Solucionario parte 2 Matemáticas Avanzadas para Ingeniería - 2da Edición - Glyn James
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2 Laplace Transforms Exercises 2.2.6 1(a)
1 s 1 1 1 + = 2 , Re(s) > 2 L{cosh 2t} = L{ (e2t + e−2t ) = 2 2 s−2 s+2 s −4
1(b)
L{t2 } =
1(c)
L{3 + t} =
1 3 3s + 1 + 2 = , Re(s) > 0 s s s2
1(d)
L{te−t } =
1 , Re(s) > −1 (s + 1)2
2(a) 5 (g) 0
2 , Re(s) > 0 s3
(b) -3 (h) 0
(c) 0 (i) 2
(d) 3 (j) 3
(e) 2
(f ) 0
3 5 5s − 3 − 2 = , Re(s) > 0 s s s2
3(a)
L{5 − 3t} =
3(b)
L{7t3 − 2 sin 3t} = 7.
3(c)
L{3 − 2t + 4 cos 2t} =
3(d)
L{cosh 3t} =
3(e)
L{sinh 2t} =
42 6 3 6 = 4 − 2 , Re(s) > 0 − 2. 2 4 s s +9 s s +9 2 3s − 2 3 s 4s − 2 + 4. 2 = , Re(s) > 0 + 2 2 s s s +4 s s +4
s , Re(s) > 3 s2 − 9
s2
2 , Re(s) > 2 −4 c Pearson Education Limited 2004
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3(f )
L{5e−2t + 3 − 2 cos 2t} =
3(g)
L{4te−2t } =
3(h)
L{2e−3t sin 2t} =
3(i)
L{t2 e−4t } =
3 s 5 + − 2. 2 , Re(s) > 0 s+2 s s +4
4 , Re(s) > −2 (s + 2)2 4 4 = , Re(s) > −3 (s + 3)2 + 4 s2 + 6s + 13
2 , Re(s) > −4 (s + 4)3
3(j) 36 6 4 2 − 3+ 2− 4 s s s s 36 − 6s + 4s2 − 2s3 = , Re(s) > 0 s4
L{6t3 − 3t2 + 4t − 2} =
3(k)
3(l)
L{2 cos 3t + 5 sin 3t} = 2.
s2
3 2s + 15 s + 5. 2 = 2 , Re(s) > 0 s +9 s +9 +9
s +4 s2 − 4 d s = L{t cos 2t} = − ds s2 + 4 (s2 + 4)2 L{cos 2t} =
s2
3(m) 6s d 3 = 2 2 ds s + 9 (s + 9)2 (s2 + 9)2 6 − 6s(s2 + 9)2 4s 6s d = − L{t2 sin 3t} = − ds (s2 + 9)2 (s2 + 9)4 18s2 − 54 , Re(s) > 0 = 2 (s + 9)3 L{t sin 3t} = −
3(n)
L{t2 − 3 cos 4t} =
2 3s , Re(s) > 0 − 2 3 s s + 16
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 3(o) 3 2 (s + 1) + + 3 2 (s + 2) (s + 1) + 4 s s+1 3 2 + 2 + , Re(s) > 0 = 3 (s + 2) s + 2s + 5 s
L{t2 e−2t − e−t cos 2t + 3} =
Exercises 2.2.10
4(a)
L−1
4(b)
L−1
4(c)
L−1
4(d)
L−1
1 14 1 1 = L−1 − 4 = [e−3t − e−7t ] (s + 3)(s + 7) 4 s+3 s+7
−1 2 s+5 = L−1 + = −e−t + 2e3t (s + 1)(s − 3) s+1 s−3
1 4 s−1 4 4 1 4 3 9 −1 9 = − t − e−3t = L − − 2 2 s (s + 3) s s s+3 9 3 9
2s + 6 2 s = L−1 2. 2 + 3. 2 = 2 cos 2t + 3 sin 2t 2 2 s +4 s +2 s + 22
4(e) −1
L
4(f )
L−1
1 1 1 −1 0 16 16 + = L − s2 (s2 + 16) s s2 s2 + 16 1 1 1 t− sin 4t = [4t − sin 4t] = 16 64 64
s+8 −1 (s + 2) + 6 = L = e−2t [cos t + 6 sin t] s2 + 4s + 5 (s + 2)2 + 1
4(g) L−1
1 − 18 s + 12 s+1 −1 8 = L + s2 (s2 + 4s + 8) s (s + 2)2 + 22 1 1 1 (s + 2) − 3(2) = L−1 . − 8 s 8 (s + 2)2 + 22 1 = [1 − e−2t cos 2t + 3e−2t sin 2t] 8 c Pearson Education Limited 2004
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4(h) L−1
4(i)
L−1
1 1 2 4s −1 − + = L (s − 1)(s + 1)2 s − 1 (s + 1) (s + 1)2 = et − e−t + 2tet
s+7 −1 (s + 1) + 3(2) = L = e−t [cos 2t + 3 sin 2t] s2 + 2s + 5 (s + 1)2 + 22
4(j) −1
L
4(k) L−1
1 12 3 s2 − 7s + 5 −1 =L − + 2 (s − 1)(s − 2)(s − 3) s−1 s−2 s−3 1 11 = et − 3e2t + e3t 2 2
−2 5s − 7 2s − 1 −1 = L + (s + 3)(s2 + 2) s + 3 s2 + 2 √ √ 1 = −2e−3t + 2 cos 2t − √ sin 2t 2
4(l) −1
L
4(m)
L−1
15 1 s−2 s −1 = L − (s − 1)(s2 + 2s + 2) s − 1 5 s2 + 2s + 2 15 1 (s + 1) − 3 = L−1 − s − 1 5 (s + 1)2 + 1 1 1 = et − e−t (cos t − 3 sin t) 5 5
(s + 1) − 2 s−1 −1 = L = e−t (cos 2t − sin 2t) s2 + 2s + 5 (s + 1)2 + 22
4(n) L−1
3 12 2 s−1 = L−1 − + 2 (s − 2)(s − 3)(s − 4) s−2 s−3 s−4 1 3 = e2t − 2e3t + e−4t 2 2
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4(o)
3s 3s −1 = L (s − 1)(s2 − 4) (s − 1)(s − 2)(s + 2) 3 1 −1 = L−1 + 2 − 2 s−1 s−2 s+2 3 1 = −et + e2t − e−2t 2 2
9 1 36 −1 4 2s 2s = L − + 2 2 2 2 s(s + 1)(s + 9) s s +1 s +9 1 9 = 4 − cos t + cos 3t 2 2
L−1
4(p) L−1
4(q) L−1
9 7s + 9 2s2 + 4s + 9 = L−1 − 3 2 2 (s + 2)(s + 3s + 3) s + 2 (s + 2 ) + 3/4 √ √ 9 7(s + 32 ) − 3. 3/2 −1 √ − =L s+2 (s + 32 )2 + ( 3/2)2 √ √ √ 3 3 3 e2t − 3 sin t = 9e−2t − e− 2 t 7 cos 2 2
4(r) L−1
1 1 1 19 1 −1 10 90 s + 9 = L − − 2 2 (s + 1)(s + 2)(s + 2s + 10) s + 1 s + 2 s + 2s + 10 1 1 1 s + 10 −1 9 10 =L − − s + 1 s + 2 90 (s + 1)2 + 32 1 1 1 (s + 1) + 3(3) −1 9 10 =L − − s + 1 s + 2 90 (s + 1)2 + 32 1 1 1 = e−t − e−2t − e−t (cos 3t + 3 sin 3t) 10 90 9
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Exercises 2.3.5 5(a) 1 2s + 5 = s+2 s+2 1 1 2s + 5 = + X(s) = (s + 2)(s + 3) s+2 s+3
(s + 3)X(s) = 2 +
x(t) = L−1 {X(s)} = e−2t + e−3t 5(b) s2 + 6 2 = s2 + 4 s2 + 4 35 3 4 s2 + 6 26 26 s + 26 = − X(s) = (3s − 4)(s2 + 4) 3s − 4 s2 + 4 2 35 4 t 3 e 3 − (cos 2t + sin 2t) x(t) = L−1 {X(s)} = 78 26 3 (3s − 4)X(s) = 1 +
5(c) (s2 + 2s + 5)X(s) =
1 s
1 1 s+2 1 5 − = · X(s) = s s(s2 + 2s + 5) 5 s2 + 2s + 5 1 1 (s + 1) + 12 (2) = 5 − s 5 (s + 1)2 + 22 1 1 x(t) = L−1 {X(s)} = (1 − e−t cos 2t − e−t sin 2t) 5 2
5(d) 4s 2s2 + 4s + 8 = s2 + 4 s2 + 4 2s2 + 4s + 8 X(s) = (s + 1)2 (s2 + 4) 12 6 1 12s − 32 25 5 − = + (s + 1) (s + 1)2 25 s2 + 4 16 12 −t 6 −t 12 e + te − cos 2t + sin 2t x(t) = L−1 {X(s)} = 25 5 25 25 (s2 + 2s + 1)X(s) = 2 +
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 5(e) (s2 − 3s + 2)X(s) = 1 +
X(s) =
x(t) = L−1 {X(s)} =
2 s+6 = s+4 s+4
1 7 4 s+6 = 15 − 5 + 3 (s + 4)(s − 1)(s − 2) s+4 s−1 s−2
1 −4t 7 t 4 2t e − e + e 15 5 3
5(f ) (s2 + 4s + 5)X(s) = (4s − 7) + 16 +
X(s) =
3 s+2
3 (s + 2) + 1 4s2 + 17s + 21 = + 2 (s + 2)(s + 4s + 5) s + 2 (s + 2)2 + 1
x(t) = L−1 {X(s)} = 3e−2t + e−2t cos t + e−2t sin t
5(g) (s2 + s − 2)X(s) = s + 1 +
5(2) (s + 1)2 + 4
s3 + 3s2 + 7s + 15 X(s) = (s + 2)(s − 1)(s2 + 2s + 5) 13 1 5 − 13 12 4s − 4 + + = s + 2 s − 1 s2 + 2s + 5 13 − 13 1 (s + 1) − 3(2) 12 = + + s + 2 s − 1 4 (s + 1)2 + 22
1 13 1 3 x(t) = L−1 {X(s)} = − e−2t + et + e−t cos 2t − e−t sin 2t 3 12 4 4 c Pearson Education Limited 2004
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5(h) (s2 + 2s + 3)Y (s) = 1 +
3 s2
2 4 − 23 1 s2 + 3 3s + 3 = + 2+ 2 Y (s) = 2 2 s (s + 2s + 3) s s s + 2s + 3 √ 1 − 23 1 2 (s + 1) − √2 ( 2) √ + 2+ = s s 3 (s + 1)2 + ( 2)2
√ √ 2 2 1 y(t) = L−1 {Y (s)} = − + t + e−t (cos 2t + √ sin 2t) 3 3 2
5(i) (s2 + 4s + 4)X(s) = X(s) =
= x(t) = L−1 {X(s)} =
2 1 1 s+2+ 3 + 2 s s+2 s5 + 6s4 + 10s3 + 4s + 8 2s3 (s + 2)3 3 8
s
−
1 2 s2
+
1 2 s3
+
1 8
s+2
+
3 4
(s +
2)2
+
1 (s + 2)3
1 3 1 1 3 1 − t + t2 + e−2t + te−2t + t2 e−2t 8 2 4 8 4 2
5(j) (9s2 + 12s + 5)X(s) = X(s) =
= x(t) = L−1 {X(s)} =
1 s 1 1 4 1 5 5 s + 15 − = s 9s(s2 + 43 s + 59 ) (s + 23 )2 + 1 5
s
−
1 [(s + 23 ) + 23 ] 5 (s + 23 )2 + ( 13 )2
1 1 1 −2t 1 − e 3 (cos t + 2 sin t) 5 5 3 3
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5(k) 4 −s3 − 6s2 − 16s + 32 1 = (s2 + 8s + 16)X(s) = − s + 1 − 4 + 16· 2 2 s + 16 2(s2 + 16) −s3 − 6s2 − 16s + 32 X(s) = 2(s + 4)2 (s2 + 16) 1 1 0 2s + − = 2 2 s + 4 (s + 4) s + 16 1 x(t) = L−1 {X(s)} = te−4t − cos 4t 2 5(l) (9s2 + 12s + 4)Y (s) = 9(s + 1) + 12 +
1 s+1
9s2 + 30s + 22 Y (s) = (3s + 2)2 (s + 1) 0 18 1 + + = s + 1 3s + 2 (3s + 2)2 2
y(t) = L−1 {Y (s)} = e−t + 2te− 3 t 5(m) 2 1 + 2 s s 3 s − 2s2 + 2s + 1 X(s) = 2 s (s − 1)(s − 2)(s + 1)
(s3 − 2s2 − s + 2)X(s) = s − 2 +
5 4
1 2 s2
5 2 1 12 + − 3 − = + s s−1 s−2 s+1 5 1 5 2 x(t) = L−1 {X(s)} = + t − et + e2t − e−t 4 2 12 3
5(n) s +9 9 3 2 1 s+9 s + 2s + 10s + 18 1 7s − 25 20 = − X(s) = 2 − s + 1 16 s2 + 1 (s + 9)(s + 1)(s2 + 1) 80 s2 + 9 25 1 3 9 −t 7 e − cos t + sin t − cos 3t − sin 3t x(t) = L−1 {X(s)} = 20 16 16 80 80 (s3 + s2 + s + 1) = (s + 1) + 1 +
s2
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6(a) 1 1 2sX(s) − (2s + 9)Y (s) = − + 2 s+2 (2s + 4)X(s) + (4s − 37)Y (s) = 1 Eliminating X(s) 1 1 )(2s + 4) − 2s = −3s [−(2s + 9)(2s + 4) − 2s(4s − 37)]Y (s) = (− + 2 s+2 3s 1 s Y (s) = = · 12s2 − 48s + 36 4 (s − 3)(s − 1) 3 1 1 2 = − 2 4 s−3 s−1 1 3 3t 1 t 3 3t 1 t y(t) = L−1 {Y (s)} = e − e = e − e 4 2 2 8 8 Eliminating
dx from the two equations dt
dy + 4x − 28y = −e−2t dt 1 dy 1 −2t 21 3t 7 t 27 3t 3 t = −e x(t) = −e−2t + 28y − 6 + e − e − e + e 4 dt 4 4 2 3 4 1 1 15 3t 11 t e − e − e−2t , y(t) = (3e3t − et ) i.e. x(t) = 4 4 4 8 6
6(b) 5 +1 1 (2s + 1)X(s) + (3s − 1)Y (s) = s−1 (s + 1)X(s) + (2s − 1)Y (s) =
s2
Eliminating X(s) s+1 5 (2s + 1) − +1 s−1 s+1 10s + 5 − Y (s) = 2 s(s + 1)(s − 2) s(s − 1)(s − 2) 5 3 − 52 5 12 2 − + 2 = + 2 − 2 − s s−2 s +1 s s−1 s−2 5 5 1 3 y(t) = L−1 {Y (s)} = − + e2t − 5 sin t − + 2et − e2t 2 2 2 2 2t t = −3 + e + 2e − 5 sin t [(2s − 1)(2s + 1) − (3s − 1)(s + 1)]Y (s) =
s2
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition Eliminating
dx dt
from the original equations
dy + x − y = 10 sin t − et dt x(t) = 10 sin t − et − 3 + e2t + 2et − 5 sin t − 2e2t − 2et + 5 cos t = 5 sin t + 5 cos t − 3 − et − e2t
6(c) (s + 2)X(s) + (s + 1)Y (s) = 3 +
3s + 10 1 = s+3 s+3
5X(s) + (s + 3)Y (s) = 4 +
4s + 13 5 = s+2 s+2
Eliminating X(s) [5(s + 1) − (s + 2)(s + 3)]Y (s) =
−4s2 − 10s + 11 15s + 50 − (4s + 13) = s+3 s+3
9 7 − 12 4s2 + 10s − 11 2s − 2 Y (s) = = + 2 s+3 s +1 (s + 3)(s2 + 1)
7 1 9 y(t) = L−1 {Y (s)} = − e−3t + cos t − sin t 2 2 2 From the second differential equation 21 3 3 27 5x = 5e−2t + e−3t − cos t + sin t − e−3t 2 2 2 2 +
7 9 sin t + cos t 2 2
x(t) = 3 sin t − 2 cos t + e−2t
6(d) (3s − 2)X(s) + 3sY (s) = 6 + sX(s) + (2s − 1)Y (s) = 3 +
6s − 5 1 = s−1 s−1 3s + 1 1 = s s
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Eliminating X(s) [3s2 − (3s − 2)(2s − 1)]Y (s) = Y (s) =
s(6s − 5) (3s − 2)(3s + 1) − s−1 s 6s2 − 5s 9s2 − 3s − 2 − s(3s − 1)(s − 2) (s − 1)(3s − 1)(s − 2)
18 14 1 5 + 5 = − + s 3s − 1 s − 2 9 14 − 12 10 − − + 5 s − 1 3s − 1 s − 2 1 1 =− + 2 s s−1 1 y(t) = L−1 {Y (s)} = −1 + et + 2
Eliminating
+
9 2
3s − 1
3 t e3 2
dx from the original equations dt 1 3 9 t 3 3 t 3 − et − 3 + et + e 3 − et − e 3 2 2 2 2 2 3 t 1 = e 3 − et 2 2
x(t) =
6(e) (3s − 2)X(s) + sY (s) = −1 +
5s −s2 + 5s + 2 3 + = s2 + 1 s2 + 1 s2 + 1
2sX(s) + (s + 1)Y (s) = −1 +
s −s2 + s 1 + = s2 + 1 s2 + 1 s2 + 1
Eliminating Y (s) [(3s − 2)(s + 1) − 2s2 ]X(s) = X(s) = =
s2
1 [(−s2 + 5s + 2)(s + 1) − (−s2 + s)s] +1
3s + 1 3s2 + 7s + 2 = 2 (s + 2)(s − 1)(s + 1) (s − 1)(s2 + 1) 2s − 1 2 − 2 s−1 s +1
x(t) = L−1 {X(s)} = 2et − 2 cos t + sin t c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition
Eliminating
dy from the original equation dt dx dt = −2 sin t − 4 cos t − 4et + 4 cos t − 2 sin t + 2et + 2 sin t + cos t
y(t) = −2 sin t − 4 cos t − 2x +
i.e. y(t) = −2et − 2 sin t + cos t, x(t) = 2et − 2 cos t + sin t
6(f ) 1 s2 + 1 = s2 s2 s+1 1 (s + 1)X(s) + 4sY (s) = 1 + = s s sX(s) + (s + 1)Y (s) = 1 +
Eliminating Y (s) s2 + 1 (s + 1)2 3s2 − 2s + 3 = [4s2 − (s + 1)2 ]X(s) = 4s − s2 s s 2 −3 1 9 3s − 2s + 3 = − + X(s) = s(s − 1)(3s + 1) s s − 1 3s + 1 t
x(t) = L−1 {X(s)} = −3 + et + 3e− 3 Eliminating
dy from the original equation dt 1 dx 4t − 1 + x + 3 4 dt t t 1 = 4t − 1 − 3 + et + 3e− 3 − 3et + 3e− 3 4 t 1 3 t i.e. y(t) = t − 1 − et + e− 3 , x(t) = −3 + et + 3e− 3 2 2 y=
6(g) 14 + 7s 12 7 = + s2 s s2 14 − 14s 14 14 = (5s + 4)X(s) − (3s − 6)Y (s) = 2 − s s s2 (2s + 7)X(s) + 3sY (s) =
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Eliminating Y (s) 1 [(3s − 6)(14 + 7s) + 3s(14 − 14s)] s2 21 21(s2 + s − 2)X(s) = 21(s + 2)(s − 1)X(s) = 2 (−s2 + 2s − 4) s 2 −s + 2s − 4 X(s) = 2 s (s + 2)(s − 1) 1 0 2 −1 + + + 2 = s−1 s+2 s s −1 x(t) = L {X(s)} = −et + e−2t + 2t
[(2s + 7)(3s − 6) + (5s + 4)(3s)]X(s) =
Eliminating
dy from the original equations dt
dx dt t = 28t − 7 + 7e + 14e−2t − 14 + 11et − 11e−2t − 22t 1 7 giving y(t) = t − + 3et + e−2t , x(t) = −et + e−2t + 2t. 2 2 6y = 28t − 7 − 11x − 7
6(h) (s2 + 2)X(s) − Y (s) = 4s −X(s) + (s2 + 2)Y (s) = 2s Eliminating Y (s) [(s2 + 2)2 − 1]X(s) = 4s(s2 + 2) + 2s (s4 + 4s2 + 3)X(s) = 4s3 + 10s 3s s 4s3 + 10s = 2 + 2 2 2 (s + 1)(s + 3) s +1 s +3 √ −1 x(t) = L {X(s)} = 3 cos t + cos 3t X(s) =
From the first of the given equations √ √ d2 x = 6 cos t + 2 cos 3t − 3 cos t − 3 cos 3t 2 dt √ √ i.e. y(t) = 3 cos t − cos 3t, x(t) = 3 cos t − cos 3t y(t) = 2x +
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6(i)
35 + 12 = (5s2 + 6)X(s) + 12s2 Y = s 4 35 2 2 + 16 = 5s X(s) + (16s + 6)Y (s) = s 4 Eliminating X(s)
83 s 4 99 s 4
s [83(5s2 ) − 99(5s2 + 6)] 4 s 4 2 [−20s − 126s − 36]Y (s) = [−80s2 − 594] 4
[60s4 − (5s2 + 6)(16s2 + 6)]Y (s) =
s(40s2 + 297) 4(s2 + 6)(10s2 + 3) 25 − 14 s 2 s + = 2 s + 6 10s2 + 3 √ 3 1 5 t y(t) = L−1 {Y (s)} = − cos 6t + cos 4 4 10
Y (s) =
Eliminating
d2 x from the original equations dt2
√ 3 d2 y 15 3 3 cos 3x = 3y + 3 2 = − t + − + 3 cos 6t dt 4 4 10 4 √ √ 3 1 3 3 5 t + cos 6t, x(t) = cos t − cos 6t. i.e. x(t) = cos 10 4 4 10 4 6(j)
(2s2 − s + 9)X(s) − (s2 + s + 3)Y (s) = 2(s + 1) − 1 = 2s + 1 (2s2 + s + 7)X(s) − (s2 − s + 5)Y (s) = 2(s + 1) + 1 = 2s + 3
Subtract (−2s + 2)X(s) − (2s − 2)Y (s) = −2 ⇒ X(s) + Y (s) =
1 s−1
⇒ x(t) + y(t) = et
(i)
Add (4s2 + 16)X(s) − (2s + 8)Y (s) = 4(s + 1) 2(s + 1) ⇒ 2x(t) − y(t) 2X(s) − Y (s) = 2 s +4 = 2 cos 2t + sin 2t c Pearson Education Limited 2004
(ii)
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Then from (i) and (ii) x(t) =
1 2 1 2 1 t 2 e + cos 2t + sin 2t, y(t) = et − cos 2t − sin 2t 3 3 3 3 3 3
Exercises 2.4.3 7
1µF = 10−6 F so 50µ = 5.105 F
Applying Kirchhoff’s second law to the left hand loop 1 5.105
di1 di2 − = E. sin 100t i1 dt + 2 dt dt
Taking Laplace transforms 2.104 100 I1 (s) + 2s[I1 (s) − I2 (s)] = E. 2 s s + 104 50s (104 + s2 )I1 (s) − s2 I2 (s) = E. 2 s + 104
(i)
Applying Kirchhoff’s law to the right hand loop 100i2 (t) − 2
di1 di2 − =0 dt dt
which on taking Laplace transforms gives sI1 (s) = (50 + s)I2 (s) Substituting in (i) 50s2 s2 + 104 Es2 2 4 (s + 200s + 10 )I2 (s) = 2 s + 104 s2 I2 (s) = E 2 (s + 104 )(s + 100)2 s(50 + s) then from (ii) I1 (s) = E 2 (s + 104 )(s + 100)2
(104 + s2 )(50 + s)I2 (s) − s2 I2 (s) = E.
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(ii)
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Expanding in partial functions 1 1 1 − 200 2 200 s + + I2 (s) = E s + 100 (s + 100)2 s2 + 104
1 −100t 1 −100t 1 e cos 100t + te + i2 (t) = L−1 {I2 (s)} = E − 200 2 200
8
Applying Kirchhoff’s second law to the primary and secondary circuits
respectively gives 2i1 +
di1 di2 +1 = 10 sin t dt dt
2i2 + 2
di1 di2 + =0 dt dt
Taking Laplace transforms (s + 2)I1 (s) + sI2 (s) =
10 +1
s2
sI1 (s) + 2(s + 1)I2 (s) = 0 Eliminating I1 (s) [s2 − 2(s + 1)(s + 2)]I2 (s) =
10s s2 + 1
I2 (s) = −
(s2
10s 10s =− 2 2 + 1)(s + 7s + 6) (s + 1)(s + 6)(s + 1)
12 25 −1 s + 35 + 37 + 37 2 37 =− s+1 s+6 s +1
i2 (t) = L−1 {I2 (s)} = e−t −
9
35 12 −6t 25 e cos t − sin t − 37 37 37
Applying Kirchhoff’s law to the left and right hand loops gives
d (i1 + i2 ) + (i1 + i2 ) + 1 i1 dt = E0 = 10 dt
di2 − 1 i1 dt = 0 i2 + dt c Pearson Education Limited 2004
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Applying Laplace transforms 1 10 (s + 1)I1 (s) + (s + 1)I2 (s) + I1 (s) = s s 1 (s + 1)I2 (s) − I1 (s) = 0 ⇒ I1 (s) = s(s + 1)I2 (s) s Substituting back in the first equation s(s + 1)2 I2 (s) + (s + 1)I2 (s) + (s + 1)I2 (s) = (s2 + s + 2)I2 (s) = I2 (s) =
10 s 10 s(s + 1) 10 s(s + 1)(s2 + s + 2)
Then from (i) 10 10 = 1 +s+2 (s + 2 )2 + √ 7 20 1 t i1 (t) = L−1 {I1 (s)} = √ e− 2 t sin 2 7 I1 (s) =
10
s2
Applying Newton’s law to the motion of each mass x ¨1 = 3(x2 − x1 ) − x1 = 3x2 − 4x1 x ¨2 = −9x2 − 3(x2 − x1 ) = −12x2 + 3x1
giving x ¨1 + 4x1 − 3x2 = 0, x1 (0) = −1, x2 (0) = 2 x ¨2 + 12x2 − 3x1 = 0 Taking Laplace transforms (s2 + 4)X1 (s) − 3X2 (s) = −s −3X1 (s) + (s2 + 12)X2 (s) = 2s c Pearson Education Limited 2004
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Eliminating X2 (s) [(s2 + 4)(s2 + 12) − 9]X1 (s) = −s(s2 + 12) + 6s (s2 + 13)(s2 + 3)X1 (s) = −s3 − 6s 3 7 − 10 s −s3 − 6s 10 s = − (s2 + 13)(s2 + 3) s2 + 3 s2 + 13 √ √ 3 7 cos 13t x1 (t) = L−1 {X1 (s)} = − cos 3t − 10 10
X1 (s) =
From the first differential equation 3x2 = 4x1 + x ¨1 √ √ √ √ 6 14 9 91 cos 13t + cos 3t + cos 13t = − cos 3t − 5 5 10 10 √ √ 1 [21 cos 13t − cos 3t] x2 (t) = 10 √ √ √ √ 1 1 (3 cos 3t + 7 cos 13t), x2 (t) = [21 cos 13t − cos 3t] 10 10 √ √ Natural frequencies are 13 and 3. Thus x1 (t) = −
11
The equation of motion is Mx ¨ + bx˙ + Kx = M g ; x(0) = 0 , x(0) ˙ =
2gh
The problem is then an investigative one where students are required to investigate for different h values either analytically or by simulation.
12 By Newton’s second law of motion ¨2 = −K2 x2 − B1 (x˙ 2 − x˙ 1 ) + u2 M2 x M1 x ¨1 = B1 (x˙ 2 − x˙ 1 ) − K1 x1 + u1 Taking Laplace transforms and assuming quiescent initial state (M2 s2 + B1 s + K2 )X2 (s) − B1 sX1 (s) = U2 (s) −B1 sX2 (s) + (M1 s2 + B1 s + K1 )X1 (s) = U1 (s) c Pearson Education Limited 2004
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Eliminating X1 (s) [(M1 s2 + B1 s + K1 )(M2 s2 + B1 s + K2 ) − B12 s2 ]X2 (s) = (M1 s2 + B1 s + K1 )U2 (s) + B1 sU1 (s) B1 s (M1 s2 + B1 s + K1 ) U1 (s) + U2 (s) ∆ ∆ B1 s (M1 s2 + B1 s + K1 ) U1 (s) + U2 (s) and x2 (t) = L−1 {X2 (s)} = L−1 ∆ ∆ i.e. X2 (s) =
Likewise eliminating X2 (t) from the original equation gives x1 (t) = L−1 {X1 (s)} = L−1
(M1 s + B1 s + K2 ) B1 s U1 (s) + U2 (s) ∆ ∆
Exercises 2.5.7 13 f (t) = tH(t) − tH(t − 1) = tH(t) − (t − 1)H(t − 1) − 1H(t − 1) Thus, using theorem 2.4 L{f (t)} =
1 1 1 −s −s 1 −s −s − e − e = (1 − e ) − e s2 s2 s2 s
14(a) f (t) = 3t2 H(t) − (3t2 − 2t + 3)H(t − 4) − (2t − 8)H(t − 6) = 3t2 H(t) − [3(t − 4)2 + 22(t − 4) + 43]H(t − 4) − [2(t − 6) + 4]H(t − 6) Thus 6 − e−4s L[3t2 + 22t + 43] − e−6s L[2t + 4] s3 6 22 43 −4s 2 4 6 e − 2 + e−6s = 3− 3+ 2 + s s s 5 s s
L{f (t)} =
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14(b) f (t) = tH(t) + (2 − 2t)H(t − 1) − (2 − t)H(t − 2) = tH(t) − 2(t − 1)H(t − 1) − (t − 2)H(t − 2) Thus
1 − 2e−s L{t} + e−2s L{t} 2 s 1 = 2 [1 − 2e−s + e−2s ] s
L{f (t)} =
e−5s 1 and by the first = L−1 {e−5s F (s)} where F (s) = 4 (s − 2) (s − 2)4 1 shift theorem f (t) = L−1 {F (s)} = t3 e2t . 6 Thus by the second shift theorem 15(a)
L−1
L−1
15(b)
L−1
L−1
3e−2s = L−1 {e−2s F (s)} where (s + 3)(s + 1) 3 − 23 3 F (s) = = + 2 (s + 3)(s + 1) s+3 s+1 3 3 f (t) = L−1 {F (s)} = e−t − e−3t 2 2 −2s 3e = f (t − 2)H(t − 2) L−1 (s + 3)(s + 1) 3 = [e−(t−2) − e−3(t−2) ]H(t − 2) 2
so
15(c)
e−5s = f (t − 5)H(t − 5) (s − 2)4 1 = (t − 5)3 e2(t−5) H(t − 5) 6
s+1 −s e = L−1 {e−s F (s)} where s2 (s2 + 1) F (s) =
1 1 s+1 s+1 = + 2− 2 + 1) s s s +1
s2 (s2
f (t) = L−1 {F (s)} = 1 + t − cos t − sin t c Pearson Education Limited 2004
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so L−1
s+1 e−s = f (t − 1)H(t − 1) + 1)
s2 (s2
= [1 + (t − 1) − cos(t − 1) − sin(t − 1)]H(t − 1) = [t − cos(t − 1) − sin(t − 1)]H(t − 1)
15(d)
L−1
s+1 −πs = L−1 {e−πs F (s)} where e s2 + s + 1 √ √1 ( 3 ) 3 2 √ 1 2 3 2 ) + ( 2 2 )
(s + 12 ) +
s+1 = + s + 1) (s + √ √ 1 1 3 3 t + √ sin t f (t) = e− 2 t cos 2 2 3
F (s) =
(s2
so L−1
√ √ √ 1 3 3 s+1 1 − (t−π) −πs 2 √ e (t − π) + sin (t − π) .H(t − π) 3 cos = e s2 + s + 1 2 2 3
15(e)
L−1
s −4πs/5 e = L−1 {e−4πs/5 F (s)} where s2 + 25 F (s) =
so L−1
15(f )
L−1
s2
s ⇒ f (t) = L−1 {F (s)} = cos 5t s2 + 25 4π s 4π e−4πs/5 = f t − H t− + 25 5 5 4π = cos(5t − 4π)H t − 5 4π = cos 5t H t − 5
e−s (1 − e−s ) = L−1 {(e−s − e−2s )F (s)} where s2 (s2 + 1) F (s) = −1
f (t) = L
1 s2 (s2
+ 1)
=
1 1 − 2 2 s s +1
{F (s)} = t − sin t
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so L−1 {(e−s − e−2s )F (s)} =f (t − 1)H(t − 1) − f (t − 2)H(t − 2) =[(t − 1) − sin(t − 1)]H(t − 1) − [(t − 2) − sin(t − 2)]H(t − 2)
16
1 dx + x = f (t), L{f (t)} = 2 (1 − e−s − se−s ) dt s
Taking Laplace transforms with x(0) = 0 1 (1 + s) − e−s 2 s s2 1 1 − e−s 2 X(s) = 2 s (s + 1) s 1 1 1 − e−s L{t} =− + 2 + s s s+1
(s + 1)X(s) =
Taking inverse transforms x(t) = −1 + e−t + t − (t − 1)H(t − 1) = e−t + (t − 1)[1 − H(t − 1)] or x(t) = e−t + (t − 1) for t ≤ 1 x(t) = e−t for t ≥ 1 Sketch of response is
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17
d2 x dx + x = g(t), x(0) = 1, x(0) ˙ =0 + dt2 dt
with L{g(t)} =
1 (1 − 2e−s + e−2s ) s2
Taking Laplace transforms (s2 + s + 1)X(s) = s + 1 +
1 (1 − 2e−s + e−2s ) s2
1 s+1 + 2 2 (1 − 2e−s + e−2s ) + s + 1) s (s + s + 1) 1 1 s (s + 1) + − + 2+ 2 [1 − 2e−s + e−2s ] = 2 (s + s + 1) s s s +s+1 √ √
1 √1 ( 3 ) (s + 12 ) + √13 ( 23 ) 1 (s + ) − 1 2 3 2 √ √ = [1 − 2e−s + e−2s ] + − + 2+ s s 3 3 )2 )2 (s + 12 )2 + ( (s + 12 )2 + ( 2 2 √ √ 1 1 3 3 −2t −1 t + √ sin t cos x(t) = L {X(s)} = e 2 2 3 √ √ 1 1 3 3 + t − 1 + e− 2 t cos t − √ sin t 2 2 3 √ 1 3 − 2 (t−1) − 2H(t − 1) t − 2 + e (t − 1) cos 2 √
1 3 − √ sin (t − 1) 2 3 √ 1 3 − 2 (t−2) (t − 2) cos + H(t − 2) t − 3 + e 2 √
1 3 − √ sin (t − 2) 2 3
X(s) =
(s2
i.e. 1
x(t) = 2e− 2 t cos
√ 3 2 t
+t−1 √ √
1 1 3 3 − 2 (t−1) (t − 1) − √ sin (t − 1) − 2H(t − 1) t − 2 + e cos 2 2 3 √ √
1 1 3 3 − 2 (t−2) (t − 2) − √ sin (t − 2) cos + H(t − 2) t − 3 + e 2 2 3 c Pearson Education Limited 2004
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101
π π π = cos t − H t− f (t) = sin tH t − 2 2 2 π since cos t − = sin t. 2
Taking Laplace transforms with x(0) = 1, x(0) ˙ = −1 π π (s2 + 3s + 2)X(s) = s + 2 + L cos t − H t− 2 2 π
= s + 2 + e− 2 s L{cos t} π s = s + 2 + e− 2 s · 2 s +1 π s 1 + e− 2 s X(s) = s+1 (s + 1)(s + 2)(s2 + 1) 2 π −1 1 1 s+3 2 = + e− 2 s + 5 + · 2 s+1 s + 1 s + 2 10 s + 1 π 1 1 2 1 = + e− 2 s L − e−t + e−2t + (cos t + 3 sin t) s+1 2 5 10 π π π 1 2 1 so x(t) = L−1 {X(s)} = e−t + − e−(t− 2 ) + e−2(t− 2 ) + (cos t − 2 5 10 2
π π + 3 sin t − ) H t− 2 2 π π 1 sin t − 3 cos t + 4eπ e−2t − 5e 2 e−t H t − = e−t + 10 2
19
f (t) = 3H(t) − (8 − 2t)H(t − 4) = 3H(t) + 2(t − 4)H(t − 4) 2 3 3 L{f (t)} = + 2e−4s L{t} = + 2 e−4s s s s
Taking Laplace transforms with x(0) = 1, x(0) ˙ =0 2 3 + 2 e−4s s s 3 2 s + + 2 2 e−4s X(s) = 2 2 s + 1 s(s + 1) s (s + 1) 1 3 3 s 1 −4s + − 2 +2 2 − 2 e = 2 s +1 5 s +1 s s +1 2 3 + 2e−4s L{t − sin t} = − 2 5 s +1
(s2 + 1)X(s) = s +
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Thus taking inverse transforms x(t) = 3 − 2 cos t + 2(t − 4 − sin(t − 4))H(t − 4)
20 θ¨0 + 6θ˙0 + 10θ0 = θi θi (t) = 3H(t) − 3H(t − a) 3 3 3 so L{θi } = − e−as = (1 − e−as ) s s s Taking Laplace transforms in (1) with θ0 = θ˙0 = 0 at t = 0 3 (1 − e−as ) s Φ0 (s) = 3(1 − e−as )
(s2 + 6s + 10)Φ0 (s) =
1 s(s2 + 6s + 10) 1 (s + 3) + 3 3 (1 − e−as ) − = 10 s (s + 3)2 + 1 3 = (1 − e−as )L 1 − e−3t cos t − 3e−3t sin t 10
Thus taking inverse transforms θ0 (t) =
3 [1 − e−3t cos t − 3e−3t sin t]H(t) 10 3 − [1 − e−3(t−a) cos(t − a) − 3e−3(t−a) sin(t − a)]H(t − a) 10
If T > a then H(T ) = 1, H(T − a) = 1 giving 3 −3T [e cos T − e−3(T −a) cos(T − a)] 10 3 − [3e−3T sin T − 3e−3(T −a) sin(T − a)] 10 3 −3T {cos T + 3 sin T − e3a [cos(T − a) + 3 sin(T − a)]} =− e 10
θ0 (T ) = −
21
θi (t) = f (t) = (1 − t)H(t) − (1 − t)H(t − 1) = (1 − t)H(t) + (t − 1)H(t − 1) c Pearson Education Limited 2004
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so
1 1 − 2 + e−s L{t} s s 1 1 s−1 1 1 + 2 e−s = − 2 + 2 e−s = 2 s s s s s Then taking Laplace transforms, using θ0 (0) = θ˙0 (0) = 0 L{θi (t)} =
(s2 + 8s + 16)Φ0 (s) =
s−1 1 −s + e s2 s2
s−1 1 −s + e s2 (s + 4)2 s2 (s + 4)2 3 1 1 3 2 10 e−s 3 2 2 + + = 2 − 2− − + + s s s s + 4 (s + 4)2 32 s s2 s + 4 (s + 4)2
Φ0 (s) =
which on taking inverse transforms gives 1 [3 − 2t − 3e−4t − 10te−4t ] 32 1 + [−1 + 2(t − 1) + e−4(t−1) + 2(t − 1)e−4(t−1) ]H(t − 1) 32 1 [3 − 2t − 3e−4t − 10te−4t ] = 32 1 + [2t − 3 + (2t − 1)e−4(t−1) ]H(t − 1) 32
θ0 (t) = L−1 {Φ0 (s)} =
22
e(t) = e0 H(t − t1 ) − e0 H(t − t2 ) e0 L{e(t)} = (e−st1 − e−st2 ) s c Pearson Education Limited 2004
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By Kirchhoff’s second law current in the circuit is given by 1 Ri + C
idt = e
which on taking Laplace transforms RI(s) +
I(s) =
e0 1 I(s) = (e−st1 − e−st2 ) Cs s
e0 C (e−st1 − e−st2 ) RCs + 1
=
e0 /R −st1 − e−st2 ) 1 (e s + RC
=
e0 /R −st2 e0 /R −st1 − 1 e 1 e s + RC s + RC
then i(t) = L−1 {I(s)} e0 −(t−t1 )/RC e = H(t − t1 ) − e−(t−t2 )/RC H(t − t2 ) R
23
Sketch over one period as shown and readily extended to 0 ≤ t < 12 .
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f1 (t) = 3tH(t) − (3t − 6)H(t − 2) − 6H(t − 4) = 3tH(t) − 3(t − 2)H(t − 2) − 6H(t − 4) 3 3 6 L{f1 (t)} = F1 (s) = 2 − 2 e−2s − e−4s s s s Then by theorem 2.5 1 F1 (s) 1 − e−4s 1 (3 − 3e−2s − 6se−4s ) = 2 s (1 − e−4s )
L{f (t)} = F (s) =
24
Take K t, T = 0,
f1 (t) =
then f1 (t) =
0 b y(x) = −
W 1 1 x2 x3 (x − b)3 − x3 + y2 (0) + y3 (0) EI 6 6 2 6
y2 (x) = −
W (x − b) − x + y2 (0) + y3 (0)x EI
y3 (x) = −
W 1 − 1 + y3 (0) ⇒ y3 (0) = 0 EI
Using the boundary condition y2 () = 0 0=−
Wb W (−h) + y2 (0) ⇒ y2 (0) = − EI EI
giving (x − b)3 bx2 W x3 − H(x − b) − EI 6 6 2 W x2 − (3b − x), 0 < x ≤ b 6EI = 2 − W b (3x − b), b < x ≤ 6EI
y(x) =
Exercises 2.6.5 34(a) gives
Assuming all the initial conditions are zero taking Laplace transforms (s2 + 2s + 5)X(s) = (3s + 2)U (s)
so that the system transfer function is given by
G(s) =
3s + 2 X(s) = 2 s + 2s + 5 U (s)
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34(b)
The characteristic equation of the system is s2 + 2s + 5 = 0
and the system is of order 2.
34(c)
The transfer function poles are the roots of the characteristic equation s2 + 2s + 5 = 0
which are s = −1 ± j . That is, the transfer function has single poles at s = −1 + j and s = −1 − j . The transfer function zeros are determined by equating the numerator polynomial 2 to zero; that is, a single zero at s = − . 3
35
Following the same procedure as for Exercise 34
35(a)
The transfer function characterising the system is
G(s) =
s3 + 5s + 6 s3 + 5s2 + 17s + 13
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35(b)
The characteristic equation of the system is s3 + 5s2 + 17s + 13 = 0
and the system is of order 3.
35(c)
The transfer function poles are given by s3 + 5s2 + 17s + 13 = 0 i.e. (s + 1)(s2 + 4s + 13) = 0
That is, the transfer function has simple poles at s = −1, s = −2 + j3, s = −2 − j3 The transfer function zeros are given by s2 + 5s + 6 = 0 (s + 3)(s + 2) = 0 i.e. zeros at s = −3 and s = −2 .
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36(a)
Poles at (s + 2)(s2 + 4) = 0 ; i.e. s = −2, s = +2j, s = −j.
Since we have poles on the imaginary axis in the s-plane, system is marginally stable.
Poles at (s + 1)(s − 1)(s + 4) = 0 ; i.e. s = −1, s = 1, s = −4 .
36(b)
Since we have the pole s = 1 in the right hand half of the s-plane, the system is unstable.
36(c)
Poles at (s + 2)(s + 4) = 0 ; i.e. s = −2, s = −4 .
Both the poles are in the left hand half of the plane so the system is stable. √ 3 1 . 36(d) Poles at (s + s + 1)(s + 1) = 0 ; i.e. s = −1 (repeated), s = − ± j 2 2 Since all the poles are in the left hand half of the s-plane the system is stable. 2
2
√ 39 1 . 36(e) Poles at (s + 5)(s2 − s + 10) = 0 ; i.e. s = −5, s = ± j 2 2 Since both the complex poles are in the right hand half of the s-plane the system is unstable.
37(a)
s2 − 4s + 13 = 0 ⇒ s = 2 ± j3 .
Thus the poles are in the right hand half s-plane and the system is unstable.
37(b) 5s3 + 13s2 + 31s + 15 = 0 a2 a3 a1 a0 Routh–Hurwitz (R-H) determinants are: 13 5 > 0, ∆3 = 15∆2 > 0 ∆1 = 13 > 0, ∆2 = 15 31 so the system is stable.
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s3 + s2 + s + 1 = 0
R-H determinants are 1 ∆1 = 1 > 0, ∆2 = 1
1 = 0, ∆3 = 1∆2 = 0 1
Thus system is marginally stable. This is readily confirmed since the poles are at s = −1, s = ±j
37(d)
24s4 + 11s3 + 26s2 + 45s + 36 = 0
R-H determinants are 11 24 0, ∆2 = 45 26 so the system is unstable.
37(e)
s3 + 2s2 + 2s + 1 = 0
R-H determinants are 2 ∆1 = 2 > 0, ∆2 = 1
3 = 1 > 0, ∆3 = 1∆2 > 0 2
√ 1 3 confirming the and the system is stable. The poles are at s = −1, s = − ± j 2 2 result. d3 x d2 x dx + c +K + Krx = 0; m, K, r, c > 0 3 2 dt dt dt R-H determinants are 38 m
∆1 = c > 0 c m = cK − mKr > 0 provided r < c ∆2 = Kr K m ∆3 = Kr∆2 > 0 provided ∆2 > 0 Thus system stable provided r <
c m
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39
s4 + 2s2 + (K + 2)s2 + 7s + K = 0 a3 a2 a1 a0
R-H determinants are ∆1 =| a3 |= 9 > 0 a3 a4 = 2 ∆2 = 7 a1 a2 a3 a4 0 ∆3 = a1 a2 a3 = 0 a0 a1
3 1 = 2K − 3 > 0 provided K > K + 2 2 2 1 0 7 K + 2 2 = 10K − 21 > 0 provided K > 2 0 K 7
∆4 = K∆3 > 0 provided ∆3 > 0 Thus the system is stable provided K > 2.1 . 40 s2 + 15Ks2 + (2K − 1)s + 5K = 0, K > 0 R-H determinants are ∆1 = 15K > 0 15K 1 = 30K 2 − 20K ∆2 = 5K (2K − 1) ∆3 = 5K∆2 > 0 provided ∆2 > 0 Thus system stable provided K(3K − 2) > 0 that is K >
41(a)
2 , since K > 0 . 3
Impulse response h(t) is given by the solution of dh d2 h + 56h = 3δ(t) + 15 2 dt dt
with zero initial conditions. Taking Laplace transforms (s2 + 15s + 56)H(s) = 3 H(s) =
3 3 3 = − (s + 7)(s + 8) s+7 s+8
so h(t) = L−1 {H(s)} = 3e−7t − 3e−8t Since h(t) → 0 as t → ∞ the system is stable. c Pearson Education Limited 2004
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121
Following (a) impulse response is given by (s2 + 8s + 25)H(s) = 1 1 (s + 4)2 + 32 1 so h(t) = L−1 {H(s)} = e−4t sin 3t 3 H(s) =
Since h(t) → 0 as t → ∞ the system is stable. 41(c)
Following (a) impulse response is given by (s2 − 2s − 8)H(s) = 4 4 2 1 2 1 = − (s − 4)(s + 2) 3s−4 3s+2 2 so h(t) = L−1 {H(s)} = (e4t − e−2t ) 3 H(s) =
Since h(t) → ∞ as t → ∞ system is unstable. 41(d)
Following (a) impulse response is given by (s2 − 4s + 13)H(s) = 1 H(s) = so h(t) = L−1 {H(s)} =
s2
1 1 = − 4s + 13 (s − 2)2 + 32
1 2t e sin 3t 3
Since h(t) → ∞ as t → ∞ system is unstable. 7 dx 2 = e−t − 3e−2t + e−4t 42 Impulse response h(t) = dt 3 3 System transfer function G(s) = L{h(t)} ; that is 3 2 7 − + 3(s + 1) s + 2 3(s + 4) s+8 = (s + 1)(s + 2)(s + 4)
G(s) =
Note The original unit step response can be reconstructed by evaluating 1 L−1 G(s) . s c Pearson Education Limited 2004
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f (t) = 2 − 3 cos t , F (s) =
43(a)
3s 2 − 2 s s +1
sF (s) = 2 −
3 3s2 =2− 2 s +1 1 + s12
Thus lim (2 − 3 cos t) = 2 − 3 = −1 t→0+
and lim sF (s) = 2 − s→∞
3 = −1 so confirming the i.v. theorem. 1
43(b) f (t) = (3t − 1)2 = 9t2 − 6t + 1, lim f (t) = 1 t→0+
18 6 6 1 18 − F (s) = 3 − 2 + so lim sF (s) = lim + 1 =1 s→∞ s→∞ s2 s s s s thus confirming the i.v. theorem.
43(c) f (t) = t + 3 sin 2t , lim = 0 t→0+
F (s) =
1 6 1 6 =0 − 2 so lim sF (s) = lim + 2 s→∞ s→∞ s s s +4 s + 4s
thus confirming the i.v. theorem.
44(a) f (t) = 1 + 3e−t sin 2t , lim f (t) = 1 t→∞
6 6s 1 and lim sF (s) = lim 1 + =1 F (s) = + 2 2 s→0 s→0 s (s + 1) + 4 (s + 1) + 4 thus confirming the f.v. theorem. Note that sF (s) has its poles in the left half of the s-plane so the theorem is applicable.
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 44(b)
123
f (t) = t2 3e−2t , lim f (t) = 0 t→∞
2s 2 and lim sF (s) = lim =0 F (s) = s→0 s→0 (s + 2)3 (s + 2)3 thus confirming the f.v. theorem. Again note that sF (s) has its poles in the left half of the s-plane. 44(c)
f (t) = 3 − 2e−3t + e−t cos 2t , lim f (t) = 3 t→∞
3 2 (s + 1) F (s) = − + s s + 3 (s + 1)2 + 4 s(s + 1) 2s + =3 lim sF (s) = lim 3 − s→0 s→0 s + 3 (s + 1)2 + 4 confirming the f.v. theorem. Again sF (s) has its poles in the left half of the s-plane. 45
For the circuit of Example 2.28 I2 (s) =
1.22 4.86 3.64 + − s s + 59.1 s + 14.9
Then by the f.v. theorem 1.22s 4.86s − lim i2 (t) = lim sI2 (s) = lim 3.64 + t→∞ s→0 s→0 s + 59.1 s + 14.9 = 3.64 which confirms the answer obtained in Example 2.28. Note that sI2 (s) has all its poles in the left half of the s-plane. 46
For the circuit of Example 2.29 sI2 (s) =
28s2 (3s + 10)(s + 1)(s2 + 4)
and since it has poles at s = ±j2 not in the left hand half of the s-plane the f.v. theorem is not applicable. c Pearson Education Limited 2004
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47
Assuming quiescent initial state taking Laplace transforms gives 1 4 + +2 s s+3 4 1 2 Y (s) = + + s(7s + 5) (s + 3)(7s + 5) 7s + 5 2s s 4 + + sY (s) = 7s + 5 (s + 3)(7s + 5) 7s + 5
(7s + 5)Y (s) =
By the f.v. theorem lim y(t) = lim sF (s) = lim
t→∞
s→0
s→0
=
s 2s 4 + + 7s + 5 (s + 3)(7s + 5) 7s + 5
4 5
By the i.v. theorem lim+ y(t) = y(0+) = lim sF (s) = lim s→∞
t→0
s→∞
=
s 4 2 + + 7s + 5 (1 + 3s )(7s + 5) 7 + 5s
2 7
2 Thus jump at t = 0 = y(0+) − y(0−) = 1 . 7
Exercises 2.6.8 48(a)
f ∗ g(t) =
0
t
τ cos(3t − 3τ )dτ
t 1 1 = − τ sin(3t − 3τ ) + cos(3t − 3τ ) 3 9 0 1 = (1 − cos 3t) 9
t (t − τ ) cos 3τ dτ g ∗ f (t) = 0
=
t t τ 1 1 sin 3τ − sin 3τ − cos 3τ = (1 − cos 3t) 3 3 9 9 0 c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 48(b)
f ∗ g(t) =
t
(τ + 1)e−2(t−τ ) dτ
0
1 t 1 = (τ + 1)e−2(t−τ ) − e−2(t−τ ) 2 4 0 1 1 1 = t + − e−2t 2 4 4
t g ∗ f (t) = (t − τ + 1)e−2τ dτ 0
t 1 1 = − (t − τ + 1)e−2τ + e−2τ 2 4 0 1 1 −2t 1 = t+ − e 2 4 4 48(c)
Integration by parts gives
t
0
2
τ sin 2(t − τ )dτ =
t
0
(t − τ )2 sin 2τ dτ
1 1 1 = cos 2t + t2 − 4 2 4 48(d)
Integration by parts gives
t
−τ
e 0
sin(t − τ )dτ =
t
e−(t−τ ) sin τ dτ
0
1 = (sin t − cos t + e−t ) 2
49(a)
Since L−1 −1
L
1 1 2 −3t 1 = 1 = f (t) and L−1 = t e s (s + 3)3 2
1 1 · = 3 s (s + 3)
t
0
f (t − τ )g(τ )dτ
t
1 1. τ 2 e−3τ dτ 2 0 t 1 2 2 = −τ 2 e−3τ − τ e−3τ − e−3τ 4 3 9 0 1 = [2 − e−3t (9t2 + 6t + 2)] 54
=
c Pearson Education Limited 2004
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Directly L−1
L−1
49(b)
−1
L
2 1 6 2 1 18 −1 1 − − = L · − s (s + 3)3 54 s (s + 3)3 (s + 3)2 (s + 3) 1 [2 − e−3t (9t2 + 6t + 2)] = 54
1 1 = te2t = f (t), L−1 = te−3t = g(t) 2 2 (s − 2) (s + 3)
1 1 · = 2 2 (s − 2) (s + 3)
0
t
(t − τ )e2(t−τ ) .τ e−3τ dτ
−2t
t
=e
0
(tτ − τ 2 )e−5τ dτ
1 1 2 −5τ t e = e2t − (tτ − τ 2 )e−5τ − (t − 2τ )e−5τ + 5 25 125 0 2 2 t t e−5t + e−5t + − = e2t 25 125 25 125 1 2t e (5t − 2) + e−3t (5t + 2) = 125 Directly −2 1 2 1 1 125 25 125 25 + + = + (s − 2)2 (s + 3)2 s − 2 (s − 2)2 (s + 3) (s + 3)2 1 −2 2t 1 2 −3t 1 e + te2t + e = ∴ L−1 + te−3t 2 2 (s − 2) (s + 3) 125 25 125 25 1 2t [e (5t − 2) + e−3t (5t + 2)] = 125
49(c)
L−1
1 1 = e−4t = g(t) = t = f (t), L−1 2 s (s + 4) −1
L
1 1 = · s2 s + 4
0
t
(t − τ )e−4t dτ
1 t 1 = − (t − τ )e−4τ + e−4τ 4 16 0 1 1 −4t 1 = + t− e 16 4 16 c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition Directly L−1
50
1 1 1 1 1 1 −1 1 = L · − · + · s2 (s + 4) 16 s + 4 16 s 4 s2 1 1 −4t 1 e + t = − 16 16 4
Let f (λ) = λ and g(λ) = e−λ so F (s) =
1 1 and G(s) = s2 s+1
Considering the integral equation
y(t) =
t
λe−(t−λ) dλ
0
By (2.80) in the text −1
L
{F (s)G(s)} =
t
0
=
t
f (λ)g(t − λ)dλ λe−(t−λ) dλ = y(t)
0
so
1 + 1) 1 1 1 = L−1 − + 2 + s s s+1 −t = (t − 1) + e
y(t) = L−1 {F (s)G(s)} = L−1
51
s2 (s
Impulse response h(t) is given by the solution of d2 h 7dh + 12h = δ(t) + dt2 dt
subject to zero initial conditions. Taking Laplace transforms (s2 + 7s + 12)H(s) = 1 H(s) =
1 1 1 = − (s + 3)(s + 4) s+3 s+4
giving h(t) = L−1 {H(s)} = e−3t − e−4t c Pearson Education Limited 2004
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Response to pulse input is
[e−3(t−τ ) − e−4(t−τ ) ]dτ H(t) 0
t −3(t−τ ) [e − e−4(t−τ ) ]dτ H(t − T ) −A T 1 1 1 −3t 1 −4t =A H(t) + e − − e 3 4 3 4 1 1 1 −3(t−T ) 1 −4(t−T ) − − − e − e H(t − T ) 3 4 3 4 1 = A 1 − 4e−3t + 3e−4t − (1 − 4e−3(t−T ) + 3e−4(t−T ) )H(t − T ) 12
x(t) = A
t
or directly u(t) = A[H(t) − H(t − T )] so U (s) = L{u(t)} =
A [1 − e−sT ] s
Thus taking Laplace transforms with initial quiescent state A [1 − e−sT ] s 1 1 1 1 1 1 · − · + · (1 − e−sT ) X(s) = A 12 s 3 s + 3 4 s + 4 A [1 − 4e−3t + 3e−4t − (1 − 4e−3(t−T ) + 3e−4(t−T ) )H(t − T )] x(t) = L−1 {X(s)} = 12
(s2 + 7s + 12)X(s) =
52
Impulse response h(t) is the solution of dh d2 h ˙ + 5h = δ(t), h(0) = h(0) =0 +4 2 dt dt
Taking Laplace transforms (s2 + 4s + 5)H(s) = 1 H(s) =
s2
1 1 = + 4s + 5 (s + 2)2 + 1
so h(t) = L−1 {H(s)} = e−2t sin t. c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition By the convolution integral response to unit step is
t
e−2(t−τ ) sin(t − τ ).1dτ 0
t −2t e2τ sin(t − τ )dτ =e
θ0 (t) =
0
which using integration by parts gives t e−2t 2τ e [2 sin(t − τ ) + cos(t − τ )] 0 θ0 (t) = 5 1 1 −2t = − e (2 sin t + cos t) 5 5 Check Solving
d2 θ0 dθ0 + 5θ0 = 1 , θ˙0 (0) = θ0 (0) = 0 + 4 dt2 dt
gives 1 s 1 1 s+4 1 = − · Φ0 (s) = 2 s(s + 4s + 5) 5s 5 (s + 2)2 + 1 1 1 so θ0 (t) = L−1 {Φ0 (s)} = − [cos t + 2 sin t]e−2t . 5 5 (s2 + 4s + 5)Φ0 (s) =
Review Exercises 2.8 1(a)
d2 x dx + 5x = 8 cos t, x(0) = x(0) ˙ = 0 Taking Laplace transforms +4 2 dt dt (s2 + 4s + 5)X(s) =
8s +1
s2
8s + 1)(s2 + 4s + 5) s+5 s+1 − 2 = 2 s + 1 s + 4s + 5 s 1 (s + 2) + 3 = 2 + 2 − s + 1 s + 1 (s + 2)2 + 1
X(s) =
(s2
giving x(t) = L−1 {X(s)} = cos t + sin t − e−2t [cos t + 3 sin t] c Pearson Education Limited 2004
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Glyn James: Advanced Modern Engineering Mathematics, Third edition
dx d2 x − 2x = 6, x(0) = x(0) ˙ =1 1(b) 5 2 − 3 dt dt Taking Laplace transforms 6 5s2 + 2s + 6 (5s − 3s − 2)X(s) = 5(s + 1) − 3(1) + = s s 5s2 + 2s + 6 X(s) = 5s(s + 25 )(s + 1) 2
13 15 3 =− + 7 + 7 2 5 s−1 s+ 5
giving x(t) = L−1 {X(s)} = −3 +
13 t 15 − 2 t e + e 5 7 7
2(a)
Thus L−1
2(b)
1 1 1 1 1 s+2 = − · − · 2 2 (s + 1)(s + 2)(s + 2s + 2) s + 1 2 s + 2 2 s + 2s + 2 1 1 1 1 (s + 1) + 1 = − · − · s + 1 2 s + 2 2 (s + 1)2 + 1 1 1 1 = e−t − e−2t − e−t (cos t + sin t) 2 (s + 1)(s + 2)(s + 2s + 2) 2 2
From equation (2.26) in the text the equation is readily deduced.
Taking Laplace transforms (s2 + 3s + 2)I(s) = s + 2 + 3 + V.
1 (s + 1)2 + 1
1 s+5 +V 2 (s + 2)(s + 1) (s + 2)(s + 1)(s + 2s + 2) 3 4 − + V extended as in (a) = s+1 s+2
I(s) =
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition Thus using the result of (a) above 1 1 i(t) = L−1 {I(s)} = 4e−t − 3e−2t + V e−t − e−2t − e−t (cos t + sin t) 2 2 3
Taking Laplace transforms 1 s2 2 −2sX(s) + (s2 − 4)Y (s) = − s (s2 − 1)X(s) + 5sY (s) =
Eliminating Y (s) [(s2 − 1)(s2 − 4) + 2s(5s)]X(s) =
s2 − 4 11s2 − 4 + 10 = s2 s2
11s2 − 4 s2 (s2 + 1)(s2 + 4) 4 1 5 − 2 =− 2 + 2 s s +1 s +4
X(s) =
giving x(t) = L−1 {X(s)} = −t + 5 sin t − 2 sin 2t From the first differential equation 1 d2 x dy = t+x− 2 dt 5 dt 1 = [t − t + 5 sin t − 2 sin 2t + 5 sin t − 8 sin 2t] 5 = (2 sin t − 2 sin 2t) then y = −2 cos t + cos 2t + const. and since y(0) = 0 , const. = 1 giving y(t) = 1 − 2 cos t + cos 2t x(t) = −t + 5 sin t − 2 sin 2t
4
Taking Laplace transforms (s2 + 2s + 2)X(s) = sx0 + x1 + 2x0 + c Pearson Education Limited 2004
s2
s +1
131
132
Glyn James: Advanced Modern Engineering Mathematics, Third edition s sx0 + x1 + 2x0 + s2 + 2s + 2 (s2 + 1)(s2 + 2s + 2) 1 s+4 x0 (s + 1) + (x1 + x0 ) 1 s + 2 + · 2 − · = 2 (s + 1) + 1 5 s + 1 5 (s + 1)2 + 1
X(s) =
giving
x(t) = L−1 {X(s)} 1 = e−t (x0 cos t + (x1 + x0 ) sin t) + (cos t + 2 sin t) 5 1 −t − e (cos t + 3 sin t) 5
i.e. x(t) =
1 3 1 (cos t + 2 sin t) + e−t (x0 − ) cos t + (x1 + x0 − ) sin t 5 5 5 ↑ ↑ steady state
transient
1 2 Steady state solution is xs (t) = cos t + sin t ≡ A cos(t − α) 5 5 1 1 2 2 2 having amplitude A = ( 5 ) + ( 5 ) = √ 5 and phase lag α = tan−1 2 = 63.4◦ . 5 Denoting the currents in the primary and secondary circuits by i1 (t) and i2 (t) respectively Kirchoff’s second law gives di1 di2 + = 100 dt dt di1 di2 + =0 20i2 + 3 dt dt 5i1 + 2
Taking Laplace transforms 100 s sI1 (s) + (3s + 20)I2 (s) = 0 (5 + 2s)I1 (s) + sI2 (s) =
Eliminating I1 (s) [s2 − (3s + 20)(2s + 5)]I2 (s) = 100 −100 20 = − 5s2 + 55s + 100 s2 + 11s + 20 20 1 20 √ = − =− 11 2 41 11 (s + 2 ) − 4 41 (s + 2 −
I2 (s) =
√ 41 2 )
c Pearson Education Limited 2004
−
1 (s +
11 2
+
√ 41 2 )
Glyn James: Advanced Modern Engineering Mathematics, Third edition
133
giving the current i2 (t) in the secondary loop as √ 20 −(11+√41)t/2 i2 (t) = L−1 {I2 (s)} = √ − e−(11− 41)t/2 e 41
6(a) (i) L{cos(wt + φ)} = L{cos φ cos wt − sin φ sin wt} s w = cos φ 2 − sin φ s + w2 s2 + w2 2 = (s cos φ − w sin φ)/(s + w2 ) (ii) L{e−wt sin(wt + φ)} = L{e−wt sin wt cos φ + e−wt cos wt sin φ} w s+w = cos φ + sin φ 2 2 (s + w) + w (s + w)2 + w2 = [sin φ + w(cos φ + sin φ)]/(s2 + 2sw + 2w2 )
6(b)
Taking Laplace transforms s +4 3 2 2s + 9s + 9s + 36 = 2 (s + 4)(s2 + 4s + 8) 1 39s + 172 1 s+4 · 2 + · 2 = 20 s + 4 20 s + 4s + 8 1 39(s + 2) + 47(2) 1 s+4 · 2 + · = 20 s + 4 20 (s + 2)2 + (2)2
(s2 + 4s + 8)X(s) = (2s + 1) + 8 +
giving x(t) = L−1 {X(s)} =
7(a) L−1
s2
1 1 (cos 2t + 2 sin 2t) + e−2t (39 cos 2t + 47 sin 2t). 20 20
s−4 −1 (s + 2) − 2(3) = L s2 + 4s + 13 (s + 2)2 + 32 = e−2t [cos 3t − 2 sin 3t] c Pearson Education Limited 2004
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7(b)
Taking Laplace transforms (s + 2)Y (s) = −3 +
2s 4 4 + 2 + 2 s s +1 s +1
Y (s) =
−3s3 + 6s2 + s + 4 s(s + 2)(s2 + 1)
=
5 2 2 − + 2 s s+2 s +1
∴ y(t) = L−1 {Y (s)} = 2 − 5e−2t + 2 sin t
8
Taking Laplace transforms (s + 5)X(s) + 3Y (s) = 1 +
5X(s) + (s + 3)Y (s) =
2s s2 − 2s + 6 5 − = s2 + 1 s2 + 1 s2 + 1
3s 6 − 3s 6 − = s2 + 1 s2 + 1 s2 + 1
Eliminating Y (s) [(s + 5)(s + 3) − 15]X(s) = (s2 + 8s)X(s) =
X(s) =
(s + 3)(s2 − 2s + 6) 3(6 − 3s) − 2 s2 + 1 s +1 s3 + s2 + 9s s2 + 1 s2 + s + 9 1 1 = + (s + 8)(s2 + 1) s + 8 s2 + 1
so x(t) = L−1 {X(s)} = e−8t + sin t From the first differential equation 3y = 5 sin t − 2 cos t − 5x −
dx = 3e−8t − 3 cos t dt
Thus x(t) = e−8t + sin t, y(t) = e−8t − cos t . c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 9
135
Taking Laplace transforms 100 + 104 2 (s + 100)(s + 200)Q(s) = 104 · 2 s + 104 2.104 Q(s) = (s + 100)(s + 200)(s2 + 104 ) 1 2 1 1 3s − 100 1 · − · − · = 100 s + 100 500 s + 200 500 s2 + 104
(s2 + 300s + 2 × 104 )Q(s) = 200·
giving q(t) = L−1 {Q(s)} = i.e. q(t) =
s2
1 −100t 2 −200t 1 − − e e (3 cos 100t − sin 100t) 100 500 500
1 1 [5e−100t − 2e−200t ] − [3 cos 100t − sin 100t] 500 500 ↑ ↑ transient
steady state
1 3 sin 100t + cos 100t ≡ A sin(100t + α) 5o 5 where α = tan−1 51 18 12 . o Hence the current leads the applied emf by about 18 12 .
Steady state current =
10 dx + 6x + y = 2 sin 2t dt d2 x dy = 3e−2t +x− 2 dt dt 4
dx = −2 when t = 0 so from (i) y = −4 when t = 0 . dt Taking Laplace transforms
Given x = 2 and
8s2 + 36 4 = s2 + 4 s2 + 4 2s2 + 6s + 7 3 = (s2 + 1)X(s) − sY (s) = 2s − 2 + 4 + s+2 s+2 (4s + 6)X(s) + Y (s) = 8 +
Eliminating Y (s) [s(4s + 6) + (s2 + 1)]X(s) =
8s2 + 36 2s2 + 6s + 7 + s2 + 4 s+2
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(i) (ii)
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X(s) = = =
8s2 + 36 2s2 + 6s + 7 + s(s2 + 4)(s + 1)(s + 15 ) 5(s + 2)(s + 1)(s + 15 ) 11 5
s+1 29 20
s+1
− +
227 505
s+ 1 3
1 5
s+2
1 3 49 1 76s − 96 3 4 · 2 + − + 60 1 − 505 s + 4 s+2 s+1 s+ 5
+
445 1212 s + 15
−
1 76s − 96 505 s2 + 4
giving x(t) = L−1 {X(s)} =
11(a)
29 −t 1 −2t 445 − 1 t 1 + e + e e 5 − (76 cos 2t − 48 sin 2t) 20 3 1212 505
Taking Laplace transforms (s2 + 8s + 16)Φ(s) =
s2
2 +4
Φ(s) =
2 4)2 (s2
(s + + 4) 1 1 1 1 1 2s − 3 · + · = − · 2 2 25 s + 4 10 (s + 4) 50 s + 4
1 −4t 1 1 + · te−4t − e (4 cos 2t − 3 sin 2t) so θ(t) = L−1 {Φ(s)} = 25 10 100 1 (4e−4t + 10te−4t − 4 cos 2t + 3 sin 2t) i.e. θ(t) = 100
11(b)
Taking Laplace transforms (s + 2)I1 (s) + 6I2 (s) = 1 I1 (s) + (s − 3)I2 (s) = 0
Eliminating I2 (s) [(s + 2)(s − 3) − 6]I1 (s) = s − 3 I1 (s) = giving i1 (t) = L−1 {I1 (s)} =
1 6 s−3 = 7 + 7 s−4 s+3 (s − 4)(s + 3)
1 4t (e + 6e−3t ) 7
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition Then from the first differential equation 6 di1 6 = − e4t + e−3t dt 7 7 1 −3t 1 giving i2 (t) = (e − e4t ), i1 (t) = (e4t + 6e−3t ). 7 7 6i2 = −2i1 −
12
The differential equation LCR
di d2 i + L + Ri = V 2 dt dt
follows using Kirchhoff’s second law. Substituting V = E and L = 2R2 C gives 3
2R C
which on substituting CR =
2d
2
i
dt2
+ 2R2 C
di + Ri = E dt
1 leads to 2n E 1 d2 i 1 di +i= + 2 2 2n dt n dt R
and it follows that
E di d2 i + 2n + 2n2 i = 2n2 2 dt dt R
Taking Laplace transforms (s2 + 2ns + 2n2 )I(s) =
2n2 E 1 · R s
2n2 E 2 2 R s(s + 2ns + 2n ) s + 2n E 1 − = 2 2 R s (s + n) + n
I(s) =
so that i(t) =
E [1 − e−nt (cos nt + n sin nt)] R
c Pearson Education Limited 2004
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13
The equations are readily deduced by applying Kirchhoff’s second law to the
left and right hand circuits. Note that from the given initial conditions we deduce that i2 (0) = 0 . Taking Laplace transforms then gives E s −RI1 (s) + (sL + 2R)I2 (s) = 0 (sL + 2R)I1 (s) − RI2 (s) =
Eliminating I2 (s) E (sL + 2R) s E (sL + 3R)(sL + R)I1 (s) = (sL + 2R) s [(sL + 2R)2 − R2 ]I1 (s) =
I1 (s) =
s + 2R E L L s(s + R L )(s +
3R L ) 1 6
1 E 23 − 2R − R s s+ L s + 3R L R 3R E 1 4 − 3e− L t − e− L t giving i1 (t) = L−1 {I1 (s)} = 6R
=
For large t the exponential terms are approximately zero and i1 (t)
2E 3R
From the first differential equation Ri2 = 2Ri1 + L
di1 −E dt
Ignoring the exponential terms we have that for large t i2
E 1E 4E − = 3R R 3R
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition
14
139
Taking Laplace transforms (s2 + 2)X1 (s) − X2 (s) =
s2
2 +4
−X1 (s) + (s2 + 2)X2 (s) = 0 Eliminating X1 (s) [(s2 + 2)2 − 1]X2 (s) =
X2 (s) = = so x2 (t) = L−1 {X2 (s)} =
s2
2 +4
2 (s2 + 4)(s2 + 1)(s2 + 3) s2
2 3
+4
+
s2
1 3
+1
−
s2
1 +3
√ 1 1 1 sin 2t + sin t − √ sin 3t 3 3 3
Then from the second differential equation √ √ √ 2 2 1 d2 x2 2 4 √ sin 2t + sin t − sin 2t − sin t + = 3t − 3 sin 3t sin dt2 3 3 3 3 3 √ 1 1 2 or x1 (t) = − sin 2t + sin t + √ sin 3t 3 3 3
x1 (t) = 2x2 +
15(a) (i) L−1
s2
(s + 1) + 3 s+4 = L−1 + 2s + 10 (s + 1)2 + 32 = e−t (cos 3t + sin 3t)
(ii) L−1
1 2 s−3 1 = L−1 + − 2 2 (s − 1) (s − 2) (s − 1) (s − 1) s−2 = et + 2tet − e2t c Pearson Education Limited 2004
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15(b)
Taking Laplace transforms (s2 + 2s + 1)Y (s) = 4s + 2 + 8 + L{3te−t } 3 (s + 1)2 Y (s) = 4s + 10 + (s + 1)2 4s + 10 3 Y (s) = + 2 (s + 1) (s + 1)4 6 4 3 + = + 2 s + 1 (s + 1) (s + 1)4 1 giving y(t) = L−1 {Y (s)} = 4e−t + 6te−t + t3 e−t 2 1 −t i.e. y(t) = e (8 + 12t + t3 ) 2
16(a) 2 5 5 = · − 14s + 53 2 (s − 7)2 + 22 5 ∴ f (t) = L−1 {F (s)} = e7t sin 2t 2
F (s) =
16(b)
s2
d2 θ dθ n2 i 2 ˙ + n , θ(0) = θ(0) = 0, i const. + 2K θ = dt2 dt K
Taking Laplace transforms n2 i (s + 2Ks + n )Φ(s) = Ks 2
2
∴ Φ(s) =
n2 i Ks(s2 + 2Ks + n2 )
For the case of critical damping n = K giving 1 1 K12 Ki K2 K = Ki Φ(s) = − − s s+K (s + K)2 s(s + K)2
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition Thus θ(t) = L−1 {Φ(s)} =
i [1 − e−Kt − Kte−Kt ] K
17(a) (i) L{sin tH(t − α)} = L{sin[(t − α) + α]H(t − α)} = L{[sin(t − α) cos α + cos(t − α) sin α]H(t − α)} cos α + s sin α −αs .e = s2 + 1 (ii) L−1
17(b)
se−αs (s + 1) − 1 = L−1 e−αs 2 s + 2s + 5 (s + 1)2 + 4 1 = L−1 eαs L[e−t (cos 2t − sin 2t)] 2 1 = e−(t−α) [cos 2(t − α) − sin 2(t − α)]H(t − α) 2
Taking Laplace transforms
−e−sπ 1 − by (i) above in part (a) s2 + 1 s2 + 1 1 + e−πs = 2 s +1 −πs 1 s−2 1 s 1+e = − + (1 + e−πs ) Y (s) = 2 (s + 1)(s2 + 2s + 5) 10 s2 + 1 10 s2 + 2s + 5 (s2 + 2s + 5)Y (s) =
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giving 1 [2 sin t − cos t + e−(t−π) [2 sin(t − π) − cos(t − π)]H(t − π)] 10 1 + e−t (cos 2t − sin 2t) + e−(t−π) [cos 2(t − z) 2 1 − sin 2(t − π)]H(t − π)] 2 1 −t 1 e (cos 2t − sin 2t) + 2 sin t − cos t = 10 2 1 + [e−(t−π) (cos 2t − sin 2t) + cos t − 2 sin t]H(t − π) 2
y(t) = L−1 {Y (s)} =
18
By theorem 2.5
T 1 e−st v(t)dt L{v(t)} = V (s) = 1 − e−sT 0 T /2
T 1 −st −st = e dt − e dt 1 − e−sT 0 T /2 1 −st T /2 1 −st T 1 − e − − e = 1 − e−sT s s 0 T /2 1 1 (e−sT − e−sT /2 − e−sT /2 + 1) = · s 1 − e−sT (1 − e−sT /2 )2 1 1 − e−sT /2 1 = = s (1 − e−sT /2 )(1 + e−sT /2 ) s 1 + e−sT /2
Equation for current flowing is 1 250i + (q0 + C
t
i(τ )dτ ) = v(t), 0
q0 = 0
Taking Laplace transforms 250I(s) +
1 1 − e−sT /2 1 1 · I(s) = V (s) = · 10−4 s s 1 + e−sT /2 1 1 − e−sT /2 (s + 40)I(s) = 250 1 + e−sT /2 1 − e−sT /2 1 · or I(s) = 250(s + 40) 1 + e−sT /2
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3 1 (1 − e−sT /2 )(1 − e−sT /2 + e−sT − e− 2 sT + e−2sT . . .) 250(s + 40) 3 1 [1 − 2e−sT /2 + 2e−sT − 2e− 2 sT + 2e−2sT . . .] = 250(s + 40) 1 −40t 1 = using the second shift theorem gives Since L−1 e 250(s + 40) 250
I(s) =
1 T −40(t−T /2) e−40t − 2H t − e i(t) = + 2H(t − T )e−40(t−T ) 250 2
3T −40(t−3T /2) e −2H t − + ... 2 If T = 10−3 s then the first few terms give a good representation of the steady 1 of the circuit is large compared to the period T . state since the time constant 4 19
The impulse response h(t) is the solution of d2 h 2dh + 2h = δ(t) + dt2 dt
˙ subject to the initial conditions h(0) = h(0) = 0 . Taking Laplace transforms (s2 + 2s + s)H(s) = L{δ(t)} = 1 1 H(s) = (s + 1)2 + 1 i.e. h(t) = L−1 {H(s)} = e−t sin t. Using the convolution integral the step response xs (t) is given by
xs (t) =
t
0
h(τ )u(t − τ )dτ
with u(t) = 1H(t) ; that is
xs (t) =
t
1.e−τ sin τ dτ
0
1 = − [e−τ cos τ + e−τ sin τ ]t0 2 1 i.e. xs (t) = [1 − e−t (cos t + sin t)]. 2 c Pearson Education Limited 2004
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Solving
2dxs d2 xs + 2xs = 1 directly we have taking Laplace transforms + dt2 dt (s2 + 2s + 2)Xs (s) =
1 s
1 s(s2 + 2s + 2) s+2 1 1 1 = · − 2 2 s 2 (s + 1) + 1
Xs (s) =
giving as before xs (t) =
1 1 −t − e (cos t + sin t) 2 2
20
d4 y = 12 + 12H(x − 4) − Rδ(x − 4) dx4 y(0) = y (0) = 0, y(4) = 0, y (5) = y (5) = 0 EI
With y (0) = A, y (0) = B taking Laplace transforms 12 12 −4s + e − Re−4s s s A B 12 1 12 1 −4s R 1 −4s · 5+ · 5e · e Y (s) = 3 + 4 + − s s EI s EI s EI s4
EIs4 Y (s) = EI(sA + B) +
giving A B 1 4 1 x + (x − 4)4 H(x − 4) y(x) = L−1 {Y (s)} = x2 + x3 + 2 6 2EI 2EI R (x − 4)3 H(x − 4) − 6EI c Pearson Education Limited 2004
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145
or EIy(x) =
1 1 1 1 R A1 x2 + B1 x3 + x4 + (x − 4)4 H(x − 4) − (x − 4)3 H(x − 4) 2 6 2 2 6
32 B1 + 128 ⇒ 3A1 + 4B1 = −48 3 y (5) = 0 ⇒ 0 = A1 + 5B1 + 6(25) + 6 − R ⇒ A1 + 5B1 − R = −156 y(4) = 0 ⇒ 0 = 8A1 +
y (5) = 0 ⇒ 0 = B1 + 12(5) + 12 − R ⇒ B1 − R = −72 which solve to give A1 = 18, B1 = −25.5, R = 46.5 Thus
1 4 x − 4.25x3 + 9x2 , 0 ≤ x ≤ 4 2 y(x) = 1 x4 − 4.25x3 + 9x2 + 1 (x − 4)4 − 7.75(x − 4)3 , 4 ≤ x ≤ 5 2 2 R0 = −EIy (0) = 25.5kN, M0 = EIy (0) = 18kN.m Check
√
R0 + R = 72kN, Total load = 12 × 4 + 24 = 72kN
Moment about x = 0 is 12 × 4 × 2 + 24 × 4.5 − 4R = 18 = M0
21(a)
f (t) = H(t − 1) − H(t − 2) e−2s e−s − and L{f (t)} = F (s) = s s c Pearson Education Limited 2004
√
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Taking Laplace transforms throughout the differential equation 1 −s (e − e−2s ) s 1 (e−s − e−2s ) X(s) = s(s + 1) 1 1 −s 1 1 −2s − e − − e = s s+1 s s+1
(s + 1)X(s) =
giving x(t) = L−1 {X(s)} = [1 − e−(t−1) ]H(t − 1) − [1 − e−(t−2) ]H(t − 2)
21(b) I(s) =
E s[Ls + R/(1 + Cs)]
(i) By the initial value theorem E =0 s→∞ Ls + R/(1 + Cs)
lim i(t) = lim sI(s) = lim
t→0
s→∞
(ii) Since sI(s) has all its poles in the left half of the s-plane the conditions of the final value theorem hold so lim i(t) = lim sI(s) =
t→∞
22
s→0
E R
We have that for a periodic function f (t) of period T 1 L{f (t)} = 1 − e−sT
T
e−sT f (t)dt
0
Thus the Laplace transform of the half-rectified sine wave is
π 1 L{v(t)} = e−sT sin tdt −2πs 1−e 0
π 1 (j−s)t = Im e dt 1 − e−2πs 0 π e(j−s)t 1 = Im 1 − e−2πs j − s 0 (−e−πs − 1)(−j − s) 1 + e−πs 1 = = Im 1 − e−2πs (j − s)(−j − s) (1 − e−2πs )(1 + s2 ) 1 i.e. L{v(t)} = 2 (1 + s )(1 − e−πs ) c Pearson Education Limited 2004
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Applying Kirchoff’s law to the circuit the current is determined by di + i = v(t) dt which on taking Laplace transforms gives (s + 1)I(s) =
1 s2 )(1
(1 + − e−πs ) 1 1 s+1 1 · I(s) = − 1 − e−πs s + 1 s2 + 1 2 s + 1 1 1 − 2 1 + e−πs + e−2πs + . . . = 2 s+1 s +1
Since L−1
1 1 s + 1 1 − 2 = (sin t − cos t + e−t )H(t) = f (t) 2 s+1 s +1 2
we have by the second shift theorem that i(t) = f (t) + f (t − π) + f (t − 2π) + . . . =
∞
f (t − nπ)
n=0
The graph may be plotted by computer and should take the form
1 1 , L{te−t } = 2 s (s + 1)2 −t taking f (t) = t and g(t) = te in the convolution theorem 23(a)
Since L{t} =
L−1 [F (s)G(s)] = f ∗ g(t) c Pearson Education Limited 2004
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gives −1
L
1 1 = · 2 2 s (s + 1)
t
0
= 0
t
f (t − τ )g(τ )dτ (t − τ )τ e−τ
t = −(t − τ )τ e−τ − (t − 2τ )e−τ + 2e−τ 0
i.e. L−1
23(b)
1 1 · = t − 2 + 2e−t + te−t . 2 2 s (s + 2)
y(t) = t + 2
t 0
y(u) cos(t − u)du
s giving on taking Taking f (t) = y(t), g(t) = cos t ⇒ F (s) = Y (s), G(s) = 2 s +1 transforms 1 s + 2Y (s) 2 2 s s +1 2 s +1 (s2 + 1 − 2s)Y (s) = s2 1 2 s2 + 1 2 2 + or Y (s) = 2 = + 2− 2 s (s − 1) s s s − 1 (s − 1)2 Y (s) =
and y(t) = L−1 {Y (s)} = 2 + t − 2et + 2tet . Taking transforms (s2 Y (s) − sy(0) − y (0))(sY (s) − y(0)) = Y (s) or (s2 Y (s) − y1 )(sY (s)) = Y (s) 1 y1 giving Y (s) = 0 or Y (s) = 2 + 3 s s which on inversion gives y(t) = 0 or y(t) =
1 2 t + ty1 2
In the second of these solutions the condition on y (0) is arbitrary.
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 24
Equation for displacement is EI
d4 y = −W δ(x − ) dx4
with y(0) = 0, y(3) = 0, y (0) = y (3) = 0 with y (0) = A, y (0) = B then taking Laplace transforms gives EIs4 Y (s) = EI(sA + B) − W e−s −W −s A B Y (s) = e + 3+ 4 4 EIs s s −W A B (x − )3 .H(x − ) + x2 + x3 giving y(x) = 6EI 2 6 −3W B (x − )2 + Ax + x2 6EI 2 so y (3) = 0 and y(3) = 0 gives For x > , y (x) =
2 2W 2 + 3A + 9B EI 2 3 9 4W 9 + A2 + B3 0=− 3EI 2 2 20 W 4W and B = giving A = − 9EI 27 EI 0=−
Thus deflection y(x) is y(x) = −
2 W 2 10 W 3 W x + x (x − )3 H(x − ) − 6EI 9 EI 81 EI c Pearson Education Limited 2004
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With the added uniform load the differential equation governing the deflection is EI
25(a)
d4 y = −W δ(x − ) − w[H(x) − H(x − )] dx4
Taking Laplace transforms (s2 − 3s + 3)X(s) =
1 −as e s
1 1 −as 16 1 −as 6s − 2 · e − ·e = s(s2 − 3s + 3) s s2 − 3s + 3 √ √ 1 1 (s − 32 ) − 3( 23 ) −as √ − e = 6 s (s − 3 )2 + ( 3 )2
X(s) =
2
2
√ √ √ 3 3 3 e−as −2t L 1−e t − 3 sin t = cos 6 2 2 giving −1
x(t) = L
√ √
√ 3 3 3 1 − 2 (t−a) 1−e cos {X(s)} = (t − a) − 3 sin (t − a) H(t − a) 6 2 2
25(b) X(s) = G(s)L{sin wt} = G(s) =
s2
w + w2
w G(s) (s + jw)(s − jw)
Since the system is stable all the poles of G(s) have negative real part. Expanding in partial fractions and inverting gives x(t) = 2Re
F (jw)w jwt ·e + terms from G(s) with negative exponentials 2jw
Thus as t → ∞ the added terms tend to zero and x(t) → xs (t) with xs (t) = Re
ejwt F (jw) j
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151
In the absence of feedback the system has poles at s = −3 and s = 1
and is therefore unstable.
26(b) G1 (s) =
26(c)
1 1 G(s) = = 2 1 + KG(s) (s − 1)(s + 3) + K s + 2s + (K − 3)
Poles G1 (s) given by s = −1 ±
√
4−K.
These may be plotted in the s-plane for different values of K . Plot should be as in the figure
26(d) Clearly from the plot in (c) all the poles are in the left half plane when K > 3 . Thus system stable for K > 3 . a1 a0 a2 2 1s + 2s + (K − 3) = 0 Routh–Hurwitz determinants are 26(e)
∆1 = 2 > 0 a1 a2 = 2 ∆2 = 0 0 a0
1 = 2(K − 3) > 0 if K > 3 K − 3
thus confirming the result in (d). 27(a)
Closed loop transfer function is G1 (s) =
2 G(s) = 2 1 + G(s) s + αs + 5
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2 = h(t) = 2e−2t sin t Thus L−1 2 s + αs + 5 α 2 −2t −1 i.e. L sin (5 − = 2e 2 (s + α2 )2 + (5 − α4 ) giving α = 4 27(b)
α2 4 )t
= 2e−2t sin t
Closed loop transfer function is G(s) = 1
10 s(s−1) − (1+Ks)10 s(s−1)
=
s2
10 + (10K − 1)s + 10
Poles of the system are given by s2 + (10K − 1)s + 10 = 0 which are both in the negative half plane of the s-plane provided (10K − 1) > 0 ; 1 that is, K > 10 . Thus the critical value of K for stability of the closed loop system is K = 28(a)
1 10
.
Overall closed loop transfer function is G(s) =
28(b)
1+
K s(s+1) K s(s+1) (1 +
K1 s)
=
s2
K + s(1 + KK1 ) + K
Assuming zero initial conditions step response x(t) is given by X(s) = G(s)L{1.H(t)} =
K s[s2 + s(1 + KK1 ) + K]
wn + 2ξwn s + wn2 ] s + 2ξwn 1 = − 2 s s + 2ξwn s + wn2
(s + ξwn ) + ξwn 1 = − s (s + ξwn )2 + [wn2 (1 − ξ 2 )]
1 (s + ξwn ) + ξwn = − s (s + ξwn )2 + wd2 ξ sin wd t , t ≥ 0. giving x(t) = L−1 {X(s)} = 1 − e−ξwn t cos wd t + 1 − ξ2 =
s[s2
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 28(c)
The peak time tp is given by the solution of
dx =0 dt t=tp
ξ 2 wn ξwd dx −ξwn t =e ξwn − + wd sin wd t cos wd t dt 1 − ξ2 1 − ξ2 wn sin wd t = e−ξwn t 1 − ξ2 Thus tp given by the solution of wn
e−ξwn tp
1 − ξ2
sin wd tp = 0
i.e. sin wd tp = 0 Since the peak time corresponds to the first peak overshoot wd tp = π or tp =
π wd
The maximum overshoot Mp occurs at the peak time tp . Thus ξw π − wn d cos π
Mp = x(tp ) − 1 = e
ξw π − wn d
=e
ξ + sin π 1 − ξ2
= e−ξπ/
√
1−ξ 2
π
We wish Mp to be 0.2 and tp to be 1 s, thus e−ξπ/
√
1−ξ 2
= 0.2 giving ξ = 0.456
and π = 1 giving wd = 3.14 wd wd Then it follows that wn = = 3.53 from which we deduce that 1 − ξ2 tp =
K = wn2 = 12.5 and K1 =
2wn ξ − 1 = 0.178. K
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28(d)
The rise time tr is given by the solution of ξ x(tr ) = 1 = 1 − e−ξwn tr cos wd tr + sin wd tr 1 − ξ2
Since e−ξwn tr = 0 ξ
cos wd tr + 1 − ξ2 giving tan wd tr = − ξ or
1 tan−1 − tr = wd
1 − ξ2
sin wd tr = 0
1 − ξ 2 π − 1.10 = = 0.65s. ξ wd
The response x(t) in (b) may be written as e−ξwn t x(t) = 1 − sin wα t + tan−1 1 − ξ2
1 − ξ2 ξ
e−ξwn t so the curves 1 ± are the envelope curves of the transient response to a 1 − ξ2 1 . The settling time ts may unit step input and have a time constant T = ξwn be measured in terms of T . Using the 2% criterion ts is approximately 4 times the time constant and for the 5% criterion it is approximately 3 times the time constant. Thus 4 = 2.48s 2% criterion : ts = 4T = ξwn 3 5% criterion : ts = 3T = = 1.86s ξwn Footnote
This is intended to be an extended exercise with students being encouraged to carry out simulation studies in order to develop a better understanding of how the transient response characteristics can be used in system design.
29 As for Exercise 28 this is intended to be an extended problem supported by simulation studies. The following is simply an outline of a possible solution. c Pearson Education Limited 2004
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Figure 2.63(a) is simply a mass-spring damper system represented by the differential equation d2 x dx + K1 x = sin wt M1 2 + B dt dt Assuming that it is initially in a quiescent state taking Laplace transforms X(s) =
M1
s2
w 1 · 2 + Bs + K1 s + w2
The steady state response will be due to the forcing term and determined by the αs + β term in the partial fractions expansion of X(s) . Thus, the steady state s2 + w2 response will be of the form A sin(wt + δ) ; that is, a sinusoid having the same frequency as the forcing term but with a phase shift δ and amplitude scaling A . In the situation of Figure 2.63(b) the equations of motion are d2 x dx + K2 (y − x) + sin wt = −K1 x − B 2 dt dt d2 y M2 2 = −K2 (y − x) dt
M1
Assuming an initial quiescent state taking Laplace transforms gives [M1 s2 + Bs + (K1 + K2 )]X(s) − K2 Y (s) = w/(s2 + w2 ) −K2 X(s) + (s2 M2 + K2 )Y (s) = 0 Eliminating Y (s) gives X(s) =
w(s2 M2 + K2 ) (s2 + w2 )p(s)
where p(s) = (M1 s2 + Bs + K1 + K2 )(s2 M2 + K2 ) . Because of the term (s2 + w2 ) in the denominator x(t) will contain terms in sin wt and cos wt . However, if (s2 M2 + K2 ) exactly cancels (s2 + w2 ) this will be avoided. Thus choose K2 = M2 w2 . This does make practical sense for if the natural frequency of the secondary system is equal to the frequency of the applied force then it may resonate and therefore damp out the steady state vibration of M1 . It is also required to show that the polynomial p(s) does not give rise to any undamped oscillations. That is, it is necessary to show that p(s) does not possess purely imaginary roots of the form jθ, θ real, and that it has no roots with a positive real part. This can be checked using the Routh–Hurwitz criterion. c Pearson Education Limited 2004
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To examine the motion of the secondary mass M2 solve for Y (s) giving Y (s) =
(s2
K2 w + w2 )p(s)
Clearly due to the term (s2 + w2 ) in the denominator the mass M2 possesses an undamped oscillation. Thus, in some sense the secondary system has absorbed the energy produced by the applied sinusoidal force sin wt .
30
Again this is intended to be an extended problem requiring wider exploration
by the students. The following is an outline of the solution.
30(a) Students should be encouraged to plot the Bode plots using the steps used in example 2.62 of the text and using a software package. Sketches of the magnitude and phase Bode plots are given in the figures below. 30(b) With unity feedback the amplifier is unstable. Since the −180◦ crossover gain is greater than 0dB (from the plot it is +92dB) .
30(c)
Due to the assumption that the amplifier is ideal it follows that for 1 must be 92dB (that is, the plot is effectively marginal stability the value of β lowered by 92dB) . Thus 20 log
1 = 92 β 92 1 ⇒ β 2.5 × 10−5 = antilog β 20
30(d) From the amplitude plot the effective 0dB axis is now drawn through the 100dB point. Comparing this to the line drawn through the 92dB point, corresponding to marginal stability, it follows that Gain margin = −8dB and Phase margin = 24◦ .
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30(e) G(s) =
K (1 + sτ1 )(1 − sτ2 )(1 + sτ3 )
Given low frequency gain K = 120dB so 20 log K = 120 ⇒ K = 106 1 where fi is the oscillating frequency in cycles per second of the pole. fi Since 1MHz = 10 cycles per second Ti =
1 1 = 6 since f1 = 1MHz f1 10 1 1 τ2 = = since f2 = 10MHz f2 10.106 1 1 = since f3 = 25MHz τ3 = f3 25.106 τ1 =
Thus G(s) = =
106 (1 +
s 106 )(1
(s +
106 )(s
+
s 10.106 )(1 24
+
s 25.106 )
250.10 + 107 )(s + 52 .107 )
The closed loop transfer function G1 (s) is G(s) =
30(f )
G(s) 1 + βG(s)
The characteristic equation for the closed loop system is (s + 106 )(s + 107 )(s + 52 .107 ) + β25.1025 = 0
or
s3 + 36(106 )s2 + (285)1012 s + 1019 (25 + 25β106 ) = 0 ↓ ↓ ↓ A1 A2 A3
By Routh–Hurwitz criterion system stable provided A1 > 0 and A1 A2 > A3 . If β = 1 then A1 A2 < A3 and the system is unstable as determined in (b). For marginal stability A1 A2 = A3 giving β = 1.40−5 (compared with β = 2.5.10−5 using the Bode plot). c Pearson Education Limited 2004
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Glyn James: Advanced Modern Engineering Mathematics, Third edition Magnitude vs Frequency Plot
120
Data margin – 8 dB
100 80 60 Gain dB
1/β = 92 dB 40 20 0 Corresponds to 180° phase lag
To phase plot
1
10
25
Log freq. MHz
Phase vs Frequency Plot 0°
– 90°
– 180°
Phase margin 24°
16 MHz – 270°
1
10
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25
ln freq. MHz
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