hw 1 solutions - EE 3‘!“ “V l Shiv-Haas 3.1-1 Using...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE 3‘!“ “V l Shiv-Haas 3.1-1 Using Euler‘s identity, we have on W W F{g(t)}=] g(t}e'i”‘dt= / g(t)cosutdt—-j/ g(t}sinwtdt -M 'W ~90 If g{t) is an even function of t, 9(t) sin nut is an odd function of t, and the second integral vanishes. Moreover, g(t) cos wt is an even function of t. and the first integral is twice the integral over the interval 0 to 00. Thus when 9(t) is even, ' Gm = 2/: g{£) ceswtdt = 2/: g(t)cos[21rft)dt Similar argument shows that when 9(t} is odd, cu) = —2;' f: g{t) sin 2'fl'ftdfi If gm is also real (in addition to being even). the integral from GU) = 2ft?” g(t) cos wtdt real. Moreover. G(—f) = 2 ngstflfl—ffildt == fomgmcoswwm dt = Gm Hence GU) is a real and even function of f. Similar arguments can be used to prove the rest of the preper'ties. 3.1-7 (3) B a I (t) = e—jitnftoejfivrfl df = f eflxflt—lo} if 9' ‘B _8 - s . " fl = sm his“ — to) =- 23 sine [27rB(t — to” j21r(t +- to) _B «(t — to) (b) D ' a '2 ft mt) = ] J's-Wm; —/0 is: * df we I ej2nft ‘1 812:}: 5 1_ E—nwst _ ej‘ZsrBt +1 1_ ms 27”?“ _. = _.._._———-‘ = = 2111 _3 2111 o 2111 ' art 3.3-2 1 91 (t) = 9(‘0- Then G1 = (3—.i2'l'f + jgflfe—jihrf __ (2mg ('3) 920) =g(t-— 1) +91 {t— 1). Then 6—12"; I _ . 8‘12” _-2«,r - am! _. = e;w_ 2“. 913*.f—1 + e 3 +J2'I'FJre 1) 020') (2"“2 ( 3 f ) (27m2 ( I (d) Qaltl=9[t-1)+91(t+1}. Then _ e—J‘m ed; .M_ 21'2" 42,7; ‘2 —:'2«J_1 GaU} — affine mre’ 1)+(2m2(e Hm ) _ 1 _ -'21r_f___ '2»; z 1 _1 2 ' We” 8’ 8’ ’ W“ m 134-2 (9) 94itl=9{t-%)+gl [t+%). Then _ e_jwf 1r 'n' 9:"! — p . _ 04 (n — (QM-A 2 ’ r32frfe’2 ’ — 1) + (W); (e ’2 f Hare 3w — 1) 1 , .w 1 _.fi_ _’ 1r _.fi I = (21rf)2(—327rfe’ ’+:2=rre J J“) = fi (‘9 "e ’ f) = (m (f) 95 (t) = 1.59% (t — 2)). Then 38—1-1"! _ I 3 I G = 4n! w v 4n; _ _ # . _ f, W (3" J4wfe’ 1) — {4”“2 [I Jdfl'f—e 3" f) 3.3—6 Note cos angt = cos IOt =r fo = Ill/211' = 5/71: (a) We take g(t) = (Mg-1;). From pair 19 of Table 3.1 and the modulation property, we obtain A cos 10¢ <=> g [sinc2 (ngf —- 5n) + Silica (nzf + 511)] (b) The signal 9(t) here is the same as the signal in Fig. 53.3-6 delayed by 2n. From the time-shifting property, its Fourier transform is the same as in part (a) multiplied by rim”. Therefore, 0”) = 1;— [sir‘ic2 (nzf — 5n) + sing:2 (11.2}- + 511)] 8—1413} The Fourier transform in this case is the same as that in part (a) multiplied by e‘jhzf. This multiplying factor l‘flpresents a linear phase spectrum "-41r2f. Thus we have an amplitude spectrum {same as in part (a)] as well as a linear phwe spectrum AGO») = —4n2f as shown in Fig. 53.3-6b. Note: In the foregoing solution, we first multiplied the triangle pulse A(%) by cos It}: and then delayed the result by 2a. This means the signal in Fig. S3.3-6b is expressed as IAN—333,1) cos 10(t w 2n). We could have interchanged the operation in this particular case; that is, we could have first delayed the triangle pulse A(%) by 211 and then multiplied the result by cos 10:. In this alternate procedure, the signal in Fig. 83.3-6b would be expressed as A(—-"2':"Jcos 10L This interchange of operation is permissible here only because the sinusoid cos IDt executes an integral number of cycles in the interval 211‘. Thus both expressions are equivalent, since cos 10(t e- 211') = cos Iflt. “K (c) In this case the signal is identical to that in Fig. 33.3-6b, except that the basic pulse is the gate pulse TI instead of a triangle pulse Ab—‘fl-J. F‘rom pair 19 of Table 3.1 and the modulation property, we obtain GU} = % [$110973]- — 101T) + sinc (21r2f + 101)] e-jh’; Fig. 53.3-fib er (I) Fig. See—ea 3,6»! (3) '9 '[—21rft —ksin2rrfT) -— 'ksanirfT — 'Eirft Huggllqurnea Mf)._.e.1 u =8} 65" o Recognizing the second term as a time-shift and employing the hint e‘iksl" 3"” z 1 ~ 3']: sin 27rfT for small it, we obtain I H (I) m {1 ujk sin 27rfT}e-J31rfto Rewriting the sine function in terms of complex exponentials, we obtain H U) = (1 __ £612wa + Eeijrro)e—3'21rftg : e—jmrfto _ Ee—jzxmo-Ti + Ee—jzwmowi 2 2 2 2 Thus the time-domain impulse response consists of time-shifted impulses, that is, k ' k h“) =5(t—to) - aduwto-FT) +§5(t—tD—T) By the sampling property of convolution, we obtain in yo) = homo) =go—to)+ Elgii‘to—Tl —g(z— tom} 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 to :l:T could prove problematic for a TDM system if T is large (is, 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 (Le. the signal copies would overlap the original signal). FDM systems are not affected because the magnitude response of the system is ideal. 3.6-3 (a) :r (t) = % sinc(1000t). Then 3; (t) 2 a- (t) + 0.001x2{t) = 9:9 sinc(1000t} + 19? sine2 (1000:) - Using pairs 18 and 20 from Table 3.1. we have * Ff fif Hf) — l1(1000) +A(2000) .(b) The Fourier transform of the input signal is X(f) = 1'! (fi), which has bandwidth %9 Hz. The Fourier transform of the output contains the same gate pulse with a superimposed triangle pulse of bandwidth % Hz. Thus the resulting signal has total bandwidth £39 Hz = 2 x % Hz as claimed. Signal squaring in the time domain is equivalent to convolving the signal with itself in the frequency domain, which as we have learned. doubles the signal duration. The superposition of the original signal, because it is lower in bandwidth than the squared signal and is low-pass, does not affect the bandwidth of the output. (c) The superposition of the squared pulse causes distortion analogous to that shown in Figure 3.3613 (Le.I low- pass filtering can suppress the extra "tails" but not the superposed signal). If the original signal is low-pass, then the superposed squared signal will also be low-pass, and this distortion cannot be compensated without knowledge of 2: (t). (d) Signal squaring has the effect of widening the bandwidth required for transmission by a factor of 2. Thus the FDM band would have to be wider or would suffer interference. (e) As explained in Example 3.18, the interleaved TDM signal could be recovered without interference, but with distortion. 2.6-1 We shall use Eq. {2.51) to compute pH for each of the four cases. Let us first compute the energies of all the signals: . l E... z / sin2 2nd: 2 0.5 0 In the same way, we find 891 = E9, 2 E9: = E9,, = 0.5. From Eq. (2.51), the correlation coefficients for four cases are found as follows: (1) ft,1 sin 2st sin 41rtdt = 0 (2) 7m ffsin 2st) (u Sin 2st] at! = -1 (3) fol 0.707 sin 2m! dt = 0 (4) [ :5 0.70mi: 2ntdt — [E15 0.7mm was] = 2328/1: = 0.9 Signals 2(t) and 9:0) provide the maximum protection against noise. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern