11.53: A.8.4- Section 8.4 Answers
( \newcommand{\kernel}{\mathrm{null}\,}\)
selected template will load here
This action is not available.
( \newcommand{\kernel}{\mathrm{null}\,}\)
1. 1+u(t−4)(t−1);1s+e−4s(1s2+3s)
2. t+u(t−1)(1−t);1−e−ss2
3. 2t−1−u(t−2)(t−1);(2s2−1s)−e−2s(1s2+1s)
4. 1+u(t−1)(t+1);1s+e−s(1s2+2s)
5. t−1+u(t−2)(5−t);1s2−1s−e−2s(1s2−3s)
6. t2(1−u(t−1));2s3−e−s(2s3+2s2+1s)
7. u(t−2)(t2+3t);e−2s(2s3+7s2+10s)
8. t2+2+u(t−1)(t−t2−2);2s3+2s−e−s(2s3+1s2+2s)
9. tet+u(t−1)(et−tet);1−e−(s−1)(s−1)2
10. e−t+u(t−1)(e−2t−e−t);1−e−(s+1)s+1+e−(s+2)s+2
11. −t+2u(t−2)(t−2)−u(t−3)(t−5);−1s2+2e−2ss2+e−3s(2s−1s2)
12. [u(t−1)−u(t−2)]t;e−s(1s2+1s)−e−2s(1s2+2s)
13. t+u(t−1)(t2−t)−u(t−2)t2;1s2+e−s(2s3+1s2)−e−2s(2s3+4s2+4s)
14. t+u(t−1)(2−2t)+u(t−2)(4+t);1s2−2e−ss2+e−2s(1s2+6s)
15. sint+u(t−π/2)sint+u(t−π)(cost−2sint);1+e−π2ss−e−πs(s−2)s2+1
16. 2−2u(t−1)t+u(t−3)(5t−2);2s−e−s(2s2+2s)+e−3s(5s2+13s)
17. 3+u(t−2)(3t−1)+u(t−4)(t−2);3s+e−2s(3s2+5s)+e−4s(1s2+2s)
18. (t+1)2+u(t−1)(2t+3);2s3+2s2+1s+e−s(2s2+5s)
19. u(t−2)e2(t−2)={0,0≤t<2,e2(t−2),t≥2
20. u(t−1)(1−e−(t−1))={0,0≤t<1,1−e−(t−1),t≥1
21. u(t−1)(t−1)22+u(t−2)(t−2)={0,0≤t<1,(t−1)22,1≤t<2,t2−32,t≥2
22. 2+t+u(t−1)(4−t)+u(t−3)(t−2)={2+t,0≤t<1,6,1≤t<3,t+4,t≥3
23. 5−t+u(t−3)(7t−15)+32u(t−6)(t−6)2={5−t,0≤t<3,6t−10,3≤t<6,44−12t+32t2,t≥6
24. u(t−π)e−2(t−π)(2cost−5sint)={0,0≤t<π,e−2(t−π)(2cost−5sint)t≥π
25. 1−cost+u(t−π/2)(3sint+cost)={1−cost,0≤t<π2,1+3sint,t≥π2
26. u(t−2)(4e−(t−2)−4e2(t−2)+2e(t−2)={0,0≤t<2,4e−(t−2)−4e2(t−2)+2e(t−2),t≥2
27. 1+t+u(t−1)(2t+1)+u(t−3)(3t−5)={t+1,0≤t<1,3t+2,1≤t<3,6t−3,t≥3
28. 1−t2+u(t−2)(−t22+2t+1)+u(t−4)(t−4)={1−t2,0≤t<2,−3t22+2t+2,2≤t<4,−3t22+3t−2,t≥4
29. e−τss
30. For each t only finitely many terms are nonzero.
33. 1+∑∞m=1u(t−m);1s(1−e−s)
34. 1+2∑∞m=1(−1)mu(t−m);1s;1−e−s1+e−s
35. 1+∑∞m=1(2m+1)u(t−m);e−s(1+e−s)s(1−e−s)2
36. ∑∞m=1(−1)m(2m−1)u(t−m);1s(1−es)(1+es)2