CS3282 Exam Summer 2006 1 25/08/2006 BMGC CS3282 Digital Communications Examination June’06 Electronic calculators ALLOWED. Answer THREE questions in TWO hours. Attached: Appendix 1: Graph of complementary error function Q(z) against z. Appendix 2: Fourier transform properties Appendix 3: Trigonometric formulae ------- ------- ------ ------ ----- ----- ---- ----- ---- ---- ---- --- 1(a) What causes ‘fading’ in a radio channel and how does it affect digital transmissions? Explain the difference between 'flat' fading and 'frequency-selective' fading and state what is meant by the ‘coherence bandwidth’ of a channel. [5 marks] (b) Compared to analogue techniques, what do you consider to be the three main advantages of digital voice transmission in wired and wireless telephony? [3 marks] (c) Explain the difference between ‘waveform coding’ and ‘parametric coding’ for speech in telephony. What features of speech and its perception are exploited by the G711 64 kb/s standard coder for wired telephony to achieve acceptable speech quality at the required bit-rate.? [7 marks] (d) Speech is digitised at 64 kb/s. How could this bit-stream be efficiently transmitted over a channel of 48 kHz bandwidth centred on 100 kHz? According to the Shannon-Hartley Law, what signal-to-noise ratio would be required to ensure that arbitrarily low bit-error rates are achievable for this transmission? [5 marks] 2 (a)Define the energy and power of an analogue signal x(t). State what is meant by (i) an energy signal and (ii) a power signal. Define the ‘1-sided energy spectral density’ of an energy signal and the ‘1-sided power spectral density’ of a power signal. [5 marks] (b) If an energy signal with ‘1-sided energy spectral density’ E(f) is passed through a filter with frequency response H(f), what is the energy of the resulting signal? [2 marks] (c) If a power signal with ‘1-sided power spectral density’ P(f) is passed through a filter with frequency response H(f), what is the power of the resulting signal? [2 marks] (d) A single symbol equal to s(t) for logic ‘1’ or zero for all t for ‘0’ is received with additive white Gaussian noise of ‘1-sided power spectral density’ N0 watts/Hz. The energy signal s(t) has Fourier transform S(f). To distinguish s(t) from zero, the receiver employs a matched filter with frequency-response

H(f) = S*(f)e-2πjfT and passes its output to a threshold detector. Show that the voltage at the input to the threshold detector at time t = T is:

∫∞

∞

+=-

20 |)(| = E ere wh

logic'0'for : 0logic'1'for :

)( dttsE

nTz

and n0 is a Gaussian random variable of zero mean and variance EN0/2. [8 marks] Assuming that a ‘1’ or a ‘0’ is equally likely, show that with an appropriate choice of threshold, the probability of making a correct decision is 1 − PB where:

CS3282 Exam Summer 2006 2 25/08/2006 BMGC

=

0B N2

EQP

and Q(z) is the ‘complementary error function’ as plotted in the graph attached. [3 marks] 3 (a) What is the purpose of the ‘physical layer’ in a digital communication system? Draw a block-diagram of a ‘single carrier’ digital transmitter and corresponding receiver for the physical layer transmission of a bit-stream over a wired or wireless channel. As used in this block diagram, state the purpose of

(i) a transmitter pulse shaping filter (ii) a matched filter and (iii) an adaptive equaliser

Explain why a Nyquist frequency-response is required between appropriate points in the block-diagram and indicate where these points occur. `[7 marks] (b) Using the Fourier transform properties in Appendix B, show that inter-symbol interference (ISI) is eliminated by shaping each symbol, s(t), transmitted such that, when it is received at the threshold decision circuit, its Fourier Transform S(f) satisfies: S(f) = 0 for |f| > 1/T S(f) + S(f – 1/T) = Ts(0) for 0 ≤ f ≤ 1/T Sketch the general shape of an “100r% raised cosine” spectrum which satisfies this property, and explain why ‘root raised cosine (RRC)’ spectrally shaped symbols are generated by the transmitter. [7 marks] (c) An appropriately shaped symbol with zero-crossings at t = ±T, ±2T, ±3T, etc. relative to the centre is distorted by the frequency-response of a wired channel. It is estimated, by averaging over several training symbols, that when the received symbol, x(t) say, is normalised to 1 volt at its centre (assumed to occur at t=0), its voltages at t = ±T and ±2T relative to the centre would be as follows in the absence of channel noise:

x(−2T) = 0.2, x(−T) = −0.3, x(0) = 1.0, x(T) = 0.4, x(2T) = −0.1 Inter-symbol interference is to be reduced by a 3-term ‘zero-forcing’ transversal equaliser. Explain how this may be achieved and give a diagram of the equaliser. Given that the zero-forcing equaliser has coefficients C0, C1, C2, expressed as a column vector c, show that the required zero-crossings are achieved if Ac = b , where

−−

−=

14.01.03.014.0

2.03.01A

and b is a suitably chosen column-vector. Given that the inverse of matrix A is approximately as follows:

−−

−=−

87.029.020.030.080.029.086.030.087.0

1A

calculate the coefficients C0, C1, and C2. [6 marks]

CS3282 Exam Summer 2006 3 25/08/2006 BMGC 4. .(a) Explain the term asynchronous as applied to digital transmission. Describe one commonly used asynchronous technique and its main applications. [6 marks] (b) What are the main requirements for PCM wave-forms used for the synchronous transmission of

digital signals over wired links in telephone networks. Show how the stream of eight bits: 1 1 0 0 0 0 0 1 would be transmitted at base-band using (i) NRZ-HDB3 coding and (ii) Manchester coding. What are the advantages and disadvantages of Manchester coding as compared to NRZ-HDB3? [7 marks] (c) What is a regenerative repeater and why are such repeaters used in wired telephony? A synchronous transmission scheme uses cable with 20 dB attenuation per km and identical regenerative repeaters every 2 km. Each repeater is affected by the same level of additive white Gaussian noise and has been set to transmit at the same power level. It is observed that each repeater increases the bit-error rate by 1 bit in 1000. Calculate the increase in power level that would be required at the output of each repeater to reduce the increase in bit-error-rate per repeater to 1 bit in 109 . If the output power of the repeaters could not be increased for practical reasons, how must the repeater spacing be modified to achieve the same error-rate reduction at each repeater ? If the total transmission distance is 34 km, how will the overall bit-error-rate be affected by the change in repeater spacing. [7 marks] 5.(a) What are the advantages of orthogonal frequency division multiplexing (OFDM) for digital transmission over channels subject to fading, and what is the main disadvantage of OFDM? [8 marks] (b) State what is meant by the ‘bandwidth-efficiency’ of a digital communication system. What is the maximum-bandwidth efficiency achievable, in theory, by a base-band binary transmission system? Why would this maximum be difficult to achieve in practice? State the bandwidth efficiency of (i) Binary PSK with 100% RC pulse shaping (ii) QPSK with 50% RRC pulse-shaping (iii) 8-PSK with 50% RRC pulse-shaping (iv) OFDM at 250 kbaud with 64 sub-carriers spanning a 20 MHz bandwidth. Assume that only 48 of the 64 sub-carriers are modulated (according to the IEEE802.11a standard) and that QPSK is used for each sub-carrier. [9 marks] (c) Give a constellation diagram for the 8-PSK scheme. Label each symbol and explain why a form of ‘Gray’ coding is used. [3 marks]

CS3282 Exam Summer 2006 4 25/08/2006 BMGC

Graph of Complementary Error function, ∫∞

−=

zdttzQ

2exp

21)(

2

π

CS3282 Exam Summer 2006 5 25/08/2006 BMGC

Appendix 2: Properties of the Fourier Transform (1/2/03) Property Signal x(t) Fourier Transform X(f) Transform & inverse: x(t) = X(f) = ∫

∞

∞−dfefX jftπ2)( ∫

∞

∞−

− dtetx jftπ2)(

Similarly let the Fourier transforms of y(t), y1(t) & y2(t) be Y(f), Y1(f) & Y2(f) respectively. Rect & sinc rectA(t) = rect(t/A) Asinc1/A(f)=A sinc(Af) sincA(t) = sinc(t/A) Arect1/A(f) = A rect(Af) Delay y(t-τ) e-2πjτY(f) Frequency shift e2πjFt y(t) Y(f-F) Amplitude scaling: Ay(t) AY(f) Time-reversal: y(-t) Y(-f) Superposition: Ay1(t)+By2(t) AY1(f)+BY2(f) Constant: A Aδ(f) Impulse: Aδ(t) A Gaussian ( π /α) exp(−π2 t2 / α2) exp(−α2 f 2 ) Time-scaling: y(At) (1 / |A|) Y(f/A) Differentiation: dmy(t)/dtm (2πjf)m Y(f) Product: y1(t)y2(t) Y1(f) ⊗ Y2(f) = Y Y f d1 2( ) ( )θ θ−

−∞

∞

∫ θ

τ

τ

τ

Convolution: Yy y t d1 2( ) ( )τ τ−−∞

∞

∫ 1(f) Y2(f)

Cross-correlation: Yy y t d1 2( ) ( )τ τ+−∞

∞

∫ 1(f) Y2*(f)

Auto-correlation: |Y(f)|y y t d( ) ( )τ τ+−∞

∞

∫ 2

Repeat: repeatPy(t) (1/P)sample1/PY(f) Sample: sampleTy(t) (1/T)repeat1/TY(f) Properties for real signals:- For all real signals: x*(t)=x(t) X(−f) = X*(f) i.e. |X(−f)| = |X(f)| & φ(−f) = −φ(f) Real and even: x(t) = x(−t) X(f) is purely real & X(−f) = X(f) Real and odd: x(t) = −x(−t) X(f) is purely imaginary & X(−f) = −X(f) Formulae:-

sinc

rect( ) =1 :| |< 0.50.5:| |= 0.5

sinc ( ) sinc( / A) rect ( ) rect( / AA A( )

sin( )( ) :

: :| | .)x

xx x

xx

xxx

x x x x= ≠=

>

= =π

π 01 0 0 05

repeat sample repeatT TPn n

x t x t nP x t x nT t nT x t t ( ) ( ) ( ) ( ) ( ) ( ) ( )= − = − ==−∞

∞

=−∞

∞

∑ ∑ δ δ

Fourier series for repeatPx(t):-

( )C e M k P t A A k P t B k P tkjkt P

k k k kkkk

20

102 2π π θ π π/ cos(( / ) ) cos(( / ) ) sin(( / ) )= + = + +

=

∞

=

∞

=−∞

∞

∑∑∑ 2

k k k

C P X k P M C M C C A C A C B Ck k k k k k= = = = = = =( / ) ( / ) ; | |; | |; arg( ); ; Re ; Im( 1 2 20 0 0 0 2θ

CS3282 Exam Summer 2006 6 25/08/2006 BMGC

Appendix 3: Trigonometric formulae sin(A ± B) = sin(A) cos(B) ± cos(A)sin(B) cos(A ± B) = cos(A) cos(B) sin(A)sin(B) mtan(A ± B) = ( tan(A) ± tan(B) ) / (1 m tan(A)tan(B) ) sin(2A) = 2 sin(A) cos(A) cos(2A) = 2 cos2(A) – 1 = 1 – 2 sin2(A) tan (2A) = 2 tan(A) / (1 – tan2(A) ) 2cos(A) cos(B) = cos(A + B) + cos(A - B) 2 sin(A) cos(B) = sin(A + B) + sin(A - B) 2 sin(A) sin(B) = cos(A - B) – cos(A + B) cos(A) + cos(B) = 2cos( (A + B)/2 ) cos( (A – B)/2 ) sin(A) + cos(B) = sin(A) + sin(B+π/2) sin(A) + sin(B) = 2sin( (A + B)/2 ) cos( (A - B)/2 ) cos(θ) = (ejθ + e-jθ) / 2 sin(θ) = (ejθ - e-jθ) / (2j) λ cos(θ) + µ sin(θ) = R cos (θ + φ) where R2 = λ2 + µ2 and φ = tan-1(µ/λ) + π if λ<0

CS3282 Exam Summer 2006 7 25/08/2006 BMGC

Solution methods Question 1(a):

Fading is due to the effect of multi-path propagation since the radio transmission from a given base-station or a given mobile user will be reflected by buildings and thus reach its intended target by a multitude of different routes. In fact, direct "line-of-sight" paths between base-stations and mobiles are rare in large cities with tall buildings. [1] The different routes introduce different phase shifts, and so when the reflected signals reach their target some of them will add in phase and reinforce each other, and some will be out of phase and can cancel each other out. The cancellation is referred to as "fading", and it will be frequency dependent, i.e. the same phase differences will cause reinforcement at some frequencies and cancellation at other frequencies. [1] If we restrict transmissions to a relatively narrow band of frequencies, the gain and phase differences across this narrow band may not be too serious, and the main problem will be "flat fading", i.e. the gain and phase-delay across the whole band being affected more or less equally due to cancellation. [1] However, if we wish to increase the bit-rate of our transmissions by using a wider radio-frequency band, "frequency selective" fading will become a problem where some frequencies within the wider band will be severely attenuated while others are enhanced. [1] The "coherence bandwidth", BC Hz say, of the radio channel is the largest bandwidth we can use without having to worry about frequency selective fading. For a channel whose bandwidth is less than BC Hz, any fading may be considered "flat fading". [1] (b) Advantages of digital voice networks in telephony (Choose three): 1. Ease of multiplexing: - TDM (time-division multiplexing) equipment is simpler and less expensive

than FDM (frequency-division multiplexing) equipment even when the cost of digitising speech is taken into account.

2. Ease of signalling: - Control information (e.g. on/off hook, dialling, coin deposits, charging) is inherently digital and can therefore be transmitted in exactly the same way as digitised speech.

3. Use of modern technology: - May be applied to switching, multiplexing and signal processing. Digital I/Cs are easier to manufacture than analogue components especially as the latter are usually required to be linear.

4. Integration of transmission and switching: - The multiplexing operations of a transmission system may be easily integrated into the switching equipment.

5. Operability at low signal-to-noise-ratios: - Analogue noise and interference is most noticeable during speech pauses or when the amplitude is low. Noise in digital systems is mostly quantisation noise produced at the A-to-D converter and since companding and adaptive gain control (mirrored at the receiver) may be used to encode low level signals with the same signal to quantisation noise ratio as high level signals (approx.), we do not need an excessively high ratio for high level signals just to ensure that the ratio for low level signals is acceptable. Also, crosstalk in analogue systems is particularly annoying as it tends to be intelligible. Even if crosstalk in digital systems is bad enough to cause bit-errors, the resulting noise will be unintelligible and therefore not as disturbing.

6. Signal regeneration possible: - Distortion in analogue transmission systems cannot be corrected. In a digital system, the probability of transmission errors can be made arbitrarily small by inserting regenerative repeaters, which reconstruct the original pulse shape at intermediate points in the transmission link.

7. Accommodation of other services: - A transmission link can be totally indifferent to the nature of the traffic it carries - it is just a bit-stream. Hence data, video, music etc., can be accommodated in a totally integrated system.

CS3282 Exam Summer 2006 8 25/08/2006 BMGC 8. Performance monitoring: - Possible, for example by recording parity errors. 9. Ease of encryption: - Scrambling and unscrambling of the transmitted bit stream for security is much

easier. [3]

(c) Waveform coding, parametric coding & G711 Waveform coding techniques such as PCM, and differential PCM try to preserve the exact shape of the speech waveform as far as possible with the allowed bit-rate. They are relatively simple to understand and to implement, but cannot achieve very low bit-rates. [2] Parametric coding techniques such as ‘linear predictive coding’ (LPC) do not aim to preserve the exact wave-shape, but instead represents features of the speech signal which are expected to be perceptually significant by sets of parameters such as predictor coefficients which characterise the short term spectral envelope. Parametric coding is considerably more complicated to understand and implement than waveform coding, but can achieve much lower bit-rates. [2] G711 (64 kb/s log-pcm) is essentially a “waveform coding” approach, but it also exploits the nature of sound perception by humans. It relies mainly on one characteristic of speech waveforms and two properties of sound perception by humans:

(i). In speech waveforms, lower amplitude sample values are more common than higher ones and these are quantised more accurately than is possible with 8-bit uniform quantisation. This reduces the average quantisation noise power. [1] (ii). G711 exploits the fact that speech may be band-limited to the frequency range 300 Hz to 3.4kHz without loss of intelligibility. (The non-sensitivity of mono hearing to phase is also exploited to a small degree since the band-limiting filters introduce phase distortion). [1] (iii). Since ‘A-law’ companding (G711) quantises the lower level samples more accurately than the higher ones, this tends to lower the quantisation noise when the signal is quiet, and allows it to increase in amplitude when the speech gets louder. Hence it exploits perception by relying on quantisation noise being masked by a higher energy signal. [1]

(d) Calculation Speech is digitised at 64 kb/s. How could this bit-stream be efficiently transmitted over a channel of 48 kHz bandwidth centred on 100 kHz? According to the Shannon-Hartley Law, what signal-to-noise ratio would be required to ensure that arbitrarily low bit-error rates are achievable for this transmission. Required bandwidth efficiency is 64/48 = 4/3. Greater than one so cannot be achieved with BPSK. Can use QPSK requiring minimum of 32kHz bandwith. Excess bandwidth available is 16 kHz. Use this for RC pulse shaping. (1+r) * 32 = 48 Therefore 1+r = 1.5 and r = .5 Use QPSK with 50% RC pulse shaping. [2] By S-H law Capacity = 48000 log2(1 + S/R) must be greater than 64 kb/s [1] Therefore 48 log2(1+S/R) > 64 log2(1 + S/R) > 4/3 1 + S/R > 21.33333 = 2.52 Therefore S/R > 1.52 10 log10(1.52) = 1.8 dB Therefore signal to noise ratio must be greater than 1.8 dB. [2]

CS3282 Exam Summer 2006 9 25/08/2006 BMGC Question 2 (a)

∫∞

∞-

2|x(t)| = x(t)ofEnergy dt

(Total energy, in Joules, that would be delivered to a 1 Ohm resistor if a voltage of value x(t) were applied to it for all time). [1] Let xD(t) be defined as follows for any constant D:

≤≤−

=otherwise

DtDtxtxD : 0

2/2/:)()(

then,

dt (t)][xD1 lim = x(t)ofPower

-

2DD ∫

∞

∞∞→

(Average power in Watts that would be dissipated (probably as heat) by a 1 Ω resistor when a voltage x(t) is applied for all time.) [1] ‘Energy-signal’ has finite energy and zero average power. ‘Power-signal’: has finite average power and therefore infinite energy; e.g. a sine wave .. [1] |X(ƒ)|2 is the 1-sided “energy spectral density” (ESD) of the energy signal x(t) whose FT is X(f). Its units are joules per Hz. [1] For a power signal x(t), its 1-sided power spectral density (PSD) is

2)(1lim)( fXD

fP DD ∞→=

where XD(f) is the FT of xD(t) as defined above. The units of P(ƒ) are Watts per Hz. [1] (b) Since E(f) = |X(ƒ)|2 and the FT of the filter output will be H(f)X(ƒ), the 1-sided ESD of the filter output will be ESD(f) = |H(f)X(ƒ)|2 = |H(f)|2 E(f) [1] [1] ∫

∞

∞−= dffEfH )(|)(|outputfilter ofEnergy 2

(c) By a similar argument, 1-sided PSD of filter output is PSD(f) = H(f)|2 P(f) [1] [1] ∫

∞

∞−= dffPfH )(|)(|outputfilter ofPower 2

2 (d) Response of H(f) to s(t) has frequency response: S0(f) = S(f) S*(f) e-2πjfT = |S(f)|2 e-2jπfT [1] By the inverse FT, the time-domain response is:

∫∞

∞−

−= dfeefSts jftjfT ππ 2220 |)(|)( [1]

and it follows that at the sampling instant t=T, by Parseval’s Theorem. [1] dffSdffSTs ∫∫

∞

∞−

∞

∞−== 22

0 |)(||)(|)(

= E as defined [1] Response of H(f) to 0 is clearly 0 for all t.

CS3282 Exam Summer 2006 10 25/08/2006 BMGC What about the noise? Can't predict its value at t=T, but can estimate its power. Its '2-sided' PSD is N0 /2 therefore the noise power is:

∫∫∞

∞−

∞

∞−= dffHNdffHN 2

02

0 |)(|)2/( |)(|)2/( [1]

∫ ∫∞

∞−

∞

∞−−== dttTsNdtthN 2

02

0 |)4(|)2/( |)(|)2/( [1]

∫∞

∞−== ENdttsN )2/( |)(|)2/( 0

20 [1]

Under the assumption that the noise is a signal whose samples may be considered samples of a zero mean random variable, the variance of this random variable, n0 say, is equal to the power (N0 /2)E. [1] Therefore when the input is s(t) or zero plus AWGN of '2-sided' PSD N0/2 the filter, the output at t=T is:

∫∞

∞

+=-

20 |)(| = E ere wh

logic"0"for : 0logic"1"for :

)( dttsE

nTz

Taking a threshold at E/2, the bit-error probability PB = Q( (E/2)/σ ) [1] where σ = standard deviation of noise which is square root of variance. Therefore σ = √(EN0/2) and it follows that: [1] PB = Q( (E/2)/ √(EN0/2)) = Q( √(E/(2N0) ) [1] End of question 2

CS3282 Exam Summer 2006 11 25/08/2006 BMGC

Question 3(a) The physical layer controls the method by which "raw" binary digits are transmitted over the real channel and received. It must try to ensure that when a stream of bits are sent, we receive them and also the correct number of bits. The received bit-stream will sometimes contain bit-errors. [1]

systemnoise

s(t)

r (t)

t

t

**

Sync symbol sampling points

Symboldetect

Equal-iser

H(f)

Derive local carrier

De- mod-ulate

Channel

Carrier

Mod -ulate

Map to bb with pulse shaping

[1]

The channel distorts the waveform by having non-flat gain-response and group-delay characteristics which vary with frequency and, especially with mobile, with time. Thermal noise, modelled as AWGN (system noise) added at the point shown in the block-diagram, is generated by electronic components at various stages in the receiver system.

(i) Map to base-band with transmitter pulse-shaping filter: converts a bit stream to a sequence of suitably shaped symbols or pulses of voltage. This process is achieved by exciting a pulse-shaping filter at the transmitter with impulses indicative of the binary data. [1] (ii) Matched filter H(f): minimises the effect of AGWN on the detection process. [1] (iii) Adaptive channel equaliser aims to cancel out the gain and phase distortion introduced by the filtering effect of the channel which will have changed the shape of the signal launched into it. [1]

A Nyquist frequency response: must exist between points marked "*" in the block-diagram. This ensures that the centre of a given symbol coincides with zero-crossings of previous and subsequent symbols. The product, HN(f) say. of the transmitting pulse shaping filter's frequency response and the matched filter's frequency response must be a Nyquist frequency-response, assuming that the equaliser has cancelled out the filtering effect of the channel. [2] HN(f) should be chosen such that its impulse response away as quickly as possible.

CS3282 Exam Summer 2006 12 25/08/2006 BMGC 3(b) Use definition of sampleT(s(t)) and repeat1/T( S(f) ) and the FT relationships given in Appendix B as attached to this examination paper. If a time-domain waveform s(t) say has the required zero crossings, sampleT(s(t)) = s(0)δ(t) [1] since sampling s(t) at t= ±T, ±2T, ±3T, gives zero in each case. (Remember that sampleT(s(t)) represents each sample it obtains as the strength of (an infinitely high) impulse).

By FT properties, Fourier Transform of sampleT(s(t)) is (1/T) repeat1/T( S(f) )

)/(1 ∑∞

−∞=

−k

TkfST

= [1]

And the FT of δ(t) is 1, so the FT of s(0)δ(t) is s(0) for all f. Therefore, [1]

)/(1 ∑∞

−∞=

−k

TkfST

= s(0) constant for any value of f. [1]

If S(ƒ) is band-limited between ƒ = ±1/T Hz, only terms with k = 0 & 1 can be non-zero for f in range 0 ≤ ƒ ≤ 1/T.

∴ S(ƒ) + S(ƒ-1/T) = T s(0) which is constant for 0 ≤ ƒ ≤ 1/T [1] (Similarly, for -1/T ≤ ƒ ≤ 0, only terms with k = 0 & -1 can be non-zero ∴ .S(f) + S(f+k/T) = Ts(0). ) 100% RC spectrum (note symmetry about f = 1/(2T) :-

1/T-1/T

t

1/(2T)-1/(2T) 01/4T 3/4T

0.5

1

|H((f))|

[1] We need to make the overall frequency response H((f)) satisfy Nyquist’s first criterion by being a raised cosine spectrum. With a bandlimiting pulse shaping filter at the transmitter and its matched filter equivalent at the receiver, assuming the frequency response of the channel is cancelled out by an equaliser, H((f)) will have magnitude spectrum equal to the square of that of the pulse-shaping filter. Hence we make the magnitude spectrum of the pulse shaping filter RRC. [1] 3 (b) A zero-crossing equaliser is a simple way of cancelling ISI. More sophisticated adaptive FIR digital filtering schemes are more successful for radio communications. [1] The zero-forcing equaliser is as follows:

CS3282 Exam Summer 2006 13 25/08/2006 BMGC

Delay T s

Delay T s x(t)

C0 C1 C2 y(t)

[1] When the input is a single signalling pulse of the appropriate shape but distorted by the channel’s frequency response, and it is centred on t=0 as detected by the timing circuitry, the output must be forced to be zero at t=0 and t=2T. The output at t=T is arbitrary and may be taken to be 1. The output at t= 0, T and 2T is as follows: 0 = y(0) = x(0)C0 + x(-T)C1 + x(-2T) C2 1 = y(T) = x(T)C0 + x(0)C1 + x(-T) C2 0 = y(2T) = x(2T)C0 + x(T)C1 + x(0) C2 [1] In matrix form:

−−

−=

2

1

0

14.01.03.014.0

2.3.01

010

CCC

[1]

b = Ac [1] Therefore c = A-1 b and it follows that C0 = 0.34 ; C1 = 0.8 ; C2 = -0.29; [1]

CS3282 Exam Summer 2006 14 25/08/2006 BMGC Question 4(a) Solution: Asynchronous transmission: A digital transmitter must apply suitably shaped symbols to the channel at times specified by a timing reference or 'clock'. For short block-length transmissions as used for transmitting 8-bit binary numbers between computers and peripherals over short distances the transmitter and receiver clocks need only be approximately matched and they may resynchronise at the beginning of each short block. [1] This is 'asynchronous' transmission and is the basis of the well known RS232 standard.. Data is sent in short words, say 8 bits long, with synchronising start and stop bits. The receiver clock resynchronises itself at each start-bit. [1] Transmission of 8-bit ASCII characters according to the RS232 protocol: When idle, the line remains high at voltage V1. A “start-bit” signifies the start of a transmission. This bit is always “0”. The eight bits of data are then transmitted using “non-return to zero” (NRZ) pulses and finally a number of “1” stop-bits (in this case two) are transmitted to ensure that the next character is not sent immediately. [1] The receiver waits for a transition from the 'idle state' "1" to "0" indicating a 'start bit'. It delays for half a symbol period according to its own free running clock having approximately the same frequency as that of the transmitter, and then samples the channel eleven times at intervals of T seconds. The samples will hopefully lie in or close to the centre of each symbol, but the timing will drift over the eleven samples. The drift is acceptable because of the frequent resynchronisation. [1] [1]

V1 V0

0 1 0 1 1 0 0 0 1 1 1 Start-bit Data Stop-bits

t

Advantage: simplicity of transmitter and especially the receiver. Disadvantage: inefficient in its utilisation of the channel capacity. [1] 4(b) Synchronous techniques are used for the efficient transmission of continuous data for long periods of time, often at data rates close to the maximum possible over a channel of specified bandwidth. A synchronising code (say 10101010) is sent at start of transmission, and thereafter, the receiver clock must be kept synchronised in frequency and symbol-timing from the transmission itself. When considering how to synchronously transmit digital information over wires, two factors must be borne in mind:- [1] (i) We would like to keep the average voltage level as close as possible to zero since any voltage offset carries no data and just wastes power. In many cases, the DC component of a signal is lost over wire lines sometimes because of AC coupling, the use of transformers and/or because the line is used to carry power as well as the data. This is certainly the case with telephone lines. [1] (ii) For synchronous transmission, we need to ensure that the signal always has a frequency component at the signalling rate (or an exact multiple or sub-multiple of the signalling rate) to allow a timing waveform to be extracted at the receiver for synchronising the detection process. [1] HDB3 coding: (high density bipolar, order 3): uses ternary coding to send binary coded data, as described above for AMI, but places an incorrectly signed pulse in place of any 4th consecutive. zero. E.g. for ' 1 1 0 0 0 0 0 1 … ' we send: ' +V -V 0 0 0 -V 0 +V … '

CS3282 Exam Summer 2006 15 25/08/2006 BMGC The incorrectly phased pulses are included only for clock synchronisation. They are taken to be a “0” at the receiver. The average voltage is no longer zero in the short block above, but over a longer time-span the average will still remain zero since incorrect +V pulses and -V pulses will occur equally often. [1] Bi-phase-L is better known as “Manchester coding”, and represents a “one” by a pulse of width T/2 positioned during the first half bit-interval. A zero has a pulse of width T/2 in the second half interval.

t

−V

T T

'one'

−V

+V +V Manchester coding

'zero' [1]

1 1 0 0 0 0 0 1 −V

V

t

[1] Manchester coding has the advantage of absolutely guaranteeing zero dc level and is easy for the receiver clock to synchronise itself to. Its disadvantage in comparison to NRZ-HDB3 is that the bandwidth requirement is considerably higher. [1] Question 4(c) Solution: Each regenerative repeater has within it a receiver and a re-transmitter. At each sampling point, the receiver receives a +V or a –V corrupted by noise and possibly other distortion. It must decide whether a “0” or a “1” was intended and then the re-transmitter reconstructs a perfectly shaped symbol of appropriate amplitude and sends it on to the next repeater. [1] Assume that, at a receiver, the symbols are expected to be +V and –V at the sampling points to signal “1” and 0 respectively. We could take the threshold to be 0 volts, and decide a +V symbol was intended if the voltage is greater than 0 and a –V symbol was intended if the received voltage is negative at the sampling point. To produce an error, the noise must exceed +V when –V (logic “0”) is sent, or be less than –V when +V (logic “1”) is sent. Hence the probability of an error is: (Prob of a “0”)x (Prob of noise sample being greater than +V) + (Prob of “1”) x (Prob of noise sample being less than –V) )

CS3282 Exam Summer 2006 16 25/08/2006 BMGC We assume an equal probability of “1”s and “0”s. Therefore probability of an error is: 0.5 x Q(V/σ) + 0.5 x Q(V/σ) where σ is the standard deviation (σ2 = variance) of the noise. The probability of a noise sample being less than –V is same as the probability of a noise sample being greater than V. Q(z) as plotted in these notes is the probability of a noise sample being greater than z for Gaussian noise of zero mean and standard deviation equal to one. Therefore Q(z/σ) is the probability of a noise sample being greater than z/σ for Gaussian noise of zero mean and standard deviation equal to one. But, most importantly, Q(z/σ) is also the probability of a Gaussian noise sample being greater than z when the noise has zero mean and standard deviation equal to σ. It is also the probability of a Gaussian noise sample being less than –z when the noise has zero mean and standard deviation equal to σ. If Q(V/σ) = 10-3, from the graph we find that (V/σ) = 3 Therefore σ = V/3. This is the standard deviation of the zero mean Gaussian noise that is causing the errors. With the same noise level, to produce an error probability of 10-9 rather than 10-3, we need the regenerative repeaters to receive higher voltage levels for the symbols. Assume they are raised from ±V to amplitude ±U at the sampling points. Then Q(U/σ) = 10-9 and from the graph, we find that U/σ = 6. This means that U = 6σ = (6/3)V = 2 V. 20log10(2) = 6 dB. Therefore we need to arrange that the voltages as received at the receiver are raised by 6 dB. [2] If we cannot raise the voltages transmitted by the regenerative repeaters because of cross-talk, we can only reduce the distance between the repeaters so that less attenuation occurs between one regenerative repeater and the next. The attenuation over 2 km must be reduced from 40 dB to 34 dB. 20 x distance = 34 Distance between repeaters must be reduced to 34/20 = 1.7 km instead of 2 km. [2] Originally, need 34/2 =17 repeaters therefore error rate (approx adds): 17 in 1000, i,e, 1 in 143. With reduced spacing, need 34/1.7 = 20 repeaters, therefore error-rate is 20 in 109 i.e. 1 in 5 x 107. [2]

CS3282 Exam Summer 2006 17 25/08/2006 BMGC Question 5 (a) OFDM is very good for channels affected by frequency selective fading for several reasons. (1) info can be spread out across sub-carriers so that when some are lost, others will compensate. [1] (2) guard-band allows for ISI; if OFDM symbol rings on, it only affects beginning of next symbol, repeated at end. (Can have cyclic "prefix"). So no pulse-shaping necessary! [1] (3) equalisation much easier than with single carrier systems. OFDM equalisation done by multiplication in frequency-domain after FFT. [1] Easier than adaptive filtering used for single carrier. [1] Works because of cyclic extension. [1] Disadvantage of OFMD is "peak to mean" ratio of symbols which can be very large by nature of FFT & its inverse. Shape of each OFDM symbol is very complex & must be sent & received accurately. [1] Amplitudes can become very large in comparison to the mean. Definitely not "constant envelope". [1] Transmitter & receiver must be linear to preserve shape. Therefore, need ‘class A’ amplifiers: less power efficient than those for constant envelope transmissions. [1] (b) Bandwidth efficiency is the number of bits/second per Hz of bandwidth. It is sometimes quoted in symbols/second per Hz of bandwidth [1] Max achievable bandwidth efficiency at baseband with zero ISI is 2 bit/second per Hz. [1] This is difficult to achieve in practice because it requires a ‘brick wall’ pulse shaping filter whose frequency response is as shown below:

f

HN(f)

1/(2T) The pulses generated by exciting this filter with impulses are ‘pure sinc’ functions with zero-crossings at t = ±T, ±2T, ±3T, ……Such pure sinc functions in theory give zero ISI, but they do not die away quickly enough. Hence we need a high order pulse shaping filter. And a slight timing error in the sampling point will give a significant non-zero sample value. [1] Bandwidth efficiency of

(i) Binary PSK with 100% RC pulse shaping 0.5 b/s per Hz [1]

(ii) QPSK with 50 % RRC pulse-shaping QPSK with 0% pulse shaping gives 2 b/s /Hz QPSK with 50% pulse shaping gives 2 b/s / (1.5 Hz) i.e. 4/3 b/s /Hz [1] (iii) 8-PSK with 50% RRC pulse-shaping 8/3 b/s /Hz [1]

CS3282 Exam Summer 2006 18 25/08/2006 BMGC (iv) OFDM at 250 kbaud with 64 sub-carriers spanning a 20 MHz bandwidth. Only 48 sub-carriers are modulated by QPSK (count FEC bits as data) 2 * 48 = 96 bits per 4 us in 20 MHz [1] 24 Mb/s in 20 MHz [1] 24/20 b/s per Hz = 1.2 b/s per Hz [1] (e) Gray code used to make a symbol error cause just one bit error as far as possible(except where noise is very large). [1] This makes bit- error probability = symbol error probability divided by number of bits per symbol. [1]

100

101

111

110

010

011

001

000

R

‘8-PSK’

Q

I

[1] End of all answers