bookstylehw9solution

# bookstylehw9solution - 9.9 Using partial fraction expansion...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 9.9. Using partial fraction expansion 4 2 3+4 3+3. X(s) = Taking the inverse Laplace transform, 30,) m tie—“um - 2e“3‘u(t). 9.29. (3.) Using Table 9.2, we obtain : “R > —1 Ms) 3+1, e{s} and 1 c = R > ——2 III-(3) 8+2, e{s} (1)) Since y(t) = \$05} * Mt), we may use the convolution property to obtain 1 The ROG of Y(s) is 72:2{3} > ~1. (c) Performing partial fraction expansion on Y(s), we obtain 1 1 3+1 3+2' Taking the inverse Laplace transform, we get Y(s) 2 y(t) = e—tu(t) — e‘2tu(t). (d) Explicit convolution of z(t) and h(t) gives us y(t) = fie h('r)x(t —— ﬂair 00 00 2 / e‘gTe"(t"T}u(t—'r)dv- 0 H a (ft/- (Td'r for t > 0 0 [6“ w 6‘2‘}u(t). H (C) Us ‘3 15(4) Figure 89.37 9.40. Taking the unilateral Laplace transform of both sides of the given differential equation, we get 3339(8) H 3211(0“) - Sy'(0") - ENG") + 63231(3) - 63940") —6y(0—) + 11sy{s) — 11y(0‘) + 6343) = “ml {3940—1) (a) For the zero state response, assume that all the initial conditions are zero. Furthermore, from the given :r(t) we may determine 1 (15(8) 2 m, R€{S} > —4. From eq. (89.40-1), we get 1 r3 2 . r + +11 +6 3 . 32(9)“; 63 s ] 3+4 Therefore, 1 (s+4)(s3 +632+113+6)' ' Taking the inverse unilateral Laplace transform of the partial fraction expansion of the above equation, we get y(t) = ée”‘u(t) — éeﬂuuﬁ) + 32(3) = ée‘ztuﬁ) — —e—3tu[t). (b) For the zero—input response, we assume that X(.9) : 0. Assuming that the initial conditions are as given, we obtain from (89.404) 32+53+6 1 s3+632~§~lls+6 = 3+1‘ Taking the inverse unilateral Laplace transform of the above equation, we get y(t) = e“u(t). (c) The total response is the sum of the zero-state and zero-input responses. 1 1 1 y(t) = gee—tum — E‘s—“1413 + ifs-”um — 56—311136). NS) = ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern