16 Sampling. Solutions to Recommended Problems. a WYE~k. P(co) 2r =o -2rk S16.1

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Christopher Mills
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1 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. he sampling function p(t) = (t - n), = 13, has a spectrum given by P(co) r =o -rk a WYE~k = 67r ( (w - 61rk), shown in Figure S P(W) 67] -17r -67r r Figure S16.-1 cos(wot) has a spectrum given by rb(o - wo) + rb(o + wo), shown in Figure S16.-. cos (CO 0 t) F r S 1. Figure S16.- S16-1

2 Signals and Systems S16- From the convolution theorem 1 = P(w) * [irb(w wr Hence, it is straightforward to find X,(x). (a) (i) For wo = 7r: wo) + gr(w + wo)] 37r -7n -6n -5r - n Figure S (ii) For wo = r: 37r -81r r -In 7 4 1r 67r 87r Figure S16.-4 (iii) For wo = 31r:

3 Sampling / Solutions S16-3 (iv) For wo = 5r: Figure S16.-6 (b) From part (a), it is clear that (i) and (iv) are identical. S16.3 he signal x(t) = cos(wet + 0), where wo = wfo, can be written as and the spectrum of x(t) is given by he spectrum of p(t) is given by x(t) = lei'eiot + le -' 0 e -jwot X(w) = rej'"(w - wo) + re ~j' 0 (w + wo) rk) P(W)= k -0, herefore, the spectrum of x,(t) is X,(W) = / L e 6 e k - Wo+ e -job (W - rk + ) and the spectrum of X,(W) is given by X,(w) = H(w)X,(w) wr (a) wo = r X 50, 0 =-) = 10-, 4, X,,(w) = r [e'b(w - w X 10'k - 1 X 50) k= - + e -j"s(w - 1 X 103k + 7 X 50)] Hence, only the k = 0 term is passed by the filter: and X,(w) = w[ej'5(w - w X 50) + e -i"s(w + 1 X 50)] X,(t) = 1 ei'ei x 5 ot + 1 e -'e -rx5o = cos (w X 50t + 0) = cos (1 X 50t +

4 Signals and Systems S16-4 (b) wo = w X 750 Hz, = 10-, X,(w) = >" [eb(w - w X 10 3 k - r X 750) Only the k = + e -j'b(w - r X 10 3 k + x X 750)] 1 term has nonzero contribution: X,(w) = r [ej"b(w + 7 X 50) + e ~ib(w - r X 50)] Hence, X,(t) = cos (w X 50t - 0) = cos (7 X 50t - (c) wo 7 X X,(w) =~ 500, =, = 10-, [ej 0 b(w - r X 10 3 k - r X 500) + e -j"b(w - 7 X 10 3 k + r X 500)] Since H(w) = 0 at w = r X 500, the output is zero: x,(t) = 0. S16.4 (a) x,(t) = x(t) 6(t - An) - x(t) 6(t - A - An) = x(t) :b(t - An) - Ln =- o n -o t - A - An)I By the convolution theorem, 1 wr X,(w) = X(w) 6 ( 1 * X(w) * - 7 An _o = X(W)* - -n ) e w,, 1(An)fl) - n7 X,(w) is sketched in Figure S and Y(w) is sketched in Figure S16.4-.

5 Sampling / Solutions S16-5 X,(w) _3ff -r 7r 3L7r 3nn n +, n 3n Figure S Y(W) (b) Fi S1.4ff Figure S16.4- xp(t) X x(t) -WM CJM cos("f) Figure S (c) yt X x(t) A -CJM lim cos(z) Figure S (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.

6 Signals and Systems S16-6 S16.5 (a) he transform of the sampled function appears as in Figure S Passband -- IW n n_ 0 x1 Figure S (b) Hence, (a) matches (i). Passband " "- "- 0 I 1r Figure S16.5- (c) (d) Hence, (b) does not match any. Matches (ii). M MKY7 M Hence, (d) does not match any. Figure S 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. herefore, Figure P is a possible candidate, but Figure P16.6- is not.

7 Sampling / Solutions S16-7 Solutions to Optional Problems S16.7 P(w) A _ 1 0 w 4r NA NA Xp(w) NA NAa N A W n W Figure S Note that as increases, (1r/) - W approaches w = 0. Also, there is aliasing when xo - C < (r/) - w < C. If w 1 - W >---0 (as given in the problem), then it is easy to see that there is no aliasing when 0! (1r/) - W! oi - C. For maximum, we choose a minimum allowable value of 1r/: - = x max = 1, ax W which is sampling at half the Nyquist rate. X,(w) for this case is given in Figure S XP(W) _ 0 - Figure S16.7- Hence, A =, wb = 7r/, Wa = WI

8 Signals and Systems S16-8 S16.8 We are given the system shown in Figure S x(t) - )(- xix(t) H, (c) X (t) X, (t) eiwot p(t) Figure S (a) X(o) and Xi(o) are as shown in Figure S X (w) is as S16.8-3, and X,(o) is therefore as given in Figure S shown in Figure X(W) I W- W, X1(Go) Wl w (W 1 - W) (W - 1 ) Figure S16.8- X (o) 1W 0) (W -- W1) Figure S Xp(w) 1 ' Figure

9 Sampling / Solutions S16-9 (b) x/max equals the Nyquist rate for X ((o): (c) Hence, (W - Wi = W - max 7 (W - wi) WI xp(t) x Re g xh1(o) X(t) e -'Ot Figure S S16.9 he composite waveform spectrum is given in Figure S We can alias the noise region to get maximum. his corresponds to the aliased spectrum, shown in Figure S W -W 0 W W 3W 4W 5W Figure S16.9- he value of is given by he value of A is max for y(t) = x(t). w-= 3W- ax = "'' 3W

10 Signals and Systems S16-10 S16.10 he spectra of x, (t), where = have generated x,(t): ir/w, given in Figures S and S16.10-, could X 1 (O) - -w W W W 6 6 W 6w 6 Figure S X (O) Figure S S16.11 (a) From the sampling theorem, 7r/ >_W. Hence, Since 7r 7r W max W X,(W) = k X k, we require A = for x,(t) = x(t). he minimum value of We is W so that we do not lose any information, and the maximum value of W is (7r/) - W to avoid periodic spectral contribution. (b) (i) X(w) = 0 for IwI > W Hence, (ii) m.ax = I, W X(w) = 0 for coi > W. Hence, A =, W< W < - W. =, A =, W< W < - W W

11 Sampling / Solutions S16-11 (iii) (iv) X(w) = 0 for coi > 3W Hence, max =, 3W' X(w) = 0 for w I > W/10. Hence, ax - "W A =, 3W< We < A=, W - '10 7r - 3W W -1

12 MI OpenCourseWare Resource: Signals and Systems Professor Alan V. Oppenheim he following may not correspond to a particular course on MI OpenCourseWare, but has been provided by the author as an individual learning resource. For information about citing these materials or our erms of Use, visit:

8 Continuous-Time Fourier Transform

8 Continuous-Time Fourier Transform Recommended Problems P8. Consider the signal x(t), which consists of a single rectangular pulse of unit height, is symmetric about the origin, and has a total width