Part 2 - Properties of Laplace Transform

DIFFERENTIAL EQUATIONS 2.2

PROPERTIES OF LAPLACE TRANSFORMATION



Laplace Transforms of Discontinuous Functions Jump discontinuities often occur naturally in physical problems. For example, an electric circuit with on/off switches. To handle such behavior, an English mathematician, Oliver Heaviside introduced the unit step function or the Heaviside function, given as follows.



Definition 2.2.1: Unit Step Function The unit step function U(t) is defined by 0 t  0 U (t )   1 t  0

By shifting the argument of U(t), we obtain 0 t  a  0 0 t  a U (t  a)     1 t  a  0 1 t  a

(1)

U(ta)

Figure 2.2.1: The unit step function U(t – a)

It has a discontinuity at t = a. By multiplying by a constant M, the height of the jump can be modified.

0 MU (t  a)   M

ta ta

Figure 2.2.2: Two step functions expressed using the unit step function 

In Laplace transform, we are only concern with t  0 in the step function.



Example 2.2.1 When a function f (t) = 2t – 3 defined for t  0 is multiplied by U(t – 1), the unit step function “turns off” a portion of the graph of f for 0  t < 1. Refer to Figure 2.2.3.

Figure 2.2.3: Graph of f(t) = (2t – 3)U(t – 1)



Example 2.2.2 Write the following function f(t) in terms of unit step functions. Refer to Figure 2.2.4.

 3,  1,  f (t )    t, t 2 10,

t2 2t 5 5t 8 t 8

Figure 2.2.4: Graph of function f(t) in Example 2.2.2

Solution: From Figure 2.2.4, the function f(t) equals 3 until t reaches 2, at which point it jumps by 2 units to the value 1. We can express at this jump with the formula 3 – 2U(t – 2), since U(t – 2) is 0 until t reaches 2, after which it has the value 1.

At t = 5, the function changes from constant value 1 to the value t. Thus, effectively we subtract the constant 1 from f(t) at t = 5 and add t. Thus, we write this condition with the formula (t – 1)U(t – 5).

Similarly, we express the formula for f(t) from t to t2/10 at t = 8 by (t2/10 – t)U(t – 8).

Therefore, the function f(t) can be written in terms of step functions as, f (t )  3  2U (t  2)  (t  1)U (t  5)  (t 2 10  t )U (t  8)



Example 2.2.3 Express the function f(t) as shown in Figure 2.2.5 in terms of step functions.

Figure 2.2.5: Graph of function f(t) Solution: Function f(t) begins at 2 then there is a jump by 3 units to the value 1 at t = 2. We express this by 2 – 3U(t – 2). Then U(t – 3) increases by +1 at t  3. Thus, f(t) = 2 – 3U(t – 2) + U(t – 3).



The general piecewise-defined function of the type  f1 t ,  f (t )   f 2 t ,  f t ,  3

0t a at b t b

= f1 t U (t  0)  U (t  a)  f 2 t U t  a   U t  b  f 3 t U t  b = f1 t U t    f 2 t   f1 t U t  a    f 3 t   f 2 t U t  b 

(2)

Example 2.2.4

20t , 0  t  5 Express f (t )   in terms of unit step functions. Refer to Figure 2.2.6. t 5  0,

Figure 2.2.6: Graph of f(t) in Example 2.2.5.

Solution: Function f(t) can be written as in (2). Thus referring to (3), with a = 5, g(t) = 20t and h(t) = 0, we can write f(t) as f(t) = 20t – 20t U(t – 5).



Consider a function y = f(t) defined for t  0.

The piecewise-defined function

Figure 2.2.7(a): y = f(t), t  0

0t a  0, f (t  a)U (t  a)   ta  f (t  a),

is shown in Figure 2.2.7.

Figure 2.2.7(b): y = f(t – a)U(t – a)

The graph of y = f(t – a)U(t – a) coincides with the graph of y = f(t – a) for t  a (which is the entire graph of y = f(t), t  0 shifted a units to the right on the t-axis), but zero for 0  t < a.



Second Translation Theorem Theorem 2.2.1: Second-Shift Property If F(s) = ℒ{f(t)} exists and a > 0, then ℒ{f(t  a)U(t – a)} = eas ℒ{f(t)}

=

eas F(s).

Proof: By definition of Laplace transform, 

e

ℒ{f(t  a)U(t – a)} =

 st

f (t  a)U (t  a)dt

0

a



0

a

 st  e f (t  a)U (t  a)dt +

=

e

 st

f (t  a)U (t  a)dt

Since U(t – a) = 0 for 0  t < a and U(t – a) = 1 for t  a 

e

ℒ{f(t  a)U(t – a)} =

 st

f (t  a)dt

a

Let v = t – a then dv = dt, thus 

ℒ{f(t  a)U(t – a)} =  e s ( va ) f (v)dv a



= e as  e sv f (v)dv a

=

eas ℒ{f(v)}

change to variable t, we obtain

= eas ℒ{f(t)} = eas F(s).



In Theorem 2.2.1, if f(t – a) = 1, F(s) = ℒ{1} =



Example 2.2.5

1 e  as . So ℒ{U(t – a)} = . s s

Find the Laplace transform of f(t) = 2 – 3U(t – 2) + U(t – 3).

Solution: ℒ{f(t)} = 2ℒ{1}  3ℒ{U(t – 2)} + ℒ{U(t – 3)} 2 e 2 s e 3s = .  3  s s s

(5)



In practice it is more common to be faced with problem of computing the Laplace transform of g(t)U(t – a) rather than f(t – a)U(t – a).

Let v = t – a, so f(t – a) = g(t) becomes f(v) = g(v+a).

Change the function to variable t we write f(t) = g(t + a).

Theorem 2.2.1 then gives ℒ{g(t)U(t – a)} =



e

 st





g (t )dt

a



e

s ( va )

g (v  a)dv = eas ℒ{g(t+a)}

0

Example 2.2.6 Determine the Laplace transform of t2U(t – 1).

Solution: To apply equation (6), we take g(t) = t2 and a = 1. Then g (t  a)  g (t  1)  (t  1) 2  t 2  2t  1

Now the Laplace transform of g(t + a) is 2 2 1  2  .

3 s s s

ℒ{g(t+a)} = ℒ{ t 2  2t  1} = So, by equation (6), we have

2 1 2 ℒ{t2U(t – 1)} = e s  3  2   . s s s



Example 2.3.7 Determine ℒ{(cos t)U(t – )}.

Solution: Here g(t) = cos t and a = . Hence,

g (t  a)  g (t  π)  cos(t  π)   cos t , So the Laplace transform of g(t + a) is ℒ{g(t+a)} = ℒ{cos t} = 

s . s 1 2

(6)

Thus, from equation (6), we get

 s  ℒ{(cos t)U(t – )} =  e π s  2 .  s  1

Exercises 2.3 In problems 1-4, sketch the graph of the given function and determine its Laplace transform.

1)

f (t )  t  1 U t  1

2)

f (t )  U (t  1)  U (t  4)

3)

f (t )  t 2U t  2

4)

f (t )  tU t  1

2

In Problems 5-10, express the given function using unit step functions and compute its Laplace transform.

5)

0, 2,  g (t )   1, 3,

0  t 1 1 t  2

6)

2t 3 3t

7)

8)

9)

10)

 0, g (t )   t  1,

0t 2 2t

Answers 2.3 In problems 1-4, determine the Laplace transform of each function.

1)

ℒ{ t  12U t  1 } =

3)

ℒ{ t 2U t  2 } =

2e  s s3



e 2 s 4s 2  4s  2 s3



2)

ℒ{ U t  1  U t  4 } =

4)

ℒ{ tU t  1 } =

e  s  e 4 s s

e  s s  1 s2

In Problems 5-10, express the given function using unit step functions and compute its Laplace transform.

5)

0,   2 , ℒ  1, 3, 

6)

 0, ℒ  t  1,

7)

ℒ{g(t)} =

0  t  1  1  t  2  2e  s  e 2 s  2e 3s   = 2  t  3 s 3  t 

0  t  2 e 2 s 3s  1 =  2t  s2

e



8)

ℒ{g(t)} =

9)

ℒ{g(t)} =

10)

ℒ{g(t)} =



 e 2 s s  1 s2

s

s

e 2s s2  1

e

s

2e  s s3

 2e 2 s  3e 3s s2

