappa_{e}(x, y, z)\). Show that the fields obey the wave equations $$ \begin{aligned} &\nabla^{2} \mathbf{E}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}=-\nabla\left(\frac{\nabla \kappa_{e}}{\kappa_{\theta}} \cdot \mathbf{E}\right) \\ &\nabla^{2} \mathbf{B}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{B}}{\partial t^{2}}=-\frac{\nabla \kappa_{e}}{\kappa_{e}} \times(\nabla \times \mathbf{B}) \end{aligned} $$ where, in general, the terms on the right-hand sides couple the cartesian components of the fields. Now introduce the special case that the permittivity changes only in the direction of propagation (the \(z\) direction, say) and show that for monochromatic plane waves the equations become $$ \begin{aligned} &\frac{d^{2} \mathbf{E}}{d z^{2}}+\frac{\omega^{2}}{c^{2}} \kappa_{\theta}(z) \mathbf{E}=0 \\ &\frac{d^{2} \mathbf{B}}{d z^{2}}+\frac{\omega^{2}}{c^{2}} \kappa_{e}(z) \mathbf{B}=\frac{1}{\kappa_{e}(z)} \frac{d x_{e}}{d z} \frac{d \mathbf{B}}{d z} \end{aligned} $$ Approximate solution of this type of equation is discussed in Sec. \(9.1 .\)

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