3015_Homework_11_Solutions

3015_Homework_11_Solutions - E.2T. Determine the bilateral...

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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 j-l—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) = mama-rose} (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 yin-2| = 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 + 29-29) u[t] {iii} with R00 —2 < 123(5) < —1 [two sided) m; = 33-°u(_t]+23-2‘um {c1 XisJ = 5 1 {i} with R00 RHES] -:: _1 {left-sided] 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 = 2-way. ya) = fl mamas) m] = (.33.. Hts] = = 1 Mt] = sm—(s-Q‘cssu)+s-9*sis(s)}s{c] ...
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