8.4 The Unit Step Function

In the next section we’ll consider initial value problems

\[ ay''+by'+cy=f(t),\quad y(0)=k_0,\quad y'(0)=k_1, \]

where \(a\), \(b\), and \(c\) are constants and \(f\) is piecewise continuous. In this section we’ll develop procedures for using the table of Laplace transforms to find Laplace transforms of piecewise continuous functions, and to find the piecewise continuous inverses of Laplace transforms.

Example 8.4.1

Use the table of Laplace transforms to find the Laplace transform of

\begin{equation} f(t)=\left\{\begin{array}{cl} 2t+1,&0\le t<2,\\[3pt] 3t,&t\ge2 \end{array}\right. \tag{8.4.1}\end{equation}

(Figure 8.4.1).

Solution Since the formula for \(f\) changes at \(t=2\), we write

\begin{equation} \begin{array}{ccl} {\cal L}(f)&=&\dst \int_0^\infty e^{-st}f(t)\,dt\\[6pt] &=&\dst\int_0^2 e^{-st}(2t+1)\,dt+\int_2^\infty e^{-st}(3t)\,dt. \end{array} \tag{8.4.2}\end{equation}

To relate the first term to a Laplace transform, we add and subtract

\[ \int_2^\infty e^{-st}(2t+1)\,dt \]

in (8.4.2) to obtain

\begin{equation} \begin{array}{ccl} {\cal L}(f)&=&\dst\int_0^\infty e^{-st}(2t+1)\,dt+ \int_2^\infty e^{-st}(3t-2t-1)\,dt\\[6pt] &=&\dst\int_0^\infty e^{-st}(2t+1)\,dt+ \int_2^\infty e^{-st}(t-1)\,dt\\[6pt] &=&\dst{\cal L}(2t+1)+\int_2^\infty e^{-st}(t-1)\,dt. \end{array} \tag{8.4.3}\end{equation}

To relate the last integral to a Laplace transform, we make the change of variable \(x=t-2\) and rewrite the integral as

\begin{eqnarray*} \int_2^\infty e^{-st}(t-1)\,dt&=&\int_0^\infty e^{-s(x+2)}(x+1)\,dx \\[6pt] &=&e^{-2s}\int_0^\infty e^{-sx}(x+1)\,dx. \end{eqnarray*}

Since the symbol used for the variable of integration has no effect on the value of a definite integral, we can now replace \(x\) by the more standard \(t\) and write

\[ \int_2^\infty e^{-st}(t-1)\,dt =e^{-2s}\int_0^\infty e^{-st}(t+1)\,dt=e^{-2s}{\cal L}(t+1). \]

This and (8.4.3) imply that

\[ {\cal L}(f)={\cal L}(2t+1)+e^{-2s}{\cal L} (t+1). \]

Now we can use the table of Laplace transforms to find that

\[ {\cal L}(f)={2\over s^2}+{1\over s} +e^{-2s}\left({1\over s^2}+{1\over s}\right).\bbox \]
The piecewise continuous function eq:8.4.1
Figure 8.4.1. The piecewise continuous function (8.4.1)
y=u(t-)
Figure 8.4.2. \(y=u(t-\tau)\)

Laplace Transforms of Piecewise Continuous Functions

We’ll now develop the method of Example 8.4.1 into a systematic way to find the Laplace transform of a piecewise continuous function. It is convenient to introduce the unit step function, defined as

\begin{equation} u(t)=\left\{\begin{array}{rl} 0,&t<0\\ 1,&t\ge0. \end{array}\right. \tag{8.4.4}\end{equation}

Thus, \(u(t)\) “steps” from the constant value \(0\) to the constant value \(1\) at \(t=0\). If we replace \(t\) by \(t-\tau\) in (8.4.4), then

\[ u(t-\tau)=\left\{\begin{array}{rl} 0,&t<\tau,\\ 1,&t\ge\tau \end{array}\right.; \]

that is, the step now occurs at \(t=\tau\) (Figure 8.4.2).

The step function enables us to represent piecewise continuous functions conveniently. For example, consider the function

\begin{equation} f(t)=\left\{\begin{array}{rl} f_0(t),&0\le t<t_1,\\[3pt] f_1(t),&t\ge t_1, \end{array}\right. \tag{8.4.5}\end{equation}

where we assume that \(f_0\) and \(f_1\) are defined on \([0,\infty)\), even though they equal \(f\) only on the indicated intervals. This assumption enables us to rewrite (8.4.5) as

\begin{equation} f(t)=f_0(t)+u(t-t_1)\left(f_1(t)-f_0(t)\right). \tag{8.4.6}\end{equation}

To verify this, note that if \(t<t_1\) then \(u(t-t_1)=0\) and (8.4.6) becomes

\[ f(t)=f_0(t)+(0)\left(f_1(t)-f_0(t)\right)=f_0(t). \]

If \(t\ge t_1\) then \(u(t-t_1)=1\) and (8.4.6) becomes

\[ f(t)=f_0(t)+(1)\left(f_1(t)-f_0(t)\right)=f_1(t). \]

We need the next theorem to show how (8.4.6) can be used to find \({\cal L}(f)\).

Theorem 8.4.1

Let \(g\) be defined on \([0,\infty).\) Suppose \(\tau\ge0\) and \({\cal L}\left(g(t+\tau)\right)\) exists for \(s>s_0.\) Then \({\cal L}\left(u(t-\tau)g(t)\right)\) exists for \(s>s_0\), and

\[ {\cal L}(u(t-\tau)g(t))=e^{-s\tau}{\cal L}\left(g(t+\tau)\right). \]

Proof By definition,

\[ {\cal L}\left(u(t-\tau)g(t)\right)=\int_0^\infty e^{-st} u(t-\tau)g(t)\, dt. \]

From this and the definition of \(u(t-\tau)\),

\[ {\cal L}\left(u(t-\tau)g(t)\right)=\int_0^\tau e^{-st}(0)\,dt+\int_{\tau}^\infty e^{-st}g(t)\,dt. \]

The first integral on the right equals zero. Introducing the new variable of integration \(x=t-\tau\) in the second integral yields

\[ {\cal L}\left(u(t-\tau)g(t)\right)=\int_0^\infty e^{-s(x+\tau)}g(x+\tau)\,dx =e^{-s\tau}\int_0^\infty e^{-sx} g(x+\tau)\,dx. \]

Changing the name of the variable of integration in the last integral from \(x\) to \(t\) yields

\[ {\cal L}\left(u(t-\tau)g(t)\right) =e^{-s\tau}\int_0^\infty e^{-st} g(t+\tau)\,dt=e^{-s\tau}{\cal L}(g(t+\tau)).\bbox \]

Example 8.4.2

Find

\[ {\cal L}\left(u(t-1)(t^2+1)\right). \]

Solution Here \(\tau=1\) and \(g(t)=t^2+1\), so

\[ g(t+1)=(t+1)^2+1=t^2+2t+2. \]

Since

\[ {\cal L}\left(g(t+1)\right)={2\over s^3}+{2\over s^2}+{2\over s}, \]

Theorem 8.4.1 implies that

\[{\cal L}\left(u(t-1)(t^2+1)\right) =e^{-s}\left({2\over s^3}+{2\over s^2}+{2\over s}\right). \]

Example 8.4.3

Use Theorem 8.4.1 to find the Laplace transform of the function

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

from Example 8.4.1.

Solution We first write \(f\) in the form (8.4.6) as

\[ f(t)=2t+1+u(t-2)(t-1). \]

Therefore

\begin{eqnarray*} {\cal L}(f)&=&{\cal L}(2t+1) +{\cal L}\left(u(t-2)(t-1)\right) \\ &=&{\cal L}(2t+1) +e^{-2s}{\cal L}(t+1)\quad\mbox{ (from Theorem~\ref{thmtype:8.4.1}}) \\ &=&{2\over s^2}+{1\over s}+e^{-2s}\left({1\over s^2}+{1\over s}\right), \end{eqnarray*}

which is the result obtained in Example 8.4.1.∎

Formula (8.4.6) can be extended to more general piecewise continuous functions. For example, we can write

\[ f(t)=\left\{\begin{array}{rl} f_0(t),&0\le t<t_1,\\[3pt] f_1(t),&t_1\le t<t_2,\\[3pt] f_2(t),&t\ge t_2, \end{array}\right. \]

as

\[ f(t)=f_0(t)+u(t-t_1)\left(f_1(t)-f_0(t)\right)+ u(t-t_2)\left(f_2(t)-f_1(t)\right) \]

if \(f_0\), \(f_1\), and \(f_2\) are all defined on \([0,\infty)\).

Example 8.4.4

Find the Laplace transform of

\begin{equation} f(t)=\left\{\begin{array}{cl} 1,&0\le t<2,\\[3pt] -2t+1,&2\le t<3,\\[3pt] 3t,&3\le t<5,\\[3pt] t-1,&t\ge5 \end{array}\right. \tag{8.4.7}\end{equation}

(Figure 8.4.3).

The piecewise contnuous function eq:8.4.7
Figure 8.4.3. The piecewise contnuous function (8.4.7)

Solution In terms of step functions,

\begin{eqnarray*} f(t)&=&1+u(t-2)(-2t+1-1)+u(t-3)(3t+2t-1) \\ &&+u(t-5)(t-1-3t), \end{eqnarray*}

or

\[ f(t)=1-2u(t-2)t+u(t-3)(5t-1)-u(t-5)(2t+1). \]

Now Theorem 8.4.1 implies that

\begin{eqnarray*} {\cal L}(f)&=&{\cal L}(1)-2e^{-2s}{\cal L}(t+2)+e^{-3s}{\cal L}\left(5(t+3)-1\right)-e^{-5s}{\cal L}\left(2(t+5)+1\right) \\[3pt] &=&{\cal L}(1)-2e^{-2s}{\cal L}(t+2)+e^{-3s}{\cal L}(5t+14)-e^{-5s}{\cal L}(2t+11) \\[3pt] &=&{1\over s}-2e^{-2s}\left({1\over s^2}+{2\over s}\right)+ e^{-3s}\left({5\over s^2}+{14\over s}\right)-e^{-5s}\left({2\over s^2}+{11\over s}\right).\bbox \end{eqnarray*}

The trigonometric identities

\begin{eqnarray} \sin (A+B)&=&\sin A\cos B+\cos A\sin B \tag{8.4.8} \\ \cos (A+B)&=&\cos A\cos B-\sin A\sin B \tag{8.4.9}\end{eqnarray}

are useful in problems that involve shifting the arguments of trigonometric functions. We’ll use these identities in the next example.

Example 8.4.5

Find the Laplace transform of

\begin{equation} f(t)=\left\{\begin{array}{cl} \sin t,&0\le t<\dst{\pi\over 2},\\[6pt] \cos t-3\sin t,&\dst{\pi\over 2}\le t<\pi,\\[6pt] 3\cos t,&t\ge\pi \end{array}\right. \tag{8.4.10}\end{equation}

(Figure 8.4.4).

The piecewise continuous function eq:8.4.10
Figure 8.4.4. The piecewise continuous function (8.4.10)

Solution In terms of step functions,

\[ f(t)=\sin t+u(t-\pi/2) (\cos t-4\sin t)+u(t-\pi) (2 \cos t+3\sin t). \]

Now Theorem 8.4.1 implies that

\begin{equation} \begin{array}{ccl} {\cal L}(f)&=&{\cal L}(\sin t)+e^{-{\pi\over 2}s}{\cal L} \left(\cos\left(t+{\pi\over2}\right)-4\sin\left(t+{\pi\over2}\right)\right)\\ [3pt]&&\qquad+e^{-\pi s}{\cal L}\left(2\cos(t+\pi)+3\sin(t+\pi)\right). \end{array} \tag{8.4.11}\end{equation}

Since

\[ \cos\left(t+{\pi\over 2}\right)-4\sin\left(t+{\pi\over 2}\right)=-\sin t-4\cos t \]

and

\[ 2\cos (t+\pi)+3\sin (t+\pi)=-2\cos t-3\sin t, \]

we see from (8.4.11) that

\begin{eqnarray*} {\cal L}(f)&=&{\cal L}(\sin t)-e^{-\pi s/2}{\cal L}(\sin t+4\cos t) -e^{-\pi s}{\cal L}(2\cos t+3\sin t) \\[3pt] &=&{1\over s^2+1}-e^{-{\pi\over 2}s}\left({1+4s\over s^2+1}\right) -e^{-\pi s}\left({3+2s\over s^2+1}\right).\bbox \end{eqnarray*}

The Second Shifting Theorem

Replacing \(g(t)\) by \(g(t-\tau)\) in Theorem 8.4.1 yields the next theorem.

Theorem 8.4.2

If \(\tau\ge0\) and \({\cal L}(g)\) exists for \(s>s_0\) then \({\cal L}\left(u(t-\tau)g(t-\tau)\right)\) exists for \(s>s_0\) and

\[ {\cal L}(u(t-\tau)g(t-\tau))=e^{-s\tau}{\cal L}(g(t)), \]

or, equivalently,

\begin{equation} \mbox{if } g(t)\leftrightarrow G(s),\mbox{ then }u(t-\tau)g(t-\tau)\leftrightarrow e^{-s\tau}G(s). \tag{8.4.12}\end{equation}

Remark

Recall that the First Shifting Theorem (Theorem 8.1.3 states that multiplying a function by \(e^{at}\) corresponds to shifting the argument of its transform by \(a\) units. Theorem 8.4.2 states that multiplying a Laplace transform by the exponential \(e^{-\tau s}\) corresponds to shifting the argument of the inverse transform by \(\tau\) units.

Example 8.4.6

Use (8.4.12) to find

\[ {\cal L}^{-1}\left(e^{-2s}\over s^2\right). \]

Solution To apply (8.4.12) we let \(\tau=2\) and \(G(s)=1/s^2\). Then \(g(t)=t\) and (8.4.12) implies that

\[ {\cal L}^{-1}\left(e^{-2s}\over s^2\right)=u(t-2)(t-2).\bbox \]

Example 8.4.7

Find the inverse Laplace transform \(h\) of

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

and find distinct formulas for \(h\) on appropriate intervals.

Solution Let

\[ G_0(s)={1\over s^2},\quad G_1(s)={1\over s^2}+{2\over s},\quad G_2(s)={4\over s^3}+{1\over s}. \]

Then

\[ g_0(t)=t,\; g_1(t)=t+2,\; g_2(t)=2t^2+1. \]

Hence, (8.4.12) and the linearity of \({\cal L}^{-1}\) imply that

\begin{eqnarray*} h(t)&=&{\cal L}^{-1}\left(G_0(s)\right)-{\cal L}^{-1}\left(e^{-s}G_1(s)\right)+{\cal L}^{-1}\left(e^{-4s}G_2(s)\right) \\[3pt] &=&t-u(t-1)\left[(t-1)+2\right]+u(t-4)\left[2(t-4)^2+1\right] \\[3pt] &=&t-u(t-1)(t+1)+u(t-4)(2t^2-16t+33), \end{eqnarray*}

which can also be written as

\[ h(t)=\left\{\begin{array}{cl} t,&0\le t<1,\\[3pt] -1,&1\le t<4,\\[3pt] 2t^2-16t+32,&t\ge4. \end{array}\right.\bbox \]

Example 8.4.8

Find the inverse transform of

\[ H(s)={2s\over s^2+4}-e^{-{\pi\over 2}s} {3s+1\over s^2+9}+e^{-\pi s}{s+1\over s^2+6s+10}. \]

Solution Let

\[ G_0(s)={2s\over s^2+4},\quad G_1(s)=-{(3s+1)\over s^2+9}, \]

and

\[ G_2(s)={s+1\over s^2+6s+10}={(s+3)-2\over (s+3)^2+1}. \]

Then

\[ g_0(t)=2\cos 2t,\quad g_1(t)=-3\cos 3t-{1\over 3}\sin 3t, \]

and

\[ g_2(t)=e^{-3t}(\cos t-2\sin t). \]

Therefore (8.4.12) and the linearity of \({\cal L}^{-1}\) imply that

\begin{eqnarray*} h(t)&=&2\cos 2t-u(t-\pi/2)\left[3\cos 3(t-\pi/2)+{1\over 3}\sin 3\left(t-{\pi\over 2}\right)\right] \\[6pt] &&+u(t-\pi)e^{-3(t-\pi)}\left[\cos (t-\pi)-2\sin (t-\pi)\right]. \end{eqnarray*}

Using the trigonometric identities (8.4.8) and (8.4.9), we can rewrite this as

\begin{equation} \begin{array}{rcl} h(t)&=&2\cos 2t+u(t-\pi/2)\left(3\sin 3t- {1\over 3}\cos 3t\right)\\[6pt] &&-u(t-\pi)e^{-3(t-\pi)} (\cos t-2\sin t) \end{array} \tag{8.4.13}\end{equation}

(Figure 8.4.5).

The piecewise continouous function eq:8.4.13
Figure 8.4.5. The piecewise continouous function (8.4.13)

8.4 Exercises

In Exercises 1–6 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 Eqn. (8.4.6), and use Theorem 8.4.1 to find \({\cal L}(f)\). Where indicated by C/G , graph \(f\).

  1. \(\dst{f(t)=\left\{\begin{array}{cl} 1,&0 \le t<4,\\[6pt] t,&t\ge4.\end{array}\right.}\)

    Show answer

    \( 1+u(t-4)(t-1)\);  \(\dst{{1\over s}+e^{-4s}\left({1\over s^2}+{3\over s}\right)}\)

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

    Show answer

    \(t+u(t-1)(1-t)\);  \(\dst{1-e^{-s}\over s^2}\)

  3. C/G \(\dst{f(t)=\left\{\begin{array}{cl} 2t-1,& 0\le t<2,\\[6pt] t,&t\ge2.\end{array}\right.}\)

    Show answer

    \(2t-1-u(t-2)(t-1)\);  \(\dst{\left({2\over s^2}-{1\over s}\right)-e^{-2s}\left({1\over s^2} +{1\over s}\right)}\)

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

    Show answer

    \(1+u(t-1)(t+1)\);  \(\dst{{1\over s} +e^{-s}\left({1\over s^2}+{2\over s}\right)}\)

  5. \(\dst{f(t)=\left\{\begin{array}{cl} t-1,& 0\le t<2,\\[6pt] 4,&t\ge2.\end{array}\right.}\)

    Show answer

    \(t-1+u(t-2)(5-t)\);  \(\dst{{1\over s^2}-{1\over s}-e^{-2s}\left({1\over s^2}-{3\over s}\right)}\)

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

    Show answer

    \(t^2\left(1-u(t-1)\right)\);  \(\dst{ {2\over s^3}-e^{-s}\left({2\over s^3}+{2\over s^2}+{1\over s}\right)}\)

In Exercises 7–18 express the given function \(f\) in terms of unit step functions and use Theorem 8.4.1 to find \({\cal L}(f)\). Where indicated by C/G , graph \(f\).

  1. \(\dst{f(t)=\left\{\begin{array}{cl} 0, &0\le t<2,\\[6pt] t^2+3t,&t\ge2.\end{array}\right.}\)

    Show answer

    \(u(t-2)(t^2+3t)\);  \(\dst{ e^{-2s}\left({2\over s^3}+{7\over s^2}+{10\over s}\right)}\)

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

    Show answer

    \(t^2+2+u(t-1)(t-t^2-2)\);  \(\dst{{2\over s^3} +{2\over s}-e^{-s}\left({2\over s^3}+{1\over s^2}+{2 \over s}\right)}\)

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

    Show answer

    \(te^t+u(t-1)(e^t-te^t)\);  \(\dst{1-e^{-(s-1)}\over (s-1)^2}\)

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

    Show answer

    \(e^{-t}+u(t-1)(e^{-2t}-e^{-t})\) ;  \(\dst{{1-e^{-(s+1)}\over s+1}+{e^{-(s+2)}\over s+2}}\)

  5. \(\dst{f(t)=\left\{\begin{array}{cl} -t,&0 \le t<2,\\[6pt] t-4,&2\le t<3,\\[6pt] 1,&t\ge3. \end{array}\right.}\)

    Show answer

    \(-t+2u(t-2)(t-2)-u(t-3)(t-5)\);  \(\dst{-{1\over s^2}+ {2e^{-2s}\over s^2}+e^{-3s}\left({2\over s}-{1\over s^2}\right)}\)

  6. \(\dst{f(t)=\left\{\begin{array}{cl} 0,&0 \le t<1,\\[6pt] t,&1\le t<2,\\[6pt] 0,&t\ge2.\end{array}\right.}\)

    Show answer

    \(\left[u(t-1)-u(t-2)\right]t\) ;  \(\dst{ e^{-s}\left({1\over s^2}+{1\over s}\right)-e^{-2s}\left({1\over s^2} +{2\over s}\right)}\)

  7. \(\dst{f(t)=\left\{\begin{array}{cl} t,&0 \le t<1,\\[6pt] t^2,&1\le t<2,\\[6pt] 0,&t\ge2. \end{array}\right.}\)

    Show answer

    \(t+u(t-1)(t^2-t)-u(t-2)t^2\);  \(\dst{{1\over s^2}+ e^{-s}\left({2\over s^3}+{1\over s^2}\right) -e^{-2s}\left({2\over s^3}+{4\over s^2}+{4\over s}\right)}\)

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

    Show answer

    \(t+u(t-1)(2-2t)+u(t-2)(4+t)\);  \(\dst{{1\over s^2} -2{e^{-s}\over s^2}+e^{-2s}\left({1\over s^2}+{6\over s}\right)}\)

  9. C/G \(\dst{f(t)=\left\{\begin{array}{cl} \phantom{2} \sin t,&0\le t<\dst{\pi\over 2},\\[6pt] 2\sin t,& \dst{\pi\over 2}\le t<\pi,\\[6pt]\phantom{2}\cos t, &t\ge\pi.\end{array}\right.}\)

    Show answer

    \(\dst{\sin t+ u(t-\pi/2)\sin t+u(t-\pi) (\cos t-2\sin t)}\);  \(\dst{{1+ e^{-{\pi\over 2}s}s-e^{-\pi s}(s-2)\over s^2+1}}\)

  10. C/G \(f(t)=\dst{\left\{\begin{array}{cl}\phantom{-} 2,&0\le t<1,\\[6pt]-2t+2,&1\le t<3,\\[6pt]\phantom{-}3t,&t\ge 3.\end{array}\right.}\)

    Show answer

    \(\dst{2-2u(t-1)t+u(t-3)(5t-2)}\);  \(\dst{{2\over s}-e^{-s}\left({2\over s^2}+{2\over s}\right)+e^{-3s}\left({5\over s^2}+{13\over s}\right)}\)

  11. C/G \(f(t)=\dst{\left\{\begin{array}{cl}3,&0\le t<2,\\[6pt]3t+2,&2\le t<4,\\[6pt]4t,&t\ge 4.\end{array}\right.}\)

    Show answer

    \(\dst{3+u(t-2)(3t-1)+u(t-4)(t-2)}\);  \(\dst{{3\over s}+e^{-2s}\left({3\over s^2}+{5\over s}\right)+e^{-4s}\left({1\over s^2}+{2\over s}\right)}\)

  12. C/G \(\dst{f(t)=\left\{\begin{array}{ll}(t+1)^2,&0\le t<1, \\[6pt](t+2)^2,&t\ge1.\end{array}\right.}\)

    Show answer

    \(\dst{(t+1)^2+u(t-1)(2t+3)}\);  \(\dst{{2\over s^3 }+{2\over s^2}+{1\over s}+e^{-s}\left({2\over s^2}+{5\over s}\right)}\)

In Exercises 19–28 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. Where indicated by C/G , graph the inverse transform.

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

    Show answer

    \(u(t-2)e^{2(t-2)}= \dst{\left\{\begin{array}{cl} 0,&0\le t<2, \\[6pt] e^{2(t-2)},&t\ge2.\end{array}\right.}\)

  2. \(\dst{H(s)={e^{-s}\over s(s+1)}}\)

    Show answer

    \(u(t-1)\left(1-e^{-(t-1)}\right)= \dst{\left\{\begin{array}{cl} 0,&0\le t<1,\\[6pt] 1-e^{-(t-1)},&t\ge1.\end{array}\right.}\)

  3. C/G \(\dst{H(s)={e^{-s}\over s^3}+ {e^{-2s}\over s^2}}\)

    Show answer

    \(\dst{u(t-1){(t-1)^2 \over 2}+u(t-2)(t-2)}= \dst{\left\{\begin{array}{cl} 0,&0 \le t< 1,\\[6pt]\dst{(t-1)^2\over 2},&1\le t<2,\\[6pt] \dst{t^2-3\over 2},&t\ge2.\end{array}\right.}\)

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

    Show answer

    \(2+t+u(t-1)(4-t)+u(t-3)(t-2)= \dst{ \left\{\begin{array}{cl} 2+t,& 0\le t<1,\\[6pt] 6,&1\le t<3,\\[6pt] t+4,&t\ge 3.\end{array}\right.}\)

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

    Show answer

    \(\dst{5-t+u(t-3)(7t-15)+ {3\over 2} u(t-6)(t-6)^2}= \dst{\left\{\begin{array}{cl} 5-t,&0\le t<3,\\[6pt] 6t-10,&3\le t<6,\\[6pt] 44-12t+{3\over2}t^2, &t\ge6. \end{array}\right.}\)

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

    Show answer

    \(u(t-\pi)e^{-2(t-\pi)} (2\cos t- 5\sin t)= \dst{\left\{\begin{array}{cl} 0, &0\le t<\pi,\\[6pt] e^{-2(t-\pi)}(2\cos t-5\sin t),&t\ge\pi. \end{array}\right.}\)

  7. C/G \(\dst{H(s)=\left({1\over s}-{s\over s^2+1}\right)+e^{-{\pi\over 2}s}\left({3s-1\over s^2+1}\right)}\)

    Show answer

    \(\dst{1-\cos t+ u(t-\pi/2) (3\sin t+\cos t)}= \dst{\left\{\begin{array}{cl} 1-\cos t,&0\le t<\dst{\pi\over 2},\\[6pt] 1+3\sin t,&t\ge \dst{\pi\over 2}.\end{array}\right.}\)

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

    Show answer

    \(\dst{u(t-2)\left (4e^{-(t-2)}-4e^{2(t-2)}+2e^{(t-2)}\right)}= \dst{\left\{\begin{array}{cl} 0,&0\le t<2,\\[6pt]4e^{-(t-2)}-4e^{2(t-2)}+2e^{(t-2)},&t\ge 2.\end{array}\right.}\)

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

    Show answer

    \(\dst{1+t+u(t-1)(2t+1)+u(t-3)(3t-5)}= \dst{ \left\{\begin{array}{cl} t+1,&0\le t<1,\\[6pt] 3t+2,&1\le t<3,\\[6pt]6t-3,&t\ge3.\end{array}\right.}\)

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

    Show answer

    \(\dst{1-t^2+u(t-2)\left(-{t^2\over2}+2t+1\right) +u(t-4)(t-4)}= \dst{\left\{\begin{array}{cl} 1-t^2,&0\le t<2\\[6pt]-\dst{3t^2\over2} +2t+2,&2\le t<4,\\[6pt]-\dst{3t^2\over2}+3t-2,&t\ge 4.\end{array}\right.}\)

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

    Show answer

    \(\dst{e^{-\tau s}\over s}\)

  12. 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 \([0,\infty)\)?

    Show answer

    For each \(t\) only finitely many terms are nonzero.

  13. In addition to the assumptions of Exercise 30, assume that

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

    and that the series

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

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

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

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

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

    1. Use (A) and Theorem 8.1.6 to show that

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

      is defined for \(s>s_0\).

    2. Show that (D) can be rewritten as

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

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

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

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

    4. Show that (E) can be rewritten as

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

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

    5. Complete the proof of (C).

  14. Suppose \(\{t_m\}_{m=0}^\infty\) and \(\{f_m\}_{m=0}^\infty\) satisfy the assumptions of Exercises 30 and 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 31 converges for any \(\rho>0\), and conclude from this that (C) of Exercise 31 holds for \(s>s_0\).

In Exercises 33–36 find the step function representation of \(f\) and use the result of Exercise 32 to find \({\cal L}(f)\). Hint: You will need formulas related to the formula for the sum of a geometric series.

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

    Show answer

    \(\dst{1+\sum_{m=1}^\infty u(t-m);\; {1\over s(1-e^{-s})}}\)

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

    Show answer

    \(\dst{1+2\sum_{m=1}^\infty (-1)^mu(t-m);\; {1\over s};\; {1-e^{-s}\over 1+e^{-s}}}\)

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

    Show answer

    \(\dst{1+\sum_{m=1}^\infty(2m+1)u(t-m);\; {e^{-s}(1+e^{-s})\over s(1-e^{-s})^2}}\)

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

    Show answer

    \(\dst{\sum_{m=1}^\infty(-1)^m(2m-1)u(t-m);\; {1\over s}{(1-e^s)\over(1+e^s)^2}}\)