8.4E: The Unit Step Function (Exercises)

Q8.4.1

In Exercises 8.4.1-8.4.6 find the Laplace transform by the method of Example 8.4.1. Then express the given function \(f\) in terms of unit step functions as in Equation 8.4.8, and use Theorem 8.4.1 to find \({\cal L}(f)\). Graph \(f\) for Exercises 8.4.3 and 8.4.4.

1. \(f(t)=\left\{\begin{array}{cl} {1,}&{0 \le t<4,}\\ {t,} & {t\ge4.} \end{array}\right.\)

2. \(f(t)=\left\{\begin{array}{cl} t,&0 \le t<1,\\[4pt] 1,& t\ge1.\end{array}\right.\)

3. \(f(t)=\left\{\begin{array}{cl} 2t-1,& 0\le t<2,\\[4pt] t,&t\ge2.\end{array}\right.\)

4. \(f(t)=\left\{\begin{array}{cl}1, &0\le t<1,\\[4pt] t+2,&t\ge1.\end{array}\right.\)

5. \(f(t)=\left\{\begin{array}{cl} t-1,& 0\le t<2,\\[4pt] 4,&t\ge2.\end{array}\right.\)

6. \(f(t)=\left\{\begin{array}{cl} t^2,& 0\le t<1,\\[4pt] 0,&t\ge1.\end{array}\right.\)

Q8.4.2

In Exercises 8.4.7-8.4.18 express the given function \(f\) in terms of unit step functions and use Theorem 8.4.1 to find \(\cal{L} (f)\). Graph \(f\) for Exercises 8.4.15-8.4.18.

7. \(f(t)=\left\{\begin{array}{cl} 0, &0\le t<2,\\[4pt] t^2+3t,&t\ge2.\end{array}\right.\)

8. \(f(t)=\left\{\begin{array}{cl} t^2+2, &0\le t<1,\\[4pt] t,&t\ge1.\end{array}\right.\)

9. \(f(t)=\left\{\begin{array}{cl} te^t,& 0\le t <1,\\[4pt] e^t,&t\ge1.\end{array}\right.\)

10. \(f(t)=\left\{\begin{array}{cl} e^{\phantom{2}-t}, &0\le t<1,\\[4pt] e^{-2t},&t\ge1.\end{array}\right.\)

11. \(f(t)=\left\{\begin{array}{cl} -t,&0 \le t<2,\\[4pt] t-4,&2\le t<3,\\[4pt] 1,&t\ge3. \end{array}\right.\)

12. \(f(t)=\left\{\begin{array}{cl} 0,&0 \le t<1,\\[4pt] t,&1\le t<2,\\[4pt] 0,&t\ge2.\end{array}\right.\)

13. \(f(t)=\left\{\begin{array}{cl} t,&0 \le t<1,\\[4pt] t^2,&1\le t<2,\\[4pt] 0,&t\ge2. \end{array}\right.\)

14. \(f(t)=\left\{\begin{array}{cl} t,&0\le t<1,\\[4pt] 2-t,&1\le t<2,\\[4pt] 6,&t > 2. \end{array}\right.\)

15. \(f(t)=\left\{\begin{array}{cl} {\sin t,}&{0\leq t<\frac{\pi }{2}}\\{2\sin t,}&{\frac{\pi }{2}\leq t<\pi }\\{\cos t,}&{t\geq \pi } \end{array} \right.\)

16. \(f(t)=\left\{\begin{array}{cl}\phantom{-} 2,&0\le t<1,\\[4pt]-2t+2,&1\le t<3,\\[4pt]\phantom{-}3t,&t\ge 3.\end{array}\right.\)

17. \(f(t)=\left\{\begin{array}{cl}3,&0\le t<2,\\[4pt]3t+2,&2\le t<4,\\[4pt]4t,&t\ge 4.\end{array}\right.\)

18. \(f(t)=\left\{\begin{array}{ll}(t+1)^2,&0\le t<1, \\[4pt](t+2)^2,&t\ge1.\end{array}\right.\)

Q8.4.3

In Exercises 8.4.19-8.4.28 use Theorem 8.4.2 to express the inverse transforms in terms of step functions, and then find distinct formulas the for inverse transforms on the appropriate intervals, as in Example 8.4.7. Graph the inverse transform for Exercises 8.4.21, 8.4.22, and 8.4.25.

19. \(H(s)={e^{-2s}\over s-2}\)

20. \(H(s)={e^{-s}\over s(s+1)}\)

21. \(H(s)={e^{-s}\over s^3}+ {e^{-2s}\over s^2}\)

22. \(H(s)=\left({2\over s}+{1\over s^2}\right) +e^{-s}\left({3\over s}-{1\over s^2}\right)+e^{-3s}\left({1\over s}+{1\over s^2}\right)\)

23. \(H(s)=\left({5\over s}-{1\over s^2}\right) +e^{-3s}\left({6\over s}+{7\over s^2}\right)+{3e^{-6s}\over s^3}\)

24. \(H(s)={e^{-\pi s} (1-2s)\over s^2+4s+5}\)

25. \(H(s)=\left({1\over s}-{s\over s^2+1}\right)+e^{-{\pi\over 2}s}\left({3s-1\over s^2+1}\right)\)

26. \(H(s)= e^{-2s}\left[{3(s-3)\over(s+1)(s-2)}-{s+1\over(s-1)(s-2)}\right]\)

27. \(H(s)={1\over s}+{1\over s^2}+e^{-s}\left({3\over s}+{2\over s^2}\right) +e^{-3s}\left({4\over s}+{3\over s^2}\right)\)

28. \(H(s)={1\over s}-{2\over s^3}+e^{-2s}\left({3\over s}-{1\over s^3}\right) +{e^{-4s}\over s^2}\)

Q8.4.4

29. Find \({\cal L}\left(u(t-\tau)\right)\).

30. Let \(\{t_m\}_{m=0}^\infty\) be a sequence of points such that \(t_0=0\), \(t_{m+1}>t_m\), and \(\lim_{m\to\infty}t_m=\infty\). For each nonnegative integer \(m\), let \(f_m\) be continuous on \([t_m,\infty)\), and let \(f\) be defined on \([0,\infty)\) by

\[f(t)=f_m(t),\,t_m\le t<t_{m+1}\quad (m=0,1,\dots).\nonumber \]

Show that \(f\) is piecewise continuous on \([0,\infty)\) and that it has the step function representation

\[f(t)=f_0(t)+\sum_{m=1}^\infty u(t-t_m)\left(f_m(t)-f_{m-1}(t)\right),\, 0\le t<\infty.\nonumber\]

How do we know that the series on the right converges for all \(t\) in \([0,\infty)\)?

31. In addition to the assumptions of Exercise 8.4.30, assume that

\[|f_m(t)|\le Me^{s_0t},\,t\ge t_m,\,m=0,1,\dots, \tag{A}\]

and that the series

\[\sum_{m=0}^\infty e^{-\rho t_m} \tag{B}\]

converges for some \(\rho>0\). Using the steps listed below, show that \({\cal L}(f)\) is defined for \(s>s_0\) and

\[{\cal L}(f)={\cal L}(f_0)+\sum_{m=1}^\infty e^{-st_m}{\cal L}(g_m) \tag{C}\]

for \(s>s_0+\rho\), where

\[g_m(t)=f_m(t+t_m)-f_{m-1}(t+t_m).\nonumber\]

1. Use (A) and Theorem 8.1.6 to show that \[{\cal L}(f)=\sum_{m=0}^\infty\int_{t_m}^{t_{m+1}}e^{-st}f_m(t)\,dt \tag{D}\] is defined for \(s>s_0\).
2. Show that (D) can be rewritten as \[{\cal L}(f)=\sum_{m=0}^\infty\left(\int_{t_m}^\infty e^{-st}f_m(t)\,dt -\int_{t_{m+1}}^\infty e^{-st}f_m(t)\,dt\right). \tag{E}\]
3. Use (A), the assumed convergence of (B), and the comparison test to show that the series \[\sum_{m=0}^\infty\int_{t_m}^\infty e^{-st}f_m(t)\,dt\quad \text{and} \quad \sum_{m=0}^\infty\int_{t_{m+1}}^\infty e^{-st}f_m(t)\,dt\nonumber\] both converge (absolutely) if \(s>s_0+\rho\).
4. Show that (E) can be rewritten as \[{\cal L}(f)={\cal L}(f_0)+\sum_{m=1}^\infty\int_{t_m}^\infty e^{-st} \left(f_m(t)-f_{m-1}(t)\right)\,dt\nonumber \] if \(s>s_0+\rho\).
5. Complete the proof of (C).

32.  Suppose \(\{t_m\}_{m=0}^\infty\) and \(\{f_m\}_{m=0}^\infty\) satisfy the assumptions of Exercises 8.4.30 and 8.4.31, and there’s a positive constant \(K\) such that \(t_m\ge Km\) for \(m\) sufficiently large. Show that the series (B) of Exercise 8.4.31 converges for any \(\rho>0\), and conclude from this that (C) of Exercise 8.4.31 holds for \(s>s_0\).

In Exercises 8.4.33-8.4.36 find the step function representation of \(f\) and use the result of Exercise 8.4.32 to find \(\cal{L}(f)\). HINT: You will need formulas related to the formula for the sum of a geometric series.

33. \(f(t)=m+1,\,m\le t<m+1\; (m=0,1,2,\dots)\)

34. \(f(t)=(-1)^m,\,m\le t<m+1\; (m=0,1,2,\dots)\)

35. \(f(t)=(m+1)^2,\,m\le t<m+1\; (m=0,1,2,\dots)\)

36. \(f(t)=(-1)^mm,\,m\le t<m+1\; (m=0,1,2,\dots)\)