Chapter 15, Solution 1.
e at + e - at
(a) cosh(at ) =
2
1 1 1 s
L [ cosh(at ) ] = + = s2 − a2
2s − a s + a
e at − e - at
(b) sinh(at ) =
2
1 1 1 a
L [ sinh(at ) ] = − = s2 − a2
2s − a s + a
Chapter 15, Solution 2.
(a) f ( t ) = cos(ωt ) cos(θ) − sin(ωt ) sin(θ)
F(s) = cos(θ) L [ cos(ωt ) ] − sin(θ) L [ sin(ωt ) ]
s cos(θ) − ω sin(θ)
F(s) =
s 2 + ω2
(b) f ( t ) = sin(ωt ) cos(θ) + cos(ωt ) sin(θ)
F(s) = sin(θ) L [ cos(ωt ) ] + cos(θ) L [ sin(ωt ) ]
s sin(θ) − ω cos(θ)
F(s) =
s 2 + ω2
Chapter 15, Solution 3.
s+2
(a) L [ e -2t cos(3t ) u ( t ) ] =
(s + 2 ) 2 + 9
4
(b) L [ e -2t sin(4 t ) u ( t ) ] =
(s + 2) 2 + 16
s
(c) Since L [ cosh(at ) ] =
s − a2
2
s+3
L [ e -3t cosh(2 t ) u ( t ) ] =
(s + 3 ) 2 − 4
a
(d) Since L [ sinh(at ) ] =
s − a2
2
1
L [ e -4t sinh( t ) u ( t ) ] =
(s + 4) 2 − 1
2
(e) L [ e - t sin( 2t ) ] =
(s + 1) 2 + 4
If f (t) ←
→ F(s)
-d
t f (t) ←
→ F(s)
ds
Thus, L [ t e - t sin(2 t ) ] =
-d
ds
[
2 ( (s + 1) 2 + 4)
-1
]
2
= ⋅ 2 (s + 1)
((s + 1) 2 + 4) 2
4 (s + 1)
L [ t e -t sin( 2t ) ] =
((s + 1) 2 + 4) 2
Chapter 15, Solution 4.
s 6se −s
(a) G (s) = 6 e −s =
s2 + 42 s 2 + 16
2 e −2s
(b) F(s) = +5
s2 s+3
Chapter 15, Solution 5.
s cos(30°) − 2 sin(30°)
(a) L [ cos(2t + 30°) ] =
s2 + 4
d 2 s cos(30°) − 1
L [ t 2 cos(2t + 30°) ] =
ds 2 s 2 + 4
d d 3 -1
= s − 1 (s 2 + 4)
ds ds 2
d 3 2 3 -2
(s + 4) − 2s s − 1 (s 2 + 4)
-1
= 2
ds 2
3 3 3
3 2
(- 2s ) 2 s − 1 2s
(8s 2 )
2 s − 1
= 2 − − 2 +
(
s +4
2 2
) s +4
2 2
(
s +4
2 2
) s +4
2
( 3
) ( )
3
(8s 2 )
2 s − 1
- 3s − 3s + 2 − 3s
= +
s +4
2 2
( )
s +4
2 3
( )
(-3 3 s + 2)(s 2 + 4) 4 3 s3 − 8 s 2
= +
(s 2
+ 4)
3
(s 2
+ 4)
3
8 − 12 3 s − 6s 2 + 3s 3
L [ t 2 cos(2t + 30°) ] =
( s 2 + 4) 3
4!
L [ 30 t 4 e - t ] = 30 ⋅
720
(b) 5 =
(s + 2) (s + 2 ) 5
d 2 2
(c) L 2t u ( t ) − 4 δ( t ) = 2 − 4(s ⋅ 1 − 0) = 2 − 4s
dt s s
(d) 2 e -(t-1) u ( t ) = 2 e -t u ( t )
2e
L [ 2 e -(t-1) u ( t ) ] =
s+1
(e) Using the scaling property,
1 1 1 5
L [ 5 u ( t 2) ] = 5 ⋅ ⋅ = 5⋅ 2⋅ =
1 2 s (1 2) 2s s
6 18
(f) L [ 6 e -t 3 u ( t ) ] = =
s + 1 3 3s + 1
(g) Let f ( t ) = δ( t ) . Then, F(s) = 1 .
dn dn
L n δ( t ) = L n f ( t ) = s n F(s) − s n −1 f (0) − s n − 2 f ′(0) −
dt dt
dn dn
L n δ( t ) = L n f ( t ) = s n ⋅ 1 − s n −1 ⋅ 0 − s n − 2 ⋅ 0 −
dt dt
dn
L n δ( t ) = s n
dt
Chapter 15, Solution 6.
(a) L [ 2 δ( t − 1) ] = 2 e -s
10 - 2s
(b) L [ 10 u ( t − 2) ] = e
s
1 4
(c) L [ ( t + 4) u ( t ) ] = +
s2 s
2 e -4s
(d) L[ 2e -t
u ( t − 4) ] = L [ 2 e -4
e -(t - 4)
u ( t − 4) ] =
e 4 (s + 1)
Chapter 15, Solution 7.
s
(a) Since L [ cos(4t ) ] = , we use the linearity and shift properties to
s + 42 2
10 s e - s
obtain L [10 cos(4 ( t − 1)) u ( t − 1) ] = 2
s + 16
2 1
(b) Since L [ t 2 ] = , L [ u ( t )] = ,
s3 s
2 e -3s
L [ t 2 e -2 t ] = , and L [ u ( t − 3)] =
(s + 2) 3 s
2 e -3s
L [ t 2 e -2 t u ( t ) + u ( t − 3) ] = +
(s + 2 ) 3 s
Chapter 15, Solution 8.
(a) L [ 2 δ(3t ) + 6 u (2t ) + 4 e -2 t − 10 e -3 t ]
1 1 1 4 10
= 2⋅ + 6⋅ ⋅ + −
3 2 s 2 s+2 s+3
2 6 4 10
= + + −
3 s s+2 s+3
(b) t e -t u ( t − 1) = ( t − 1) e -t u ( t − 1) + e -t u ( t − 1)
t e -t u ( t − 1) = ( t − 1) e -(t-1) e -1 u ( t − 1) + e -(t-1) e -1 u ( t − 1)
e -1 e -s e -1 e -s e -(s+1) e -(s+1)
L [ t e - t u ( t − 1) ] = + = +
(s + 1) 2 s + 1 (s + 1) 2 s + 1
s e -s
(c) L [ cos(2 ( t − 1)) u ( t − 1) ] =
s2 + 4
(d) Since sin(4 ( t − π)) = sin(4t ) cos(4π ) − sin( 4π) cos(4t ) = sin(4t )
sin(4t ) u ( t − π ) = sin(4 ( t − π )) u ( t − π )
L[ sin( 4 t ) [ u ( t ) − u ( t − π )] ]
= L[ sin( 4 t ) u ( t ) ] − L[ sin( 4( t − π )) u ( t − π) ]
4 4 e - πs 4
= − 2 = 2 ⋅ (1 − e -πs )
s + 16 s + 16 s + 16
2
Chapter 15, Solution 9.
(a) f ( t ) = ( t − 4) u ( t − 2) = ( t − 2) u ( t − 2) − 2 u ( t − 2)
e -2s 2 e -2s
F(s) = 2 − 2
s s
(b) g( t ) = 2 e -4t u ( t − 1) = 2 e -4 e -4(t -1) u ( t − 1)
2 e -s
G (s) = 4
e (s + 4)
(c) h ( t ) = 5 cos(2 t − 1) u ( t )
cos(A − B) = cos(A) cos(B) + sin(A) sin(B)
cos(2t − 1) = cos(2t ) cos(1) + sin(2t ) sin(1)
h ( t ) = 5 cos(1) cos(2 t ) u ( t ) + 5 sin(1) sin(2t ) u ( t )
s 2
H(s) = 5 cos(1) ⋅ + 5 sin(1) ⋅ 2
s +4
2
s +4
2.702 s 8.415
H(s) = +
s2 + 4 s2 + 4
(d) p( t ) = 6u ( t − 2) − 6u ( t − 4)
6 - 2s 6 -4s
P(s) = e − e
s s
Chapter 15, Solution 10.
(a) By taking the derivative in the time domain,
g ( t ) = (-t e -t + e -t ) cos( t ) − t e -t sin( t )
g ( t ) = e -t cos( t ) − t e -t cos( t ) − t e -t sin( t )
s +1 d s +1 d 1
G (s) = + +
(s + 1) + 1 ds (s + 1) + 1 ds (s + 1) + 1
2 2 2
s +1 s 2 + 2s 2s + 2 s 2 (s + 2)
G (s) = − 2 − 2 = 2
s 2 + 2s + 2 (s + 2s + 2) 2 (s + 2s + 2) 2 (s + 2s + 2) 2
(b) By applying the time differentiation property,
G (s) = sF(s) − f (0)
where f ( t ) = t e -t cos( t ) , f (0) = 0
- d s +1 (s)(s 2 + 2s) s 2 (s + 2)
G (s) = (s) ⋅ = 2 = 2
ds (s + 1) 2 + 1 (s + 2s + 2) 2 (s + 2s + 2) 2
Chapter 15, Solution 11.
s
(a) Since L [ cosh(at ) ] =
s − a2
2
6 (s + 1) 6 (s + 1)
F(s) = = 2
(s + 1) − 4 s + 2s − 3
2
a
(b) Since L [ sinh(at ) ] =
s − a2
2
(3)(4) 12
L [ 3 e -2t sinh(4t ) ] = = 2
(s + 2) − 16 s + 4s − 12
2
-d
F(s) = L [ t ⋅ 3 e -2t sinh(4t ) ] = [ 12 (s 2 + 4s − 12) -1 ]
ds
24 (s + 2)
F(s) = (12)(2s + 4)(s 2 + 4s − 12) -2 =
(s + 4s − 12) 2
2
1
(c) cosh( t ) = ⋅ (e t + e - t )
2
1
f ( t ) = 8 e -3t ⋅ ⋅ (e t + e - t ) u ( t − 2)
2
= 4 e-2t u ( t − 2) + 4 e-4t u ( t − 2)
= 4 e-4 e-2(t - 2) u ( t − 2) + 4 e-8 e-4(t - 2) u ( t − 2)
L [ 4 e -4 e -2(t -2) u ( t − 2)] = 4 e -4 e -2s ⋅ L [ e -2 u ( t )]
4 e -(2s+ 4)
L [ 4 e -4 e -2(t -2) u ( t − 2)] =
s+2
4 e -(2s+8)
Similarly, L [ 4 e -8 e -4(t -2) u ( t − 2)] =
s+4
Therefore,
4 e -(2s+ 4) 4 e -(2s+8) e -(2s+ 6) [ (4 e 2 + 4 e -2 ) s + (16 e 2 + 8 e -2 )]
F(s) = + =
s+2 s+4 s 2 + 6s + 8
Chapter 15, Solution 12.
f ( t ) = te −2( t −1) e −2 u ( t − 1) = ( t − 1)e −2 e −2( t −1) u ( t − 1) + e −2 e −2( t −1) u ( t − 1)
e−2 e −s e − (s + 2) 1 s + 3 − (s + 2)
f (s) = e − s + e−2 = 1 + = e
(s + 2) 2 s+2 s + 2 s + 2 (s + 2) 2
Chapter 15, Solution 13.
d
(a) tf (t ) ←
→ − F (s)
ds
s d ( s 2 + 1)(1) − s (2s + 1)
If f(t) = cost, then F ( s )= and F ( s )=
s2 +1 ds ( s 2 + 1) 2
s2 + s −1
L (t cos t ) =
( s 2 + 1) 2
(b) Let f(t) = e-t sin t.
1 1
F (s) = = 2
( s + 1) + 1 s + 2s + 2
2
dF ( s 2 + 2s + 2)(0) − (1)(2s + 2)
=
ds ( s 2 + 2s + 2) 2
dF 2( s + 1)
L (e −t t sin t ) = − = 2
ds ( s + 2s + 2) 2
∞
f (t )
(c )
t
←
→ ∫ F (s)ds
s
β
Let f (t ) = sin βt , then F ( s ) =
s +β2
2
∞
sin βt β 1 −1 s ∞ π s β
L = ∫ s 2 + β 2 ds = β β tan β = − tan −1
β
= tan −1
t s s 2 s
Chapter 15, Solution 14.
5t 0 < t Chapter 15, Solution 15.
f ( t ) = 10 [ u ( t ) − u ( t − 1) − u ( t − 1) + u ( t − 2)]
1 2 e -2s 10
F(s) = 10 − e -s + = (1 − e -s ) 2
s s s s
Chapter 15, Solution 16.
f ( t ) = 5 u ( t ) − 3 u ( t − 1) + 3 u ( t − 3) − 5 u ( t − 4)
1
F(s) = [ 5 − 3 e -s + 3 e - 3 s − 5 e - 4 s ]
s
Chapter 15, Solution 17.
0 t3
f ( t ) = t 2 [ u ( t ) − u ( t − 1)] + 1[ u ( t − 1) − u ( t − 3)]
= t 2 u ( t ) − ( t − 1) 2 u ( t − 1) + (-2t + 1) u ( t − 1) + u ( t − 1) − u ( t − 3)
= t 2 u ( t ) − ( t − 1) 2 u ( t − 1) − 2 ( t − 1) u ( t − 1) − u ( t − 3)
2 2 e -3s
F(s) = (1 − e -s ) − 2 e -s −
s3 s s
Chapter 15, Solution 18.
(a) g ( t ) = u ( t ) − u ( t − 1) + 2 [ u ( t − 1) − u ( t − 2)] + 3 [ u ( t − 2) − u ( t − 3)]
= u ( t ) + u ( t − 1) + u ( t − 2) − 3 u ( t − 3)
1
G (s) = (1 + e -s + e - 2s − 3 e - 3s )
s
(b) h ( t ) = 2 t [ u ( t ) − u ( t − 1)] + 2 [ u ( t − 1) − u ( t − 3)]
+ (8 − 2 t ) [ u ( t − 3) − u ( t − 4)]
= 2t u ( t ) − 2 ( t − 1) u ( t − 1) − 2 u ( t − 1) + 2 u ( t − 1) − 2 u ( t − 3)
− 2 ( t − 3) u ( t − 3) + 2 u ( t − 3) + 2 ( t − 4) u ( t − 4)
= 2t u ( t ) − 2 ( t − 1) u ( t − 1) − 2 ( t − 3) u ( t − 3) + 2 ( t − 4) u ( t − 4)
2 2 - 3s 2 - 4 s 2
H(s) = 2 (1 − e ) − 2 e
-s
+ 2 e = 2 (1 − e -s − e - 3s + e -4s )
s s s s
Chapter 15, Solution 19.
1
Since L[ δ( t )] = 1 and T = 2 , F(s) =
1 − e - 2s
Chapter 15, Solution 20.
Let g 1 ( t ) = sin(πt ), 0 < t < 1
= sin( πt ) [ u ( t ) − u ( t − 1)]
= sin(πt ) u ( t ) − sin(πt ) u ( t − 1)
Note that sin(π ( t − 1)) = sin(πt − π) = - sin(πt ) .
So, g1 ( t ) = sin( πt) u(t) + sin( π( t - 1)) u(t - 1)
π
G 1 (s) = (1 + e -s )
s + π2
2
G 1 (s) π (1 + e -s )
G (s) = =
1 − e -2s (s 2 + π 2 )(1 − e - 2s )
Chapter 15, Solution 21.
T = 2π
t
Let f 1 ( t ) = 1 − [ u ( t ) − u ( t − 1)]
2π
t 1 1
f1 ( t ) = u ( t ) − u(t) + ( t − 1) u ( t − 1) − 1 − u ( t − 1)
2π 2π 2π
1 1 e -s 1 1 [ 2π + (-2π + 1) e -s ] s + [ - 1 + e -s ]
F1 (s) = − + + - 1 + e -s ⋅ =
s 2πs 2 2πs 2 2π s 2πs 2
F1 (s) [ 2π + (-2π + 1) e -s ] s + [ - 1 + e -s ]
F(s) = =
1 − e -Ts 2πs 2 (1 − e - 2 πs )
Chapter 15, Solution 22.
(a) Let g1 ( t ) = 2t, 0 < t < 1
= 2 t [ u ( t ) − u ( t − 1)]
= 2t u ( t ) − 2 ( t − 1) u ( t − 1) + 2 u ( t − 1)
2 2 e -s 2 -s
G 1 (s) = 2 − 2 + e
s s s
G 1 (s)
G (s) = , T =1
1 − e -sT
2 (1 − e -s + s e -s )
G (s) =
s 2 (1 − e -s )
(b) Let h = h 0 + u ( t ) , where h 0 is the periodic triangular wave.
Let h 1 be h 0 within its first period, i.e.
2t 0 < t 2 4 -s 2 e -2s 2
H 1 (s) = 2 − 2 e − 2 = 2 (1 − e -s ) 2
s s s s
2 (1 − e -s ) 2
H 0 (s) = 2
s (1 − e -2s )
1 2 (1 − e -s ) 2
H(s) = + 2
s s (1 − e - 2s )
Chapter 15, Solution 23.
1 0 < t Chapter 15, Solution 24.
10s 4 + s
(a) f (0) = lim sF(s) = lim
s →∞ s + 6s + 5
2
s →∞
1
10 +
s3 10
= lim = =∞
s →∞ 1 6 5 0
+ +
s 2 s3 s 4
10s 4 + s
f (∞) = lim sF(s) = lim =0
s →0 s → 0 s 2 + 6s + 5
s2 + s
(b) f (0) = lim sF(s) = lim =1
s →∞ s − 4s + 6
2
s →∞
The complex poles are not in the left-half plane.
f (∞) does not exist
2s 3 + 7s
(c) f (0) = lim sF(s) = lim
s →∞ (s + 1)(s + 2)(s + 2s + 5)
2
s →∞
2 7
+
s s3 0
= lim = =0
s →∞ 1 2 2 5 1
1 + 1 + 1 + + 2
s s s s
2s 3 + 7s 0
f (∞) = lim sF(s) = lim = =0
s → 0 (s + 1)(s + 2)(s + 2s + 5)
2
s →0 10
Chapter 15, Solution 25.
(8)(s + 1)(s + 3)
(a) f (0) = lim sF(s) = lim
s →∞ s →∞ (s + 2)(s + 4)
1 3
(8) 1 + 1 +
s s
= lim =8
s →∞ 2 4
1 + 1 +
s s
(8)(1)(3)
f (∞) = lim sF(s) = lim =3
s →0 s → 0 ( 2)( 4)
6s (s − 1)
(b) f (0) = lim sF(s) = lim
s →∞ s − 1
4
s →∞
1 1
6 2 − 4
s s 0
f (0) = lim = =0
s →∞ 1 1
1− 4
s
All poles are not in the left-half plane.
f (∞) does not exist
Chapter 15, Solution 26.
s 3 + 3s
(a) f (0) = lim sF(s) = lim =1
s →∞ s →∞ s 3 + 4s 2 + 6
Two poles are not in the left-half plane.
f (∞) does not exist
s 3 − 2s 2 + s
(b) f (0) = lim sF(s) = lim
s →∞ (s − 2)(s + 2s + 4)
2
s →∞
2 1
1−+
= lim s s2 =1
s →∞ 2 2 4
1 − 1 + + 2
s s s
One pole is not in the left-half plane.
f (∞) does not exist
Chapter 15, Solution 27.
(a) f ( t ) = u(t ) + 2 e -t
3 (s + 4) − 11 11
(b) G (s) = = 3−
s+4 s+4
g( t ) = 3 δ(t ) − 11 e -4t
4 A B
(c) H(s) = = +
(s + 1)(s + 3) s + 1 s + 3
A = 2, B = -2
2 2
H(s) = −
s +1 s + 3
h ( t ) = 2 e -t − 2 e -3t
12 A B C
(d) J (s) = = + 2 +
(s + 2) (s + 4) s + 2 (s + 2)
2
s+4
12 12
B= = 6, C= =3
2 (-2) 2
12 = A (s + 2) (s + 4) + B (s + 4) + C (s + 2) 2
Equating coefficients :
s2 : 0= A+C → A = -C = -3
s1 : 0 = 6A + B + 4C = 2A + B → B = -2A = 6
s0 : 12 = 8A + 4B + 4C = -24 + 24 + 12 = 12
-3 6 3
J (s) = + 2 +
s + 2 (s + 2) s+4
j( t ) = 3 e -4t − 3 e -2t + 6 t e -2t
Chapter 15, Solution 28.
(a)
2(−2) 2(−4)
−2 4
F(s) = 2 + − 2 = +
s+3 s+5 s+3 s+5
f ( t ) = (−2e − 3t + 4e − 5t )ut ( t )
(b)
3s + 11 A Bs + C
H(s) = = +
(s + 1)(s 2 + 2s + 5) s + 1 s 2 + 2s + 5
3s + 11 = A(s 2 + 2s + 5) + (Bs + C)(s + 1) = (A + B)s 2 + (2A + B + C)s + 5A + C
5A + C = 11; A = −B; − B + C = 3, B = C − 3 → A = 2; B = −2; C = 1
H(s) =
2
+
− 2s + 1
s + 1 s 2 + 2s + 5
( )
→ h ( t ) = 2e − t − 2e − t cos 2t + 1.5e − t sin 2t u ( t )
Chapter 15, Solution 29.
2 As + B
V(s) = + ; 2s 2 + 8s + 26 + As 2 + Bs = 2s + 26 → A = −2 and B = −6
2
s (s + 2) + 3 2
2 2(s + 2) 2 3
V(s) = − −
s (s + 2) 2 + 3 2 3 (s + 2) 2 + 3 2
2
v(t) = 2u ( t ) − 2e − 2 t cos 3t − e − 2 t sin 3t , t ≥ 0
3
Chapter 15, Solution 30.
2(s + 2) + 2 2(s + 2) 2 3
(a) H1 (s) = = +
(s + 2) 2 + 32 (s + 2) 2 + 3 2 3 (s + 2) 2 + 32
2 −2 t
h1 ( t ) = 2e −2 t cos 3t + e sin 3t
3
s2 + 4 A B Cs + D
(b) H 2 (s) = = + +
(s + 1) 2 (s 2 + 2s + 5) (s + 1) (s + 1) 2 (s 2 + 2s + 5)
s 2 + 4 = A(s + 1)(s 2 + 2s + 5) + B(s 2 + 2s + 5) + Cs(s + 1) 2 + D(s + 1) 2
or
s 2 + 4 = A(s 3 + 3s 2 + 7s + 5) + B(s 2 + 2s + 5) + C(s 3 + 2s 2 + s) + D(s 2 + 2s + 1)
Equating coefficients:
s3 : 0= A+C
→ C = −A
s2 : 1 = 3A + B + 2C + D = A + B + D
s: 0 = 7 A + 2B + C + 2D = 6A + 2B + 2D = 4A + 2
→ A = −1 / 2, C = 1 / 2
constant : 4 = 5A + 5B + D = 4A + 4B + 1
→ B = 5 / 4, D = 1 / 4
1 −2 5 2s + 1 1 − 2 5 2(s + 1) − 1
H 2 (s) = + + = + +
4 (s + 1) (s + 1) 2
(s 2 + 2s + 5) 4 (s + 1) (s + 1) 2
(s + 1) 2 + 2 2 )
Hence,
h 2 (t) =
1
4
( )
− 2e − t + 5te − t + 2e − t cos 2t − 0.5e − t sin 2t u ( t )
(s + 2)e − s A B 1 −s 1 1
(c ) H 3 (s) = = e−s + = e +
(s + 1)(s + 3) (s + 1) (s + 3) 2
(s + 1) (s + 3)
h 3 (t) =
2
e(
1 −( t −1)
)
+ e −3( t −1) u ( t − 1)
Chapter 15, Solution 31.
10s A B C
(a) F(s) = = + +
(s + 1)(s + 2)(s + 3) s + 1 s + 2 s + 3
- 10
A = F(s) (s + 1) s= -1 = = -5
2
- 20
B = F(s) (s + 2) s= -2 = = 20
-1
- 30
C = F(s) (s + 3) s= -3 = = -15
2
-5 20 15
F(s) = + −
s +1 s + 2 s + 3
f ( t ) = - 5 e -t + 20 e -2t − 15 e -3t
2s 2 + 4s + 1 A B C D
(b) F(s) = 3 = + + 2 +
(s + 1)(s + 2) s + 1 s + 2 (s + 2) (s + 2) 3
A = F(s) (s + 1) s= -1 = -1
D = F(s) (s + 2) 3 s = -2 = -1
2s 2 + 4s + 1 = A(s + 2)(s 2 + 4s + 4) + B(s + 1)(s 2 + 4s + 4)
+ C(s + 1)(s + 2) + D(s + 1)
Equating coefficients :
s3 : 0= A+B → B = -A = 1
s2 : 2 = 6A + 5B + C = A + C → C = 2 − A = 3
s1 : 4 = 12A + 8B + 3C + D = 4A + 3C + D
4 = 6+A+ D → D = -2 − A = -1
s0 : 1 = 8A + 4B + 2C + D = 4A + 2C + D = -4 + 6 − 1 = 1
-1 1 3 1
F(s) = + + 2 −
s + 1 s + 2 (s + 2) (s + 2) 3
t 2 -2t
f(t) = -e - t + e -2t + 3 t e -2t − e
2
t2
f ( t ) = - e -t + 1 + 3 t − e - 2t
2
s +1 A Bs + C
(c) F(s) = = + 2
(s + 2)(s + 2s + 5) s + 2 s + 2s + 5
2
-1
A = F(s) (s + 2) s= -2 =
5
s + 1 = A (s 2 + 2s + 5) + B (s 2 + 2s) + C (s + 2)
Equating coefficients :
1
s2 : 0= A+B
→ B = -A =
5
s1 : 1 = 2A + 2B + C = 0 + C → C = 1
s0 : 1 = 5A + 2C = -1 + 2 = 1
-1 5 1 5⋅ s +1 -1 5 1 5 (s + 1) 45
F(s) = + 2 = + 2 +
s + 2 (s + 1) + 2
2
s + 2 (s + 1) + 2
2
(s + 1) 2 + 2 2
f ( t ) = - 0.2 e -2t + 0.2 e -t cos( 2t ) + 0.4 e -t sin( 2t )