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3 years ago

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In class we learned that in order to uniquely identify one of N equally likely symbols, ceiling (〖log〗_2 N) bits of information

must be communicated. Answer the questions below:
(a) How many bits are necessary to encode an integer in the range of 0 to 512 (inclusive)?
(b) It must require 10 bits at least, not 9, because 2^9 = 512, however this is inclusive so it is actually 513 numbers and therefore 10 bits. How many bits are necessary to encode an integer in the range of 0 to 75 (inclusive)?
(c) It would require at least 7 bits. 2^6 = 64 which is not enough, so it must be 2^7 = 128 bits. How many bits are necessary to encode an integer in the range of -20 to 13 (inclusive)?

1 answer:

AleksAgata [21]3 years ago

6 0

Answer:

a. 10bits

b. 7 bits

c. 6 bits

Explanation:

a. for 0 to 512

# of numbers = 512 - 0 + 1 =513

[log ₂513] = 9 bits

we actually need 10 bits

b. for 0 to 75

# of numbers = 75 - 0 + 1 =76

[log ₂76] = 6 bits

we actually need 7 bits

c. for -20 to 13

# of numbers = 13 - (-20) + 1 =34

[log ₂34] = 5 bits

we actually need 6 bits

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a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

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$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

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$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

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$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

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