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The Laplace Transform The Laplace transform of a function f (t) is defined as: Z ∞ F (s) = L{f (t)} = f (t)e−st dt. 0

The transformed function F (s) provides an alternative representation of the original function f (t). The key feature of the Laplace transform is that it replaces the real-valued variable t, which you can think of as representing time, with the complex-valued variable s, which really does not represent anything – concrete, that is. Despite its abstract and obtuse nature, the Laplace transform provides an extremely powerful method for solving differential equations, because it allows one to transform differential equations into algebraic ones. It also provides a convenient mathematical structure for representing feedback control loops. For this reason, elementary control theory uses the language of Laplace transforms. I would go as far as to say that it provides the simplest mathematical framework for understanding control. Let me be clear: no one likes the Laplace transform. Many have tried to explain control without it. However, the result is a dumbed-down version of control that misses many critical elements. No, you do not need the Laplace transform to configure or tune a PID controller (in fact, it would probably be detrimental to employ it in such practice). But if you even want to understand control concepts beyond a superficial level and consider more advanced applications, then you really need to understand the Laplace transform. The best way to understand the Laplace transform is by example: Z ∞ L{f (t) = 1} = e−st dt, 0  −1 −st ∞ = e  , s 0 −1 (0 − 1) , (implicitly assuming that Re(s) > 0) = s 1 = . s

Z

∞

te−st dt 0 ∞ Z ∞ −st e −te−st  dt, + = s s 0 0 ∞ e−st  =− 2  , s 0 1 = 2. s

L{f (t) = t} =

Z

(using integration by parts)

∞

t2 e−st dt 0  Z ∞ −t2 e−st ∞ e−st = dt, + 2t  s s 0 0

L{f (t) = t2 } =

1

(using integration by parts)

Z 2 ∞ −st te dt, s 0 2 (using result from above) = 2. s =

L{f (t) = eat } =

Z

∞

eat e−st dt,

Z 0∞

e(a−s)t dt, ∞ e(a−s)t  =  , a−s  =

0

0

1 = . s−a L{f (t) = sin(ωt)} = =

Z

∞

Z 0∞

sin(ωt)e−st dt, sin(ωt)e−st dt,

Z 0∞

eiωt − e−iωt −st e dt, (using complex numbers) 2i 0 Z ∞ (iω−s)t e − e−(iω+s)t dt, = 2i 0 ∞ e−(iω+s)t  e(iω −s)t + =  , 2i(iω + s)  2i(iω − s) =

0

= =

= =

−1 −1 , + 2i(iω − s) 2i(iω + s) −1 −1 , + 2(−ω − si) 2(−ω + si) ω + si ω − si + , 2 2 2(ω + s ) 2(ω 2 + s2 ) ω . 2 ω + s2

We rarely need to evaluate these integrals in practice because others have already done so. These transforms are available in tables, which can be found in your textbook, mathematical handbooks, or on the web. When applying these tables to a specific problem, the following properties of the Laplace transform are useful: 1. L{af (t)} = aL{f (t)}, where a is a constant. 2. L{af (t) + bg (t)} = aL{f (t)} + bL{g(t)}, where a and b are constants. You can also calculate the inverse Laplace transform: Z γ+iT 1 est F (s)ds. lim f (t) = L−1 {F (s)} = 2πi T →∞ γ−iT 2

However, direct calculation of the inverse transform involve integration in the complex domain, which typically requires the use of the Cauchy Residue Theorem. In practice, we never directly calculate the inverse transform but rather take advantage of the following property: f (t) = L−1 {L{f (t)}} . The main utility of the Laplace transform derives from the following property:   Z ∞ df df −st L = e dt, dt dt 0 ∞ Z ∞ f (t)se−st dt, (using integration by parts) = −f (t)e−st 0 + 0

= sF (s) − f (0).

The same approach can also be used to determine the Laplace transform for derivatives of arbitrary order, which is given by  n      df n−2  df n−1  d f n n−1 n−2 df  − . . . − s n−2  − n−1  = s F (s) − s f (0) − s L . dtn dt t=0 dt dt t=0 t=0

Note, the Laplace transform converts a derivative with respect to t into a polynomial in s. As we demonstrate with the examples below, this property enables easily us to solve linear differential equations. Example 1. Solve the differential equation dy + ay = 0, dt

y(0) = b.

Solution: We begin by taking the Laplace transform of both sides of the differential equation: b ✯+ ✟✟ y(0) aY (s) = 0. sY (s) − ✟ Collecting terms and using the initial condition, we have Y (s) =

b . s+a

Using the result derived above, the inverse Laplace transform of Y (s) is: y(t) = be−at , which is the solution to the differential equation. Example 2: Solve the differential equation dy + ay = b, dt

y(0) = c.

Solution: We first take the Laplace transform of the both sides of the differential equation: c b ✯+ ✟✟ sY (s) − ✟ y(0) aY (s) = . s

3

Collecting terms and rearranging yields: Y (s) =

c b . + s(s + a) s+a

We simplify this expression by considering the partial fraction expansion of the last term in the above expression:   b 1 1 b = − . s(s + a) s+a a s Using the above result, we obtain the simplified expression   b 1 1 c + − . Y (s) = s+a a s s+a Taking the inverse transform yields y(t) = ce−at +

 b  1 − e−at . a

Example 3. Solve the differential equation dy + ay = sin(ωt), dt

y (0) = 0.

Solution: We again begin by taking the Laplace transform of both sides of the differential equation: 0 ✯+ ✟✟ sY (s) − ✟ y(0) aY (x) =

ω . s2 + ω 2

Collecting terms, we have Y (s) =

ω 1 . 2 s + a s + ω2

Few tables will include such an expression. However, we need to simplify this expression into terms that are present in tables using a partial fraction expansion: Bs + C 1 ω A . + 2 = s + a s2 + ω 2 s+a s + ω2 Collecting terms, we obtain the following equation for the coefficients: A(s2 + ω 2 ) + (Bs + C)(s + a) = ω. We can further simply as follows: (A + B)s2 + (aB + C )s + (ω 2 A + aC) = ω. By matching the coefficients of the different powers of s, we have s2 : A = −B,

s1 : aB = −C,

s0 : ω 2 A + aC = ω. 4

Solving these equations yields: ω , a2 + ω 2 −ω , B= 2 a + ω2 aω . C= 2 a + ω2

A=

We then have Y (s) =

ω a2 + ω 2



a s 1 + 2 − 2 2 s+a s +ω s + ω2



Taking the inverse Laplace transform yields:   ω a −at sin( ωt) − cos( ωt) . y(t) = 2 e + ω a + ω2 We can further simplify as follows:   a 1 ω ω −at cos(ωt) √ e + √ 2 y(t) = 2 sin(ωt) − √ 2 a + ω2 a + ω2 a + ω2 a2 + ω 2 (using the trigometric identity: sin(ωt + φ) = cos(φ) sin(ωt) + sin(φ) cos(ωt)) 1 ω e−at + √ sin(ωt + φ) = 2 2 2 a +ω a + ω2 where φ = − arctan

ω a

.

As we will demonstrate later, this form is more convenient for analysis. Observe in the all examples above that we transform a differential equation in the variable t to an algebraic one in the variable s. This is the key feature of the Laplace transform, namely that it eliminates the differentials. The Laplace transform also has a few additional convenient features that we will use in the course: • Transform of an integral.  Z Z t f (u)du = L 0

∞ 0

Z

t



f (u)du e−st dt, 0

(using integration by parts)  Z t  ∞ Z ∞ e−st 1 −st  f (t) f (u)du e  + =− dt, s s 0 0 0 Z ∞ e−st f (t) = dt, s 0 1 = L {f (t)} . s

5

• Transform of the step function S(t). The step function is defined as  0 : t...