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)}
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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)
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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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