This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 = 5872498. (c) Using the results from 3.23 and H.2b,
(1 + ;)(—4 + 33} = (ﬁwf‘xseﬂm) = 7.071633“.
2 2
'114 —'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) 131 + 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 = 1.755e11”. (f) Using the results from 3.23. and B.2b, 1 +3 __ vie7"“
—4+ 33 ‘ 5312493 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’itﬁ‘rﬂiﬂtg the denominator in standard polar finmI % = 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“: = 2414" = 2410735. 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 = 1.414 431.414. (a) Mom above, 1H]. = TLC”1 = 5510327. {In} From above,
113‘: = 4:; +331; = 1.414 +31.414.
(c) I‘m1”: 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 (ﬁgg) =
1.021. Thus,
«.41 — w: = r339 = 3.93991“1. (f) 191w; = rieJalrge"J = 10331713. Converting to Certain form.1 9: =
19 4490.713) = —1.414 and y = 19 511241.713) = 9.399. Thus, 19119: = .r +Jy = —1.414 439.399. (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 ﬁHz, 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'); ...
View
Full
Document
This note was uploaded on 02/06/2012 for the course EE 2120 taught by Professor Aravena during the Fall '08 term at LSU.
 Fall '08
 ARAVENA

Click to edit the document details