Seat Number Problem 1: Pulse amplitude modulation...

Seat Number

King Mongkut’s University of Technology Thonburi Final Examination

Semester 2 -- Academic Year 2015

Subject: EIE 467 Digital Communications For: Electrical Communication and Electronic Engineering, 3rd Yr (Inter. Program) Exam Date: Tuesday May 10, 2016 Time: 9.00am-12.00pm Instructions:-

1. This exam consists of 5 problems with a total of 24 pages, including the cover. 2. This exam is closed-book. 3. You are not allowed to use a written A4 note for this exam. 4. Answer each problem on the exam itself. 5. A calculator compiling with the university rule is allowed. 6. A dictionary is not allowed. 7. Do not bring any exam papers and answer sheets outside the exam room. 8. The exam has no correction. If you think that you find a mistake, explain it for a possibility

of a partial credit. 9. Open Minds … No Cheating! GOOD LUCK!!!

Remarks:- • Raise your hand when you finish the exam to ask for a permission to leave the exam

room.

• Students who fail to follow the exam instruction might eventually result in a failure of the class or may receive the highest punishment within university rules.

• Carefully read the entire exam before you start to solve problems. Before jumping into the mathematics, think about what the question is asking. Investing a few minutes of thought may allow you to avoid twenty minutes of needless calculation!

Question No. 1 2 3 4 5 TOTAL

Full Score 22 25 16 23 14 100

Graded Score

Name ______________________________________________Student ID___________________

This examination is designed by Watcharapan Suwansantisuk; Tel: 9069

This examination has been approved by the committees of the ENE department.

_____________________________________ (Asst. Prof. Suwat Pattaramalai, Ph.D.)

Acting Head of Electronic and Telecommunication Engineering Department

Name . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Student ID 560705038 .. . . . . Seat Number . . . . . . . . . . . .

Problem 1: Pulse amplitude modulation (PAM) [22 points]

modulatorsignalencoder

pass filtersignaldecoder

ideal low−

baseband to passband

passband to baseband

demodulatorbaseband

baseband1 1 0 u(t)

2 cos(2πfct)

x(t)

cos(2πfct)

g(t)v(t)

A PAM modulator is connected directly to a PAM demodulator through a noiseless

channel, as shown in the above figure. The binary input b0 b1 b2 is

1 1 0

The amplification factor is a = 5. The symbol interval is T = 2 sec, and the pulse is the

square root of raised cosine filter with the roll-off factor α = 0.4. The carrier frequency

is fc = 0.5 Hz.

Fill in the blanks below. You do not need to justify your answer (2 points each).

(a) The output from the signal encoder is .

(b) The output from the baseband demodulator is .

(c) The output from the signal decoder is .

2


Map a figure to each signal in the table below. The available figures appear in

pages 4–7. You do not need to justify your answer (2 points each).

Signal Figure Signal Figure

(d) u(t) . . . . . . . . . . . . . . . . . (h) g(t) . . . . . . . . . . . . . . . . .

(e) u(f) . . . . . . . . . . . . . . . . . (i) g(f) . . . . . . . . . . . . . . . . .

(f) x(t) . . . . . . . . . . . . . . . . . (j) v(t) . . . . . . . . . . . . . . . . .

(g) x(f) . . . . . . . . . . . . . . . . . (k) v(f) . . . . . . . . . . . . . . . . .

3


−10 −5 0 5 10 15−10

−5

0

5

10

time t (sec)

valu

e

Figure 1

−10 −5 0 5 10 15−5

0

5

time t (sec)

valu

e

Figure 2

4


−10 −5 0 5 10 15−10

−5

0

5

10

time t (sec)

valu

e

Figure 3

−2 −1 0 1 20

5

10

15

20

frequency f (Hz)

mag

nitu

de

−2 −1 0 1 2−4

−2

0

2

4

frequency f (Hz)

angl

e (ra

dian

)

Figure 4

5


−2 −1 0 1 20

5

10

15

20

frequency f (Hz)

mag

nitu

de

−2 −1 0 1 2−4

−2

0

2

4

frequency f (Hz)

angl

e (ra

dian

)

Figure 5

−2 −1 0 1 20

5

10

15

20

frequency f (Hz)

mag

nitu

de

−2 −1 0 1 2−4

−2

0

2

4

frequency f (Hz)

angl

e (ra

dian

)

Figure 6

6


−2 −1 0 1 20

5

10

15

20

frequency f (Hz)

mag

nitu

de

−2 −1 0 1 2−4

−2

0

2

4

frequency f (Hz)

angl

e (ra

dian

)

Figure 7

−2 −1 0 1 20

5

10

15

20

frequency f (Hz)

mag

nitu

de

−2 −1 0 1 2−4

−2

0

2

4

frequency f (Hz)

angl

e (ra

dian

)

Figure 8

7


Problem 2: Shift-orthonormal pulse [25 points]

a

p(f)

f (Hz)10−1

A baseband real-valued signal p(t) has the Fourier transform p(f) as shown in the

above picture. The amplitude or height of the Fourier transform is a, which is a positive

real number.

Questions (a)–(d) are unrelated and can be answered separately.

(a) [5 points] Find the value of a that makes the norm of p equal one:∫∞−∞ p2(t)dt = 1.

8


(b) [8 points] Suppose the symbol interval is T = 12 sec. Show that p(t) is orthogonal

to its shifts:∫ ∞

−∞p(t)p(t−mT )dt = 0, for each m = 1, 2, 3 . . . .

9


For questions (c)–(e), let h(t) denote a passband signal: h(t) = p(t) sin(2πt). Let

h(f) denote the Fourier transform of h(t).

(c) [3 points] Plot the real part of h(f) as a function of the frequency f .

(d) [3 points] Plot the imaginary part of h(f) as a function of the frequency f .

10


(e) [6 points] Find the norm ∥h∥. Leave your answer in terms of a.

11


Problem 3: Random processes [16 points]

A random process is given by

X(t) =

⎧

⎨

⎩

A sin(t), 0 ≤ t ≤ 2π

0, otherwise,

where A is a random amplitude, equally likely to be 1, 2, and 3:

P {A = 1} = P {A = 2} = P {A = 3} =1

3.

Answer the following questions. You do not need to justify your answer (4 points each).

(a) Which graphs in the next page are the sample paths of X(t)? List all correct

answers.

Answer: Graphs

(b) Which graph in the next page is the mean function E {X(t)}?

Answer: Graph

12


−π 0 π 2π 3π

1

2

3

0

−1

−2

−3

t

Graph 1

Graph 2

Graph 3

−π 0 π 2π 3π

1

2

3

0

−1

−2

−3

t

Graph 4

Graph 5

Graph 6

−π 0 π 2π 3π

0

23

13

−23

−13

t

Graph 7

Graph 8 Graph 9

−π 0 π 2π 3π

0

23

13

−23

−13

t

Graph 10

Graph 11

Graph 12

−π 0 π 2π 3π

1

2

3

0

t

Graph 13

Graph 14

Graph 15

Graph 16

13


(c) For 0 ≤ s ≤ 2π and 0 ≤ t ≤ 2π, what is the value of E {X(s)X(t)}? Circle the

correct answer.

(A) 0

(B) 113 sin(s)

(C) 113 sin(t)

(D) 113 sin(s) sin(t)

(E) 143 sin(s)

(F) 143 sin(t)

(G) 143 sin(s) sin(t)

(H) None of the above

(d) For s /∈ [0, 2π] or t /∈ [0, 2π], what is the value of E {X(t)X(s)}? Circle the correct

answer.

(A) 0

(B) 113 sin(s)

(C) 113 sin(t)

(D) 113 sin(s) sin(t)

(E) 143 sin(s)

(F) 143 sin(t)

(G) 143 sin(s) sin(t)

(H) None of the above

14


Problem 4: Detection [23 points]

demodulator

modulatorX(t)

Z(t)

Y (t)

B ∈ {0, 1}

C ∈ {0, 1}

In a certain communication system, a bit B, which is equally likely to be 0 and 1, is

modulated into a waveform X(t), where

X(t) = p(t−B),

and p(t) is the square wave:

p(t) =

⎧

⎨

⎩

1, 0 ≤ t ≤ 1

0, otherwise.

The received waveform is

Y (t) = X(t) + Z(t),

where Z(t) is the additive white Gaussian noise (AWGN) with the power spectral density

of σ2, and the noise {Z(t) : t ∈ R} is independent of the transmitted bit B.

To decode (demodulate) B, the receiver uses the following decision rule:

∫ 1

0Y (t)dt

C=0≷

C=1

∫ 2

1Y (t)dt.

15


−2 −1 0 1 2 3

0

1

time t

a sample path of Y (t)

(a) [4 points] Suppose the received waveform Y (t) is given by the above figure. What

is the value of the decoded bit C? Why?

16


For each question (b)–(c) below, circle a correct answer. You do not need to justify

your answer (4 points each).

(b) Given that bit B = 0 was transmitted, what is the probability distribution of∫ 10 Y (t)dt?

(A) Normal N (−1,σ2) distribution

(B) Normal N (−1, 2σ2) distribution

(C) Normal N (0,σ2) distribution

(D) Normal N (0, 2σ2) distribution

(E) Normal N (1,σ2) distribution

(F) Normal N (1, 2σ2) distribution

(c) Given that bit B = 0 was transmitted, what is the probability distribution of∫ 21 Y (t)dt?

(A) Normal N (−1,σ2) distribution

(B) Normal N (−1, 2σ2) distribution

(C) Normal N (0,σ2) distribution

(D) Normal N (0, 2σ2) distribution

(E) Normal N (1,σ2) distribution

(F) Normal N (1, 2σ2) distribution

17


(d) [7 points] Find the conditional probability P {C = 1 | B = 0}. Write your answer

in terms of the Q-function and σ.

18


−5 −4 −3 −2 −1 0 1 2 3 4 510−1

100

SNR (dB)

Probability of errorProbability of error

(e) [4 points] The probability of error P {C = B} as a function of the signal-to-noise

ratio (SNR) in decibel (dB) is shown in the above figure. Here, the SNR equals

SNR =1

σ2(no unit).

If the power spectral density of the noise is σ2 = 2, what is the approximate value

of the probability of error? Justify your answer.

19


Problem 5: Channel coding [14 points]

A channel encoder uses a linear block code with the following generator matrix:

G =

⎡

⎢

⎣

1 0 0 1 0 1

0 1 0 1 1 1

0 0 1 0 1 1

⎤

⎥

⎦

.

Questions (a)–(c) are unrelated and can be answered separately. You do not need to

justify your answer.

(a) [4 points] Fill in the table below, which lists some codewords.

input to channel encoder output from channel encoder (codeword)

0 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

0 0 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

0 1 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20


Suppose the modulator is the PAM with the amplification factor a = 2, and the

channel decoder uses the minimum-distance rule.

Circle a correct answer for each question (5 points each).

(b) If a soft-decisioned input to the channel decoder is

−2 − 1 0 0 1 2

what is the output from the channel decoder?

(A) 0 0 0

(B) 0 0 1

(C) 0 1 0

(D) 0 1 1

(c) If a hard-decisioned input to the channel decoder is

0 0 0 1 1 1

what is the output from the channel decoder?

(A) 0 0 0

(B) 0 0 1

(C) 0 1 0

(D) 0 1 1

21


Formula Sheet

E {X} =∑

x∈XxP {X = x}

E {aX + bY + c} = aE {X}+ bE {Y }+ c

u(f) =

∫ ∞

−∞u(t)e−2πiftdt

u(t) =

∫ ∞

−∞u(f)e2πiftdf

|z|2 = zz∗

⟨u,v⟩ =∫ ∞

−∞u(t)v∗(t)dt =

∫ ∞

−∞u(f)v∗(f)df

p(t) =

⎧

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎩

1√T(1− α+ 4α

π ), t = 0

α√2T

[

(1 + 2π ) sin

(

π4α

)

+ (1− 2π ) cos

(

π4α

)]

, t = ± T4α

√T

πt!

1−( 4αt

T)2

"

{

sin[

π(1−α)tT

]

+ 4αtT cos

[

π(1+α)tT

]}

, otherwise

p(f) =

⎧

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎩

√T , |f | ≤ 1−α

2T√T cos

[

πT2α

(

|f |− 1−α2T

)]

, 1−α2T < |f | < 1+α

2T

0, |f | ≥ 1+α2T

Bandwidth =1 + α

2T

u(t) =N∑

k=0

ukp(t− kT )

u(f) =

[

N∑

k=0

uke−2πifkT

]

p(f)

vk =

∫ ∞

−∞v(t)p(t− kT )dt

ck =

⎧

⎨

⎩

1, if vk ≥ 0

0, if vk < 0

x(t) = ag(t) + bh(t) ↔ x(f) = ag(f) + bh(f)

x(t) = g(t− τ) ↔ x(f) = g(f)e−2πifτ

g(t) = h(t) cos(2πf0t) ↔ g(f) = 12 h(f + f0) +

12 h(f − f0)

g(t) = h(t) sin(2πf0t) ↔ g(f) = i2 h(f + f0)− i

2 h(f − f0)

22


fZ(z) =1√2πσ2

e−1

2σ2(z−µ)2 , −∞ < z < ∞

KX(s, t) = E

{

X(s)X(t)}

− E

{

X(s)}

· E{

X(t)}

Q (x) = P {Z > x} , where Z ∼ N (0, 1)

=

∫ ∞

x

1√2π

e−t2/2dt

SNR in dB = 10 log10 SNR

u(v) = argmaxm

P {U = am | V = v} (MAP rule)

fV |U (v|a0)fV |U (v|a1)

U=a0≷

U=a1

P {U = a1}P {U = a0}

(Likelihood-ratio test)

ln

[

fV |U (v|a0)fV |U (v|a1)

]

U=a0≷

U=a1

ln

[

P {U = a1}P {U = a0}

]

(Log likelihood-ratio test)

u(v) = argminm

|v − am| (Minimum-distance rule)

C = W log2

(

1 +P

WN0

)

bit/sec

uG = u[

I | P]

=[

u | uP]

• If random variables X and Y are independent, then E {XY } = E {X}E {Y }.

• If[

XY

]

is a Gaussian vector and if E {XY } = 0, then random variables X and Y

are independent.

• For every real-valued L2-function f(t), the integral

∫ ∞

−∞f(t)Z(t)dt

is a Gaussian random variable of mean 0 and variance σ2∫∞−∞ f2(t)dt.

23


• Suppose {Z(t) : t ∈ R} is an AWGN process with the power spectral density of σ2.

Suppose f(t) and g(t) are real-valued L2-functions. Let random variables F and

G be defined by the integrals

F =

∫ ∞

−∞f(t)Z(t)dt and G =

∫ ∞

−∞g(t)Z(t)dt.

Then,[

FG

]

is a Gaussian vector, and the expectation of FG is

E {FG} = σ2∫ ∞

−∞f(t)g(t)dt.

In particular, F and G are independent if and only if∫ ∞

−∞f(t)g(t)dt = 0

• If W is a Gaussian N (µ,σ2) random variable and if x is a constant, then

P {W ≥ x} = Q

(

x−mean

standard deviation

)

= Q

(

x− µ

σ

)

.

• Suppose Z ∼ N (µ,σ2). Let W = bZ + c for constants b and c. Then, W is a

Gaussian random variable with mean and variance

E {W} = E {bZ + c} = bµ+ c

V {W} = V {bZ + c} = b2σ2,

respectively.

• Suppose X ∼ N (µX,σ2X) and Y ∼ N (µY,σ2

Y) are independent. Let W = X + Y .

Then, W is a Gaussian random variable with mean and variance

E {W} = E {X + Y } = µX + µY

V {W} = V {X + Y } = σ2X + σ2

Y,

respectively.

24

Seat Number Problem 1: Pulse amplitude modulation...

Documents

Transcript of Seat Number Problem 1: Pulse amplitude modulation...