hw1soln - BME 3H l . - b 2 )(m): 35m [NW/o) (Mm/Mme...

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Unformatted text preview: BME 3H l . - b 2 )(m): 35m [NW/o) (Mm/Mme Age/rtoot N , Sin RNA) 0 IL XCh)= N 0T VQ/rtaw‘c C . L PW Sta/mi / I. WW SFg/mJL , ’9' I m): 35m (ZN/o) PQ/VFGIOUC (I U Hl/Jfirl % % problem “ Clear; nl : n2 = 2 . c. 11- W94 wig/wk Pov=° Ew= J‘” WWW”? )ar 0 20 = 5 Nmmwks’l = [—15:15,; [zeros(1,15) 0:10 zeros(l,5)]; X = 3*sin(2*pi*nl/10); Y = sin(2*pi*n2/20); subplot(3,2,;),stem(n1,X), Xlabel('n’), ylabel(‘x(n)'), title(‘la ‘); subplot(3,2,2),Stem(n1,Y), Xlabel('n‘), ylabel('x(n)'), title(‘la ); % problem 2 t1 = [~15:O.l:15]; t2 = [—15:l:15]; X1 = 3*sin(2*pi*tl/10); X2 = 3*sin(2*pi*t2/10); Y1 = sin(2*pi*tl/20).*[zeros(l,150) ones(l,lOl) zeros(l,50)]; Y2 = sin(2*pi*t2/20).*[zeros(l,15) ones(l,ll) zeroS(l,5)]; subplot(3,2,3),plot(tl,Xl), Xlabel(‘t'), ylabel('x(t)‘), title(‘2 ) dt=0.l'); subplot(3,2,5),plot(t2,X2), Xlabel(‘t'), ylabel('x(t)‘), title('2 ) dtil'); subplot(3,2,4),plot(tl,Yl), xlabel('t'), yLabel('X(t)'), title('2 dt=0.l'); subplot(3,2,6),plot(t2,Y2), Xlabel('t'), yLabel(‘X(t)'), title(‘2 dt=l‘); 1.21. The signals are sketched in Figure 81.21. 7:6: 4) 2. ’43-‘15) 2. l 4 2. t o L 2. 3 * oi ~l [z(+)+x(—t)]u1+) 3 .0“! 3c .. O ' 3/2. a. a. a . a . %%JXAMrOMS 11d AUMMJ QPFYGmcflb in :1. X(n) = 35in(2\pin/10) x(n) = Sin(2\pin/20)' X(t) = 35in(2\pit/10 X(t) = 3sin(2\pit/10 X(t) = sin(2\pit/20) X(t) = sin(2\pit/20) K K 1 j 1 s l i 1a. x(n) = 33in(21m/1O) -. - II‘I n 2a. x(t)=331n(2m/10) dt=O.1 -15 -10 -5 O 5 1O 15 t 2a. X(t)=351n(2n:t/10) dt=1 1a. x(n) = sin(21m/2O) -15 ~10 -5 O 5 10 15 n 2a. x(t) = sin(2m/20) dt=0.1 -15 -1O —5 O 5 10 15 104—4] xfanJ V» (/1. 1 o 1 n 4 0 vi -l/L 4 (0L) adorn] 1fn]-u[M-fl = 1M , , , i l i , o I 2. n (e) (h) 1.34. (a) Consider 2 min] = 2401+ Zorn] + cc[~n]} n=—oo 17.21 i If :13[n] is odd, x[n] + a:[— ] == 0. Therefore, the given summation evaluates to zero. E (b) Let y[n] = m1[n]xg[n]. Then y[—“] = wli-niwzi—ni = —$1[ni$2ini = *yini- This implies that y[n] is odd. ((3) Consider Z + main]? M HM g H n:——oo n=—oo i 00 oo oo = 2 $311.] + Z + 2 Z $e[n]:ro[n]. n=—oo 122—00 722—00 Using the result of part (b), we know that $e[n]a:o[n] is an odd signal. Therefore, using V the result of part (a) we may conclude that 2 Z me[n]:1;a[n] 2: 0. n=—oo Therefore, CD 00 00 Z m2[n]== Z $2[n]+ Z “3’00 713*00 112—00 ((1) Consider 00 CO 00 00 / m2(t)dt= / z3(t)dz+ / x§(t)dt+2/ moment. ._00 -00 —oo ’00 Again, since xe(t):z:o(t) is odd, Therefore, ...
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hw1soln - BME 3H l . - b 2 )(m): 35m [NW/o) (Mm/Mme...

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