Problem Sheets: Communication Systems

Problem Sheets: Communication Systems

Professor A. ManikasChair of Communications and Array ProcessingDepartment of Electrical & Electronic Engineering

Imperial College London

2011-2012

Introductory Concepts

1. Sketch and mathematicaly represent the pdfs of the following signals:

(a) 4 rep3T{

Λ(tT

)}−1 (10%)

i.e.pdfg(g) =1

2δ(g + 1) +

1

6rect

{g − 1

4

}(b) rep2T

{2rect

(tT

)+ 4Λ

(2tT

)}(10%)

i.e.pdfg(g) =1

2δ(g) +

1

8rect

{g − 4

4

}

(c) rep2T

{5 rect

(t

T

)− rect

(t− TT

)}(10%)

1

E303: Comm Systems (Problem Sheets) 2

pdfg(g) =1

2δ(g + 1) +

1

2δ(g − 5)

(d) rep6T

{4 rect

(t

5T

)− rect

(t− 3T

T

)}(5%)

pdfg(g) =1

6δ(g + 1) +

5

6δ(g − 4)

(e) N+1N repNT

{Λ

(t

T

)}− 1

Nwith N ∈ Z+ > 2 (15%)

i.e.

pdfg(g) =N − 2

Nδ(g +

1

N) +

2

N + 1rect

{g − N−1

2NN+1N

}

=N − 2

Nδ(g +

1

N) +

2

N + 1rect

{2Ng −N + 1

2(N + 1)

}(f) 3rep2 {Λ (t)} − 2 (5%)

pdfg(g) =1

3rect

{g − 0.5

3

}2. Evaluate:

(a)∞∫−∞

(t4 − 3t+ 1).δ(t− 2) .dt (10%)

∞∫−∞

(t4 − 3t+ 1).δ(t− 2) .dt = (t4 − 3t+ 1)|t=2 = 24 − 3× 2 + 1 = 11

(b)∞∫−∞

(cos(4πt) ∗ δ(t+ 1

4 )).δ(t− 1

8 ).dt (10%)

∞∫−∞

(cos(4πt) ∗ δ(t+ 1

4 )).δ(t− 1

8 ).dt =∞∫−∞

cos(4π(t+ 1

4 )).δ(t− 1

8 ).dt =

= cos(4π(t+ 1

4 ))|t= 1

8= cos

(4π( 18 + 1

4 ))

= cos( 32π) = 0

Imperial College London Coursework 2011-12


(c)∞∫−∞

(t3 − 3t2 − 11).δ(t− 1) .dt (5%)

∞∫−∞

(t3 − 3t2 − 11).δ(t− 1) .dt = (t3 − 3t2 − 11)|t=1 = 13 − 3× 12 − 11 = −13

(d)∞∫−∞

{(sin(4πt) ∗ δ(t+ 1

4 )}.δ(t− 1

4 ).dt (5%)

∞∫−∞

{(sin(4πt) ∗ δ(t+ 1

4 )}.δ(t− 1

4 ).dt =

∞∫−∞

(sin(4π

(t+ 1

4

)).δ(t− 1

4 ).dt =

sin(4π(t+ 1

4

))|t= 1

4= sin

(4π(14 + 1

4

))= sin (2π) = 0

(e)∞∫−∞

(t3 − 2t2 + 1).δ(t− 2) .dt (5%)

∞∫−∞

(t3 − 2t2 + 1).δ(t− 2) .dt = (t3 − 2t2 + 1)|t=2 = 23 − 2× 22 + 1 = 1

(f)∞∫−∞

{(cos(2πt) ∗ δ(t− 1

4 )}.δ(t− 1

12 ).dt (5%)

∞∫−∞

(cos(2πt) ∗ δ(t− 1

4 )).δ(t− 1

12 ).dt =∞∫−∞

cos(2π(t− 1

4 )).δ(t− 1

12 ).dt =

= cos(2π(t− 1

4 ))|t= 1

12= cos

(2π( 112 −

14 ))

= cos(− 13π) = 12

(g) h(3) where h(t) =(t.rect

{t8

})∗ δ(t+ 3) (5%)

h(t) =(t.rect

{18

})∗ δ(t+ 3) = (t+ 3) .rect

{t−38

}⇒ h(3) = 6

(h) h(3) where h(t) =(t.rect

{18T

})∗ δ(t− 2) (10%)

h(t) =(t.rect

{18T

})∗ δ(t− 2) = (t− 2) .rect

{t−28T

}h(3) = (3− 2).rect

{3−28T

}⇒ h(t) =

{t− 2 if −0.5 < t−2

8T < 0.50 otherwise

⇒ h(3) =

{1 if −0.5 < t−2

8T < 0.50 otherwise

=

{1 if T > 1

40 if T ≤ 1

4

(i) h(3.5) where h(t) =(t.rect

{18T

})∗ δ(t− 3) (10%)

3. The waveform below shows the autocorrelation function Rbb(τ) of what is called in commu-nications a pseudo-random (PN) signal b(t).

(a) Write a mathematical expression, using Woodward’s notation, to describe the aboveautocorrelation function. (15%)



(b) Find the power spectral density PSDb(f) of b(t). (20%)

Solution

(a) Rbb(τ) = N+1N repNTc

{Λ(τTc

)}− 1

N

(b) PSD(f) =FT{Rbb(τ)} = N+1N2 comb 1

NTc

{sinc2(fTc)

}− 1

N δ(f)

4. At the input of a filter there is white Gaussian noise of power spectral density PSDni(f) =3210−6. If the transfer function of the filter is

H(f) = Λ

{f

106

}exp(−jφ(f)

calculate the power of the signal at the output of the filter. (10%)

Solution

5. For the following differential circuitfind:

(a) the impulse response and (5%)

(b) frequency response (5%)



Solution

(a) h(t) = δ(t)− δ(t− T )

(b) H(f) =FT{h(t)}= 1− exp(−j2πfT )

= {exp(jπfT )− exp(−jπfT )} exp(−jπfT )

= 2 sin(πfT ) exp(−jπfT )

⇒ |H(f)| = 2 |sin(πfT )| =

6. Consider the filter with impulse response

h(t) = sinc2{

106(t− 3)}

and assume that the input signal ni(t) is white Gaussian noise with double-sided powerspectral density PSDni(f) = 1.5× 10−6 W/Hz.

For the signal n(t) at the output of the filter

(a) find and plot its power spectral density PSDn(f); (10%)

(b) calculate its power Pn (5%)

Solution

(a) H(f) =FT{h(t)}= 1106 Λ

(f106

)exp

(−j2πf106 × 3

)⇒PSDn(f) =PSDni(f). |H(f)|2 =

1.5× 10−6(1106

)2Λ2(

f106

)= 1.5× 10−18Λ2

(f106

)

(b) Pn =106∫−106

PSDn(f).df = 2106∫0

PSDn(f).df =

= 2106∫0

(1.5× 10−18 ×

(1− f

106

)2).df =

= 2× 1.5× 10−18 ×106∫0

(1− f

106

)2.df =

= 3× 10−18 ×106∫0

(1− 2 f

106 + f2

1012

).df



= 3× 10−18 ×(

106 − 2 1012

2×106 + 1018

3×1012

)= 3× 10−18 ×

(106 − 106 + 1

3106)

= 10−12

7. Consider a bandpass filter with impulse response

h(t) = 8× 103sinc{

4× 103t}. cos(2π104t)

and assume that at the input of this filter there is white Gaussian noise ni(t) of power spectraldensity PSDni(f) = 10−6.

For the signal n(t) at the output of the filter

(a) find and plot its power spectral density PSDn(f); (10%)

(b) calculate its power Pn (5%)

Solution



Information Sources

8. The signal at the output of an analogue information source x(t) having a uniform pdf be-tween ±2Volts, is passed through a half-wave and a full-wave rectifier circuits. Sketch andmathematically represent the pdfs of:

(a) the original analogue information source, 5%

(b) the output from the half-wave rectifier, 10%

(c) the output from the full-wave rectifier. 10%

(d) Determine

• the mean value, and 15%• the rms value 15%

of the signals in cases (a),(b) and (c) above.

N.B.: Assume ideal diodes

Solution

(a) pdfx(x)= 14 rect

{x4

}

mean: E {x(t)} =∫ +∞−∞ x.pdfx(x).dx =

∫ +2−2 x.

14 .dx = 0 Volts

power:Px = E{x2(t)

}=∫ +∞−∞ x2.pdfx(x).dx =

∫ +2−2 x

2. 14 .dx = 43

rms=√Px = 2√

3= 1.1547

(b) pdfn(n) = 12δ(n) + 1

4 rect{n−12

}

mean: E {n(t)} =∫ +∞−∞ n.pdfn(n).dn =

∫ +20

n. 14 .dn = 12 Volts

power:Pn = E{n2(t)

}=∫ +∞−∞ n2.pdfn(n).dn =

∫ +20

n2. 14 .dx = 23

rms=√Px =

√23 = 0.81650

(c) pdfn(n) = 12 rect

{n−12

}



mean: E {n(t)} =∫ +∞−∞ n.pdfn(n).dn =

∫ +20

n. 12 .dn = 1 Volts

power:Pn = E{n2(t)

}=∫ +∞−∞ n2.pdfn(n).dn =

∫ +20

n2. 12 .dx = 43

rms=√Px = 2√

3= 1.1547

9. Consider an analogue signal source x(t) having a uniform amplitude probability densityfunction

pdfx(x) =1

2arect

{ x

2a

}(a) Estimate the average power Px of the signal x(t). 10%

(b) Find the differential entropy Hx of the signal source x(t) 10%

(c) Find Hy −Hx 10%where Hy denotes the differential entropy of an analogue signal source y(t) having aGaussian amplitude probability density function with mean µy and σy =

√Px

(d) What is the entropy power of the signal x(t). 10%

Solution

(a) average power (source with uniform pdf):Px = E

{x2(t)

}=∫ +∞−∞ x2.pdfx(x).dx =

∫ +3−3 x

2. 16 .dx = 93 = 3

(b) Hx = −∫ +∞−∞ pdfx(x). log2 (pdfx(x)) dx = −

∫ +3−3

16 . log2

(16

)dx =

= − 16 . log2(16

) ∫ +3−3 dx = log2 6 = 2.585

(c) If y(t)=Gaussian with mean=µ and rms=√Px then

Hy = log2√

2πePx = log2√

2πe3 = 2.8396

Hy −Hx = 2.8396− 2.585 = 0.2546

(d) Nx = 1πe22Hx = 1

πe22×2.585 = 4.2158

10. A signal g(t) having the pdf shown in Fig.1 is bandlimited to 4 kHz. The signal is sampled atthe Nyquist rate and is fed through a 2-level quantizer. The transfer function of the quantizeris shown in Fig.2.

Fig. 1 Fig. 2

Consider the output of the quantizer as the output of a discrete information source (X, p).Calculate:

(a) the symbol rate rX of the source (X, p). 10%

(b) the amplitude pdf of the signal at the quantizer’s output. Sketch this pdf. 10%

(c) the rms value of the signal at the output of the quantizer. 10%

(d) the entropy HX 10%

(e) the entropy of the source (X ×X, J) (10%)

Solution



(a) rx = 2× Fg = 8k levelssec

(b) Pr(−2V ) = 34 ; Pr(+2V ) = 1

4 ⇒ p = [ 34 ,14 ]T

pdf:

(c) rms=√

(−2)2 Pr(−2V ) + 22 Pr(+2V ) =√

(−2)2 34 + 22 14 = 2V

(d) HX = − 34 log2(34

)− 1

4 log2(14

)= 0.81128 bits

level

(e) (X ×X, J) =

{ (x1x1,

916

) (x2x1,

316

)(x1x2,

316

) (x2x2,

116

) }HX×X = − 9

16 log2(916

)− 3

16 log2(316

)− 3

16 log2(316

)− 1

16 log2(116

)= 1.6226 bits

double level

11. A signal g(t) having the probability density function (pdf) shown below is sampled and fedthrough an 4-level quantizer.

Consider the output of the quantizer as the output of a discrete information source (X, p).

(a) Calculate and sketch the pdf of the signal at the output of the quantizer. 10%

(b) Calculate the rms value of the signal at the output of the quantizer. 10%

(c) What is the ensemble of the source (X ×X, J)? 10%

(d) Calculate the entropy HX×X 10%

Solution

(a) Quantiser:"end"-points Quantisation levelsb0 = −2V (or −∞)b1 = −1V x1 = −1.5Vb2 = 0V x2 = −0.5Vb3 = 1V x3 = 0.5Vb4 = 2V (or +∞) x4 = 1.5V

input pdf:



Therefore, power of g(t) = Pg =+∞∫−∞

g2.pdfg(g).dg

= 20∫−2g2.(116 + 3

16 (g + 2)).dg = 2

0∫−2

(116g

2 + 316g

3 + 616g

2).dg = 2

0∫−2

(716g

2 + 316g

3).dg

= 2×716

g3

3 ]0−2 + 2×3

16g4

4 ]0−2 = 5

6 = 0.8333

input pdf:

output pdf:

(b) rms:

i.e. A3 = A2 = 1132 and A4 = A1 = 5

32

Pgout = 2 ×((−1.5)2 × 0.1563 + (−0.5)2 × 0.3438

)= 0.8752 ⇒ rms =

√0.8752 =

0.9355

(c) (X, p) ={(−1.5V, 532

),(−0.5V, 1132

),(+0.5V, 1132

),(+1.5V, 532

)}(X ×X, J) =

(x1x1,

251024

) (x2x1,

551024

) (x3x1,

551024

) (x4x1,

251024

)(x1x2,

551024

) (x2x2,

1211024

) (x3x2,

1211024

) (x4x2,

551024

)(x1x3,

551024

) (x2x3,

1211024

) (x3x3,

1211024

) (x4x3,

551024

)(x1x4,

251024

) (x2x4,

551024

) (x3x4,

551024

) (x4x4,

251024

)

(d) Entropy:HX×X = 1T4 (J� log2 J) 14 = 3.7921 bits

double level



Communication Channels

12. If one binary source and two binary channels are connected in cascade as shown below

ChannelNo.1

ChannelNo.2

Binary Source

where both channels have the following forward transition probability diagram

0.7

0.7

find the bit-error-rate pe at the output of the second channel. 10%

Solution

• The matrix F is:F =

[0.7 0.30.3 0.7

]• ∴ the overall matrix Fcsc is as follows:

Fcasc = F.F =

[0.7 0.30.3 0.7

] [0.7 0.30.3 0.7

]=

[0.7× 0.7 + 0.3× 0.3, 0.7× 0.3 + 0.3× 0.70.3× 0.7 + 0.7× 0.3, 0.7× 0.7 + 0.3× 0.3

]=

[0.58 0.420.42 0.58

]=

[Pr(z0|x0), Pr(z0|x1)Pr(z1|x0) Pr(z1|x1)

]∴

pe = Pr(z1|x0) Pr(x0) + Pr(z0|x1) Pr(x1)

= 0.42× 0.4 + 0.42× 0.6 = 0.42

13. A digital communication system, operating at 100 bits/sec in the presence of additive whiteGaussian noise of power spectral density PSDn(f) = N0

2 , is represented in the energy utiliza-tion effi ciency (EUE) - bandwidth utilization effi ciency (BUE) plane, as follows:



What is the capacity C of the channel in bits/sec? 20%

Solution

C = B log 2(1 + SNRin︸︷︷︸=30/2

) = B log2(1 + 15) = 4B (1)

However, BUE= Brb⇒ B = BUE︸︷︷︸

=2

× rb︸︷︷︸=100

= 200 (2)

Therefore: (1) ∧ (2)⇒ C = 4× 200 = 800 bitssec

14. A digital communication system having an energy utilisation effi ciency (EUE) equal to 30operates in the presence of additive white Gaussian noise of double-sided power spectraldensity PSDn(f)= 0.5× 10−6W/Hz. If the channel capacity C is 16 kbits/s and the channelbandwidth B is 4 kHz, estimate

(a) the bit rate rb 10%

(b) the noise power at the channel output 10%

Solution

(a)

(b) Pn = N0B = 10−6 × 4k = 4mW

15. A discrete channel is modelled as follows:

Estimate:

(a) The probability of error at the output of the channel 5%

(b) The amount of information delivered at the output of the channel 10%



Solution

(a) pe = Pr(H2, D1) + Pr(H1, D2) = Pr(D1|H2) Pr(H2) + Pr(D2|H1) Pr(H1)

= 0.04× 23 + 0.018× 1

3 = 0.032667

(b) any expression of mutual information Hmut can be used.

For instance: Hmut =

2∑m=1

2∑k=1

Jkm log2

(pmqmJkm

)where

q = F.p⇒[q1q2

]=

[0.982 0.040.018 0.96

]︸︷︷︸

F

[p1p2

]

p1 = Pr(H1) = 13 ,

p2 = Pr(H2) = 23

J11 = Pr(H1, D1) = Pr(D1|H1) Pr(H1) = 0.982× 13 = −0.3273

J12 = Pr(H2, D1) = Pr(D1|H2) Pr(H2) = 0.04× 23 = −0.0267

J21 = Pr(H1, D2) = Pr(D2|H1) Pr(H1) = 0.018× 13 = −0.006

J22 = Pr(H2, D2) = Pr(D2|H2) Pr(H2) = 0.96× 23 = −0.64

or J =

[J11 J12J21 J22

]= Fdiag

(p)

=

[0.982 0.040.018 0.96

]︸︷︷︸

F

.

[13 00 2

3

]

⇒ Hmut = 0.7326

16. Consider a binary Communication System that uses the following two equally probable energysignals:

0 7→ s0(t) = 2Λ

{t

10µs

}1 7→ s1(t) = −2Λ

{t

10µs

}The channel is assumed additive white Gaussian noise of double-sided power spectral densityPSDn(f) = 10−6W/Hz. Find:

(a) the bandwith B of the channel; 5%

(b) the channel symbol rate rcs (baud rate) & data bit rate; 5%

(c) the Energy Utilisation Effi ciency (EUE); 10%

(d) the channel capacity C in bits/sec. 10%

Solutions

(a) B = rcs2 = 1

2Tcs= 1

2×20×10−6 = 25kHz

(b) rcs = 1Tcs

= rb = 50kbits/ sec

(c) EUE=EbN0,

E1 =

10µ∫−10µ

4Λ2{

t10µ

}dt = 2

0∫−10µ

4(t+10µ10µ

)2dt

= 8(10µ)2

0∫−10µ

(t2 + 20µt+ (10µ)

2)dt = 1

380µ = E2 = Eb

i.e. Eb = 803 × 10−6 = 26.667× 10−6

N0

2 = 10−6 ⇒ N0 = 2× 10−6

Thus EUE=EbN0

= 26.667×10−62×10−6 = 13.334



(d) C = B log2(1 + SNRin) = B log2(1 + EUE

BUE

)=

25× 103 × log2

(1 + 13.334

25×10350×103

)= 25× 103× log2 (1 + 26.668)

⇒ C = 119.75× 103

17. Consider a binary Communication System that operates with a bit rate 100kbits/sec anduses the following two equally probable energy signals:

0 7→ s0(t) = 3

(Λ

{t

5µs

}+ rect

{t

10µs

})1 7→ s1(t) = −3

(Λ

{t

5µs

}+ rect

{t

10µs

})The channel is assumed additive white Gaussian noise of double-sided power spectral densityPSDn(f) = 0.5× 10−6W/Hz. Find:

(a) the bandwith B of the channel; 5%

(b) the channel symbol rate rcs (baud rate); 5%

(c) the Energy Utilisation Effi ciency (EUE); 20%

(d) the channel capacity C in bits/sec. 15%

Solution

(a) Tcs = 10µs⇒ B = 12Tcs

= 12×10µs = 105

2 = 50kHz

(b) rcs = 1Tcs

= 105 symbols/sec (=bit rate rb)

(c) N0 = 2× 0.5× 10−6 ⇒EUE= EbN0

= Eb10−6

However,

Eb =

5µ∫−5µ

9

(Λ

{t

5µ

}+ rect

{t

10µ

})2dt

= 2× 9

5µ∫0

(Λ

{t

5µ

}+ rect

{t

10µ

})2dt

note that Λ

{t

5µ

}|5µ0 is equal to

−t+ 5µ

5µ

and Λ

{t

5µ

}+ rect

{t

10µ

}= 1 +

−t+ 5µ

5µ= 2− 2× 105t

and(

Λ

{2t

Tcs

}+ rect

{t

Tcs

})2= 4 + 4× 1010t2 − 8× 105t

= 2× 9

5×10−6∫0

(4 + 4× 1010t2 − 8× 105t

)dt

= 18

5µ∫0

4dt+

5µ∫0

4× 1010t2dt−5µ∫0

8× 105tdt

= 18×

(20 +

4

3× 10−2 × 125− 4× 10−1 × 25

)× 10−6

= 210× 10−6

EUE= 210×10−610−6 = 210



(d) EUE=210; BUE= Brb

= 50×103105 = 0.5

C = 50× 103 log2(1 + 2100.5 ) = 435.88 kbits/s

18. Consider a binary digital communication system in which a binary sequence is transmitted

as a signal s(t) with a one being sent as 6Λ{

tTcs/2

}and a zero being sent as −6Λ

{t

Tcs/2

}.

The source at the input to the system provides a binary sequence of ones and zeros, withthe number of ones being twice the number of zeros. The transmitted signal is corruptedby channel noise n(t) of bandwidth B and has an amplitude probability density functiondescribed by the following expression:

pdfn(n) =1

6.rect

{n6

}Find a bound on the ratio C/B 20%

where C denotes the capacity of the channel in bits/s.

Solution

Answer

Ps = E{s(t)2

}=

+∞∫−∞

s2pdfs(s)ds =

0∫−6

s2 118ds+

6∫0

s2 19ds = 118s3

3

∣∣∣0−6

+ 19s3

3

∣∣∣60

= 4 + 8 = 12W

Since noise 6=white Gaussian⇒ B log2

(Ps+Nn

Nn

)≤ C ≤ B log2

(Ps+PnNn

)However, Pn =

+3∫−3

n2 16dn = 16n3

3

∣∣∣+3−3

= 118 (27 + 27) = 3W

Nn =entropy power= 12πe22H

where H = −+∞∫−∞

pdfn(n) log2 (pdfn(n)) dn = − 16 log216

+3∫−3

dn = − log216 = log2 6 = 2.5850

Thus, Nn = 12πe22×2.5850 = 36. 002

2πe = 2.1079

log2(12+2.10792.1079

)≤ C

B ≤ log2(12+32.1079

)2.7426≤ C

B ≤ 2.8311

19. Consider a binary digital communication system in which the transmitted signal is corruptedby channel noise of bandwidth B having an amplitude probability density function describedby the following expression:

pdfn(n) =1

6.rect

{n6

}If the power of the received signal is 12W then

(a) find the entropy power of the noise; 10%



(b) find an upper and a lower bound on the ratio C/B where C denotes the capacity of thecommunication channel. 10%

Solution

(a)

Pn = E{n(t)2} =

∫ ∞−∞

n.pdfn(n).dn

=

∫ 3

−3n.16.dn = 3

Nn = (entropy power) =1

2πe22H

where H = −∫ ∞−∞

pdfn(n) log2 (pdfn(n)) .dn

= −1

6log2

(1

6

)∫ 3

−3dn

= − log2

(1

6

)= log2(6) = 2.585

∴ Nn = (entropy power) =1

2πe22H

=36

2πe= 2.1078

(b)

B log2

(Ps +NnNn

)≤ C ≤ B log2

(Ps + PnNn

)bitssec

log2

(Ps +NnNn

)≤ C

B≤ log2

(Ps + PnNn

)log2

(12 + 2.1078

2.1078

)≤ C

B≤ log2

(12 + 3

2.1078

)2.7427 ≤ C

B≤ 2.8312


Estimate:



Solution



(a) pe = Pr(y2|x1).Pr(x1)︸︷︷︸Pr(y2,x1)

+ Pr(y1|x2).Pr(x2)︸︷︷︸Pr(y1,x2)

= 0.1× 0.25 + 0.2× 0.75 = 0.175

(b) F =

[0.9 0.20.1 0.8

]p =

[0.25 0.75

]TJ = F.diag(p) =

[0.9 0.20.1 0.8

]diag

([0.250.25

])=

[0.225 0.150.025 0.6

]q = Fp =

[0.9 0.20.1 0.8

] [0.250.25

]=

[0.3750.625

]HR = qT . log2

(q)

= 0.9544

HR|M = −1T2 (J� log2(F)) 12 = 0.658695

Hmut = HR −HR|M = 0.9544− 0658695 = 0.295705 bits/symbol


Estimate:



Solution

(a)

pe = Pr(y1, x2) + Pr(y2, x1)

= Pr(y1|x2).Pr(x2) + Pr(y2|x1).Pr(x1)

= 0.04× 2

3+ 0.018× 1

3= 0.032667

(b)

J =

[J11 = Pr(x1, y1) = 0.3267 J12 = Pr(x1, y2) = 0.006J21 = Pr(x2, y1) = 0.02667 J22 = Pr(x2, y2) = 0.64

]T

p =

[p1 = Pr(x1) = 1

3p2 = Pr(x2) = 2

3

],F =

[F11 = 0.982 F12 = 0.04F21 = 0.018 F22 = 0.96

]q = Fp =

[q1 = Pr(y1) = 0.982× 1

3 + 0.04× 23 = 0.354

q2 = Pr(y2) = 0.018× 13 + 0.96× 2

3 = 0.646

]



Hmut , Hmut(p,F) = (using any of the following expression)

= −M∑m=1

K∑k=1

Fkm.pm log2

(qkFkm

)

= −M∑m=1

K∑k=1

Jkm log2

(pm.qkJkm

)

= −1TK

J� log2

=q︷︸︸︷F.p .pT

∅J

︸︷︷︸K×M matrix

1Mbits

symbol

= 0.7327

22. A signal g(t) bandlimited to 4kHz is sampled at the Nyquist rate and is fed through a 2-levelquantizer. A Huffman encoder is used to encode triples of successive output quantizationlevels as follows:

symbols probs Huffmanm1m1m1 27/64 1m1m1m2 9/64 001m1m2m1 9/64 010m2m1m1 9/64 011m1m2m2 3/64 00000m2m1m2 3/64 00001m2m2m1 3/64 00010m2m2m2 1/64 00011

while the binary sequence at the output of the Huffman encoder is fed to a Binary on-offKeyed Communication System which employs the following two energy signals of durationTcs

s0 (t) = 0

s1 (t) =

√3

8Λ

(t

0.5Tcs

)

The transmitted signals are corrupted by additive white Gaussian channel noise having adouble-sided power spectral density of 10−3 W/Hz. The figure below shows a modelling ofthe whole system where the output of the Huffman encoder is modelled as the output of abinary discrete information source (X, p) with

X = {x1 = 1, x2 = 0},p = [Pr(x1),Pr(x2)]

T

while the binary on-off Keyed system is modelled as a discrete channel as shown below.



(a) Find the entropy of the information source (X, p), the information rate and the bit datarate (symbol rate) at the channel input. 15%

(b) Estimate the bit-error probability of the system. 10%

(c) Estimate the energy utilization effi ciency (EUE) and bandwidth utilization effi ciency(BUE) using the bit data rate as well as the information rate. 15%

(d) Represent the communication system, as a point on the (EUE,BUE) parameter plane.In this plane show also the locus of the system properly labelled. 10%

(e) Is the system a ‘realizable’communication system? 5%

(f) What is the signal-to-noise ratio, SNRin, at the receiver’s input? 5%

Solution



23. A signal g(t) having the pdf shown in Figure 1 is bandlimited to 4 kHz. The signal is sam-pled at the Nyquist rate and fed through a 2-level quantizer. The transfer function of thequantizer is shown in Figure 2.

Figure-1 Figure-2

A Huffman encoder is used to encode triples of successive output quantization levels whilethe binary sequence at the output of the Huffman encoder is fed to a Binary on-off KeyedCommunication System which employs the following two energy signals

s1(t) = 0; s2(t) = 0.5 cos

(2π

5

Tcst

);with 0 < t < Tcs

The whole system is modelled as follows

where the binary information source represents the system up to the output of the Huffmanencoder. The discrete channel models the binary on-off keyed Transmitter/Receiver (withx1 = 1 and x2 = 0) and the additive white Gaussian noisy channel with noise having adouble-sided power spectral density of 10−3 W/Hz.

(a) Estimate the bit-error probability of the system. 5%

(b) Find the information rate and the bit data rate (symbol rate) at the channel input. 10%

(c) Estimate the data point (EUE,BUE), where EUE denotes the energy utilization effi -ciency and BUE represents the bandwidth utilization effi ciency of the system. 10%

(d) Estimate the information point (EUE,BUE), where EUE denotes the information en-ergy utilization effi ciency and BUE represents the information bandwidth utilizationeffi ciency of the system. 15%

(e) Is the system a ’realizable’communication system? 5%

(f) What is the signal-to-noise ratio SNR, at the receiver’s input? 5%

Solution


a) Hz; Hz; J œ % ‚ "! J œ # ‚ J œ ) ‚ "! U œ #1 = 1$ $

From Figures 1 and 2 we get: VPrÐ # Ñ œ $Î% VPrÐ # Ñ œ "Î%

R œ # ‚ "!!$

symbols probabilities Huffman bits)

33355

6 ÐB B B #(Î'% " "B B B *Î'% !!"B B B *Î'% !"!B B B *Î'% !""B B B $Î'% !!!!!B B B $Î'% !!!!"B B B

3

" " "

" "

" "

" "

"

"

2

2

2

2 2

2 2

2 2 " $Î'% !!!"!B B B "Î'% !!!""

552 2 2

6 œ " ‚ #(Î'% $ ‚ *Î'% $ ‚ *Î'% $ ‚ *Î'% & ‚ $Î'% & ‚ $Î'% bits/ triple-level

or 3-samples)

& ‚ $Î'% & ‚ "Î'% œ #Þ%')(&

Ð

Alphabet: \ œ B œ "

B œ !ÐB Ñ ÐB ÑÑ"

#" #(since Pr Pr

with probabilities: : œ œ œ: ÐB Ñ !Þ'$%%: ÐB Ñ !Þ$'&' " "

# #

PrPr

Note:PrÐB Ñ ‚ ‚ ‚ ‚ ‚ ‚ ‚#

# * # * " * & $ % $ % $ $ "$ '% $ '% $ '% & '% & '% & '% & '%=

œ !Þ$'&'

: œ ÐC ß B Ñ ÐC ß B Ñ/ # " " #Pr Pr œ ÐC lB Ñ ÐB Ñ ÐC lB Ñ ÐB ÑPr Pr Pr Pr# " " " # #

œ !Þ!& ‚ !Þ'$%% !Þ# ‚ !Þ$'&' œ !Þ"!%)

b) HB 7 # 7 #7œ"

#Xœ : Þ : œ : Þ : œ !Þ*%($ log log bits

symbol

data rate: 6583 3 bits/sec< œ < œ J 6 œ Þdata , =

"$

information rate: H 6 bits/sec< œ < ‚ œ < ‚ œ #$'Þ%inf , ,B !Þ*%($

c) Q œ # i.e. binary CS Therefore: 1.5190 secX œ œ ‚ "!-=

"<

-=

4

1.I œ X ‚ œ #!%' ‚ "!, -=!Þ&#

#

PrÐB Ñ"5

EUE data EUEœ œ 'Þ!##) ‚ "! Ð ÑIR

$,

!

BUE (data BUE with denoting the baseband bandwidth)œ œ œ FF F "< F‚ Q #-= #2 log

data point (EUE,BUE)œ œ Ð'Þ!##) ‚ "! ß Ñ$ "#



d) CS inf.point (EUE BUE (data point)œ œ ß Ñ œ ‚380 380ÐQÑlog#

Hmut

Therefore we have to estimate the mutual information Hmut

H H H H H Hmut mutœ œ ] \] l\ \l]or i.e.

: œ œ ; œ Þ: œ!Þ'$%% !Þ*&ß !Þ# !Þ'(&)!Þ$'&' !Þ!&ß !Þ) !Þ$#%# … …

…œ Ð;Ñ Þ Þ Ð:Ñ œ!Þ)*")ß !Þ"!)#!Þ!*()ß !Þ*!##

diag diag" ‰ œ … Þ Ð:Ñ œ Ð;ÑÞ œ

!Þ'!#(ß !Þ!($"!Þ!$"( !Þ#*#&

diag diag

H\ œ : : œ : : œ7œ"

#

7 7X . 0.9473 log log# # bits

symbol

H] œ : : œ ; ; œ5œ"

#

5 5X . 0.9089 log log# # bits

symbol

H\‚ 57 577œ" 5œ" "‡

] œ N N œ 2 2bits

symbol . = 1.3929 log log# # ‰ ‰

H H] l\ ] l\œ Ð Ñ ´ N‰ 7œ"

# #

5œ"57

N: .log# 57

7

=0.4456 œ Þ :

‰ ‰

…

log diag( )#"

"*

bitssymbol

H H HÊ mut œ œ !Þ%'$$] ] l\

Therefore CS inf.point 1.0792œ œ Ð!Þ!"$ß Ñ

e) CS is not realizable ( since EUE 0.693)Ê œ !Þ!"$ 380

f) SNR =0.012 SNR = 19.2082dBin in

EUEBUEœ Ê




Wireless Channels

24. Find the minimum channel symbol rate needed by a digital communication system to resolvea multipath, with an additional path length of 30m compared to the direct path. 10%

Solution

Tspread = 303×108 = 10−7 sec = Tc,max

In this case L =⌊TspreadTc

⌋+ 1 = 1 + 1 = 2

chip rate= 1Tc

= 110−7 = 10Mchips/ sec

25. If BD = 8MHz denotes the Doppler spread, Bcoh represents the coherent bandwidth andTcs is the channel symbol period, then in a frequency selective fast fading channel which ofthe following is correct? 10%

(a) Tc = 61n sec and Bcoh = 3MHz.

(b) Tc = 61n sec and Bcoh = 100MHz.

(c) Tc = 244n sec and Bcoh = 3MHz.

(d) Tc = 244n sec and Bcoh = 100MHz.

(e) None of the above.

Solution

(a) Tc = 61ns; Bcoh = 3MHz Tc = 61ns < Tspread = 1Bcoh

= 13×106 = 333ns

and Tc = 61ns < Tcoh = 1BD

= 18×106 = 125ns

(b) Tc = 61ns; Bcoh = 100MHz Tc = 61ns > Tspread = 1Bcoh

= 1100×106 = 10ns

and Tc = 61ns < Tcoh = 1BD

= 18×106 = 125ns

(c) Tc = 244ns⇒ Bcoh = 3MHz Tc = 244ns < Tspread = 1Bcoh

= 13×106 = 333ns and

Tc = 244ns > Tcoh = 1BD

= 18×106 = 125ns

(d) Tc = 244ns ⇒ Bcoh = 100MHz Tc = 244ns > Tspread = 1Bcoh

= 1100×106 = 10ns

and Tc = 244ns > Tcoh = 1BD

= 18×106 = 125ns


Thus, the answer is (e)

26. The minimum chip rate needed by a DS-BPSK spread spectrum system to resolve a multi-path, with an additional path length of 30m compared to the direct path, is

(a) 10 Mchips/second

(b) 20 Mchips/second

(c) 40 Mchips/second

(d) 60 Mchips/second

(e) none of the above.

Solution

Tspread = 303×108 = 10−7 sec = Tc,max

In this case L =⌊TspreadTc

⌋+ 1 = 1 + 1 = 2

chip rate= 1Tc

= 110−7 = 10Mchips/ sec

That is, the answer is (a)



Digital Modulators & Line Codes

27. The next figure illustrates the signal constellation points of twoM -ary signals si(t) and sj(t)of equal energy.

The energy of each of these two signals is

(a) 25,

(b) 50,

(c) 75,

(d) 100,


Solution

(a) d2ij = Esi + Esj − 2ρij√EsiEsj with Esi = Esj = E

ρij = −1

102 = 2E + 2E = 4E ⇒ E = 25\That is, the correct answer is (a)

28. Consider a random binary sequence of 0’s and 1’s. This binary sequence is transmitted as arandom signal with 1’s and 0’s being sent using the pulses s1(t) and s0(t) described below:

0 7−→ s0(t) = 3rect{

t

1ms

}and

1 7−→ s1(t) = −3rect{

t

1ms

}If 1’s and 0’s are statistically independent with Pr(1) = Pr(0) = 0.5, find the Power SpectralDensity of the transmitted signal.

Solution

A = 3mV

Tcs = 1ms

= 9× 10−9 sinc2(f10−3

)Imperial College London Coursework 2011-12


29. A binary PSK signal is decoded coherently in the presence of white noise having a doublesided power spectral density 0.5×10-6 Watts/Hz. If Pr(1) = Pr(0) and the bit rate is 220kbits/sec, what is the average received signal power at which a probability of error of 10-5

can be achieved? (10%)

Solution

30. Consider a Biphase Shift-keyed digital modulator/demodulator operating the presence ofadditive white Gaussian noise with double-sided power spectral density 2×10−9W/Hz. Thedigital modulator maps zeros and ones as follows:

0 7−→ s0(t) = 3 cos(2πFct− 300)

1 7−→ s1(t) = 3 cos(2πFct+ 300)

for 0 ≤ t ≤ Tcs

where Pr(0) = Pr(1), Tcs = 4ns and Fc = 5Tcs. Find

(a) the Energy Utilisation Effi ciency (EUE); [5 marks]

(b) the bit error rate pe at the demodulator’s output. [5 marks]

Solution

(a) EUE=EbN0

=32

2 4×10−9

10−9 = 18

(b) Initially you have to prove that pe =T{√

2EUE sin2(300)

}pe =T

{√2× 18 sin2(300)

}=T{√

2× 18× 14

}=T{3} ' 10−3

31. The following HDB3 encoded signal

represents the binary sequence:



(a) 1 0 0 1 0 0 0 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 1

(b) 1 0 0 1 0 0 0 1 1 1 1 0 0 1 1 0 0 1 0 0 1 1 1

(c) 1 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1

(d) 1 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 1 1


Solution

The correct answer is (c)



SSS: PN-Codes, Direct Sequence and Frequency Hopping

32. A pseudo random (PN) signal b(t) is generated by using a maximal length shift register ofm-stages and has the following double-sided Power Spectral Density.

PSDb(f) :

(a) Find the number m of shift register stages. 5%

(b) Find ∆F . 5%

Solution

(a) Nc = 31⇒ Nc = 2m − 1

⇒ m = log2(Nc + 1)

⇒ m = log2(31 + 1) =5

(b) Tc = 10−8

∆f = 1NcTc

= 131×10−8 =3.2258× 106

33. Sketch the feedback shift register whose feedback connections are represented by the primitivepolynomial x24 + x7 + x2 + x+ 1 and find the length N of this sequence. [6 marks]

If the clock rate is 2.7 chips/s, find the period of this sequence in minutes. [4 marks]

Solution

N = 2m − 1

m = 24

Thus N = 224 − 1 = 16777215

Tc = 12.7

schip = 0.37037 s

chip

NTc =(224 − 1

)× 0.37037 = 6.21377712× 106s

= 6.21377712×10660 min = 1.0356× 105 min



34. Sketch the feedback shift register whose feedback connections are represented by the primitivepolynomial D24 +D7 +D2 +D + 1 and operates with a clock rate 1Mb/sec. 5%

Find the period of the output sequence in minutes. 15%

Solution

N = 2m − 1m = 24

}⇒ N = 224 − 1 == 16.77 7× 106(16777215)

Tc = 1106 s/bit = 10−6s/bit

NTc = (224 − 1)10−6/60 = 0.27962minutes

35. A ‘short-code’BPSK DS/SSS uses an m-sequence and a data rate 9.6 kbits/sec. If it isrequired that the spread spectrum signal will have bandwidth no larger than 25MHz, whatis the largest period of the m-sequence that can be used?

(a) 255

(b) 511

(c) 1023

(d) 2047

(e) None of the above

Solution

rb = 9.6 kbits⇒ Tcs = 19.6×103

PG = N = TcsTc

Bss ≤ 25 MHz⇒ 1Tc≤ 25× 106 ⇒ Tc ≥ 1

25×106

⇒ TcsN ≥

125×106 ⇒ N ≤ 25× 106Tcs

⇒ N ≤ 25×1069.6×103 = 2604. 2

However, N = 2m − 1⇒ N = 211 − 1 = 2047 ≤ 2604.2

That is, the correct answer is (d)

36. Consider a binary message signal of rate 8 kbits/s at the input of a fully synchronizedBPSK direct sequence spread spectrum system (DS/SSS-BPSK). The system operates in thepresence of both additive white noise, n(t), and a broadband noise jammer, j(t), of power1 Watt. The double sided power spectral density of the noise is 10-12 Watts/Hz and theprocessing gain of the system is 105. The bit error probability at the output of the receiveris equal to 4×10−6 while the protection probability is equal to 4× 10−2.

(a) What is the amplitude A of the sinewaves which are used by the binary PSK modulator? 15%

(b) What is the bit error probability if the jammer switches to a "pulse jammer" mode,which is “on”for 40% and “off”for 60% of the time? 10%



(c) What is the Anti-jam Margin, in dBs, when the jammer switches to the above-mentionedmode? 10%

Solution

(a) rcs = 8× 103bits/ secN0

2 = 10−12 ⇒ N0 = 2× 10−12

Pj = 1W ;PG=105; pe = 4× 10−6; pe,PR = 4× 10−2

PG=TcsTc⇒ Bss = 1

Tc=PG× rcs︸︷︷︸

= 1Tcs

= 105 × 8× 103 = 800× 106Hz

Baseline Performance: Bj = Bss

⇒ pe =T{√(1− ρ)EUEequ

}; ρ = −1

⇒ 4× 10−6 =T{√2EUEequ

}

⇒√

2EUEequ = 4.5⇒EUEequ = 4.52

2 = 10.125

⇒ Eb

N0+PjBss

= 10.125⇒ Eb = 10.125×(N0 +

PjBss

)= 10.125×

(2× 10−12 + 1

800×106

)=

12.677× 10−9

Ps = EbTcs

= Ebrcs = 12.677× 10−9 × 8× 103 = 101.42× 10−6

A =√

2Ps =√

2× 101.42× 10−6 ⇒A = 14.242mV

(b) pulse jammer with q = 0.4 "on"

pe = (1− q)T{√

2EbN0

}︸︷︷︸

jammer=”off”

+ qT{√

2Eb

N0 + PJqBss

}︸︷︷︸

jammer=”on”

pe = 0 + 0.4×T{√2× 4.05

}= 0.4×T{2.8474} = 0.4× 2.2× 10−3

i.e. pe = 8.8× 10−4

(c) pe,PR = qT{√2qEUEPR

}⇒ 4× 10−2 = 0.4T{√

2× 0.4× EUEPR}

⇒ 10−1 =T{√0.8× EUEPR

}⇒√

0.8× EUEPR = 1.25⇒EUEPR = 1.252

0.8 = 1.9531

AJM= 10 log10EUEequ − 10 log10EUEpr =7.1467

(or AJM = 10 log10 (10.125)− 10 log10 (0.7813) = 11.126)

37. A speech signal having a maximum frequency of 4kHz is sampled at twice the Nyquist rateand then fed through an 8-bit uniform quantizer. The generated binary sequence is then fedthrough a binary PSK direct sequence spread spectrum system which operates in the presenceof a broadband jammer of power 1.6 Watts and in the presence of additive white Gaussiannoise with double-sided power spectral density 0.5× 10−12 Watts/Hz. The amplitude of theBPSK signal is 0.5V.

For this system, the spread spectrum bandwidth Bss is 32 MHz and the system is fullysynchronised.

Find:



(a) the power of the code noise, 5%

(b) the power of the noise at the output of the correlator, 5%

(c) the power of the jammer at the output of the correlator. 10%

Solution

(a) fully synchronised system ⇒code noise = zero(b) Fs = 2× 2× 4k = 16kHz ⇒bit-rate= rb = rcs = 8× 16k = 128k

⇒ Tcs = 1rcs

= 1128×103 = 7. 812 5× 10−6

N0 = 2× 0.5× 10−12 = 10−12

Pn,out = N0

Tcs= 10−12

7. 812 5×10−6 =1.28× 10−7

(c) Tc = 1Bss

= 132×106 = 3.125× 10−8 sec

PG= TcsTc

= 7. 812 5×10−63.125×10−8 = 250

Pj,out =PjPG = 1.6

25 =0.0064

38. Two m-sequence PN-signals, generated by two 3-stage shift registers, are shown below.

Construct a Gold code signal from these two PN-signals. 10%

Solution

(or, other valid codes - for various delays)

39. Consider a feedback shift register whose feedback connections are represented by the primitivepolynomial D4+D1+ 1. Give one period of its output sequence - starting with all 1’s (initialcondition). 15%

Solution

1 1 1 1

0 1 1 1

0 0 1 1

0 0 0 1

1 0 0 0

0 1 0 0

0 0 1 0

1 0 0 1

1 1 0 0

0 1 1 0

1 0 1 1



0 1 0 1

1 0 1 0

1 1 0 1

1 1 1 0

1 1 1 1

o/p = 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0

40. A DS/SSS uses an m-sequence for spreading the spectrum with a processing gain equal toone period of the m-sequence. If the data rate is 28 kbits/sec and it is required that thespread-spectrum signal has a bandwidth no larger than 25 MHz, what is the largest periodof the m-sequence that can be used? 15%

Solution

rb = 28kbits/ sec⇒ Tcs = 128×103

Note: PG=N = TcsTc

Bss 5 25MHz ⇒ 1Tc5 25× 106 ⇒ Tc = 1

25×106 = 40× 10−9 sec

⇒ TcsN = 40× 10−9 sec⇒ N 5 25× 106Tcs ⇒ N 5 25×106

28×103 = 892

However,N = 2m − 1 5 892 (1)

This implies that m = 9⇒ N = 29 − 1 = 511 that satisfies Equation 1 given above.



41. A pseudo random (PN) signal b(t) is generated by using a maximal length shift register ofm-stages and has the following double-sided Power Spectral Density.

PSDb(f) :

f

Find the number m of shift register stages. 15%

Solution

1Tc

= 10M ⇒ Tc = 10−7 sec1

NTc= 39.2k ⇒ NTc = 25.5µ sec

}⇒ N = 255

N = 2m − 1⇒ m = log2 (N + 1) = 8

42. An analogue message signal having a maximum frequency of 4kHz is sampled at the Nyquistrate and then is fed through a 4-level quantizer where each level is encoded using 2 bitcodewords. The binary sequence is then fed through a fully synchronized Binary PSK DirectSequence Spread Spectrum System (BPSK/DS-SSS) of processing gain 108. The systemoperates in the presence of white Gaussian noise having a double-sided power spectral densityof 10−12 W/Hz and its Energy Utilization Effi ciency is 40 (i.e. EUE= Eb

N0=40). What would

be the power PJ of a jammer which, if it was distributed over 50% of the spread spectrumsignal bandwidth, would provide a bit error probability pe of 3× 10−5 ? 25%

Solution



N.B.: EUEequ = 8;Tc = 116 × 10−11;Tcs = 1

16 × 10−3



PCM & PSTN

43. For a speech signal of 4 kHz bandwidth transmitted using a uniform quantiser of 256 levelsthe bit rate at the output of the source encoder is

(a) 8 kbits/s

(b) 16 kbits/s

(c) 32 kbits/s

(d) 64 kbits/s

(e) 128 kbits/s

Solution

Q = 2γ ⇒ rb = 2Fgγ = 4× 103 × 2× log2(256) = 64k

That is, the correct answer is (d)

44. Consider the mse-differential quantizer shown in the following figure

A B C

D

E

which employs a 6-level non-uniform quantizer, having the following input and output levels

I/P (volts) O/P (volts)+8 ≤input≤+255 +23+4≤input≤+7 +60≤input≤+3 +1−3 ≤input≤ −1 −1−7 ≤input≤ −4 −6−255 ≤input≤ −8 −23

If the input is a step signal of amplitude 0V→ 41V, and assuming “E”=0 as an initial value,then find the data sequence that is read from point ’D’

Solution

point ’D’data sequence: 23V, 46V, 40V, 41V, 42V

45. An analogue message signal g(t) with amplitude probablity density function 0.5Λ(g2

)and

a bandwidth of 10kHz, is applied to a 256-levels uniform PCM system (i.e. PCM systemwhich employs a uniform quantizer of 256 levels). For this system calculate the Signal-to-Quantization-Noise ratio (SNRq).

Solution

PCM using a 256-level uniform quantizer:

Q = 256⇒ 2γ = 256⇒ γ = 8 bitslevel

CF=crest factor of the signal= peakrms

peak= g = 2V



rms= σg =√Pg ⇒ Pg = 2

∫ 20g2 pdfg(g)dg = 2

∫ 20g2 1

2Λ(g2

)dg

= 2∫ 20g2 1

22−g2 dg =

∫ 20

(g2 − 1

2g3)dg

=(g3

3 −12g4

4

)]20 = 2

3

i.e. rms=√

23

SNRq = 4.77 + 6× γ − 20 log102√23︸︷︷︸

20 log10(√6)=7.7815

= 4.77 + 6× 8− 7.7815 = 44.989dB

46. A high quality music signal with a Crest Factor of 4.4668, having a maximum frequency18 kHz, is applied to a uniform PCM system (i.e. PCM with a uniform quantizer). If itis specified that the Signal-to-Quantisation-Noise ratio (SNRq) should be better than 50dB,find the minimum data bit rate required [6 marks]

Solution

SNRq >50 dB;CR=ˆVσ = 4.4668

SNRq = 4.77 + 6γ − 20 log10

ˆ

V

σ︸︷︷︸=13dB

dB

SNRq = 4.77 + 6γ − 13

i.e.

⇒ 4.77 + 6γ − 13 > 50

⇒ γ > 50 + 13− 4.77

6

⇒ γ > 10⇒ log2Q > 10⇒ Q > 210 = 1024

Therefore,rb = γFs = γ2Fg = 10× 2× 18k = 360kbits/ sec

47. In a binary PCM communication system prove that bandwidth expansion factor β is equalto the average number of bits γ per quantization level, i.e. [4 marks]

β = γ

Solution

B= channel symbol rate2 = bit rate

2 = γFs2 = γFg ⇒ B = γFg ⇒ B

Fg= γ ⇒ β = γ

48. Consider a PCM system where its quantizer consists of a µ-law compander (with µ = 100)followed by a uniform quantizer with “end points”bi, and “output levels”mi. The maximumvalue of the input signal is 10V olts and the input/output characteristics of the uniformquantizer are given in the following tables:

b0 b1 b2 b3 b4 b5 b6 b7- 1 0V -8 .7 5V -7 .5 V -6 .2 5V -5V -3 .7 5V -2 .5 V -1 .2 5V

b8 b9 b1 0 b1 1 b1 2 b1 3 b1 4 b1 5 b1 60V 1 .2 5V 2 .5V 3 .7 5V 5V 6 .2 5V 7 .5V 8 .7 5V 1 0V

m1 m2 m3 m4 m5 m6 m7 m8

- 9 .3 7 5V -8 .1 2 5V -6 .8 7 5V -5 .6 2 5V -4 .3 7V -3 .1 2 5V -1 .8 7 5V -0 .6 2 5V



m9 m1 0 m1 1 m1 2 m1 3 m1 4 m1 5 m1 6

0 .6 2 5V 1 .8 7 5V 3 .1 2 5V 4 .3 7 5V 5 .6 2 5V 6 .8 7 5V 8 .1 2 5V 9 .3 7 5V

Note that µ-law compression is defined as follows:

output= ln(1+µ. |x|)ln(1+µ) .sign(x) where x = input value in V olts

maximum input value in V olts

(a) If the signal at the output of the sampler at time kTs is equal to 2.4V olts, what is thecorresponding output level of the uniform quanitizer? [5 marks]

(b) Find the corresponding value of the signal at the output of the expander. [5 marks]

(c) Estimate the instantaneous quantisation noise nq(kTs) [1 marks]

Note: Ts is the sampling period and k is an integer.

Solution

(a) ggmax

= 2.410 = 0.24

gc = ln(1+100×0.24)ln(1+100) × gmax = 3.218

4.615 × 10 = 6.974

⇒ b13 < gc < b14 ⇒ gc = 6.875V = m14

(b) gq,out = 1µ

(exp

(m14

gmaz× ln (1 + µ)

)− 1)× 10 = 2.287

(c) nq = 2.4− 2.28 = 0.12V (or − 0.12V )

49. The CCITT standards 32kbits/second Differential PCM are

(a) for speech signals with bandwidth 3.2 kHz.

(b) for audio signals with bandwidth 7 kHz

(c) specifying a sampling frequency 16 ksamples/second

(d) specifying an 8 levels quantizer

(e) none of the above

Solution

The correct answer is (a)

50. The first TDMA multiplexing level of a 30-channel PCM Telephone system uses

(a) an AMI line code;

(b) a polar RZ line code;

(c) a Manchester line code;

(d) an HDB3 line code;


Solution

The correct answer is (d)

END


Problem Sheets: Communication Systems

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