Problem Set Solution 16: Sampling

16 Sampling Solutions to Recommended Problems S16.1 If wo = 7r X 10', then cos(won X 10-3) = cos(irn) Similarly, for wo = 31 X 10-3 = and wo = 57 X 10-3, cos((on X 10-3) = (-1)" S16.2 The sampling function p(t) = (t - nT), T = 13, has a spectrum given by =o 2r P(co) 67r = -2rk a WYE~k ( (w - 61rk), shown in Figure S16.2-1. P(W) 67] -127r 67T 0 -67r 121r Figure S16.2-1 cos(wot) has a spectrum given by rb(o S16.2-2. wo) + rb(o + wo), shown in Figure cos (CO0 t) F r S 1. 2 Figure S16.2-2 S16-1 Signals and Systems S16-2 From the convolution theorem 1 = 2wr P(w) * [irb(w ­ wo) + gr(w + wo)] Hence, it is straightforward to find X,(x). (a) (i) For wo = 7r: 37r -7n -6n -5r - 57 nT 67 77 Figure S16.2-3 (ii) For wo = 2r: 37r -81r - 6 7T 4 - 7r -2In 2 7T Figure S16.2-4 (iii) For wo = 31r: 4 1r 67r 87r Sampling / Solutions S16-3 (iv) For wo = 5r: Figure S16.2-6 (b) From part (a), it is clear that (i) and (iv) are identical. S16.3 The signal x(t) = cos(wet + 0), where wo = 2wfo, can be written as x(t) = 0 lei'eiot + le -' e -jWOt and the spectrum of x(t) is given by X(w) rej'"(w - wo) + re ~j'0 (w + wo) = The spectrum of p(t) is given by 2rk) P(W)= 2 T k T, -0 Therefore, the spectrum of x,(t) is 2Tk - X,(W) = / Wo+ e6e L e -job (W - 2 rk+ and the spectrum of X,(W) is given by = H(w)X,(w) X,(w) (a) wr wo = 2r X 250, X,,(w) = Tr 0 =-) 4, [e'b(w Tk= - T = 10- , 2w X 10'k - 21 X 250) - + e -j"S(w - 2 1 X 103k + 27 X 250)] Hence, only the k = 0 term is passed by the filter: X,(w) = w[ej'5(w - 2w X 250) + e -i"S(w + 21 X 250)] and X,(t) = 1 ei'ei2 x 25 ot + 1 e -'e 2 = cos (2w X 250t + 0) 2 = cos (21 X 250t + -22rX25o ) Signals and Systems S16-4 (b) wo = 2w X 750 Hz, T = 10-, 2 >" [eb(w - 2w X 103 k - X,(w) = + e -j'b(w Only the k = 2 r X 750) r X 103 k + 2 x X 750)] 1 term has nonzero contribution: 2 X,(w) = r [ej"b(w + 27 X 250) + e ~ib(w T r X 250)] Hence, X,(t) = cos (2w X 250t - (27 = cos (c) 27 X 500, wo [ej0b(w - 2 X,(w) = ~ r + e -j"b(w - X 250t - T = 10- , = 0) , X 10 3 k - 2 r X 500) 2 7 X 10 3k + 2 r X 500)] Since H(w) = 0 at w = 2r X 500, the output is zero: x,(t) = 0. S16.4 (a) x,(t) = = x(t) 6(t - x(t) :b(t L n =- 2An) - x(t) 6(t - 2An) - o n -o A - A - 2An) 2An)I t By the convolution theorem, X,(w) = 1 X(w) 1 27 2wr 6 ( 2 X(w) * - * T 2An _o = X(W)* - -n 2-7 -- ) e w,, 12(An)fl) - n7 X,(w) is sketched in Figure S16.4-1 and Y(w) is sketched in Figure S16.4-2. Sampling / Solutions S16-5 X,(w) _3ff -r 3L7r 7r n 3n2n + , 2n 3n Figure S16.4-1 Y(W) Fi S1.4ff­ Figure S16.4-2 (b) xP(t) X x(t) CJM -WM cos("f) Figure S16.4-3 (c) 2A yt x(t) X -CJM liM cos(Z) Figure S16.4-4 (d) A is maximum when ir/A is minimum. From part (a) we see that aliasing is avoided in X,(w) if wm zr/A. Hence, Amax = */Wu. Signals and Systems S16-6 S16.5 (a) The transform of the sampled function appears as in Figure S16.5-1. Passband -2n __n_ T 2T IW x21T 0 2T T Figure S16.5-1 Hence, (a) matches (i). (b) Passband 2" T "T "2T 0 I 2T 21r T T T Figure S16.5-2 Hence, (b) does not match any. (c) Matches (ii). (d) M MKY7 M 2T 2T Figure S16.5-3 Hence, (d) does not match any. S16.6 Since the input x,(t) cannot be distinguished for certain values of w,the output also should not be distinguishable for certain values of w. Hence, Q(w) must be periodic in w. Therefore, Figure P16.6-3 is a possible candidate, but Figure P16.6-2 is not. Sampling / Solutions S16-7 Solutions to Optional Problems S16.7 P(w) ... 4-A _ 21T .. 4r 2w 0 T T T T Xp(w) NA 4 T 0 NA NA NAa N A 1 W2 2T 2n T W2 Figure S16.7-1 Note that as T increases, (21r/T) - W2 approaches w = 0. Also, there is aliasing when 2xo - C 2 < (2r/T) - w2 < C 2. If 2 w1 - W2 >---0 (as given in the problem), then it is easy to see that there is no aliasing when 0 ! (21r/T) - W2 ! 2oI - C 2 . For maximum T, we choose a minimum allowable value of 21r/T: = - x2 Tmax = Tax 21, W2 which is sampling at half the Nyquist rate. X,(w) for this case is given in Figure S16.7-2. XP(W) _ 2 T 0 2T -T 2 2T T Figure S16.7-2 Hence, A = T, wb = 27r/T, Wa = WI Signals and Systems S16-8 S16.8 We are given the system shown in Figure S16.8-1. xix(t) x(t) - X2 (t) )(- H, (c) X, (t) p(t) eIwot Figure S16.8-1 (a) X(o) and Xi(o) are as shown in Figure S16.8-2. X 2(w) is as shown in Figure S16.8-3, and X,(o) is therefore as given in Figure S16.8-4. X(W) X1(Go) I W- 2 Wl W ________W, w2 (W1 - W2) 2 (W 2 - 1 ) 2 Figure S16.8-2 X 2 (o) 1W­ 02) (W2 2 -- W1) 2 Figure S16.8-3 Xp(w) 1 T T T' Figure 516.8-4 Sampling / Solutions S16-9 (b) 2x/Tmax equals the Nyquist rate for X 2 ((o): 2(W2 - Wi = 2 W2 - WI Hence, 27 Tmax (W2 - wi) (c) g X(t) x2TH1(o) Re x xP(t) e -'Ot Figure S16.8-5 S16.9 The composite waveform spectrum is given in Figure S16.9-1. We can alias the noise region to get maximum T. This corresponds to the aliased spectrum, shown in Figure S16.9-2. -2W -W 4W 3W 2W W Figure S16.9-2 0 The value of T is given by w-= 3W- T The value of A is Tmax for y(t) = x(t). Tax =­ "'' 3W 5W Signals and Systems S16-10 S16.10 The spectra of x, 2(t), where T = ir/W, given in Figures S16.10-1 and S16.10-2, could have generated x,(t): X 1 (O) T-­ -w 6 W W W W 2 6 6w 6 2 Figure S16.10-1 X 2 (O) 22 2 2 Figure S16.10-2 S16.11 (a) From the sampling theorem, 27r/T >_2W. Hence, 7r W 7Tr Tmax W Since X,(W) = Tk X k , we require A = T for x,(t) = x(t). The minimum value of We is W so that we do not lose any information, and the maximum value of W is (27r/T) - W to avoid periodic spectral contribution. (b) (i) X(w) = 0 for IwI > W Hence, T.ax (ii) TmT = I , A = T, W W< W < T - W X(w) = 0 for coI > 2W. Hence, T. = , A = T, 2W 2W< W < T - 2W Sampling / Solutions S16-11 (iii) X(w) = 0 for |coI > 3W Hence, Tmax = (iv) 3W' , 3W< We < A = T, T - 3W X(w) = 0 for w I > W/10. Hence, Tax - "W A=T, '10 W - 27r T W -1