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hw 1 solutions

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

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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. ...
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