chapter 98 the laplace transform of the heaviside function
CHAPTER 98 THE LAPLACE TRANSFORM OF THE HEAVISIDE FUNCTION EXERCISE 357 Page 1042
1. A 6 V source is switched on at time t = 4 s. Write the function in terms of the Heaviside step function and sketch the waveform.
The function is shown sketched below The Heaviside step function is:
V(t) = 6 H(t – 4)
2 for 0 〈 t 〈 5 2. Write the function V (t ) = in terms of the Heaviside step function and sketch 0 for t 〉 5 the waveform.
The voltage has a value of 2 up until time t = 5; then it is turned off The function is shown sketched below The Heaviside step function is:
V(t) = 2 H(t) – H(t – 5)
3. Sketch the graph of: f(t) = H(t – 2)
A function H(t – 2) has a maximum value of 1 and starts when t = 2, as shown in the sketch below
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4. Sketch the graph of: f(t) = H(t)
A function H(t) has a maximum value of 1 and starts when t = 0, as shown in the sketch below
5. Sketch the graph of: f(t) = 4 H(t – 1)
A function 4H(t – 1) has a maximum value of 4 and starts when t = 1, as shown in the sketch below
6. Sketch the graph of: f(t) = 7H(t – 5)
A function 7H(t – 5) has a maximum value of 7 and starts when t = 5, as shown in the sketch below
π 7. Sketch the graph of: f(t) = H t − . cos t 4
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π Below shows a graph of H t − . cos t where the graph of cos t does not ‘switch on’ until t = π/4 4
π π 8. Sketch the graph of: f(t) = 3H t − .cos t − 2 6 π π Below shows a graph of f(t) = 3H t − .cos t − where the graph of 3 cos(t – π/6) does not 2 6 ‘switch on’ until t = π/2
9. Sketch the graph of: f(t) = H ( t − 1) . t 2 Below shows a graph of f(t) = H ( t − 1) . t 2 where the graph of t 2 does not ‘switch on’ until t = 1
10. Sketch the graph of: f(t) = H(t – 2). e
−
t 2
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Below shows a graph of f(t) = H(t – 2). e
−
t 2
−
t
where the graph of e 2 does not ‘switch on’ until t = 2
11. Sketch the graph of: f(t) = [H(t – 2) – H(t – 5)]. e
−
t 4
Below shows the graph of f(t) = [H(t – 2) – H(t – 5)]. e
−
t 4
−
t
where the graph of e 4 does not ‘switch
on’ until t = 2, but then ‘switches off’ at t = 5
π π 12. Sketch the graph of: f(t) = 5 H t − .sin t + 3 4
π π Below shows a graph of f(t) = 5 H t − .sin t + where the graph of 5 sin(t + π/4) does not 3 4 ‘switch on’ until t = π/3
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EXERCISE 358 Page 1044
1. Determine ℒ{H(t – 1)} where in this case, F(s) = ℒ {1} and c = 1
ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
1 ℒ{H(t – 1)} = e − s s =
from (i) of Table 95.1, page 1023
e− s s
2. Determine ℒ{7 H(t – 3)} ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {7} and c = 3
7 ℒ{7 H(t – 3) } = e − 3 s s =
from (ii) of Table 95.1, page 1023
7 e −3 s s
3. Determine ℒ{H(t – 2).(t – 2) 2 } ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {t 2 } and c = 2
2! ℒ{H(t – 2).f(t – 2) 2 } = e − 2 s s3 =
from (vii) of Table 95.1, page 1023
2 e− 2 s s3
4. Determine ℒ{H(t – 3).sin(t – 3)} ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {sin t} and c = 3
1 ℒ{H(t – 3).sin(t – 3) } = e − 3 s s 2 + 12 =
from (iv) of Table 95.1, page 1023
e− 3s s2 + 1
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5. Determine ℒ{H(t – 4). et − 4 } where in this case, F(s) = ℒ {et } and c = 4
ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
1 ℒ{H(t – 4). et − 4 } = e − 4 s s −1 =
from (iii) of Table 95.1, page 1023
e− 4 s s −1
6. Determine ℒ{H(t – 5).sin 3(t – 5)} where in this case, F(s) = ℒ {sin 3t} and c = 5
ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
3 ℒ{H(t – 5).sin 3(t – 5) } = e − 5 s s 2 + 32 =
from (iv) of Table 95.1, page 1023
3e − 5 s s2 + 9
7. Determine ℒ{H(t – 1).(t – 1) 3 } ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {t 3 } and c = 1
3! ℒ{H(t – 1).(t – 1) 3 } = e − s s 3+1 =
from (viii) of Table 95.1, page 1023
6 e− s s4
8. Determine ℒ{H(t – 6).cos 3(t – 6)} ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {cos 3t} and c = 6
s ℒ{H(t – 6).cos 3(t – 6)} = e − 6 s s 2 + 32 s e− 6 s = s2 + 9
from (v) of Table 95.1, page 1023
9. Determine ℒ{5 H(t – 5).sinh 2(t – 5)} 1480
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ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {sinh 2t} and c = 5
2 ℒ{5 H(t – 5).sinh 2(t – 5) } = 5 e − 5 s s 2 − 22
from (x) of Table 95.1, page 1023
10 e − 5 s s2 − 4
=
π π 10. Determine ℒ{ H t − .cos 2 t − } 3 3
where in this case, F(s) = ℒ {cos 2t} and c =
ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
π s − s π π ℒ{ t − . cos 2 t − } = e 3 3 3 s 2 + 22 −
π
π 3
from (v) of Table 95.1, page 1023
s
se 3 = s2 + 4 11. Determine ℒ{2 H(t – 3). et − 3 } ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {et } and c = 3
1 ℒ{2 H(t – 3). et −3 } = 2 e −3 s s −1 =
from (iii) of Table 95.1, page 1023
2 e− 3 s s −1
12. Determine ℒ{3 H(t – 2).cosh(t – 2)} ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {cosh t} and c = 2
s ℒ{3 H(t – 2).cosh(t – 2) } = 3 e − 2 s s 2 − 12 =
from (ix) of Table 95.1, page 1023
3 s e− 2 s s2 −1
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EXERCISE 359 Page 1045
e− 9 s 1. Determine ℒ −1 s
Part of the numerator corresponds to e − c s where c = 9. This indicates H(t – 9) Then
1 = F(s) = ℒ{1} from (i) of Table 97.1, page 1033 s
Hence,
e− 9 s ℒ −1 = H(t – 9) s
4 e− 3 s 2. Determine ℒ −1 s
Part of the numerator corresponds to e − c s where c = 3. This indicates H(t – 3) Then
4 = F(s) = ℒ{4} from (ii) of Table 97.1, page 1033 s
Hence,
4 e− 3 s ℒ −1 = 4 H(t – 3) s
2 e− 2 s 3. Determine ℒ −1 s2
The numerator corresponds to e − c s where c = 2. This indicates H(t – 2) 1 = F(s) = ℒ{t} from (vii) of Table 97.1, page 1033 s2 Then
2 e− 2 s ℒ −1 = 2 H(t – 2).(t – 2) s2
5e − 2 s 4. Determine ℒ −1 s2 + 1
Part of the numerator corresponds to e − c s where c = 2. This indicates H(t – 2)
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5 s2 + 1
Then
Hence,
1 may be written as: 5 s 2 + 12
1 5 = F(s) = ℒ{5 sin t} from (iv) of Table 97.1, page 1033 s 2 + 12 5e − 2 s ℒ −1 = H(t – 2).5 sin(t – 2) = 5 H(t – 2).sin(t – 2) s2 + 1
3 s e− 4 s 5. Determine ℒ −1 s 2 + 16
Part of the numerator corresponds to e − c s where c = 4. This indicates H(t – 4) 3s s + 16 2
Then
Hence,
s may be written as: 3 s 2 + 42 s 3 = F(s) = ℒ{3 cos 4t} from (v) of Table 97.1, page 1033 s 2 + 42 3 s e− 4 s ℒ −1 =H(t – 4).3 cos 4(t – 4) = 3 H(t – 4).cos 4(t – 4) s 2 + 42
6 e− 2 s 6. Determine ℒ −1 s2 −1
Part of the numerator corresponds to e − c s where c = 2. This indicates H(t – 2) 6 s2 −1
Then
Hence,
1 may be written as: 6 s 2 − 12 1 6 = F(s) = ℒ{6 sinh t} from (x) of Table 97.1, page 1033 s 2 − 12 6 e− 2 s ℒ −1 =H(t – 2).6 sinh (t – 2) = 6 H(t – 2).sinh (t – 2) s2 −1
3e − 6 s 7. Determine ℒ −1 s3
The numerator corresponds to e − c s where c = 6. This indicates H(t – 6)
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1 1 = F(s) = ℒ t 2 from (viii) of Table 97.1, page 1033 s3 2
Then
1 3e − 6 s ℒ −1 = 3 H(t – 6). (t – 6) 2 = 1.5 H(t – 6). (t – 6) 2 2 s2
2 s e− 4 s 8. Determine ℒ −1 s 2 − 16
Part of the numerator corresponds to e − c s where c = 4. This indicates H(t – 4) 2s s − 16 2
Then
Hence,
s may be written as: 2 s 2 − 42 s 2 = F(s) = ℒ{2 cosh 4t} from (ix) of Table 97.1, page 1033 s 2 − 42 2 s e− 4 s ℒ −1 =H(t – 4).2 cosh 4(t – 4) = 2 H(t – 4).cosh 4(t – 4) s 2 − 42
1 − s 2 2 s e 9. Determine ℒ −1 2 s + 5
1 1 . This indicates H t − 2 2
Part of the numerator corresponds to e − c s where c = 2s s2 + 5
Then
Hence,
s may be written as: 2 2 s + 5
( )
s 2 s2 + 5
( )
1 − s 2 2 s e ℒ −1 s2 + 5
2
= F(s) = ℒ 2 cos 5 t 2
( )
{
}
from (v) of Table 97.1, page 1033
1 1 1 1 = H t − .2 cos 5 t − = 2 H t − .cos 5 t − 2 2 2 2 2
4 e− 7 s 10. Determine ℒ −1 s −1 Part of the numerator corresponds to e − c s where c = 7. This indicates H(t – 7) 1484
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Then
1 = F(s) = ℒ{ et } from (iii) of Table 97.1, page 1033 s −1
Hence,
4 e− 7 s ℒ −1 = 4 H(t – 7). et −7 s −1
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1. A 6 V source is switched on at time t = 4 s. Write the function in terms of the Heaviside step function and sketch the waveform.
The function is shown sketched below The Heaviside step function is:
V(t) = 6 H(t – 4)
2 for 0 〈 t 〈 5 2. Write the function V (t ) = in terms of the Heaviside step function and sketch 0 for t 〉 5 the waveform.
The voltage has a value of 2 up until time t = 5; then it is turned off The function is shown sketched below The Heaviside step function is:
V(t) = 2 H(t) – H(t – 5)
3. Sketch the graph of: f(t) = H(t – 2)
A function H(t – 2) has a maximum value of 1 and starts when t = 2, as shown in the sketch below
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4. Sketch the graph of: f(t) = H(t)
A function H(t) has a maximum value of 1 and starts when t = 0, as shown in the sketch below
5. Sketch the graph of: f(t) = 4 H(t – 1)
A function 4H(t – 1) has a maximum value of 4 and starts when t = 1, as shown in the sketch below
6. Sketch the graph of: f(t) = 7H(t – 5)
A function 7H(t – 5) has a maximum value of 7 and starts when t = 5, as shown in the sketch below
π 7. Sketch the graph of: f(t) = H t − . cos t 4
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π Below shows a graph of H t − . cos t where the graph of cos t does not ‘switch on’ until t = π/4 4
π π 8. Sketch the graph of: f(t) = 3H t − .cos t − 2 6 π π Below shows a graph of f(t) = 3H t − .cos t − where the graph of 3 cos(t – π/6) does not 2 6 ‘switch on’ until t = π/2
9. Sketch the graph of: f(t) = H ( t − 1) . t 2 Below shows a graph of f(t) = H ( t − 1) . t 2 where the graph of t 2 does not ‘switch on’ until t = 1
10. Sketch the graph of: f(t) = H(t – 2). e
−
t 2
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Below shows a graph of f(t) = H(t – 2). e
−
t 2
−
t
where the graph of e 2 does not ‘switch on’ until t = 2
11. Sketch the graph of: f(t) = [H(t – 2) – H(t – 5)]. e
−
t 4
Below shows the graph of f(t) = [H(t – 2) – H(t – 5)]. e
−
t 4
−
t
where the graph of e 4 does not ‘switch
on’ until t = 2, but then ‘switches off’ at t = 5
π π 12. Sketch the graph of: f(t) = 5 H t − .sin t + 3 4
π π Below shows a graph of f(t) = 5 H t − .sin t + where the graph of 5 sin(t + π/4) does not 3 4 ‘switch on’ until t = π/3
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EXERCISE 358 Page 1044
1. Determine ℒ{H(t – 1)} where in this case, F(s) = ℒ {1} and c = 1
ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
1 ℒ{H(t – 1)} = e − s s =
from (i) of Table 95.1, page 1023
e− s s
2. Determine ℒ{7 H(t – 3)} ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {7} and c = 3
7 ℒ{7 H(t – 3) } = e − 3 s s =
from (ii) of Table 95.1, page 1023
7 e −3 s s
3. Determine ℒ{H(t – 2).(t – 2) 2 } ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {t 2 } and c = 2
2! ℒ{H(t – 2).f(t – 2) 2 } = e − 2 s s3 =
from (vii) of Table 95.1, page 1023
2 e− 2 s s3
4. Determine ℒ{H(t – 3).sin(t – 3)} ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {sin t} and c = 3
1 ℒ{H(t – 3).sin(t – 3) } = e − 3 s s 2 + 12 =
from (iv) of Table 95.1, page 1023
e− 3s s2 + 1
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5. Determine ℒ{H(t – 4). et − 4 } where in this case, F(s) = ℒ {et } and c = 4
ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
1 ℒ{H(t – 4). et − 4 } = e − 4 s s −1 =
from (iii) of Table 95.1, page 1023
e− 4 s s −1
6. Determine ℒ{H(t – 5).sin 3(t – 5)} where in this case, F(s) = ℒ {sin 3t} and c = 5
ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
3 ℒ{H(t – 5).sin 3(t – 5) } = e − 5 s s 2 + 32 =
from (iv) of Table 95.1, page 1023
3e − 5 s s2 + 9
7. Determine ℒ{H(t – 1).(t – 1) 3 } ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {t 3 } and c = 1
3! ℒ{H(t – 1).(t – 1) 3 } = e − s s 3+1 =
from (viii) of Table 95.1, page 1023
6 e− s s4
8. Determine ℒ{H(t – 6).cos 3(t – 6)} ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {cos 3t} and c = 6
s ℒ{H(t – 6).cos 3(t – 6)} = e − 6 s s 2 + 32 s e− 6 s = s2 + 9
from (v) of Table 95.1, page 1023
9. Determine ℒ{5 H(t – 5).sinh 2(t – 5)} 1480
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ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {sinh 2t} and c = 5
2 ℒ{5 H(t – 5).sinh 2(t – 5) } = 5 e − 5 s s 2 − 22
from (x) of Table 95.1, page 1023
10 e − 5 s s2 − 4
=
π π 10. Determine ℒ{ H t − .cos 2 t − } 3 3
where in this case, F(s) = ℒ {cos 2t} and c =
ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
π s − s π π ℒ{ t − . cos 2 t − } = e 3 3 3 s 2 + 22 −
π
π 3
from (v) of Table 95.1, page 1023
s
se 3 = s2 + 4 11. Determine ℒ{2 H(t – 3). et − 3 } ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {et } and c = 3
1 ℒ{2 H(t – 3). et −3 } = 2 e −3 s s −1 =
from (iii) of Table 95.1, page 1023
2 e− 3 s s −1
12. Determine ℒ{3 H(t – 2).cosh(t – 2)} ℒ{H(t – c).f(t – c)} = e − c s F(s) Hence,
where in this case, F(s) = ℒ {cosh t} and c = 2
s ℒ{3 H(t – 2).cosh(t – 2) } = 3 e − 2 s s 2 − 12 =
from (ix) of Table 95.1, page 1023
3 s e− 2 s s2 −1
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EXERCISE 359 Page 1045
e− 9 s 1. Determine ℒ −1 s
Part of the numerator corresponds to e − c s where c = 9. This indicates H(t – 9) Then
1 = F(s) = ℒ{1} from (i) of Table 97.1, page 1033 s
Hence,
e− 9 s ℒ −1 = H(t – 9) s
4 e− 3 s 2. Determine ℒ −1 s
Part of the numerator corresponds to e − c s where c = 3. This indicates H(t – 3) Then
4 = F(s) = ℒ{4} from (ii) of Table 97.1, page 1033 s
Hence,
4 e− 3 s ℒ −1 = 4 H(t – 3) s
2 e− 2 s 3. Determine ℒ −1 s2
The numerator corresponds to e − c s where c = 2. This indicates H(t – 2) 1 = F(s) = ℒ{t} from (vii) of Table 97.1, page 1033 s2 Then
2 e− 2 s ℒ −1 = 2 H(t – 2).(t – 2) s2
5e − 2 s 4. Determine ℒ −1 s2 + 1
Part of the numerator corresponds to e − c s where c = 2. This indicates H(t – 2)
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5 s2 + 1
Then
Hence,
1 may be written as: 5 s 2 + 12
1 5 = F(s) = ℒ{5 sin t} from (iv) of Table 97.1, page 1033 s 2 + 12 5e − 2 s ℒ −1 = H(t – 2).5 sin(t – 2) = 5 H(t – 2).sin(t – 2) s2 + 1
3 s e− 4 s 5. Determine ℒ −1 s 2 + 16
Part of the numerator corresponds to e − c s where c = 4. This indicates H(t – 4) 3s s + 16 2
Then
Hence,
s may be written as: 3 s 2 + 42 s 3 = F(s) = ℒ{3 cos 4t} from (v) of Table 97.1, page 1033 s 2 + 42 3 s e− 4 s ℒ −1 =H(t – 4).3 cos 4(t – 4) = 3 H(t – 4).cos 4(t – 4) s 2 + 42
6 e− 2 s 6. Determine ℒ −1 s2 −1
Part of the numerator corresponds to e − c s where c = 2. This indicates H(t – 2) 6 s2 −1
Then
Hence,
1 may be written as: 6 s 2 − 12 1 6 = F(s) = ℒ{6 sinh t} from (x) of Table 97.1, page 1033 s 2 − 12 6 e− 2 s ℒ −1 =H(t – 2).6 sinh (t – 2) = 6 H(t – 2).sinh (t – 2) s2 −1
3e − 6 s 7. Determine ℒ −1 s3
The numerator corresponds to e − c s where c = 6. This indicates H(t – 6)
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1 1 = F(s) = ℒ t 2 from (viii) of Table 97.1, page 1033 s3 2
Then
1 3e − 6 s ℒ −1 = 3 H(t – 6). (t – 6) 2 = 1.5 H(t – 6). (t – 6) 2 2 s2
2 s e− 4 s 8. Determine ℒ −1 s 2 − 16
Part of the numerator corresponds to e − c s where c = 4. This indicates H(t – 4) 2s s − 16 2
Then
Hence,
s may be written as: 2 s 2 − 42 s 2 = F(s) = ℒ{2 cosh 4t} from (ix) of Table 97.1, page 1033 s 2 − 42 2 s e− 4 s ℒ −1 =H(t – 4).2 cosh 4(t – 4) = 2 H(t – 4).cosh 4(t – 4) s 2 − 42
1 − s 2 2 s e 9. Determine ℒ −1 2 s + 5
1 1 . This indicates H t − 2 2
Part of the numerator corresponds to e − c s where c = 2s s2 + 5
Then
Hence,
s may be written as: 2 2 s + 5
( )
s 2 s2 + 5
( )
1 − s 2 2 s e ℒ −1 s2 + 5
2
= F(s) = ℒ 2 cos 5 t 2
( )
{
}
from (v) of Table 97.1, page 1033
1 1 1 1 = H t − .2 cos 5 t − = 2 H t − .cos 5 t − 2 2 2 2 2
4 e− 7 s 10. Determine ℒ −1 s −1 Part of the numerator corresponds to e − c s where c = 7. This indicates H(t – 7) 1484
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Then
1 = F(s) = ℒ{ et } from (iii) of Table 97.1, page 1033 s −1
Hence,
4 e− 7 s ℒ −1 = 4 H(t – 7). et −7 s −1
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