6.3: test
- Page ID
- 154469
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Graph \(u_c(t)= \begin{cases}0 & t<c \\ 1 & t \geq c\end{cases}\).
- Answer
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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()
Graph \(g(t) =\sin(t)\).
- Answer
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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()
Graph \(h(t)=u_\pi(t) \sin (t) \)
- Answer
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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()
Graph \(f(t)=2 t+u_\pi(t)[\sin (t)-2 t]= \begin{cases} t<\pi \\ t \geq \pi\end{cases}\)
- Answer
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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()
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
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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: \(f(t)= \begin{cases}f_1, & \text { if } t<4 \\ f_2, & \text { if } 4 \leq t<5 \\ f_3, & \text { if } 5 \leq t<10 \\ f_4, & \text { if } t \geq 10\end{cases}\)
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 \(t=9: f(9)=f_1(9)+1\left[f_2(9)-f_1(9)\right] +1\left[f_3(9)-f_2(9)\right]+0\left[f_4(9)-f_3(9)\right]=f_3(9) \)
\(f(t)=\left\{\begin{array}{ll}0 & 0 \leq t<2 \\ t^2 & t \geq 2\end{array} \quad\right.\) implies \(\quad f(t)=\)
Solution
\(f(t) = u_2(t)*t^2\)
\(g(t)=\left\{\begin{array}{ll}t^2 & 0 \leq t<3 \\ 0 & t \geq 3\end{array} \quad\right.\) implies \(g(t)=\)
Solution
\(g(t) = t^2 - u_3(t)*t^2\)
\(j(t)=\left\{\begin{array}{ll}t & 0 \leq t<5 \\ 2 & 5 \leq t < 8 \\ e^t & t \geq 8\end{array} \quad\right.\) implies \(j(t)=\).
Solution
\(j(t) = t -u_5(t) *t +2*u_5(t) -2*u_8(t) + e^t * u_8(t)\)
\[ \notag \mathcal{L}\left(u_c(t) f(t-c)\right)=e^{-c s} \mathcal{L}(f(t))\]
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 \(t+c\).
Find the LaPlace transform of \(\mathcal{L}\left(u_3(t)\left(t^2-2 t+1\right)\right)\)
Solution
\[\begin{aligned}
& \mathcal{L}\left(u_3(t)\left(t^2-2 t+1\right)\right)=e^{-3 s}\left(\frac{2}{s^3}+\frac{4}{s^2}+\frac{4}{s}\right) \\
& \left.\mathcal{L}\left(u_3(t)\left(t^2-2 t+1\right)\right)=e^{-3 s} \mathcal{L}\left((t+3)^2-2(t+3)+1\right)\right) \\
& \left.=e^{-3 s} \mathcal{L}\left(t^2+6 t+9-2 t-6+1\right)\right) \\
& =e^{-3 s} \mathcal{L}\left(t^2+4 t+4\right)=e^{-3 s}\left(\frac{2}{s^3}+\frac{4}{s^2}+\frac{4}{s}\right) \\
&
\end{aligned}\]
Find the LaPlace transform of \(\mathcal{L}\left(u_4(t)\left(e^{-8 t}\right)\right)\)
Solution
\[\notag \begin{aligned}
& \mathcal{L}\left(u_4(t)\left(e^{-8 t}\right)\right)=e^{-4 s-32}\left(\frac{1}{s+8}\right) \\
& \left.\mathcal{L}\left(u_4(t)\left(e^{-8 t}\right)\right)=e^{-4 s} \mathcal{L}\left(e^{-8(t+4)}\right)=e^{-4 s} \mathcal{L}\left(e^{-8 t} e^{-32}\right)\right) \\
& =e^{-4 s} e^{-32} \mathcal{L}\left(e^{-8 t}\right)=e^{-4 s-32}\left(\frac{1}{s+8}\right)
\end{aligned}\]
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 \(F(s-3)=\frac{2}{(s-3)^3}.\)
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
\[ \notag
\mathcal{L}\left(u_3(t) e^{t-3}\right)=e^{-3 s} \mathcal{L}\left(e^t\right)=\frac{e^{-3 s}}{s-1}
\]
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}
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 \(f(t)=\mathcal{L}^{-1}(F(s))\)
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}
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 \(f(t)=\frac{1}{\sqrt{3}} \mathcal{L}^{-1}\left(\frac{\sqrt{3}}{s^2-3}\right)=\frac{1}{\sqrt{3}} \sinh (\sqrt{3} t)\)
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 \(f(t)=\frac{5}{6} \mathcal{L}^{-1}\left(\frac{3!}{(s-3)^4}\right)=\frac{5}{6} t^3 e^{3 t}\)
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 \(f(t-1)=1\)
Find the inverse LaPlace transform of \(\mathcal{L}^{-1}\left(e^{-s}\right)\)
Solution
\(\mathcal{L}^{-1}\left(e^{-s}\right)=\underline{\delta(t-1)}\)
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 \(f(t)=\frac{1}{2} \mathcal{L}^{-1}\left(\frac{2}{(s-3)^2+4}\right)=\frac{1}{2} e^{3 t} \sin (2 t)\)
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}
\(\frac{2 s-5}{s^2+6 s+13}=\frac{2 s-5}{s^2+6 s+9-9+13}=\frac{2 s-5}{(s+3)^2+4}=\frac{2(s+3)-6-5}{(s+3)^2+4}\)
\(\begin{aligned} \mathcal{L}^{-1}\left(\frac{2 s-5}{s^2+6 s+13}\right) & =\mathcal{L}^{-1}\left(\frac{2(s+3)-11}{(s+3)^2+4}\right) \\ & =2 \mathcal{L}^{-1}\left(\frac{s+3}{(s+3)^2+4}\right)-11 \mathcal{L}^{-1}\left(\frac{1}{(s+3)^2+4}\right) \\ & =2 \mathcal{L}^{-1}\left(\frac{s+3}{(s+3)^2+4}\right)-\frac{11}{2} \mathcal{L}^{-1}\left(\frac{2}{(s+3)^2+4}\right) \\ & =2 e^{-3 t} \cos (2 t)-\frac{11}{2} e^{-3 t} \sin (2 t)\end{aligned}\)
\(\begin{aligned} & \mathcal{L}^{-1}\left(e^{-s} \frac{2 s-5}{s^2+6 s+13}\right)=u_1(t) f(t-1) \\ & \quad=u_1(t)\left[2 e^{-3(t-1)} \cos (2(t-1))-\frac{11}{2} e^{-3(t-1)} \sin (2(t-1))\right] \\ & \quad=u_1(t) e^{-3 t+1}\left[2 \cos (2 t-2)-\frac{11}{2} \sin (2 t-2)\right]\end{aligned}\)