Test 1 Solutions - Operational Methods | MATH 4564, Exams of Production and Operations Management

Test 1 Solutions Material Type: Exam; Professor: Kim; Class: Operational Methods; Subject: Mathematics; University: Virginia Polytechnic Institute And State University; Term: Fall 2007;

Typology: Exams

Pre 2010

Uploaded on 09/30/2007

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Test I, Math 4564 Solution
[1]
1
(s+1)(s2โˆ’1) =A
sโˆ’1+Bs +C
(s+1)
2
A=1
4,B=โˆ’1
4,C=โˆ’3
4
Thus,
f(t)=1
4etโˆ’1
4eโˆ’tโˆ’1
2teโˆ’t
[2]
s+1
s(s2โˆ’2s+2) =A
s+Bs +C
s2โˆ’2s+2
A=1
2,B=โˆ’1
2,C=2
Thus,
f(t)=1
2โˆ’1
2etcos(t)+3
2etsin(t)
[3]
s2Yโˆ’1โˆ’2sY +Y=1
sโˆ’1
Y(s)= 1
(sโˆ’1)2+1
(sโˆ’1)3
Thus,
y(t)=te
t+1
2t2et
[4]
L{u1(t)g(tโˆ’1)}=eโˆ’sL{g(t)}
If g(tโˆ’1) = et,then g(t)=et+1,and
L{g(t)}=e
sโˆ’1
L{u2(t)h(tโˆ’2)}=eโˆ’2sL{h(t)}
If h(tโˆ’2) = t2,then h(t)=t2+4t+4,and
L{h(t)}=2
s3+4
s2+4
s
Thus,
F(s)= ee
โˆ’s
sโˆ’1+eโˆ’2s๎˜2
s3+4
s2+4
s๎˜‚
pf2

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Test I, Math 4564 Solution

[1] 1 (s + 1)(s^2 โˆ’ 1)

A

s โˆ’ 1

Bs + C (s + 1)^2

A =

, B = โˆ’

, C = โˆ’

Thus,

f (t) =

et^ โˆ’

eโˆ’t^ โˆ’

teโˆ’t

[2] s + 1 s(s^2 โˆ’ 2 s + 2)

A

s

Bs + C s^2 โˆ’ 2 s + 2

A =

, B = โˆ’

, C = 2

Thus,

f (t) =

et^ cos(t) +

et^ sin(t)

[3]

s^2 Y โˆ’ 1 โˆ’ 2 sY + Y =

s โˆ’ 1

Y (s) =

(s โˆ’ 1)^2

(s โˆ’ 1)^3

Thus,

y(t) = t et^ +

t^2 et

[4] L{u 1 (t)g(t โˆ’ 1)} = eโˆ’sL{g(t)}

If g(t โˆ’ 1) = et, then g(t) = et+1, and

L{g(t)} = e s โˆ’ 1

L{u 2 (t)h(t โˆ’ 2)} = eโˆ’^2 sL{h(t)}

If h(t โˆ’ 2) = t^2 , then h(t) = t^2 + 4t + 4, and

L{h(t)} =

s^3

s^2

s

Thus,

F (s) = e eโˆ’s s โˆ’ 1

  • eโˆ’^2 s

s^3

s^2

s

[5]

L{f (t)} =

1

eโˆ’stet^ dt = eโˆ’(sโˆ’1) s โˆ’ 1

So

Y (s) = eโˆ’(sโˆ’1) (s โˆ’ 1)(s^2 + 1)

= e eโˆ’s

A

s โˆ’ 1

Bs + C s^2 + 1

A =

, B = โˆ’

, C = โˆ’

Thus, y(t) = e 2 u 1 (t) etโˆ’^1 โˆ’ e 2 u 1 (t) cos(t โˆ’ 1) โˆ’ e 2 u 1 (t)sin(t โˆ’ 1)

[6]

s^2 Y โˆ’ 1 โˆ’ Y =

1 โˆ’ eโˆ’^2 s

1 โˆ’ eโˆ’s s

s

1 + eโˆ’s

Y (s) =

s^2 โˆ’ 1

s

s s^2 โˆ’ 1

โˆ‘^ โˆž

k=

(โˆ’1)keโˆ’ks

y(t) = sinh(t) โˆ’ 1 + cosh(t) +

โˆ‘^ โˆž

k=

(โˆ’1)kuk(t)(cosh(t โˆ’ k) โˆ’ 1)