Problem Set 5

(7.1)

Cables designed to carry signals with minimum pickup of interference often consist of a twisted pair of conductors surrounded by a grounded shield. Why the twist? Why the shield?

Let’s answer this from the ground up. The simplest way to transfer a signal is to use a single wire, where a voltage is applied at one end and measured at the other. In practice this requires at least some current flow, so there must be a return path which for now I’ll assume is literal ground. The voltage we measure will be different than the voltage we apply if and only if the path integral of the electric field along our wire is nonzero.

A stationary electric field won’t have any effect, since it will quickly induce a charge distribution that cancels itself (i.e. the field inside a conductor is zero). But a quickly changing one can induce a voltage, since the rate of charge transfer within the conductor is finite. This is capacitive coupling.

Similarly, a stationary magnetic field will only apply a net force on electrons perpendicular to the current. This induces a Hall effect but won’t impact our measurement. But a varying magnetic field (or moving the wire through a fixed one) can induce a voltage because it causes curl in the electric field. This is inductive coupling.

How can we mitigate these effects? If we surround our signal wire with another conductor, the capacitively induced voltages will mostly appear there. This is a shield. The effectiveness of the shield depends on its conductivity, its thickness, and the frequency of the interference (recall that oscillating electric fields penetrate conductors up to a skin depth). It doesn’t necessarily need to be grounded to do its job, but doing so ensures it can source enough charge to counteract even the strongest external fields, and dissipate static buildup.

The inductive coupling is minimized by minimizing the magnetic flux between our signal wire and the current return path. So if we use a second wire for our return path, and keep it next to the first, there’s much less area for flux to penetrate. By twisting the pair of wires, we create alternating regions where the flux will have opposite sign. This causes inductively coupled signals to cancel themselves out. If we want to take this even further, we could use star quad wire.

Continuing with the theme of noise cancellation, if the drive, line, and load impedances are matched in the two wires, then any noise signals that do couple will tend to cause equal voltage fluctuations. This is why differential signaling is common. Note that this works even if we don’t put equal and opposite voltages on the wires: we just need matching electrical characteristics (and the assumption of equal noise patterns). But applying equal and opposite voltages to the wires does help in another way: it causes the fields generated by our wires to cancel each other out except at small distances, so it helps our wires not induce noise in any other signal lines.

(7.2)

Salt water has a conductivity ∼4 S/m. What is the skin depth at $10^4$ Hz?

$\begin{align*} \delta &= \frac{1}{\sqrt{\pi \nu \mu \sigma}} \\ &= \frac{1}{\sqrt{\pi \cdot 10^4 \si{Hz} \cdot \num{1.3e-6} \si{H/m} \cdot 4 \si{S/m}}} \\ &= 2.5 \si{m} \end{align*}$

(7.3)

Integrate Poynting’s vector $P = E \times H$ to find the power flowing across a cross-sectional slice of a coaxial cable, and relate the answer to the current and voltage in the cable.

We saw that the magnetic field has a magnitude of $I/(2 \pi r)$ and wraps around the inner conductor according to the right hand rule. The electric field has magnitude $Q/(2 \pi \epsilon r)$ and points radially from the inner conductor toward the outer. So $P = E \times H$ points along the direction of current in the inner conductor, and has magnitude

$| P | = \frac{I Q}{(2 \pi r)^2 \epsilon}$

Integrating over the area between the conductors, we find that the cross-sectional power transfer is

$\begin{align*} \int_0^{2 \pi} \int_{r_i}^{r_o} \frac{I Q}{(2 \pi r)^2 \epsilon} r \mathrm{d} r \mathrm{d} \theta &= \frac{I Q}{2 \pi \epsilon} \int_{r_i}^{r_o} r^{-1} \mathrm{d} r \mathrm{d} \theta \\ &= \frac{I Q}{2 \pi \epsilon} \ln \left( \frac{r_o}{r_i} \right) \end{align*}$

But the charge is

$Q = CV = \frac{2 \pi \epsilon}{\ln (r_0 / r_i)}$

so we can write the total power as $IV$ .

(7.4)

Find the characteristic impedance and signal velocity for a transmission line consisting of two parallel strips with a width $w$ and a separation $h$ . You can ignore fringing fields by assuming that they are sections of conductors infinitely wide.

Take the plates to have length one. Suppose the bottom plate has a charge $Q$ , and the top plate a charge of $-Q$ . Then as we found last week the field between the plates will have magnitude $Q/(\epsilon w)$ (since the charge density is the total charge divided by area $1 \cdot w$ ), and point from the bottom plate toward the top plate. Integrating from the bottom plate to the top one, we find that the voltage difference is $Qh/(\epsilon w)$ . Thus the capacitance (charge per voltage) per length is

$C = \frac{\epsilon w}{h}$

By symmetry the magnetic field must point perpendicular to the direction of current travel and parallel to the surfaces of the plates. Consider a square perpendicular to the current, with a width of $w$ , bisected by the lower plate. Ampère’s Law tells us that the integral of the magnetic field along the boundary of this square is $I$ . So we deduce that the magnetic field has magnitude $I/w$ . Integrating over the surface perpendicular to the field between the plates, we find that the flux is $\mu_0 I h / w$ . Thus the inductance (flux per current) per length is

$L = \frac{\mu_0 h}{w}$

From here we can use the results derived in the chapter.

$v = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{\mu_0 \epsilon}}$ $Z = \sqrt{\frac{L}{C}} = \frac{\mu_0 h^2}{\epsilon w^2}$

(7.5)

The most common coaxial cable, RG58/U, has a dielectric with a relative permittivity of 2.26, an inner radius of 0.406 mm, and an outer radius of 1.48 mm.

(a)

What is the characteristic impedance?

$\begin{align*} Z &= \sqrt{\frac{L}{C}} \\ &= \sqrt{\frac{\mu_0}{2 \pi} \ln \left( \frac{r_o}{r_i} \right) \frac{1}{2 \pi \epsilon_0 \epsilon_r} \ln \left( \frac{r_o}{r_i} \right)} \\ &= \frac{1}{2 \pi} \sqrt{\frac{\mu_0}{\epsilon_0 \epsilon_r}} \ln \left( \frac{r_o}{r_i} \right) \\ &= 51.6 \si{\ohm} \end{align*}$

(b)

What is the velocity?

$\begin{align*} v &= \frac{1}{\sqrt{L C}} \\ &= \frac{1}{\sqrt{\frac{\mu_0 \ln(r_o / r_i)}{2 \pi} \frac{2 \pi \epsilon_0 \epsilon_r}{\ln( r_o / r_i )}}} \\ &= \frac{1}{\sqrt{\mu_0 \epsilon_0 \epsilon}} \\ &= \num{2e8} \si{m/s} \end{align*}$

This is two thirds the speed of light.

(c)

If a computer has a clock speed of 1 ns, how long can a length of RG58/U be and still deliver a pulse within one clock cycle?

$\num{1e-9} \si{s} \cdot \num{2e8} \si{m/s} = 0.2 \si{m}$

(d)

It is often desirable to use thinner coaxial cable to minimize size or weight but still match the impedance of RG58/U (to minimize reflections). If such a cable has an outer diameter of 30 mils (a mil is a thousandth of an inch), what is the inner diameter?

Solving the formula above for $r_i$ ,

$\begin{align*} r_i &= r_0 \cdot e^{-2 \pi \cdot 51.6 \sqrt{\epsilon_0 \epsilon_r / \mu_0}} \\ &= 8.2 \si{mils} \end{align*}$

(e)

For RG58/U, at what frequency does the wavelength become comparable to the diameter?

$\begin{align*} \nu &= \frac{c}{\lambda} \\ &= \frac{\num{2e8} \si{m/s}}{2 \cdot \num{1.48e-3} \si{m}} \\ &= \num{6.76e10} \si{Hz} \end{align*}$

(7.6)

CAT6 twisted pair cable used in ethernet networks has a propagation delay of 4.6 ns/m, and an impedance of 100 ohms.

(a)

What is the physical length of a minimum size 64 byte frame?

The inverse of the propagation delay is the velocity, so the signal travels at $0.22 \si{m/ns}$ . Assuming we need one nanosecond clock cycle per bit, the length of a 64 byte frame is

$64 \si{bytes} \cdot \frac{8 \si{bits}}{1 \si{byte}} \cdot \frac{1 \si{ns}}{1 \si{bit}} \cdot \frac{1 \si{m}}{4.6 \si{ns}} = 111 \si{m}$

(b)

Now consider what would happen if a “T” connector was used to connect one CAT6 cable to two other ones. Estimate the reflection coefficient for a signal arriving at the T.

The two connected CAT6 cables will look like two parallel impedances. So we can take $Z_0$ to be 100 $\si{\ohm}$ , and $Z_L$ to be 50 $\si{\ohm}$ . Then

$\begin{align*} R &= \frac{Z_L - Z_0}{Z_L + Z_0} \\ &= -\frac{1}{3} \end{align*}$