CHAPTER 4 The Laplace Transform

Contents 4.1 Definition of the Laplace Transform 4.2 The Inverse Transform and Transforms of Derivatives 4.3 Translation Theorems 4.4 Additional Operational Properties 4.5 The Dirac Delta Function

4.1 Definition of Laplace Transform Basic Definition If f(t) is defined for t  0, then the improper integral (1) If f(t) is defined for t  0, then (2) is said to be the Laplace Transform of f. DEFINITION 4.1 Laplace Transform

Example 1 Evaluate L{1} Solution: Here we keep that the bounds of integral are 0 and  in mind. From the definition , s>0 Since e-st  0 as t , for s > 0.

Example 2 Evaluate L{t} Solution , s>0

Example 3 Evaluate L{e-3t} Solution

Example 4 Evaluate L{sin 2t} Solution

Example 4 (2) Laplace transform of sin 2t ↓

L.T. is Linear We can easily verify that (3)

(a) (b) (c) (d) (e) (f) (g) Transform of Some Basic Functions THEOREM 4.1 Transform of Some Basic Functions

A function f(t) is said to be of exponential order, DEFINITION 4.2 A function f(t) is said to be of exponential order, if there exists constants c>0, M > 0, and T > 0, such That |f(t)|  Mect for all t > T. See Fig 4.2, 4.3. Exponential Order

Fig 4.2

Examples See Fig 4.3

Fig 4.4 A function such as is not of exponential order, see Fig 4.4

If f(t) is piecewise continuous on [0, ) and of THEOREM 4.2 If f(t) is piecewise continuous on [0, ) and of exponential order, then L{f(t)} exists for s > c. Sufficient Conditions for Existence

Fig 4.1

Example 5 Find L{f(t)} for Solution

4.2 If F(s)=L(f(t)), then f(t) is the inverse Laplace transform of F(s) and f(t)=L(F(s)) THEOREM 4.3 (a) (b) (c) (d) (e) (f) (g) Some Inverse Transform

Example 1 Find the inverse transform of (a) (b) Solution (a) (b)

L -1 is also linear We can easily verify that (1)

Example 2 Find Solution (2)

Example 3: Partial Fraction Find Solution Using partial fractions Then (3) If we set s = 1, 2, −4, then

Example 3 (2) (4) Thus (5)

Uniqueness of L -1 Suppose that the functions f(t) and g(t) satisfy the hypotheses of Theorem 4.2, so that their Laplace transform F(s) and G(s) both exist. If F(s)=G(s) for all s>c (for some c), then f(t)=g(t) whenever on [0, + ) both f and g are continuous.

Transform of Derivatives (7) (8)

If are continuous on [0, ) and are of THEOREM 4.4 If are continuous on [0, ) and are of Exponential order and if f(n)(t) is piecewise-continuous On [0, ), then where Transform of a Derivative

Solving Linear ODEs Then (9) (10)

We have (11) where

Example 4: Solving IVP Solve Solution (12) (13)

Example 4 (2) We can find A = 8, B = −2, C = 6 Thus

Example 5 Solve Solution (14) Thus

4.3 Translation Theorems If f is piecewise continuous on [0, ) and of exponential order, then lims L{f} = 0. Behavior of F(s) as s →  Proof

F(s) by s-a If L{f} = F(s) and a is any real number, then THEOREM 4.6 If L{f} = F(s) and a is any real number, then L{eatf(t)} = F(s – a), See Fig 4.10. Translation on the s-axis Proof L{eatf(t)} =  e-steatf(t)dt =  e-(s-a)tf(t)dt = F(s – a): replacing all s in F(s) by s-a

Fig 4.10

Example 1 Find the L.T. of (a) (b) Solution (a) (b)

Inverse Form of Theorem 4.6 (1) where

Parttial Fraction To perform the inverse transform of R(s)=P(s)/Q(s): Rule 1: Linear Factor Partial Fractions Rule 2: Quadratic Factor Partial Fractions

Example 2 Find the inverse L.T. of (a) (b) Solution (a) we have A = 2, B = 11 (2)

Example 2 (2) And (3) From (3), we have (4)

Example 2 (3) (b) (5) (6) (7)

Example 3 Solve Solution

Example 3 (2) (8)

Example 4 Solve Solution

Example 4 (2)

The Unit Step Function U(t – a) defined for is Unit Step Function DEFINITION 4.3 The Unit Step Function U(t – a) defined for is Unit Step Function See Fig 4.11.

Fig 4.11

Fig 4.12 Fig 4.13 Fig 4.12 shows the graph of (2t – 3)U(t – 1). Considering Fig 4.13, it is the same as f(t) = 2 – 3U(t – 2) + U(t – 3),

Also a function of the type. (9) is the same as Also a function of the type (9) is the same as (10) Similarly, a function of the type (11) can be written as (12)

Example 5 Express in terms of U(t). See Fig 4.14. Solution From (9) and (10), with a = 5, g(t) = 20t, h(t) = 0 f(t) = 20t – 20tU(t – 5)

Fig 4.14

Consider the function (13) See Fig 4.15.

Fig 4.15

If F(s) = L{f}, and a > 0, then L{f(t – a)U(t – a)} = e-asF(s) THEOREM 4.7 If F(s) = L{f}, and a > 0, then L{f(t – a)U(t – a)} = e-asF(s) Second Translation Theorem Proof

Let v = t – a, dv = dt, then If f(t) = 1, then f(t – a) = 1, F(s) = 1/s, (14) eg: The L.T. of Fig 4.13 is

Inverse Form of Theorem 4.7 (15)

Example 6 Find the inverse L.T. of (a) (b) Solution (a) then (b) then

Alternative Form of Theorem 4.7 Since , then The above can be solved. However, we try another approach. Let u = t – a, That is, (16)

Example 7 Find Solution With g(t) = cos t, a = , then g(t + ) = cos(t + )= −cos t By (16),

Example 8 Solve Solution We find f(t) = 3 cos t U(t −), then (17)

Example 8 (2) It follows from (15) with a = , then Thus (18) See Fig 4.16

Fig 4.16

4.4 Additional Operational Properties Multiplying a Function by tn that is, Similarly,

If F(s) = L{f(t)} and n = 1, 2, 3, …, then THEOREM 4.8 If F(s) = L{f(t)} and n = 1, 2, 3, …, then Derivatives of Transform

Example 1 Find L{t sin kt} Solution With f(t) = sin kt, F(s) = k/(s2 + k2), then

Different approaches Theorem 4.6: Theorem 4.8:

Example 2 Solve Solution or From example 1, Thus

Convolution A special product of f * g is defined by and is called the convolution of f and g. Note: f * g = g * f

IF f(t) and g(t) are piecewise continuous on [0, ) and THEOREM 4.9 IF f(t) and g(t) are piecewise continuous on [0, ) and of exponential order, then Convolution Theorem Proof

Holding  fixed, let t =  + , dt = d The integrating area is the shaded region in Fig 4.32. Changing the order of integration:

Fig 4.32

Example 3 Find Solution Original statement = L{et * sin t}

Inverse Transform of Theorem 4.9 L-1{F(s)G(s)} = f * g (4) Look at the table in Appendix III, (5)

Example 4 Find Solution Let then

Example 4 (2) Now recall that sin A sin B = (1/2) [cos (A – B) – cos (A+B)] If we set A = k, B = k(t − ), then

Transform of an Integral When g(t) = 1, G(s) = 1/s, then

Examples:

Periodic Function f(t + T) = f(t): periodic with period T If f(t) is a periodic function with period T, then THEOREM 4.10 Transform of a Periodic Function

Proof Use change of variable

Example 7 Find the L. T. of the function in Fig 4.35. Solution We find T = 2 and From Theorem 4.10,

Fig 4.35

Example 8 The DE (13) Find i(t) where i(0) = 0, E(t) is as shown in Fig 4.35. Solution or (14) Because and

Assuming R=1ohm, L=1 henry Then i(t) is described as follows and see Fig 4.36: (15)

Fig 4.36

4.5 The Dirac Delta Function Unit Impulse See Fig 4.43(a). Its function is defined by (1) where a > 0, t0 > 0. For a small value of a, a(t – t0) is a constant function of large magnitude. The behavior of a(t – t0) as a  0, is called unit impulse, since it has the property . See Fig 4.43(b).

Fig 4.43

The Dirac Delta Function This function is defined by (t – t0) = lima0 a(t – t0) (2) The two important properties: (1) (2) , x > t0

Proof The Laplace Transform is THEOREM 4.11 For t0 > 0, Transform of the Dirac Delta Function Proof The Laplace Transform is

When a  0, (4) is 0/0. Use the L’Hopital’s rule, then (4) becomes 1 as a  0. Thus , Now when t0 = 0, we have

Example 1 Solve subject to (a) y(0) = 1, y’(0) = 0 (b) y(0) = 0, y’(0) = 0 Solution (a) s2Y – s + Y = 4e-2s Thus y(t) = cos t + 4 sin(t – 2)U(t – 2) Since sin(t – 2) = sin t, then (5) See Fig 4.44.

Fig 4.44

Example 1 (2) (b) Thus y(t) = 4 sin(t – 2)U(t – 2) and (6)

Fig 4.45