A deep space satellite transmits data to a ground station at a constant rate of 1.6 megabi… — SAT

A deep space satellite transmits data to a ground station at a constant rate of $1.6$ megabits per second. The transmitted data consists of actual scientific data and protocol overhead. The protocol overhead accounts for $25\%$ of the total transmitted bits, meaning the actual scientific data constitutes only $75\%$ of the total transmitted bits. The ground station needs to receive $5$ scientific data files, each with a size of $1.08$ gigabytes. How many hours will it take to transmit all $5$ files? (Given that $1\\text{ byte} = 8\\text{ bits}$ , $1\\text{ megabit} = 10^6\\text{ bits}$ , and $1\\text{ gigabyte} = 10^9\\text{ bytes}$ .)

Answer: 10 hours

Answer

It will take 10 hours to transmit all 5 files.

The correct answer is 10. To find this, first calculate the total size of the scientific files: 5 files * 1.08 gigabytes per file = 5.4 gigabytes. Convert this to bytes: 5.4 * 10^9 bytes. Convert bytes to bits: 5.4 * 10^9 * 8 = 43.2 * 10^9 bits. Since protocol overhead is 25%, the scientific data is 75% of the total transmitted bits, so the total bits transmitted is 43.2 * 10^9 / 0.75 = 57.6 * 10^9 bits. The transmission rate is 1.6 megabits per second, or 1.6 * 10^6 bits per second. The transmission time in seconds is 57.6 * 10^9 / (1.6 * 10^6) = 36,000 seconds. Converting seconds to hours: 36,000 / 3,600 = 10 hours.

Step-by-Step Solution

Calculate the total size of the scientific files in gigabytes.

5.4

gigabytes

To find the total amount of scientific data that must be received.

Convert the total scientific data size from gigabytes to bytes and then to bits.

43.2 \\times 10^9

bits

The transmission rate is given in megabits per second, so the data size must be converted to bits for unit consistency. Since

1\\text{ GB} = 10^9\\text{ bytes}

and

1\\text{ byte} = 8\\text{ bits}

, we have

5.4 \\times 10^9 \\times 8 = 43.2 \\times 10^9

bits.

Calculate the total number of bits transmitted, including protocol overhead.

57.6 \\times 10^9

bits

Since protocol overhead is

25\%

, the scientific data is only

75\%

of the total bits transmitted. Thus, we divide the scientific bits by

0.75

to find the total bits transmitted:

\\frac{43.2 \\times 10^9}{0.75} = 57.6 \\times 10^9

bits.

Calculate the transmission time in seconds.

36,000

seconds

Divide the total bits by the transmission rate in bits per second (

1.6\\text{ Mbps} = 1.6 \\times 10^6

bits per second):

\\frac{57.6 \\times 10^9}{1.6 \\times 10^6} = 36,000

seconds.

Convert the transmission time from seconds to hours.

10

hours

Since there are

3,600

seconds in one hour, divide the total seconds by

3,600

\\frac{36,000}{3,600} = 10

Key Concept

Multi-step dimensional analysis and compound rate conversions incorporating percentage overhead

Alternative Method

Instead of converting units step-by-step, dimensional analysis can be set up as a single product of conversion factors: 5 files * (1.08 GB / 1 file) * (10^9 bytes / 1 GB) * (8 bits / 1 byte) * (1 total bit / 0.75 scientific bits) * (1 second / 1.6 * 10^6 bits) * (1 hour / 3600 seconds) = 10 hours.

Estimated Time:3m 0s

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