diff --git a/_psets/5.md b/_psets/5.md index 48c0e094a05dab62a3fe09d7df64aeac9104f888..c8792a42eb5d42960d7d8096057e478ad4c03e5d 100644 --- a/_psets/5.md +++ b/_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 âTâ 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 + +$$ +\begin{align*} +R &= \frac{Z_L - Z_0}{Z_L + Z_0} \\ +&= -\frac{1}{3} +\end{align*} +$$