# Laplace Transform First Derivative

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```					1.9 Laplace Transforms
The Laplace transformation is essentially a tool for solving linear differential equations with constant
coefficients, i.e.

2
d y     dy
a      2
+b    +c y= f ( x )
dx      dx

The Laplace transform is of particular importance when initial conditions are given for which the solution must
satisfy. The transformation converts a problem in differential equations into one involving one or more linear
algebraic equations in which the initial conditions are already incorporated and from which the transform can be
found by elementary algebra. As the initial conditions are included from an early stage, the calculation of a
particular solution is shortened. The Laplace transform also allows discontinuous functions to be dealt with,
whereas other methods require the function to be continuous.

The Laplace transform of a function f(t) is denoted by F(s) and is given by,


L f (t )  F ( s)   f (t )e  st dt
0
It can be seen from the definition that only those functions limited to the positive real axis have a Laplace
transform and also the product f(t)e-st
upon evaluation of the integral, hence, the result is a function of s only.

1.9.1 Basic Laplace Transforms
Consider the Laplace transform of a constant,

                             
 e  st 
   0  1
k
L[k ]   ke dt  k 
 st

0             s 0      s

k
 L[k ] 
s
By a similar analysis we find,
1
L[ ek t ] =
s-k
It is possible to derive the Laplace transform of tn, where n is a positive integer, although the derivation is more
complicated and involves a reduction formulae. The result is quoted below,

n!
L[ t n ] =       n+1
s
From this result, we see that,

1
L[1] =     as before with k = 1
s

1
also L[t] =         2
s
Consider now the Laplace transforms of the trigonometric functions sine and cosine. Recall Euler=s formulae,

j k t
e           = cos k t + j sin k t

 sin k t = Im ( e j k t ) , cos k t = Re ( e j k t )

Therefore,
                       
L[sin(kt)]   e  st sin(kt)dt  Im  e  st e jkt dt
0                       0

 e            
 ( s  jk ) t
1
 Im                 Im
  ( s  jk )  0    s  jk

 Lsin(kt)  2
k
s  k2
Similarly,

s
L[cos(kt)]  Re  e  st e jkt dt 
0          s  k2
2

The hyperbolic functions can also be expressed in exponential form to evaluate their Laplace transform,

k t   -k t               k t   -k t
e -e                     e +e
sinh(kt) =                 , cosh(kt) =
2                        2
Therefore,

                                                      
1  st kt          e ( k  s )t   e ( k  s )t   1   1   1 
L[sin(kt)]   e (e  e )dt  
 kt
                 0      
20                 k  s  (k  s )  0 2   k  s   k  s  

k
 L[ sinh(kt) ] =          2 2
s -k
By a similar approach,

s
L [ cosh(kt) ] =      2  2
s -k

1.9.2 Properties of Laplace Transforms
The functions above have been rather simple, however, there are some important theorems of Laplace transforms
that allow more complicated functions to be transformed.

1.9.2.1 Linearity
If a and b are constants, then,

L[a f 1 (t)+ b f 2 (t)] = L[a f 1 (t)] + L[b f 2 (t)] = a L[ f 1 (t)] + bL[ f 2 (t)]
However,

L[ f 1 (t) f 2 (t) ]  L[ f 1 (t)]L[ f 2 (t)]

1.9.2.2 First Shift Theorem
The first shift theorem or shifting theorem states that,

if L[f(t)]= F(s)

then L[ e k t f(t)]= F(s - k)
Where k is a constant. Hence, wherever s appears in the Laplace transform of f(t), it is replaced by (s-k). For
example,

2
L[ e 3 t sin(2 t ) )] =
( s - 3 )2+4

1.9.2.3 Multiplication and Division by t

if L[f(t)]= F(s)

d
then L[tf (t)] = -           [ F(s) ]
ds
This can be expressed more generally as,

n
d
L[ t n f(t)] = (- 1 ) n        [F(s)]
d sn

For example,

d  s               2
s + 25
L[t cosh(5 t ) ] = -                        = 2
d s  s2 - 25

 ( - 25 )2
   s
The theorem for dividing by t is as follows,
If
L[ f (t )]  F (s)
Then

 f (t ) 
L           F ( s)ds
 t  s
This theorem is true if and only if the limit below exists,

 f (t)
lim 
 t   
t 0
Often it may be necessary to use l’Hopital’s rule to evaluate the limit. l’Hopital’s rule is stated below,

       d             
          f ( x)     
 f(x)                             dx
lim       = lim                                    
 g(x) x  0                       d            
x 0                             
g(x)

       dx            
The following example demonstrates the Laplace transform of a function divided by t which requires the use of
l’Hopital’s rule.


 sin(t )       1
L           2   ds
 t  s s 1
provided that

 sin(t ) 
lim t  0 
 t      
exits.

By l’Hopital’s rule,

 sin(t )               cos(t ) 
lim t  0            lim t  0  1   1
 t                             
i.e. limit exists

l’Hopital’s rule is required because the initial limit, sint/t, becomes 0/0 thus cannot be determined in this state.

1.9.2.4 Laplace Transforms of Derivatives
To enable us to solve differential equations using Laplace transforms we must be able to find the Laplace
transforms of derivatives. Consider the Laplace transform of the first derivative of f(t), by definition,


L f (t )   f (t )e  st dt
0
Integrating by parts,

                                             

L[ f (t )]  f (t )e    st 
0     f (t )(se       st
)dt   f (0)  s  f (t )e  st
0                                            0

Therefore, the Laplace transform of the first derivative is given by,

L[ f  (t)] =  f(0) + sL[f(t)]

The transform of the second derivative is again found using integration by parts,

                                             
 st

L f (t )   f (t )e dt  f (t )e                f (t )(se
 st 
0
 st
)dt
0                                             0

 L f (t )   f (0)  sL f (t )
Substituting the result for the first derivative,

L[ f (t )]   f (0)  sf (0)  s 2 L[ f (t )]

It is now possible to evaluate the Laplace transform of first and second derivatives and hence the Laplace
transform of first and second order differential equations. However, before we consider the solution of
differential equations, transforms of further functions and inverse transforms will be discussed.

1.9.3 Discontinuous Functions
The functions described so far have all been continuous functions of t. However, in practice certain functions are
discontinuous e.g. square waves. To describe these functions mathematically, two new functions are defined.
They are the Heaviside Unit Step function and the Delta Dirac or Impulse function.

1.9.3.1 Heaviside Unit Step Function

The Heaviside unit step function is denoted by,

f(t)= U(t - a)

It is a unit step occurring at t=a where,

f(t)= 0 for t < a

f(t)= 1 for t  a

The Heaviside step function can be multiplied with other functions to give discontinuities and more than one step
can be added to form other functions, for example,

f(t)= 2U(t - 1) - 2U(t - 2)

Consider the Laplace transform of U(t-a),
                       a               
LU (t  a)   U (t  a)e  st dt   (0)e  st dt   (1)e  st dt
0                      0               a
                           
 e  st 
 LU (t  a)   e  st dt          
a               s a
e  as
 LU (t  a) 
s
If the step occurs at the origin then a=0 and,

1
 L[U(t)]= = L[1]
s
Consider the Laplace transform of a product of the Heaviside step function, U(t-a), and another function, f(t-a). It
can be shown that,

L[U(t - a)f(t - a)] = e- a s L[f(t)]

1.9.3.2 Delta Dirac Function

This is an important function and is of particular use in digital signal theory and sampling. The function,
occurring at t=a, is written,

f(t)=  (t - a)

The delta function is defined by,

 (t - a)= 0 for t  a

 (t - a)=  for t = a
The function can be thought of as a rectangular pulse centred on t=a, being of width b and height 1/b, such that it
has unit area. The delta function is then the limiting case as the width tends to zero and its height tends to infinity
to maintain its unit area. The integral of the delta function is,

q

  (t  a)dt  1
p                           provided p<a<q

This again indicates the unit area of the pulse. Consider the integral of a function, f(t), that is multiplied with a
delta function,

q

 f (t ) (t  a)dt  f (t ) |
p
t a    f (a)
provided p<a<q

The result is the value of the function evaluated at the point where the delta function occurs, f(a). This
phenomenon is known as the sifting property of the delta function. For example,
4

 (2t  4) (t  1)dt  (2t  4) |
1
t 1   6
The result above enables the Laplace transform to be determined easily. The function f(t) becomes e -st and the
limits p and q become zero and infinity respectively,

L (t  a)    (t  a)e  st dt  e  st |t a  e as
0
For a delta function occurring at the origin,

L (t )    (t )e  st dt  e  st |t 0  1
0
Also we find that,

L f (t ) (t  a)   f (t ) (t  a)e  st dt  f (a)e as
0

From the definitions of the Delta function and the Heaviside step function, the two functions are related by,

t

  (t  a)dt  U (t  a)                                              (t  a) 
d
dt
U (t  a)
0                                                 also

1.9.4 Laplace Transforms of Periodic Functions
Consider a function that is periodic, for example a sinewave, of period T. Then, for an integer n,

f(t)= f(t + n T)

It can be shown that the Laplace transform of the periodic function is,

T
L f (t ) 
1
 f (t )e
 st
dt
1  e Ts   0
Note, that the Laplace transform of the periodic function involves an integral with limits from zero to the period
of the function, T, and not from zero to infinity as with other Laplace transforms.
1.9.5 Inverse Laplace Transforms
The following table lists the standard Laplace transforms,


L f (t )  F ( s)   f (t )e  st dt
0

k
L[k]=
s
1
L[ ek t ] =
s-k
n!
L[ t n ] =        n+1
s
k
L[ sin(kt) ] =
s k
2  2

s
L[ cos(kt) ] = 2
s k
2

k
L[ sinh(kt) ] = 2 2
s -k
s
L[ cosh(kt) ] = 2 2
s -k
From this table it is possible to determine simple inverse Laplace transforms, for example,

 k 
L1  2   2
 sin(kt)
s  k 
However, more complex functions may not be obvious examples from the table above, consider,
 3s  1 
L1  2
s  s  6

Expressing the function in partial fractions enables a solution to be found,

 3s  1      1  1       2       1  1        1  1 
L1  2         L  s  2  s  3   L  s  2   2L  s  3 
s  s  6                                             

 3s  1 
 L1  2             2t
  e  2e
3t

s  s  6
Evaluating the inverse Laplace transform of a periodic function is not so straight forward. Consider the
following example. Find,

 2  e 2 s  3e  s 
L1             2 s     
 s(1  e ) 
The factor (1-e-2s ) in the denominator is significant as it indicates a periodic function with a period of 2
units, recalling the Laplace transform of a periodic function,
T
L f (t ) 
1
 f (t )e
 st
dt
1  e Ts   0

Re-writing the original function with (1-e-2s ) in the numerator giving,

2 + e- 2 s - 3 e- s 1
-2s
= ( 2 + e- 2 s - 3 e- s ) ( 1 - e- 2 s )- 1
s ( 1- e         ) s

The term (1-e-2s )-1 can be expanded using the binomial theorem,

1                                            1
( 2 + e- 2 s - 3 e- s ) ( 1 - e- 2 s )- 1 = ( 2 + e- 2 s - 3 e- s ) ( 1 + e- 2 s + e- 4 s + e- 6 s + ... )
s                                            s

-s   -2s      -3s     -4 s      -5s
2    e   e        e        e        e
=     - 3 +3       -3       +3       -3       + ...
s     s    s        s        s        s
Earlier we saw that,

-as
e
L[ U ( t - a ) ] =
s
Hence,

 2 + e- 2 s - 3 e- s 
L1                        = 2U (t )  3U (t  1)  3U (t  2)  3U (t  3)  3U (t  4)  3U (t  5)....
 s ( 1- e
-2s
)  

This represents a square wave of period 2 and it is shown below,

1.9.6 Solutions to Differential Equations using Laplace Transforms
We have now established all the tools to allow us to solve differential equations using Laplace
transforms. The following steps form the method,

i)       Laplace transform each term in the differential equation.
ii)      Insert the initial conditions.
iii)     Re-write the equation algebraically to a form that can be inverse Laplace transformed.
iv)      Perform the inverse transform.
Consider the following first order differential equation, given the initial condition that x=2 at t=0,

dx
- 4 x= 8
dt
Perform the Laplace transform,

dx        
L    - 4 x  = L[ 8 ]
 dt       

8
 - x ( 0 ) + s X (s) - 4 X ( s ) =                 e
wher X ( s ) = L[ x ( t ) ]
s
Insert initial condition, x(0)=2,

8
- 2 + s X (s) - 4 X ( s ) =
s

8+2 s
 X ( s )=
s( s-4 )
Taking the inverse Laplace transform,

 8  2s        1  1        1  1 
x(t )  L1 X (s)  L1             4L  s  4   2L  s 
 s(s  4)                       
 x(t )  4e  2
4t

Where x(t) is the solution to the differential equation with the initial condition included.

Consider the second order differential equation,

2
d x
+ x = cos(t )
d t2

Given that,

dx
= x = 0 when t = 0
dt
Therefore, taking the Laplace transform,

 d2 x     
L      + x  = L[ cos(t ) ]
 dt       
2

dx                                              s
-       ( 0 ) - s x ( 0 ) + s2 X ( s ) + X ( s ) = 2                   e
wher X ( s ) = L[x (t)]
dt                                           s +1
Substituting the initial conditions,
s
X ( s ) ( s2 + 1 ) =      2
s +1

s
 X ( s )=
( s +1 )2
2

Taking the inverse Laplace transform to obtain the solution gives,

1
x(t)=     t sin(t )
2

```
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