Pset 5, Abubakar Abid (It may take a second for the LaTeX-javascript library to load...)

7.1) The twist increases the inductance of the device, which makes filters out low-frequency noise in the power and the ground. The shield prevents electric fields from forming within the cable, reducing external electrical noise.

7.2) The skin depth is $$ \frac{1}{\sqrt{\pi \nu \mu \sigma}} \approx 2.51 \text{ meters} $$

7.3) The H-field is $$\frac{I}{2 \pi r}$$ and the E-field is $$\frac{Q}{2 \pi \epsilon r}$$ They are perpendicular to each other so the Poynting vector is $$ \frac{QI}{(2 \pi)^2 \epsilon r^2} $$ Integrating that, we get $$ \frac{QI}{(2 \pi)^2 \epsilon}(\frac{1}{r_i} - \frac{1}{r_o}) $$ Replacing Q using (7.45), we get: $$ \frac{VI}{2 \pi \ln(\frac{r_o}{r_i})}(\frac{1}{r_i} - \frac{1}{r_o}) $$ 7.4) The characteristic impedance and velocity depend on the capacitance and inductance per unit length, \(l\). The capacitance would be simply \(\frac{l w \epsilon}{h}\), so the capacitance per unit length is: $$\frac{ w \epsilon}{h} $$ While the inductance, using the magnetic field calculated from Stoke's Law, would be \(\frac{l h \mu}{w} \) so the inductance per unit length would be $$\frac{ h \mu}{w} $$ So the velocity is \(\frac{1}{\sqrt{LC}}\) = $$ 1/\sqrt{LC} = c $$ And the impedance is $$ \sqrt{L/C} = \frac{h}{w} \sqrt{\frac{\mu}{\epsilon}} $$ 7.5) (a) Plugging in these numbers, we get \(91 \ln(1.48/.4)\) = 120 ohms

7.5) (b) The transmission velocity is \(c/\sqrt(2.24)\)

7.5) (c) Multiplying the velocity by 1 ns, we get 20 cm

7.5) (d) Keeping the ratio of the diameters the same, we get 8.1 mils

7.5) (e) Dividing the velocity by the outer radius gives us 135 GHz.

7.6) (a) Incredibly, the physical length of a bit is the velocity divided by (10MHz), so about 20 meters!

7.6) (b) Upon hitting two other cables, it would see a load impedance half it's own, so the reflection would be \((1/2 - 1)/(1/2 + 1) \approx 1/3\) assuming the connector does not try to do any impedance matching