Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 6.3 Step Function due 04/01/2021 at 11:59pm MST
1. (1 point)
(1) Graph the function
f(t) = 2t(u(t−5)−u(t−10))
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:
2t(u(t−5)−u(t−10)) =
if 0 ≤t<5,
if 5 ≤t<10,
if 10 ≤t<∞.
help (formulas)
(2) Evaluate f(7.5).
f(7.5) = help (formulas)
Answer(s) submitted:
•0
•1
•0
•((75)/(sˆ(2)))
(score 0.5)
Correct Answers:
•0
•2*t
•0
•2*7.5
2. (1 point) Find the Laplace transform of
f(t) = 2u(t−5)−4u(t−6) + 5u(t−9)
F(s) = .
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•2*exp(-5*s)/s - 4*exp(-6*s)/s+5*exp(-9*s)/s
3. (1 point)
Consider the function f(t) =
0,t<0
6,0≤t<3
−4,3≤t<7
6,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) = .
Answer(s) submitted:
•
•
(incorrect)
Correct Answers:
•6*u(t)-10*u(t-3)+10*u(t-7)
•6/s-10*exp(-3*s)/s+10*exp(-7*s)/s
4. (1 point) Find the Laplace transform of
f(t) = (0,t<7
(t−7)2,t≥7
F(s) = .
Solution:
f(t) = u7(t)·(t−7)2
Thus
L{f(t)}=e−7sL{t2}=e−7s2!
s3
Answer(s) submitted:
•((2eˆ(-7s))/(sˆ(3)))
(correct)
Correct Answers:
•exp(-7*s)*2/sˆ3
5. (1 point) Find the Laplace transform of
f(t) = 1+u(t−5)·(t+9)
F(s) = .
Solution:
L{f(t)}=1
s+e−5sL{t+14}=1
s+e−5s1
s2+14
s
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•1/s+exp(-5*s)(1/sˆ2+14/s)
1
6. (1 point) Find the Laplace transform of
f(t) = u(t−6)·t2
F(s) = .
Solution:
L{f(t)}=e−6sL{(t+6)2}=e−6sL{t2+12t+36}=e−6s2
s3+12
s2+36
s
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•eˆ(-6*s)*[2/(sˆ3)+12/(sˆ2)+36/s]
7. (1 point) Find the Laplace transform of
f(t) = ut−3π
2·sint
F(s) = .
Solution:
L{f(t)}=e−3πs/2Lsint+3π
2
=e−3πs/2Lsintcos3π
2+costsin3π
2
=e−3πs/2L{−cost}
=−e−3πs/2s
s2+1
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•-(eˆ[-(3*pi*s/2)]*s/(sˆ2+1))
8. (1 point) Find the Laplace transform of
f(t) = u(t−4)·e2t
F(s) = .
Solution:
L{f(t)}=e−4sL{e2(t+4)}
=e−4(s−2)
s−2
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•exp(-4*(s-2))/(s-2)
9. (1 point)
Find the Laplace transform of
f(t) = (0,t<2
t2−4t+7,t≥2
F(s) = .
Solution:
f(t) = u2(t)·(t2−4t+7) = u2(t)·[(t−2)2+3]
Thus
L{f(t)}=e−2sL{t2+3}=e−2s2
s3+3
s
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•exp(-2*s)*(2/sˆ3 + 3/s)
10. (1 point)
Find the inverse Laplace transform of
F(s) = −e−2s−e−3s−6e−5s−e−8s
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−2s
s} − 1L−1{e−3s
s} − 6L−1{e−5s
s} − 1L−1{e−8s
s}
=−1·u2(t)−1·u3(t)−6·u5(t)−1·u8(t)
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•-1*u(t-2)+-1*u(t-3)+-6*u(t-5)+-1*u(t-8)
11. (1 point)
Find the inverse Laplace transform of
F(s) = e−9s
s2−16
f(t) = .
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
Solution:
L−1e−9s
s2−16 =u(t−9)·f(t−9)
where
f(t) = L−11
s2−16 =L−1(−1
8
s+4+
1
8
s−4)=−1
8e−4t+1
8e4t
2
Thus
L−1e−9s
s2−16 =u(t−9)·−1
8e−4(t−9)+1
8e4(t−9)
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•u(t-9)*[-0.125*exp(-4*(t-9))+0.125*exp(-(-4)*(t-9))]
12. (1 point)
Find the inverse Laplace transform of
F(s) = 8e−7s
s2+49
f(t) = .
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
Solution:
L−18e−7s
s2+49 =u(t−7)·f(t−7)
where
f(t) = L−18
s2+49 =8
7L−17
s2+49 =8
7sin(7t)
Thus
L−18e−7s
s2+49 =8
7u(t−7)·sin(7(t−7))
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•u(t-7)*8/7*sin(7*(t-7))
13. (1 point)
Find the inverse Laplace transform of
F(s) = e−s(7−9s)
s2+16 .
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−s(7−9s)
s2+16 =−9e−ss
s2+16 +
7e−s1
s2+16 =−9e−ss
s2+16 +7e−s
4
4
s2+16
Since L−1s
s2+16 =cos(4t)and L−14
s2+16 =sin(4t),
we have that
L−1{F(s)}=−9cos(4(t−1))u(t−1)+ 7
4u(t−1)sin(4(t−1))
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•7*sin(4*(t-1))*u(t-1)/4-9*cos(4*(t-1))*u(t-1)
14. (1 point) Compute the inverse Laplace transform of
F(s) = 2s−1
s2−s−12 e−2s
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−s−12 = (s−4)(s+3).
•You must solve the Type I partial fractions problem:
2s−1
(s−4)(s+3)=A
s−4+B
s+3
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•u(t-2)*(eˆ[4*(t-2)]+eˆ[-3*(t-2)])
15. (1 point) Compute the inverse Laplace transform of
F(s) = −2s−4
s2+6s+9e−2s
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+9= (s+3)2.
•Solve the partial fractions problem:
−2s−4
(s+3)2=A
(s+3)2+B
s+3
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•u(t-2)*[2*(t-2)-2]*eˆ[-3*(t-2)]
3
16. (1 point) Compute the inverse Laplace transform of
F(s) = 2−2s
s2−6s+25 e−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+25 = (s−3)2+16.
•2−2s
(s−3)2+16 =−2(s−3)−4
(s−3)2+16
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•-u(t-3)*[2*cos(4*(t-3))+sin(4*(t-3))]*eˆ[3*(t-3)]
17. (1 point) Find the Laplace transform of
f(t) =
0,t<8
4sin(πt),8≤t<9
0,t≥9
and
L{f(t)}=4e−8sπ
s2+π2+4e−9sπ
s2+π2
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•4*pi*(exp(-8*s)+exp(-9*s))/(sˆ2+pi**2)
18. (1 point)
Find the inverse Laplace transform f(t) = L−1{F(s)}of the
function
F(s) = 3s−7
s2−4s+20 s>2
.
f(t) = help (formulas)
Solution: Completing the square at the denominator yields:
F(s) = 3s−7
(s−2)2+16
We rearrange the numerator:
F(s) = 3(s−2)−1
(s−2)2+16
Separating the fraction:
F(s) = 3s−2
(s−2)2+16 −11
(s−2)2+16 =3s−2
(s−2)2+16 −1
4
4
(s−2)2+16
Thus
L−1{F(s)}=3e2tcos(4t)−1
4e2tsin(4t)
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•3*eˆ(2*t)*cos(4*t)-0.25*eˆ(2*t)*sin(4*t)
F(s) = .
Solution: Rewriting the function f (t) in terms of unit step
function, yields
f (t) = 4 · u(t − 8)sin(πt) − 4 · u(t − 9)sin(πt)
Since the function sin(πt) is periodic with period T = 2, we have
that sin(πt) = sin(π(t − 8) and −sin(πt) = sin(π(t − 9)). Thus
f (t) = 4 · u(t − 8)sin(π(t − 8)) + 4 · u(t − 9)sin(π(t − 9)
4
