Jonathan Greenfield Brewer MAT 275 ONLINE B Spring 2020
Assignment Section 6.3 Step Function due 04/21/2020 at 11:59pm MST
1. (1 point)
(1) Graph the function
f(t) = 3t(u(t−5)−u(t−7))
for 0 ≤t<∞, where uis the unit step function with a
jump at 0, i.e. u(t) = 0,t<0
1,t≥0
Use your graph to write this function piecewise as fol-
lows:
3t(u(t−5)−u(t−7)) =
if 0 ≤t<5,
if 5 ≤t<7,
if 7 ≤t<∞.
help (formulas)
(2) Evaluate f(6).
f(6) = help (formulas)
Correct Answers:
•0
•3*t
•0
•3*6
2. (1 point) Find the Laplace transform of
f(t) = u(t−4) + 5u(t−5)−3u(t−7)
F(s) = .
Correct Answers:
•1*exp(-4*s)/s+5*exp(-5*s)/s - 3*exp(-7*s)/s
3. (1 point)
Consider the function f(t) =
0,t<0
−3,0≤t<2
−2,2≤t<7
−3,t≥7
;
1. Write the function in terms of unit step function
f(t) = .
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c. For example, u5(t)should be entered as
u(t−5).)
2. Find the Laplace transform of f(t)
F(s) = .
Correct Answers:
•u(t-2)-3*u(t)-u(t-7)
•[exp(-2*s)]/s-3/s-[exp(-7*s)]/s
4. (1 point) Find the Laplace transform of
f(t) = (0,t<3
(t−3)3,t≥3
F(s) = .
Solution:
f(t) = u3(t)·(t−3)3
Thus
L{f(t)}=e−3sL{t3}=e−3s3!
s4
Correct Answers:
•exp(-3*s)*6/sˆ4
5. (1 point) Find the Laplace transform of
f(t) = 4+u(t−3)·(t+5)
F(s) = .
Solution:
L{f(t)}=4
s+e−3sL{t+8}=4
s+e−3s1
s2+8
s
Correct Answers:
•4/s+exp(-3*s)(1/sˆ2+8/s)
6. (1 point) Find the Laplace transform of
f(t) = u(t−2)·t2
F(s) = .
Solution:
L{f(t)}=e−2sL{(t+2)2}=e−2sL{t2+4t+4}=e−2s2
s3+4
s2+4
s
Correct Answers:
•eˆ(-2*s)*[2/(sˆ3)+4/(sˆ2)+4/s]
7. (1 point) Find the Laplace transform of
f(t) = ut−11π
2·sint
F(s) = .
Solution:
L{f(t)}=e−11πs/2Lsin t+11π
2
=e−11πs/2Lsintcos11π
2+costsin11π
2
=e−11πs/2L{−cost}
=−e−11πs/2s
s2+1
Correct Answers:
•-(eˆ[-(11*pi*s/2)]*s/(sˆ2+1))
1
8. (1 point) Find the Laplace transform of
f(t) = u(t−4)·et
F(s) = .
Solution:
L{f(t)}=e−4sL{et+4}
=e−4(s−1)
s−1
Correct Answers:
•exp(-4*(s-1))/(s-1)
9. (1 point)
Find the Laplace transform of
f(t) = (0,t<5
t2−10t+27,t≥5
F(s) = .
Solution:
f(t) = u5(t)·(t2−10t+27) = u5(t)·[(t−5)2+2]
Thus
L{f(t)}=e−5sL{t2+2}=e−5s2
s3+2
s
Correct Answers:
•exp(-5*s)*(2/sˆ3 + 2/s)
10. (1 point)
Find the inverse Laplace transform of
F(s) = 4e−4s−e−s−3e−7s−4e−9s
s
f(t) = .
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
Solution: L−1{F(s)}=−1L−1{e−s
s}+4L−1{e−4s
s} − 3L−1{e−7s
s} − 4L−1{e−9s
s}
=−1·u1(t) + 4·u4(t)−3·u7(t)−4·u9(t)
Correct Answers:
•-1*u(t-1)+4*u(t-4)+-3*u(t-7)+-4*u(t-9)
11. (1 point)
Find the inverse Laplace transform of
F(s) = e−8s
s2+2s−8
f(t) = .
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
Solution:
L−1e−8s
s2+2s−8=u(t−8)·f(t−8)
where
f(t) = L−11
s2+2s−8=L−1(−1
6
s+4+
1
6
s−2)=−1
6e−4t+1
6e2t
Thus
L−1e−8s
s2+2s−8=u(t−8)·−1
6e−4(t−8)+1
6e2(t−8)
Correct Answers:
•u(t-8)*[-0.166667*exp(-4*(t-8))+0.166667*exp(-(-2)*(t-8))]
12. (1 point)
Find the inverse Laplace transform of
F(s) = 2e−9s
s2+4
f(t) = .
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
Solution:
L−12e−9s
s2+4=u(t−9)·f(t−9)
where
f(t) = L−12
s2+4=2
2L−12
s2+4=2
2sin(2t)
Thus
L−12e−9s
s2+4=2
2u(t−9)·sin(2(t−9))
Correct Answers:
•u(t-9)*2/2*sin(2*(t-9))
13. (1 point)
Find the inverse Laplace transform of
F(s) = −e−3s(5s+3)
s2+36 .
f(t) = help (formulas)
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
Solution: F(s) = −e−3s(5s+3)
s2+36 =−5e−3ss
s2+36 −
3e−3s1
s2+36 =−5e−3ss
s2+36 −3e−3s
6
6
s2+36
Since L−1s
s2+36 =cos(6t)and L−16
s2+36 =sin(6t),
we have that
L−1{F(s)}=−5cos(6(t−3))u(t−3)−1
2u(t−3)sin(6(t−3))
Correct Answers:
•-[5*cos(6*(t-3))*u(t-3)+3*sin(6*(t-3))*u(t-3)/6]
2
14. (1 point) Compute the inverse Laplace transform of
F(s) = −3s−10
s2+6s+8e−3s
f(t) =
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
If you don’t get this in 2 tries, you can get a hint.
Hint:
•The denominator factors as: s2+6s+8= (s+2)(s+4).
•You must solve the Type I partial fractions problem:
−3s−10
(s+2)(s+4)=A
s+2+B
s+4
Correct Answers:
•-u(t-3)*(2*eˆ[-2*(t-3)]+eˆ[-4*(t-3)])
15. (1 point) Compute the inverse Laplace transform of
F(s) = −2s−3
s2+4s+4e−3s
f(t) =
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
If you don’t get this in 2 tries, you can get a hint.
Hint:
•The denominator factors as: s2+4s+4= (s+2)2.
•Solve the partial fractions problem:
−2s−3
(s+2)2=A
(s+2)2+B
s+2
Correct Answers:
•u(t-3)*(t-3-2)*eˆ[-2*(t-3)]
16. (1 point) Compute the inverse Laplace transform of
F(s) = 2−s
s2+2s+10 e−5s
f(t) =
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
If you don’t get this in 2 tries, you can get a hint.
Hint:
•The denominator factors as: s2+2s+10 = (s+1)2+9.
•2−s
(s+1)2+9=−(s+1) + 3
(s+1)2+9
Correct Answers:
•u(t-5)*[sin(3*(t-5))-cos(3*(t-5))]*eˆ[-(t-5)]
17. (1 point) Find the Laplace transform of
f(t) =
0,t<6
4sin(πt),6≤t<7
0,t≥7
F(s) = .
Solution: Rewriting the function f(t)in terms of unit step
function, yields
f(t) = 4·u(t−6)sin(πt)−4·u(t−7)sin(πt)
Since the function sin(πt)is periodic with period T=2, we have
that sin(πt) = sin(π(t−6)and −sin(πt) = sin(π(t−7)). Thus
f(t) = 4·u(t−6)sin(π(t−6)) + 4·u(t−7)sin(π(t−7)
and
L{f(t)}=4e−6sπ
s2+π2+4e−7sπ
s2+π2
Correct Answers:
•4*pi*(exp(-6*s)+exp(-7*s))/(sˆ2+pi**2)
18. (1 point)
Find the inverse Laplace transform f(t) = L−1{F(s)}of the
function
F(s) = 8s−12
s2−4s+13 s>2
.
f(t) = help (formulas)
Solution: Completing the square at the denominator yields:
F(s) = 8s−12
(s−2)2+9
We rearrange the numerator:
F(s) = 8(s−2) + 4
(s−2)2+9
Separating the fraction:
F(s) = 8s−2
(s−2)2+9+41
(s−2)2+9=8s−2
(s−2)2+9+4
3
3
(s−2)2+9
Thus
L−1{F(s)}=8e2tcos(3t) + 4
3e2tsin(3t)
Correct Answers:
•8*eˆ(2*t)*cos(3*t)+1.33333*eˆ(2*t)*sin(3*t)
3
