DIGITAL COMMUNICATIONS IAGlover and P M Grant Prentice Hall 997 PROBLEM SOLUTIONS CHAPTER 6 6. P e erf V σ erf. 5 +. 5 0.705 [ erf (. 009)] [ 0. 999 979 ]. 0 0 5 The optimum DC level is zero. For equiprobable symbols this requires pulses with amplitudes of ±.0 V. Transmitted power before adjustment: V 0 + V 9. 5 (V ) Transmitted power after adjustment (. 5) +. 5 V 0 + V. 0 9. 0 (V ) Reduction in transmitted power 9. 5 0 log 0 9. 0 0. db 6. P e erf V σ V erf P e σ erf 0 6 P R R
erf [0. 999 998]. 8 0 R 0. 0 R V Voltage attenuation. 8 db V trans V 0. 6607.6 0 0 0. 6607 For unipolar RZ signalling each power is: 0. 0 R 0. 6607 0. 60 R V S peak ( V trans) R (0. 60 R ) R 0. 7 W for 50% duty cycle and equiprobable symbols mark voltage is only present for a quarter of time, i.e. S AV S peak 0. 7 0.09 W 6. AMbaud signal with a SER of error/s has a probability of symbol error given by: P e 0 6 For 6identical sections the individual section P e is: P e hop P e mhops m 0 6 6. 86 0 8 Thus, using equation (6.8): P em P e8. 86 0 8 8 8 erf V σ 7 erf 8 σ σ erf 8 7. 86 0 8 erf. 95 0 8 erf [ 0. 999 99995695 ] 0. 09 V
Thus the errors in the approximate analysis are only tenths of a db! 6. PSD is controlled by pulse shape and type of signalling code OOK/NRZ etc used. For a sample rate f o and period the pulse of width / implies a first null in the spectrum of f o for the RZ signal. (In comparison the NRZ signal has a first null at f o Hz.) The minimum theoretical transmission bandwidth is half the values, i.e. f o for RZ and f o / for NRZ. The disadvantages of OOK are no timing extraction, no error detection, it needs a channel with a DC response and it is not transparent. Thus AC coupled amplifier designs cannot be used. 6.5 p(t) Π t / 0. 5 Π t / / 0. 5 Π t + / / p(t).0 -T 0 -T 0 -T 0 0 T 0 6 6 -T 0 -T 0 t -.0 (a) Total area under pulse is zero therefore DC level iszero. Using equation (6.) the PSD of the binary signal is: G p ( f ) + T o 0. 5 F ( f ) F ( f ) Σ 0. 5 F (nf o ) + 0. 5 F (nf o ) δ ( f nf o ) n F ( f ) sinc f 0. 5 sinc f 0. 5 sinc f e jω sinc f sinc f T 0 e jω cos ω + e jω jω e
sinc f sin π 6 f sinc ( f ) 0-9 -6-6 9 T 0 T 0 T 0 T 0 T 0 T 0 f.0 sin ( π 6 f ) -9-6 - 0 6 9 T 0 T T T T T 0 0 0 0 0 f F (f) T 0 ( F (f) 0 ) sinc ( f ) sin ( π 6 f ) 0-9 -6-6 9 f T 0 T 0 T 0 T 0 T 0 T 0 Continuous part of G p (f) is therefore: G p (f) continuous 9 sinc ( f ) sin ( π f ) 6 0-9 -6-6 9 T 0 T 0 T 0 T 0 T 0 T 0 The discrete part of G p ( f )is: G p ( f ) discrete T o Σ n 0.5 sinc nf o sin π 6 nf o δ ( f nf o ) Σ n sinc n sin π n δ ( f nf o ) Σ n sin 6 Π n Π n δ ( f nf o )
i.e. there are spectral lines at all integer multiples of f o (the clock frequency) except n, 6, 9,,... etc. G p (f) A δ (f - f 0 ) etc A δ (f - f 0 ) B δ (f - f 0 ) C δ (f - f ) 0-9 -6-0 6 9 T 0 T 0 T 0 T 0 T 0 T 0 f (b) (c) The PSD at f 0Hziszero. This, together with the zero DC component (part (a)), makes the line code suitable for AC coupled transmission. The first null bandwidth is Hz (d) T For a given P e (using CPD) unipolar NRZ would require o 0. 5 + 0. T 5 o + T more o energy per digital than this line code. Since zero energy is required for digital 0 s inboth cases unipolar NRZ will require: 0 log 0 0. 5 + 0. 5 + db more power than the given line code. (e) (f) Since there is a spectral line at the symbol frequency, f o,the line code is potentially selfclocking. There is no error detection capability. 6.6 Unipolar NRZ Unipolar RZ Polar NRZ Dipolar OOK Bipolar NRZ Bipolar RZ All the above suffer from a loss of clock signal in the event of long strings of zero s preventing such strings being successfully transmitted.
Polar RZ - Has non-zero G(0) (i.e. power spectral density at f 0). It therefore suffers from DC accumulation (and therefore signal droop on AC coupled lines) in the presence of long strings of s or 0 s. (Unipolar NRZ, RZ and polar NRZ also suffer from this in addition to the loss of clock signal as described above.) Dipolar split phase - Has zero G(0) and therefore does not suffer from signal droop. Split phase symbols ensure recoverable clock signal even inpresence of long strings of zero s and one s. HDB - Has zero G(0) and therefore does not suffer from signal droop. Four zeros subsitution prevents loss of clock signal when long strings of zeros occur in data. CMI - Has zero G(0) and therefore does not suffer from signal droop. Split phase zeros and alternating polarity ones ensures clock signal is present even in presence of long strings of ones and zeros. 6.7 + V - 0 0 0 Space part of CMI only - V - Space part of CMI sequence can be resolved into the sum of an antipodal dipolar (Manchester code) signal and a (periodic) clock signal i.e.: +V T 0 clock signal ( + V - ) -V +V Manchester code + ( - V ) -V Using equation (6.), noting that for Manchester coding F ( f ) F (f)and assuming p then: G Man ( f ) 0. 5 F ( f ) + 0 F ( f ) FT V V Π t V Π t + sinc f jω e sinc f jω e+
V 6 V 6 i. e. G Man ( f ) V sinc f jω e e + jω sinc f j sin ω sinc f π sin The spectral density of bipolar NRZ is given inquestion as: f G BNRZ ( f ) V sinc ( f )sin (π f ) The spectrum of the clock signal component is the set of impulses (or delta functions) for a square wave of amplitude V/ volts and period seconds. Table. gives the (one sided) amplutides of impulses for a unit amplitude square wave for as π n.the one sided amplitudes for a V amplitude square is thus V π n. The power in each line in the one sided spectrum is: V π n V (n ) (V ) The power in each line in a two sided spectrum is: V ( n ) V ( n ) (V ) The position of the spectral lines is given by: f n Hz Thus: G Impulses ( f ) V Σ n ( n ) δ f n The CMI spectrum is thus given by: G CMI ( f ) G Man ( f ) + G BNRZ ( f ) + G Impulses ( f ) V sinc f π sin + V sinc ( f ) sin (π f ) + V Σ n f ( n ) δ f n
The spectra are shown in Figure 6.. 6.8 The trajectories are obtained from Figure 6. by using the equalised cosine filter impulse responses, as in Figure 8.(b). In this figure since signal is AMI the trajectories +, 0, + and, 0, are not allowed, although their trajectories effectively appear in the eye as elements of other legitimate trajectories.? Refer to Figure 6. for the diagram showing the capacitive coupling between the pairs throughout their length. Since outgoing signal is largest at point of launch into the cable and incoming signal on return path is smallest at same point (after attenuation from distant end) the dominant crosstalk occurs at this point. This point is near the receive end and the coupled noise is called near end cross-talk or NEXT. Clearly the picture is symmetrical and NEXT is also present at the receiving input at the other end of the span. Figure 6. gives the required spectral plots. 6.9 The HDB signal is bipolar with alternating positive and negative pulses. This gives a zero in the PSD at DC. To recover the clock component this signal has to be rectified to turn the negative pulses into positive pulses. Then all the pulses give the clock signal component. The rectification operation is a nonlinear process as it involves the use of a nonlinear diode rectifier. 6.0(a) First we find z where m X: z x m 5. 5 0. 889 σ Therefore probability that x 5. 5 is: + erf(0. 889) + 0. 7887 0. 89 6.0(b) P e is now the above value subtracted from, i.e. 0. 6.0(c) z x m σ 750 80 60. Probability of the clouds being higher than 750 metres is therefore:- ( + erf(. )) 0. 95 0. 0
6.0(d) (i) This question assumes zero mean, therefore for the three cases:- i) z σ σ, ii) z, iii) z This gives erf z as i) erf(0.7) 0.687, ii) erf (.) 0.95, iii) erf(.) 0.997. Therefore probability that the Gaussian random variable will exceed these values is given by:- i) ( + 0. 687) 0. 6 ii) ( + 0. 95) 0. 0 iii) ( + 0. 997) 0. 00 6.0(a) Alternative solution for Question 6.0 This is a question with Gaussian variables developed from equation (6.) as: P e x σ π Now for σ this becomes: P e π e x y e σ y dy dy which is the Q function as in footnote on page 5. In appendix A the erf tables give: erf(x) π x e y dy The relation between these is thus: P e 0 erfc x erf x Thus for X and σ we obtain the normalised value of x as: and: x norm x X σ P e 5. 5. 5 erf. ( erf 0. 88) (0. 0) 0. 05 This is the solution to Problem 6.0(b). 6.0(a). 5 P e is now the above value subtracted from, i.e. 0.898.
6.0(c) x X 750 80 σ 60 P e ( erf. ) ( 0. 9506) 0. 00 6.0(d) (i) x X σ P e erf ( 0. 686) 0. 587 (ii) (iii) x X σ P e erf x X σ P e erf 0. 00 0. 00 pmg/www/digicomms/tut6