A Wireless Protocol Transmission - Essay Example

A wireless protocol transmits over a channel 20 MHz wide over a noiseless channel. How many bits/sec can be sent if 64-level digital signals are used?
Channel bandwidth = 20 MHz = 20 * 106
Using the Nyquist theorem;
C = 2Blog2M
C = 2 * 20 *106 * log 264
=240Mbps

2) If a binary signal is sent over a 20 3KHz MHz channel whose signal-to-noise ratio is 40 20 dB, what is the maximum achievable data rate?

We know that 10log10S/N = 40dB
It follows that S/N = 100
Now from the Shannon theorem;
The maximum number of bits per second = H* log2 (1 + S/N) = 20KHz * log2(1+100)
Log2(101)= 6,658
= 20 * 6658
=133160
Now from the Nyquist theorem;
The maximum number of bits per second = 2 * H * log2V = 2 * 20 * log22
=40
The Nyquist theorem gives us a lower data rate hence it forms the maximum digital signals used.
3) Radio antennas often work best when the diameter of the antenna is equal to the wavelength of the radio wave. An antenna has a diameter of 10 cm. What frequency range does it work best for?

From the formulae λf = c
We get f = c/λ.
Now c = 3×108 m/sec and λ = 3×10-2 m/cycle
We obtain (3×108 m/sec)/(3×10-2 m/cycle) = 3 × 1010 m/cycle = 30 GHz.
When λ = 5 m/cycle we obtain (3×108 m/sec)/(5 m/cycle) = 60×106 m/cycle = 60 MHz.
Hence the frequency range is 60 MHz to 30 GHz.

4) Suppose we want to send 2-bit strings. Each codeword contains three copies of the string. If the bit-string is 01, we send the 6-bit string 010101. What is the Hamming distance of this code? How many errors is it possible to detect? How many errors is it possible to correct?
The length of the data sequence is 6
The number of parity bits is 5
The length of the code word is 11
The parity scheme is represented as
P0 = D0 + D1 + D3 + D4
P1 = D0 + D2 + D3 + D5
P2 = D1 + D2 + D3
P3 = D4 + D5
P4 = D0 + D1 + D2 + D3 + D4 + D5 + P0 + P1 + P2 + P3

The syndrome matrix is
D5 D4 D3 D2 D1 D0 P4 P3 P2 P1 P0
1 1 1 1 1 1 1 1 1 1 1
1 1 0 0 0 0 0 1 0 0 0
0 0 1 1 1 0 0 0 1 0 0
1 0 1 1 0 1 0 0 0 1 0
0 1 1 0 1 1 0 0 0 0 1
Its parity scheme is
P0 = D0 + D1 + D3 + D4 + D6
P1 = D0 + D2 + D3 + D5 + D6
P2 = D1 + D2 + D3
P3 = D4 + D5 + D6
P4 = D0 + D1 + D2 + D3 + D4 + D5 + D6 + P0 + P1 + P2 + P3
Its syndrome matric is
D6 D5 D4 D3 D2 D1 D0 P4 P3 P2 P1 P0
1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 0 0 0 0 0 1 0 0 0
0 0 0 1 1 1 0 0 0 1 0 0
1 1 0 1 1 0 1 0 0 0 1 0
1 0 1 1 0 1 1 0 0 0 0 1


Length of Data Sequence Number of Parity Bits Length of CodeWord
7 5 12

Parity Scheme
P0 = D0 + D1 + D3 + D4 + D6
P1 = D0 + D2 + D3 + D5 + D6
P2 = D1 + D2 + D3
P3 = D4 + D5 + D6
P4 = D0 + D1 + D2 + D3 + D4 + D5 + D6 + P0 + P1 + P2 + P3

Syndrome Matrix
D6 D5 D4 D3 D2 D1 D0 P4 P3 P2 P1 P0
1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 0 0 0 0 0 1 0 0 0
0 0 0 1 1 1 0 0 0 1 0 0
1 1 0 1 1 0 1 0 0 0 1 0
1 0 1 1 0 1 1 0 0 0 0 1