Topic 1 - Laplace Transforms for Solving Differential Equations

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Overview

In this section we will examine how the Laplace Transform can be used to solve a variety of engineering problems involving differential equations.

Objective

The student will examine how the Laplace Transform can be used to solve a variety of engineering problems involving differential equations.

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Many engineering problems involve relationships between functions and their derivatives. One example would be a relationship that involves both displacement (x) and its first derivative with respect to time, velocity (dx/dt).

Equations such as this are referred to as differential equations. Often these relationships are complicated enough that a direct solution using the calculus of integration is not possible.

However, by using the Laplace Transform approach, we have another tool by which such problems can be solved. This discussion will cover:

The difference between time domain and frequency domain
The mathematical concept of the Laplace Transform and Inverse Laplace Transform
The use of common transforms
Solution of engineering problems involving differential equations by use of Laplace.

Topic 1 – Time and Frequency Domains

We often refer to the Laplace Transform (named after its inventor Pierre-Simon Laplace) as taking us from the time domain into the frequency domain and the Inverse Laplace Transform as taking us from the frequency domain to the time domain. This statement in and of itself confuses many students from the outset. So, let us begin by examining what we mean by these two domains.

Here is a practical application which will help with our understanding. The figure below is a complex electrical signal (an example could be the output of an accelerometer mounted on a running engine). It is difficult to make sense of this data.

Figure of the amplitude of a random electrical signal plotted versus time

However, using a computer and the mathematical algorithms of curve fitting, it is possible to derive a time-based function that represents this signal over a small time period.

And once that function has been derived, the computer can perform operations to break that function into its components of sine and cosine (This process uses another transform, called the Fourier Transform). The computer can then produce a plot of individual discrete frequency sinusoidal waves that make up the complex signal. This concept is shown in the next figure.

Diagram of how a Fast Fourier Transform analysis convers signals from the time domain to the frequency domain

unknown origin

This can turn a time domain signal that looks like one on the upper left of the figure below into a frequency domain spectrum like that shown on the lower right of the same figure. It is then possible to relate the responses at various frequencies to their sources within the machine.

Figure showing the same signal in the time domain and the frequency domain

Original image : dplot.com

This demonstrates how the time domain and the frequency domain are two completely different ways of looking at exactly the same thing. Thus, when the Laplace Transform takes us from one domain to the other, we have not changed the problem, we are merely looking at it from a different perspective. If, in this new domain, the problem can be solved more easily, we can then Inverse Transform back to the original domain with our answer.

We live in the time domain, and that is how we like to look at things. Thus, we would like to look at a simple time domain relationship, such as the displacement of an object with time. But if we have a complicated relationship in the time domain that involves displacement, velocity (time derivative of displacement) and acceleration (time derivative of velocity or second time derivative of displacement) then we often cannot see a way to solve for that simple relationship of displacement with time. However, if we can transform the relationship to the frequency domain, solve the problem there, and then inverse transform back to the time domain, we have obtained exactly what we wanted, a time domain expression for the displacement as a function of time.

In the world of Laplace, we often refer to the frequency domain as the “S” domain, because when Laplace transformed a function with variable x in the time domain, and transformed the function to the frequency domain, he used the variable s to avoid confusion between use of the same variable name in two different domains.

Topic 2 – Laplace Transforms

Laplace transforms can be used to solve differential equations for the analysis of systems varying with time.

If we can create a simplified model of a system (i.e. functions or equations which describe the system) then we can solve dynamic problems involving

Electrical circuits
Harmonic oscillators
Optical devices
Mechanical systems

Laplace Transforms are often interpreted as a transformation from the time domain to the frequency domain (or “s” domain where s is a complex number of the form a + bi, recalling that a and b are real numbers and i is an imaginary number representing the square root of negative one)

Time Domain : Inputs and outputs are functions of time (unit is usually seconds)
Frequency Domain : Inputs and outputs are functions of complex angular frequency (units of radians/second)

The Laplace Transform is mathematically defined as the following interval

$F (s) = L \{f (t)\} = \int_{0}^{∞} e^{- s t} f (t) d t$

Where F(s) is a function F of variable s in the frequency domain and it is the Laplace Transform of f(t) which is a function f of variable t in the time domain. F(s) is a complex function, where s is a complex variable as previously discussed, L is the operational symbol for the Laplace Integral.

The Inverse Laplace Transform, returning from the frequency domain to the time domain is defined by the following:

$f (t) = L^{- 1} \{F (s)\} = \frac{1}{2 π i} \lim_{T \to ∞} \int_{γ - i T}^{γ + i T} e^{s t} F (s) d s$

Fortunately, we seldom have to perform the actual integration, because others have done the work of transforming typical functions and building tables of the transforms, thus allowing us to simply use the following tables.

	$f (t)$	$F (s)$
1	$Unit impulse δ (t)$	1
2	$Unit step 1 (t)$	$\frac{1}{s}$
3	$t$	$\frac{1}{s^{2}}$
4	$\frac{t^{n - 1}}{(n - 1)!}$ $(n = 1, 2, 3, ...)$	$\frac{1}{s^{n}}$
5	$t^{n}$ $(n = 1, 2, 3, ...)$	$\frac{n!}{s^{n + 1}}$
6	$e^{- a t}$	$\frac{1}{s + a}$
7	${t e}^{- a t}$	$\frac{1}{{(s + a)}^{2}}$
8	$\frac{1}{(n - 1)!} t^{n - 1} e^{- a t}$ $(n = 1, 2, 3, ...)$	$\frac{1}{{(s + a)}^{n}}$
9	$t^{n} e^{- a t}$ $(n = 1, 2, 3, ...)$	$\frac{n!}{{(s + a)}^{n + 1}}$
10	$sin ω t$	$\frac{ω}{s^{2} + ω^{2}}$
11	$cos ω t$	$\frac{s}{s^{2} + ω^{2}}$
12	$sinh ω t$	$\frac{ω}{s^{2} - ω^{2}}$
13	$cosh ω t$	$\frac{s}{s^{2} - ω^{2}}$
14	$\frac{1}{a} (1 - e^{- a t})$	$\frac{1}{s (s + a)}$
15	$\frac{1}{b - a} (e^{- a t} - e^{- b t})$	$\frac{1}{(s + a) (s + b)}$
16	$\frac{1}{b - a} ({b e}^{- b t} - {a e}^{- a t})$	$\frac{s}{(s + a) (s + b)}$
17	$\frac{1}{a b} [1 + \frac{1}{a - b} ({b e}^{- a t} - {a e}^{- b t})]$	$\frac{1}{s (s + a) (s + b)}$
18	$\frac{1}{a^{2}} (1 - e^{- a t} - {a t e}^{- a t})$	$\frac{1}{{s (s + a)}^{2}}$
19	$\frac{1}{a^{2}} (a t - 1 + e^{- a t})$	$\frac{1}{s^{2} (s + a)}$
20	$e^{- a t} sin ω t$	$\frac{ω}{{(s + a)}^{2} + ω^{2}}$
21	$e^{- a t} cos ω t$	$\frac{s + a}{{(s + a)}^{2} + ω^{2}}$
22	$\frac{ω_{n}}{\sqrt{1 - ζ^{2}}} e^{- ζ ω_{n} t} sin ω_{n} \sqrt{1 - ζ^{2}} t$	$\frac{ω_{n}^{2}}{s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}}$
23	$- \frac{1}{\sqrt{1 - ζ^{2}}} e^{- ζ ω_{n} t} sin (ω_{n} \sqrt{1 - ζ^{2}} t - Φ)$ $Φ = {tan}^{- 1} \frac{\sqrt{1 - ζ^{2}}}{ζ}$	$\frac{s}{s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}}$
24	$1 - \frac{1}{\sqrt{1 - ζ^{2}}} e^{- ζ ω_{n} t} sin (ω_{n} \sqrt{1 - ζ^{2}} t + Φ)$ $Φ = {tan}^{- 1} \frac{\sqrt{1 - ζ^{2}}}{ζ}$	$\frac{ω_{n}^{2}}{s (s^{2} + 2 ζ ω_{n} s + ω_{n}^{2})}$
25	$1 cos ω t$	$\frac{ω^{2}}{s (s^{2} + ω^{2})}$
26	$ω t - sin ω t$	$\frac{ω^{3}}{s^{2} (s^{2} + ω^{2})}$
27	$sin ω t - ω t cos ω t$	$\frac{2 ω^{3}}{{(s^{2} + ω^{2})}^{2}}$
28	$\frac{1}{2 ω} t sin ω t$	$\frac{s}{{(s^{2} + ω^{2})}^{2}}$
29	$t cos ω t$	$\frac{s^{2} - ω^{2}}{{(s^{2} + ω^{2})}^{2}}$
30	$\frac{1}{ω_{2}^{2} - ω_{1}^{2}} (cos ω_{1} t - cos ω_{2} t)$ $(ω_{1}^{2} \neq ω_{2}^{2})$	$\frac{1}{(s^{2} + ω_{1}^{2}) (s^{2} + ω_{2}^{2})}$
31	$\frac{1}{2 ω (sin ω t + ω t cos ω t)}$	$\frac{s^{2}}{{(s^{2} + ω^{2})}^{2}}$

Table data from Ogata

Note that for any function F(s) which can be broken into components, such that

$F(s) = F_{1} (s) + F_{2} (s) + F_{3} (s) + ........... F_{n (s)}$

Then the inverse transform is the sum of the inverses of the individual terms

$L^{- 1} [F(s)] = L^{- 1} F_{1} (s) + L^{- 1} F_{2} (s) + L^{- 1} F_{3} (s) + ........... L^{- 1} F_{n (s)}$

$= f_{1} (t) + f_{2} (t) + f_{3} (t) + ........... f_{n} (t)$

Application 1

Find the Laplace Transform of f(t)

$f(t) = 3sin(5t + 45^{o})$

(Note: recall from Trig that sin (A + B) = sinAcosB + cosAsinB)

$f(t) = 3 sin5t cos 45^{o} + 3 cos5t sin 45^{o}$

$= 3(0.707) sin5t + 3(0.707) cos5t$

$= 2.121 sin5t + 2.121 cos5t$

Transform using lines 10 and 11 of the chart

$F(s) = 2.121[5 / (s^{2} + 5^{2})] + 2.121[s/(s^{2} + 5^{2})] = 2.121[(s + 5)/ (+ 25)]$

Application 2

Find the Laplace Transform of f(t)

$f(t) = 0.03(1 – cos2t)$

Transform using lines 2 and 11 of the chart

$F(s) = 0.03[1/s] - 0.03 [s/(s^{2} + 2^{2})]$

$F(s) = 0.03[1/s][(s^{2} + 2^{2}) / (s^{2} + 2^{2})] - 0.03 [s / (s^{2} + 2^{2})] [s/s]$

$= (0.03 s^{2} + 0.12) / s (s^{2} + 2^{2}) - 0.03 s^{2} / s (s^{2} + 2^{2})$

$= (0.03 s^{2} + 0.12 - 0.03 s^{2}) / s (s^{2} + 2^{2})$

$= 0.12 / s (s^{2} + 2^{2})$

$= 0.12 / s (s^{2} + 4)$

Application 3

Find the Inverse Laplace Transform of F(s)

$F(s) = (6s + 3)/ s^{2}$

$= 6 s / s^{2} + 3 / s^{2}$

$= 6 / s + 3 / s^{2}$

Inverse transform using lines 2 and 3 of the chart

$f(t) = L^{- 1} [6 / s] + L^{- 1} [3 / s^{2}]$

$= 6 + 3 t$

Topic 3 – Inverse Laplace Transforms using Partial Fractions

When our function in the “s” domain is more complicated, we may have to use the concept of Partial Fractions to simplify the F(s) function such that we can use the terms in the table.

To do this, we need to recall zeros and poles. Consider a function like G(s)

$G(s) = \frac{K (s + 3) (s + 20)}{s (s - 4) (s + 6) {(s + 15)}^{2}}$

The poles of this function are values of s at which G(s) = infinity (i.e. the denominator of the function goes to zero). In this example the poles are = 4, 0, -6, -15 and -15

NOTE: -15 is referred to as a repeated pole, since it appears twice due to (s + 15) being squared in the denominator

The zeros of this function are values of s at which G(s) = 0 (i.e. the numerator of the function goes to zero). In this example the poles are -3 and -20. Technically, the function should have as many zeros as poles. (in which case for this example the zeros are -2, -10 and three repeated zeros at infinity)

For a function F(s) = B(s)/A(s) F(s) can be expanded using Partial Fractions and the poles of A(s)

$F (s) = \frac{a_{1}}{s + p_{1}} + \frac{a_{2}}{s + p_{2}} + \frac{a_{3}}{s + p_{3}} + ........... \frac{a_{n}}{s + p_{n}}$

Where the p_n values are the poles and each a_n (named the “residue) can be found from

$a_{n} = (s + p_{n}) \frac{B (s)}{A (s)}$ evaluated at $s = p_{n}$

When the denominator contains repeated poles, as in A(s) = (s+1)²then the expansion must use both (s+1) and (s+1)²

So the partial fraction expansion is

$F (s) = \frac{a_{1}}{s + p_{1}} + \frac{a_{2}}{{(s + p_{1})}^{2}}$

And

$a_{2} = {(s + p_{1})}^{2} \frac{B (s)}{A (s)}$ evaluated at $s = p_{n}$

But a₁ comes from the derivative

$a_{1} = \frac{d {(s + p_{1})}^{2}}{d s}$ $\frac{B (s)}{A (s)}$ evaluated at $s = p_{n}$

This may be best understood by looking at an example.

Application 1

Find the inverse Laplace Transform of F(s)

$F (s) = \frac{(5 s + 2)}{(s + 1) {(s + 2)}^{2}}$

We must use partial fractions to get F(s) into a form that will allow us to use the chart

$F (s) = a / (s + 1) + b_{1} / (s + 2) + b_{2} / {(s + 2)}^{2}$

Finding the a and b coefficients:

$a = (5 s + 2) / {(s + 2)}^{2}$ evaluated at $s = - 1$

$= (- 5 + 2) / 1^{2}$

$= - 3$

$b_{2} = (5 s + 2) / (s + 1)$ evaluated at $s = - 2$

$= (- 10 + 2) / (- 2 + 1)$

$= 8$

$b_{1} = derivative of (5 s + 2) / (s + 1)$ evaluated at $s = - 2$

$= [5 (s + 1) - (5 s + 2)] / {(s + 1)}^{2}$ evaluated at $s = - 2$

$[5 (- 1) - (- 10 + 2)] / {(- 1)}^{2}$

$[- 5 - (- 8)] / 1 = 3$

$F (s) = - 3 / (s + 1) + 3 / (s + 2) + 8 / {(s + 2)}^{2}$

Inverse transform using lines 6 and 7 of the chart

$f (t) = - 3 e^{- t} + 3 e^{- 2 t} + 8 {t e}^{- 2 t}$

Topic 4 - Laplace Transforms of derivatives

When we need to take the Laplace Transform of a time domain function which involves derivatives, we need to understand how to treat the derivative terms.

While $L f (t)$ transforms to $F (s)$

$L \frac{d f (t)}{d t}$ transforms to $s F (s) - f (0)$

where f(0) is the initial condition for the function f(t), i.e. the value of f(t) when t = 0

$L \frac{d^{2} f (t)}{{d t}^{2}}$ transforms to $s^{2} F (s) - s f (0) - f^{'} (0)$

where f’(0) is the initial condition for the derivative of the function f(t), i.e. the value of f’(t) when t = 0

and similarly for any derivative

$L \frac{d^{n} f (t)}{{d t}^{n}}$ transforms to $s^{n} F (s) - s^{n - 1} f (0) - s^{n - 2} f^{'} (0) - .......... f^{(n - 1)'} (0)$

where f^(n-1)’(0) is the n-1 derivative of f(t) evaluated at t = 0

Thus, if we have a differential equation in the time domain, we can convert the differential equation to a simple algebraic equation in terms of s in the frequency domain by using the Laplace Transform.

We can convert the differential equation by changing $\frac{d}{dt} to s$ , $\frac{d^{2}}{{dt}^{2}} to s^{2}$ , etc. per the explanation just given. We can then solve this new equation in the frequency domain for F(s) and then the time solution f(t) is obtained by taking the Inverse Laplace Transform of the expression we just derived for F(s)

Application 1

Find the solution f(t) for the differential equation x” + 3x’ + 2x = 0 with initial conditions x(0) = 1 and x’(0) = -1

Take the transform of the DE

$[s^{2} X (s) - s x (0) - x^{'} (0)] + [3 s X (s) - 3 x (0)] + 2 X (s) = 0$

$[s^{2} X (s) - s (1) + 1] + [3 s X (s) - 3 (1)] + 2 X (s) = 0$

$[s^{2} + 3 s + 2] X (s) - s + 1 - 3 = 0$

$X (s) [s^{2} + 3 s + 2] = s - 1 + 3 = s + 2$

solve for X(s)

$X (s) = \frac{s + 2}{(s + 2) (s + 1)} = 1 / (s + 1)$

Inverse Transform using line 6 of the chart

$x (t) = e^{- t}$

Topic 5 - Laplace Transforms as a Solution Approach to Differential Equations

Let us consider the general equation of motion for a simple spring-mass system shown here.

spring-mass system example

The equation of motion, derived from Newton’s Second Law, would involve a mass times acceleration term (mx”), and forces derived from the spring (spring rate times displacement = kx) and the damper (velocity times damping coefficient = cx’) and any externally applied force, f(t)

The resulting equation of motion would be mx” + cx’ + kx = f(t)

Where m is mass, c is damping, k is spring rate, x is displacement, and x’ and x” are its derivatives (velocity and acceleration) and f(t) is a forcing function that varies with time

We will now examine a couple of equations of this form

Application 1

Begin by assuming no damping (c = 0) and no forcing function, f(t) = 0

So we have $mx” + kx = 0$

Rearrange to $x” + (k/m)x = 0$ and let $k/m = ω^{2}$

So we have $x” + ω^{2} x = 0$

Laplace transform this: $s^{2} X(s) - sx(0) - x’(0) + ω^{2} X(s) = 0$

Assume an initial displacement, but no initial velocity, so the initial conditions (IC) are: $x(0) = 5 and x’(0) = 0$

Apply these IC: $X(s)[s^{2} + ω^{2}] = 5s$

Rearrange: $X(s) = 5s/(s^{2} + ω^{2})$

And inverse transform using line 10 of the chart: $x(t) = 5cosωt$

Let’s examine this solution

It is an oscillating response of amplitude 5 and frequency ω
5 was the initial displacement, so it is oscillating at the amplitude we initially gave it
And it is oscillating at the frequency ω, which you will recall we set as ω = (k/m) ^½

In vibration theory, the square root of k over m is referred to as the natural frequency, which is the frequency that any spring-mass system in nature will oscillate at if caused to vibrate on its own. So our answer is exactly what vibration theory would lead us to expect.

Application 2

Let’s return to our spring-mass system and assume m = 1, k = 1 and c = 0 (undamped). But with a forcing function which is sinusoidal, sin3t. We will choose initial conditions of zero, i.e. x(0) = 0 and x’(0) = 0

So we have $x” + x = sin3t$

Transform this equation using line 10 of the chart

$s^{2} X(s) + X(s) = 3/(s^{2} + 3^{2})$

Solve this for X(s)

$X(s) = 3/(s^{2} + 1)(s^{2} + 9)$

Manipulate this using partial fractions into a form that can be used with the chart

$X(s) = (3/8)[1/(s^{2} + 1)] - (1/8)[3/(s^{2} + 9)]$

Inverse transform using line 10 of the chart

$x(t) = (3/8)sin(t) - (1/8)sin3t$

Let’s again examine our result.

There is an aspect of this which is a sinusoidal oscillation at frequency 3t, which would correspond to the forcing function which was sin3t
There is also a sinusoidal oscillation at frequency 1t, which would correspond to the natural frequency ω = (k/m)^½ = (1/1)^½ = 1

So not surprisingly, there is a portion of the response due to the applied force and a portion at the natural frequency due to our having given it an excitation

Application 3

spring-mass system example

Let us return to our simple spring-mass system $mx” + cx’ + kx = f(t)$

And let Let k = 4 lb/ft, c = 0.4 lb-sec/ft, m = 0.1 lb-sec²/ft

$mx’’ + cx’ + kx = 0$

$0.1 x’’ + 0.4 x’ + 4x = 0$

$x’’ + 4x’ + 40x = 0$

Apply Laplace (assume x’₀ = 0, initially at rest)

$s^{2} X(s) - {sx}_{0} + 4[ sX(s) - x_{0}] + 40X(s) = 0$

Solve for X(s)

$X(s) = \frac{(s + 4) x_{0}}{{(s + 2)}^{2} + 6^{2}}$ $= \frac{2 x_{0} + ({sx}_{0} + 2 x_{0})}{{(s + 2)}^{2} + 6^{2}}$

$= \frac{x_{0}}{3}$ $\frac{6}{{(s + 2)}^{2} + 6^{2}}$ $+ x_{0} \frac{(s + 2)}{{(s + 2)}^{2} + 6^{2}}$

Inverse Laplace using lines 20 and 21

$x(t) = x_{0} / 3 e^{- 2 t} sin6t + x_{0} e^{- 2 t} cos6t$

$= e^{- 2 t} [1 / 3 sin6t + cos6t] x_{0}$

Again examine the response and find these bits which will result in the decaying sinusoidal response shown

Figure showing the solution of x as it varies with time and the plotted response

Application 4

Consider another spring-mass system which has the equation of motion $x’’ + 9x = t$ with initial conditions $x(0) = 0 and x’(0) = 2.$

Transform the equation using line 3 of the chart

$[s^{2} X(s) – sx(0) -x’(0)] + 9X(s) = 1 / s^{2}$

$[s^{2} X(s) – 0 - 2] + 9X(s) = 1 / s^{2}$

$s^{2} X(s) + 9X(s) = 1 / s^{2} + 2$

$(s^{2} + 3^{2}) X(s) = 1 / s^{2} + 2$

$X(s) = 1/[s^{2} (s^{2} + 3^{2})] + 2 / (s^{2} + 3^{2})$

Rearrange in preparation for Inverse Transform

$\frac{(3^{2}) 1}{(3^{2}) s^{2} (s^{2} + 3^{2})}$ $+$ $\frac{(3^{2}) 2}{(3^{2}) (s^{2} + 3^{2})}$

$\frac{3^{2} + s^{2} - s^{2}}{(3^{2}) s^{2} (s^{2} + 3^{2})}$ $+$ $\frac{(3^{2}) 2}{(3^{2}) (s^{2} + 3^{2})}$

$\frac{s^{2} + 3^{2}}{(3^{2}) s^{2} (s^{2} + 3^{2})}$ $-$ $\frac{s^{2}}{(3^{2}) s^{2} (s^{2} + 3^{2})}$ $+$ $\frac{(2) (3^{2})}{(3^{2}) (s^{2} + 3^{2})}$

$\frac{1}{(3^{2}) s^{2}}$ $-$ $\frac{1}{(3^{2}) (s^{2} + 3^{2})}$ $+$ $\frac{(2) (3^{2})}{(3^{2}) (s^{2} + 3^{2})}$

$\frac{1}{(3^{2}) s^{2}}$ $-$ $\frac{3}{(3^{2}) (3) (s^{2} + 3^{2})}$ $+$ $\frac{(2) (3) 3}{(3^{2}) (s^{2} + 3^{2})}$

Inverse transform using lines 3 and 10 from the table

$x(t) = (1/9)t - (1/27) sin3t + (2/3)sin3t$

$= (1/9)t - (1/27)sin3t + (18/27)sin3t$

$= (t/9) + (17/27)sin3t$

Summary

The Laplace Transform and Inverse Laplace Transform allow us to move functional relationships between the time domain and the frequency domain so as to allow us to solve difficult time domain relationships in the frequency domain and then convert our solution back into the time domain.

TASK 1

Tutorial problems will allow the student to practice the concepts shown in this discussion.

References and Bibliography

Ogata, K. (2004). System Dynamics, 4th edition. Upper Saddle River, USA: Pearson Prentice Hall.