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Unformatted text preview: Mesaﬂ (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 112 (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:
: ﬁlthti 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?Kf1tf4)+ 3erO'JU5‘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'mmt4)]
XE“) = “jeﬁiomwswm + 158 j{lt}m —5m‘6)]
de) : [eﬂlem  1:!6) + 8 31101:: e 11.36) + 2SEJUT‘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) + 156 —j(l0m  51116)] Problem 115 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 cosineconvention Phase Specnum. This destro s the odd ‘ _
part convention. y Symmetry present m the phase Spectrum "51113 the real Problem 120 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(r4) x40) = dz) — 2u(t — 1) ~ r(1’ 2) Problem 123 (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 127 (a) Using (166), this integral becomes fe3'8(r—2)d: = ODE—€31 = 9.e‘S
I' 2 —¢= (b) Again applying (166), 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 (166) we gel ﬂea: +cos(21'cr)]5(r)dr = (1)?!” +cos(2m)],:D = [—3e"' —2nsin(2wﬂ,=o = 3 Problem 138 (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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This note was uploaded on 09/15/2008 for the course EE 308 taught by Professor Atkin during the Spring '08 term at Illinois Tech.
 Spring '08
 ATKIN

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