Unformatted text preview: E.2T. Determine the bilateral Laplace transform and ROG for the following signals: [a] Iﬁfj = 19—qu + 2) H5} f ﬂips—3f Eff f 6411“ + ﬁle—5i Eff fix 6—i:1+s:] EH.
—2 £.,:2{1+al l + 3
ROC: Rel[sj : —1 {cl} :rﬁf} = sinﬁfjuEi} 5° 1 . .
X s f —_ cit—r3" e‘stdf
{J D 231 1 no fEij—s} it _ ['33 —E_¢{j+5} [it
fa 91 . s 21 1(—1 1)
ﬂj j—s jl—s 1
(1+5?)
soc: Re[sj 3:: n 6.29. Use the basic Laplace transforms and the Laplace transform properties given in Tables Ill and [1.2 to determine the unilateral Laplace transform of the following signals: {11] :rﬁf} = 1111(1)”: cosl[27ri}u{fj £1 1
11H] = tuﬁf} <—+ 51(5) = 3—2
{1(1) = mamarose} (i. m
as) = as; 1 so} 1:. my; = Areas]
1 X”) = {f} IHJ = ti {ﬁt—*GOEUJHHIIJ oEi} = :24 cosEiIIuIifJ L m5} = [s :13ng
m d
mm = the) <—+ XIISJ = —EB{5}
—52 — 43 — 2 X“) m 13.33. Determine the forced and natural responses for the LT] systems described by the following dif— ferential equations with the speciﬁed input and initial conditions:
(a) sym + 10w) = lulu}: ytﬂ‘} = 1.. rm = um I 11(3) = E
Y[s](s+lﬂj = lﬂX(5}+y[ﬂ‘} = 23:63: _ 10 _ s[s+1'} _ 1 —1 _ E+s+lﬂ
W) = [l—e—m‘hm
we = e'm‘um {d1 sins + 2st + syn} = smut yin2 = 21§iylfllt=m = ﬂ: wit} = nit} Y{5}{52 + 25 + s] sX{sj + slim—II + QyEIJ—j 1
“(33' = yfﬂ] = %e_¢sin{2fju[t]
n _ sow}
Y [33' — m 26—: cosﬁjluﬁ} “a:
a
.u.
~23  6.43. Use the method of partial Fractions to determine the time signals corresponding to the following bilateral Laplace transforms: (a) ms} = —3 2
XIES} : 5+1+ 5+2
(i) with R00 Re[s] c _2
(left—sided}
3(1) = (3e—t — 2.2”) um)
(ii) with RUG Reps} } —1
(right—sided)
mm = {—39.6 + 2929) u[t]
{iii} with R00 —2 < 123(5) < —1
[two sided)
m; = 33°u(_t]+232‘um
{c1 XisJ = 5 1 {i} with R00 RHES] :: _1 {leftsided]
1111*) = {—59.—B + iii—t} Eli—f“) {ii} with RUG Reﬁs) 2. _1 [right—sided)
mm = [5e—‘ — :34) 1.:{tj {3.45. A system has transfer Function H (5} as given below. Determine the impulse response assuming
{i} that the system is causal, and {ii} that the system is stable. 52 Es—
{CJ 3(5) = H“? = 3111+ + {i} system is causal
hm = [—s¢ + Estoosmf} + s*s1n[3sj}s{sj
[ii] system is stable
hm = —s*s[rj — {ssi msl[3fj + si sin[3t]} uE—t} 6.46. A stable system has input I“) and output y(t] as given below. Use Laplace transforms to de—
termine the transfer function and impulse response of the system.
(a) mm = 2way. ya) = ﬂ mamas) m] = (.33..
Hts] = = 1 Mt] = sm—(sQ‘cssu)+s9*sis(s)}s{c] ...
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 Fall '08
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