Solutions of problems

2.1. Energies of signals are . Correlation coefficients are . According to the cosine theorem the squared

distances between signals appear to be , and .

2.2. Correlations of with all three candidate signals are . Then according to the decision rule (2.8)

has values for respectively. Therefore decision will be that signal is received.

2.3. Bandwidth necessary to run BPSK transmission is of the order of , i.e. 10 kHz. Widening a bandwidth up to 10 MHz means involving spread spectrum signals with , which has no positive effect on bit error probability depending on only bit pulse energy.

2.4. The case (a) is better, since energy per bit for it is higher, while bandwidth has no effect on bit error probability.

2.5. All the signals have equal energies, thus pairs are different in only correlation coefficients within a pair. The third one is antipodal, i.e. having correlation coefficient . This is an optimal pair, i.e. has no energy loss. For the first and second pairs correlation coefficients are and respectively. Energies equal, squared distance is proportional to . Therefore energy loss versus the third (optimal) pair for the first pair is

(4.8 dB) and for the second (1.8 dB).

2.6. There are 4 possible configurations of transmitted pairs of consecutive pulses: . They correspond to four signal points on the plane having squared distances from the origin and from the nearest other point, where is a pulse energy. Bit error happens any time when a transmitted point is confused with the one of an adjacent quadrant. Block of successive pulses is capable to transmit bits meaning that

, being energy per bit. Hence, squared minimum distance for DBPSK is , which is close to the one of BPSK whenever

. The pulse for DBPSK is times shorter than that of BPSK, rate fixed, implying the same bandwidth broadening. Again when this is negligible and asymptotically DBPSK and BPSK are equivalent in both energy consumption and bandwidth.

2.7. Correlation coefficient between adjacent signals is zero, therefore minimum distance , where is energy of the manipulated pulse. Since every pulse transmits two bits, and (the same as for BPSK). For the optimal (simplex) four signals and

, so that the asymptotic energy loss of QPSK to the simplex set is or 1.2 dB.

2.8. This is possible for no more than eight signals.

2.9. Squared distances for orthogonal and simplex signals are respectively and . Therefore energy loss of orthogonal signals is

and drops from 3 dB tending to zero as grows.

2.10. means transmission of one bit. The best way of doing it is antipodal pair rather than orthogonal one having 3 dB energy loss versus the first.

2.11. Under the uncoded transmission of two bits minimum squared distance between two closest blocks of two bits (say 00 and 10) is . In the family of four orthogonal signals of energy minimum squared distance is . Since those signals bear two data bits , which makes minimum squared distances for both transmission modes equal.

2.12. Optimal transmission mode, no bandwidth limits imposed, uses a set of eight simplex signals. Their minimum squared distance is , while for 8-PSK with the same signal energy the closest signals have correlation coefficient and therefore squared distance

. Then energy loss of 8-PSK to the optimal set is or 5.9 dB.

2.13. For M-PSK correlation coefficient between the closest pair and with pulse energy the squared minimum distance

. With the orthogonal signalling , so that energy loss of M-PSK is

when . At the same time bandwidth for M-PSK , while for orthogonal coding bandwidth . Therefore and M-PSK may be recognized preferable if bandwidth economy is more important than energy saving.

2.14. For orthogonal signals kHz.

2.15. Since , . Solution of this equation in is orthogonal signals. When signals are bandpass this number may be doubled by involving quadrature phase-shifted copies of every of them.

2.16. Gain 6 dB is four times, therefore means and . Then necessary bandwidth kHz, i.e. 6

dB gain within the bandwidth 320 kHz is achievable with orthogonal coding.

2.17. Again , which means , and or 7 dB.

2.18. Gain 10 dB is ten times, therefore from it follows that and necessary bandwidth GHz,

which is greater than double the carrier frequency. Of course, it is not realizable.

2.19. According to the Sylvester rule

,

where matrix may be taken from Section 2.7.3.

2.20. First of all, transpose of a Hadamard matrix is again Hadamard matrix. Indeed, since rows of are orthogonal and have squared norms

equalling , , where is identity matrix of size . Hence, is

inverse to and , meaning that rows of are orthogonal. Then:

(a) Does not effect row orthogonality, the matrix remains Hadamard-type;(b) Permutation of columns of permutes rows of preserving their

orthogonality. Reverse transpose of after row permutation produces again a Hadamard matrix;

(c) Does not effect inner products of rows, hence leaves the matrix of Hadamard type;

(d) Does not effect row orthogonality leaving the matrix of Hadamard type;(e) Does not effect row orthogonality of , hence preserves the Hadamard

type of ;(f) The inner product of the first row with any other becomes nonzero; a

new matrix is no longer a Hadamard one.

2.21. The inner product of any different rows and squared norm of any row in the matrix obtained becomes and respectively, so that correlation

coefficient of any two different rows . Thus rows of the new matrix form simplex signal set. With the data rate fixed, each of signals has the same duration as any of initial orthogonal signals, containing smaller by one number of chips (first column of Hadamard matrix dropped!). Hence, simplex set occupies narrower bandwidth. Certainly, with this bandwidth gain of the simplex set is insignificant as well as energy gain.

2.22. Amplitude and duration are energy parameters, the rest being non-energy ones.

2.23. The best accuracy will be achieved in the case (b). In the case (a) correlation coefficient is smoother in dependence on , which has negative effect on estimate precision. In the case (c) the central peak of is as narrow as in the case (b), however high sidelobes make abnormal errors quite likely, i.e. confusion of a true value of with the one distant from it by a sidelobe-mainlobe space.

2.24. The best is case (c), since drops most quickly in all directions of -plane, making well distinguishable any signal copies whose

mismatch in goes beyond the shadowed circle. The case (a) is as good for measuring alone, being known. The case (b) is equally good for measuring any single of parameters, the second being known.

2.25. All three signals have the same energy , providing thereby equal variance of amplitude estimation.

2.26. Standard deviation of estimate is inversely proportional to a square root of energy , and linearly depends on duration. Therefore, to halve standard deviation signal duration has to be increased four times.

2.27. Energy is proportional to duration and squared amplitude, so halving the first and doubling the second doubles energy and reduces the standard deviation of amplitude estimate times.

2.28. Phase estimate accuracy is defined by only signal energy. Thus(a) Has no effect, since energy does not change;(b) Energy is two times more, the standard deviation becomes times

lower;(c) Energy drops four times, standard deviation grows two times;(d) Energy does not change; the accuracy remains the same.

2.29. Answer is the same as in Problem 2.25.

2.30. Among those given only amplitude and duration will affect phase estimation accuracy. Increase of will proportionally lower the standard deviation of phase estimate; increase of will lower it proportionally to the square root of . The changes pointed will not affect signal energy, i.e. the standard deviation of phase estimate.

2.31. To make it easiest way, note that ACF is piecewise-linear, so it is enough to calculate its values at the shifts being multiples of and connect them by straight lines. The results for are shown in Figure S.1, where .

2.32. Figure S.2 answers the problem.

03E

0E

0E

0E

2 Δ4Δ3Δ Δ2Δ

04E

Δ

02E

00

)(R

a)c)

Δ3Δ2

0

b)

Figure S.1 Autocorrelation functions of signals of Figure 2.34.

)(R

0E

)(R

)(R

3 3

Figure S.2 ACF of signal of Problem 2.32.

2.33. After the pulse matched filter there has to be a tapped delay-line with delay between the consecutive taps. Each tap with sign plus or minus is

connected to the adder. Signs of taps (first tap is from delay-line input) at the adder input are mirror-like to the signs of rectangular pulses of a signal. Figure S.3 presents an example for signal of Figure 2.34, c. Response of this matched filter to its own signal is shown by Figure S.1, c.

2.34. After the pulse matched filter there should be a delay-line with three taps: from input and with delay and , first connected to the adder with minus and two others with pluses. Response to the signal is shown in Figure S.2.

2.35. The best choice would be (b). For details see answer to Problem 2.23.

2.36. The standard deviation is given by Woodward formula: , which may be trusted since . This gives kHz. Since the signal is plain the order of its duration may be estimated as

s.

2.37. In order to preserve SNR after reduction of peak-power duration should be proportionally (i.e. 100 times) increased. But simultaneously to reduce 10 times standard deviation the bandwidth should be 10 times increased. Thus the time-frequency product equal to one for a plain signal should be increased 1000 times in total, i.e. made equal to 1000.

2.38. Since the standard deviation is inversely proportional to bandwidth and the latter (equalling initially frequency deviation of LFM signal) drops to the one of an unmodulated signal, i.e. 1000 times, precision of time measuring also drops 1000 times.

2.39. The standard deviation of frequency estimate is inversely proportional to the product , i.e. , where is signal peak-power. Since may be

increased only , i.e. 1.25 times, to reduce standard deviation 10 times should become 8 times greater, which is equivalent to increasing 4 times.

2.40. Since the standard deviation is inversely proportional to or with increasing four times it drops 8 times.

2.41. Precision of measuring only one parameter, the other being known, is determined by the extension of ambiguity diagram along the corresponding axis. From this angle cases (b), (c), and (d) are equally good in time-only estimation, as well as cases (a), (c), and (d) are equally good in frequency-only measuring. When both parameters are to be measured the extension of diagram in any direction has to be small, which means that the case (d) is best of all.

2.42. It is obvious that SNR remains the same, since four times increase of duration compensates fully for the 6 dB (4 times) power reduction. This means that the standard deviation of frequency estimate decreased 4 times due to duration increase. Since new time-frequency product is 100 with four times greater duration, bandwidth of the new signal is 25 times wider, entailing the same reduction of the standard deviation of time-delay estimate. Note that when both parameters are measured simultaneously these conclusions may appear not true, if ambiguity diagram of the spread spectrum signal is not “good” enough. E.g. for LFM signal estimates of both parameters are strongly dependent, which makes precision of simultaneous estimates poor (see Section 6.2).

2.43. The requirements towards the ambiguity function in time-frequency resolution are quite similar to those in time-frequency measuring. Thus, the answer repeats the one of Problem 2.41 after changing the “measuring” to “resolution”.

2.44. For a good resolution in time only the ambiguity diagram should have small extension along -axis. Thus, cases (b) and (c) are equally favourable, while cases (a) and (d) are poorer since signal ACF for them has high time sidelobes (side zones along the -axis). The same way cases (a) and (c) are equally favourable for frequency-only resolution, while (b) and (d) are worse. When signal copies should be resolved both in time and frequency, extension of the ambiguity diagram in any direction should be small. Therefore, case (c) is best, while the rest of ambiguity diagrams have side zones, corresponding to intense sidelobes, very harmful in resolution (see also Section 6.1).