ECE_308_F'_08_Sample_Problems_1-Solution

ECE_308_F'_08_Sample_Problems_1-Solution - Mesa-fl (a)...

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Unformatted text preview: Mesa-fl (a) 2111}, = 101t, so T =1rffl :15 = 0.2 s. (b) 21th = 171:, so To = Ufi, =1rs.s = 0.1176 s. (c ) 21th = 191:, so To = 1151‘Cl = U95 = 0.1053 s. (d) We have 101t = 27tme and Hit = 2nnfl), where m and n are integers and f0 is the largest constant that satisfies these equations. The largest fl, is 0.5 Hz with m = 10 and n = 1?. (e) We have 101': = 2am}; and l91t = Znnfo. where m and n are integers and jg is the largest constant that satisfies these equations. The largest fcI is 0.5 Hz with m = 10 and n = 19. (Q We have l‘ht = anfo and 19a = 21tnfi,, where m and n are integers andf0 is the largest constant that satisfies these equations. The largest flJ is 0.5 Hz with m = I? and n = 19. Problem 1-12 (3) Written as the real part of rotating phasors: XIII!) : Rel'zefllmlr ° 1116)]; x31!) : Reiseflnm - 1114)] x (I) : Rc[3ej{10m - 11:13 - 1132)] : Renemom - 51116)] t: : filth-ti o m‘fi} jtl'l'iu - 11M) . = jtl0'rt.‘ v m'fi) __i(|0m - 51116} xd(t) Re[2e + 5e ], xt(t) Re[2e + 38 ] x1“) : Re[56j(i?Kf-1tf4)+ 3erO'JU-5‘JII6J] (b) In terms of counter rotating phasors, the signals are: 1:00) = [gflmm mm) + e “mum 416)]; Ibo) = [2.561(1m - mu} + 2.581(1'J'm-mt4)] XE“) = “jefiiomwswm + 158 -j{lt}m —5m‘6)] de) : [efllem - 1:!6) + 8 31101:: e 11.36) + 2-SEJUT‘M - 11M) + 256 -j(l'?m —nt4)] xi!) : [ej{10flt ° 11:!6) + e j'tlfl'm v 1H6) + 1.581(1011: " 5111'6) + -j(|01u - 51116)] XI“) = [2.Sejtl'hu - m'd) + 2.58 —jt1'l'1u — M4) + LSejth'lu - Surfs) + 1-56 —j(l0m - 51116)] Problem 1-15 Note that sinuoa: + a) = Leieeiw _ i e -,:'B e and“: 21 2;“ _ l JIe—flfl) Jim; 1 _ _. 5.6 e t + e J‘e‘mle J'ng 2 Thus we conclude the following: (I) The amplitude spectrum does not change; (2) The phase spectrum has a -1tf2 radian phase shift with respect to the cosine-convention Phase Specnum. This destro s the odd ‘ _ part convention. y Symmetry present m the phase Spectrum "51113 the real Problem 1-20 Representations for the signals are given below (others may be possible): xaU) = 2 m — 3n)u(2 ‘ r — 3n) n =9 xbU) = E m — 4n)u(2 —: — 4n) n =0 xr_.(f) = 2 25(1’ - 2.511) n =0 de) = z %u(t - 3n)r(3 - I — 3n) n =0 fliflJ—hmlfl One possible representation fdr each is xl(r) = MU) + r(t — 1) - 2:1! - 2) + r(r - 3) - u(r - 4) x20) = u(!) * 2u(! - l) + 2u(r - 2) - u(t ~ 3) x30) = rm—m- l)—rtr'3)+r(r-4) x40) = dz) — 2u(t — 1) ~ r(1’- 2) Problem 1-23 (a) First note that the integral of the function is 1. no matter what the value for e: U ml — LE“ 1 = [640$ = {Ee- "Edi = —e “' lo =1 Second, note that the pulse becomes infinitely narrow and infinitely high as e — no. (b) Use the integral to Show that the area under the given function is 1. Then note that as o — 0, the function becomes infinitely narrow and infinitely high. Thus the properties of a delta function are satisfied. Problem 1-27 (a) Using (1-66), this integral becomes fe3'8(r—2)d: = ODE—€31 = 9.e‘S I' 2 —¢= (b) Again applying (1-66), we have m 3 fcos(2m)5(r —0.S)dr = (—1)3£—3cos(2m) = —(21I)3sin(2“rtr)|I :05 = 0 dr ' 0 r :05 (c ) Using (1-66) we gel flea: +cos(21'cr)]5(r)dr = (-1)?!” +cos(2m)],:D = -[—3e"' —2nsin(2wfl,=o = 3 Problem 1-38 (a) Power: ' lim 1 ’ 2 T lim 16(T 2) P: — 2 + 2 +43dz=0+0+ ———__* =8W a quzrfldz fsdzf Tm 2T 0 I 2 (b)Energy: I 2 Eb =f12dr+f62dr : 37 J 6' I (C)Energy: ‘” Io: "' Ec=fe"°'dt=-e =-'—J a 10 10 (d)Power: lim 1 T lim 1 T P = _ -5r 1 d = _ -10: -S:+ I d T_m2T e + F: quzrfle +22 1]d O O T - -l[} -5: = 11mi_e '_2£ +1 :_l_w T"°° 2T 10 5 2 (e) Power: similarly to (d), it can be shown that P: = 11$ W. (0 Neither: it can be shown that both the power and energy are infinite. (g) Power: P8 = 1/2 W. (h) Neither: E].l = no and P,I = 0. ...
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ECE_308_F'_08_Sample_Problems_1-Solution - Mesa-fl (a)...

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