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Chapter 2

# Chapter 2 - Chapter 2 2.1-1 Let us denote the signal in...

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Unformatted text preview: Chapter 2 # 2.1-1 Let us denote the signal in question by g(t) and its energy by By. For parts (a) and (b) 2: 1 2n 1 2w 5,: sin'rdt=— dt-— cos2tdt=1r+0=7r 0 2 0 2 0 4w 1 4w 1 (n (c) E,-:./ sin’m=—/ dt—— costht=1r+0=1r 2 2 2 2 1! 2w 2n 1 2x 1 21 (d) E, =/ ('2 sin f)? dt = 4 [3/ df. - 3/ cos 2t dz] = 4[1r + 0] = 41r o o " o ‘3 Sign change and time shift do not affect the signal energy. Doubling the signal quadruples its energy. In the same way we can show that the energy of 1:90) is 1:215”. 2.1-2 (3) a, = f:(l)2df = 2. E, = fo‘nﬁdmfﬂ—nzdz = 2 1 2 Ew=f (2)’dt=4. EH (2)”m=4 0 1 Therefore E”, = E, . Ey. .w 27: «[2 a 37/2 I 27 (ME. =/ (1)2dr+/ (4)7.” = 2n. E, =/ (1)’m+/ (—1)’dr+/ (l)’dr+/ t-l)zo’1 = ‘21r ‘ 0 t' 0 - w/2 '1 . 173/2 7/2 rig/2 2r Ew=/ (2)7dt+/ (0)2dr+/ (-1)2d!=47r 0 1/2 31!,1'3 Similaxly. we can show that 51-, = 411’ Therefore E”, = E“- + 5,. We are tempted to conclude that Eris! = E, - 15y in general. Let us see. arr/d w w (c) E,=j (1)2(It+/ (—1)2dt=1r £,=/ (wt/mu O n 0 l4 n/d 1.- 1/4 1: 51+, =/ (2):." +/ (07.1: = 1: EN, =/ (0W: +/ (4)2.“ = 3n 0 «,‘4 0 1/4 Therefore. in general E,“ 76 E, + E, 2.1-3 1 To 2 a 02 To = — ' t 9 z = — 1 2' t 29 J: P, To 0 C cos (we + )d 27.0 o l +cos( do + )] C2 To To C2 C2 = —- ~ _ ' . 0 = --—- = —— 2T0 [/0 (I! -t A coe (2‘20! + ‘2 )dt 2T0 [To + 0] 2 2.1-4 This problem is identical to Example 2.2b. except that on at m. In this case. the third integral in 1’, (See p. 19 is not zero. This integral is given by 2 ‘ a, 1/2 1:; = lim Cl ‘ cos (all + 01)cos (wit + 92”! 7—0» T _T/= C c 1/2 172 = lim coslal — 02):]! 4-] cos(2.u41r + 0, + 92):]! 472 -1‘/2 C1C2 T—m [7‘ cos(01 — 02)} -'r 0 = (THC; cos(0i - 02) ii if (4 Therefore (:2 2 p” r; —,)i + 95“- + am 605(91 - 92) 2.1-5 l 2 ~ 1 2 p9 =—. _ (1")2dr = 64/7 (8)119 = — (—13)2dt = 64/7 4 «2 4 -2 2 2 (b) P2y = % / (2r“)’dt = 4(64/7) = 256/7 ((1) PC, = % / (ct-"W: = 643/7 2 -2 Sign change of a signal does not affect its power. Multiplication of a signal by a conStant r increases the power by a factor I". 2.1-6 1 * —c/2 2 1 qr 4 1 ~1r (a) Pg=- (r )df=— v dt=-—[l—n ] 1.- 0 n o n 1 1r 2 1 1/2 (1,) Pg = .—- w (t) (I! = —— dt = 0-5 21r __" {Zr _*/2 , To/z 70/2 (c) Pr = 71- u-ﬁitwt = -T‘— I dr=1 ° -To/2 0 J—ro/z (d) P,= 1/ (i1)2dt=l 4 —2 ZR '7 1 l ‘ 1 (e) 'Pv’27 ‘0 (E) ""5 2-1-7 7/2 H TI'? 19 _ ‘ l l ‘r _ ' l ' 1(I'L'b'r)‘ Py — 7hr); T mug r011! — Z Z DkD rr df Ta ‘77: k=m r=m rns (when k aé r) are ﬁnite because the integrands are periodic signals The integrals of the cross-product ter d zero. The remaining terms (I: = 1‘) yield (made up of sinusoids). These terms. when divided by T —- oo. yiel 172 n n _ . 1 2 __ 2 Py-rli_r.rLT/ END“ 41.2 mil 'T/z ksm k=m 2.1-8 (3) Power of a sinusoid of amplitude C‘ is (“/2 [Eq. (2.63)] regardless of its frequency (a) 59 0) and phase. Therefore. in this case P = (10)’/2 = 50. (b) Power of a sum of sinusoids is equal to the sum 0 this case P= £313 + =17s. (c) (10 + 2 sin 3!)cos 101 = lOcos 10! -1- sin 13! — sin 3!. Hence from Eq. (2.6b) P = 10 2 + % + g = 51. (d) 10 cos 5! cos 10! = 5(cos 5! + cos 15t. Hence from Eq. (2.6b) P = (-22):- + 59g: = 25. (c) 10 sin Stcos 10! = 5(sin 15! — sin 5!. Hence from Eq. (2.6b) P = \$33: + '5‘: = 25. g [r-"°+~°>' + «Na-WW Using the result in Prob. 2.1.7. we obtain P = (1/4) + (1/4) = 1/2. " ‘0» E, =/ (e "Ml? dt =/ (2'2"! 4!! = 00 -- a. may l T/'-’ ' 2 1 7/2 2 I P = l' - ._‘ It = Ii —- (t— a dt = r 9 TE“ T /_T/2(r ) 4 73"” T [—172 x f the powers of the sinusoids [Eq. (2.6b)]. Therefore. in (f) r-’°' c0540! = 2.2-1 For a real It For imaginary a. let a = jrr. Then 1/2 1 m - 1 =I1i'anmx—/ (01“)(0-lﬂ)dt lim - / (It: 9 T -772 7—“: T . -172 Fig. 82.3-2 Clearly. if a is real. r‘" is neither energy not power signal. Hawever. if a is imaginary, it is a p0wer signal with power 1. 2.3-1 gg(r) = g(f — 1) + g;(f - 1). 9.10) = g(t - l) + 91(f+1), 94(1) = g(t - 0.5) + 910 —2 0.5) The signal 950) can be obtained by (i) delaying gli) by 1 second (replace 1 with t - 1). (ii) then time-expanding by a factm' 2 (replace f with #2). (iii) then multiply with 1.5. Thus 95(t) = 1.59% — l). 2.3-2 All the signals are shown in Fig. 52.3-2. 2.3-3 All the signals are shown in Fig, 52.3-3 2. Flg. 52-3-3 2.3.4 1 'N m m £_y=/ [—g(:)]’m=/ g2(t)dt=E,. E,(-.,=/ [g(—t))2dt=/ 92(m)d.‘t=E, . nix, -t‘; 'N ‘W X ‘ "‘4 I'll l 1. Est.-n= / {nu—“rim = / g’(:n)dr= 1-2.. Eg(.,)= / [9(at)l’dt= 3 / 99(1):!” Eg/a. Ey(a.,.o).=/ [9(01 — h)]2df = % / 92(T)(LT = 59/0. Egg/.3 =/ [,q(t/a)]2 dt = a/ 92(1) d! = aEg Eu“) =/ [(190)]? (If = 122/ 92(t)dt = 025,, 2.4-1 Using the fact that g(.7)ﬁ(:r) = g(0)h(:r). we ha‘vc (a) 0 (b) gm) (c) %h(1) (d) —§b(r — 1) (e) 5:11-55(J+3) (r) k6(w) (use L' Hopital‘s rule) 2.4-2 In these problems remember that impulse Mm) is located at z = 0. Thus. an impulse 6(t — r) is located at 'r = i. and so on. (a) The impulse is located at r = I and 9!?) at 'r = f is 9(t). Therefore 3 / g(r)b(t -— T)tl7' = g(t) (b) The impulse 5(7) is at ‘r = 0 and 9(1— 7) at r = 0 is 9(1). Therefore f“. 6(r)_q(t — 1')dr = 90) —m Using similar arguments. we obtain (c)1 (d) 0 (e) r” (f) 5 (z) 9(-1) 2.4-3 Letting n! = :r. we obtain (fora > 0) / 0(1)5(at)dt f.- i / ¢(%)6(1)d1=i—¢(0) may (h) -e2 —’D Similarly for a < 0, we show that this integral is —£~¢(0). Therefore [-~a>(f)b(al)dt = man) = m -0“ ¢(t)6(t)d1 Therefore Mat) = I‘ll—Ian) 2.5-1 Trivial. Take the derivative of le:2 with respect to r: and equate it to zero. 2.5-2 (8) [n this case E, =fo‘dr=1.and 1 ' 1 1 l r: gif}.’r(t)dt=;‘/o td1=0.5 (b) Thus. qlt) z 0.5.10). and the error r(t) = t — 0.5 over (0 g t S l). and zero outside this interval Also E, and E. {the energy of the error) are l l 1 Eg=/ 92(’)'H=/ '24t=1/3 and E.-=/ (1—0.5)2dt=1/12 0 0 D The error (I — 0.5) is orthogonal to .r(t) because 1 / (r - 0.5)(1m = o 0 Note that E, - ("£1 + [5,. 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 ax. Because of this orthogonality. the length-square of 3 {energy of 9(1)] is equal to the sum of the square of the lengths of ex and e {sum of the energies of {27(1) and (0)]. 2.5.3 In this case Eg = fo‘ 92(1)!” = fo‘ Fm. = 1/3. and l 1 “if :r(t)g(t)df=3/ rm=r5 E9 0 0 Thus. .r(!) z 1.590). and the error r-(t) = 1(1) — 1.590) = 1 -- 1.5t over (0 S f g 1). and zero outside this interval. Also E, (the energy of the error) is Ee = fol(1 — 1.5!)2 dt = 1/4. 2.5-4 (a) In this case 5, = [0‘ sinz21rhlt = 0.5, and l 1 (~=.._... 1 1 t..f1=— t' 2tdt=-—1 E: o g( )r( )d 0.5/0 srn 1r [r (b) Thus. 9(1) 3 —(1/1t):r(r). and the error 0(1) = t+ (l/n)sirr 27:! over (0 s t g 1). and zero outside this interval. Also E9 and E. (the energy of the error) are 4 .‘,,.~. l 1 l l 1 E,=/ 92(1).u=/ 12:11:1/3 and E.=/ [r—(l/r)sin21rt]2di=§--2-;5 O .o 0 The error If + (l/ir) sin 21H} is orthogonal to .10) because i / sin 21rflt + (1/1r)sin 21f} dt = 0 0 Note that E, = (‘25, + 113.. 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 ax. Because of this orthogonality. the length 0‘ f {energy 0‘ 9(1)] is equal to the sum of the square of the lengths of «:x and e [sum of the energies of nz(t) and «(1)1. (a) If 10) and Mt) are orthogonal. then we can show the energy of1:(t) :l: p/(t) is E: + 8,. m a; nu [unmiwwm=fnmwm+jiwwwi/ wwmmt/ Hmwmt =[wnmrm+/mmmﬁw The last result follows from the fact that because of orthogonality, tli :(r)y'(r) and :r'(t)y(r) are zero [see Eq. (2.40)]. Thus the energy of 1(1) .10) and 11(1) are orthogonal. (b) Using similar argument. we can show that the energy of c): n-(t) and 11-3) are orthogonal. This energy is given by 1c;leI + l('2|2£y- (c) If 2(r) = :r(t) i g/(f). then it follows from Eq. (1) in the above derivation that 2.5-5 (1) (2) e two integrals of the cross products + t/(t) is equal to that of 1(1) — y/(t) if (t) + (tap/(t) is equal to that of 1'11“) - (at/(1) if E: = E, + By :l: (E:y + £141) 2-5-6 gr(2.-1)r z:(—1.2). 83(0.—2). g¢(l.2). 30.1) and “(3.0). From Fig. 52.5-6. we see that pairs (1.33.35). ‘ and (52. gal are orthogonal. We can verify this also analytically. (81.34.: 4 G “a ch 2 gig. 516-8 gn-g5=(0x3)+(-2x0)=0 81'84=(2X1)+(-1x2)=0 gg.gs=(—1X2)+(2Xl)=0 We can show that the corresponding signal pairs are also orthogonal. /WMMWWW=/VFMWWHMW=° ,x, -1, fat. 91(t)g4(t:lllf = f» [2.710) - 12(f)][:n(1) + 212(1)] d! = 0 .‘ .. q. -m /‘ ﬂ2(’)95(ill1l = fixl‘11(’)+ 212(1)ll211(’)+ millldt = 0 In deriving these results, we used the fact that [Tm zfdt = f“ §(£)dt = 1 and mi(?)zg(t) d: = 0 I -ou 2.6-1 We shall compute «-,. using Eq. (2.48) for each of the 4 cases. Let us ﬁrst compute the energies of all the signals. 1 E,=/ sin’2nm=o.5 0 In the same way we find I5“,l = E“ = E9, = Eg4 = 0.5. Using Eq. (2.48). the correlation coefﬁcients for four cases are found as 1 1 x u 9 . = ' _. ' t = - (1) m l/o‘ sm -1rt sm 41ml: 0 (2) Wei/O (sm 21H)( sm 2M)df 1 l l 0.5 (3) j; 0.707sin 2mar=o (4) f0 0.707sin 21rtdt—‘/°-°0.707sm 21am] =MI4/7r Signals 1(t) and 920) provide the maximum protection against noise. 2.8-1 Here To = 2. so that .uo = 21r/2 = 11’. and "N g(t) = (1.0 + 2a,. cosmrt +bnsinmrt -152\$] n=l where l 1 \n 1 no:1 izdt=l. a,.=Z Izcosnmdr=i-—I-)—. b,.=?- tzsinnmdt=0 ‘2 _l 3 2 _l #1172 2 _1 Therefore 1 4 "” (—1)" = -— V —— I , - < < gh‘) 34—1-4: n2 cosmrt 1_t_l The power of g(r) is Moreover. from Parseval's theorem {Eq. (2.90)] "~ 2 o 00‘ n 2 m _2 Cn_1*l 4(—1) _1 8 1_18_1 Pa‘C°*Z?'(§) +§Z(7572‘ “9+:4Z—n4-g"’90“5 1 1:31 (b) If the N-term Fourier series is denoted by .770). then 1 4 N—l( l)" 1(I)=§+F; n2 cosmrt —-15_151 7|: The power P, is required 10 be 99%)”, = 0.198. Therefore 2 -1 8 1 F 1—): -O.198 a l .31 -14, ‘9 6 For N = 1. P, = 0.1111; for N = 2. P, = 0.19323. For N = 3. P, = 0.19837. which is greater than 0.198. Thus. N = 3. 2.8-2 Here To = 2'». so that can = 27r/21r = Land (14 g(r)=ao+Zancosnt+b,.sinnt -1r\$tg1r n=l where 217 _' l I 2 1r 2 W ao=—- “11:0. a..=— tcosntdt=0, bn=— tsinntdt: 21: _, 2 Therefore W 9(r)=2(—1)"“Z%sinm —1r5t\$1r as! Figure 52.8-2 shows g(!) = t for all I and the corresponding Fourier series to represent 9(1) over (—1r, 1r). Fig. 52.8-2 The power of QM) is P —1 *War—"2 9 m -17 - 3 (b) if the N-tetm Fourier series is denoted by .r(f). then N :r.(!)=2(-])"“Z%sinnm —1r5!'_<_7r n=1 The power P: is required to be 0.90 x = 0.3113. Therefore N l 4 2 F’s-~2- 21:13—03! 7" at) For .'\' = 1. P, = 2; for A' = 2. P3 = 2.5. for N = 5. P, = 2.927. which is less than 0.31r2. For N = 6. P: = 2.9825. which is greater than 0.312. Thus. N = 6. 2.8-3 Rer‘ail that 1 70/2 no = — (1(1)!!! To -To/2 ,2 To]. a, = — g(f)cos nuot dt To -To/2 2 To/2 h" = — g(l)sin 17.40! (H To —To/2 .y I (la) (lb) (1‘?) Recall also that cos mm! is an even function and sin nwot is an odd function of t. It g(t) is an even function of t. then g(f)cos 77.20! is also an even function and g(t)sin man! is an odd function of 2. Therefore (see hint) 7h/2 2 = — t 1 2 no To 0 g( )d ( a) 4 7b/2 a... = — g(l)cos "wot df. (2b) To . o b“ = 0 (2“) Similarly. if _q(t) is an odd function of t, then 9(1) cos nwot is an odd function of t and g(t) sin mast is an even function of I. Therefore no = n... = 0 (3a) 4 70/2 1),. = — 9(1) sin twat dt (3b) To 0 Observe that. because of symmetry. the integration required to compute the coefficients need be performed over only half the period. 2.8-4 (3) To = 4. .u0 = % = Because of even symmetry, all sine terms are zero. 9(1)= no + 2a,. cos us] no = 0 (by inspection) 4 1 rm 2 mt 4 _ mt - cos (—t) d! — cos (—t) dt = —-— stn —— 4 0 2 1 2 mr 2 Therefore. the Fourier series for git) is "n (t' -£(cosx—'—lco E312+'1coss—1rt-—loos-211 ‘ 9"} 3 s2 5 2 7 2" Here I». = 0. and we allow C" to take negative values. Figure \$2.843 shows the plot of C... (b) To = l07r. .co = a}? = Because of even symmetry. all the sine terms are zero. .0”) = no + :0" cos + bu sin n=) no = E (by inspection) a -—Z- scos(£t)lt——1-(5)' (IE-t),r --—2—sin(ﬂ) “‘10:: _” 5 ‘ _51r n 5|“ 5 -,,-1m 5 2 w n = —— ‘ — . = ' ‘ ' ft 1;" 10" qsm ( 5!) d1 0 (integrand IS an odd function 0 ) Here 1),, = 0. and we allow C,I to take negative values. Note that C,I = a.| for n = 0, 1. 2. 3, - - -. Figure 52.8-4h shows the plot of Cn- (c) To = 21:. “:0 =1. m g(t) = no + :0" cos m + h" sin m with no = 0.5 (by inspection) nzl ' 21v 21 f a,.=l —cosnrdt=0. bn=lf .t—sinnidt=—-1— 7r 0 21 1r 0 2n 1m and 9(1) = 0.5 — (sin! + ésin‘Zt + §sin3t + 1Zsintlt + 1 ? =0.5+-1— [cos(t+1)+lcos(2f+1)+lcos(3t+1)+:--] -r 2 2 2 3 ’7 8 The reason f0 Fig. 82.8-4 r vanishing of the cosines terms is that when 0.5 (the dc component) is subtracted from 90). the remaining function has odd symmetry. Hence, the Fourier series would contain dc and sine terms only. Figure 52.8-4c shows the plot of (7.I and 9... (d) To = 1r. «)0 = 2 and g(!) = \$1. (10:0 an=0 (11> (by inspection). 0) because of odd symmetry. all h" = 5- itsin 2ntdt = -2— sing-l — cos E1-) 1r 0 1r 1m 1m 2 2 4 l 4 1 =_."_n.-__v f-.._' _.__' 1r: 5111 x mm: 9W2 sm 6! 2‘” Sin 8t + 1 . 1 =\$cos<2t—§)+;cos(4t—g)+§%5cos(6t+%)+;cos(81+%)+~~ Figure S2.8»4d shows the plot of C‘n and 9". (e) To = a. .00 = 21r/3, l 1 1 11:» £11:— 0 6 ’2 1 21m 3 27m 21m 21m n = ' 1 -—-t = __ __ ‘ .— . 1 a 3 [0 cos 3 dt 2W2": [cos 3 + 3 sm 3 l ‘7 1 2mr 3 91m 21m 27m In = z . i — , = ' ‘ .__—’ .. — — I 3 /o ism 3 id! 212”, [Sm 3 3 cos 3 3 432712 21m 41m 21m V 2' C05 212 " 5in 23—" \) C = _ _ . = ’ -l . 3 .4 r: 212”: 2 + 9 2cos —3 ———-3 sin —3 and 0,. tan cos Tm + gusin I? _ 1, (f) To = 6. .40 = 7r/3. uo = 0‘5 (by inspection). Even symmetry; bn == 4 '1 mr a]; g\_t)cos—3—tlt a": ‘2 1 mt 2 m: =3 A cos—3—dt+ 1 (2-i)c08-3—tllﬂ - 6 [cos mr cos 21”] _ #2712 A 3 3 . _, 6 1r ‘ 2 ‘ 1 51! l ‘74-. -'/l1}'-=0.a~,-;‘2'(C053!—§COS1H+ECOSTITECOS?I+---) Observe that even harmonics vanish. The reason is that if the dc (0.5) is subtracted from "(9). the resulting function has hall-wave symmetry. (See Prob. 2.8-6). Figure 518-41" shows the plot of Cu. 2.8-5 An even function 9!”) and an odd function golf) have the property that .Cnalll = ski") and ‘30“) = ’goi") ill Every signal 9U) can be expressed as a sum of even and odd components because 9“) = 2 in“) + 9(4)} + % lgU) - 9(4)} W W oven ed A From the deﬁnitions in Eq. (1). it can be seen that the ﬁrst component on the right-hand side is an even function. while the second component is odd. This is readily seen from the fact that replacing t by —I in the first component yields the same function. The same maneuver in the second component yields the negative of that component. To ﬁnd the odd and the even components of g(l) = u(t). we have W) = Mt) + you) where {from Eq. (1)] gulf-l = yum + “(—01 = and am) = g M!) — "(—01 = éssntr) The even and odd components of the signal 11(0) are shown in Fig. 82.8-53. Similarly. to ﬁnd the odd and the even components of 9(1) = e“°‘u(t). we have Elf) = (M!) +.<Io(i) where .09“) = 5:; [F_”u(f) + c“u(-t)] and 10 where and 2.8-6 and and The even and odd components of the signal r""n(I) are shown in Fig. S2.8-5b. For 9(1) = r". we have (a) For half wave symmetry Let .1“ = i - To/2 in the second integral. This gives 9 To/2 I 7.0/2, To g(l.)cosnuotdt + g (.1: + -2- cosvwo o o 70/2 Tia/2 [/ g(f)cosn.uot d! +/ —g(m)[— cosnworn} 4.1:] o o To/2 [/ g(t) cos 71th d1] 0 In a similar way we can show that I! (b) To = 8. ..:o = (-b D Fig. 52.8-5 9,0) = % [rfﬂnﬂj —- cu‘u(—t)] \$1" = 9.0) + gm.) 9:“) IJ|-‘ [n1'+ 0'11]: cos I you) = g [M - n"'] = jsin t 70/2 Too g(!)cos nwot dt g(!)cosn.uoi d1 = g(t)cosn.uof dt +/ To/2 1+7 °) dz] _To To]? g(f) sin 11on (It 11,:- Too 1. no = 0 (by inspection). Half wave symmetry. Hence a a I! QIA 4 mr mr . mr "a": (cos? + -2—sm T — 1) (11 odd) 4 mr , mr "2R2 sm 7 — 1) (71 odd) ll ll Therefore 71: l,5.9.13.--- a ={m‘a("—'-1) 2 ‘n‘:7 (4:?r +1) n=3.7.11,15,--- Similarly 2 2 ‘2 and 'N 9(1) = Z n=1.3.5.-~ mt , mr aﬂ cos-4-t -+- (1,. sm —l 4 (ii) To = 27. .40 = 1. no = 0 (by inspection). Half wave symmetry. Hence 'L 2 an cos nt +1». sin nt n=1,3,5. ~ .00) = 2 r "n = -/ 0—H", cosntllr o 11' * (—0.1cosnf+nsinn!)] (n 0 2 p-t/IO .1? [n2 + 0.01 —1r/10 3[' (0.1)— 1 (41.1)] F H I! 1-:2 + 0.01 in,2 + 0.01 ’ 2 (P—R/XO 101r(n'-‘ +0.01) 7 / r"“°sinntdt o 2 p—t/IO ; [n2 + 0.01 -_ I) _ 0.0465 ’ n2 + 0.01 and all.) I)” ‘l' (—0.1:innt -— n. cos 110] 0 1.46111 2" "/10 _ (P 1) " "2 +0.01 (1')2 + 0.01) 2.9-1 (a): To = 4.40 = 77/2. Also Do = 0 (by inspection). l H —- 1 n""‘-"""2)' dr -/ (-"(Mny d! = —2— sin 2!- l "—21: _ «n- 1 (b) To = 101.40 = 217/10" =1/5 9(')= i Dnr’é'. 7’=»'I.a where 4 mr 1 2t f _. =_ .. ﬁnd; U. W 4 ‘4'] 2U. i 2 f . . _ mt b,=1/ -5inﬂm= 4 (sinﬂ_ﬂ.cos£1)= 2425,“ o nvr 2 odd) (11 odd) 2 . . 717T . Sln— =——sm l ” —J"-¢ ( = — I .1 It = —— .. D" 103/; r ( 27rn 2" 5 ) (n odd) |n321 (ﬂ (c) "J 90) = D0 + Z Dnr'i". where. by inspection .00 = 0.5 n=~1 ‘ 2-. >. ' n > 0 D,I = -1— -’—r'-’"'dt = —J—. so that ID"! = -]-, and AD.I = {3 27r 0 21r 21m 21m :5! n < 0 (d) To=7r..vo=28nd Dn=0 1 ’r/4 . . 1 4f - «a —] 2 . 1m 1m , = "1211!. , n=_ _" J-’\’dt=_(_-_‘_— __) .0( ) X D r‘ where D W [in «I 1m 1m 5m 2 cos 2 n=—'). 2.9-2 (e) ‘10:??? m, It?! 1 1 _ 3 {gin ﬂu" 9“) = z D" p} ‘ whole Dn = -/ tP-st‘hdt'; 2 2 [r J ( 1 1) -1] rum, o 477 1') Therefore 3 4 2 9 mcoszn_ - m '9” = [—27 2+ 1' " ~2coszﬂ — mun—L" and 40,. = ta _:,___r_—T—- m in 3": a..- ﬁn 9 3 3 3 COST+Tsm—jﬂ_1 (f) To = 6. do = 7r/3 Do = 0.5 Fig. 52.9-2 _q(I) = 3cost +siu (5t - — Zone (8! — For a compact trigonometric form. all terms must have cosine form and amplitudes must be positive. For this reason. we rewrite 9U) as .00) = 3cost +cos (5r. — 7% .. 21r I < ——> cos +cos 5t 3 Figure S2.9-2a shows amplitude and phase spectra. (b) By inSpection of the trigonometric spectra in Fig. 82.9-2a. we plot the exponential spectra as shown in Fig. 52.9-2b. By inspection of exponential spectra in Fig. 52.9-23\ we obtain W) = gu" + r' ”) 4-; MW?" + p"i‘5“33‘)] + [AW-‘3") +p-j(at—5§-)] = + 14 2.9-3 (0.) gm = 2 + 2cos(2t _ it) + mm - g) = 2 — 2cosZt +sin3¢ (b) The exponential speCtra are shown in Fig. 52.9-3. (c) By inspection of exponential spectra 9(t) = 2 + [v""*’ + c’j‘2‘"*’] + -1-' [H‘“"*’ + c""’“9)] =2+2cos(2t—1r)+cos(3t—%) ) are equivalent. (d) Observe that the two expressions (trigonometric and exponential Fourier series Fig. 52.9-3 2.9-4 To/Q 1 To/2 Du = — f(f)cos nuofdt — j/ f(t)sin nuotdl] T“ —Tn/2 43/2 because its integrand is an odd function of t. If 9H) is even. the second term on the right-hand side is zero he right-hand side is zero because its integrand Hence. 0,, is real. In contrast. if git) is odd. the ﬁrst term on t is an odd function of f. Hence. DH is imaginary. ...
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Chapter 2 - Chapter 2 2.1-1 Let us denote the signal in...

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