Question

Do: Find the Laplace transform, F(s), for each f(t) given below in parts a) and b). Express F(s) polynomials ins where the de
Part a) NNNNA VNNN a 8 b 3 sec (-1) c 2 rad/sec d (5/9)*pi rad Part b) a = -4 b -7 c 4 sec^ (-1) NNNN NNNNy
Do: Find the Laplace transform, F(s), for each f(t) given below in parts a) and b). Express F(s) polynomials ins where the denominator polynomial, A(s) Le., it has the value "1" (one) Monic Rational Form (MRF). This means that the result is a ratio of polynomials, and the coefficient, a, in the denominator polynomial, A(s) below is a, 1 as a ratio of =s"+a-1s"- + a28 +a1s +a0, is monic as the leading coefficient on the highest power of s. We will call this format Your final answers to parts a) and b) must be in the following form: bm &m+bm-1sm-1+ .+bis+bo + azs2 B(s) A(s) F(s) + a18 + a0 s an-18 + and must be devoid of imaginary numbers or else lose at least 50% of the total points available in that part. -bt. =(e cos (ct + d) fort>0 a) f(t) +bt e-et fort>0 b) f(t) a6(t) Use the Laplace Transform Pairs and Properties found below: httpa://clcee.net/clc_ece/ECE2020/Supplementa/Laplace_simple_table . pdf Write each Pair or Property as it is used in your solution development. For example, if you use Pair 4) at some stage of your solution, then write something like, be t Any work beyond the Pairs and Properties shown in the table, must be shown using the Laplace Transform definition: 8+a -N f(t)e#dt F(s) The Laplace transform of a time function f(t), is a definite integral with respect to (w.r.t.) t. The result, F(s), must not contain the time variable, t. If your final answer contains the variable t it indicates that you have little or no understanding of what you are doing, and the grade for the entire offending part, a) or b), will be 0 points.
Part a) NNNNA VNNN a 8 b 3 sec (-1) c 2 rad/sec d (5/9)*pi rad Part b) a = -4 b -7 c 4 sec^ (-1) NNNN NNNNy
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Answer #1

Part (a)

a(t)e co Cet d) coet) cood Smfet) sm d LT cood. S sc2 cSmd c, d in Gls) we et Cop s2+4 - 6.173) s 2l0.98) GIS) = 0.173 s- 1.9

Part (b)

(i) aSlt) i) hti b.21 S3 bte-ct 2b (Ste)3 f(t) F(s) a t 2b (S+93 alsre)26 Src) - a (s3++3sc(s+ c)) + 26 3+c+3s2 c+3sc as+3as2

in patlb) Fls)- BCS) -4s3-48s-192s-270 + 12 s +48S +64 A(S)

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