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SCHOOL OF ENGINEERING & BUILT ENVIRONMENT
Mathematics
Laplace Transforms 1. Introduction ……………………………………. 1 2. Definition ………………………………………. 2 3. Properties of the Laplace Transform …………… 4 4. Inverse Laplace Transforms …………………… 6 5. Using Laplace Transforms to Solve ODEs ……. 9 6. The Step Function ……………………………... 11 7. The Dirac Delta Function ……………………… 20 8. Convolution ……………………………………. 25 9. A Note on Control Theory …………………….. 26 Tutorial Exercises ……………………………… 27 Answers to Tutorial Exercises …………………. 33
Dr Derek Hodson
1
LTNOTES_Rev/D Hodson Laplace Transforms 1) Introduction The idea of transforming a “difficult” problem into an “easier” problem is one that is used widely in mathematics. Diagrammatically we have
Problem Transformed
Problem
Solution Transformed Solution
Transform
Solve
Inverse Transform
Difficult
There are many types of transforms available to mathematicians, engineers and scientists. We are going to examine one such transformation, the Laplace transform, which can be used to solve certain types of differential equations and also has applications in control theory. 2) Definition The Laplace transform operates on functions of t . Given a function we define its Laplace transform as
)( tf
∫∞ −=0
)(])([ dtetftfL ts
where s is termed the Laplace variable. We sometimes denote the Laplace transform by or )( sF f . We start off with a function of t and end up with a function of s ; s is, in fact, a complex variable, but this need not concern us too much. Because the function of t is often some form of time signal, we often talk about moving from the time domain to the Laplace domain when we perform a Laplace transformation. Note: Laplace transforms are only concerned with functions where . 0≥t
2
Examples (1) : 1)( =tf
∞=
=
−
∞ −
⎥⎦⎤
⎢⎣⎡ −=
= ∫t
t
ts
ts
es
dteL
0
0
1
.1]1[
Strictly speaking, we can’t set t equal to ∞, but we can “take the limit” as t heads
towards infinity. Providing , thereby ensuring that we have a negative exponential, the limit of the inside of the square brackets as t tends to infinity will be zero. Also, since , this leaves us with
0>s
10 =e
ss
L 110]1[ =⎟⎠⎞
⎜⎝⎛ −−= .
So → 1)( =tfs
L 1]1[ =
(2) ttf =)( :
∫∞ −=
0.][ dtettL ts
This integral requires integration by parts to complete the process, but with the same
assumptions regarding s as before, it can readily be shown that
2
1][s
tL = .
So → ttf =)( 2
1][s
tL =
Very quickly the integrations required to complete the Laplace transformation become difficult and messy. For this reason, we generally work from a table of pre-determined Laplace transforms (see Appendix).
3
Examples Using Table of Laplace Transforms (3) (a) Determine . ][ 3tL isn’t in the table explicitly, but is: 3t 1−nt
nn
sn
tL)!1(
][ 1 −=−
For we require : 3t 4=n
4443 6123!3][
ssstL =
××==
(b) Determine . ][ 2 teL −
From table:
α
α
+=−
seL t 1][ .
Set 2=α :
2
1][ 2
+=−
seL t
(c) Determine . ])4(sin[ tL From table:
22])(sin[ω
ωω+
=s
tL .
Set 4=ω :
16
44
4])4(sin[ 222 ++ ss==tL
4
3) Properties of the Laplace Transform The Laplace transform has several special properties that make it a useful mathematical tool. We consider some of these now. a) Linearity Suppose we have two functions along with their respective Laplace transforms:
)(])([ sFtfL = )(])([ sGtgL = . The property of linearity means that )()(])()([ sGbsFatgbtfaL +=+ providing a and b are constants. This makes the transformation of a string of functions straightforward. Example
(4) 222222
3232])(sin3)(cos2[
ωω
ωω
ωωω
++
=+
++
=+ss
sssttL
Warning: The Laplace transform of a product is NOT EQUAL TO the product of
the individual Laplace transforms. We have to invoke other properties of the Laplace transform to deal with such.
b) The First Shifting Theorem Suppose a function has the Laplace transform . It is easily demonstrated that )( tf )( sF . )(])([ αα +=− sFtfeL t
Example (5) From tables and Example (3)(a) we have
43 6][
stL = . [1]
By the first shifting property
432
)2(6][+
=−
steL t . [2]
To obtain [2] from [1] we merely replace by s )2( +s .
5
c) Transformation of Derivatives As before, denote the Laplace transform (LT) of by . Now consider the LT of the derivative of , denoted by :
)( tf )( sF)( tf )( tf&
. dtetftfL ts−∞
∫= )(])([0
&&
Integrate by parts (integrating and differentiating ): )( tf& tse −
[ ] ∫∞ −∞− −−=
00 )()()( dtestfetf tsts
∫∞ −+−=
0)()0(0 dtetfsf ts
)()0( sFsf +−= . So . )
)
0()(])([ fsFstfL −=&
We have expressed the Laplace transform of a derivative in terms of the Laplace transform of the undifferentiated function. In effect, the Laplace transform has converted the operation of differentiation into the simpler operation of multiplication by s . In a similar fashion, using repeated integration by parts, we can show that . 0()0()(])([ 2 ffssFstfL &&& −−= This is one of the most important properties of the Laplace transform. The Laplace transform “gets rid of” derivatives; just the thing for solving differential equations! When we come to solve differential equations using Laplace transforms we shall use the following alternative notation: xxL =][ )0(][ xxsxL −=& )0()0(][ 2 xxsxsxL &&& −−= . However, before we can solve differential equations, we need to look at the reverse process of finding functions of t from given Laplace transforms.
6
4) Inverse Laplace Transforms So far, we have looked at how to determine the LT of a function of t , ending up with a function of s . The table of Laplace transforms collects together the results we have considered, and more. When we apply Laplace transforms to solve problems we will have to invoke the inverse transformation. That is, given a Laplace transform we will want to determine the corresponding . In general we have
)( sF)( tf
∫∞+
∞−
− =j
j
ts dsesFj
sFLγ
γπ)(
21])([1 ,
where the evaluation of the integral requires a knowledge of complex analysis, which is too difficult to consider here. Instead, we shall rely on the table of Laplace transforms used in reverse to provide inverse Laplace transforms. This will mean manipulating a given Laplace transform until it looks like one or more entries in the right of the table. The inverse is then determined from the left of the table. The following examples illustrate the main algebraic techniques required. These include completing the square, factorisation and the formation of partial fractions. See separate documents for the details of completing the square and partial fractions. Examples
(6) Invert the Laplace transform 6
3s
.
The closest entry in the table gives:
1)!1( −→− n
n ts
n .
Setting 6=n :
56
!5 ts
→ . [Note: 12012345!5 =××××= ], so 56
120 ts
→ .
In the given Laplace transform there is a 3 on the top; we would like there to be a 120 to
match the table entry. We can re-write the transform providing we don’t alter its “net value”:
⎥⎦⎤
⎢⎣⎡=== 6666
120401120
1203133
ssss .
The term in the square brackets is now exactly the table entry so we can invert that and
simply multiply by the fraction in front:
56s
→4013 t .
7
(7) Invert 52
12 ++ ss
.
This requires the technique of “completing the square” and a little bit of fine tuning to
re-write it in a form that can be inverted from tables. Note that the closest entry in the table gives:
)(sin)( 22 te
st ω
ωαω α−→
++ .
Now complete the square in the denominator: . 4)1(51)1(52 222 ++=+−+=++ ssss The given transform becomes:
2222 2)1(1
4)1(1
521
++=
++=
++ ssss .
This is now very close to the table entry with 1=α and 2=ω . We would like there
to be a 2 on top so “fine tuning” gives
⎥⎦
⎤⎢⎣
⎡++
=++ 2222 2)1(
221
2)1(1
ss ,
where the term in the square brackets is exactly the table entry with 1=α and 2=ω .
Inverting this and multiplying by the fraction gives
)2(sin2)1(
221
21
22 tes
t−→⎥⎦
⎤⎢⎣
⎡++
.
(8)(a) Invert 32
52 −−
+ss
s .
Factorising the denominator and splitting the result into its partial fractions deals with
this one. Note that the details of the partial fraction expansion have been omitted.
1
13
2)1()3(
532
52 +
−−
=+−
+=
−−+
sssss
sss
Use tes
α
α−→
+ )(1 twice with 3−=α and 1+=α to give
tt eess +− 13
−−→− 3212
8
(8)(b) Invert 32
52 −−
+ss
s by completing the square.
If we failed to notice that the denominator factorised, them completing the square is still
an option:
22
2
22
2)1(5
4)1(5
31)1(5
325
−−+
=
−−+
=
−−−+
=−−
+
ss
ss
ss
sss
The minus sign in the denominator is very significant since it no longer conforms to . 22)( ωα ++s Instead we need the following results:
22)(])(sinh[
βαββα
−+=−
steL t 22)(
])(cosh[βα
αβα
−++
=−
ssteL t
where sinh (pronounced “shine”) and cosh (pronounced “cosh”) are the so-called
hyperbolic functions. Fine tuning the Laplace transform a little bit more gives
.2)1(
232)1(
)1(2)1(6)1(
2)1(5
22222222 ⎥⎦
⎤⎢⎣
⎡−−
+⎥⎦
⎤⎢⎣
⎡−−
−=
−−+−
=−−
+ss
ss
ss
s
Inverting now gives
.])2(sinh3)2(cosh[
)2(sinh3)2(cosh2)1(
232)1(
)1(2222
tte
tetess
s
t
tt
+=
+→⎥⎦
⎤⎢⎣
⎡−−
+⎥⎦
⎤⎢⎣
⎡−−
−
+
++
By applying the following mathematical identities: ][)(cosh 2
1 tt eet βββ −+ += and ][)(sinh 21 tt eet βββ −+ −=
we can easily show that the two versions of the answer are equivalent.
9
5) Using Laplace Transforms to Solve ODEs We have seen how the Laplace transform of the derivative of a function can be expressed in terms of the Laplace transform of the undifferentiated function. We can use this property to derive solutions to certain types of differential equations. The process is broken down into the following steps:
• Transform both sides of the ODE; • Substitute initial values; • Solve for x ; • Manipulate into a form that can be inverted from tables; • Invert to give the solution of the ODE.
The method is best illustrated by example and we shall need the results from earlier: xxL =][ )0(][ xxsxL −=& )0()0(][ 2 xxsxsxL &&& −−= . Example (9) Consider the ODE texxx 3852 −=++ &&&
Subject to the initial conditions . 0)0()0( == xx &
Transform both sides of the equation:
3
85])0([2])0()0([ 2
+=+−+−−
sxxxsxxsxs & .
Substitute initial values:
3
85]0[2]00[ 2
+=+−+−−
sxxssxs
3
8)52( 2
+=++
sxss .
10
Solve for x :
)52()3(
82 +++
=xss
x .
Manipulate into a form that can be inverted from tables. In this case we form partial
fractions, complete the square on the quadratic denominator and finish up with a little “fine tuning” of the last term:
⎥⎦
⎤⎢⎣
⎡++
+⎥⎦
⎤⎢⎣
⎡++
−⎥⎦
⎤⎢⎣
⎡+
= 2222 2)1(2
21
2)1(31
sss
sx .
Finally invert term-by-term and tidy-up:
.])2(sin)2(cos[
)2(sin])2(sin)2(cos[)(3
21
213
ttee
tetteetxtt
ttt
−−=
+−−=−−
−−−
Further Example (10) Solve subject to 125 =+ xx&& 10)0( =x and 0)0( =x& . Transform:
s
xxxsxs 125])0()0([ 2 =+−− &
Substitute initial conditions:
s
xsxs 125]010[ 2 =+−−
Solve for x :
ss
xxs 101252 +=+
ss
xs 101)25( 2 +=+
)25(
10)25(
122 +
++
=s
sss
x
11
Manipulate and invert:
⎥⎦
⎤⎢⎣
⎡+
+⎥⎦
⎤⎢⎣
⎡+
=)25(
10)25(
25251
22 ss
ssx
)5(cos10])5(cos1[)( 25
1 tttx +−= Using Laplace transforms to solve ODEs of the form
)( tfxcxbxa =++ &&& allows us to tackle problems where a solution by the method of undetermined coefficients (say) is rendered difficult or impossible because of the specific type of function on the right-hand-side. We shall now look at two such functions for which this is the case. 6) The Step Function The step function is defined as
. ⎩⎨⎧
><
=−TtTt
Ttu,1,0
)(
Its graph is shown below in Figure 1:
)( Ttu −
Ot
1
T
Figure 1 T is called the critical value of the step function; it is where the function changes value. The value of the step function at is not defined above. Some authors define the value as 1, others define it as 0 or 0.5. We shall not worry about it!
Tt =
Note: In the notation for the step function, the u is NOT multiplying the bracket. The u is the short-hand name for the function (full name: unit step function) and the bracket contains information regarding the function variable (usually t ) and the location of the step.
12
The step function is like a mathematical switch. It can be used to model situations where something occurs suddenly, such as when a circuit is switched on, or a voltage is instantaneously modified, or a force is suddenly applied or removed. This may be done by taking combinations of step functions with different critical values, or even combining them with more conventional functions, as the following examples illustrate. Note: Since Laplace transforms are only concerned with functions where , if the step
occurs at then 0≥t
0=t 1)0( ≡−tu . Examples (11) Graph )15()10(3)5(2)( −+−−−= tutututf . Thinking of the step functions as switches, for 5<t all step functions are “off” and so
are all zero; for the first step function is “on” and has the value 1; for the second step function now switches on to 1 also; and for all step
functions are “on” and equal to 1. This gives
105 << t1510 << t 15>t
⎪⎪⎩
⎪⎪⎨
⎧
><<−<<<<
=
15,01510,1105,250,0
)(
tttt
tf
and the graph
f ( t )
tO 5 10 15
2
−1
(12) / Over the page . . .
13
(12) Graph ])4()2([)4()2()( −−−−−= tututttf . We do this in two parts. Without the step functions, we would have . 86)4()2()( 2 −+−=−−= tttttf This graph is a parabola that crosses the horizontal axis at 2=t and and has a
maximum turning point at 4=t
)1,3( . The step functions take the following values:
. ⎪⎩
⎪⎨
⎧
>=−<<=−
<=−=−−−
4,01142,101
2,000)4()2(
tt
ttutu
When the two parts are multiplied together we get
⎪⎩
⎪⎨
⎧
><<
<=
4,042,parabola
2,0)(
tt
ttf
and the graph
f ( t )
tO 2 4
1
We can think of a difference of two step functions
)()( btuatu −−− like a mathematical window whose frame obscures the graph of any multiplying function to the sides, i.e. and . at < bt >
14
(13) Determine from the following graph: )( tf
f ( t )
tO 1 2 3
4
−2
2
The jumps in the graph can be represented by step functions. The location of the jump
is determined by the critical value of the step function, the size of the jump can be achieved by multiplying the step function be a suitably sized constant, the direction of the jump by the sign of the constant. Working from left to right:
1st jump: up by 4 at )1(4 −tu 1=T 2nd jump: down by 6 at )2(6 −− tu 2=T 3rd jump: up by 4 at )3(4 −tu 3=T . Adding these together gives us the function: )3(4)2(6)1(4)( −+−−−= tutututf . (14) / Over the page . . .
15
(14) Determine from the following graph: )( tf
f ( t )
tO 1 2
3
Using basic coordinate geometry, the equation of the oblique straight line is given by . 33 −= ty We are only seeing the straight line b en etwe 1=t and 2=t ; this gives a
“mathematica
l window” of [ ])2()1( −−− tutu . Multiplying the two parts together gives us our function: ])2()1([)33()( −−−−= tututtf . For more general piece-wise functions we can break them down into fragments, each with their own window, and add the parts together. Example
(15) ⎪⎩
⎪⎨
⎧
><<−<<
=2,0
21,210,
)(t
tttt
tf
])2()1([)2(])1()0([)( −−−−+−−−= tututtututtf As far as Laplace transforms are concerned 1)0( ≡−tu so this can be re-written as )2( −t . )2()1()1(2)( −+−−−= uttutttf
16
There are two results that we will need in order to solve ODEs containing step functions. Result 1
Tses
TtuL −=−1])([
Proof
.)0(1
10
10
)(])([
0
0
>=
⎥⎦⎤
⎢⎣⎡ −+=
+=
−=−
−
∞−
∞ −−
∞ −
∫∫
∫
ses
es
dtedte
dteTtuTtuL
Ts
T
ts
T
tsT ts
ts
Result 2 – The Second Shifting Theorem Given a function with Laplace transform , it can be shown that )( tf )( sF TsesFTtuTtfL −=−− )(])()([ . The proof of this is omitted. The implication of this result is demonstrated graphically below in Figure 2:
O Ot t
T
f Shifted and truncated f
)()( tfsF → )()()( TtuTtfesF Ts −−→−
Figure 2
17
We shall use modified versions of the following diagram to help us apply the 2nd Shifting Theorem:
TsesFTtuTtf
sFtf
−↔−−
↔
)()()(
)()(bb
Examples
(16) Invert the Laplace transform sess
4
)2(1 −
+.
For this we are starting at the bottom-right of the diagram and moving round anti-
clockwise:
TsesFTtuTtf
sFtf
−−−
↑↓
←
)()()(
)()(
Step 1: Ignore exponential to give ⎥⎦
⎤⎢⎣
⎡+
=+
=)2(
221
)2(1)(
sssssF .
Step 2: Invert to give )( sF ]1[)( 2
21 tetf −−= .
Step 3: Shift and truncate with 4=T , i.e. replace by t )4( −t and multiply by
)4( −tu , to give the completed inversion )4(]1[)()( )4(2
21 −−=−− −− tueTtuTtf t .
18
(17) Solve the differential equation )3(4 −=+ tuxx&& subject to . 0)0()0( == xx &
Transform: ses
xxxsxs 32 14])0()0([ −=+−− &
Substitute initial conditions and solve for x :
ses
xs 32 1)4( −=+
sess
x 32 )4(1 −
+=
Invert with the help of the 2nd Shifting Theorem: Again, we are starting at the bottom-right of the diagram and moving round anti-
clockwise:
TsesFTtuTtf
sFtf
−−−
↑↓
←
)()()(
)()(
Step 1: Ignore exponential to give ⎥⎦
⎤⎢⎣
⎡+
=+
=)2(
241
)4(1)( 22
2
2 sssssF .
Step 2: Invert to give )( sF ])2(cos1[)( 4
1 ttf −= . Step 3: Shift and truncate with 3=T , i.e. replace by t )3( −t and multiply by
)3( −tu , to give )(( Tttf )uT −− and the completed solution )3(]))3(2(cos1[)( 4
1 −−−= tuttx . These two examples illustrate the use of the 2nd Shifting Theorem to invert Laplace transforms that contain an exponential factor. The next example shows the theorem working in the other direction.
19
(18) Determine . ])10([ 2 −tutL This time we go round the diagram the other way:
TsesFTtuTtf
sFtf
−−−
↓↑
→
)()()(
)()(
We have, however, a slight problem in that the given function of t does not quite conform to the structure bottom-left. We need to re-write the multiplying function so that the variable t always has a “minus T ”. For this example so we introduce this as follows:
2t10=T
. 100)10(20)10(]10)10([ 222 +−+−=+−= tttt The original function becomes which
now does conform. Next we use the theorem to transform: )10(]100)10(20)10([ 2 −+−+− tutt
Step 1: Cover up the step function and the “minus 10s” to leave . 10020)( 2 ++= tttf
Step 2: Transform to give sss
sF 100202)( 23 ++= .
Step 3: Everything we covered up in Step 1 is now accounted for by multiplying
by and so completes the transformation, )( sF se 10−
sesss
tutL 1023
2 100202])10([ −⎥⎦⎤
⎢⎣⎡ ++=− .
Now we shall look at another new type of function, one that is closely related to the step function.
20
7) The Dirac Delta Function Consider the graph of a function something like the following: )( tD
TT − ε T + εt
D ( t )
O
Figure 3
This graph might represent a force applied and then removed, or a voltage switched on and then off. We can then define another function as
∫=t
dzzDtU0
)()(
which records the area under the graph from 0 to the variable t . Assume that the function
is such that the total area under the graph is 1 unit)( tD 2. This means that
⎪⎩
⎪⎨
⎧
+≥+≤≤−
−≤=
εεε
ε
TtTtT
TttU
,1,1 toincreasing0,0
)(
and so the graph of will look like )( tU
TT − ε T + εt
U ( t )
O
1
Figure 4
21
Now suppose that the interval over which is non-zero is reduced but the height of the graph is increased in such a way that the total area under the graph is still 1 unit
)( tD2:
Tt
D ( t )
O
Figure 5
We get what might be termed a "blip". The corresponding graph of looks like )( tU
Tt
U ( t )
O
1
Figure 6
Notice that is beginning to look like the step function. If we let the interval about T shrink further towards zero and the height increase towards infinity (again, in such a way that the area under the graph is always 1 unit
)( tU
t =2), then, in the limit, we obtain an "instantaneous,
infinite blip" at . This "infinite blip" at T Tt = is called the Dirac delta function and is denoted by
)( Tt −δ ; it has the graph as shown in Figure 7 on the next page.
22
Tt
δ ( t − T )
O
Figure 7
The Dirac delta function has the property that
.)(,1,0
)(0
TtuTtTt
dzTzt
−=⎩⎨⎧
><
=−∫ δ
Another way of expressing this is that
)()( TtudtdTt −=−δ .
That is, the Dirac delta function is the derivative of the step function. The Dirac delta function also has the property that
)
.
()()(0
TfdtTttfM
=−∫ δ
for any function f , providing . From this result we can determine the Laplace transform of the Dirac delta function:
TM >
)(])([0
Tsts edteTtTtL −∞ − =−=− ∫ δδ
The Dirac delta function is often used the model actions or events that occur over a short period of time; reality is compressed into an instant of time for mathematical convenience. It is sometimes referred to as the impulse function.
23
Examples (19) Solve the differential equation 5)0(,)2(34 =−=+ xtxx δ& . Transform: sexxxs 234])0([ −=+− Substitute initial condition and solve for x : 53)4( 2 +=+ − sexs
)4(
5)4(
3 2
++
+= −
se
sx s
[A] [B] Invert to give solution:
[A]: teInverts
43)4(
3 −
+
By 2nd Shifting Theorem,
)2(3)4(
3 )2(42 −+
−−− tueInvertes
ts
[B]: teInverts
45)4(
5 −
+
tt etuetx 4)2(4 5)2(3)( −−− +−= (20) In a branch of an electronic circuit, the variation of current with time is modelled by the
differential equation
dtdVi
dtid
=+ 362
2
,
where )( is an input voltage. Suppose tV ])10()5([240)( −−−= tututV .
Assuming zero initial conditions, determine i as a function of t .
24
Referring back to the introduction of the Dirac delta function, we can write
])10()5([240 −−−= ttdtdV δδ
and so the differential equation becomes
])10()5([240362
2
−−−=+ ttidt
id δδ .
The Laplace transformation process results in
][36
240 1052
ss ees
i −− −+
= ,
which can be inverted with the help of the 2nd Shifting Theorem:
)10(])10(6[sin40)5(])5(6[sin40)( −−−−−= tuttutti . Try and fill in the details for yourselves. You may recall this warning from earlier in the notes: The Laplace transform of a product is NOT EQUAL TO the product of the
individual Laplace transforms. We shall now look at a kind of product rule for Laplace transforms.
25
8) Convolution The convolution of two functions and is another function of t denoted by
and defined by )( tf )( tg
gf ∗
[a] ∫ −=∗t
dzzgztfgf0
)()(
or
. [b] ∫ −=∗t
dzztgzfgf0
)()(
It can be shown that these two integrals are equal. If and are the Laplace transforms of and respectively, then )( sF )( sG )( tf )( tg
)()(][ sGsFgfL =∗ . Example
(21) Invert the Laplace transform 2)4(1+ss
.
This could be done using partial fractions (try it!), but convolution could be used as an
alternative:
Set s
sF 1)( = and 2)4(1)(+
=s
sG and invert individually from tables:
. 1)( =tf tettg 4)( −= Form the convolution of f and g using form [a] : 1)( =− ztf zezzg 4)( −=
.
)()(
0
4
0
∫
∫−=
−=∗
t z
t
dzez
dzzgztfgf
Omitting the details, integration by parts gives 16
141614 . 4
1 −−=∗ −− t eetgf +t
26
9) A Note on Control Theory As mentioned earlier, Laplace transforms are an important tool in control theory. Keeping things basic, in a simple control system there may be an input and an output. Control theory is concerned with the relationship between the input and the output within the system. Often a control system can be modelled by a differential equation that relates input to output in what may be referred to as the time domain. For example, a differential equation like
)( tvxkdtdx
=+
might relate an input to an output . For a given input, the equation is solved to give the corresponding output. However, in control theory, options regarding the input are often left open. Without a specific input, we can’t determine a corresponding output, but valuable information about the systems controllability can be established by leaving the input unspecified, and continuing to apply the solution process to the differential equation in any case.
)( tv )( tx
For the above differential equation, denote the following Laplace transforms: )(])([ sXtxL = )(])([ sVtvL = . Now take the Laplace transform of the differential equation; assume that the initial condition of is zero: )( tx . )()()( sVsXksXs =+ Now re-arrange: )()()( sVsXks =+
)(
1)()(
kssVsX
+= .
What we have here is the ratio of the output of the system to its input in what is called the Laplace domain. This ratio is called the system’s transfer function. In general, for a system with a single input and a single output we have
)()()( sH
sVsX
= .
The transfer function is itself a Laplace transform and its form will depend on the structure of the differential equation modelling the system. The transfer function plays a huge role in control theory as much information can be derived from it. But that is another story for another module!
)( sH
27
Tutorial Exercises (1) With the aid of the table of Laplace transforms, transform the following functions : (i) 2 (ii) a t b5t − + (iii) 2 (iv) ( 3 12t t− + )a b t+ 2
(v) 3 t e (vi) t e t t2 2−
(vii) e (viii) e tt2 1− t− cos( )2 (ix) sin ( )ω θt + (x) cos( )ω θt + (xi) sin (xii) cos 2 t 2 t (xiii) e (xiv) tt− 2 5sin ( ) e tt− +2
45sin ( )π . (2) Invert each of the following Laplace transforms :
(i) 53s +
(ii) 2162s +
(iii) ss++
112 (iv)
ss−−
242
(v) s
s−−
442 (vi) 1
9 2( )s +
(vii) s
s+
+ +2
2 12( ) (viii)
ss
++ +
42 12( )
(ix) 14s
(x) 45s
(xi) ss( )− −1 42 (xii)
5 11 2s
s s+
− +( )( )
(xiii) s
s s+
+ −5
1 3( )( ) (xiv) 32
162s s( )+ .
28
(3) Using Laplace transforms, solve the following ordinary differential equations : (i) &x x+ =2 0 ; x ( )0 1= (ii) &x x+ =2 1 ; x ( )0 0= (iii) ; & sinx x− =3 10 t x ( )0 0= (iv) &&x x− =4 0 ; x x( ) , & ( )0 0 0 6= = − (v) ; &&x x+ =ω 2 0 x A x B( ) , & ( )0 0= = (vi) && &x x x+ + =4 4 0 ; x x( ) , & ( )0 2 0 3= = − (vii) 9 6 0&& &x x x− + = ; x x( ) , & ( )0 3 0 1= = (viii) && &x x+ = 0 ; x x( ) , & ( )0 0 0 4= = (4) Sketch the graphs and determine the Laplace transforms of each of the following : (i) 2 (ii) 5u t( − ) − −4 3u t( ) (iii) 3 (iv) u t2 4[ ( ) ( ) ]u t u t− − − u t u t( ) ( ) ( )− − − + −1 2 2 3 (v) u t (vi) u t( ) (− −2 )4 − − −2 3u t u t( ) ( 6)
)
)
2
)
(vii) ( ) (viii) ( (t u t− −3 3 ) ( )t u t− −5 52
(ix) t u (x) t u t( − 3 t2 5( )− (xi) [ ( ) ( ) ] ( )u t u t t− − − −2 4 2
(xii) [ ( ) ( ) ] ( ) [ ( ) ( ) ] (u t u t t u t u t t− − − − + − − − −1 2 1 2 3 32
29
(5) Using combinations of step-functions, determine single-line expressions for each of the following functions and (hence) their Laplace transforms :
(i)
5
2 4O
f(t)
t
(ii)
2 4O
f(t)
t
3
1
3
(iii)
2 4O
f(t)
t
3
1
3
(iv)
O
f(t)
t
5
5 10
30
(v)
O
f(t)
t
5
5
(vi)
O
f(t)
t
5
5 10 15
(vii)
O
f(t)
t
5
5
(6) Invert each of the following Laplace transforms :
(i) 1s
e s− (ii) 1 2
se s− (iii) 1
2se s−
(iv) 12
2
se s− (v) 1
23
se s
+− (vi) s
se s
25
9+−
(vii) 442
10
se s
+− (viii) s
s se s
24
2 10+ +− (ix)
225
2 4
2
( )e es
s s− −−+
31
(7) The ODE below models the behaviour of a time-varying current i(t) in a circuit :
didt
i v+ = ( )t .
The function on the RHS, v(t) , is a source voltage. Below are various source voltages.
For each, graph v(t) and write as a single-line expression using step-functions. Then determine and graph i(t) .
(i) v t , v t
t( )
,,
=≤ ≤
≥⎧⎨⎩
0 0 10 1
i v( )0 0=
(ii) v t , i t t
t( )
,,
=≤ ≤≥
⎧⎨⎩
0 10 1
( )0 0=
(8) The ODE and initial conditions m x c x k x f t&& & ( )+ + = , x x( ) &( )0 0 0= = models the mass displacement of a spring-mass-dashpot system, where m is the mass ,
c is the damping constant, k is the spring stiffness and f(t) is a forcing term. For each of the following forcing terms graph f(t) and express it as a single-line expression . Setting m = 1, c = 2 and k = 5, determine x in terms of t (i.e. solve the ODE).
(i) f t t
t( )
,,
=≤ ≤
≥⎧⎨⎩
1 0 10 1
00
00
dt dt dt
t t t
(ii) f t t t
t( )
,,
=≤ ≤
≥⎧⎨⎩
0 10 1
(9) Evaluate the following integrals :
(i) (ii) δ ( (iii) δ ( δ ( )t −∫ 10
2)t −∫ 3
0
2)t −∫ 1
2
4
(iv) t t (v) e t (vi) e t
d2
0
32δ ( )−∫ dt− −∫ δ ( )3
0
5dt− −∫ δ ( )3
0
2
32
(10) Write down the Laplace transforms of (i) 2 1δ ( t − ) (ii) 4 2δ ( t )− (iii) 6 δ ( )t (11) Invert the following Laplace transforms :
(i) s
s+ 1
(ii) s
se s+ −1 2
(iii) ss
e s++
−21
4 (iv) s ss
e s2
252
9++
−
(12) Solve the following ODEs : (i) & ( )x x+ t=4 2δ , x( )0 0= (ii) & (x x t+ )= −4 2 1δ , x( )0 0= (iii) && (x x t+ = − −4 4 )δ π , x x( ) , &( )0 0 0 4= = (iv) && & (x x x t+ )+ = −4 5 1δ , x x( ) &( )0 0 0= = (13) The ODE and initial condition
didt
i v+ = &( )t , i( )0 0=
models the behaviour of a time-varying current in a circuit. The function v(t) is a
source voltage. For each of the source voltages below, write down ; hence determine and graph i(t) :
&( )v t
(i) v t v u t( ) ( )= −0 1 (ii) v t v u t u t( ) [ ( ) ( ) ]= − − −0 1 2 .
33
Answers
(1) (i) 22s s
−5 (ii) a
sbs2 +
(iii) 4 33 2s s s− +
1 (iv) as
abs
bs
2
2
2
3
2 2+ +
(v) 31 2( )s −
(vi) 22 3( )s +
(vii) es
−
−
1
2 (viii)
ss
++ +
11 42( )
(ix) s
ssin cosθ ω θ
ω++2 2 (x)
ss
cos sinθ ω θω
−+2 2
(xi) 242s s( )+
(xii) s
s s
2
2
24
++( )
(xiii) 52 22( )s + + 5
(xiv) ( ) sin cos
( )s
s+ +
+ +2 5
2 254 42
π π
(2) (i) (ii) 5 3e t− 1
2 4sin ( )t (iii) cos (iv) e or cosh ( sint + t t− 2 ) sinh ( )2 2t t− (v) 3
22 1
22e t− − e t (vi) t e t− 9
or cosh ( ) sinh ( )2 2 2t t− (vii) e (viii) e t t tt− 2 cos t− +2 2( cos sin ) (ix) 1
63t (x) 1
64t
(xi) e tt [ cosh ( ) sinh ( ) ]2 1
2+ t2
t ) ]
(xii) 2 3 2e et t+ −
(xiii) 2 (xiv) 23e et − − 1 4[ cos(− t
34
(3) (i) (ii) x e t= − 2 x e t= − −12
21( ) (iii) x e (iv) x ttt= − −3 3cos sin t = − 3 2sinh ( ) or x e et t= − − −3
22 2( )
(v) x A t B t= +cos( ) sin ( )ωω
ω (vi) x t e t= + −( )2 2
(vii) x et= 3 3 (viii) x e t= − −4 1( )
(4) (i) 2 5
se s− (ii) − −4 3
se s
(iii) 3 2 4
se es( − −− s ) (iv) 1 2 2 3
se e es s( )− − −− + s
(v) 1 4
se s− (vi) − −2 6
se s
(vii) 12
3
se s− (viii) 2
35
se s−
(ix) 1 32
3
s se s+⎛
⎝⎜⎞⎠⎟
− (x) 2 10 253 2
5
s s se s+ +⎛
⎝⎜⎞⎠⎟
−
(xi) 2 2 4 43
23 2
4
se
s s ses s− −− + +⎛
⎝⎜⎞⎠⎟
(xii) 2 2 3 13 3 2
22
3
se
s se
ses s− −− +⎛
⎝⎜⎞⎠⎟
+ s−
35
(5) (i) 5 2 4[ ( ) ( ) ]u t u t− − −
5 2 4
se es s( )− −−
(ii) u t u t u t( ) ( ) ( )− + − − −2 2 3 3 4
1 2 32 3
se e es s( )− − −+ − 4 s
(iii) 3 2 2 3 4u t u t u t( ) ( ) ( )− − − − −
1 3 22 3
se e es s( )− − −− − 4 s
]
(iv) ( ) [ ( ) ( )t u t u t− − − −5 5 10
1 1 52
52
10
se
s ses s− −− +⎛
⎝⎜⎞⎠⎟
(v) t u t[ (1 5− − ) ]
1 1 52 2
5
s s se s− +⎛
⎝⎜⎞⎠⎟
−
(vi) ( ) ( ) ( ) ( ) ( ) (t u t t u t t u t− − − )− − + − −5 5 2 10 10 15 15
1 225 10
se e es s( )− − −− + 15 s
(vii) ( ) [ ( ) ]5 1 5− − −t u t
5 1 12 2
5
s s se s−⎛
⎝⎜⎞⎠⎟
+ −
36
(6) (i) (ii) u tu t( − 1) ( )− 2 (iii) ( ) (iv) ( )(t u t− −1 )1 ( )t u t− −2 2 (v) e u (vi) ctt− − −2 3 3( ) ( ) )os[ ( ) ] (3 5 5t u t− − (vii) 2 2 10 10sin[ ( ) ] ( )t u t− − (vii) [ ]e t t ut− − − − − −( ) cos[ ( ) ] sin[ ( ) ] ( )4 1
33 4 3 4 4t (viii) { }2
5 5 2 2 5 4 4sin[ ( ) ] ( ) sin[ ( ) ] ( )t u t t u t− − − − − (7) (i) i t v v e u tt( ) [ ] ( )( )= − − −− −
0 011 1
(ii) i t e t t u tt( ) ( ) ( ) ( )= + − − − −− 1 1 1 (8) (i) x t f t f t u t( ) [ ( ) ( ) ( ) ]= − − −1
5 1 1 10 10 where f t e t e tt t
1121 2( ) cos( ) sin ( )= − −− − 2
(ii) x t g t g t u t g t u t( ) ( ) ( ) ( ) ( ) ( )= − − − − − −1 1 210 10 2 10 10 where [ ]g t t e t e tt t
1125
325 2 2 2 2( ) cos( ) sin ( )= − + −− −
g t e t e tt t
2121 2( ) cos( ) sin ( )= − −− − 2
37
(9) (i) 1 (ii) 0 (iii) 0 (iv) 4 (v) e − (vi) 0 3
(10) (i) (ii) 4 (iii) 6 2 e s− 2e s−
(11) (i) δ ( )t + 1 (ii) δ ( ) ( )t u t− + −2 2 (iii) δ ( ) ( )( )t e u tt− + −− −4 44
(iv) δ ( ) { cos[ ( ) ] sin[ ( ) ]} ( )t t t− u t+ − − − −5 2 3 5 3 3 5 5 (12) (i) x t e t( ) = −2 4
(ii) x t e u tt( ) ( )( )= −− −2 14 1
(iii) x t t u t( ) sin ( ) [ ( ) ]= − −2 2 1 π (iv) x t e t u tt( ) sin ( ) ( )( )= −− −2 1 1 1−
−
(13) (i) i t v e u tt( ) ( )( )= −− −
01 1
(ii) i t v e u t e u tt t( ) [ ( ) ( ) ]( ) ( )= − −− − − −
01 21 2
38
39
TABLE OF LAPLACE TRANSFORMS
])([ tfL is defined by dtetf ts∫∞ −
0)(
)( tf ])([)( tfLsF =
1 s1
t 2
1s
2t 3
2s
1−nt nsn )!1( −
te α− )(1α+s
tet α− 2)(1α+s
te α−−1 )( αα+ss
)(sin tω 22 ωω+s
)(cos tω 22 ω+ss
)(cos1 tω− )( 22
2
ωω+ss
)(sin tt ωω 222
2
)(2
ωω+s
s
Over . . . .
40
)(cos)(sin ttt ωωω − 222
3
)(2
ωω+s
)(sin φω +t 22
cossinω
φωφ++
ss
)(sin te t ωα− 22)( ωαω
++s
)(cos te t ωα− 22)( ωαα++
+s
s
⎟⎠⎞
⎜⎝⎛ −− )(sin)(cos tte t ω
ωαωα 22)( ωα ++s
s
)(cos)(sin tte t ωωωαα −+−
)()( 22
22
ωαωα++
+ss
)(sinh tβ 22 ββ−s
)(sinh te t βα− 22)( βαβ
−+s
)(cosh tβ 22 β−ss
)(cosh te t βα− 22)( βαα−+
+s
s
)( tfe tα− )( α+sF
x x
x& )0(xxs −
x&& )0()0(2 xxsxs &−−
Over . . . .
41
Unit Step Function
⎩⎨⎧
><
=−TtTt
Ttu,1,0
)( Tses
−1
)()( TtuTtf −− TsesF −)(
Dirac Delta Function
)( tδ 1
)( Tt −δ Tse −
Convolution Integral
∫
∫
−
−
t
t
dzzgztf
dzztgzf
0
0
)()(
OR
)()(
)()( sGsF