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7.3 Operational Properties I: Class Notes

7.3a

Theorem: First Translation Theorem

If $ℒ\{f(t)\}=F(s)$ and $a$ is any real number, then $ℒ\{e^{at}f(t)\}=F(s-a)$.

(This is also called a shifting theorem.)

Proof

By Definition 7.1.1,

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{e^{at}f(t)\}}& {\displaystyle ={\int }_0^{\infty }{e}^{-st}{e}^{at}f(t)dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle ={\int }_0^{\infty }{e}^{-(s-a)t}f(t)dt=F(s-a)}\hfill \end{array}}$$

$-(s-a)$ is playing the role of $-s$.

Notation: $ℒ\{e^{at}f(t)\}=ℒ{\{f(t)\}}_{s\to s-a}$

We also get an inverse rule: ${ℒ}^{-1}\{F(s-a)\}=e^{at}f(t)$

Example

Find either $F(s)$ or $f(t)$, as indicated.

$ℒ\left\{t^{10}{e}^{-7t}\right\}$

Solution

$t^{10}$ is what we’re transforming, and $e^{-7t}$ shifts $s$ to $s+7$. Since ${\displaystyle ℒ\left\{t^{10}\right\}=\frac{10!}{s^{11}}}$,

$ℒ\left\{e^{-2t}\cos 4t\right\}$

Solution

Transform $\cos 4t$, then shift $s$ to $s+2$:

${\displaystyle {ℒ}^{-1}\left\{\frac{1}{s^2+2s+5}\right\}}$

Solution

Complete the square: $s^2+2s+5={(s+1)}^2+4$. The $s+1$ is a shift, and $4=k^2$ gives $k=2$ and sine (no $s$ in the numerator). Multiply and divide by 2:

$${\displaystyle {ℒ}^{-1}\left\{\frac{1}{s^2+2s+5}\right\}={ℒ}^{-1}\{\frac{1}{2}·\frac{1·2}{{(s+1)}^2+4}\}}$$

Tip: Recall from partial fractions: an irreducible quadratic factor gets a linear numerator, as in ${\displaystyle \frac{Ax+B}{x^2+4}}$.

${\displaystyle {ℒ}^{-1}\left\{\frac{(s+1{)}^2}{(s+2{)}^4}\right\}}$

Solution

Expand the numerator. The denominator is a repeated linear factor, so the partial fractions are

$${\displaystyle \frac{s^2+2s+1}{(s+2{)}^4}=\frac{A}{s+2}+\frac{B}{(s+2{)}^2}+\frac{C}{(s+2{)}^3}+\frac{D}{(s+2{)}^4}}$$

$$s^2+2s+1=A(s+2{)}^3+B(s+2{)}^2+C(s+2)+D$$

Compare coefficients and substitute $s=-2$:

$$\begin{array}{cc}\hfill {\displaystyle s^3:}& {\displaystyle A=0}\hfill \\ \hfill {\displaystyle s=-2:}& {\displaystyle 1=D}\hfill \\ \hfill {\displaystyle s^2:}& {\displaystyle 1=B}\hfill \\ \hfill {\displaystyle \text{const}:}& {\displaystyle 1=8A+4B+2C+D=4+2C+1\Rightarrow C=-2}\hfill \end{array}$$

Each term is a power of $s+2$, a shift: use part (b) with $n=1$, $n=2$, and $n=3$. The middle numerator is already $2!=2$, and the last needs $3!=6$, so multiply and divide by 6:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {ℒ}^{-1}\{\frac{1}{{(s+2)}^2}-\frac{2}{{(s+2)}^3}+\frac{1}{6}·\frac{1·3!}{{(s+2)}^4}\}}& {\displaystyle =t{e}^{-2t}-t^2{e}^{-2t}+\frac{1}{6}{t}^3{e}^{-2t}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-2t}(t-t^2+\frac{1}{6}{t}^3)}\hfill \end{array}}$$

7.3b

Example

Use the Laplace transform to solve the initial-value problem.

$$y\prime -y=1+t{e}^t,y(0)=0$$

Solution

In $t{e}^t$, $f(t)=t$ and $e^t$ is a shift, so ${\displaystyle ℒ\left\{t{e}^t\right\}=\frac{1}{(s-1{)}^2}}$. With $y(0)=0$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle sY(s)-Y(s)}& {\displaystyle =ℒ\left\{1\right\}+ℒ\left\{t{e}^t\right\}=\frac{1}{s}+\frac{1}{(s-1{)}^2}}\hfill \\ \hfill {\displaystyle Y(s)}& {\displaystyle =\frac{1}{s(s-1)}+\frac{1}{(s-1{)}^3}}\hfill \end{array}}$$

Partial fractions: ${\displaystyle \frac{1}{s(s-1)}=\frac{A}{s}+\frac{B}{s-1}}$, so $1=A(s-1)+Bs$, giving $A=-1,B=1$. The last term needs $2!=2$ in the numerator ($n=2$):

$${\displaystyle Y(s)=-\frac{1}{s}+\frac{1}{s-1}+\frac{1}{2}·\frac{1·2}{(s-1{)}^3}}$$

$$2y″+20y\prime +51y=0,y(0)=2,y\prime (0)=0$$

Solution

Transform each term (colors match the equation); $y\prime (0)=0$ and $y(0)=2$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle 2[s^{2}Y(s)-sy(0)-y\prime (0)]+20[sY(s)-y(0)]+51Y(s)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle 2s^{2}Y(s)-4s+20sY(s)-40+51Y(s)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (2s^2+20s+51)Y(s)}& {\displaystyle =4s+40}\hfill \\ \hfill {\displaystyle Y(s)}& {\displaystyle =\frac{4s+40}{2s^2+20s+51}}\hfill \end{array}}$$

Multiply the numerator and denominator by $\frac{1}{2}$, then complete the square:

$${\displaystyle s^2+10s+\frac{51}{2}=s^2+10s+25+\frac{1}{2}={(s+5)}^2+\frac{1}{2}}$$

$${\displaystyle Y(s)=\frac{2s+20}{s^2+10s+25+\frac{1}{2}}=\frac{2s+20}{{(s+5)}^2+\frac{1}{2}}}$$

The shift $s+5$ has to show up in the numerator too: $2s+20=2(s+5)+10$. For sine and cosine, $k^2=\frac{1}{2}$, so $k=\frac{1}{\sqrt{2}}$; the sine term needs $\frac{1}{\sqrt{2}}$ in its numerator, and $10=10\sqrt{2}·\frac{1}{\sqrt{2}}$:

$${\displaystyle Y(s)=2·\frac{s+5}{{(s+5)}^2+\frac{1}{2}}+10\sqrt{2}·\frac{1/\sqrt{2}}{{(s+5)}^2+\frac{1}{2}}}$$

Example

Solve the boundary-value problem.

$$y″+8y\prime +20y=0,y(0)=0,y\prime (\pi )=0$$

Solution

The value $y\prime (0)$ is not given, so call it $c_1=y\prime (0)$. The terms with $y(0)=0$ drop out:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle s^{2}Y(s)-sy(0)-y\prime (0)+8sY(s)-8y(0)+20Y(s)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (s^2+8s+20)Y(s)-c_1}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle Y(s)=\frac{c_1}{s^2+8s+16+4}}& {\displaystyle =\frac{1}{2}·\frac{2c_1}{(s+4{)}^2+4}}\hfill \end{array}}$$

Here $k=2$ (sine) with the shift $s+4$. Writing $c=c_1/2$ for the constant,

$$y(t)=c{e}^{-4t}\sin 2t$$

Now use $y\prime (\pi )=0$:

$$\begin{array}{cc}\hfill {\displaystyle y\prime (t)}& {\displaystyle =-4c{e}^{-4t}\sin 2t+2c{e}^{-4t}\cos 2t}\hfill \\ \hfill {\displaystyle y\prime (\pi )}& {\displaystyle =2c{e}^{-4\pi }=0\Rightarrow c=0}\hfill \end{array}$$

(At $t=\pi$, $\sin 2\pi =0$ and $\cos 2\pi =1$.)

7.3c

The First Translation Theorem showed how to translate on the $s$-axis. A translation on the $t$-axis is essentially an adjustment for moving the lower limit in the definition of a Laplace transform. This is accomplished via the Heaviside function or unit step function.

Definition

The unit step function is

$$𝒰(t-a)=\{\begin{array}{cc}{\displaystyle 0,}\hfill & {\displaystyle 0\le t<a}\hfill \\ {\displaystyle 1,}\hfill & {\displaystyle t\ge a}\hfill \end{array}$$

Graph of the unit step function $𝒰(t-a)$. It is 0 (“off”) for $0\le t<a$, with an open circle at $(a,0)$; it is 1 (“on”) for $t\ge a$, with a filled dot at $(a,1)$.

To make it first on and then off:

$$1-𝒰(t-a)=\{\begin{array}{cc}{\displaystyle 1-0,}\hfill & {\displaystyle 0\le t<a}\hfill \\ {\displaystyle 1-1,}\hfill & {\displaystyle t\ge a}\hfill \end{array}=\{\begin{array}{cc}{\displaystyle 1,}\hfill & {\displaystyle 0\le t<a}\hfill \\ {\displaystyle 0,}\hfill & {\displaystyle t\ge a}\hfill \end{array}$$

Consider

$$\sin t𝒰(t-\pi /2)=\{\begin{array}{cc}{\displaystyle 0,}\hfill & {\displaystyle 0\le t<\pi /2}\hfill \\ {\displaystyle \sin t,}\hfill & {\displaystyle t\ge \pi /2}\hfill \end{array}$$

Graph of $\sin t𝒰(t-\pi /2)$: 0 for $0\le t<\pi /2$ (open circle at $(\pi /2,0)$), then the sine curve from $(\pi /2,1)$ onward, crossing 0 at $t=\pi$ and reaching −1 at $t=3\pi /2$. The sine wave is cut off, not shifted.

Shifting the sine yields

$$\sin (t-\pi /2)𝒰(t-\pi /2)=\{\begin{array}{cc}{\displaystyle 0,}\hfill & {\displaystyle 0\le t<\pi /2}\hfill \\ {\displaystyle \sin (t-\pi /2),}\hfill & {\displaystyle t\ge \pi /2}\hfill \end{array}$$

Graph of $\sin (t-\pi /2)𝒰(t-\pi /2)$: 0 for $0\le t<\pi /2$, then a full sine wave that starts at $(\pi /2,0)$, rises to 1 at $t=\pi$, and returns to 0 at $t=3\pi /2$. It is the graph of $\sin t$ moved $\pi /2$ units to the right.

Theorem: Second Translation Theorem

If $F(s)=ℒ\{f(t)\}$ and $a>0$, then $ℒ\{f(t-a)𝒰(t-a)\}=e^{-as}F(s)$.

Watch the signs: compare with ${\displaystyle ℒ\left\{e^{at}\right\}=\frac{1}{s-a}}$. Here the shift $t-a$ comes with $e^{-as}$.

Proof

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{f(t-a)𝒰(t-a)\}}& {\displaystyle ={\int }_0^{\infty }{e}^{-st}f(t-a)𝒰(t-a)dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle ={\int }_0^a{e}^{-st}f(t-a)·0dt+{\int }_a^{\infty }{e}^{-st}f(t-a)dt}\hfill \end{array}}$$

The first integral is 0. In the second, let $v=t-a$; then $dv=dt$ and $t=v+a$. The limits change: $t=a$ becomes $v=0$, and $t=\infty$ stays $v=\infty$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle }& {\displaystyle ={\int }_0^{\infty }{e}^{-s(v+a)}f(v)dv}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-as}{\int }_0^{\infty }{e}^{-sv}f(v)dv=e^{-as}F(s)}\hfill \end{array}}$$

This is the shift: the lower limit moved from $a$ to $0$.

If we let $f(t)=1$, then we have ${\displaystyle ℒ\{𝒰(t-a)\}=\frac{e^{-as}}{s}}$.

In inverse form, ${\displaystyle {ℒ}^{-1}\left\{\frac{e^{-as}}{s}\right\}=𝒰(t-a)}$.

Example

Find either $F(s)$ or $f(t)$, as indicated.

$ℒ\{(3t+1)𝒰(t-1)\}$

Solution

The theorem needs the form $ℒ\{f(t-a)𝒰(t-a)\}$, but $3t+1$ is not written in terms of $t-1$: these don’t match. Call $g(t)=3t+1$ and use the alternative form below.

Alternative form of the second translation theorem

$ℒ\{g(t)𝒰(t-a)\}=e^{-as}ℒ\{g(t+a)\}$

Why it works. Start from the original theorem:

$$ℒ\{f(t-a)𝒰(t-a)\}=e^{-as}ℒ\{f(t)\}$$

Let $g(t)=f(t-a)$; then $f(t)=g(t+a)$. So

$$\begin{array}{cc}\hfill {\displaystyle ℒ\{g(t)𝒰(t-a)\}}& {\displaystyle =ℒ\{f(t-a)𝒰(t-a)\}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-as}ℒ\left\{f(t)\right\}=e^{-as}ℒ\left\{g(t+a)\right\}}\hfill \end{array}$$

Back to the example, with $g(t)=3t+1$ and $a=1$: replace $t$ by $t+1$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{(3t+1)𝒰(t-1)\}}& {\displaystyle =e^{-s}ℒ\{3(t+1)+1\}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-s}ℒ\{3t+4\}=e^{-s}(\frac{3}{s^2}+\frac{4}{s})}\hfill \end{array}}$$

7.3d

Inverse form of the Second Translation Theorem

If $f(t)={ℒ}^{-1}\{F(s)\}$, then ${ℒ}^{-1}\{e^{-as}F(s)\}=f(t-a)𝒰(t-a)$.

You have to make the $t-a$ happen: replace $t$ by $t-a$ everywhere in $f(t)$.

Example

Find either $F(s)$ or $f(t)$, as indicated.

${\displaystyle {ℒ}^{-1}\left\{\frac{(1+e^{-2s}{)}^2}{s+2}\right\}}$

Solution

Expand the square and split into three inverse transforms. Each uses ${\displaystyle f(t)={ℒ}^{-1}\left\{\frac{1}{s+2}\right\}=e^{-2t}}$, shifted by $a=2$ or $a=4$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {ℒ}^{-1}\left\{\frac{(1+e^{-2s}{)}^2}{s+2}\right\}}& {\displaystyle ={ℒ}^{-1}\left\{\frac{1+2e^{-2s}+e^{-4s}}{s+2}\right\}}\hfill \\ \hfill {\displaystyle }& {\displaystyle ={ℒ}^{-1}\left\{\frac{1}{s+2}\right\}+{ℒ}^{-1}\left\{\frac{2e^{-2s}}{s+2}\right\}+{ℒ}^{-1}\left\{\frac{e^{-4s}}{s+2}\right\}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-2t}+2e^{-2(t-2)}𝒰(t-2)+e^{-2(t-4)}𝒰(t-4)}\hfill \end{array}}$$

${\displaystyle {ℒ}^{-1}\left\{\frac{e^{-2s}}{s^2(s-1)}\right\}}$

Solution

Set the $e^{-2s}$ aside for now and find partial fractions for the rest:

$${\displaystyle \frac{1}{s^2(s-1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s-1}}$$

Tip: Or use ${\displaystyle \frac{As+B}{s^2}+\frac{C}{s-1}}$; it gives the same result.

$$1=As(s-1)+B(s-1)+C{s}^2$$

$$\begin{array}{cc}\hfill {\displaystyle s^2:}& {\displaystyle A+C=0}\hfill \\ \hfill {\displaystyle s:}& {\displaystyle -A+B=0}\hfill \\ \hfill {\displaystyle \text{const}:}& {\displaystyle -B=1}\hfill \end{array}$$

So $B=-1$, then $A=B=-1$, and $C=-A=1$. Bring back $e^{-2s}$ and use the inverse form with $a=2$:

$${\displaystyle {ℒ}^{-1}\left\{e^{-2s}(-\frac{1}{s}-\frac{1}{s^2}+\frac{1}{s-1})\right\}=𝒰(t-2)[-1-(t-2)+e^{t-2}]}$$

Example

Use the Laplace transform to solve the given initial-value problem.

$y\prime +y=f(t)$,   $y(0)=0$, where

$$f(t)=\{\begin{array}{cc}{\displaystyle 1,}\hfill & {\displaystyle 0\le t<1}\hfill \\ {\displaystyle -1,}\hfill & {\displaystyle t\ge 1}\hfill \end{array}$$

Solution

Graph of $f(t)$: a horizontal segment at height 1 for $0\le t<1$ (open circle at $(1,1)$), then a horizontal ray at height −1 for $t\ge 1$ (filled dot at $(1,-1)$).

First express $f(t)$ using the unit step function. Write it as a sum of two pieces:

$$f(t)=\{\begin{array}{cc}{\displaystyle 1,}\hfill & {\displaystyle 0\le t<1}\hfill \\ {\displaystyle 0,}\hfill & {\displaystyle t\ge 1}\hfill \end{array}+\{\begin{array}{cc}{\displaystyle 0,}\hfill & {\displaystyle 0\le t<1}\hfill \\ {\displaystyle -1,}\hfill & {\displaystyle t\ge 1}\hfill \end{array}$$

$$f(t)=1-𝒰(t-1)-𝒰(t-1)=1-2𝒰(t-1)$$

Check: $f(t)=\{\begin{array}{cc}{\displaystyle 1-0,}\hfill & {\displaystyle 0\le t<1}\hfill \\ {\displaystyle 1-2,}\hfill & {\displaystyle t\ge 1}\hfill \end{array}$ ✓

Now solve $y\prime +y=1-2𝒰(t-1)$ with $y(0)=0$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle sY(s)-y(0)+Y(s)}& {\displaystyle =\frac{1}{s}-\frac{2e^{-s}}{s}}\hfill \\ \hfill {\displaystyle Y(s)}& {\displaystyle =\frac{1}{s(s+1)}-\frac{2e^{-s}}{s(s+1)}}\hfill \end{array}}$$

Partial fractions: ${\displaystyle \frac{1}{s(s+1)}=\frac{A}{s}+\frac{B}{s+1}}$, so $1=A(s+1)+Bs$, giving $A=1,B=-1$.

$${\displaystyle Y(s)=\frac{1}{s}-\frac{1}{s+1}-2e^{-s}(\frac{1}{s}-\frac{1}{s+1})}$$

For the last term, use the inverse form with $a=1$: replace $t$ by $t-1$ in $1-e^{-t}$.