Communications Exam Deck

Question 1

What are the advantages and disadvantages of binary digital signalling with respect to M-ary digital signalling? Justify

Answer 1

Question 2

Do question 2

Answer 2

Question 3

Question 3

Answer 3

Note that for the signalling scheme in question 1 Sx(f = 0) > 0 but for the signalling scheme in question 2 Sx(f = 0) = 0. Therefore, the first signalling scheme is appropriate for channels that let a DC component go through whereas the second signalling scheme is appropriate for channels that block a DC component.

Question 4

What do we define bandwidth to be?

Answer 4

We define bandwidth to be equal to the width of the interval from frequency equal to zero to the very first frequency where the power density spectrum is equal to zero

Question 5

Comment on possible differences between the power density spectrum associated with unipolar binary transmission and polar binary transmission. Justify your comments.

Answer 5

The most relevant difference between the power density spectrum associated with the unipolar case compared to the polar case relates to the fact that the first exhibits a Dirac delta impulse centred at zero frequency, whereas the second does not. This is due to the fact that (on average) a unipolar signal exhibits a DC component whereas a polar signal does not. The Dirac delta impulse captures this DC term.

Question 6

Determine the average energy conveyed per bit Eb;

Answer 6

Question 7

Error probability: do a) iii)

Answer 7

Question 8

Difference between error probability equations for unipolar and polar transmission?

Answer 8

The attached is for polar.

Unipolar has an extra factor of 2 in the denominator with No

Question 9

Comment on the differences between the error probability associated with unipolar binary digital transmission and polar binary digital transmission. In particular, justify why one scheme may perform better than the other. [10 Marks]

Answer 9

The average energy per bit required to achieve a certain error probability for the unipolar scheme is twice that for the polar scheme.

The reason has to do with the fact that – for a fixed Eb – the separation between the constellation points in the unipolar case is smaller than that for the polar case. In particular, the constellation points are separated by 2 √𝐸_b in the polar case and are separated by √(2𝐸_b) in the unipolar case.

Question 10

𝑃(𝑓) for binary amplitude shift keying

Answer 10

Question 11

𝑃(𝑓) for binary phase shift keying

Answer 11

Question 12

What is the bandwidth efficiency for binary phase-shift keying?

Answer 12

Question 13

What is the bandwidth efficiency for binary amplitude-shift keying?

Answer 13

Question 14

Comment on how the bandwidth efficiency of an M-ary passband digital signalling scheme compares to that of a binary passband digital signalling scheme. Justify your comments.

Answer 14

Question 15

What is the expression for the bandwidth efficiency of an M-ary passband transmission scheme?

Answer 15

Question 16

What is the bandwidth efficiency for a binary passband transmission scheme?

Answer 16

Question 17

What is the engineering relevance of the power density spectrum of a digital signal?

Answer 17

The power density spectrum of a digital signal measures the distribution of power over frequency. It can be used to determine the bandwidth occupancy of the digital signal hence the bandwidth efficiency of the signal.

Question 18

The technique known as ‘spectral shaping by precoding’ can be used to shape the power density spectrum of a digital signal. Describe this technique.

Answer 18

Question 19

What is the difference between a maximum likelihood (ML) detector and a maximum a posteriori (MAP) detector? What are the advantages and disadvantages of one detector in relation to the other?

Answer 19

• These detectors deliver estimates of the transmit symbols based on the output of a matched filter or a correlation and dump filter. The difference between the detectors is that the maximum likelihood detector does not take into account the prior probabilities of the transmit symbols to deliver the decision whereas the maximum a posteriori detector does take these into account.
• maximum a posteriori detector
  
  • minimizes the probability of decision error,
  • but it requires extra knowledge about the transmission process
    
    • (the prior probabilities of the transmit symbols).
• The maximum likelihood detector
  
  • does not in general minimize the probability of decision error
  • but it also does not require extra knowledge about the transmission process.
• `The two detectors are equivalent provided that the transmit symbols are equally likely.

Question 20

What is an essential expression to remember for error probability calculations?

Answer 20

In calculations the integral of this expression is taken

Question 21

Assume a quaternary digital transmission where each symbol in the sequence of symbols is drawn from a quaternary alphabet A={-3A,-A,+A,+3A} with equal probability. Determine the error probability as a function of the ratio Es/N0 where Es represents the average energy conveyed per symbol.

Answer 21

he error probability can be written as follows:

𝑃𝑟(𝑒) = 𝑃𝑟(𝑋𝑘 = −3𝐴) ∙ 𝑃𝑟(𝑒|𝑋𝑘 = −3𝐴) + 𝑃𝑟(𝑋𝑘 = −𝐴) ∙ 𝑃𝑟(𝑒|𝑋𝑘 = −𝐴) +𝑃𝑟(𝑋𝑘 = +𝐴) ∙ 𝑃𝑟(𝑒|𝑋𝑘 = +𝐴) + 𝑃𝑟(𝑋𝑘 = +3𝐴) ∙ 𝑃𝑟(𝑒|𝑋𝑘 = +3𝐴)

(Calculations for -3A can be deduced from others)

Question 22

Assume that one wishes to convey information with an error probability equal to 0.001. Determine how much more energy per symbol one requires by using the quaternary transmission scheme in relation to the binary one.
[You can use the fact that 1/2 x erfc (2.18) ≈ 0.001 and 3/4 x erfc (2.27) ≈ 0.001.]

Answer 22

Note how 2.18 is equated to sqrt(Es/No)

Note the change to dB to compare the values.