Name: Student No.:

UNIVERSITY OF TORONTO

FACULTY OF APPLIED SCIENCE AND ENGINEERING

FINAL EXAMINATION, April 17, 2017

Third Year Electrical and Computer Engineering/Erigi neering Science

ECE316H1 S - COMMUNICATION SYSTEMS

Calculabor Type: 2

Exam Type: C

Examiner Deepa Kundur

Each question is worth 10 points in total. Answer all questions in the space provided in this question booklet. You may use the backs of pages if needed, but please specify the question you are answering clearly. There are 10 pages in total to this exam.

Allowable aids: (1) a 8.5"x11 double-sided handwritten aid sheet and (2) Type 2 (nonpro-grammable) calculator. Please note that important tables are given at the end of the test.

If a particular question seems unclear, please explicitly state any reasonable assumptions and proceed with the problem.

Please properly label all points of interest 011 sketches and graphs that you are requested to draw. so that there is no ambiguity.

For full marks, show all steps and present results clearly.

Question 1 2 3 4 5 Total Value of Ques. 10 10 10 10 10 50

Earned

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1. (10 marks)

(2 points) Distinguish the concepts of coherent and non-coherent demodulation. How are they different and what is an advantage and disadvantage of each?

(1 point) What is an advantage of quadrature amplitude modulation (QAIVI) over double-sideband suppressed carrier amplitude modulation?

(2 points) List four properties of angle modulation discussed in class?

(1 point) What is the main objective of a regenerative repeater?

(2 points) Suppose we have a binary phase shift keying system (BPSK) for which Tb =

where k is a positive integer, Tb is the hit duration and f. is the corresponding carrier frequency. Show that the BPSK pulses will have k cycles within hit period Tb.

(1 point) What is meant by conjugate symmetry of a Fourier Transform C(f)?

(1 point) Why would one use a Costas receiver instead of an envelope detector for AM demodulation?

2. (10 marks) A compact disk (CD) records audio signals digitally by using pulse code modu-lation (PCM). Assume that the audio signal bandwidth equals 12.5 kHz.

(a) (3 points) If the Nyquist samples are uniformly quantized into L = 65,536 levels and then binary coded, determine the number of bits required to encode one sample.

(h) (2 points) Determine the number of bits/second required to encode the audio signals.

(3 points) For practical reasons, signals are sampled at a rate above the Nyquist rate. Practical CDs use 44, 100 samples /second. If L = 65, 536, determine the number of bits/second required to encode the signal and the minimum bandwidth required to trans-mit the encoded signal.

(2 points) If a digital audio system designer wanted to improve the fidelity of the encoded CD data, state one way in which this can be achieved and justir it.

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3. (10 marks) Consider the frequency demodulation scheme shown in the following figure in which the incoming frequency modulation (FM) wave s(t) is passed through a delay line that produces a phase shift of —7r/2 radians at the carrier frequency f. The delay-line output is subtracted from s(t), and the resulting composite wave is then envelope-detected. Assuming that:

s(t) = cos [27rft +,3 sin(27rfmt)]

analyze the operation of this demodulator when the modulation index 3 < 1 and the de- lay T < 1 such that the following approximations can be made: cos(27rf 1T) 1 and sin(2irfrnT) 27, f rnT.

Modulating SiYIIdl E~4T +

Envelope O Delay Line Output

Ctor

wave

5

{bk}

4. (10 marks) Consider the following baseband digital transmission system.

Mod Output

Adder

ulo 2 I r --------, Sequence Decision-making xo: {d}

______

{a} f6k j

1 I Delay

Tb att= kTh

Precoder Duobinary conversion fitter [1(f)

A precoder stage takes the input binary sequence {bk} and produces a new binary sequence {dk} defined by:

dk = hk T ('k—i

where denotes the exclusive OR (XOR) operation. The precoded binary sequence {dk} is applied to a pulse-amplitude modulator that produces the line encoded sequence {ak } E

{-1, 11 (such that ak = 1 for d = 1 and ak = —1 for d,, = 0) followed by a duohinary conversion filter as shown in the figure where

H f -- for B( <f < B0

0 otherwise

The output sequence {ck} has three possible levels and enables the decoded result {c} ac- cording to the decision rule stated in the figure above.

(a) (3 points) Consider the input data sequence: {hk} = 0011100 for k = 1,2, , 7. Let the initial bit d0 = 1. Determine the precoded sequence {d} and then sketch the two-level sequence {ak) produced by the pulse-amplitude modulator.

(h) (3 points) Determine the duobinary code output {ck}.

(2 points) Applying {ck} to the decision-making device above, determine time resulting binary sequence {b}. What can you say about the sequence {bk} in relation to {bk}?

(2 points) What main advantage does the precoding stage provide for system operation? Please explain.

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5. (10 marks) In this question, we will extend what we have learned in class on binary amplitude

shift keying (BASK) to a simple M-ary amplitude shift keying scheme. Suppose Al = 4 and

the transmitted pulses for different bit-pairs are given by:

s(t) = a/2-cos(2irft), 0 <t < T. VTb

The following table defines how, the different bit pairs are communicated as symbol pulses:

Index i bit pair ai s(t)

0 < t < T 0 < t < T

1 00 3 3Jcos(27, fct)

2 0 1 1 /cos(2mrft)

3 1 1 —1 /cos(fct)

4 10 —3 —3/cos(2irft)

(3 points) Draw a corresponding transmitter for this system. Hint: Consider viewing:

s(t) = b(t). J,cos(2ft).

(2 points) Draw s(t) for the following bit sequence carefully labelling your sketch:

0110110010. Assume T =

(3 points) Draw a coherent receiver for the system in this question showing how signal

detection is possible.

(2 points) Draw a mien-coherent receiver for the system in this question and explain how

signal detection is possible.

M.

Fourier Transform Properties and Theorems

Property/Theorem Time Domain Frequency Domain Notation: g(t) G(f)

91(t) Gi(f) 92 (t) G2 Y)

Linearity: c191(t)+c292(t) c1G1(f)+c2G2(f)

Dilation: g(at) lal (fl Conjugation: g*(t) G*(_f)

Duality: G(t) g( —f) Time Shifting: g(t - to) C(f)eJ2ft0

Frequency Shifting: &?2tg(t) G(f - fc)

Area Under G(f): g(0) = fG(f)df

Area Under g(t): fg(t)dt = G(0)

Time Differentiation: dt

g(t) j2irfG(f)

Time integration: f t_g(T)dT 13 f G(f) Modulation Theorem: 91(t)92(t) f G)G2(f -

Convolution Theorem: fgj(r)g2 (t —r)d G1(f)G2(f) Correlation Theorem: fg1(t)g(t - r)dt G1 (f)G.(f)

Rayleigh's Energy Theorem: g(t)12dt = f G(f)Vdf

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Fresnel Reflection & Transm

o1 HI TE-case H7 \ k TM-case

E. n. cos O—n, cos O, E. E. —n,cos+n.cos&

TE =

iF = n:cos6+n,coso, r

IM = TM = n,

O. cos01+n1cos0,

-

E,0 -

E, -

2n, cos -

E10 -

E1 -

2n, cos 0,

E,0 E, n, cos 0, + nt cos 0, E,0 TM

E, n, cos 0, + n, cos 0,

R = r2

T= n1cos0, 2

ni Cos 01

Total Internal Reflection TE-case:

ii, cos 0,, -in, V(ni sin 0,. /n, )2 -

rJE=

n1 cos 0 + in, J(n, sin 0, /n, ) —1

TM-case:

- fl, CO S (9, + in1 ,j (ni, sin q.1 n1 )2 —1 rJf =

n, cos q. + in, (n, sin 01 In, ) —1

- ..i' 11 — e

n 7E = —2 tan ' —i

(ni sin q. In, Y —I )

n1 cost )

I / n1 S In 01. / n ) —1 cbJM = —2 tan'

ii' f

J + a ii, cos O,

Fabry-Pe'rot Interferometer

TFI = F 1 _4R

)2 5r_ ___

i + Fsin2 (6/2) - 2

Airy Function coefficient of Finesse Finesse

Fraunhofer (far-field) diffraction -, _,

.v. +v• 00

E1 (x1 ,y) = 2

ff E,(x,,y,)expl-ik Lx, +LYO )]dxO dY()

Far-field diffraction of a circular aperture

-1.22 Azi - 0 1.22- 2a 2a p1

-1.22-- 1.22-- sinG 2i 2a

Approximations

2 3 X x e =1+x+—+—+... foralix

2! 3! 3

X x 5 sinx=x--+--... foralix

3! 5! 2 4

cosx=1—x —+—

x —... for all x

2! 4! 5 X 2x Jr

tan x=x+—+ +... for x<- 315 2

=1+x+x2 +13 +... for x<i 1-x

(I _ I X)2 = 1+2x+3x2 +4x3 +... forx< 1

I 1 1 2 1 3 i1+x =1+—x----x +—x -... for all x

2 8 16

2 3 =1--x+ - -x - - --x +... for all x ,fl +—x 2 8 16

Depending on the situation, you may choose to use 0th, 1st, or 2nd order approximations.