numerical methods - 17.31

17.11 PRACTICE PROBLEM SOLUTIONS

1.

5

l)

0.5

0.5

a)

--

s – 3

– s + 3

s

1

m)

–4

16s

2

=

12s2 – 16

b)

--------------------

+ --------------------

----------------------

s-----------+ 3

( s2 + 4) 2

( s2 + 4) 3

( s2 + 4) 3

5

n)

2s

c)

------------------

-----------

( s + 3) 3

s + 3

d)

5

o)

( s

2

x – 7s – 8) + 5( sx – 7) + 3x

(------------------s + 3) 2

6s

5

p)

e)

----------------

----

s

2

+ 36

s2

f)

8

q)

2

----

s3

s

r)

y

g)

----------------

2

--

s

+ 25

s

72

18

1

h)

3

3

e–1

s)

-----

–

-----

-----------

–

+

s

5

s

4 + s + 5

----

--s

s-----------+ 1

s

2

e

–s

e

–2s

5( s + 3)

t)

–

i)

------------------------------

------s

--------

s

( s + 3) 2 + 52

e4 – 2s

j)

2.5 1

2.5 –1

u)

s + 3 – 5j + s + 3 + 5j

-------------s + 2

=

s(

) +

v)

e

2 – s

s----------------------------2 + 6s + 34

-----------

s + 1

k)

5

w)

5

e–s

e–2s

s----------------2 + 25

s + 3

+ s –

s

x)

cos ( 2) s – sin ( 2) 7

e–3

– 0.416s – 6.37

e–3

-------------------------------------------s

2

+ 49

+ s----------– 1 =

------------------------------------s

2

+ 49

+ s----------– 1

y)

s cos 1 – 5 sin 1

----------------------------------

s2 + 25

3 3

3 –3

s( ) +

z)

s-------------------------------+ 2.7 – 9.2j + s-------------------------------+ 2.7 + 9.2j

= ----------------------------------------

2

+ 5.4s + 91.93

s

2.

a)e–t

b)5e–t

c)6t

d)3t2

e)– e–3t + 2e–4t

f)6 sin ( 6t)

3.

numerical methods - 17.32

g)5 – 5u( t – 4.5)

() –( 1) t 3

h)2 5 e cos 2t + atan –--4

i)t3 + 2 sin ( 3t)

j)6e–2t – 6e–3t

k)1.5e–2t – 1.5e–3t

a)	– e–3t + 2e–4t		g)
b)			h)
c)			i)
d)			j)	δ ( t)	– e–2t
e)	1.2 – 1.2e	–5t	k)	δ ( t)
				--------- + 0.834 cos ( t + 0.927)
				6
f)	8.34e–4.4t + 2( 0.99) e–0.3t cos ( 1.13t + 1.23)		l)	δ ( t)	+ 2te–t

4.

y( t) = 5 + 6 sin ( 2t) + 2( 3) e–2t cos ( 3t – 0)

5.

a)	x	=			5
	--		----------------------------
	F		5s	2	+ 6s + 2
b)	y	=	3
	-		-----------
	x		s + 8
c)	y	=	–5
	-		----------
	x		s – 1

numerical methods - 17.33

6.

y( t) = 40 – 40e–2t

7.

a)	5	5	–3t	–	5	–4t
	--	+ --e			--e
	6	3			2
b)	( 5δ ( t) + 30 – 5e–2t) N

8.

a)

–4t

–α t

α = 4

β

= 3

L[ 5e

cos ( 3t + 2) ] = L[ 2

A

e

cos ( βt + θ ) ]

A

= 2.5

θ

= 2

A = 2.5 cos 2 + 2.5j sin 2 =

– 1.040 + 2.273j

A

+

Acomplex conjugate

=

– 1.040 + 2.273j

+

– 1.040 – 2.273j

s-----------------------+ α

–β j

-------------------------------------s + α +β j

---------------------------------------

s + 4 – 3j

--------------------------------------s + 4 + 3j

b)

L[ e–2t + 5t( u( t – 2) – u( t) ) ] = L[ e–2t] + L[ 5tu( t – 2) ] – L[ 5tu( t) ]
	1	+ 5L[ tu( t – 2) ] –			5				1			+ 5L[ ( t – 2) u( t – 2) + 2u( t – 2) ] –			5
=	-----------				----			=	-----------						----
	s + 2				s	2			s + 2						s	2
	1												5
		+ 5L[ ( t – 2) u( t – 2) ] + 10L[ u( t – 2) ] –
=	-----------												----
	s + 2												s	2
	1										5
		–2s			–2s
=	-----------	+ 5e L[ t] + 10e L[ 1] –								----
	s + 2									s		2
	1	5e–2s		10e–2s		5
=					–
	-----------	+ ------------		+ ---------------		----
	s + 2	s	2	s		s	2

c)

d

3

s

3

y + 1s

2

+ 2s

1

+ 3s

0

----

y =

dt

d

y

=

s

1

y + s

0

1

----

dt

d

3

d

3

2

L

----

y + 2

----

=

( s y + 1s + 2s + 3) + ( sy + 1) + ( y)

dt

y + y

dt

= y( s3 + s + 1) + ( s2 + 2s + 4)

numerical methods - 17.34

d)

1 + j

1 – j

A

A

complex conjugate

L–1

----------------------

+

= L–1

-----------------------

+

s + α +β j

s + 3 + 4j

s + 3 – 4j

s + α –β

j

2

2

–1

π

α = 3

β = 4

-----

--

A

= 1

+ 1

= 1.141

θ = atan 1

= –4

= 2

A

e

–α t

cos ( βt + θ ) = 2.282e

–3t

cos

4t –

π

--

4

e)

–1

1

3

= L[ s] + L

1

3

L

-----------

+

----------------------------

-----------

+ L

----------------------------

s + s + 2

s

2

+ 4s + 40

s + 2

s

2

+ 4s + 40

d

–2t

3

d

δ ( t) + e

–2t

6

=

----

+ e

+ L

------------------------------

=

----

+ 0.5L

------------------------------

dtδ ( t)

( s + 2) 2 + 36

dt

( s + 2) 2 + 36

d

–2t

–2t

=

----

δ ( t)

+ e

+ 0.5e sin ( 6t)

dt

9.

a)s3 + 4s2 + 4s + 4

----------------------------------------

s3 + 4s

	= 1 +	4s2 + 4				=
		----------------s3	+ 4s
	= 1 +	1-- +	-------------	3s
		s	s	2	+ 4

							1
						s3 + 4s			s3 + 4s2 + 4s + 4
									–( s3 + 4s)
											4s2 + 4
1 +	A	+	Bs + C			= 1 +	s2( A + B) + s( C) + ( 4A)
	--s		----------------s	2	+ 4		-----------------------------------------------------------		s	3	+ 4s
						= δ ( t)	+ 1 + 3 cos ( 2t)

A = 1

C = 0

B = 3