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EE314HW1

# EE314HW1 - EE314 Homework 1 Chapter 2 21-1 2.1-2...

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Unformatted text preview: EE314 Homework 1 Chapter 2; 21-1, 2.1-2, 2.1-3 (a,b,c), 2.3-2, 2.4-1 (d,e.f), 2.4-3 (e)(f)(g)(h), 2.5-2, 2.5-4, 2.6-1, 2.3-2 (b,c,e), 2.9-1 (b,c,e), 2.9-2 4+10+6+8+6+8+8+1U*5=100pt5 2.1-1 Both tp(t) and wok) are periodic. The average power of <p{t) is P = %— 0719026) dt = i o” (e"/2)2 d; = 1’“frm”. The average power of 1120(3) is P5 = fj" wﬂt) (it : Tl" ff" 1 -dt = 1. 2.1-2 (3.) Since is a real signal, Ex = 162?:2ﬁ) dt. Solving for Fig. 32.1—2(a), we have ' E... = f§(1)2dt = 2, 3;, = f01(1)2dt + f12(,—1}2dt = 2 . Em = fo‘(2)2dt = 4, Ex... 2 L212? dz = 4 _ Therefore, Em =' Ex + E... Solving for Fig. 32.1—2(b), we have 139; =fo"(1)2dt+ff"(-1)2dt =21, E3, = D"’2(1)2dt+f:/2(—1)th+ffmuﬁdt +f3ij2(-1)2dt= 21? EM = 0"/2(2)2dz + [jg/2(0)? dt +f321f/20—2J2dt .—. 417 E24, : 7 0mm)2 dt + ﬁne? dt + fjfz’2(—2)2dt + fir/2(0)th = 473 Therefore, Emiy = E, + Ev. (a) (a) . (b) (c) X(!)-y(!) Fig. 32.1-2 EE314 Homework 1 (b) . '13: =fa"/4(1)2dt +' f;,4{~1)2dt:?r, Ev =f0”(1)2dt= 1r _ 3/4 2 7r ' 4 EH3, 3 G (2) emf,”4 (0)2031: = 71', EM, = 07‘” (0}2 dt +f7f/q1(—2)2 dt = 371- Therefore, Em, aé Eac +Ey. and Eiig = 3,3 2|: E9 are not true in genera}. ' 2.1-8 (a) ham Eq. [2.5a), the power of a. Sigma! of ampiitude C is P9 = regardless of phase and frequency; therefore, P9 = 100/2 = 50; the rms yaiue is_ ‘/P = 5J5. (b) Horn Eq. (2.5%)), the power of the sum of two sinusoids of different frequencies is the sum of the power of 2 2 individual sinusoids, regardless of the phase, 92’- + 951. therefore, Pg 2 100/2 + 256/2 2 50 + 128 = 178; the rms value is JP 2 V178. (c) gﬁ) = (10 + 2 sin (3t))cos {10:}:10 cos (101i) + 2 sin (30003 (10¢) = 10 cos(10t)+ sin (13\$) — cos(7t) Therefore, P9 : 100/2 + 1/2 + 1/2 = 50 + 0.5 + 0.5 = 51; the rms vaine is , [Pg = m. " 2.3-2 0!) Fig. 52.323 EE314 Homework 1 (b) Energy of 90:), °° 2 . 151 2 2“ 1 2 9 3 E9:]_Wg(t)di:/ﬁ [Eu—12)] dt+f15 [Ts—(344)] dt=E+4—=3 Since [see Problem 2.3-5], tim‘e shifting ortime inversion does not change the Signal energy. EM—z) = 3904-12) = Ego.) = 3 On the other hand, a. scaling of g(at) will change the signal energy to Eg/a [Problem 2.3—5], the energy of g(3t) is 1 and energy of 9(6 — 2t) is . 2.4-1 (d) 0 re) (5925-1953) 6m + a) = 5(2) = o (f) krﬂw) (use L’Hésﬁitél’s rule limeD = 1) 2.4-3 (8} 0 (f) 5 (g) 9(4) ' (h) ‘ on 00 . .3 ‘ /; cos — 5)6(2\$ w 3)d:c = 10° 603%(33 —— 5)6 (2(z — d3: 00 3 5 = é‘fwcosgﬁmmdcc— )dz=%cos-g (3—5)':\b< “‘00 Mag i Here, we used the fact 6(at) = [711T5(t)(see Probfem 2.3-2) 2.5-? (a) In this case Em : f0} di = 1, and 1 1 1 3d =— i 3di=-/ii=0.5 c EI/ngasu In EE314 Homework 1 (b) Thus, g(t) % 0.5::(tj. and the error e(t) :- t —0.5 over (0 s t S I), and zero outside this interval. Also E9 end E3 (the energy of the error) are ' l 1 1 E9 =/ ﬁnd: =/ 13sz =1/3 and Ee =f (z_0.5)2dr = 1/12 0 o o The error (E —' 0.5) is orthogonal to z(t) because ‘ f[(t—0.5}(l)dt=0 CI Note that ES. = 2EI + E2. To explain these results in terms of vector concepts, we observe from Fig. 2.15 that the error vector e is orthogonal to the component cx. Because of this orthogonality, the length~square of g [energy of g(t)] is equal to the sum of the square of the lengths of c -x and e {sum of the energies of ca:(t) and eft)]. 2.5-4 (3) In this case Es = f0l sin2 21rtdt = 0.5, and l 1 c = FEE-fa g{tJm(t)dt = tsin 21rtdt= --I/7r (b) Thus, 9(a) z -(1/7r)a:(t). and the error e(t) 2 13+ (l/vr) sin 2r: over (0 S t g i}. and zero outside this interval. Also E9 and EQ: (the energy of the error) are 1 1 E£7 =f 92(t)dt=f tzdt=1/3 0 0 and 1 e E6 =/ [i + (1/17) sin 2m]2 cit = % — mim- o 2712 The error [t +. f i /-:r) sin 2711] is orthogonal to x(t) because I / sin 2mi[t +(1/n)sin 2nt]dt = 0 o . Note that E9 = 2EE + E6. To explain these results in terms of vector concept we observe from Fig: 2.15 that the error vector 6 is orthogonal to the component c- or. Because of this orthogonality, the square of length of g [energy of g(t)] is equal to the sum of squares .of the lenghs ex and f [sum of the energies of ca:(t) and em} ' EE314 Homework 1 2.6-i We shall use Eq. (2.51) to compute pn for each of the four cases. Let us ﬁrst compute the energies of all lshe signals: 1 E: = f sin22wtdt = 0.5 a In the same way, we ﬁnd Es.1 = Eg2 = E93 2 Em = 0.5. From Eq. (2.51), the correlation coefﬁcients for four cases are found as follows: (1) (0 gm 5)- f01 sin 2m: sin 41rtdt= 0 (2) W f(5lﬂ 27ft) (—Slﬂ 27Tﬂdt = -1 (3) W f; 0.7mm Zntdt = 0 r- (4) ‘ . Van—anus) Signals :r(t) and 5320:) provide the maximum protectidn against noise. [ 350.707sm 2ntdt — fofs 0.707sin 211ml] = 2328/7: = 0.9 2.8-2 0,) To = 1077 , wo = 31—: I Because of even symmetry, all the sine terms are zero. 9(i) = qo+:%cos(§t)+bnsin(gt) an an 2 % _:cos(gt) dz ﬂﬁlﬂnﬁoi bn sin dz “3’ , 0 (integrand is an odd function of t) , 1 g (by inspection) El ﬁsh} I! Here b11 = D, and we all0w On to take negative values. Note that. On = an for n = O, 1, 2, 3, Fig. 328-20)} shows the plot of Cu. o 1 2 3 4 (b) 5 6 7 3 9 1o EE314 Homework 1 (c) To = 211', we = l, and with and 00 9(1) : (131+ Zan cos m. + bu Sinni. 11:] an = 0.5 {by inspection of the dc or average) 1 21ft l 2." t I 1 anz— —cosntdt=0. and bn=— —smmdi=—— ,T o 2” _ 1r 0 2r mm (a) — 05 I 'a 1' 2t+1'3t 15mm» g — .mﬂ smn +251n 35m +4_ - 05+1 cos(t+7r)+1cos(2t+ﬂ)+lcos(3t+w)+ _ ' 1r 2 2 2 3 2 The cosine terms vanish because when 0.5 (the dc component) is subtracted from 905), the remaining function has odd symmetr 0" and an. O5 IJ_.L.LL..L .1 .4 d .L 2 4 5 8 10 (c (e) To = 3, wo : 21r/3, and 1 1 1 a=— nit:— " 311A 6 8 2 1 21m 3 21m ern _ 21r%‘:l/ an _' E]; tcos Twit :- 21r2n2 [COS T + Tin-Sin T 1 ' 2 ,2 'bnx—w/ tSiﬂ—Etdt: 3 [sing—gq—mcosm] 0 Therefore, Cg : é and and Lg" cos Lg” — sin L” = m: 3 9“ ta” 5""+2—§—.’lsin-2"T"——l) COS T Figure S2.8—2(e) shows the plots of On and 6“. y. Hence, the Fourier series would contain dc and sine terms only. Figure 823—202) shews th plots of "} iii'iﬁiijﬂfiﬂ 10 EE314 Homework 1 2.9—1 Since Dn = 0.5(an —jbn), from an and 19,, calcufated in 2.8-2. jut (b)Tu =10ﬂ',0)0 =1/5, 903) = 2?:4. D918? 1 . mt where D0 = g; D" = *jbn) = 0.50.]1 = .71. St 0. m ((1)1?) : 273.3690 :1; 90:) = 21:02—00 Dnejnt where DO = 0.5; Dn = 0.5(071—1'13“): —jo.5bn = Ln: at 0. ' 27m (9) T5 : 3:500 = 9(3) = ZZZ—m Dada-m”3 where D0 = 1/6; Dn = 0.5(an —jbn) and £171 . 1311 are given in 2.8-2 (e). Or solving JDTl directly from the analysis equation , 1 1 . , . Dn : te"32“”‘/3dt : 3 [e—jQHn/S + 1) _ l] 47r2n2 EE314 Homework 1 0.2 Inn] 3 —1 0 -5 0 5 10 2.9—2 . . 211' 1r g(t} = 351nt + cos (3t -— + 2005 (8: + For a compact trigonometric form, all terms must have cpsine form and amplitudes must be positive. F03 mascm'I we rewrite g(t) as follows: 211' ﬁt) = 3coé (t — +cos (St a?) +2003 (815 + ...
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EE314HW1 - EE314 Homework 1 Chapter 2 21-1 2.1-2...

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