periodic functions, the unit step function and the second shifting theorem

Copyright Rene Barrientos

PERIODIC FUNCTIONS AND THE UNIT STEP FUNCTION The Unit Step Function

The First Shifting Theorem tells us how to compute the transform of the product . Now we introduce a very valuable tool called the Unit Step Function1, which allows us to find transforms of piecewise-defined functions.

Graphically, this function looks like a shifted version of the unit function 1, 0:

For example, u3(t) is a function that is 0 until 3 and 1 thereafter:

The idea is this: turns a function “off” prior to and “on” after that and so whereas the function is a horizontal translation of , · is a new function that equals only

on the interval :

1 Also known as Heaviside’s function after the self-taught English mathematician, physicist, and electrical engineer Oliver Heaviside (1850 – 1925).

The Unit Step Function, is defined by

This function is also denoted by .

1

u(t)

t

Figure 1

ua(t)

a

1

t

Figure 2 3

u3(t)

1

t

a

–a


Example 1 Graph the function 3

Solution Since 3 0 before 3 and 1 afterward, we can write

0 3 3

Thus, the effect of 3 is to kill the function 2 prior to 3. The graph of is shown below.

Now suppose we want to describe the piecewise defined function

1 3 2 3 whose

graph is How can we write this in terms of unit step functions? Idea: “turn off” the function “1” after 3 and “turn on” the function “2” in its place. We know that turns a function “off” until and “on” after that. Therefore, the function does the opposite: it turns a function “on” until we get to

and “off after that”:

Figure 3

0 1

1 0

In other words, 1 leaves a function alone until and kills it after that so the function

1 3 2 3

May be written as follows:

3

g(t) = t2 f(t) = t2∙u(t – 3)

3

2

1

3

a

1 1 0 1

10


· ·

This can easily be verified by noting that if 3, 0 so 1 · 1 2 · 0 1 and if 3, then 1 · 0 2 · 1 2.

Example 2 Express the function 0 2 1 2

Solution We wish the function 2 to be “on” until 2 and “off” after that. This is accomplished by multiplying by 1 − u2(t). After that, we want to bring the function 1 to life. This is accomplished by multiplying by u1(t):

· 1 1 ·

In terms of notation,

· 1 2 1 · 2

Collect like terms and write: ·

The key to solving initial value problems whose forcing function is piecewise defined is to first write the forcing function in terms of unit step functions.

It is sometimes easier to use the pulse function instead, but the method described in the previous examples is just as good.

The Pulse Function

What does , look like? It is the difference of two step functions as shown below. Algebraically it is this:

,

0 0 1 0

,

Let 0. Define the pulse function , where 0 ∞ is defined by

We also use the notation ,

where , and ,∞ . Alternatively, we may define and ∞ .

Turn “on” until 2, then “off”

Turn 1 “on” after 2

Turn “1” off after t = 3

Turn “2” on after t = 3

ua

a b

a b

ub

a b


You can clearly see why , is called a pulse function; that is exactly what it does and its effect on other functions is to “turn them off” outside the interval (a, b), which is what we want to do when we are dealing with piecewise defined functions.

Example 3 Plot the pulse 2,3.

Solution

Algebraically,

2,3

0 ∞, 21 2,3 0 3,∞

We can also write 2,3 2 3 . This will be the preferred way of expressing the pulse when we study the Laplace Transform of these functions, and particularly when we compute inverse transforms.

How do we use the pulse function to express piecewise defined functions un terms of the unit step function? Here is how: suppose that f is given by

0 .. ∞

We can write f in terms of pulse functions:

· , · , · , · ,∞

Since , with 0, 1 and ,∞ , we can write

· 1 · · ·

· · ·

Example 4 Write the function 0 2

2 51 5 ∞

in terms of unit step functions.

Solution We first write it in terms of the pulse function:

· , · , 1 · ,∞

Now in terms of unit step functions:

· · 1 · ∞

Finally, since and ∞ , we have:

· 1 · 1 · 0

· · · 1 ·

· 1 ·

2 3

1


In terms of the more formal notation,

· 1 ·

Verify that this function reduces to what it should on the different intervals in which it is defined.

Laplace Transform of the Unit Step Function By definition,

∞

But since 0 for and 1 thereafter,

0

· 1∞

Evaluating this integral is a simple exercise, but by using the appropriate substitution we can avoid the limit process associated with the improper integral altogether.

Let . Then and . Also when and ∞ as ∞. Thus:

· 1∞

· 1∞

0

· 1∞

This last integral is just 1 with variable replaced by . Therefore,

Example 5 Find 3

Solution

3 ·1

Example 6 Find 5 · 1

Solution 5 · 1 5 · 1

5 ·

We may generalize formula (1) for the product · :

· ·∞

∞

The substitution ; when transforms the last integral as follows: ∞ ∞

0

∞

0

∞

0

But this last integral is just .

;


Thus,

This looks very clumsy and we will fix it, but first notice that formula (2) is a special case of formula (1): let 1. Then 1 and applying (2) gives us

1 · 1 ·

Example 7 Find 2

Solution Let and a = 2 in (3). Then 2 2 . Thus,

2 2

6 12 8 ·

3!

6 ·2!

12 ·1

8 ·1

·

·

Example 8 Find 4

Solution Let and 4. Then 4 . Applying (2),

4 ·

· ·

·12 ·

·

Formula (2) is adequate when computing a transform, but essentially useless when we try to use it to find inverse transforms.

We remedy this via the following observation: Suppose that we first express as function of , that is, . Then,

· · ·∞

Using the same substitution used earlier results in the more useful formula which we call Second Shifting Theorem:

Letting ,

· ·

This a much useful version when used to compute an inverse transform because it states that

· ·

The Second Shifting Theorem

· ·


· · We will come back to this thought later; first let us look at examples of formula (3).

Example 9 Find 2 .

Solution In order to use (3), we need to express as a function of 2. This is how it is done:

3 2 2 3

2 2

2 6 2 12 2 8 Therefore,

2 2 6 2 12 2 8 · 2

Applying (4) with 2:

2 6 12 8 ·

3!

6 ·2!

12 ·1

8 ·1

Example 10 Find · sin using both formulas.

Solution Method I using formula (2) with sin :

· sin sin

sin

sin

·

Method II using formula (3), we need to express sin as a function of :

sin sin

sin Thus,

· sin sin

sin

·

Note: The identity sin sin cos cos sin in both parts. Example 11 Compute cos 4 · 1

Solution We have cos 4 and 1. applying formula (2)

cos 4 · 1 1 ·

cos 4 4 ·

where


cos 4 · cos 4 sin 4 · sin 4 ·

cos 4 · cos 4 sin 4 · sin 4 ·

cos 4 · 16 sin 4 ·416

Note: and are just numbers.

Using formula (3) instead requires that we express cos 4 as a function of t − 1:

cos 4 cos 4 1 1

cos 4 1 4

cos 4 1 · cos 4 sin 4 1 · sin 4

Thus,

cos 4 · 1 cos 4 1 · cos 4 sin 4 1 · sin 4 · 1

cos 4 · cos 4 1 · 1 sin 4 · sin 4 1 · 1

cos 4 · cos 4 · sin 4 · sin 4 ·

cos 4 · 16 sin 4 ·416

Example 12 Compute where 0 0 2cos 2 2

Solution We have 0 · 0,2 cos 2 · , cos 2 · 2 . Thus,

cos 2 · 2

In order to use formula (3), we must express cos 2 as a function of 2 :

cos 2 cos 2 2 2 cos 2 2 4

cos 2 2 cos 4 sin 2 2 sin 4

cos 2 2 Therefore,

cos 2 2 · 2

cos 2 ·

The Inverse Transform of ·

From the transform of it follows that

Example 13 Find 14

cos 4 1 where

sin 4 1 where


Solution

14

4

Let us now find inverse transforms of function of the form · . Using formula (4),

· ·

A friendlier version of this is

which says: to find · ,

1) find

2) evaluate it at ,

3) and multiply the result by . In plain English: the inverse transform of · is the inverse transform of F(s) evaluated at t − a and the result multiplied by u(t − a).

Example 14 Find

Solution We can write this as · . Thus, using (5) with and 3:

·416 3

416

3 · sin 4 |

3 · sin 4 3 Example 15 Find

Solution Here we have · and we can identify and 1. Thus,

· 1 1 · 1

1 · cos |

1 · cos 1 Example 16 Find

Solution Again, · Hence, we identify and 2. But is not an entry on the table of transforms. However, using partial fractions we can write

1 11

11

1

Thus, 1

·11

1

2 ·11

1

· · |


2 · 1 |

·

Note: 1 so if we evaluate this function at we still get 1 for any .

Example 17 (both shifting theorems) Find Solution

2 5 2 1 4

1 4

1 4

1 4 1 4

Thus,

2 5 2 ·1

1 411 4

2 ·1

1 2 · 1 2

2 · cos 212 sin 2

·

Applications to Initial Value Problems Let us solve some differential equations using the tools so far developed.

Example 18 Solve the IVP 4 9 0 13 1 ; 0 1

Solution The traditional approach to this problem would be to break it into two separate problems: an initial value problem defined on 0,1 and a differential equation with boundary conditions defined on

1,∞ . This can be avoided by expressing the forcing function 9 0 13 1 in terms

of the unit step function: 9 · 3 ·

9 9 · 1 3 · 1

9 6 · 1 Thus,

4 9 6 · 1 Applying the transform,

4 · 9 6 · 1

0 49

6

Shifted Shifted


4 19

6 Solving for :

14

94 6

14

Last two terms will require partial fractions:

14 4 1/4, 1/4

14

1/4 1/44

Thus,

4/ /

4 614

144 .

Simplifying, 94 ·

1·

14

321 1

4 .

94

1 54

14

32

1 14

94

54

32 1 · 1 |

·

Notice that the last term is 0 for 1. This is exactly what happens when one has a forcing term that kicks in after certain amount of time has elapsed.

Example 19 Solve the IVP ; 0 1 where 1 0 2 2 40 4

Solution In order to solve this equation, we would have to solve three separate ones and match their solutions at the boundaries - a lengthy process. Instead, we first use the pulse function to express f(t) in a more workable way:

1 · , · , 0 · ,

1 · · 0 ·

1 · 1 ·

1 · ·

1 1 ·

Thus, the differential equation we have to solve is

1 1 · ; 0 1

1 1 ·

1 1 · ·

1


Thus, 1

· ·

1 1

1 · 4 ·

11 1 1

·1 4

·

Solving for :

1 11 1 1

·1 4

·

11

11

11

11 ·

11

41 ·

This looks horribly long, but notice that many of the terms are identical, except for the exponential factors whose only job is to shift our final answer. Applying partial fractions to the expressions

and we get the following result:

11

1 1 11

and

11

1 11

Thus,

11

1 11

1 1 11

1 11 ·

1 1 11

4 41 ·

Simplifying,

1 1·

3 1 31 ·

We may now apply :

1 2 · | 4 · 3 3

1 2 · 2 4 · 3 4 3

· ·

Periodic Functions The functions listed in that table below are familiar to you. They are the trigonometric functions whose graphs exhibit a high degree of regularity which we call periodicity. For example, we learn that the sine function has period 2 and this means, as we also learned, that sin 2 sin for all real numbers . We call the number 2 the period of the sine function because no other real number smaller than it has this property.

Periodic functions are very important not only because they describe a multitude of natural phenomena but also because they are mathematically beautiful and worthy of study. Of course, the trigonometric functions

f(t) Period

sin 2

cos 2

tan


are not the only periodic functions and the aim of this lecture is to study their general properties in the context of the Laplace Transform.

Example 20 (the square wave) Often it is easier to describe periodic functions by their graph The square wave is shown below:

This wave has period 2 and amplitude B where a and B are positive real numbers. We may describe this function algebraically as follows:

0 0 2

but we need to indicate that this pattern repeats indefinitely. We do so by stating the additional condition:

which conveys the information that the function does not end at 2 but repeats the same pattern indefinitely.

Example 21 The saw-tooth wave shown below. Give an algebraic description of this wave.

Solution

Observe that the function is a straight line of slope 1/2 on the interval 0,2 and this line repeats itself every two units. Therefore, 2 and

12 0 2

2

Example 22 (Half-wave rectification of sin ). Give an algebraic formula that describes it. (this wave is called the half-wave rectification of sin .

Solution

Definition A function f defined on 0,∞ is periodic with period 0 if for all 0,∞ .

2a

B

3a 4a . . a

6

1

8 4 . . 2

2

3

1

. . .


Rectifying means getting rid of the portion on which the wave has negative values. Since the period of sin is 2 / the wave has negative values on the interval , . In the figure, that portion has been removed thus leaving only the positive portions. Clearly the wave repeats 2 / so this is the period and one period is described by

sin 0 / 0 / 2 /

Adding the condition 2 / completes the description of the wave.

Example 23 Show that is a periodic function with meriod then if is any positive integer, then is also a period in the sense that .

Solution (proof)

We proceed by induction. For 1 we have since is the period. Assume that

. Then

1

QED How do we find transforms of periodic functions? Let us consider the square wave of example 1. In order to find its transform, we apply the definition:

Notice that the integral on the intervals , 2 , 3 , 4 , … is zero since the function is 0. As you can see, this is not very practical even in this simple case of the square wave. We need another, more efficient, way of doing things. In the optional section of these notes we establish the following result:

Example 24 Find the transform of the square wave with 1 and 4

Solution This wave has amplitude 4 and period 2 2. Thus,

11

1

1 · 4 · 0

41

41

1

if f is a periodic function with period p, then its transform is given by


4

1 11

Note: 1 1 is a difference of squares.

Example 25 Find the transform of the saw-tooth wave shown below:

Solution The function is described by

12 0 2

2

Applying (6) 1

1 ·12

1

2 1 ·

This requires integration by parts, which may be accomplished though tabular integration (review a calculus text for this procedure as it is encountered often – the table for this particular example appears at the end of these notes). The integral is

·1 1

1 1

1 2 1

Thus,

1

2 1 ·1

2 1 ·1 2 1

Therefore,

·

Observe that 1 2 1 and that 1 1 · 1 .

Hence, 1

2 1 1 · 1

Finally, 1

2 1

Example 26 (Half-wave rectification of sin ). Find the transform of the half-wave rectification of sin

Solution One period if this wave is given by

sin 0 / 0 / 2 /

Using 2 / ,


1

12 ·

2 /

Since 0 on / 2 / ,

1

12 · sin

/

The integral may be evaluated by two applications of integration by parts. Instead, we will use the help of MATHEMATICA for this purpose. The output is

· sin/ 1 ·

Thus,

1

12 ·

1 ·

The denominator is again a difference of squares. Hence:

Example 27 Evaluate cos .

Solution First observe that the period of cos is , not 2 .

Thus,

cos1

1 cos

Solving for cos :

cos 1 · cos

Using the reduction formula cos cos 2 ,

cos 1 ·12

12 cos 2

·

Example 28 Here is an interesting thought: the constant function 1 can be thought of as a wave with infinite period. What happens if we apply formula (6) to this function?

Solution Le ∞ in formula (6):

1 2 3 4 5 6

0.2

0.4

0.6

0.8

1.0


Replacing ∞ gives us

1

1 ∞· · 1∞

1 ; 0

Periodic Forcing Functions Example 29 (The spring-mass revisited) suppose a spring mass system has equation

4 sin ; 0 0, 0 0

If we had to solve this equation with the methods already developed, we would follow these steps: 1) Find the complementary solution by solving the homogeneous equation

4 0. 2) Find a particular solution which can be done by using the variation of parameters method or

undetermined coefficients. 3) The general solution is then 4) Apply the initial conditions to determine the arbitrary constants.

Instead, we will use the Laplace transform. Taking the transform of both sides:

4 sin

4 sin

0 0 411

where . Since 0 0, 0 0 we have

411

Solving for , 1

1 4

Finally, we can reproduce using the inverse transform:

11 4

At this point, we could use partial fractions, the convolution integral or, even better, use a table of transforms in which you will find the following entry:

Thus, 1 sin sin

Identifying 1 and 2, sin 2 2 sin2 1 2

16 sin 2

13 sin

sin sin

1


Using the Convolution Integral The convolution of two functions f and is defined by

and can be used to find inverse transforms of products. The Convolution Theorem states that

·

where f and g are the inverse transforms of F and G, respectively. We can use this in the previous example:

1·

Thus, 1

1 4

12

11 ·

24

12 sin ·

12 sin 2

Example 30 (The RC circuit) suppose you have an RC circuit with 3.5Ω and 0.06 f. However, instead of having a steady source of power it has one that turns off after a few seconds:

3 0 20 2

If 0 0 find the charge on the capacitor at time 0.

Solution The differential equation that governs this circuit is

1; 0 0

Since is given piecewise, we either break down the differential equation into different time intervals or we use the Laplace Transform. If we opt for the second choice, we need first to express

in terms of the unit step function and also the equation in terms of the charge only:

3 · 1 2

does the trick. Also note that . Thus,

13 1 2 ; 0 0

Applying the transform,

·1· 3 · 1 2

01

31

Substituting the initial condition 0 0 and solving for Q(s)

13

1


31

1 1

The term 1

11·

1

0 1

has the form with / . Consulting a basic table of transforms we find that

Hence,

31

1 1

3·

11 1

3· 2 ·

3· · 2

where / . Substituting these in the expression above,

3·

0 1 0 1 · 2

3 · 1 / 1 · 2

Since 3.5Ω and 0.06 ,

. · . . ·

Example 31 (periodic forcing function) Consider the equation " " with 00, 0 0 [the square wave is shown below]. Find its solution.

Solution

This square wave is a periodic function with period 4. Accordingly, we apply the transform to obtain

0 0 01

1 · " "

Applying the initial conditions and transforming the periodic function:

1

2 4 6


11 · 1

1

1 1

Simplifying the right hand side and Solving for :

1

1 1

Thus, 1

1 1

How de we handle the term? We use the geometric series expansion:

11 1

11 1

Therefore, 11 ·

11

Let and . Then

·

We may obtain using partial fractions: 11

1

However, observe that is an infinite sum of unit step functions:

11

1 1 1 1

1 2 4 6

Thus, it behooves us to use the commutative property of convolution and write instead

·

From which it follows that

1 2 4 6 · 1

1 2 4 6

1 2 4 6

PROOF OF THE PERIODIC FUNCTION FORMULA Theorem Let f be a periodic function with period p defined on the interval 0,∞ which has a Laplace Transform. Then


11

Proof By definition,

Applying the properties of integration,

Let In the second integral. Then

Since f is a periodic with period p, . Therefore,

Solving for :

Thus,

TABULAR INTEGRATION Tabular integration is an organized way of doing integration by parts. In example 2 is as follows:

Differentiate these Integrate these Products

1 1

0 1

·

· ·

periodic functions, the unit step function and the second shifting theorem

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