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

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

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.

Page1 / 6

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online