CHAPTER 97 INVERSE LAPLACE TRANSFORMS
CHAPTER 97 INVERSE LAPLACE TRANSFORMS EXERCISE 354 Page 1035
1. Determine the inverse Laplace transforms of : (a)
7 s
(b)
2 s −5
7 1 (a) ℒ −1 = 7ℒ −1 = 7(1) = 7 s s 2 1 (b) ℒ −1 = 2ℒ −1 = 2 e5t s − 5 s − 5
2. Determine the inverse Laplace transforms of : (a)
3 2s + 1
(b)
2s s +4 2
1 3 −1 1 3 − 1t 3 −1 (a) ℒ −1 = 3 ℒ = ℒ = e 2 2 2 2s + 1 2 s + 1 s + 1 2 2
2s s (b) ℒ −1 = 2ℒ −1 = 2 cos 2t s 2 + 22 s2 + 4
3. Determine the inverse Laplace transforms of : (a)
1 s + 25
(b)
4 s +9
5s 2 s + 18
(b)
6 s2
2
2
1 1 5 1 (a) ℒ −1 = ℒ −1 = sin 5t 5 5 s 2 + 25 s 2 + 52 4 4 4 −1 3 (b) ℒ −1 = ℒ = sin 3t 3 s2 + 9 3 s 2 + 32
4. Determine the inverse Laplace transforms of : (a)
2
s 5 5s s 5 (a) ℒ −1 = 5ℒ −1 = ℒ −1 2 2 = cos 3t 2 2 2 2 s + 18 s +3 2 2 ( s + 9 ) 1464
© 2014, John Bird
6 1 (b) ℒ −1 = 6 ℒ −1 = 6t s2 s2
5. Determine the inverse Laplace transforms of : (a)
5 s3
(b)
8 s4
5 5 2! 5 (a) ℒ −1 = ℒ −1 = t 2 3 2 2! s s3 8 8 3! 4 (b) ℒ −1 = ℒ −1 = t 3 4 s 3! s4 3
6. Determine the inverse Laplace transforms of : (a)
3s 1 2 s −8 2
(b)
7 s − 16 2
3 s s s = 3ℒ −1 = 6ℒ −1 (a) ℒ −1 = 6 cosh 4t 2 − 42 1 1 s 2 2 s −8 ( s − 16 ) 2 2
1 7 7 4 7 (b) ℒ −1 = 7ℒ −1 = ℒ −1 = sinh 4t 4 4 s 2 − 42 s 2 − 16 ( s 2 − 42 )
7. Determine the inverse Laplace transforms of : (a)
15 3s − 27 2
(b)
4 ( s − 1)3
1 5 5 15 1 3 (a) ℒ −1 = 15ℒ −1 = 5ℒ −1 = ℒ −1 2 2 = sinh 3t 2 2 2 3 3 s −3 s −3 3s − 27 3 s 2 − 27 3
4 2! 1 4 −1 2! −1 −1 (b) ℒ −1 = 4 ℒ = ℒ = 2 ℒ = 2 et t 2 3 2 +1 2 +1 2 +1 2! − − − − s 1 s 1 s 1 s 1 ( ) ( ) ( ) ( )
8. Determine the inverse Laplace transforms of : (a)
1465
1 ( s + 2) 4
(b)
3 ( s − 3)5 © 2014, John Bird
1 1 3! 1 −1 3! 1 − 2t 3 1 (a) ℒ −1 = ℒ −1 = ℒ −1 = ℒ = e t 4 3+1 3+1 3+1 3! 6 6 s + 2 s + 2 s + 2 s 2 + ( ) ( ) ( ) ( )
3 1 3 −1 4! 1 −1 4! 1 3t 4 −1 (b) ℒ −1 = 3 ℒ = ℒ = ℒ = e t 5 4 +1 4 +1 4 +1 4! ( s − 3) ( s − 3) 8 ( s − 3) ( s − 3) 8
9. Determine the inverse Laplace transforms of : (a)
s +1 s 2 + 2 s + 10
3 s 2 + 6 s + 13
(b)
s +1 s +1 (a) ℒ −1 = ℒ −1 = e − t cos 3t 2 2 + 2 s + 10 2 s ( s + 1) + 3 3 3 1 2 3 −1 −1 = 3 ℒ = ℒ (b) ℒ −1 = e −3t sin 2t 2 2 2 2 2 2 2 s + 6 s + 13 ( s + 3) + 2 ( s + 3) + 2
10. Determine the inverse Laplace transforms of : (a)
2( s − 3) s − 6 s + 13 2
(b)
7 s − 8s + 12 2
2 ( s − 3) s −3 = 2ℒ −1 (a) ℒ −1 = 2 e3t cos 2t 2 2 − 6 s + 13 2 s − + s 3 2 ( ) 7 7 7 1 2 −1 (b) ℒ −1 = 7ℒ −1 = ℒ = e 4 t sinh 2t 2 2 2 2 2 2 s 2 − 8s + 12 s 4 2 − − − − s 4 2 ) ) ( (
11. Determine the inverse Laplace transforms of : (a)
2s + 5 s 2 + 4s − 5
(b)
3s + 2 s 2 − 8s + 25
2 ( s + 2 ) 1 2s + 5 (a) ℒ −1 = ℒ −1 + 2 2 s 2 + 4s − 5 ( s + 2 ) − 32 ( s + 2 ) − 32 1 s+2 3 1 −1 = 2ℒ −1 + ℒ = 2 e −2t cosh 3t + e −2t sinh 3t 2 2 3 3 ( s + 2 ) − 32 ( s + 2 ) − 32
3 ( s − 4 ) + 14 3s + 2 3s + 2 (b) ℒ −1 = ℒ −1 = ℒ −1 2 2 2 2 s 2 − 8s + 25 ( s − 4 ) + 3 ( s − 4 ) + 3
1466
© 2014, John Bird
14 3 s−4 14 = 3ℒ −1 ℒ −1 = 3e 4t cos 3t + e 4t sin 3t + 2 2 2 2 3 3 ( s − 4 ) + 3 ( s − 4 ) + 3
1467
© 2014, John Bird
EXERCISE 355 Page 1036
1. Use partial fractions to find the inverse Laplace transforms of:
11 − 3s = s + 2s − 3 2
Hence,
2 5 − ( s − 1) ( s + 3)
from Problem 1, page 188 of the textbook
2 5 11 − 3s ℒ −1 = ℒ −1 − = 2 e − t − 5e − 3t 2 + 2s − 3 s s 1 s 3 + + ( ) ( )
2. Use partial fractions to find the inverse Laplace transforms of:
2 s 2 − 9 s − 35 4 3 1 = − + ( s + 1)( s − 2 )( s + 3) ( s + 1) ( s − 2 ) ( s + 3)
Hence,
2 s 2 − 9 s − 35 4 3 1 ℒ −1 − + = ℒ −1 = 4 e − t − 3e 2t + e − 3t s 1 s 2 s 3 s 1 s 2 s 3 + − + + − + ( )( )( ) ( ) ( ) ( )
5s 2 − 2 s − 19 2 3 4 = + − ( s + 3)( s − 1) 2 ( s + 3) ( s − 1) ( s − 1)2
Note:
2 s 2 − 9 s − 35 ( s + 1)( s − 2)( s + 3)
from Problem 2, page 188 of the textbook
3. Use partial fractions to find the inverse Laplace transforms of:
Hence,
11 − 3s s + 2s − 3 2
5s 2 − 2 s − 19 ( s + 3)( s − 1) 2
from Problem 6, page 190 of the textbook
5s 2 − 2 s − 19 3 4 2 −1 ℒ −1 = ℒ = 2 e − 3t + 3et − 4 et t + − 2 2 + − s s 3 1 ( ) ( ) − s 1 + − s 3 s 1 ( ) )( ) ( 4 1 −1 ℒ −1 = 4 ℒ = 4 et t 2 1+1 ( s − 1) ( s − 1)
4. Use partial fractions to find the inverse Laplace transforms of:
3s 2 + 16 s + 15
( s + 3)
3
3 2 6 − − 2 ( s + 3 ) ( s + 3 ) ( s + 3 )3
=
3s 2 + 16 s + 15 ( s + 3)3
from Problem 7, page 191 of the textbook 1468
© 2014, John Bird
Hence,
3 3s 2 + 16 s + 15 2 6 ℒ −1 − − = ℒ −1 2 3 3 ( s + 3) ( s + 3) ( s + 3) ( s + 3)
1 1 1 −1 = 3ℒ −1 – 6 ℒ – 2ℒ −1 3 2 ( s + 3) ( s + 3) ( s + 3) 1 6 −1 2! 1 −1 = 3ℒ −1 – 2 ℒ – ℒ 2 +1 1+1 ( s + 3) ( s + 3) 2! ( s + 3)
= 3e −3t − 2 e −3t t − 3e −3t t 2
or
e −3t (3 − 2t − 3t 2 )
5. Use partial fractions to find the inverse Laplace transforms of:
7 s 2 + 5s + 13 = ( s 2 + 2)( s + 1) and
2s + 3 5 + 2 ( s + 2 ) ( s + 1)
2s + 3 5 + = ( s 2 + 2 ) ( s + 1)
Hence,
from Problem 8, page 192 of the textbook
2s 3 5 + + ( s 2 + 2 ) ( s 2 + 2 ) ( s + 1)
2 s 7 s 2 + 5s + 13 3 5 = ℒ −1 ℒ −1 + + 2 2 2 ( s + 2 ) ( s + 2 ) s + 1 ( s + 2 )( s + 1) s = 2ℒ s2 + 2 −1
( )
= 2 cos 2 t +
3 2 − 1 + ℒ 2 2 s2 + 2
3 + 6s + 4s 2 − 2s3 2 1 3 − 4s = + + s 2 ( s 2 + 3) s s2 s2 + 3
( )
1 + 5ℒ −1 2 s + 1
3 sin 2 t + 5e − t 2
6. Use partial fractions to find the inverse Laplace transforms of:
Hence,
7 s 2 + 5s + 13 ( s 2 + 2)( s + 1)
3 + 6s + 4s 2 − 2s3 s 2 ( s 2 + 3)
from Problem 9, page 192 of the textbook
3 + 6 s + 4 s 2 − 2 s 3 2 1 3 − 4s ℒ −1 = ℒ −1 + 2 + 2 2 2 s +3 s ( s + 3) s s 3 − 4s 1 1 − 1 − 1 = 2ℒ + ℒ + ℒ 2 s s s2 + 3 −1
( )
1469
2 © 2014, John Bird
3 4s 1 1 − 1 − 1 − 1 = 2ℒ + ℒ + ℒ –ℒ 2 2 s s s2 + 3 s2 + 3 −1
( )
( )
2
3 1 1 3 ℒ −1 – = 2ℒ −1 + ℒ −1 + 2 2 3 s s 2 s + 3
( )
s 4ℒ −1 2 s2 + 3
( )
= 2 + t + 3 sin 3 t − 4 cos 3 t
7. Use partial fractions to find the inverse Laplace transforms of:
Let
26 − s 2 s ( s 2 + 4 s + 13)
A ( s 2 + 4 s + 13) + ( Bs + C ) s A Bs + C 26 − s 2 ≡ + = s ( s 2 + 4 s + 13) s s 2 + 4 s + 13 s ( s 2 + 4 s + 13) 2 26 − s = A ( s 2 + 4 s + 13) + Bs 2 + Cs
Hence, When s = 0:
26 = 13A + 0 + 0
Equating s 2 coefficients:
–1 = A + B
Equating s coefficients:
0 = 4A + C
Thus,
Hence,
i.e. A = 2 i.e. B = –3 i.e. C = –8
26 − s 2 2 −3s − 8 ≡ + s ( s 2 + 4 s + 13) s s 2 + 4 s + 13 26 − s 2 2 3s + 8 ℒ −1 = ℒ −1 – ℒ −1 s 2 + 4 s + 13 s s ( s 2 + 4 s + 13) 3s + 8 1 = 2ℒ −1 – ℒ −1 2 2 s ( s + 2 ) + 3 3 ( s + 2 ) 2 1 −1 = 2ℒ −1 – ℒ −1 – ℒ 2 2 2 2 s ( s + 2 ) + 3 ( s + 2 ) + 3 ( s + 2 ) 2 3 1 −1 = 2ℒ −1 – 3ℒ −1 – ℒ 2 2 s ( s + 2 ) + 32 3 ( s + 2 ) + 32
1470
© 2014, John Bird
2 = 2 − 3e −2t cos 3t − e −2t sin 3t 3
1471
© 2014, John Bird
EXERCISE 356 Page 1038
1. Determine for the transfer function: R(s) =
50 ( s + 4) s ( s + 2)( s 2 − 8s + 25)
(a) the zero and (b) the poles. Show the poles and zeros on a pole–zero diagram.
(a) For the numerator to be zero, (s + 4) = 0 hence, s = –4 is a zero of R(s) (b) For the denominator to be zero, s = 0 or s = –2 or s 2 − 8s + 25 = 0
i.e.
s=
− −8±
( −8)
2
− 4(1)(25)
2(1)
=
8 ± −36 −8 ± j 6 = = 4 ± j3 2 2
Hence, poles occur at s = 0, s = –2, s = 4 + j3 and 4 – j3 A pole–zero diagram is shown below
2. Determine the poles and zeros for the function: F(s) =
( s − 1)( s + 2) and plot them on a ( s + 3)( s 2 − 2 s + 5)
pole–zero map.
For the numerator to be zero, (s – 1) = 0 hence, s = +1 is a zero of F(s) and
(s + 2) = 0 hence, s = –2 is a zero of F(s)
For the denominator to be zero, s = –3 or s 2 − 2 s + 5 = 0
1472
© 2014, John Bird
i.e.
s=
−−2±
( −2 )
2
− 4(1)(5)
2(1)
=
2 ± −16 2 ± j 4 = = 1± j2 2 2
Hence, poles occur at s = –3, s = 1 + j2 and 1 – j2 A pole–zero diagram is shown below
3. For the function G(s) =
s −1 , determine the poles and zeros and show them on a ( s + 2)( s 2 + 2 s + 5)
pole–zero diagram. For the denominator to be zero, s = –2 or s 2 + 2 s + 5 = 0
i.e.
s=
−2 ±
( 2)
2
− 4(1)(5)
2(1)
−2 ± −16 −2 ± j 4 = = =−1 ± j 2 2 2
Hence, poles occur at s = –2, s = –1 + j2 and –1 – j2 For the numerator to be zero, (s – 1) = 0 hence, s = 1 is a zero of G(s) A pole–zero diagram is shown below
1473
© 2014, John Bird
4. Find the poles and zeros for the transfer function: H(s) =
s 2 − 5s − 6 and plot the results in the ss ( s 2 + 4)
plane. For the denominator to be zero, s = 0 or s 2 + 4 = 0 i.e.
s2 = − 4
and s =
and
−4 = ± j2
Hence, poles occur at s = 0, s = + j2 and –j2 For the numerator to be zero, s 2 − 5s − 6 = 0 i.e.
( s − 6)( s + 1) = 0 hence, s = 6 is a zero of H(s)
and
s = –1 is a zero of H(s)
A pole–zero diagram is shown below
1474
© 2014, John Bird
1. Determine the inverse Laplace transforms of : (a)
7 s
(b)
2 s −5
7 1 (a) ℒ −1 = 7ℒ −1 = 7(1) = 7 s s 2 1 (b) ℒ −1 = 2ℒ −1 = 2 e5t s − 5 s − 5
2. Determine the inverse Laplace transforms of : (a)
3 2s + 1
(b)
2s s +4 2
1 3 −1 1 3 − 1t 3 −1 (a) ℒ −1 = 3 ℒ = ℒ = e 2 2 2 2s + 1 2 s + 1 s + 1 2 2
2s s (b) ℒ −1 = 2ℒ −1 = 2 cos 2t s 2 + 22 s2 + 4
3. Determine the inverse Laplace transforms of : (a)
1 s + 25
(b)
4 s +9
5s 2 s + 18
(b)
6 s2
2
2
1 1 5 1 (a) ℒ −1 = ℒ −1 = sin 5t 5 5 s 2 + 25 s 2 + 52 4 4 4 −1 3 (b) ℒ −1 = ℒ = sin 3t 3 s2 + 9 3 s 2 + 32
4. Determine the inverse Laplace transforms of : (a)
2
s 5 5s s 5 (a) ℒ −1 = 5ℒ −1 = ℒ −1 2 2 = cos 3t 2 2 2 2 s + 18 s +3 2 2 ( s + 9 ) 1464
© 2014, John Bird
6 1 (b) ℒ −1 = 6 ℒ −1 = 6t s2 s2
5. Determine the inverse Laplace transforms of : (a)
5 s3
(b)
8 s4
5 5 2! 5 (a) ℒ −1 = ℒ −1 = t 2 3 2 2! s s3 8 8 3! 4 (b) ℒ −1 = ℒ −1 = t 3 4 s 3! s4 3
6. Determine the inverse Laplace transforms of : (a)
3s 1 2 s −8 2
(b)
7 s − 16 2
3 s s s = 3ℒ −1 = 6ℒ −1 (a) ℒ −1 = 6 cosh 4t 2 − 42 1 1 s 2 2 s −8 ( s − 16 ) 2 2
1 7 7 4 7 (b) ℒ −1 = 7ℒ −1 = ℒ −1 = sinh 4t 4 4 s 2 − 42 s 2 − 16 ( s 2 − 42 )
7. Determine the inverse Laplace transforms of : (a)
15 3s − 27 2
(b)
4 ( s − 1)3
1 5 5 15 1 3 (a) ℒ −1 = 15ℒ −1 = 5ℒ −1 = ℒ −1 2 2 = sinh 3t 2 2 2 3 3 s −3 s −3 3s − 27 3 s 2 − 27 3
4 2! 1 4 −1 2! −1 −1 (b) ℒ −1 = 4 ℒ = ℒ = 2 ℒ = 2 et t 2 3 2 +1 2 +1 2 +1 2! − − − − s 1 s 1 s 1 s 1 ( ) ( ) ( ) ( )
8. Determine the inverse Laplace transforms of : (a)
1465
1 ( s + 2) 4
(b)
3 ( s − 3)5 © 2014, John Bird
1 1 3! 1 −1 3! 1 − 2t 3 1 (a) ℒ −1 = ℒ −1 = ℒ −1 = ℒ = e t 4 3+1 3+1 3+1 3! 6 6 s + 2 s + 2 s + 2 s 2 + ( ) ( ) ( ) ( )
3 1 3 −1 4! 1 −1 4! 1 3t 4 −1 (b) ℒ −1 = 3 ℒ = ℒ = ℒ = e t 5 4 +1 4 +1 4 +1 4! ( s − 3) ( s − 3) 8 ( s − 3) ( s − 3) 8
9. Determine the inverse Laplace transforms of : (a)
s +1 s 2 + 2 s + 10
3 s 2 + 6 s + 13
(b)
s +1 s +1 (a) ℒ −1 = ℒ −1 = e − t cos 3t 2 2 + 2 s + 10 2 s ( s + 1) + 3 3 3 1 2 3 −1 −1 = 3 ℒ = ℒ (b) ℒ −1 = e −3t sin 2t 2 2 2 2 2 2 2 s + 6 s + 13 ( s + 3) + 2 ( s + 3) + 2
10. Determine the inverse Laplace transforms of : (a)
2( s − 3) s − 6 s + 13 2
(b)
7 s − 8s + 12 2
2 ( s − 3) s −3 = 2ℒ −1 (a) ℒ −1 = 2 e3t cos 2t 2 2 − 6 s + 13 2 s − + s 3 2 ( ) 7 7 7 1 2 −1 (b) ℒ −1 = 7ℒ −1 = ℒ = e 4 t sinh 2t 2 2 2 2 2 2 s 2 − 8s + 12 s 4 2 − − − − s 4 2 ) ) ( (
11. Determine the inverse Laplace transforms of : (a)
2s + 5 s 2 + 4s − 5
(b)
3s + 2 s 2 − 8s + 25
2 ( s + 2 ) 1 2s + 5 (a) ℒ −1 = ℒ −1 + 2 2 s 2 + 4s − 5 ( s + 2 ) − 32 ( s + 2 ) − 32 1 s+2 3 1 −1 = 2ℒ −1 + ℒ = 2 e −2t cosh 3t + e −2t sinh 3t 2 2 3 3 ( s + 2 ) − 32 ( s + 2 ) − 32
3 ( s − 4 ) + 14 3s + 2 3s + 2 (b) ℒ −1 = ℒ −1 = ℒ −1 2 2 2 2 s 2 − 8s + 25 ( s − 4 ) + 3 ( s − 4 ) + 3
1466
© 2014, John Bird
14 3 s−4 14 = 3ℒ −1 ℒ −1 = 3e 4t cos 3t + e 4t sin 3t + 2 2 2 2 3 3 ( s − 4 ) + 3 ( s − 4 ) + 3
1467
© 2014, John Bird
EXERCISE 355 Page 1036
1. Use partial fractions to find the inverse Laplace transforms of:
11 − 3s = s + 2s − 3 2
Hence,
2 5 − ( s − 1) ( s + 3)
from Problem 1, page 188 of the textbook
2 5 11 − 3s ℒ −1 = ℒ −1 − = 2 e − t − 5e − 3t 2 + 2s − 3 s s 1 s 3 + + ( ) ( )
2. Use partial fractions to find the inverse Laplace transforms of:
2 s 2 − 9 s − 35 4 3 1 = − + ( s + 1)( s − 2 )( s + 3) ( s + 1) ( s − 2 ) ( s + 3)
Hence,
2 s 2 − 9 s − 35 4 3 1 ℒ −1 − + = ℒ −1 = 4 e − t − 3e 2t + e − 3t s 1 s 2 s 3 s 1 s 2 s 3 + − + + − + ( )( )( ) ( ) ( ) ( )
5s 2 − 2 s − 19 2 3 4 = + − ( s + 3)( s − 1) 2 ( s + 3) ( s − 1) ( s − 1)2
Note:
2 s 2 − 9 s − 35 ( s + 1)( s − 2)( s + 3)
from Problem 2, page 188 of the textbook
3. Use partial fractions to find the inverse Laplace transforms of:
Hence,
11 − 3s s + 2s − 3 2
5s 2 − 2 s − 19 ( s + 3)( s − 1) 2
from Problem 6, page 190 of the textbook
5s 2 − 2 s − 19 3 4 2 −1 ℒ −1 = ℒ = 2 e − 3t + 3et − 4 et t + − 2 2 + − s s 3 1 ( ) ( ) − s 1 + − s 3 s 1 ( ) )( ) ( 4 1 −1 ℒ −1 = 4 ℒ = 4 et t 2 1+1 ( s − 1) ( s − 1)
4. Use partial fractions to find the inverse Laplace transforms of:
3s 2 + 16 s + 15
( s + 3)
3
3 2 6 − − 2 ( s + 3 ) ( s + 3 ) ( s + 3 )3
=
3s 2 + 16 s + 15 ( s + 3)3
from Problem 7, page 191 of the textbook 1468
© 2014, John Bird
Hence,
3 3s 2 + 16 s + 15 2 6 ℒ −1 − − = ℒ −1 2 3 3 ( s + 3) ( s + 3) ( s + 3) ( s + 3)
1 1 1 −1 = 3ℒ −1 – 6 ℒ – 2ℒ −1 3 2 ( s + 3) ( s + 3) ( s + 3) 1 6 −1 2! 1 −1 = 3ℒ −1 – 2 ℒ – ℒ 2 +1 1+1 ( s + 3) ( s + 3) 2! ( s + 3)
= 3e −3t − 2 e −3t t − 3e −3t t 2
or
e −3t (3 − 2t − 3t 2 )
5. Use partial fractions to find the inverse Laplace transforms of:
7 s 2 + 5s + 13 = ( s 2 + 2)( s + 1) and
2s + 3 5 + 2 ( s + 2 ) ( s + 1)
2s + 3 5 + = ( s 2 + 2 ) ( s + 1)
Hence,
from Problem 8, page 192 of the textbook
2s 3 5 + + ( s 2 + 2 ) ( s 2 + 2 ) ( s + 1)
2 s 7 s 2 + 5s + 13 3 5 = ℒ −1 ℒ −1 + + 2 2 2 ( s + 2 ) ( s + 2 ) s + 1 ( s + 2 )( s + 1) s = 2ℒ s2 + 2 −1
( )
= 2 cos 2 t +
3 2 − 1 + ℒ 2 2 s2 + 2
3 + 6s + 4s 2 − 2s3 2 1 3 − 4s = + + s 2 ( s 2 + 3) s s2 s2 + 3
( )
1 + 5ℒ −1 2 s + 1
3 sin 2 t + 5e − t 2
6. Use partial fractions to find the inverse Laplace transforms of:
Hence,
7 s 2 + 5s + 13 ( s 2 + 2)( s + 1)
3 + 6s + 4s 2 − 2s3 s 2 ( s 2 + 3)
from Problem 9, page 192 of the textbook
3 + 6 s + 4 s 2 − 2 s 3 2 1 3 − 4s ℒ −1 = ℒ −1 + 2 + 2 2 2 s +3 s ( s + 3) s s 3 − 4s 1 1 − 1 − 1 = 2ℒ + ℒ + ℒ 2 s s s2 + 3 −1
( )
1469
2 © 2014, John Bird
3 4s 1 1 − 1 − 1 − 1 = 2ℒ + ℒ + ℒ –ℒ 2 2 s s s2 + 3 s2 + 3 −1
( )
( )
2
3 1 1 3 ℒ −1 – = 2ℒ −1 + ℒ −1 + 2 2 3 s s 2 s + 3
( )
s 4ℒ −1 2 s2 + 3
( )
= 2 + t + 3 sin 3 t − 4 cos 3 t
7. Use partial fractions to find the inverse Laplace transforms of:
Let
26 − s 2 s ( s 2 + 4 s + 13)
A ( s 2 + 4 s + 13) + ( Bs + C ) s A Bs + C 26 − s 2 ≡ + = s ( s 2 + 4 s + 13) s s 2 + 4 s + 13 s ( s 2 + 4 s + 13) 2 26 − s = A ( s 2 + 4 s + 13) + Bs 2 + Cs
Hence, When s = 0:
26 = 13A + 0 + 0
Equating s 2 coefficients:
–1 = A + B
Equating s coefficients:
0 = 4A + C
Thus,
Hence,
i.e. A = 2 i.e. B = –3 i.e. C = –8
26 − s 2 2 −3s − 8 ≡ + s ( s 2 + 4 s + 13) s s 2 + 4 s + 13 26 − s 2 2 3s + 8 ℒ −1 = ℒ −1 – ℒ −1 s 2 + 4 s + 13 s s ( s 2 + 4 s + 13) 3s + 8 1 = 2ℒ −1 – ℒ −1 2 2 s ( s + 2 ) + 3 3 ( s + 2 ) 2 1 −1 = 2ℒ −1 – ℒ −1 – ℒ 2 2 2 2 s ( s + 2 ) + 3 ( s + 2 ) + 3 ( s + 2 ) 2 3 1 −1 = 2ℒ −1 – 3ℒ −1 – ℒ 2 2 s ( s + 2 ) + 32 3 ( s + 2 ) + 32
1470
© 2014, John Bird
2 = 2 − 3e −2t cos 3t − e −2t sin 3t 3
1471
© 2014, John Bird
EXERCISE 356 Page 1038
1. Determine for the transfer function: R(s) =
50 ( s + 4) s ( s + 2)( s 2 − 8s + 25)
(a) the zero and (b) the poles. Show the poles and zeros on a pole–zero diagram.
(a) For the numerator to be zero, (s + 4) = 0 hence, s = –4 is a zero of R(s) (b) For the denominator to be zero, s = 0 or s = –2 or s 2 − 8s + 25 = 0
i.e.
s=
− −8±
( −8)
2
− 4(1)(25)
2(1)
=
8 ± −36 −8 ± j 6 = = 4 ± j3 2 2
Hence, poles occur at s = 0, s = –2, s = 4 + j3 and 4 – j3 A pole–zero diagram is shown below
2. Determine the poles and zeros for the function: F(s) =
( s − 1)( s + 2) and plot them on a ( s + 3)( s 2 − 2 s + 5)
pole–zero map.
For the numerator to be zero, (s – 1) = 0 hence, s = +1 is a zero of F(s) and
(s + 2) = 0 hence, s = –2 is a zero of F(s)
For the denominator to be zero, s = –3 or s 2 − 2 s + 5 = 0
1472
© 2014, John Bird
i.e.
s=
−−2±
( −2 )
2
− 4(1)(5)
2(1)
=
2 ± −16 2 ± j 4 = = 1± j2 2 2
Hence, poles occur at s = –3, s = 1 + j2 and 1 – j2 A pole–zero diagram is shown below
3. For the function G(s) =
s −1 , determine the poles and zeros and show them on a ( s + 2)( s 2 + 2 s + 5)
pole–zero diagram. For the denominator to be zero, s = –2 or s 2 + 2 s + 5 = 0
i.e.
s=
−2 ±
( 2)
2
− 4(1)(5)
2(1)
−2 ± −16 −2 ± j 4 = = =−1 ± j 2 2 2
Hence, poles occur at s = –2, s = –1 + j2 and –1 – j2 For the numerator to be zero, (s – 1) = 0 hence, s = 1 is a zero of G(s) A pole–zero diagram is shown below
1473
© 2014, John Bird
4. Find the poles and zeros for the transfer function: H(s) =
s 2 − 5s − 6 and plot the results in the ss ( s 2 + 4)
plane. For the denominator to be zero, s = 0 or s 2 + 4 = 0 i.e.
s2 = − 4
and s =
and
−4 = ± j2
Hence, poles occur at s = 0, s = + j2 and –j2 For the numerator to be zero, s 2 − 5s − 6 = 0 i.e.
( s − 6)( s + 1) = 0 hence, s = 6 is a zero of H(s)
and
s = –1 is a zero of H(s)
A pole–zero diagram is shown below
1474
© 2014, John Bird
