The string to transmit consists of 20 characters. With ASCII encoding, there are 7 bits per character plus the 1 parity bit. Therefore, \((7+1)20=160\) symbols must be transmitted. At 9600 baud this will take \(160/9600=0.017\ \text{s}\).

Drawing the circuit described results in the following diagram:

1. This circuit is a standard voltage divider. Therefore, \(V_I=V_\text{dd}\frac{R_D}{R_D+R_U}\).
2. For the pin to float in the correct direction, the voltage should be \(V_I<\frac{1}{2}V_\text{dd}\). Using the equation from the previous part, \(V_text{dd}\frac{R_D}{R_D+R_U}<\frac{1}{2}V_\text{dd}\). Simplifying results in the expression \(\frac{R_D}{R_D+R_U}<\frac{1}{2}\). Solving for \(R_D\) shows \(R_D<R_U\).
3. Assuming that the DI has a high impedance, The power required is the power flowing through the two resistors. These resistors have a combined resistance of \(R=R_U+R_D\). Therefore, the power is \(\mathcal{P}=\frac{V_\text{dd}^2}{R}=\frac{V_\text{dd}^2}{R_U+R_D}\).
4. In terms of \(V_I\), \(R_D\) can be solved using the answer to part a. First, solving the equation for \(R_D\) results in the equation \(R_D=R_U\frac{V_I}{V_\text{dd}-V_I}\). Substituting in the given values, \(R_D=40\ \text{k}\Omega\frac{V_I}{3.3\ \text{V}-V_I}\). Similarly, the power is, \(\mathcal{P}=\frac{10.89\ \text{V}^2}{R_D+40\ \text{k}\Omega}\), so \(R_D=\frac{10.89\ \text{V}^2}{\mathcal{P}}-40\ \text{k}\Omega\) TODO.
5. There is no way to pull \(V_I\) too low; therefore, the minimum value for \(R_D\) is zero. The upper limit for \(R_D\) can be calculated using the \(3.3\) V CMOS low output voltage level of \(0.5\) V. Putting this value into the equation from the preview problem, \(R_D=40\ \text{k}\Omega\frac{0.5\ \text{V}}{3.3\ \text{V}-0.5\ \text{V}}=7.14\ \text{k}\Omega\).
7. For a battery powered application, we should try to limit the power consumed. To do so, we should use the maximum resistance of \(R_D=7.14\ \text{k}\Omega\). However, resistors are not available in every value, but are instead commonly manufactured in what is called the E-series. The closest common value below \(7.14\) k\(\Omega\) is \(6.8\) k\(\Omega\). For a grid powered application, the power is less important. Therefore, a lower resistance can be used. However, a resistance of \(0\ \Omega\) would not would well, as it would make it nearly impossible to pull the input back up. Therefore, a non-zero resistance should be used. Therefore, a resistor lower in the normal range such as \(1\) k\(\Omega\) could be a good choice.

The circuit described in this problem is:

1. The total resistance of the pullup resistor circuit is \(R=\frac{1}{\frac{1}{R_U}+\frac{1}{R_{UE}}}\). Since the desired resistance is \(R=\gamma R_U\), the equations, \(\gamma R_u=\frac{1}{\frac{1}{R_U}+\frac{1}{R_{UE}}}\), can be solved for \(R_{UE}\). This results in the expression \(R_{UE}=\frac{1}{\frac{1}{\gamma R_U}-\frac{1}{R_U}}\).
2. \(R_{UE}=\frac{1}{\frac{1}{40\ \text{k}\Omega / 20}-\frac{1}{40\ \text{k}\Omega}}=2.1\ \text{k}\Omega\).
3. If the attached circuit is high impedance, the resistance to ground of the total circuit is \(R+\infty\). Therefore, no current or power flows (\(\mathcal{P}_f=0\)).
4. In this case the total circuit only consists of the two parallel resistors. Therefore, \(\mathcal{P}_s=\frac{V_\text{dd}^2}{R}=V_\text{dd}^2\left(\frac{1}{R_U}+\frac{1}{R_{UE}}\right)\).