Answer 7.6

_psets/5.md

@@ -199,8 +199,26 @@ impedance of 100 ohms.
 {:.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
+
+$$
+\begin{align*}
+R &= \frac{Z_L - Z_0}{Z_L + Z_0} \\
+&= -\frac{1}{3}
+\end{align*}
+$$