Bird J. Electrical Circuit Theory and Technology
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918 Electrical Circuit Theory and Technology
Problem 10. Use Table 45.1 to determine the Laplace transforms
of the following waveforms:
(a) a step voltage of 10 V which starts at time t D 0
(b) a step voltage of 10 V which starts at time t D 5s
(c) a ramp voltage which starts at zero and increases at 4 V/s
(d) a ramp voltage which starts at time t D 1 s and increases at
4V/s
(a) From 2 of Table 45.1, Lf10gD10Lf1gD10
1
s
D
10
s
The waveform is shown in Figure 45.10 (a).
(b) From 4 of Table 45.1, a step function of 10 V which is delayed by
t D 5 s is given by:
10
e
sT
s
D 10
e
5s
s
D
10
s
e
−5s
This is, in fact, the function starting at t D 0 given in part (a), i.e.,
10/s multiplied by e
sT
, where T is the delay in seconds.
The waveform is shown in Figure 45.10 (b).
(c) From 7 of Table 45.1, the Laplace transform of the unit ramp,
LftgD1/s
2
Hence the Laplace transform of a ramp voltage increasing at 4 V/s
is given by:
4
LftgD
4
s
2
The waveform is shown in Figure 45.10(c).
(d) As with part (b), for a delayed function, the Laplace transform is
the undelayed function, in this case 4/s
2
from part (c), multiplied
by e
sT
where T in this case is 1 s. Hence the Laplace transform is
given by: .4=s
2
/e
−s
The waveform is shown in Figure 45.10 (d).
Problem 11. Determine the Laplace transforms of the following
waveforms:
(a) an impulse voltage o
f 8 V which starts at time
t D 0
(b) an impulse voltage o
f 8 V which starts at time
t D 2s
(c) a sinusoidal current of 4 A and angular frequency 5 rad/s
which starts at time t D 0
0
10
V
t
(a)
0
V
(c)
4
0
V
(d)
4
1
t
1
t
0
10
V
(b)
5
t
2
Figure 45.10
Transients and Laplace transforms 919
V
8
0
t
(a)
V
8
0
t
(b)
2
0
4
−4
π
t
5
2π
5
(c)
i
Figure 45.11
(a) An impulse is an intense signal of very short duration. This function
is often known as the Dirac function.
From 1 of Table 45.1, the Laplace transform of an impulse starting
at time t D 0 is given by
LfυgD1, hence an impulse of 8 V is
given by: 8
LfυgD8
This is shown in Figure 45.11 (a).
(b) From part (a) the Laplace transform of an impulse of 8 V is 8.
Delaying the impulse by 2 s involves multiplying the undelayed
function by e
−sT
where T D 2s.
Hence the Laplace transform of the function is given by: 8 e
2s
This is shown in Figure 45.11(b).
(c) From 5 of Table 45.1,
LfsinωtgD
ω
s
2
C ω
2
When the amplitude i
s4Aand
ω D 5, then
Lf4sinωtgD4
5
s
2
C 5
2
D
20
s
2
Y 25
The waveform is shown in Figure 45.11 (c).
Problem 12. Find the Laplace transforms of (a) 2t
4
e
3t
(b) 4e
3t
cos5t
(a) From 12 of Table 45.1,
Lf2t
4
e
3t
gD2Lft
4
e
3t
g
D 2
4!
s 3
4C1
D
24 ð 3 ð 2 ð 1
s 3
5
D
48
.s − 3/
5
(b) From 14 of table 45.1,
Lf4e
3t
cos5tgD4Lfe
3t
cos5tg
D 4
s 3
s 3
2
C 5
2
D
4s 3
s
2
6s C 9 C25
D
4.s
− 3/
s
2
− 6s Y 34
Problem 13. Determine the Laplace transforms of (a) 2cosh3t
(b) e
2t
sin3t
(a) From 10 of Table 45.1,
Lf2 cosh3tgD2Lcosh3t D 2
s
s
2
3
2
D
2s
s
2
− 9
920 Electrical Circuit Theory and Technology
(b) From 13 of Table 45.1,
Lfe
2t
sin3tgD
3
s C 2
2
C 3
2
D
3
s
2
C 4s C 4 C9
D
3
s
2
Y 4s Y 13
Laplace transforms of derivatives
Using integration by parts, it may be shown that:
(a) for the first derivative:
Lff
.t/gDsLff .t/g − f .0/
or
L
dy
dx
D sLfyg − y.0/ 45.33
where y0 is the value of y at x D 0
(b) for the second derivative:
Lff
.t/gDs
2
Lff .t/g − sf .0/ f
.0/
or
L
d
2
y
dx
2
D s
2
Lfyg − sy.0/ − y
.0/ 45.34
where y
0
0 is the value of dy/dx at x D 0
Equations (45.33) and (45.34) are used in the solution of differential equa-
tions in Section 45.6.
The initial and final value theorems
The initial and final value theorems can often considerably reduce the
work of solving electrical circuits.
(a) The initial value theorem states: limit
t!0
[ft] D limit
s!1
[sLfftg]
Thus, for example, if ft D
v D Ve
t
CR
and if, say,
V D 10 and CR D 0.5, then
ft D
v D 10e
2t
LfftgD10
1
s C 2
from 3 of Table 45.1
s
LfftgD10
s
s C 2
From the initial value theorem, the initial value of ft is given by:
10
1
1C2
D 101 D 10
Transients and Laplace transforms 921
(b) The final value theorem states: limit
t!1
[ft] D limit
s!0
[sLfftg]
In the above example of ft D 10e
2t
the final value is given by:
10
0
0 C 2
D 0
The initial and final value theorems are used in pulse circuit applica-
tions where the response of the circuit for small periods of time, or the
behaviour immediately the switch is closed, are of interest. The final value
theorem is particularly useful in investigating the stability of systems
(such as in automatic aircraft-landing systems) and is concerned with the
steady state response for large values of time t, i.e., after all transient
effects have died away.
Further problems on Laplace transforms may be found in Section 45.10,
problems 9 to 19, page 953.
45.6 Inverse Laplace
transforms and the
solution of differential
equations
Since from 2 of Table 45.1, Lf1gD
1
s
then
L
1
1
s
= 1
where
L
1
means the inverse Laplace transform.
Similarly, since from 5 of Table 45.1,
LfsinωtgD
ω
s
2
C ω
2
then L
1
!
s
2
Y !
2
= sin!t
Thus finding an inverse transform involves locating the Laplace transform
from the right-hand column of Table 45.1 and then reading the function
from the left-hand column. The following worked problems demonstrate
the method.
Problem 14. Find the following inverse Laplace transforms:
(a)
L
1
1
s
2
C 9
(b) L
1
5
3s 1
(a) L
1
1
s
2
C 9
D L
1
1
s
2
C 3
2
D
1
3
L
1
3
s
2
C 3
2
and from 5 of Table 45.1,
1
3
L
1
3
s
2
C 3
2
D
1
3
sin3t
(b)
L
1
5
3s 1
DL
1
5
3s
1
3
D
5
3
L
1
1
s
1
3
D
5
3
e
1
3
t
from 3 of Table 45.1
922 Electrical Circuit Theory and Technology
Problem 15. Determine the following inverse Laplace transforms:
(a)
L
1
6
s
3
(b) L
1
3
s
4
(a) From 8 of Table 45.1, L
1
2
s
3
D t
2
Hence L
1
6
s
3
D 3L
1
2
s
3
D 3t
2
(b) From 9 of Table 45.1, if s is to have a power of 4 then n D 3.
Thus
L
1
3!
s
4
D t
3
, i.e., L
1
6
s
4
D t
3
Hence L
1
3
s
4
D
1
2
L
1
6
s
4
D
1
2
t
3
Problem 16. Determine (a) L
1
7s
s
2
C 4
(b) L
1
4s
s
2
16
(a) L
1
7s
s
2
C 4
D 7L
1
s
s
2
C 2
2
D 7cos2t from 6 of Table 45.1
(b)
L
1
4s
s
2
16
D 4L
1
s
s
2
4
2
D 4 cosh 4t from 10 of Table 45.1
Problem 17. Find L
1
2
s 3
5
From 12 of Table 45.1, L
1
n!
s a
nC1
D e
at
t
n
Thus L
1
1
s a
nC1
D
1
n!
e
at
t
n
and comparing with L
1
2
s 3
5
shows that n D 4anda D 3.
Hence
L
1
2
s 3
5
D 2L
1
1
s 3
5
D 2
1
4!
e
3t
t
4
D
1
12
e
3t
t
4
Transients and Laplace transforms 923
Problem 18. Determine
(a)
L
1
3
s
2
4s C 13
(b) L
1
2s C 1
s
2
C 2s C 10
(a) L
1
3
s
2
4s C 13
D L
1
3
s 2
2
C 3
2
D e
2t
sin3t from 13 of Table 45.1
(b)
L
1
2s C 1
s
2
C 2s C 10
D L
1
2s C 1
s C 1
2
C 3
2
D 2e
−t
cos3t from 14 of Table 45.1
Note that in solving these examples the denominator in each case has
been made into a perfect square.
Use of partial fractions for inverse Laplace transforms
Sometimes the function whose inverse is required is not recognisable as a
standard type, such as those listed in Table 45.1. In such cases it may be
possible, by using partial fractions, to resolve the function into simpler
fractions which may be inverted on sight.
For example, the function Fs D
2s 3
ss 3
cannot be inverted on sight
from Table 45.1. However, using partial fractions:
2s 3
ss 3
A
s
C
B
s 3
D
As 3 C Bs
ss 3
from which, 2s 3 D As 3 C Bs
Letting s D 0 gives: 3 D3A from which A D 1
Letting s D 3 gives: 3 D 3B from which B D 1
Hence
2s 3
ss 3
1
s
C
1
s 3
Thus
L
1
2s 3
ss 3
D L
1
1
s
C
1
s 3
D 1 Y e
3t
from 2 and 3 of Table 45.1
Partial fractions are explained in Engineering Mathematics and Higher
Engineering Mathematics. The following worked problems demonstrate
the method.
Problem 19. Determine L
1
4s 5
s
2
s 2
924 Electrical Circuit Theory and Technology
4s 5
s
2
s 2
4s 5
s 2s C 1
A
s 2
C
B
s C 1
D
As C 1 C Bs 2
s 2s C 1
Hence 4s 5 D As C 1 C Bs 2
When s D 2, 3 D 3A from which, A D 1
When s D1, 9 D3B from which, B D 3
Hence
L
1
4s 5
s
2
s 2
L
1
1
s 2
C
3
s C 1
D L
1
1
s 2
C L
1
3
s C 1
D e
2t
Y 3e
−t
from 3 of Table 45.1
Problem 20. Find L
1
3s
3
C s
2
C 12s C 2
s 3s C 1
3
3s
3
C s
2
C 12s C 2
s 3s C 1
3
A
s 3
C
B
s C 1
C
C
s C 1
2
C
D
s C 1
3
D
As C 1
3
C Bs 3s C 1
2
CCs 3s C 1 C Ds 3
s 3s C 1
3
Hence 3s
3
C s
2
C 12s C 2 D As C 1
3
C Bs 3s C 1
2
C Cs 3s C 1 C Ds 3
When s D 3, 128 D 64A from which, A D 2
When s D1, 12 D4D from which, D D 3
Equating s
3
terms gives: 3 D A C B from which, B D 1
Equating s
2
terms gives: 1 D 3A B C C from which, C D4
Hence
L
1
3s
3
C s
2
C 12s C 2
s 3s C 1
3
L
1
2
s 3
C
1
s C 1
4
s C 1
2
C
3
s C 1
3
D 2e
3t
Y e
−t
− 4e
−t
t Y
3
2
e
−t
t
2
from 3 and 12 of Table 45.1
Transients and Laplace transforms 925
Problem 21. Determine L
1
5s
2
C 8s 1
s C 3s
2
C 1
5s
2
C 8s 1
s C 3s
2
C 1
A
s C 3
C
Bs C C
s
2
C 1
D
As
2
C 1 C Bs C Cs C 3
s C 3s
2
C 1
Hence 5s
2
C 8s 1 D As
2
C 1 C Bs C Cs C 3
When s D320D 10A from which, A D 2
Equating s
2
terms gives: 5 D A C B from which, B D 3
Equating s terms gives: 8 D 3B C C from which, C D1
Hence
L
1
5s
2
C 8s 1
s C 3s
2
C 1
L
1
2
s C 3
C
3s 1
s
2
C 1
D L
1
2
s C 3
C L
1
3s
s
2
C 1
L
1
1
s
2
C 1
D 2e
−3t
Y 3cost − sint
from3,6and5ofTable45.1
Procedure to solve differential equations by using Laplace
transforms
(i) Take the Laplace transform of both sides of the differential equation
by applying the formulae for the Laplace transforms of derivatives
(i.e., equations (45.33) and (45.34) on page 920) and, where neces-
sary, using a list of standard Laplace transforms, such as Table 45.1
on page 917.
(ii) Put in the given initial conditions, i.e. y0 and y
0
0
(iii) Rearrange the equation to make
Lfyg the subject
(iv) Determine y by using, where necessary, partial fractions, and taking
the inverse of each term by using Table 45.1.
This procedure is demonstrated in the following problems.
Problem 22. Use Laplace transforms to solve the differential
equation
2
d
2
y
dx
2
C 5
dy
dx
3y D 0, given that when x D 0,y D 4and
dy
dx
D 9
926 Electrical Circuit Theory and Technology
(i) 2L
d
2
y
dx
2
C 5L
dy
dx
3LfygDLf0g
2[s
2
Lfygsy0 y
0
0] C 5[sLfygy0] 3LfygD0
from equations (45.33) and (45.34)
(ii) y0 D 4andy
0
0 D 9
Thus 2[s
2
]Lfyg4s 9] C5[sLfyg4] 3LfygD0
i.e., 2s
2
Lfyg8s 18 C 5sLfyg20 3LfygD0
(iii) Rearranging gives: 2s
2
C 5s 3LfygD8s C 38
i.e.,
LfygD
8s C 38
2s
2
C 5s 3
(iv) y D
L
1
8s C 38
2s
2
C 5s 3
Let
8s C 38
2s
2
C 5s 3
D
8s C 38
2s 1s C 3
D
A
2s 1
C
B
s C 3
D
As C 3 C B2s 1
2s 1s C 3
Hence 8s C 38 D As C 3 C B2s 1
When s D
1
2
, 42 D 3
1
2
A from which, A D 12
When s D3, 14 D7B from which, B D2
Hence y D
L
1
8s C 38
2s
2
C 5s 3
L
1
12
2s 1
2
s C 3
D L
1
12
2s
1
2
L
1
2
s C 3
Hence y= 6e
.1=2/x
− 2e
−3x
from 3 of Table 45.1.
Problem 23. Use Laplace transforms to solve the differential
equation:
d
2
y
dx
2
C 6
dy
dx
C 13y D 0, given that when x D 0,y D 3and
dy
dx
D 7
Using the above procedure:
(i)
L
d
2
y
dx
2
C 6L
dy
dx
C 13LfygDLf0g
Hence [s
2
Lfygsy0 y
0
0] C 6[sLfygy0] C13LfygD0
from equations (45.33) and (45.34)
Transients and Laplace transforms 927
(ii) y0 D 3andy
0
0 D 7
Thus s
2
Lfyg3s 7 C 6sLfyg18 C 13LfygD0
(iii) Rearranging gives: s
2
C 6s C 13LfygD3s C 25
i.e.
LfygD
3s C 25
s
2
C 6s C 13
(iv) y D
L
1
3s C 25
s
2
C 6s C 13
D L
1
3s C 25
s C 3
2
C 2
2
D L
1
3s C 3 C 16
s C 3
2
C 2
2
D L
1
3s C 3
s C 3
2
C 2
2
C L
1
82
s C 3
2
C 2
2
D 3e
3t
cos2t C 8e
3t
sin2t
from 14 and 13 of Table 45.1, page 917.
Hence y
= e
−3t
.3cos2t Y 8sin2t/
Problem 24. A step voltage is applied to a series C–R circuit.
When the capacitor is fully charged the circuit is suddenly broken.
Deduce, using Laplace transforms, an expression for the capacitor
voltage during the transient period if the voltage when the supply
is cut is V volts.
From Figure 45.1, page 902, v
R
C v
C
D 0 when the supply is cut
i.e., iR C
v
c
D 0
i.e.,
C
d
v
c
dt
R C v
c
D 0
i.e., CR
d
v
c
dt
C
v
c
D 0
Using the procedure:
(i)
L
CR
d
v
c
dt
C Lfv
c
gDLf0g
i.e., CR[s
Lfv
c
gv
0
] C Lfv
c
gD0
(ii)
v
0
D V, hence CR[sLfv
c
gV] C Lfv
c
gD0
(iii) Rearranging gives: CRs
Lfv
c
gCRV C Lfv
c
gD0
i.e., CRs C 1
Lfv
c
gDCRV
hence
Lfv
c
gD
CRV
CRs C 1