diff --git a/_psets/5.md b/_psets/5.md
new file mode 100644
index 0000000000000000000000000000000000000000..40f3eb62afb1852ffd56beb11bd1a9d9320cfd78
--- /dev/null
+++ b/_psets/5.md
@@ -0,0 +1,250 @@
+---
+title: Problem Set 5
+---
+
+## (7.1)
+
+{:.question}
+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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Star_quad_cable) 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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+CAT6 twisted pair cable used in ethernet networks has a propagation delay of 4.6 ns/m, and an
+impedance of 100 ohms.
+
+### (a)
+
+{:.question}
+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)
+
+{:.question}
+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
+
+$$
+\begin{align*}
+R &= \frac{Z_L - Z_0}{Z_L + Z_0} \\
+&= -\frac{1}{3}
+\end{align*}
+$$