# hw9soln - 4. (a) The Nyquist rate for the given signal is 2...

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Unformatted text preview: 4. (a) The Nyquist rate for the given signal is 2 x 5000a = 100007r. Therefore, in order to be able to recover a:(t) from 3pm, the sampling period must at most be Tm“ Tﬁr—Oﬁ— = 2 x 10‘4 sec. Since the sampling period used is T = 10'“1 < Tm”, :r(t) can be recovered ' from xp(t). (b) The Nyquist rate for the given signal is 2 x 150001r :1 300001r. Therefore, in order to be able to recover a:(t) from :rp(t), the sampling period must at most be Tm” gig—35; = 0.66 x 10““4 sec. Since the sampling period used is T = 10"4 > Tm“, z(t) cannot be recovered from 22,,(t). (c) Here, Im{X(jw)} is not speciﬁed. Therefore, the Nyquist rate for the signal ac(t) is indeterminate. This implies that one cannot guarantee that $(t) would be recoverable from 23,, ((1) Since z(t) is real, we may conclude that X (jw) = 0 for [w] > 5000. Therefore, the answer to this part is identical to that of part (a). (e) Since a:(t) is real, X (jw) = 0 for M > 150001r. Therefore, the answer to this part is identical to that of part (h). (f) If X (jw) : 0 for |w| > m, then X (jw)=l=X (jw) = 0 for lwl > 2%. Therefore, in this part, X (jw) = 0 for [w] > 750011: The Nyquist rate for this signal is 2 x 75007r = 150001r. Therefore, in order to be able to recover a:(t) from 2:1,“), the sampling period must at most be Tm 133-355 = 1.33 x 10“4 sec. Since the sampling period used is T = 10“4 < Tm”,$(t) can be recovered from 1:1,(t). (g) If |X(jw)| = 0 for w > 5000s, then X (jw) = 0 for w > 500011: Therefore, the answer to this part is identical to the answer of part (3.). Using the properties of the Fourier transform, we obtain YOU) = X1 (jlezle- Therefore, Y(jw) = 0 for [w] > 1000s. This implies that the Nyquist rate for y(t) is 2 x 10001r = 20001r. Therefore, the sampling period T can at most be 21r/(20007r) 2: 10“3 sec. Therefore we have to use T < 10"3 sec in order to be able to recover y(t) from yp(t). 7.29. From Section 7.1.1 we know that Xpuw) =% 2 mo» — law/T». kzz—oo X(ej‘“), Y(ej“’), Yp(jw), and Yc(jw) are as shown in Figure 87.29. IN“) = TPUWT) "I OTT K mils" 3mm“L ("I’M —' ‘4 Limo“ If/lxlo Figure $7.29 7.35. (a) The signals 17,,[n] and uh] are sketched in Figure 87.35. “1) [H] , X48“) ‘ﬁ‘ “.5 -_1l'_ 0 w {I Q Wt. .15} Ir. Figure$7.35 H: \Private\BME499\I—IW7__6 .m Page 1 March 25, 2004 2:31:17 PM % Hw solution for BME 499.098/311 HW #7 96 . 95 problem 6 let T = l % Adequately sampled, using A = 2 clear; ' A = 2; n = [v32z31]; w = n*2*pi/64; x = (l/A)*sinc(n/A); y = fftshift(fft(x)); sub lot(2,2,l),plot(w,abs(y)), xlabel(‘w (from ~\pi to \pi)'), ylabel(‘lx(w)l'), title('6 1 . Magnitiude of X(w), using A = 2 (adequately sampled)'); subplot(2,2,2),plot(w,angle(y)), xlabel('w (from —\pi to \pi)‘), ylabel('angle of X(w)'), 1 title('6 Angle of X(w), using A = 2 (adequately sampled)'); % Adequately sampled, using A = 0.6 clear; A=0.6; n = [-32:31]; w = n*2*pi/64; X = (l/A)*sinc(n/A); y = fftshift(fft(x)); subplot(2,2,3),plot(w,abs(y)), Xlabel('w (from —\pi to \pi)‘), ylabel('1x(w)|‘), title('6 t . Magnitiude of X(w), using A = 0.6 (inadequately sampled)'); subplot(2,2,4),plot(w,angle(y)), xlabel('w (from ~\pi to \pi)‘), ylabel('angle of X(w)'), 1 title(‘6 Angle of X(w), using A = 0.6 (inadequately sampled)'); X(u) 6 (Q) ws~$>1§ ’3? § ”$ ~F < 1% 042%uab%j.a¢m¢4%%j MT=\ 6. Magnitiude of X(w), using A = 2 (adequately sampled) 6 Angle of X(w), using A = 2 (adequately sampled) 1.4 4 , —,—— __._ 1.2 1 2 E 2 0.8 x -- 0.6 2:72) % 0.4 -2 0.2 O l _4 l 1 _.L. -4 -2 O 2 4 -4 —2 0 2 4 w (from ms to n) w (from -n to 1:) 6. Magnitiude of X(w), using A = 0.6 (inadequately sampled) 6 Angle of X(w), using A z 0.6 (inadequately sampled) 2‘2"““" 4"’*O 2 1.8 A 2 E z 1.5 x E ‘5 o 2‘— 1.4 #3, 5% 1.2 -2 1 0.8 L”— _‘ 1 -4 x _._ -4 -2 0 2 4 -4 -2 0 2 4 w (from ~75 to 7:) w (from m: to n) s; C:\Documents and Settings\Douglas Noll\My Documents\Cl. . .\hw7.m Page 1 March 22, 2004 9:26:26 AM M hw7, #7 = 7.5; = [-64:63]; = abs(n) < A/z; X5330“ Z KNEW) = A ~ SCnc H omc [-pi:.01:pi]; XC = A*sinc(A*omc./(2*pi)); Xd = real(ft(x)); omd = n*2*pi/128; Xdapprox = Xc + A*sinc(A*(omc+2*pi)./(2*pi)) + A*sinc(A*(omc-2*pi)./(2*pi)); plot(omc,Xc,‘-.',omd,xd,':',omc,Xdappr0x,‘-') legend('cont FT','disc FT','sum of cont FT') —-—‘~ cont FT disc FT -—~ sum of cont FT ...
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## This note was uploaded on 04/30/2008 for the course BIOMEDE 311 taught by Professor Steel during the Winter '06 term at University of Michigan.

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hw9soln - 4. (a) The Nyquist rate for the given signal is 2...

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