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# 8.4E: The Unit Step Function (Exercises)

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