1. Most students' solution show a total misunderstanding what happens in the investigated receiver. Independently of the dualband trick, the following must be fulfilled. We want the (LO+IF) to cover whole one band (e.g whole band II 1850-1910). If RF filter is fixed, it must pass the whole 1850-1910 band. If it is tuned, it should be tunable from 1850 to 1910. With fixed IF (the whole idea of heterodyne rcvr is to keep IF fixed...), LO must be tuned so that e.g. LO+IF covers whole RF. The dualband solution was to set LO and IF so that you can exploit the "mirror band" effect. If we set LO+IF=F1 and LO-IF=F2, we have an equation system with 2 unknowns - it should be solvable :-). One solution is: LO=(F1+F2)/2, IF=(F1-F2)/2, with use of (F1min and F2min) or (F1max and F2max) -> to keep IF constant another solution is: IF=(F1+F2)/2, LO=(F1-F2)/2 with use of (F1max and F2min) or v.v. to keep IF constant Filter F_I is of course covering one of RF bands and F_II the other one. (ver A) LO= tunable from (1850+1710)/2=1880 to 1880+60=1940 MHz IF fixed at 30 MHz. (remark: band IV is 45 MHz wide, band II is 60 MHz, so we have a bit of freedom - we can shift our choices by up to 15 MHz) Note that this design is not easy (probably not feasible in this case), as LO is partially in the RF band. (ver B) LO= tunable from (2110+1930)/2=2020 to 2020+60=2080 MHz IF fixed at 90 MHz. ...and the second solution with IF big and LO small. 2. Balanced mixer is more complicated, harder to construct on very high freqencies (poor design may result in unbalanced). However, (when ideally balanced) it has some nice properties: - no DC component at the output - no 2xLO or 2xRF or other spurious components - no LO leakage to the input or output. The text about "better noise immunity" is about audio mixers, not RF mixers. 3. on (A-B) road We have a wave from our front and another from our back. Our v=10 (verA) or 20 (verB) m/s, so the Doppler shift is 10/3e8*1.8e9=60 Hz (or 20/3e8*0.9=60Hz). Front wave (one from R1) is shifted to Fc+60Hz, back wave to Fc-60Hz. At the point D, two waves hit us from the back at 30 degrees - so both are shifted down, by 10/3e8*1.8e9*cos(30deg) (verA) which is 60Hz*sqrt(3)/2=52 Hz; (both verA and verB) The constructive interference will be whenever the distances are equal or differ by \lambda. Assuming R1 and R2 reflect exactly equal power: strongest interference will be when distances are equal, because the waves will be equal amplitude. So, middle between A and B is a candidate (on C-D road we are far from source -> interference is 100%, but waves are weaker) The destructive interference is when waves are 180deg off, so it will be e.g at \lambda/4 from the midpoint - as the distances (and amplitudes) via both paths are almost equal, so the result is almost zero. 4. What really counts in detection, is the ENERGY level at the MF output. E1=const*(Pxmit*Stgt*B*Ti)/(r^4) (verA) If we want the same energy from 16 times longer Ti, we can increase r by 2 so new Ti/(r^4) is the same as before. (verB) If we want the same energy from 16 times shorter Ti, we must decrease r by 2. 5. Again, the energy counts. For PDR, energy emitted in one pulse is 20e-6 [s] *500 [W] =0.01 [Joule]. One pulse is integrated in MF. For FMCW, energy emitted in one modulation cycle is 1e-3 [s] * 20 [W] = 0.02 [Joule]. One cycle is integrated in range FFT. The integration from pulse to pulse or from cycle to cycle will give additional improvement, but the improvements are equal in both radars (under e.g. the same antenna design and rotation parameters), so we may cancel this effect. (verA): FMCW is stronger. (verB): Pulse: 10e-6*500=0.005, FMCW: 1e-3*1=0.0001 -> Pulse is stronger. 6. If the signal x(t) is gaussian pulse, then the MF h(t) is x(-t); with symmetrical pulse it is the same. So, output is x(t) convolved with itself. A gaussian pulse convolved with itself is ... again a gaussian pulse! (but 2x longer one). 7. (verA) When antenna moves past the object, the distance first diminishes, then starts to increase - in approximately quadratic manner. Thus, the phase of the reflected signal is also quadratic function of time. Frequency is the derivative of phase -> it is linearly changing, giving a chirp in the demodulated signal. (verB) When antenna moves past the object, it first approaches it, then retreats. Thus, the Doppler frequency changes from positive, then negative. This gives us a chirp in the demodulated signal. (bothA&B) Chirp is perfect for a MF processing, so it gives a sharp peak -> this way we get sharp image in along-track dimension. 8. Multipath defines minimum GI. (verA&B) 5e3 [m]/3e8 [m/s] = 17 us Doppler defines minimum subcarrier distance - we may have a situation as in problem 3, with two paths of opposite Doppler shifts, so they induce Inter (sub)Carrier Interference. Doppler shift: (verA) 100[m/s]/3e8[m/s]*4.8e9[Hz]=1600Hz (verB) 100[m/s]/3e8[m/s]*2.4e9[Hz]=800Hz SC spacing: We choose as narrow spacing of subcarriers as possible (to pack many of them into our B-width). Doppler however limits us: (verA) 1600 Hz is to be < 5% of spacing -> spacing >32 KHz -> max 15e6/32e3=468 subcarriers (verB) 800 Hz is to be <10% of spacing -> spacing >8 kHz -> max 15e6/8e3 = 1875 subcarriers With 32 (or 8) kHz spacing, the symbol time must be 1/16 (1/8) ms. Add the GI of 1/60 ms, you'll get how many seconds we use per symbol. Subtract 10% for pilots and sync. (verA) 16QAM - 4bits*468subcarriers/(1/32000+1/60000)*90%=21.3 Mbit/s (verB) 4QAM - 2bits*1875subcarriers/(1/8000+1/60000)*90%=23.8 Mbit/s NB: There is a WRONG solution hanging in the web, where the author tries to divide Fc over B, i.e. packs many B-width channels between f=0 and f=Fc. I just wonder - how would you transmit in such a wide band? And what would FCC, ITU, and UKE say about eating whole band which should serve LW, MW, FM, TV and GSM? In OFDM we pack many B/N subcarriers INTO the band (Fc-B/2 to Fc+B/2), not many B into (what?). 9. near-far occurs in uplink of CDMA. Far user transmission is detected with a MF (matched filter) for his pseudonoise, but the near (strong) transmission is not fully orthogonal - so it "leaks" through the MF appearing as noise at the QPSK or QAM decoder. If it is strong enough, it disturbs the decoding (demodulation). In downlink, powers are equal (one transmitter). In other MA methods, users are spaced in time (TDMA) or frequency (FDMA, OFDMA etc.) (or both - e.g. GSM). 10. Primary need for GI is because of multipath propagation (this is important both in uplink and downlink). The most delayed echo should not cause InterSymbolInterference - so symbols must be spaced by GI. A secondary problem (only in uplink) is lack of synchronization between mobile stations. They derive timing from downlink, but downlink is delayed depending on the distance - so ther is some sync, but not very exact. This is strong when beginning the conversation (RACH channel - thus very long GI needed here ), later it is made better by feedback from base station - anyway, GI between uplink slots accomodates nonidealities in ME sync. 11. One pair indicates an ellipsoid. Two pairs - intersection of two ellipsoids (circle or ellipse). Third pair may give 2 points, four pairs are needed for full unambiguous 3D location. 12. CDMA codes are not 100% orthogonal because there is a limited number of orthogonal codes with given length, nonsensitive to delay. If we need more, we choose "almost orthogonal" ones. Multipath and Doppler spoil the codes again. The downside is the ICI (where "C" is "a code channel"), resulting in MAI, which gets more annoying due to near-far effect.