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# HW1 - 2006 Spring Quarter EE 113 HW#1 Solution Problem 1(a...

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Unformatted text preview: 2006 Spring Quarter EE 113 HW #1 Solution Problem 1. (a) ' I] 5 m 15 20 25 an 35 an time[msec} (b) x01):xa(nr)=xa[ij:2cos[2ooﬂi+£]:2ms[ﬂ+gj 600 600 4 3 4 x(n+N)= ZCO{W+£): Zcos[ﬂ+_‘Nﬂ+£] 3 4 3 3 4 . . . . N7: For x(n) to be penodlc 111 N, ( 1.3.x(n + N) = x(n) ), ——3— = 2?: => N = 6 The sam lin time T *— i - i(sec) p g ‘ _ F3 _ 600 . 6 l The perlod of x[n) : —(sec) = 10(m sec) "666100 Problem 2. (a) xa = cos(2t): f 2 —, TP 2 3r , periodic 1r (b) x(n.)=1+ 2cos[2¢m + Np : 1, periodic, although it is DC. (0) x(n) = sin(2n w 9:): N p = 7: , not integer, not periodic (d) 2 2 1 + cos[? + 27: x(n] = 3 oos[7m)+ sin[—m + E} — cos2 [E + it] = 3 00501113) 1 sin m 1 ﬂ] 5 6 4 5 6 2 1 N 2 N2 5 => Np = 2cm(2,5,4)= 20 N 4 H 3 Problem 3. (a) n 0.2 £1.21 13. £1.13 1 1.2 1.4 1. 1.13 2 1ima(see] (b) _ Wuém Level Level Value 2’s complement _ 3 10 on 2 6.67 010 1 ﬂ __ __ __3.33 001_ 0 o 000 ' -1 —3.33 -2 -6.66 110 —3 | -10 | 101 —4 | -1333 \ 100 anntized AID 2’s complement 6.67 \ 010 011 _ D 0.2 0.4 115 I13 1 1.2 1.4 1.5 1.8 2 timeisec) Problem 4. (a) y(t)310-cos[27r - 200: + E] - 003(221 - 200: + E] = 5-[1+ co{8007z -t + 5]] 4 4 2 (5 + 5 cos[8007r - t + g—jﬂt = 10 + 802:? [sin[8007rT + — sin[§D ---- -- (1) T: 6 :2}. =8.8004 T: 8 :21 =9.6563 T=16 32:9.5501 T: 26 :2} =9.7232 T=126 :21 =9.9429 (b) Because the second term in equation (I) -> 0 when T -> inﬁnity, Therefore A = 10. (c) We want sin[8007rT + = sing] = 1 :> 80032" = 2m, k = 0,1,2,3,. r = iJc = 0,12,“. 400 Using T = 2.5msec, the above = 10 Problem 5. ﬂ" 1"? (a) y1(n)=10-eos[0.4m +§ -cos(0.4m) = 5 005(2) + S 005(0‘8731 + y2(n):10-cos[0.4m +-si11(0.47m)= —5 \$1163] 55in[0.8m + N 2: L p 2N—l _ 1' z:—~ yn,z=— n =>'A= 2+2“ I N; 1() 2 Ngyx) 21 2 N25234:“) N:11:>31=10.0412 N=i9:>}31:9.7 (b) 9"(0'8”’+’3’i) = cos(0.8m + %)+ j - sin(0.8m + %) NJ] 1_ ejﬂﬁﬂ' 11185:: ":0 1 _ e . 2 “‘1 7r . 10 yyg 1—63-08”W :2: 21 ~ IleaEZy1(n)—10c03[2~]+1£1\$—§Re[e —10005 E 4 New N 1__e_;0.85r I 2 N—I A ’ ﬂ _ 10.3w I 2;: = 11111 b ht?) = —10\$1n[£] + hm l—qlm[e“% = —lOsm[— (c) If N=muitiple of 5, then 1 — ew's’w : 0 is always true. This cause 21 : 10003615}:2 = ~103in[%] :> = 212 + z: = 10 to be exact. So N=5 is the smallest N which will make estimate of A to be exact. Problem 6. EMU-87m + %l=1m[Ee‘“‘-W%ll=Im[e~r% = haw “=0 5:0 E 005(082171 + %)= Re[§€j{0'sm+%)] 2 Ree-31% -Eej'0'gm] = Re[ef% _ 1— ejaw] n=0 “=0 l [1(a) x4(nj imaginary 1 9 l E 1 0&0 I : ...
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HW1 - 2006 Spring Quarter EE 113 HW#1 Solution Problem 1(a...

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