3.1. According to (3.1) with no care on jammer power SIR at the matched filter output is , where is matched filter SNR without jammer and is AWGN power within signal bandwidth. With band-elimination filtering the resulting power SNR (see (3.2)). Then:(a) , and both strategies are equivalent

as for SIR/SNR;(b) , and ignoring jammer is

preferable;(c) , and band-elimination filtering is

better.

3.2. Signal spectrum has identical maximums at the frequencies . Accordingly matched filter will pass components of these frequencies with maximal amplification. So when jammer is ignored its most harmful central frequency is any of . With jammer having any of central frequencies

the band-elimination filter will remove the greatest part of signal energy, maximally reducing output SNR. Therefore, with band-elimination filtering the most harmful jammer central frequency remains to be again any of .

3.3. From it follows that if , . To have one needs , i.e. 100 times wider signal bandwidth. Then

(a) Signal duration should be 100 times reduced with simultaneous increasing signal peak-power 100 times;

(b) Spread spectrum signal of former duration should be used having processing gain 100.

3.4. From it follows that , while to have only 2 percent SNR degradation it should be that . Then(a) Signal duration should be reduced 25 times with simultaneous

increasing amplitude 5 times;(b) Spread spectrum signal of the former duration with processing gain 25

solves the same problem with no power increase.

3.5. The SNR should be increased 20 times. Then(a) Signal duration should be 20 times decreased with simultaneous

increasing power 20 times;(b) Without increasing peak power there is no possibility to improve SNR

using only a plain signal: ;

(c) Spread spectrum signal of the former duration and power with processing gain 20 solves the problem;

(d) Take a spread spectrum signal with 10 times wider bandwidth than before and 2 times greater duration (processing gain is 20). Then its energy increases 2 times and .

3.6. , where subscripts “1” and “2” correspond to situations before and after parameter changing. If , then

and after parameter changing SNR becomes five times higher meaning gain of the intended system.

3.7. Let be a fraction of total jammer power assigned to the first subband. Then jammer power spectrum densities within the first and second subbands are and respectively. Weighting the signal and jammer within the first subband by equalizes spectrum densities in both subbands, so that we have matched filtering of the signal with total energy

, being energies of the initial signal in each subband, against the white noise with power spectrum density

. Therefore, the total matched filter SNR is proportional to

or to . The derivative of this function with respect to

is , so that minimizing the function is defined by the equation , leading to . Hence, the total jammer power should be divided between subbands as .

3.8. Power SNR at the intended matched filter output is . Let denote signal power spectrum density perceived by an interceptor.

Then (a) (–15 dB);(b) (–4.1 dB).

3.9. To provide bit voltage SNR should be no smaller than 3 (see (2.19)). Then means .

3.10. MHz (12 dB corresponds to 16).

3.11. Numbers of keys of one bit duration for two systems equal respectively and . Since the first system is more immune

to a cracking.

3.12. in decibels looks as . Substitution of figures gives for both systems the same result dB. Hence the two are nearly equivalent as for the cracking immunity.

3.13. Since (20 dB) and , the necessary processing gain .

3.14. From and (–7 dB) it follows that for the first system and for the second (17 dB corresponds to 50).

3.15. Noise spectrum density is found as , being Boltzmann constant. Hence, V2/Hz. From (3.10)

.

This typically is represented in decibels as

In its turn to provide a necessary power SNR the received power should be or in decibels

dB.

Finally dB. Converting decibels back to watts gives approximately mW. If hypothetically only the distance attenuation exponent grew up to 3.84, no other change were necessary in the model, the transmit power should be increased by dB, i.e. up to about a hundred kilowatts. Fortunately, in practical scenarios not only distance exponent grows but other model parameters change, so that the resulting increase of the transmit power is not that catastrophic.

3.16. Four times drop of voltage means dB drop of power. Then according to the lognormal law of the long-term fading with dB, the probability of the system failure

.

3.17. Denote all distances as shown in Figure S.4. Then the propagation difference

, and phase difference between signals of two paths is . The period of the standing wave, i.e. distance between two

adjacent drops (or peaks) is determined as an increment of , at which changes by or equivalently changes by wavelength . Thus, the equation may be written

.

Using the fact that it is possible to skip in the Taylor series of the left-hand side all powers of but first, coming to the equation

,

and finally to the solution

.

After substituting here the values given in the problem

m.

Then at the velocity km/h time interval between the wave drops will be s.

3.18. If phase of the fading signal changes from one pulse (bit) to the next significantly, reliable reference recovery is impossible and BPSK becomes unavailable. One of possible solutions is FSK.

3.19. The lognormal fading with deviation 12 dB around 27 dB means that probability of dropping SNR below 15 dB . Bit error probability under SNR 15 dB in the Rayleigh channel according to (3.13) or Figure 3.16 . Hence, bit error probability cannot be lower than

Transmitter Receiver

Reflector

h2/D

Figure S.4 Two-path scenario.

.

3.20. For the maximal-ratio combining the resulting power SNR according to (3.15) , being the power SNR per a branch. For the selection of maximum-SNR-branch the output combiner amplitude is the greatest of branch amplitudes , i.e. if and otherwise. Let

, and be PDFs of amplitude at the combiner output and in the first and second branches respectively. Then

.Since both branches have identical Rayleigh fading

,

and PDF of the output combiner amplitude.

Then average power SNR is proportional to average squared amplitude

.

Therefore, gain of the selection combiner in average power SNR is 1.5 versus 2 of the maximal-ratio combiner, i.e. loss of the first to the second is 4/3 or 1.25 dB.

3.21. At first sight it may seem that the frequency space between the branches is fit, which is equal to the coherence bandwidth, i.e. kHz, providing 6 branches. This, however, will produce overlapping of branch spectra, i.e. violation of branch independence. To eliminate it frequency space should be no smaller than signal bandwidth, i.e. 60 kHz. Hence, maximum number of branches is 300/60=5 rather than 6.

3.22. The minimum mutual time delay is between the second and third paths: s. To resolve these paths signal bandwidth has to be

about MHz. With rate 20 kbps data symbol duration is no smaller than s, while delay spread s, meaning that RAKE arrangement is reasonable. Then the processing gain will be no smaller than .

3.23. Signal correlation spread has to be within s corresponding to bandwidth about MHz. Since only signals with mutual delay no smaller than are resolved, number of RAKE fingers is within . With QPSK transmission data symbol duration s, corresponding to the processing gain .

3.24. With being average signal power at the RAKE input signal power per finger will be , since the input signal is just sum of independent “finger” signals. Then power SNR per branch , where is input power SNR. The combiner will increase power SNR times, returning it to the initial quantity of average power SNR. So, RAKE-diversity gain is not explained by SNR gain. Its nature is similar to the one of transmit diversity: coherent summation of paths (preliminarily separated) reduces the signal scattering range, thereby decreasing error probability (see also Section 10.3).