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Mathematics LibreTexts

8.4E: The Unit Step Function (Exercises)

  • Page ID
    18294
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    In Exercises [exer:8.4.1}– [exer:8.4.6} find the Laplace transform by the method of Example [example:8.4.1}. Then express the given function \(f\) in terms of unit step functions as in Equation \ref{eq:8.4.6}, and use Theorem [thmtype:8.4.1} to find \({\cal L}(f)\). Where indicated by , graph \(f\).

    [exer:8.4.1] \(\{f(t)=\left\{\begin{array}{cl} 1,&0 \le t<4,\\[4pt] t, & t\ge4.\end{array}\right.}\)

    [exer:8.4.2] \(\{f(t)=\left\{\begin{array}{cl} t,&0 \le t<1,\\[4pt] 1,& t\ge1.\end{array}\right.}\)

    [exer:8.4.3] \(\{f(t)=\left\{\begin{array}{cl} 2t-1,& 0\le t<2,\\[4pt] t,&t\ge2.\end{array}\right.}\)

    [exer:8.4.4] \(\{f(t)=\left\{\begin{array}{cl}1, &0\le t<1,\\[4pt] t+2,&t\ge1.\end{array}\right.}\)

    [exer:8.4.5]\(\{f(t)=\left\{\begin{array}{cl} t-1,& 0\le t<2,\\[4pt] 4,&t\ge2.\end{array}\right.}\)

    [exer:8.4.6] \(\{f(t)=\left\{\begin{array}{cl} t^2,& 0\le t<1,\\[4pt] 0,&t\ge1.\end{array}\right.}\)

    [exer:8.4.7]\(\{f(t)=\left\{\begin{array}{cl} 0, &0\le t<2,\\[4pt] t^2+3t,&t\ge2.\end{array}\right.}\)

    [exer:8.4.8] \(\{f(t)=\left\{\begin{array}{cl} t^2+2, &0\le t<1,\\[4pt] t,&t\ge1.\end{array}\right.}\)

    [exer:8.4.9] \(\{f(t)=\left\{\begin{array}{cl} te^t,& 0\le t <1,\\[4pt] e^t,&t\ge1.\end{array}\right.}\)

    [exer:8.4.10] \(\{f(t)=\left\{\begin{array}{cl} e^{\phantom{2}-t}, &0\le t<1,\\[4pt] e^{-2t},&t\ge1.\end{array}\right.}\)

    [exer:8.4.11] \(\{f(t)=\left\{\begin{array}{cl} -t,&0 \le t<2,\\[4pt] t-4,&2\le t<3,\\[4pt] 1,&t\ge3. \end{array}\right.}\)

    [exer:8.4.12] \(\{f(t)=\left\{\begin{array}{cl} 0,&0 \le t<1,\\[4pt] t,&1\le t<2,\\[4pt] 0,&t\ge2.\end{array}\right.}\)

    [exer:8.4.13] \(\{f(t)=\left\{\begin{array}{cl} t,&0 \le t<1,\\[4pt] t^2,&1\le t<2,\\[4pt] 0,&t\ge2. \end{array}\right.}\)

    [exer:8.4.14] \(\{f(t)=\left\{\begin{array}{cl} t,&0\le t<1,\\[4pt] 2-t,&1\le t<2,\\[4pt] 6,&t > 2. \end{array}\right.}\)

    [exer:8.4.15] \(\{f(t)=\left\{\begin{array}{cl} \phantom{2} \sin t,&0\le t<\{\pi\over 2},\\[4pt] 2\sin t,& \{\pi\over 2}\le t<\pi,\\[4pt]\phantom{2}\cos t, &t\ge\pi.\end{array}\right.}\)

    [exer:8.4.16] \(f(t)=\{\left\{\begin{array}{cl}\phantom{-} 2,&0\le t<1,\\[4pt]-2t+2,&1\le t<3,\\[4pt]\phantom{-}3t,&t\ge 3.\end{array}\right.}\)

    [exer:8.4.17] \(f(t)=\{\left\{\begin{array}{cl}3,&0\le t<2,\\[4pt]3t+2,&2\le t<4,\\[4pt]4t,&t\ge 4.\end{array}\right.}\)

    [exer:8.4.18] \(\{f(t)=\left\{\begin{array}{ll}(t+1)^2,&0\le t<1, \\[4pt](t+2)^2,&t\ge1.\end{array}\right.}\)

    [exer:8.4.19] \(\{H(s)={e^{-2s}\over s-2}}\)[exer:8.4.20] \(\{H(s)={e^{-s}\over s(s+1)}}\)

    [exer:8.4.21] \(\{H(s)={e^{-s}\over s^3}+ {e^{-2s}\over s^2}}\)

    [exer:8.4.22] \(\{H(s)=\left({2\over s}+{1\over s^2}\right) +e^{-s}\left({3\over s}-{1\over s^2}\right)+e^{-3s}\left({1\over s}+{1\over s^2}\right)}\)

    [exer:8.4.23] \(\{H(s)=\left({5\over s}-{1\over s^2}\right) +e^{-3s}\left({6\over s}+{7\over s^2}\right)+{3e^{-6s}\over s^3}}\)

    [exer:8.4.24] \(\{H(s)={e^{-\pi s} (1-2s)\over s^2+4s+5}}\)

    [exer:8.4.25] \(\{H(s)=\left({1\over s}-{s\over s^2+1}\right)+e^{-{\pi\over 2}s}\left({3s-1\over s^2+1}\right)}\)

    [exer:8.4.26] \(\{H(s)= e^{-2s}\left[{3(s-3)\over(s+1)(s-2)}-{s+1\over(s-1)(s-2)}\right]}\)

    [exer:8.4.27] \(\{H(s)={1\over s}+{1\over s^2}+e^{-s}\left({3\over s}+{2\over s^2}\right) +e^{-3s}\left({4\over s}+{3\over s^2}\right)}\)

    [exer:8.4.28] \(\{H(s)={1\over s}-{2\over s^3}+e^{-2s}\left({3\over s}-{1\over s^3}\right) +{e^{-4s}\over s^2}}\)

    [exer:8.4.29] Find \({\cal L}\left(u(t-\tau)\right)\).

    [exer:8.4.30] 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

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

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

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

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

    [exer:8.4.31] In addition to the assumptions of Exercise [exer:8.4.30}, assume that

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

    and that the series

    \[\sum_{m=0}^\infty e^{-\rho t_m} \eqno{\rm (B)}\]

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

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

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

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

    Use (A) and Theorem 8.1.6 to show that

    \[{\cal L}(f)=\sum_{m=0}^\infty\int_{t_m}^{t_{m+1}}e^{-st}f_m(t)\,dt \eqno{\rm (D)}\]

    is defined for \(s>s_0\).

    Show that (D) can be rewritten as

    \[{\cal L}(f)=\sum_{m=0}^\infty\left(\int_{t_m}^\infty e^{-st}f_m(t)\,dt -\int_{t_{m+1}}^\infty e^{-st}f_m(t)\,dt\right). \eqno{\rm (E)}\]

    Use (A), the assumed convergence of (B), and the comparison test to show that the series

    \[\sum_{m=0}^\infty\int_{t_m}^\infty e^{-st}f_m(t)\,dt\mbox{\quad and \quad} \sum_{m=0}^\infty\int_{t_{m+1}}^\infty e^{-st}f_m(t)\,dt\]

    both converge (absolutely) if \(s>s_0+\rho\).

    Show that (E) can be rewritten as

    \[{\cal L}(f)={\cal L}(f_0)+\sum_{m=1}^\infty\int_{t_m}^\infty e^{-st} \left(f_m(t)-f_{m-1}(t)\right)\,dt\]

    if \(s>s_0+\rho\).

    Complete the proof of (C).

    [exer:8.4.32] Suppose \(\{t_m\}_{m=0}^\infty\) and \(\{f_m\}_{m=0}^\infty\) satisfy the assumptions of Exercises [exer:8.4.30} and [exer: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 [exer:8.4.31} converges for any \(\rho>0\), and conclude from this that (C) of Exercise [exer:8.4.31} holds for \(s>s_0\).

    [exer:8.4.33] \(f(t)=m+1,\,m\le t<m+1\; (m=0,1,2,\dots)\)

    [exer:8.4.34] \(f(t)=(-1)^m,\,m\le t<m+1\; (m=0,1,2,\dots)\)

    [exer:8.4.35] \(f(t)=(m+1)^2,\,m\le t<m+1\; (m=0,1,2,\dots)\)

    [exer:8.4.36] \(f(t)=(-1)^mm,\,m\le t<m+1\; (m=0,1,2,\dots)\)