Introduction to Digital Communications

Midterm

3:30AM ~ 6:30PM, 4/26/2016

1. (2+3+2+3=10 points)

Given the carrier frequency fc , symbol energy E, and symbol duration T, write

down the mathematical formats of the transmitted signals of

A. M-ary PSK.

B. M-ary QAM square constellations (given the in-phase symbol ai and

quadrature symbol bi, and the energy E0 of the signal with the lowest

amplitude.)

C. M-ary FSK (supposing that fc = nc /2T).

D. Carrierless amplitude and phase modulation (CAP) (given a pulse shaping

function of g(t), with g(t) containing no symbol energy)

[A]:

A. 𝑠𝑖(𝑡) = √2𝐸

𝑇cos (2𝜋𝑓𝑐𝑡 +

2𝜋

𝑀(𝑖 − 1)), 0 ≤ 𝑡 < 𝑇, 𝑖 = 1,… ,𝑀

B. 𝑠𝑖(𝑡) = √2𝐸0

𝑇𝑎𝑖 cos 2𝜋𝑓𝑐𝑡 −√

2𝐸0

𝑇𝑏𝑖 sin 2𝜋𝑓𝑐𝑡 , 0 ≤ 𝑡 ≤ 𝑇 with ai and bi are

integers. Define 𝐿 = √𝑀, then we have

{𝑎𝑖, 𝑏𝑖} = [

(−𝐿 + 1, 𝐿 − 1) (−𝐿 + 3, 𝐿 − 1)(−𝐿 + 1, 𝐿 − 3) (−𝐿 + 3, 𝐿 − 3)

… (𝐿 − 1, 𝐿 − 1)

… (𝐿 − 1, 𝐿 − 3)⋮ ⋮

(−𝐿 + 1,−𝐿 + 1) (−𝐿 + 3,−𝐿 + 1)⋱ ⋮… (𝐿 − 1,−𝐿 + 1)

]

C. 𝑠𝑖(𝑡) = {√2𝐸

𝑇cos (

𝜋

𝑇(𝑛𝑐 + 𝑖)𝑡) , 0 ≤ 𝑡 ≤ 𝑇

0, elsewhere

, i = 1,2,…,M,

D. 𝑠(𝑡) = Re{∑ (𝑎𝑘 + 𝑗𝑏𝑘)𝑔(𝑡 − 𝑘𝑇)exp(𝑗2𝜋𝑓𝑐𝑡) ∞𝑘=−∞ }

Remember to write down your id number and your name.

Please provide detailed explanations/derivations in your answers. Correct answer without any explanations

will not be given any credits. However, wrong answers with correct reasoning will have partial credits.

2. (2+3+5=10 points)

Following the previous problem, given the carrier frequency fc , symbol energy E,

and symbol duration T, provide the basis functions and the message points of the

following modulations

A. M-ary PSK.

B. BFSK, also called Sunde’s FSK, (supposing that fc = (f1 + f2)/2).

C. Carrierless amplitude and phase modulation (CAP) (given a pulse shaping

function of g(t)).

[A]:

A. The orthonormal bases for M-ary PSK ae

𝜙1(𝑡) = √2

𝑇cos 2𝜋𝑓𝑐𝑡 0 ≤ 𝑡 ≤ 𝑇

𝜙2(𝑡) = −√2

𝑇sin 2𝜋𝑓𝑐𝑡 0 ≤ 𝑡 ≤ 𝑇

The message points are given by.

𝑥𝐼 = √𝐸cos (2𝜋

𝑀(𝑖 − 1)) 𝑥𝑄 = √𝐸sin (

2𝜋

𝑀(𝑖 − 1))

B. The orthonormal bases functions for Sunde’s FSK are

𝜙𝑖(𝑡) = {√

2

𝑇𝑏cos(2𝜋𝑓𝑖𝑡), 0 ≤ 𝑡 < 𝑇𝑏

0, elsewhere

The two message points on are

𝑺1 = [√𝐸𝑏0

] and 𝒔𝟐 = [0

√𝐸𝑏],

C. The passband in-phase pulse is 𝑝(𝑡) = 𝑔(𝑡) cos(2𝜋𝑓𝑐𝑡), and

the passband quadrature pulse is �̂�(𝑡) = 𝑔(𝑡) sin(2𝜋𝑓𝑐𝑡).

The message points are {𝑎𝑘, 𝑏𝑘}

3. (4+8+4+4=20 points)

Given the carrier frequency fc , symbol energy Eb, and symbol duration Tb, for

Minimum shift keying (MSK).

A. Provide the mathematical format of the transmitted signal

B. Provide the basis functions and the message points

C. Show the phase trellis of the MSK signals for a sequence of binary waves:

1 1 0 1 0 0 1 0

D. For Gaussian-filtered MSK (GMSK), given the pulse shaping function

𝑔(𝑡) =1

2[erfc (𝜋√

2

log2𝑊𝑇𝑏 (

𝑡

𝑇𝑏−

1

2)) − erfc (𝜋√

2

log2𝑊𝑇𝑏 (

𝑡

𝑇𝑏+

1

2))]

with W being the 3 dB baseband bandwidth of the pulse shaping filter.

Explain how GMSK might affect the phase trellis for the cases of 𝑊𝑇𝑏=

and 𝑊𝑇𝑏

[A]:

A. 𝑠(𝑡) = √2𝐸𝑏

𝑇𝑏cos 𝜃(𝑡) cos(2𝜋𝑓𝑐𝑡) − √

2𝐸𝑏

𝑇𝑏sin 𝜃(𝑡) sin(2𝜋𝑓𝑐𝑡)

With 𝜃(𝑡) = 𝜃(0) ±𝜋

2𝑇𝑏𝑡, 0 ≤ 𝑡 ≤ 𝑇𝑏

B. The in-phase term for −𝑇𝑏 ≤ 𝑡 ≤ 𝑇𝑏 is

𝜙1(𝑡) = √2

𝑇𝑏𝑐𝑜𝑠 (

𝜋

2𝑇𝑏𝑡) 𝑐𝑜𝑠 2𝜋𝑓𝑐𝑡

The message point is 𝑠1 = ∫ 𝑠(𝑡)𝜙1(𝑡)𝑇𝑏−𝑇𝑏

𝑑𝑡 = √𝐸𝑏cos 𝜃(0).

On the other hand, the quadrature term for 0 ≤ 𝑡 ≤ 2𝑇𝑏 is

𝜙2(𝑡) = √2

𝑇𝑏𝑠𝑖𝑛 (

𝜋

2𝑇𝑏𝑡) 𝑠𝑖𝑛 2𝜋𝑓𝑐𝑡

The message point is 𝑠2 = ∫ 𝑠(𝑡)𝜙2(𝑡)2𝑇𝑏0

𝑑𝑡 = −√𝐸𝑏sin 𝜃(𝑇𝑏)

D. Phase transitions in the phase diagram become smoother.

4. (5+5+10+5=25 points)

Given that the bit energy is Eb and the noise density is N0,

A. Show the observation vectors of the received QPSK signals, and accordingly

provide the decision rule and its corresponding decision boundaries in the

signal space of QPSK

B. Suppose the input dibits are Gray-encoded, derive the bit error rate of

QPSK given that 1

√𝜋∫ exp[−𝑧2]𝑑𝑧∞

𝑥=

1

2erfc(𝑥)

C. Show the observation vectors of the received MSK signals, and accordingly

provide the detection rule and its corresponding decision boundaries in the

signal space of MSK.

D. Derive the approximate bit error rate (BER) of MSK.

[A]:

A. See lecture note

B. The average probability of symbol error is

𝑃𝑒 = 1 − 𝑃𝑐 = erfc (√𝐸

2𝑁0) −

1

4erfc2 (√

𝐸

2𝑁0)

𝑃𝑒 ≅ erfc (√𝐸

2𝑁0) = erfc (√

𝐸𝑏𝑁0) , when

𝐸

2𝑁0≫ 1

For Gray coded symbols, the most probable number of bit errors is one as a

symbol is most likely to be taken as adjacent symbols. As a result

∴ BER =𝑃𝑒2=1

2erfc (√

𝐸𝑏𝑁0)

C. For the optimum detection of 𝜃(0), we project x(t) onto 𝜙1(𝑡)

𝑥1 = ∫ 𝑥(𝑡)𝜙1(𝑡)𝑇𝑏

−𝑇𝑏

𝑑𝑡 = √𝐸𝑏cos 𝜃(0) + 𝑤1, − 𝑇𝑏 ≤ 𝑡 ≤ 𝑇𝑏

If 𝑥1 > 0, then 𝜃(0) = 0, otherwise 𝜃(0) = 𝜋.

For the optimum detection of 𝜃(𝑇𝑏), we project x(t) onto 𝜙2(𝑡)

𝑥2 = ∫ 𝑥(𝑡)𝜙2(𝑡)2𝑇𝑏

0

𝑑𝑡 = −√𝐸𝑏sin 𝜃(𝑇𝑏) + 𝑤2, 0 ≤ 𝑡 ≤ 2𝑇𝑏

If 𝑥2 > 0, then 𝜃(𝑇𝑏) = −𝜋

2, otherwise 𝜃(𝑇𝑏) =

𝜋

2.

D. The average correction I s

𝑃𝑐 = (1 − 𝑃′)2 = 1 − erfc (√𝐸𝑏𝑁0) +

1

4erfc2 (√

𝐸𝑏𝑁0)

Thus the probability of wrong decision for one message points is ≅

erfc (√𝐸𝑏

𝑁0) which yields the BER of 𝑃𝑒 =

1

2erfc (√

𝐸𝑏

𝑁0)

5. (10+5+5=20 points)

For M-ary QAM with square constellations, i.e. M=4, 16, 64,…, suppose the

energy of the signal with the lowest amplitude is E0,

A. Derive the symbol error rate of M-ary QAM.

B. Define the bandwidth efficiency as Rb/B, where Rb = 1/Tb is the bit rate, and

B=2/T with T being the symbol duration. Show the bandwidth efficiency of

M-ary QAM.

C. Compare the bandwidth efficiency of M-ary QAM with that of M-ary FSK

supposed that the adjacent frequencies of M-ary FSK are separated from

each other by 1/2T

[A]:

A. See the lecture for the details. Given the average symbol energy

𝐸𝑎𝑣 =2(𝐿2 − 1)𝐸0

3=2(𝑀 − 1)𝐸0

3

The average symbol error rate is

𝑃𝑒 ≅ 2(1 −1

√𝑀) erfc (√

𝐸0𝑁0) = 2(1 −

1

√𝑀) erfc (√

3𝐸𝑎𝑣2(𝑀 − 1)𝑁0

)

B. The symbol duration T of M-ary QAM is 𝑇𝑏 log2𝑀. Since 𝑅𝑏 =1

𝑇𝑏, and B =

2

T=

2𝑅𝑏

log2𝑀, then 𝜌 =

𝑅𝑏

𝐵=

log2𝑀

2

C. For M-ary FSK, the bandwidth is B = M / 2T , then 𝜌 =𝑅𝑏

𝐵=

2𝑙𝑜𝑔2𝑀

𝑀

6. (5+5+5=15 points)

Given that the bit energy is Eb and the noise density is N0,

A. Show the baseband power spectral density (PSD) of the BFSK signal

B. Show the baseband PSD of the MSK signal

C. Compare the baseband PSDs with that of the QPSK signal

Notice that sinc (𝑥) =sin (𝜋𝑥)

𝑥

[A]:

A. The in-phase component of BFSK is completely independent of the input

binary wave, and is equal to √2𝐸𝑏/𝑇𝑏 cos(𝜋𝑡/𝑇𝑏). The PSD of the in-phase

component is, thus, equal to

𝐸𝑏2𝑇𝑏

[𝛿 (𝑓 −1

2𝑇𝑏) + 𝛿 (𝑓 +

1

2𝑇𝑏)]

Depending on the input value, the quadrature component is equal to g(t) or

-g(t) where

g(𝑡) = {√2𝐸𝑏𝑇𝑏

sin (𝜋𝑡

𝑇𝑏) , 0 ≤ 𝑡 ≤ 𝑇𝑏

0, otherwise

The ESD of g(t) is obtained by F(g(t)) F*(g(t)), yielding

𝛹g(𝑓) =8𝐸𝑏𝑇𝑏cos

2(𝜋𝑓𝑇𝑏)

𝜋2(4𝑇𝑏2𝑓2 − 1)

2

The PSD thus equals to

𝑆𝐵(𝑓) =𝐸𝑏2𝑇𝑏

[𝛿 (𝑓 −1

2𝑇𝑏) + 𝛿 (𝑓 +

1

2𝑇𝑏)] +

8𝐸𝑏cos2(𝜋𝑓𝑇𝑏)

𝜋2(4𝑇𝑏2𝑓2 − 1)

2

B. Depending on the value of 𝜃(0), the in-phase component equals g(t) or -g(t)

where

g(𝑡) = {√2𝐸𝑏𝑇𝑏

cos (𝜋𝑡

2𝑇𝑏) , −𝑇𝑏 ≤ 𝑡 ≤ 𝑇𝑏

0, otherwise

The ESD of g(t) is obtained by F(g(t)) F*(g(t)), yielding

𝛹g(𝑓) =32𝐸𝑏𝑇𝑏cos

2(2𝜋𝑓𝑇𝑏)

𝜋2(16𝑇𝑏2𝑓2 − 1)

2

The PSD equals 𝛹g(𝑓)/2𝑇𝑏

Similarly, depending on the value of 𝜃(𝑇𝑏), the quadrature component is

equal to g(t) or –g(t) where

g(𝑡) = {√2𝐸𝑏𝑇𝑏

sin (𝜋𝑡

2𝑇𝑏) , 0 ≤ 𝑡 ≤ 2𝑇𝑏

0, otherwise

The PSD is the same to that of the in-phase component

Since the in-phase and the quadrature components of MSK are statically

independent, the baseband PSD of the MSK equals

𝑆B(𝑓) = 2𝛹g(𝑓)

2𝑇𝑏=

32𝐸𝑏cos2(2𝜋𝑓𝑇𝑏)

𝜋2(16𝑇𝑏2𝑓2 − 1)

2

C. For QPSK, the in-phase and quadrature components are statically

independent. The baseband PSD is thus equal to

𝑆𝐵(𝑓) = 2𝐸sinc2 (𝑓𝑇) = 2𝐸𝑠𝑖𝑛2 (𝜋𝑓𝑇𝑏)

(𝜋𝑓𝑇𝑏)2= 2𝐸𝑏 𝑙𝑜𝑔2𝑀𝑠𝑖𝑛𝑐2 (𝑓𝑇𝑏 𝑙𝑜𝑔2𝑀)

with M=4.