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EE314HW2

# EE314HW2 - Homework 12 Chapter 3 3.1-2 3.1-5ib 3=_1 3.2-1...

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Unformatted text preview: Homework 12 Chapter 3: 3.1-2, 3.1-5ib}, 3=_1-?, 3.2-1, 3.3-2, 3.34, 3.3-6, 3.4»2, 3.44, 3.6-1, 3.1M, 3.8-3, 3.8-5. Due: Feb. ’1 {Monday} 3.1—2 (a) Do . no G (f) = f g (t) 8—32?rft dt 2/ emzit—Evieujhrft dz. -00 _w 3 DO : / e?{t——3)e—3'22rfc dt +f e~2(£—3)e—j21rf¢dt W00 3 : e—jﬁ’rf + e—Jﬁzrf e—jﬁwf 2—j%d 2+jhf“?¥?ﬁﬁ (b) gﬂ) =[ooa(f)ej21rftdf=/w5(4ﬁf}ej2ﬂfidf__/ood(f_2)ej27rf£df Let A = 47rf in the integral involving 6 (4n f). Then . / mnneﬂwﬁdf : foo “Mew/Mi = ﬂ = 3- _m _°° A 411' 41'r 411' We also have If; 6(f — 2) ejzwf‘df = e374“. Hence 1 . t 2 __ 341:! M) E e 3.1-5. (5) Because 1,»25fg—1 2, —l< <1 em = - ”f— 1. leSQ 0, else is aneven function of I. 00 _ I 905) = /'G’(f)c232""’r“a£fw——2/0 2cos27rftdf+2f2c0327rftdf _m 1 f=2 Ii sin 21?: + sin 4m f=l 7ft 4ﬁn2wftf=1+72\$n2ﬁft 2wt fzﬂ 27ft 3.1-7 (a) 8 E g (t) = / e—j27rfioej2rrft df =/ ej2nf(t.—to) (if -—B 78 I ejzﬂa—m) 3 sin 27rB(t — to) " = .,____ =-—~—«—_—.m_:25’sinc27rBt-t jg‘ﬂ'ﬂ' — to) —B ”(If - t0) [ f ON (b) 9(6) 1! I] E / jejzﬂtdf—f jejZthdf ~18 0 {J 33'2wa ej27rft B 1_ 6—12ﬁ8£_ 63-21:.3: +1 I“ coserBt _ 27d _3 _ 21rt U - 27d, * Trt 3.2—1 See Fig. 83.2-1. _ 110/3) A(3x/100) 1 ‘I —50/ 3 0 50/3 x (b) . sin-{(brf— 10:2)15] 1 2.5 5' 7. (d) 1 sinc(m/5)n(mo) 1O Fig. 33.2—1 3.3-2 1 (2wa2 (h) 9: (i) = g(-—t). Then 01m: (€52er Jrjgﬂfe—jzn; _1) (C) 92 (t) 2 g(t — 1} + 91(i—1). Then 02 (f) - 8—3.2?” (eihf j27rf€j27rf 1) + 6.42,” ( -J27rf + ‘2 ~J'2vrf _ _ _ e _ . (2H)2 (27d)? 3 ”f9 I) 1 1— '2 *2 ‘J‘ZWI -14”; -2 —j47rf f2Wf)2( J7Tf e +e +J arje ) (d) gm) 29a — 1) +91 (12+ 1). Then e-jzwf 03 m = - (8W _ j27rfej2"! -1) + 9fo (27rf)2 . (2 ~ 2‘32"” e 81'2” ) : (x ~ ices 271' f) {3‘72"} +,j'27rfe‘j27rf w I) (2w? 2 (711)“)2 (e) 94 (t) 29' (t - %) +91 (t+ 313—). Then - _ E—jﬂ'f .21”, _ _ 7r 'Zn'f _— 53ij —j27rf . —j21rf _ G4 (f) -— (27rf)2 (ea-7 32 fe’ 1) + (BYE/f)? (e +32n'fe ‘- 1) z marina (—ﬂwfew +i27rf‘3—W) * % (W: ed”) = ﬁfe: (m (f) gsft)=1.59(%(t— 2)). Then 050) = = (4m? (1 ”'4” " 6W) 3 3—4 FY0111 the time-shifting property, g(t :h T) ¢=> C(fjcijzﬂ'r. Therefore, w + T) + at! — T) w Swami?" + awe-1'2"” = 26m cos 2an (a) Let. T = 3. Then 9(2) = 11(5) and from pair 17 ofTable 3.1. GU) = 23inc[27rf). Then 9U + 3) + 9(t — 3) {:2} claim {21rf} cos(61rf) . _ . 2 (2)) Again, let T 2 3 and than take g(t) : A (ﬁ). From palr 19 of Table 3.1, G U) — smc (17f). Then 9(1 + 3) + g(t — 3) @a" Elaine2 (Ff) cos (610') 3.3-6 Note cos 21rfgt = c0510: => fa =10/21r = 5/77. A (EEC ) <=> 1T “NE: (a) We take g(2‘.) = A{%). F‘rompaii' 19 of Table 3.1 and the modulation property, we obtain A (%) c0510: 4:)» % [Sine2 {arzf w 51:) + sine2 (Trzf + furl] ‘03) The signal g(t) here is the same as the signal in Fig. 83.3—6 delayed by 21L From the time-shifting property, its Fourier transform is the same as in part (a) multiplied by e‘jth. Therefore, C(f) 2 g [sir-1c2 (rrzf ~ 577} + sinc2 (wzf + 5w” 2“”"21 The Fourier transform in this case is the same as thet in part (a) multiplied by 2'14“”. This multiplying factor {Eprﬁems a linear phase spectrum Hirrzf. Thus we have an amplitude spectrum [same as in part. (3)] as well as a. mear phase Spectrum 4mm} = ~4rr2f as shown in Fig. 83.3-6b. at“; In the foregoing solution, we first multiplied the triangle pulse [Mi—r) by cos 101 and then delayed the rwult the Sr. This means-the signal in Fig. 33.3-6b is expressed es A(%1)c0510(t— 2w). We could have interchanged malt-Peration Hi this particular 'case; that is, we could have ﬁrst delayed the triangle pulse A0151?) by 27r and then lplred the result by cos 1011. In this alternate procedure, the signal in Fig. S3.3-6b would be expressed as 41—2024 Fig.333-6a 3.4-2 (a) See Fig. 834-203). (b) See Fig. 33.42{b). (a) See Fig. s3.4-2'(c). 40000 ﬂHﬂ ”.U) 0 10000 ﬁHa 1 Y.(f)=G.(f)H.(f) -1 0000 D 10000' KHZ) Fig. S3.3—6b (a) (b) (C) Fig. 33.4—2 KHZ} HA1) ~5ooo 0 5000 RHn Y:(f)=Gzlf)H;{f) 1 . KHZ) l0 {cl-ta, 5-501? 'Fkﬁb/ 3.4—3 To find the outpui y [i] of this system. we wiil first ﬁnd the Fourier transform X (f) of the input. Since the Fourier transform is a. linear operation. we can take the transform term by term Using Table at. we write W) m -5(f}~!~2£'ﬁ"“" --j—![a(r+ 31") +5(f~-§§ } n w... i ‘- ,f “P gﬂiﬂi +J'2tfe ﬂ 1“ Then Vii} ‘-= Hiﬂxiﬂ Ii 1+ j‘Q-rrf i .._._....._ ..WW “szjir + 24"“ i+j2nfai+ 32743 1+}2n'j ii 1.1 .,_ A———-—~ ”. J I! + 2 :E- -+ j'2irfc1—,-i- 3'2st Prom partial fraction expansion (assuming a. 95 1), m on” 1 E)_.__Lm_‘,_‘ 1+j21rja,+j2rrf“ oar-E 114217,? (ifs-ﬁn} 1 + 3122:}’ 'f‘aking the inverse Fourier transform term by .term. we: ﬁnd 2 II o; = 1. then we have the term (-————3~———) , 1 . l ‘ = ‘ —{ta—£a} . W ..- ____ ~Jwat __ .....m 'wot yit) 1+2e nit to) 2“..ng )6 2(1-4 Maia] ﬂ - A . an“ a) madam-i .. 1' Z: ai‘:1(e;-e )u(t 1) is! 0&4: S: 3! ﬂaw-mu (1% 5i} i=.i.,u‘=i 3.6-1 (t‘t') (a) H (f)— IH (Di ejﬂhU) = ej(——‘21rf1;rksin21rf7‘}_—jksin2ﬁfTeij2=rfto I Recognizing the second term as a time-shift and employing the hint 6‘3“” 2"}? m 1 — 3'): sin 2nfT for small k, we obtain _ . H (f) z {1 —jksin27rfT)e—J'~’"f‘° Rewriting the sine function in terms of complex ex'ponentiais, we obtain 2 ““7 "ﬂu .W ‘ __ ... i 6”“ i_'—+;2nfe J2 mum) Mi“ 2:) ﬂiié-jw9)6(f W0 2n 3 ‘5‘wa 2 6"?"ﬂ“'1't;t'issfii5(f "”“")+5(f ”ii ) H (j) 3 (1 _ {\$92er + femur) ewjzwm : 842nm _ Ee—jzwmn—T) + Ee—ﬂwnzum- ‘2 2 2 2 Thus the time—domain impulse response consists of time-shifted impulses, that is, hit)=5if“io)s§5(tito+T)+§6(t~to—T) By the sampling property of convolution, we obtain _ 'yinmango)=go—to)+§igu—to—T>—g(t—zo+rn as claimed. (b) The time-shift by to will not affect TDM systems because it is a constant delay experienced by all signals. In the same way, this is like a linear phase response and will also not affect FDM systems. However. the two “copies" of the signal at in :l:T could prove problematic for a. TDM system if T is large (i.e., at worst. the copied signals may overlap with a different time slot; at best, the copies cause the TDM system designer to use much larger slots). If T is small in comparison to the signal duration, it could also cause distortion (i:e. the signal copies would overlap the original signal). FDM systems are not aﬁected because the magnitude response of the system is ideal. 3.7—1 00 00 f sinc‘1 (kt) dt = / sine2 (ktlsinc:2 (kt) cit. _oo ~00 For real signals, 9 (t) = g' (15). Thus we will take g(t) = g' (t) z sinc2 (let) and apply Parseval’s theorem in the form 00 [m9(t)y(t}dt=f_le(f)lzdf to compute the desired integral, since the Fourier transform 0 (f) =' E-A (3%) is easier to handle in an integral. [We = 1.: e {a} e (as scat-new — ' .. 2"” £1.12 e _ k2 7r rr 31r F3 ll 95-121 to h) w Bx 35'. : f‘x .... | :1 eta-l.“ v I'D 53:. ll Pry-“l U to M , 3‘} "-5. :I A H | M er :3 1»,a + :1 “541‘ h: x.._./ a; 3.8- ' 3 From Example 3.16. the transfer function of the RC circuit in Fig. 3.283, is ‘ l H = —_u._ if) 1 +1.47er We also compute 2 MW: l 2 = I l+j47rf 1+161r4f2 { . or later use in the application of Equation 3.92. (a) 3 (f) = K - I ﬁnd 2 . In this case P1 = ff; 510’de —-r 00 (this signal is not realizable). Employing Eq. (3.92) we K Sno=mm123nn=1+m 4F 11' (b) S (f)= H(7rf) Then P =ff°°o SzUMf: ffﬁ; df:— Following part (a) we have 3y (f) = erst m = 1 mm 1+ 161t4f2 and 1/2" —I 2 ”2" _ P = 5y tan (4n 1') tan ‘(2 ) y 1:(f}df=/l/2W 1-i- 3511-4}? CU: 4.”? = 231-2 7? “11’2" (c) 3:(f)=r5(f+lJ+6(f—l). Then Pr=f_‘_’;.5'=(f)df=1+1=2, 8y (f) 2 IH (le2 3,, (r) =lH(f+1)f26[f+1)+lH(f—1)126(f- 1) =1—ﬁm{6(r+u+w—m Therefore, 9° 2 Py=f 3y‘{f)df= .x. 1+ E61r4 3. 8- 5 The system impulse 1s hm = e‘z‘um and the input is g C = .5 -» . ' “1613th Is a noise signal with PSD 3w (f)= 11 (H4) ( ) w J C0595“ + “/3) We are also given {a} To hm! the pawn!" of Elmmtrput F3, we ﬁrst. mad he mmpuw S, [I}-— S.” {f} + S? {I} when: S {f} dammes Lbs: \$9313 of the signal .9 {a} as» "cw-Hf [ﬁrst i1:f3}.me Problem 3 8 1 WE hm .3; ff} :2 — “a” a} + a U 4" 33; Thus 3;: U) ”1 HUN] T M5 U" J} + 5 [I + 3}}1531; Eisn- haw: the system fmqmzncy respunﬁe 1 l . -- r ‘2 Au- __.._.‘ ”LU—- 2 wk! and If; {HI ‘e1+‘m‘3j2 We can now chain Liz:- cutput PSI) 313 8y {I} in W an m -— m in (ix-211+ gs r: - 3} + is (,r + 33.] and the mmput gamx (b) This F'SD {33' Line mimic due: ta Lhc skgmﬂ is SwEfEHi-Wtfﬁg-‘Fsiﬂ ﬁgﬂ—w 33¢ l6ff+ 3)] and 15m pawcr (isn't: E6 £112 31ng i5 Pia mf 331d (13'?de 39'2275: {c} Sil'ahilariy. "ﬂan powm‘ than to the name is (a; The arm is p “*2 2 d 3 SNR=1DI- “\$.34? ”hm Pym )«B 3*“ “131 U1“- Mﬂiﬂriw 3‘“? mama“ that the power a? le sigma! is \$255 than the pawer of Liia misc. ...
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EE314HW2 - Homework 12 Chapter 3 3.1-2 3.1-5ib 3=_1 3.2-1...

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