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6.3: Step function

( \newcommand{\kernel}{\mathrm{null}\,}\)

Step function

Exercise 6.3.1

Graph uc(t)={0t<c1tc.

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 6.3.2

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 6.3.3

Graph h(t)=uπ(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 6.3.4

Graph f(t)=2t+uπ(t)[sin(t)2t]={t<πtπ

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 6.3.1

If h(t)={t0t<4ln(t)t4, 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 6.3.1

Example: f(t)={f1, if t<4f2, if 4t<5f3, if 5t<10f4, if t10

Solution

Hence
f(t)=f1(t)+u4(t)[f2(t)f1(t)]+u5(t)[f3(t)f2(t)]+u10(t)[f4(t)f3(t)]

Partial check

If t=3:f(3)=f1(3)+0[f2(3)f1(3)]+0[f3(3)f2(3)]+0[f4(3)f3(3)]=f1(3)

If t=9:f(9)=f1(9)+1[f2(9)f1(9)]+1[f3(9)f2(9)]+0[f4(9)f3(9)]=f3(9)

Example 6.3.2

f(t)={00t<2t2t2 implies f(t)=

Solution

f(t)=u2(t)t2

Example 6.3.3

g(t)={t20t<30t3 implies g(t)=

Solution

g(t)=t2u3(t)t2

Example 6.3.4

j(t)={t0t<525t<8ett8 implies j(t)=.

Solution

j(t)=tu5(t)t+2u5(t)2u8(t)+etu8(t)

 

 

Theorem 6.3.1

L(uc(t)f(tc))=ecsL(f(t))
or equivalently
L(uc(t)f(t))=ecsL(f(t+c)).

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

Example 6.3.5

Find the LaPlace transform of L(u3(t)(t22t+1))

Solution

L(u3(t)(t22t+1))=e3s(2s3+4s2+4s)L(u3(t)(t22t+1))=e3sL((t+3)22(t+3)+1))=e3sL(t2+6t+92t6+1))=e3sL(t2+4t+4)=e3s(2s3+4s2+4s)

Example 6.3.6

Find the LaPlace transform of L(u4(t)(e8t))

Solution

L(u4(t)(e8t))=e4s32(1s+8)L(u4(t)(e8t))=e4sL(e8(t+4))=e4sL(e8te32))=e4se32L(e8t)=e4s32(1s+8)

Example 6.3.7

Find the LaPlace transform of L(u2(t)(t2e3t))

Solution

L(u2(t)(t2e3t))=e2s+6(2(s3)3+4(s3)2+4(s3))L(u2(t2e3t))=e2sL([(t+2)2]e3(t+2)))=e2sL([t2+4t+4]e3t+6))=e2se6L([t2+4t+4]e3t))=e2s+6L(t2e3t+4te3t+4e3t))=e2s+6(L(t2e3t)+4L(te3t)+4L(e3t))=e2s+6(2(s3)3+4(s3)2+4(s3)) since 

Formula 14: L(ecsf(t))=F(sc). Thus L(t2e3t)=F(s3)=2(s3)3 since F(s)=L(f(t))=L(t2)=2s3 and F(s3)=2(s3)3.

Example 6.3.8

Find the LaPlace transform of
g(t)={0t<3et3t3

 

Solution

Note g(t)=u3(t)et3, thus we have
L(u3(t)et3)=e3sL(et)=e3ss1

 

Example 6.3.9

Find the LaPlace transform of f(t)={0t<353t<4t5t4

Solution

f(t)=0+u3(t)[50]+u4(t)[t55]L(f(t))=L(5u3(t)+u4(t)[t10])=5L(u3(t))+L(u4(t)[t10])=5e3s+e4sL(t+410)=5e3s+e4sL(t6)=5e3s+e4s[L(t)6L(1)]=5e3s+e4s[1s26s]=5e3s+e4s(16s)s2

Note

From the above theorem we have L(uc(t)f(tc))=ecsL(f(t)).
Let F(s)=L(f(t)).
Then L1(F(s))=L1(L(f(t)))=f(t).
Thus L1(ecsF(s))
=L1(ecsL(f(t)))=uc(t)f(tc)
where f(t)=L1(F(s))

Example 6.3.10

Find the inverse LaPlace transform of L1(e8s1s3)

Solution

 a.) L1(e8s1s3)=u8(t)e3(t8)_L1(e8s1s3)=u8(t)f(t8) where L(f(t))=1s3. Hence f(t)=L1(1s3)=e3t

Example 6.3.11

Find the inverse LaPlace transform of L1(e4s1s23)

Solution

L1(e4s1s23)=u4(t)13sinh(3(t4))
L1(e4s1s23)=u4(t)f(t4) where
L(f(t))=1s23. Hence f(t)=13L1(3s23)=13sinh(3t)

Example 6.3.12

Find the inverse LaPlace transform of L1(es5(s3)4)

Solution

L1(es5(s3)4)=u1(t)(56)(t1)3e3(t1)_
L1(es5(s3)4)=u1(t)f(t1) where
L(f(t))=5(s3)4. Hence f(t)=56L1(3!(s3)4)=56t3e3t

Example 6.3.13

Find the inverse LaPlace transform of L1(es4s)=14u1(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):
L1(es4s)=14L1(ess)=14u1(t)f(t+1) where
L(f(t))=1s. Hence f(t)=1. Thus f(t1)=1

Example 6.3.14

Find the inverse LaPlace transform of L1(es)

Solution

L1(es)=δ(t1)_

Example 6.3.15

Find the inverse LaPlace transform of L1(es1(s3)2+4)

Solution

 f.) L1(es1(s3)2+4)=12u1(t)e3(t1)sin(2(t1))_L1(es1(s3)2+4)=u1(t)f(t1) where L(f(t))=1(s3)2+4)

Hence f(t)=12L1(2(s3)2+4)=12e3tsin(2t)

Example 6.3.16

Find the inverse LaPlace transform of L1(es2s5s2+6s+13)

Solution

L1(es2s5s2+6s+13)=u1(t)e3t+1[2cos(2t2)112sin(2t2)]

2s5s2+6s+13=2s5s2+6s+99+13=2s5(s+3)2+4=2(s+3)65(s+3)2+4

L1(2s5s2+6s+13)=L1(2(s+3)11(s+3)2+4)=2L1(s+3(s+3)2+4)11L1(1(s+3)2+4)=2L1(s+3(s+3)2+4)112L1(2(s+3)2+4)=2e3tcos(2t)112e3tsin(2t)

L1(es2s5s2+6s+13)=u1(t)f(t1)=u1(t)[2e3(t1)cos(2(t1))112e3(t1)sin(2(t1))]=u1(t)e3t+1[2cos(2t2)112sin(2t2)]

 


This page titled 6.3: Step function is shared under a not declared license and was authored, remixed, and/or curated by Isabel K. Darcy.

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