Information Theory 121-152

The Hamming distance between 010111 and 010011 is

• 2

• 3

• 1

• 4

The Hamming distance between 011111 and 010011 is

• 1

• 3

• 2

• 4

The Hamming distance between 101001 and 010011 is

• 1

• 2

• 4

• 3

The length of the message is 16 symbols and the message's alphabet consists of 32 symbols. Find the amount of information in this message.

• 80

• 16

• 64

• 32

The length of the message is 6 symbols and the message's alphabet consists of 32 symbols. Find the amount of information in this message.

• 30

• 6

• 32

• 24

The redundancy of the language is 0,75 and its I average is 1 bit. Calculate the number of letters in this language's alphabet?

• 32

• 16

• 8

• 64

The string was encoded with Hamming (7,4) code using the structure (i1, i2, i3, i4, r1, r2, r3). After a channel the error syndrome is 000. Specify the position of the error.

• i1

• r1

• no error

• r3

The string was encoded with Hamming (7,4) code using the structure (i1, i2, i3, i4, r1, r2, r3). After a channel the error syndrome is 001. Specify the position of the error.

• i1

• r1

• r3

• no error

The string was encoded with Hamming (7,4) code using the structure (i1, i2, i3, i4, r1, r2, r3). After a channel the error syndrome is 010. Specify the position of the error.

• r3

• r1

• r2

• i2

The string was encoded with Hamming (7,4) code using the structure (i1, i2, i3, i4, r1, r2, r3). After a channel the error syndrome is 011. Specify the position of the error.

• r4

• i1

• i4

• r1