Moulton  Lectures

 

On

Electro-Acoustics

 

 

 

Presented by:

Dave L Moulton

 

 

 

 

Location:  Thales Acoustics Harrow UK

Date:  22-January-2003

 

Content

 

I have decided that now is the time to break off from ANR and give a couple of lectures on a very important mathematical toolset. Namely the Laplace transform. I want to explain the definition of the Laplace transform  and show how it can be used to give an insight into the stability of a system and the resulting time domain response.

 

On with the lecture

 

What is a Laplace Transform?

Before I define the transform I would first like to give a few examples of systems that can be analysed in the time domain.  By time domain I mean that the system response is a  function of time along with the input and output signals.

 

Consider the simple low pass filter and high pass filter systems shown in figure 12.1 below:

 

Figure 12.1

The time domain functions of the Low pass and High pass filters are basically differential and Integral equations. The output V_out(t)  is the solution to these equations for a given input V_in(t).  So from a simple circuit we can derive the Integral and Differential equations to describe the affect of the circuit in the time domain.

 

But what if the circuit is not quite so trivial.  In order to predict the output for a given input we may have to describe the circuit as a very complex differential equation. We would then need to find the solution to the differential equation in order to predict the output in the time domain.

 

 

Figure 12.2

 

This could be very laborious and time consuming, not to mention the possibility of making a mistake in deriving and solving the differential equations.

 

Fortunately there are ways of making the mathematics easier to manage. One such method transforms the system time variable into another temporary domain.

 

The transformation method is known as the Laplace Transform.  Named after Pierre Simon Marquis de Laplace (1749 – 1827) the French analyst, probability theorist and physicist.

 

The Laplace Transform is defined by the Integral:

F(s) is sometimes expressed as: 

 

The inverse Laplace Transform is defined by the contour integral:

 

The Laplace Transform enables us to transform a complicated differential-Integral equation for a system into a much more friendly looking polynomial.

 

For example the filter circuits shown in figure 12.1 can have their respective input and system responses transformed in terms of ‘s’ instead of  ‘t’.

 

Without going into the detail at this stage, the transforms are:

 

L{V_in(t)} = V_in(s) ………….[12.4]

 

The Low Pass Filter Transforms to:

 

The High Pass Filter Transforms to:

 

As a consequence of the above transforms the output will also be a function of ‘S’.

L{V_out(t)} = V_out(s) ………….[12.7]

 

Applying these transforms to  figure 12.1 will show how much simpler the mathematics has become. We have removed all of the Integrals and Differentials from the system and replaced them with algebraic functions of ‘S’.  Figure 12.3 illustrates the improvement:

 

 

Figure 12.3

 

Now for some simple examples of Laplace Transforms.  I will only do a few of what I consider to be the more significant transforms. 

 

Unit Step Function

 

The unit step function is easily defined as having a value of 1 for all time greater than or equal to zero:

 

Thus written mathematically we have:

 

The unit step function can be illustrated as:

Figure 12.4

 

Applying the Laplace Transform:

Thus:

Performing the Integral:

Thus

 

Exponential Function  e^at.

 

This is probably one of the most significant transforms, since it can be used to easily find the transform of other functions that are composed of exponentials, such as COS, SIN, COSH, SINH, etc…

 

Applying the Laplace Transform:

Thus

Performing the integral

 

Resulting in:

 

From this last result we can easily determine the Laplace transform of other simple exponential forms:

 

 

 

From these we can easily evaluate the Laplace transform for a sine function.

 

Using Exponentials we know that:

Thus:

Transforming:

Resulting in:

 

 

Using the same principal it is very easy to show the transforms for COS, COSH and SINH:

 

 

 

 

There are countless books and tables of Laplace transforms for many different functions.

 

I now want to focus on one more very important Laplace Transform, the one that deals with differentials. Understanding this will save you a lot of grief when trying to solve acoustical, mechanical and electrical differential equations,

 

I will express the first differential of f(t) as f’(t), The second differential as f’’(t) etc…….

 

Now lets first have a look at the Laplace transform for the first differential.

 

We have:

 

To solve this integral we need to use the method known as ‘Integration by Parts’.  Just as a reminder:

 

In this case we let:

    and   

Thus:

 

this leads to:

 

Better written as:

 

 

Using the same technique as described above it is very easy to show the Laplace transform for any Positive Integer order differentials.

 

For example :

 

Going straight to the integral:

Note that the last integral is a repeat of the process for the first differential.

 

 

Giving the final result:

 

The sequence for any differential of order ‘n’ has a Laplace transform which takes the form:

 

 

Now lets have an example.

 

Remembering back to Lecture 8, where we found a way of measuring the Compliance and Resistance of an earcushion.  At that time the method used involved creating a differential equation and then solving the equation directly, i.e staying in the time domain.  Lets now use Laplace to solve this equation.

 

Firstly a reminder of the mechanics of the problem.

 

Figure 12.5

 

The derived differential equation took the form:

 

We also had the initial condition:  t = 0, x = x₀

Now lest apply the Laplace Transform to this equation.

 

We know from the initial conditions that x(0) = x₀.

 

Hence:

 

Rearranging:

Thus:

 

 

Now we have already seen from equation [12.7] that the Inverse Laplace transform of [12.41] will take the form:

 

 

Note the ease at which we arrived at this answer. Using Laplace it only took 5 steps to reach the answer.  Whereas using the conventional time domain method in Lecture 8 it took 9 steps.

 

NOTE: The Most Important Thing to Remember when Solving Differential or Integral Equations using Laplace is that the initial conditions MUST be determined at the outset.

 

Had I not considered the initial condition for the above example, our final solution would have been very difficult to solve, because our Laplace expression would have ended up looking like:

 

The only solutions to this equation are:

  or 

 

There are no inverse Laplace transforms for these solutions except x(t)=0

 

Now to try and answer an obvious question.

 

What is ‘s’ ?

 

To answer this question we need to go back to lecture 5 where we investigated the nature of the differential equation describing the motion of a Mass M_m a spring with Compliance C _m and a Frictional Resistance R _m.  The natural response of the system could be described by a differential equation which takes the form:

 

If we make the assumption that at time t=0, f’(0) =0 and f(0) = 0.

We can take the Laplace transform for the equation:

 

A little bit of factorising:

 

 

This equation has a couple of solutions when the term in brackets equals zero.  Hence the two possible solutions for ‘s’ are:

 

If we assume that the system can oscillate (i.e has overshoot) then.

 

Lecture 5 lead to the more usual presentation for this solution in terms of the natural angular frequency w₀ and the damping coefficient z.

 

Thus giving:

 

So ‘s’ has a real and an imaginary part.  We can express ‘s’ in a more usual form.

 

 

Now for the really important bit.  The solution to the differential equation can be expressed as:

 

From the above it should be quite clear that ‘s’ tells us a lot about the  response nature of the system.  For example the sigma term ‘s’  gives a good indication of the nature of the damping in the system.  If s is positive them the system is at risk of growing in amplitude and going unstable. If s is negative then the system is damped and will eventually decay to zero.  If s is zero, then the system will oscillate indefinitely with constant amplitude.

 

Understanding the S solutions to a differential equation can give a very good indication as to what will keep the system stable and what will make it go unstable.   Stability analysis will be covered in another lecture.

 

The s-Plane

 

Since we now have an understanding of the significance of ‘s’. I will briefly introduce you to the s-plane.   The s-plane is basically a plot of all system ‘s’ solutions.  The plane is a bit like the cartesian xy plane except that the  ordinate (y) axis is complex and the abscissa (x) axis is real.

 

Just as an appetizer I have shown the S-plane plots for the solution described in equation [12.49].

 

 

Figure 12.6

 

More about the s-plane in the next lecture.

 

 

End of Lecture