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Problem 5. l

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(1)

Problem 5.

l(0) = 0, l(1) = 1.

First letter is a1:

Second letter is a1:

Third letter is a3:

Fourth letter is a2:

Fifth letter is a3:

(2)

Sixth letter is a1:

Therefore, a possible tag value is 0.0273.

Problem 6.

The tag decodes to the following sequence: a3 a2 a2 a1 a2 a1 a3 a2 a2 a3.

Problem 7.

(a) Total count = 100,

m = log2(Total_Count) + 2 = 7 + 2 = 9 bits.

(b) 00101011000111100. Note that the underlined bit patterns (i.e., EOS) is the lower limit of the final interval, l(7), you can transmit any value between l(7) and u(7). (c) Decoding of 001010110+EOS should give you abacabb.

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