6.3: Step function
( \newcommand{\kernel}{\mathrm{null}\,}\)
Step function
Graph uc(t)={0t<c1t≥c.
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Below is the answer when c=5.
Graph g(t)=sin(t).
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Graph h(t)=uπ(t)sin(t)
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Graph f(t)=2t+uπ(t)[sin(t)−2t]={t<πt≥π
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If h(t)={t0≤t<4ln(t)t≥4, then what does this imply about h(t)
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Example: f(t)={f1, if t<4f2, if 4≤t<5f3, if 5≤t<10f4, if t≥10
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)
f(t)={00≤t<2t2t≥2 implies f(t)=
Solution
f(t)=u2(t)∗t2
g(t)={t20≤t<30t≥3 implies g(t)=
Solution
g(t)=t2−u3(t)∗t2
j(t)={t0≤t<525≤t<8ett≥8 implies j(t)=.
Solution
j(t)=t−u5(t)∗t+2∗u5(t)−2∗u8(t)+et∗u8(t)
L(uc(t)f(t−c))=e−csL(f(t))
or equivalently
L(uc(t)f(t))=e−csL(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 L(u3(t)(t2−2t+1))
Solution
L(u3(t)(t2−2t+1))=e−3s(2s3+4s2+4s)L(u3(t)(t2−2t+1))=e−3sL((t+3)2−2(t+3)+1))=e−3sL(t2+6t+9−2t−6+1))=e−3sL(t2+4t+4)=e−3s(2s3+4s2+4s)
Find the LaPlace transform of L(u4(t)(e−8t))
Solution
L(u4(t)(e−8t))=e−4s−32(1s+8)L(u4(t)(e−8t))=e−4sL(e−8(t+4))=e−4sL(e−8te−32))=e−4se−32L(e−8t)=e−4s−32(1s+8)
Find the LaPlace transform of L(u2(t)(t2e3t))
Solution
L(u2(t)(t2e3t))=e−2s+6(2(s−3)3+4(s−3)2+4(s−3))L(u2(t2e3t))=e−2sL([(t+2)2]e3(t+2)))=e−2sL([t2+4t+4]e3t+6))=e−2se6L([t2+4t+4]e3t))=e−2s+6L(t2e3t+4te3t+4e3t))=e−2s+6(L(t2e3t)+4L(te3t)+4L(e3t))=e−2s+6(2(s−3)3+4(s−3)2+4(s−3)) since
Formula 14: L(ecsf(t))=F(s−c). Thus L(t2e3t)=F(s−3)=2(s−3)3 since F(s)=L(f(t))=L(t2)=2s3 and F(s−3)=2(s−3)3.
Find the LaPlace transform of
g(t)={0t<3et−3t≥3
Solution
Note g(t)=u3(t)et−3, thus we have
L(u3(t)et−3)=e−3sL(et)=e−3ss−1
Find the LaPlace transform of f(t)={0t<353≤t<4t−5t≥4
Solution
f(t)=0+u3(t)[5−0]+u4(t)[t−5−5]L(f(t))=L(5u3(t)+u4(t)[t−10])=5L(u3(t))+L(u4(t)[t−10])=5e−3s+e−4sL(t+4−10)=5e−3s+e−4sL(t−6)=5e−3s+e−4s[L(t)−6L(1)]=5e−3s+e−4s[1s2−6s]=5e−3s+e−4s(1−6s)s2
From the above theorem we have L(uc(t)f(t−c))=e−csL(f(t)).
Let F(s)=L(f(t)).
Then L−1(F(s))=L−1(L(f(t)))=f(t).
Thus L−1(e−csF(s))
=L−1(e−csL(f(t)))=uc(t)f(t−c)
where f(t)=L−1(F(s))
Find the inverse LaPlace transform of L−1(e−8s1s−3)
Solution
a.) L−1(e−8s1s−3)=u8(t)e3(t−8)_L−1(e−8s1s−3)=u8(t)f(t−8) where L(f(t))=1s−3. Hence f(t)=L−1(1s−3)=e3t
Find the inverse LaPlace transform of L−1(e−4s1s2−3)
Solution
L−1(e−4s1s2−3)=u4(t)1√3sinh(√3(t−4))
L−1(e−4s1s2−3)=u4(t)f(t−4) where
L(f(t))=1s2−3. Hence f(t)=1√3L−1(√3s2−3)=1√3sinh(√3t)
Find the inverse LaPlace transform of L−1(e−s5(s−3)4)
Solution
L−1(e−s5(s−3)4)=u1(t)(56)(t−1)3e3(t−1)_
L−1(e−s5(s−3)4)=u1(t)f(t−1) where
L(f(t))=5(s−3)4. Hence f(t)=56L−1(3!(s−3)4)=56t3e3t
Find the inverse LaPlace transform of L−1(e−s4s)=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):
L−1(e−s4s)=14L−1(e−ss)=14u1(t)f(t+1) where
L(f(t))=1s. Hence f(t)=1. Thus f(t−1)=1
Find the inverse LaPlace transform of L−1(e−s)
Solution
L−1(e−s)=δ(t−1)_
Find the inverse LaPlace transform of L−1(e−s1(s−3)2+4)
Solution
f.) L−1(e−s1(s−3)2+4)=12u1(t)e3(t−1)sin(2(t−1))_L−1(e−s1(s−3)2+4)=u1(t)f(t−1) where L(f(t))=1(s−3)2+4).
Hence f(t)=12L−1(2(s−3)2+4)=12e3tsin(2t)
Find the inverse LaPlace transform of L−1(e−s2s−5s2+6s+13)
Solution
L−1(e−s2s−5s2+6s+13)=u1(t)e−3t+1[2cos(2t−2)−112sin(2t−2)]
2s−5s2+6s+13=2s−5s2+6s+9−9+13=2s−5(s+3)2+4=2(s+3)−6−5(s+3)2+4
L−1(2s−5s2+6s+13)=L−1(2(s+3)−11(s+3)2+4)=2L−1(s+3(s+3)2+4)−11L−1(1(s+3)2+4)=2L−1(s+3(s+3)2+4)−112L−1(2(s+3)2+4)=2e−3tcos(2t)−112e−3tsin(2t)
L−1(e−s2s−5s2+6s+13)=u1(t)f(t−1)=u1(t)[2e−3(t−1)cos(2(t−1))−112e−3(t−1)sin(2(t−1))]=u1(t)e−3t+1[2cos(2t−2)−112sin(2t−2)]