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Unformatted text preview: 3’“ WILL; l" 2.57. (a) Realizing that m2[n] = y1[n], we may eliminate these from the two given difference
equations. This would give us min] = ay2[n  1] + box1[n1+ 12mm ~— 1]. This is the same as the overall difference equation. bu 'Afnl (L)
at
6 6
(d) 6 ‘ ~+—‘ w out“) ‘1 [W] “W b o '2! {I’ll if») (1°)
Figure 52.57 (b) The ﬁgures corresponding to the remaining parts of this problem are shown in the
Figure 82.57. 59. (a) Integrating the given differential equation once and simplifying, we get y(t) = —9—9 t y(r)dr + 92ft m(r)d¢ + 915cm. 0‘1 —oo 0‘1 ——oo 0'1 Therefore, A = —ao/a.1, B = bl/a1,C' = bo/al.
(b) Realizing that m2(t) = y1(t), we may eliminate these from the two given integral equa tions. This would give us t
yg(t) =: 14/: y2(7)dT + Bﬂmmlhmr + Cmﬂt). (c) The ﬁgures corresponding to the remaining parts of this problem are shown in Figure 82.59. a 3
em the) 1971i gym
I
C
) (C 3 Lb
3(4 1 £84 3 i I, I
a A
~Ld) l
2
l
Cf) E Figure 82.59 9}— 2.61. (a) (i) From Kirchoff’s voltage law, we know that the input voltage must equal the sum
of the voltages across the inductor and capacitor. Therefore, d2y(t)
at? m(t) = LC + y(t). %
E:
T.
i,
3,
f Using the values of L and C we get
d2y(t)
dt2 (ii) Using the results of Problem 2.53, we know that the homogeneous solution of the
differential equation + y(t) = d2y(t) dy(t)
dt2 dt will have terms of the form Klesot+Kzesll where so and .91 are roots of the equation + agy(t) = bx(t). +0.1 32+als+a2=0.. (It is assumed here that so # 31.) In this problem, (11 = O and a2 = 1. Therefore,
the root of the equation are so = j and 31 = —~ j .. The homogeneous solution is yh(t) = Klejt +K2€wjt. And, 021 = 1 = wz.
(iii) If the voltage and current are restricted to be real, then K1 = K2 = K. Therefore, yh(t) = 2K cos(t) = 2K sin(t + 7r/2). t+2, ~2<Zt<A1
a:(t)= 1, —1<t<1 2—m 1<t<2
T: 6, (10 = 1/2, and __ 0, k even
“"‘ ﬁﬁ$m%qmﬁ%x komi
(iii) T = 3, a0 :21, and ak = 3] [aim/3 Sin(k27r/3) + 2ejk"/3 sin(k7r/3)],
273192
ﬁﬂT=z%=—U1%=%—bﬂhk¢u
(v) T = 6, we = 7r/3, and _ cos(2k7r/3) —— cos(k7r/3)
“k" jkw/3 ' Note that (10 = 0 and (1;, mm = 0.
(vi) T = 4, mg = 7r/2, a0 = 3/4 and min/2 mam/2) + «rm/4 sinUmr/4)
ale 2 W’ km (b) T = 2, ak : mﬁgﬁjk — (3—1] for all k. % problen1 3
clear;
x : [zeros(l,lOO) ones(l,10l) zeros(l,99)];
x1 = [x —X 0]; dt = 0.01; t = [—3:0.0l:3];
for k : 0:5
a(k+l) = dt/6*sum(xl.*eXp(—i*k*2*pi*t/6)); ?
end k¢u Vk. CM: 0 h
a, :JOQSQ?” (a) Wehave
1 1 1 2
ao=—/ tdt+—/ (2—t)dt:1/2.
2 0 21 (b) The signal _q(t) : dac(t)/dt is as shown in Figure 83.24. Figure 83.24
i The F8 coefﬁcients bk of g(t) may be found as follows: 1]l 12
b:— —— dt:
0 2V/odt 2/1. 0 bk = l/le_j7”°‘(1t— 3[2:3‘3'7'Mdt
2 0 2 1 and = — 67”].
(c) Note that
g(t) = (1:?) (55—) bk = jkwak.
Therefore,
ah = j—1;bk = —;21—k—2—{1  e'j"k}. 7' 3.25. (a) The nonzero FS coefﬁcients of 33(t) are a1 2 (L1 = 1/2.
(b) The nonzero FS coefﬁcients of x(t) are bl = bi] = 1/2j. (c) Using the multiplication property, we know that 00 2a) = mm) +55 at = 2 am l=~oo Therefore, 1 1
c =a *b =——.6k—2 ———,6k+2.
k k k 4Jl l 4Jl l This implies that the nonzero Fourier series coefﬁcients of z(t) are (:2 = 6:2 2 (1 /4j ). (d) We have 1
'2 Sll’l(8t). Therefore, the nonzero Fourier series coefﬁcients of z(t) are c2 = 62 = (1/43"). z(t) = Sin(4t) cos(4t) : ...
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This note was uploaded on 04/30/2008 for the course BIOMEDE 311 taught by Professor Steel during the Winter '06 term at University of Michigan.
 Winter '06
 Steel

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