ELEC4110 Midterm Exercise 2014 With Solution (1)

ELEC 4110 Digital Communications Midterm Exercise

ELEC 4110 Digital Communications Midterm Exercise 2014

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Question 1 (30 marks)



Total Marks (100 marks)

ELEC 4110 Digital Communications Midterm Exam

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Question 1 Multiple Choices & True or False

(3 marks 10) Please select an appropriate answer to the following questions.

1. Which of the following modulation schemes is the most energy efficient under

the same BER requirement?

A. 8-PSK B. 16-PSK C. 8-FSK D. 16-FSK

Solution: D. Page 20 of lecture note 7.


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2. Which of the following modulation schemes can achieve the highest bandwidth

efficiency under the same BER requirement?

A. 8-PSK B. 16-PSK C. 8-FSK D. 16-FSK

Solution: B. Page 34 of lecture note 7.

3. Which of the following range of frequency band is the most commonly used in

cellular network?

A. 30MHz-300MHz B. 300MHz-3000MHz

C. 3GHz-30GHz D. 30GHz-300GHz

Solution: B. For example,

PCCW 3G: 1920.3 - 1935.1/2110.3 - 2125.1/ 1914.9 -1919.9 MHz

Smartone GSM: 907.5-915/952.5/960MHz

4. Which of the following modulation schemes provides 4 bits per baud?

A. QPSK B. 8-PSK C. 16-QAM D. 64-QAM

Solution: C. Using 16 possible combinations of phase and amplitude, 16-QAM can

represent any of the 16 possible combinations of 4 bits, from 0000 to 1111, with each

symbol change.


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5 Which of the following forms a set of orthonormal basis?

A.

1

2

3

1 2 0 2( )

0

1 2 0 1

( ) 1 2 1 2

0

1 2 2 3( )

0

tt

otherwise

t

t t

otherwise

tt

otherwise

B.

C.


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D.

1 2 3{ (1,0,0), (0,1,0), (0,1 2 ,1 2)}e e e

Solution: C. A is orthogonal but not normalized. B and D are normalized but not

orthogonal.

Please check the true/false of following statements. Write a T in the bracket if it is

true and a F if it is false. Briefly explain your choice if it is false.

6. Sampling makes signal discrete in amplitude and bandlimited signals can be

sampled without distortion. ( F )

Solution: Sampling makes signal discrete in time.

7. Matched filter is used to totally remove the effect of noise from the received

signal. ( F )

Solution: Although matched filter can maximize the output SNR, it cannot totally

remove the noise.


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8. Multiplying a match filter with a real constant still gets a match filter.

( T )

Solution: Page 38 of lecture note 2.


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9. The power of Additive White Gaussian Noise on whole spectrum is infinity.

( T )

Solution: AWGN is a theoretical concept with unbounded bandwidth. AWGN has a

certain power spectral density (PSD), say N0/2. 0

2

NP df

10. If two random variables are uncorrelated, they are independent.

( F )

Solution: Uncorrelated relationship cannot imply independence, except for the two

random variables are Gaussian r.v.s..


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Question 2 (35 marks) Consider the signals

1) Calculate the energy of each signal. (6 marks)

2) Find a set of orthonormal basis functions for these signals. Express them as a linear

combination of those functions. (6 marks)

3) Calculate the energy again used the coordinates in the signal space. (6 marks)

4) Find the inner product of these two signals. (5 marks)

5) Suppose all the signals are transmitted with equal probability. First draw the

corresponding constellation diagram of this modulation. Then draw the decision

region. (6 marks)

6) Try to modify only one signal to achieve a better SER performance under the same

average symbol energy . Draw the new constellation diagram. (6 marks)


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Solution

5)


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6) Answer varies. For example, make

Question 3 (35 marks) Consider a baseband digital communication system that employs an orthonormal basis

( ) shown in Figure 1, where , and a modulation scheme whose

constellation is shown in Figure 2. The ML detector is used at the receiver and

AWGN channel is assumed with power spectral density / 2.


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1) What is the bit rate? (6 marks)

2) What is the average symbol Energy in terms of ? (5 marks)

3) Find (plot) the signal transmitted for symbol . (6 marks)

4) What is the pair-wise probability of i.e., the probability of detecting

when transmitting ? Also, what is ? Express the result in terms of and

. (6 marks)

5) Assume that different symbols have equal probability of being transmitted. Derive

a union bound for the symbol error rate in the system in terms of and . (6

marks)

6) Relax the error bound derived in 5) to contain one Q function only. Then use this

relaxed error bound to determine the required to achieve a symbol error

probability of , where denotes the average bit energy. (6 marks)

You may use the Q-function table below:

x 4.7534 4.5648 4.5062 4.4172 4.2649 4.0556 3.9903 3.8906 3.7190


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Q(x)

(*10-5) 0.1 0.25 0.33 0.50 1.0 2.5 3.3 5.0 10

Solution

1) The symbol rate is 1/T = 800 symbols/s. We have log2(8) = 3 bits per symbol.

Hence the bit rate is 2400 bits/s.

2) .

3) The plotted figure is as follow:

4) ,

5)

6) . Hence, it is sufficient to have . Also,

observe that , then it is sufficient to have

From Q-function table, the required is .

ELEC4110 Midterm Exercise 2014 With Solution (1)

Documents

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t question

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f solution

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qam solution

marks total marks