Slu to ChB

# Slu to ChB - 3.2. (a) For 1 +1, 1' = 1/12 + 1 = 1/5 and s =...

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Unformatted text preview: 3.2. (a) For 1 +1, 1' = 1/12 + 1 = 1/5 and s = arctan (s) = «f4 = 0.534. Thus, 1 + 3 = View“ = 1.4141905“. (b) For —4 +33, 1' = 1f(—4}9 + 32 = 5 and 19 = arctan = —ﬂ.643 + 1r = 2.493. Thus, —4 +33 = 58-72498. (c) Using the results from 3.23 and H.2b, (1 + ;)(—4 + 33} = (ﬁwf‘xseﬂ-m) = 7.071633“. 2 2 '11-4 —'1'r4_1+ 2—2_s— _, 3 —1 __ .h (d) E: " +26 ’ " - 797% -7é- Thmrr- (:5) +61) “VP—1235 and 9 = arctah = «43.322, which yields aim + 25‘?!" = 2.236111%”. (2) 13-1 + 1 = 1203(1) +Jsin(1) + 1. Thus, 1' = 1X(ms(1) + 1]“ +(sinf1) 3 = 1.755 and a = arctan ( = 0.500, which yields 33 +1 = 1.755e-11”. (f) Using the results from 3.23. and B.2b, 1 +3 __ vie-7"“ —4+ 33 ‘ 5312-493 3.3. (3.) Using Euler‘s identity, 36"” = 3 cos[1r/4)+ 335in{1rf4) = 2.121 + 32.121. (1)} Using Euler’s identith1 1 EJ- = a“: = GOE(—1)+Jﬁin(—1) = 0.540 - 10.341. 0:) Expanding, (I +;}(—4 +33) = (-4 - 3] +:{—4+ 3) = —T—;. (d) Using Euler’s identity, 1+3l _+2_—_32 3 v’i ﬂ . -1 61am + 2e—Jri’4 = — + —-—. a "/23 (3} Using Euler’s identity, e3 + 1 = 003(1) + jsin{1)+ 1 = (cos(1)+ 1) +jsin{1}. (3 511311? 5? ex'IJI’i-tﬁ‘rﬂiﬂtg the denominator in standard polar fin-mI % = grim = 3" mm. Using Euler‘s identity, ' 5J1— : coaﬂn(2)) — gamma» = 0.759 — 30.539. BR. First, express to; in both rectangular and polar coordinates. By inspection, :91 = :1 +391 = 3+:4. Next. r1 = 9’35+4i = 5 end 91 = arotanG) = 9927 so 191 = no?“ = 5W3”. \$eoond, express to; in both reotenyﬂer and polar coordinates. By inapeotion. w: = my“: = 24-1-4" = 2410-735. Next, 4, = T1006(33) = 2coe(w,’4} = ﬂ = 1.414 and ya = r: 3111(92) = 2ain{rrj4] = '95 = 1.414. Thus, 1.92 = 1:1 +jyg = 1.414 4-31.414. (a) Mom above, 1-H]. = TLC”1 = 5510327. {In} From above, 113‘: = 4:; +331; = 1.414 +31.414. (c) I‘m-1|”: I"? E 52 = Similerl , F I'IJ'J‘QI2 = 1"; = 4. (d) 191 +99: = (:1 + 9:2) + ﬂy; + ya} = [3 + 1.414) + 3(4 + 1.414) = 4.414 + 35.414. (e) wl—wg = {191+Ig)-J(y1+yg = 3—1.414 +J(4—1.4l4} = 1.586+J2.586. Con- vertingto polar form, r = (1.536} + (2.5815)2 = 3.033 and H = arcten (ﬁg-g) = 1.021. Thus, «.41 — w: = r339 = 3.93991-“1. (f) 191w; = rieJalrge-"J = 10331-713. Converting to Certain form.1 9: = 19 4490.713) = —1.414 and y = 19 511241.713) = 9.399. Thus, 19119: = .r +Jy = —1.414 4-39.399. (9} 3.20. The general form is zit) = e'“eee(wt). At t = D. e‘“ = 1. Thus, a ﬁfty percent decrease in two seconds requires 0.5 = r”, or a. = 0.5 111(2). Te oscillate three times per second requires w = 611'. >> w - 3*2*pi; a_- D.5*1eg(2); >) t - [-2:.01:2]; x - exp(-a*t).tcus(w*t); >> plet{t.x.'k-’J; Ilahelﬂ’t'); ylabeli’xit}’}; 2 Figure 33.20: Plot of ﬁt) = e'“ meﬁut) for w = 3(21r] and e = 0.5 111(2). 13.21. (a) 1:1(t) = Re(2e{_1+-’2"l‘) = 2s" cas(21rt). This is 1H3 cosine wave that exponen- tially decays by a. factor of 1 — e‘1 = 0.532 every second. A signal peak is near 15 = D, where the signal has an amplitude of 2. See Figure 53.2%. (b) :20) = Im(3—e“‘ﬁ')*} = e‘ sin(21rt}. This is a 1H2 sine wave that exponentially grows by a. factor of e1 = 2.718 every second. A signal peak is near t = 1.34, where the signal has an amplitude of 1.284. See Figure 53.22b. (o) :36) = 3 — Im[e“‘32"}‘) = 3 + s‘sinﬂirt). This is a 1H2 sine wave that expe- nentislly grows by s. £sctor of e1 = 2.713 every second and has an offset of 3. A signal peak is near t = 1,34, where the signal has an amplitude of 4.284. See Figure 313.2%. ll U5 1 L! 2 2.! J I Figure 313.223: Plot of suit) = 2e‘t cOS(21rt)- Figure 5322‘): Plat or \$2“) = ct Singing} Figure 88.22:: Plot of 33(t) = 3 + e‘sin(2irt). B.23. Since cos{t) oscillates at ﬁ-Hz, t should cover at least 2w seconds to span one period. Since sin(20t) has a period of :-E = 0.314 seconds, the step size of t should be less than 0.0314 to ensure at least ten samples per period of this fastest component. >> 1: = DJ: 111:3]; J: = cos{t}.*sin(20*tl; >> plot (1:,1, 'ke’); xlabel(’t’): ylabel('x(t)’): 13.24. The highest frequency is IIIIle1 so the step size of t should be Dill or less to provide ten samples per period of the fastest component. The lowest frequency is 1H2, so t should span at least one second to cover one period of the slowest component. )2- 1: - [ﬂ:.005:2]; kt - {1:1ﬁl’tt; J: - sum(cos('2*pi*ktl); >> plotft,x,’k'); xlabe1(’t’); ylahellf’xlftl'); ...
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## This note was uploaded on 02/06/2012 for the course EE 2120 taught by Professor Aravena during the Fall '08 term at LSU.

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Slu to ChB - 3.2. (a) For 1 +1, 1' = 1/12 + 1 = 1/5 and s =...

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