Chapter 1, Problem P29P

a.

Explanation of Solution

The distance between the geostationary satellite and the surface of the earth is 36,000 km.

The formula used for calculating the propagation delay in terms of meters (m) and seconds (s) is ${\text{d}}_{\text{prop}}\text{=}\frac{\text{m}}{\text{s}}$ .

Substitute, “36,000×10³” for “m” and “2

b.

Explanation of Solution

The formula used for calculating the bandwidth-delay product is $\text{R}\text{×}{\text{d}}_{\text{prop}}$

The formula used for calculating the propagation delay in terms of meters (m) and seconds (s) is ${\text{d}}_{\text{prop}}\text{=}\frac{\text{m}}{\text{s}}$  seconds.

Substitute, “36,000” for “m” and “2.4 ×10⁸” for “s”

The propagation delay of the link is $\frac{\text{36000}\text{×}{\text{10}}^{\text{3}}}{\text{2}}$

c.

Explanation of Solution

Consider the following values to calculate the minimum value of x:

$\begin{array}{c}Transmissionrate\left(R\right)=10Mbps\\ =10{\text{x 10}}^{\text{6}}\\ ={10}^{7}bits\end{array}$

Whereas, “1 Mbps = ${\text{10}}^{\text{6}}$ bits”.

Assume the size of the photo as “x”.

For every one minute, the satellite takes a digital photo and sends it to the base station; hence the frequency is “1 per minute”. One photo is transmitted per 60 seconds; therefore, the interval between each photo is “60 seconds”.

Calculating minimum value of “x” by multiplying the value of transmission rate “R” and time taken for each photo to transmit