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6.3: Step function

Last updated

Jun 28, 2024
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- 6.2
- 6.4

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Step function

Exercise $\PageIndex{1}$

Graph $u_c(t)= \begin{cases}0 & t<c \\ 1 & t \geq c\end{cases}$ .

Answer

Below is the answer when $c=5$ .

import matplotlib.pyplot as plt
import numpy as np
from sympy import var, plot_implicit

x = np.linspace(1, 10, 100)
y = np.piecewise(x,[x < 5,x > 5],[0, 1])

pos = np.where(np.abs(np.diff(y)) >= 0.5)[0] #This is just to not connect the parts of the indicator function.

x[pos] = np.nan
y[pos] = np.nan

plt.plot(x, y)
plt.grid()
plt.show()

Exercise $\PageIndex{2}$

Graph $g(t) =\sin(t)$ .

Answer

Add texts here. Do not delete this text first.

import matplotlib.pyplot as plt
import numpy as np
from sympy import var, plot_implicit

x = np.linspace(0, 10, 100)
y = np.sin(x)

plt.plot(x, y)
plt.grid()
plt.show()

Exercise $\PageIndex{3}$

Graph $h(t)=u_\pi(t) \sin (t)$

Answer

import matplotlib.pyplot as plt
import numpy as np
from sympy import var, plot_implicit

x = np.linspace(1, 10, 100)
y = np.piecewise(x,[x < np.pi,x > np.pi],[0, 1]) *np.sin(x)

pos = np.where(np.abs(np.diff(y)) >= 0.5)[0]

x[pos] = np.nan
y[pos] = np.nan

plt.plot(x, y)
plt.grid()
plt.show()

Exercise $\PageIndex{4}$

Graph $f(t)=2 t+u_\pi(t)[\sin (t)-2 t]= \begin{cases} t<\pi \\ t \geq \pi\end{cases}$

Answer

Add texts here. Do not delete this text first.

import matplotlib.pyplot as plt
import numpy as np
from sympy import var, plot_implicit

x = np.linspace(1, 10, 100)
y = np.piecewise(x,[x < 5,x >= 5],[1, 1])

pos = np.where(np.abs(np.diff(y)) >= 0.5)[0]

x[pos] = np.nan
y[pos] = np.nan

plt.plot(x, y)
plt.grid()
plt.show()

Exercise $\PageIndex{1}$

If $h(t)= \begin{cases}t & 0 \leq t<4 \\ \ln (t) & t \geq 4\end{cases}$ , then what does this imply about $h(t)$

Answer

Add texts here. Do not delete this text first.

import matplotlib.pyplot as plt
import numpy as np
from sympy import var, plot_implicit

x = np.linspace(0, 10, 100)
y = np.piecewise(x, [x < 4,x >= 4], [lambda x: x, lambda x: np.log(x)])

pos = np.where(np.abs(np.diff(y)) >= 0.5)[0]

x[pos] = np.nan
y[pos] = np.nan

plt.plot(x, y)
plt.grid()
plt.show()

Example $\PageIndex{1}$

Example: $f(t)= \begin{cases}f_1, & \text { if } t<4 \\ f_2, & \text { if } 4 \leq t<5 \\ f_3, & \text { if } 5 \leq t<10 \\ f_4, & \text { if } t \geq 10\end{cases}$

Solution

Hence

$\notag f(t)=f_1(t)+u_4(t)\left[f_2(t)\right.\left.-f_1(t)\right]+u_5(t)\left[f_3(t)-f_2(t)\right] +u_{10}(t)\left[f_4(t)-f_3(t)\right]$

Partial check

If $t=3: \quad f(3) =f_1(3)+0\left[f_2(3)-f_1(3)\right] +0\left[f_3(3)-f_2(3)\right]+0\left[f_4(3)-f_3(3)\right]=f_1(3)$

If $t=9: f(9)=f_1(9)+1\left[f_2(9)-f_1(9)\right] +1\left[f_3(9)-f_2(9)\right]+0\left[f_4(9)-f_3(9)\right]=f_3(9)$

Example $\PageIndex{2}$

$f(t)=\left\{\begin{array}{ll}0 & 0 \leq t<2 \\ t^2 & t \geq 2\end{array} \quad\right.$ implies $\quad f(t)=$

Solution

$f(t) = u_2(t)*t^2$

Example $\PageIndex{3}$

$g(t)=\left\{\begin{array}{ll}t^2 & 0 \leq t<3 \\ 0 & t \geq 3\end{array} \quad\right.$ implies $g(t)=$

Solution

$g(t) = t^2 - u_3(t)*t^2$

Example $\PageIndex{4}$

$j(t)=\left\{\begin{array}{ll}t & 0 \leq t<5 \\ 2 & 5 \leq t < 8 \\ e^t & t \geq 8\end{array} \quad\right.$ implies $j(t)=$ .

Solution

$j(t) = t -u_5(t) *t +2*u_5(t) -2*u_8(t) + e^t * u_8(t)$

Theorem $\PageIndex{1}$

$\notag \mathcal{L}\left(u_c(t) f(t-c)\right)=e^{-c s} \mathcal{L}(f(t))$
or equivalently

$\notag \mathcal{L}\left(u_c(t) f(t)\right)=e^{-c s} \mathcal{L}(f(t+c)) .$

the above theorem can be restated as, replacing $t-c$ with $t$ is equivalent to replacing $t$ with $t+c$ .

Example $\PageIndex{5}$

Find the LaPlace transform of $\mathcal{L}\left(u_3(t)\left(t^2-2 t+1\right)\right)$

Solution

$\begin{aligned} & \mathcal{L}\left(u_3(t)\left(t^2-2 t+1\right)\right)=e^{-3 s}\left(\frac{2}{s^3}+\frac{4}{s^2}+\frac{4}{s}\right) \\ & \left.\mathcal{L}\left(u_3(t)\left(t^2-2 t+1\right)\right)=e^{-3 s} \mathcal{L}\left((t+3)^2-2(t+3)+1\right)\right) \\ & \left.=e^{-3 s} \mathcal{L}\left(t^2+6 t+9-2 t-6+1\right)\right) \\ & =e^{-3 s} \mathcal{L}\left(t^2+4 t+4\right)=e^{-3 s}\left(\frac{2}{s^3}+\frac{4}{s^2}+\frac{4}{s}\right) \\ & \end{aligned}$

Example $\PageIndex{6}$

Find the LaPlace transform of $\mathcal{L}\left(u_4(t)\left(e^{-8 t}\right)\right)$

Solution

$\notag \begin{aligned} & \mathcal{L}\left(u_4(t)\left(e^{-8 t}\right)\right)=e^{-4 s-32}\left(\frac{1}{s+8}\right) \\ & \left.\mathcal{L}\left(u_4(t)\left(e^{-8 t}\right)\right)=e^{-4 s} \mathcal{L}\left(e^{-8(t+4)}\right)=e^{-4 s} \mathcal{L}\left(e^{-8 t} e^{-32}\right)\right) \\ & =e^{-4 s} e^{-32} \mathcal{L}\left(e^{-8 t}\right)=e^{-4 s-32}\left(\frac{1}{s+8}\right) \end{aligned}$

Example $\PageIndex{7}$

Find the LaPlace transform of $\mathcal{L}\left(u_2(t)\left(t^2 e^{3 t}\right)\right)$

Solution

$\notag \begin{aligned} \mathcal{L}\left(u_2(t)\left(t^2 e^{3 t}\right)\right) & =e^{-2 s+6}\left(\frac{2}{(s-3)^3}+\frac{4}{(s-3)^2}+\frac{4}{(s-3)}\right)\\ \mathcal{L}\left(u_2\left(t^2 e^{3 t}\right)\right) & \left.=e^{-2 s} \mathcal{L}\left(\left[(t+2)^2\right] e^{3(t+2)}\right)\right) \\ & \left.=e^{-2 s} \mathcal{L}\left(\left[t^2+4 t+4\right] e^{3 t+6}\right)\right) \\ & \left.=e^{-2 s} e^6 \mathcal{L}\left(\left[t^2+4 t+4\right] e^{3 t}\right)\right) \\ & \left.=e^{-2 s+6} \mathcal{L}\left(t^2 e^{3 t}+4 t e^{3 t}+4 e^{3 t}\right)\right) \\ & =e^{-2 s+6}\left(\mathcal{L}\left(t^2 e^{3 t}\right)+4 \mathcal{L}\left(t e^{3 t}\right)+4 \mathcal{L}\left(e^{3 t}\right)\right) \\ & =e^{-2 s+6}\left(\frac{2}{(s-3)^3}+\frac{4}{(s-3)^2}+\frac{4}{(s-3)}\right) \text { since } \end{aligned}$

Formula 14: $\mathcal{L}\left(e^{c s} f(t)\right)=F(s-c).$ Thus $\mathcal{L}\left(t^2 e^{3 t}\right)=F(s-3)=\frac{2}{(s-3)^3}$ since $F(s)=\mathcal{L}(f(t))=\mathcal{L}\left(t^2\right)=\frac{2}{s^3}$ and $F(s-3)=\frac{2}{(s-3)^3}.$

Example $\PageIndex{8}$

Find the LaPlace transform of
$g(t)= \begin{cases}0 & t<3 \\ e^{t-3} & t \geq 3\end{cases}$

Solution

Note $g(t)=u_3(t) e^{t-3}$ , thus we have

$\notag \mathcal{L}\left(u_3(t) e^{t-3}\right)=e^{-3 s} \mathcal{L}\left(e^t\right)=\frac{e^{-3 s}}{s-1}$

Example $\PageIndex{9}$

Find the LaPlace transform of $f(t)= \begin{cases}0 & t<3 \\ 5 & 3 \leq t<4 \\ t-5 & t \geq 4\end{cases}$

Solution

$\begin{aligned} & f(t)=0+u_3(t)[5-0]+u_4(t)[t-5-5] \mathcal{L}(f(t)) & =\mathcal{L}\left(5 u_3(t)+u_4(t)[t-10]\right) \\ & =5 \mathcal{L}\left(u_3(t)\right)+\mathcal{L}\left(u_4(t)[t-10]\right) \\ & =5 e^{-3 s}+e^{-4 s} \mathcal{L}(t+4-10) \\ & =5 e^{-3 s}+e^{-4 s} \mathcal{L}(t-6) \\ & =5 e^{-3 s}+e^{-4 s}[\mathcal{L}(t)-6 \mathcal{L}(1)] \\ & =5 e^{-3 s}+e^{-4 s}\left[\frac{1}{s^2}-\frac{6}{s}\right]=5 e^{-3 s}+\frac{e^{-4 s}(1-6 s)}{s^2} \end{aligned}$

Note

From the above theorem we have $\mathcal{L}\left(u_c(t) f(t-c)\right)=e^{-c s} \mathcal{L}(f(t))$ .
Let $F(s)=\mathcal{L}(f(t))$ .
Then $\mathcal{L}^{-1}(F(s))=\mathcal{L}^{-1}(\mathcal{L}(f(t)))=f(t)$ .
Thus $\mathcal{L}^{-1}\left(e^{-c s} F(s)\right)$

$\notag =\mathcal{L}^{-1}\left(e^{-c s} \mathcal{L}(f(t))\right)=u_c(t) f(t-c)$
where

$f(t)=\mathcal{L}^{-1}(F(s))$

Example $\PageIndex{10}$

Find the inverse LaPlace transform of $\mathcal{L}^{-1}\left(e^{-8 s} \frac{1}{s-3}\right)$

Solution

$\begin{aligned} & \text { a.) } \mathcal{L}^{-1}\left(e^{-8 s} \frac{1}{s-3}\right)=\underline{u_8(t) e^{3(t-8)}} \\ & \mathcal{L}^{-1}\left(e^{-8 s} \frac{1}{s-3}\right)=u_8(t) f(t-8) \text { where } \\ & \quad \mathcal{L}(f(t))=\frac{1}{s-3} \text {. Hence } f(t)=\mathcal{L}^{-1}\left(\frac{1}{s-3}\right)=e^{3 t} \end{aligned}$

Example $\PageIndex{11}$

Find the inverse LaPlace transform of $\mathcal{L}^{-1}\left(e^{-4 s} \frac{1}{s^2-3}\right)$

Solution

$\mathcal{L}^{-1}\left(e^{-4 s} \frac{1}{s^2-3}\right)=u_4(t) \frac{1}{\sqrt{3}} \sinh (\sqrt{3}(t-4))$
$\mathcal{L}^{-1}\left(e^{-4 s} \frac{1}{s^2-3}\right)=u_4(t) f(t-4)$ where
$\mathcal{L}(f(t))=\frac{1}{s^2-3}$ . Hence $f(t)=\frac{1}{\sqrt{3}} \mathcal{L}^{-1}\left(\frac{\sqrt{3}}{s^2-3}\right)=\frac{1}{\sqrt{3}} \sinh (\sqrt{3} t)$

Example $\PageIndex{12}$

Find the inverse LaPlace transform of $\mathcal{L}^{-1}\left(e^{-s} \frac{5}{(s-3)^4}\right)$

Solution

$\mathcal{L}^{-1}\left(e^{-s} \frac{5}{(s-3)^4}\right)=\underline{u_1(t)\left(\frac{5}{6}\right)(t-1)^3 e^{3(t-1)}}$
$\mathcal{L}^{-1}\left(e^{-s} \frac{5}{(s-3)^4}\right)=u_1(t) f(t-1)$ where
$\mathcal{L}(f(t))=\frac{5}{(s-3)^4}$ . Hence $f(t)=\frac{5}{6} \mathcal{L}^{-1}\left(\frac{3!}{(s-3)^4}\right)=\frac{5}{6} t^3 e^{3 t}$

Example $\PageIndex{13}$

Find the inverse LaPlace transform of $\mathcal{L}^{-1}\left(\frac{e^{-s}}{4 s}\right)=\underline{\frac{1}{4} u_1(t)}$

Solution

In this case you can use the easier formula 12 (?), or alternatively, you can use formula 13 (but formula 12 is easier to use and applies to this case):
$\mathcal{L}^{-1}\left(\frac{e^{-s}}{4 s}\right)=\frac{1}{4} \mathcal{L}^{-1}\left(\frac{e^{-s}}{s}\right)=\frac{1}{4} u_1(t) f(t+1)$ where
$\mathcal{L}(f(t))=\frac{1}{s}$ . Hence $f(t)=1$ . Thus $f(t-1)=1$

Example $\PageIndex{14}$

Find the inverse LaPlace transform of $\mathcal{L}^{-1}\left(e^{-s}\right)$

Solution

$\mathcal{L}^{-1}\left(e^{-s}\right)=\underline{\delta(t-1)}$

Example $\PageIndex{15}$

Find the inverse LaPlace transform of $\mathcal{L}^{-1}\left(e^{-s} \frac{1}{(s-3)^2+4}\right)$

Solution

$\notag \begin{aligned} & \text { f.) } \mathcal{L}^{-1}\left(e^{-s} \frac{1}{(s-3)^2+4}\right)=\underline{\frac{1}{2} u_1(t) e^{3(t-1)} \sin (2(t-1))} \\ & \mathcal{L}^{-1}\left(e^{-s} \frac{1}{(s-3)^2+4}\right)=u_1(t) f(t-1) \text { where } \\ & \left.\mathcal{L}(f(t))=\frac{1}{(s-3)^2+4}\right) \text {. } \\ & \end{aligned}$

Hence $f(t)=\frac{1}{2} \mathcal{L}^{-1}\left(\frac{2}{(s-3)^2+4}\right)=\frac{1}{2} e^{3 t} \sin (2 t)$

Example $\PageIndex{16}$

Find the inverse LaPlace transform of $\mathcal{L}^{-1}\left(e^{-s} \frac{2 s-5}{s^2+6 s+13}\right)$

Solution

$\begin{aligned} & \mathcal{L}^{-1}\left(e^{-s} \frac{2 s-5}{s^2+6 s+13}\right) \\ & =u_1(t) e^{-3 t+1}\left[2 \cos (2 t-2)-\frac{11}{2} \sin (2 t-2)\right] \\ & \end{aligned}$

$\frac{2 s-5}{s^2+6 s+13}=\frac{2 s-5}{s^2+6 s+9-9+13}=\frac{2 s-5}{(s+3)^2+4}=\frac{2(s+3)-6-5}{(s+3)^2+4}$

$\begin{aligned} \mathcal{L}^{-1}\left(\frac{2 s-5}{s^2+6 s+13}\right) & =\mathcal{L}^{-1}\left(\frac{2(s+3)-11}{(s+3)^2+4}\right) \\ & =2 \mathcal{L}^{-1}\left(\frac{s+3}{(s+3)^2+4}\right)-11 \mathcal{L}^{-1}\left(\frac{1}{(s+3)^2+4}\right) \\ & =2 \mathcal{L}^{-1}\left(\frac{s+3}{(s+3)^2+4}\right)-\frac{11}{2} \mathcal{L}^{-1}\left(\frac{2}{(s+3)^2+4}\right) \\ & =2 e^{-3 t} \cos (2 t)-\frac{11}{2} e^{-3 t} \sin (2 t)\end{aligned}$

$\begin{aligned} & \mathcal{L}^{-1}\left(e^{-s} \frac{2 s-5}{s^2+6 s+13}\right)=u_1(t) f(t-1) \\ & \quad=u_1(t)\left[2 e^{-3(t-1)} \cos (2(t-1))-\frac{11}{2} e^{-3(t-1)} \sin (2(t-1))\right] \\ & \quad=u_1(t) e^{-3 t+1}\left[2 \cos (2 t-2)-\frac{11}{2} \sin (2 t-2)\right]\end{aligned}$

Exercise 6.3.1\PageIndex{1}

Exercise 6.3.2\PageIndex{2}

Exercise 6.3.3\PageIndex{3}

Exercise 6.3.4\PageIndex{4}

Exercise 6.3.1\PageIndex{1}

Example 6.3.1\PageIndex{1}

Solution

Example 6.3.2\PageIndex{2}

Solution

Example 6.3.3\PageIndex{3}

Solution

Example 6.3.4\PageIndex{4}

Solution

Theorem 6.3.1\PageIndex{1}

Example 6.3.5\PageIndex{5}

Solution

Example 6.3.6\PageIndex{6}

Solution

Example 6.3.7\PageIndex{7}

Solution

Example 6.3.8\PageIndex{8}

Solution

Example 6.3.9\PageIndex{9}

Solution

Note

Example 6.3.10\PageIndex{10}

Solution

Example 6.3.11\PageIndex{11}

Solution

Example 6.3.12\PageIndex{12}

Solution

Example 6.3.13\PageIndex{13}

Solution

Example 6.3.14\PageIndex{14}

Solution

Example 6.3.15\PageIndex{15}

Solution

Example 6.3.16\PageIndex{16}

Solution

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Exercise $\PageIndex{1}$

Exercise $\PageIndex{2}$

Exercise $\PageIndex{3}$

Exercise $\PageIndex{4}$

Exercise $\PageIndex{1}$

Example $\PageIndex{1}$

Example $\PageIndex{2}$

Example $\PageIndex{3}$

Example $\PageIndex{4}$

Theorem $\PageIndex{1}$

Example $\PageIndex{5}$

Example $\PageIndex{6}$

Example $\PageIndex{7}$

Example $\PageIndex{8}$

Example $\PageIndex{9}$

Example $\PageIndex{10}$

Example $\PageIndex{11}$

Example $\PageIndex{12}$

Example $\PageIndex{13}$

Example $\PageIndex{14}$

Example $\PageIndex{15}$

Example $\PageIndex{16}$