hw4_solution - Problem 6.3(a The closed loop transfer...

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Unformatted text preview: Problem 6.3 : (a) The closed loop transfer function is : : G(s)/s : C(s) : 1 1+G(s)/s s—l—G(s) 32+\/§s+1 The poles of the system are the roots of the denominator, that is — 2::w/2—4 1 1 P1,2=#=—E:J‘/§ Since the real part of the roots is negative, the poles lie in the left half plane and therefore, the system is stable. 3(8) (b) Writing the denominator in the form : D = s2 —|— Krone—kw: we identify the natural frequency of the loop as w” = 1 and the damping factor as g“ = 715 Problem 3.7 : In the non decision—directed timing recovery method we maximise the funcfion : 1'11:le = Zyif’rl with respect to 1". Thus, we obtain the condition : Suppose now that we approximate the derivative of the log—likelihood AL {1'} by the finite dilfer— ence : dym {T} air = I] sins} 1.. new} —r‘LL{—r— a} db" ”" 25 Then, if we substitute the expre-on of AL {1'} in the previous approximation, we obtain : dfiL [T] Em fiIET + '5} _ Em yin“- _ is} {IT 2:5 2—15; iUTfllgft _ mgr _ «r _ as): — (from: — mT — r +st¢)2] where gI[—t] is the impulse response of the matched filter in the receiver. However, this is the exprellon of the early—late gate synchronizer, where the lowpass filter has been substituted by the summation operator. Thus, the early—late gate synchroniser is a close approximation to the timing recovery systln. Problem 3.9 : {a} The wavelength .1 is : _99193 _9 _ 199 III‘19:“ Hence, the Doppler fiequency shift is : 9 199 9:91th 199 9:193 3-: 19 = :|:— = :l:— = :lz—H = 9:99.9999 H f” .9 13—, m 9 9: 9999 E z The plus sign holds when the vehicle travels towards the transmitter whereas the minus sign holds when the vehicle moves away from the transmitter. {b} The maximum difference in the Doppler fiequency shift, when the vehicle travels at speed lflfl kmfhr and f = 1 GHZ, is: afpm = 9,93 =195.1999 Hz This should be the bandwith of the Doppler frequency tracldng loop. {c} The maximum Doppler frequency shift is obtained when f = l GHZ —|— 1 MHz and the vehicle moves towards the transmitter. [n this case : 3 x 193 Jim, = — = 9.299? 199 + 199 m m and therefore : lfll] 1‘33 X = — = 92.9959 He f3“ 9.999? :>< 9999 Thus, the Doppler frequency spread is 3,; = Efflm = 135.3?05 Hz. Problem 3.15 : The received signal—plus—noise vector at the output of the matched filter may be represented as (see {5—2—53} for example) : r11 = x/S—sejtfln—dfl ‘l' N11 where {in = D,rr;'2,rr, Bxffl for QPSK, and 95 is the carrier phase. By raising r“ to the fourth power and neglecting all products of noise terms, we obtain : r: a (9.91)“ Ema—e + 9 [fir N, terse-mm] If the estimate is formed by averaging the received vectors {ri} over K signal intervals, we have the resultant vector U = Evie—if —|— ILEHK=1 N“. Let 9'), E £195. Then, the estimate of 95,, is : " __ _1Im[U]I ‘3'“— tfl“ Rem M.I is a complex—valued Gaussian noise component with zero mean and variance 02 = Nuffl. Hence, the pdf of 954 is given by {5—2—55} }w:here _1_E=Hv=fl .. 19(9x92] 19x99, 19m. 32 To a first approximation, the variance of the estimate is : 2 1 16 ...
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