Question

3. We want to transmit commands to a drone over a wireless channel. The commands and the probability of transmission are as follows: up (0.1), down (0.1), forward (0.4), left (0.15), right (0.15) and hover (0.1). (30) (a) What is a compact binary code to transmit these messages? (5) b) Suppose for hardware simplification reasons, a fixed 4-bit representation of each of the commands is used instead of the compact code. If a command is sent every second, how inefficient is this approach versus the compact code in terms of bps? (10) (c) If we send the 4-bit representations with BPSK modulation over an AWGN channel with an Eb/N 9.6 dB where BER-1E-5 (as shown on the figure below), what is the probability of message error? A message is in error if at least one bit in the message is incorrect. (5) (d) If we now code each message transmission with a single even parity check bit, what is the probability of an undetected message error? Assume that the energy of the transmission is no different from the uncoded case (hint: the energy per uncoded bit must be shared across the coded bits). (10) Bit error probability curve for BPSK modulation theory simulation : 10 10 6 10 Eb/No, dB

Answer 1

Answer:

We can use Huffman coding to find the compact coding. It goes as follows. List the possibilities in the order of non-decreasing probabilities from left to right. and keep combining the least two at each stage.

in all the following 'p' represents down. I intended write 'd' but didn't observe till the last moment.

List the possibilities

0.1 0.1 0.1 0.4 0.15 0.15

Combine the two least probabilites p and u

0.4 0.15 0.15 0.1 0.1 0.1 0.2

Follow this till the end

0.4 0.15 0.15 0.1 0.1 0.1 0.25 0.2

Next least are 0.2 and 0.15

0.4 0.15 0.15 0.1 0.1 0.1 0.25 0.2 0.35

Next are 0.35 and 0.25

0.4 0.15 0.15 0.1 0.1 0.1 0.25 0.2 003 0.6

Lastly

0.1 0.1 0.1 0.4 0.15 0.15 0.25 0.2 003 0.6

Now take any convention to assign the bits. i took left edge as 1 and rigth as 0

0.4 0.15 0.15 0.1 0.1 0.1 0.25 0.2 CD 003 CD 0.6 CD

Now if we have to data for an entry, start from the bottom and note the bits in the path to the corresponding entry.

so, f (forward) = 1

r (right) = 010

l(left) = 001

h(hover) = 000

p(down) = 0101

u(up) = 0100

Average bit per data . multiply bit length of each entry with probability = (0.4*1)+(0.15*3)+(0.15*3)+(0.1*3)+(0.1*4)+(0.1*4) = 2.4

But we take a fixed 4 bit representation then each entry will take 4 bits

Inefficienvcy =

4-2.4! 2.4 * 100 66.6% inefficient