Moulton Lectures

Electro-Acoustics

Lecture 13

Basic Laplace 2

Presented by:

Dave L Moulton

Location: Thales Acoustics Harrow UK

Date: 29-January-2003

Content

I will continue on from Lecture 12 and explain more about the S-plane and also the use of partial fractions to solve inverse Laplace transforms. Hopefully lectures 12 and 13 will give the budding acoustics engineer a minimum set of design tools needed to analyse acoustic problems. In particular the essential tools needed to analyse the stability of an ANR system, to be covered during another lecture in this series.

On with the lecture

Lets take a step back to previous lectures where we analysed the response of a mechanical system comprising a Mass M_m , Spring Compliance C_m and Resistance R_m. We will consider the system to be acting under the influence of an external Force F_m(t), this will result in the system undergoing a displacement x(t). You have seen this set up before in lecture 4. As a reminder of the set up figure 13.1 is a diagram of the system.

Figure 13.1

The equation for the motion of this system is shown below:

What this differential equation describes is the relationship between the Applied Force and the resulting displacement, both Force and displacement are functions of time.

Now let us analyse this equation using Laplace. We will make the assumption that at time t=0 all initial conditions are also equal zero.

Taking Laplace transforms of both sides we have:

Suppose we wish to find the displacement X(s) , we need to rearrange the equation and make X(s) the subject:

Thus:

We can write this expression in a more standard form by dividing through by M_m.

We know from lecture 12 by comparing equations [12.48] and [12.49] the quadratic equation in the denominator of [13.5] can be expressed in terms of the natural frequency w₀ and damping coefficient z as:

Factorising the quadratic gives:

Where:

and

I will also define the angular frequency w_d as

Impulse Response

Now lets consider the response of the system to an impulse, this is when the force gives the system an initial kick and is never applied again afterwards.

The Laplace transform of an un-scaled impulse is always equal to 1. I will explain more about impulses in a future lecture.

Thus for an impulse:

The nature of the impulse response tells us a lot about the natural response of the system. In fact it is very useful to plot the solutions for ‘s’ onto the s-plane and see how those solutions vary with changes in damping coefficient.

The solutions for ‘s’ take the form:

and the complex conjugate

Poles

Each of these solutions when applied to equation [13.11] will cause the function to go to infinity. Any solution of ‘s’ which has this effect is called a pole. Thus the solutions ‘s_a’ and ‘s^*_a’ are the poles of the equation.

These poles can be plotted in the S-plane as shown:

Figure 13.2

Notice that a line drawn from the origin to either of the poles has a length equal to the angular frequency w₀.

How the s-plane poles change as z changes

If we keep the values of Mass M_m and Compliance C_m constant and only increase the value of the Resistance R_m , then z will increase independent of w₀. We can plot the location of the two poles for different values of z . We will see that for the standard format of second order denominator equation [13.14] the resulting pairs of poles will sit on a semicircle in the left half s-plane.

As the value of z increases from 0 to 1 (The condition for complex poles) then the poles will map out a circular locus. The radius of this locus is equal to the natural frequency w₀.

Figure 13.3

Notice that the further the pairs of poles are from the imaginary axis, the more damped the system becomes.

Further examples of S-plane poles and the resulting time domain response.

Inverse Laplace Transforms

Now lets look at some methods for finding inverse Laplace Transforms. There are three methods that I want to cover in this lecture.

In lecture 12 we defined the inverse Laplace Transform as a rather tricky contour integral.

This is an Integral we would prefer to avoid if possible. It so happens that in most cases we can avoid having to evaluate [13.15]. The reason is simple, there are books with pages upon pages of Laplace Transforms. Thus in most cases we only need to arrange our Laplace polynomial equation into something that we know has an inverse transform. For example if we had a Laplace equation which looked like:

Then we could easily identify this as having the format of a cosine function with the shifting of an exponential multiplier.

Lets take an example of a third order Laplace transform and solve it using three different reduction methods. These reduction methods all involve breaking the Laplace function down into partial fractions and then using look up tables to find the inverse of each partial fraction component.

Consider the Laplace function:

Method 1: Traditional Partial Fractions

The first thing we need to do here is factorise the equation, this can usually be done using a calculator.

Thus:

With this particular equation only the denominator can be factorised:

Now we apply the method for breaking the function down into simple pieces. Each piece can then be easily transformed into the time domain via common look up tables of inverse Laplace Transforms.

We can already see that the inverse Laplace transform will take the form:

All we need to do is find values for the coefficients A, B and C.

This is how we do it:

Firstly multiply both sides of equation [13.20] by the denominator:

Now we expand the Right hand side.

Now we compare coefficients. Firstly lets compare all the coefficients of s²on both sides of equation [13.23]:

Coefficients of s:

Coefficients of 1.

Here we have 3 equations with 3 unknowns. This is a very easy equation to solve, which yields the results:

, ,

Thus the solution is:

Method 2: Direct Coefficient Extraction Method

This is a neat method for finding the Partial coefficients without having to expand out the whole function. We start from the form shown in equation [13.20], thus:

lets start by finding the coefficient ‘A’

Multiply both sides by (s + 1), thus:

Now evaluate the function when S =-1. This results in all the terms on the right hand side going to zero except for ‘A’, thus:

For B:

For C:

Thus we get the same result as [13.27]

Method 3: Heaviside’s Expansion Formula

This is by far the quickest method and leads directly to the inverse transforms. I will explain the derivation of Heaviside’s method and then we will apply it to our equation:

First lets consider that our Laplace function F(s) can be represented as a single polynomial fraction with a numerator function P(s) and a denominator function Q(s), thus:

Now assume that we can write this function as a series of partial fractions. The number of partial fractions will correspond to the order of the polynomial Q(s). Lets assume that Q(s) is of order ‘n’.

Now to find a general expression for establishing the k^th coefficient.

Using a similar technique to that of Method 2, we first multiply both sides of [13.31] by (s-a_k), giving:

Instead of letting s = a_k, we will instead let s approach the limit of a_k, s® a_k. As a result of this all terms on the right hand side of [13.32] will become infinitely small compared to A_k and therefore be ignored: