Moulton  Lectures

 

On

Electro-Acoustics

 

 

 

Presented by:

Dave L Moulton

 

 

 

 

Location:  Thales Acoustics Harrow UK

Date:  19-November-2003

 

Content

 

This will be the first in a series of lectures where I hope to explain the fundamental principals of wave acoustics. Acoustic wave theory is a very complex subject and to pursue it in any real depth requires a very good knowledge of mathematics, in particular vector analysis, partial differential equations and continuum mechanics. To be good at these can take many years of practice and a lot of text book reading. In order not to introduce too high a level of mathematics into this lecture I propose to adopt a simple approach to describing what an acoustic wave is and how it can be

represented mathematically.  This will be done using the following steps:

 

·        Explain wave motion created by a piston in long pipe.

·        Introduce a simplistic sinusoidal mathematical expression to describe the wave.

·        Manipulate the sinusoid to produce the simplest form of the wave equation

·        Finally, end up with an exponential solution to the wave equation.

 

The reason for the content of this lecture is to provide the audience with the most fundamental mathematical description of a moving wave. The symbols that I use will be very common in most text books and technical papers where the wave equation is discussed. It is very important that the audience become familiar with the wave equation and its symbolic format. Future lectures will often involve the wave equation.

 

On with the lecture

 

Let us start by considering a tube full of air with a piston attached to one end. Let us also have an x coordinate axis in the direction of the length of the tube starting from the front face of the stationary piston.

 

Figure 20.1

 

Now let us observe what happens when the piston is moving backwards and forwards in a sinusoidal manner.  We will consider the piston to have started its first cycle at time t=0, that is to say no other cycle has occurred before time t=0.

 

Figure 20.2

 

It is not easy to clearly represent the wave effect on a static diagram, but hopefully Figure 20.2 will give some insight into what is happening. Here is a list of events occurring during the cycle.

 

At time t=0: The piston is stationary and the air density is constant at all points within the tube.

At time t=Dt:  The piston has moved forward an amount Du resulting in the air particles in front of the piston being pushed forwards. The air particles each have mass so they must also have momentum, which results in a localized region of increased air density in front of the piston. This region extends a small distance beyond the front of the piston due the effect of the forward momentum of the air particles.

 

At time t=T/4: At this point the piston reaches it maximum extent into the tube. However the energy imparted on the air from the piston becomes detached from its source and propagates away at the speed of sound, this is the adiabatic process at work in the system. The effect looks like a localized high density region of air particles propagating away from the piston at the speed of sound C₀ .

 

NOTE 1:  It is not the air particles that are propagating down the tube at the speed of sound. It is in fact the energy given by the piston that propagates away causing a sequential chain of localized particle displacements down the length of the tube. The process is adiabatic since the energy is confined to a packet and continues to propagate away from the piston without loss.

 

NOTE 2:  For the sake of simplicity in this example using a tube, the air particles can be considered to each undergo a maximum forward displacement +u₀ equal to the original displacement of the piston

 

At time t=T/2:  The piston has now moved backwards and passes through its original rest position. Meanwhile the initial pocket of energy propagates further away from the piston at the speed of sound. The gap between the escaping energy wave front and the piston gets bigger resulting in a lower air density region starting to form at the face of the piston

 

At time t=3T/4:  The piston has now moved as far backwards as it can go in the cycle, this equates to a backwards displacement of -u_0. You will notice now that another high density region of air is forming at the front of the piston and a very low density region has formed directly between the piston and the previous energy packet. This has been caused by the fact that the volume of space between the piston and the previous energy packet is now large compared to the volume necessary to maintain the initial steady state density within the tube. The result is that a negative pressure forms in front of the piston resulting in the air particles being forced backwards towards the piston to fill the gap. This results in the particles undergoing a maximum negative displacement -u₀

 

At time t=T:  The piston has now moved forwards through it original rest position. Once again the air particles are compressed together as the piston drives forwards.

 

The complete cycle:

The positive and negative pressure regions propagate away from the piston as a complete cycle of wave energy. Producing a sequential chain of alternate high and low pressure zones which propagate down the length of the tube. The high pressure regions are called Compressions and the low pressure regions are called Rarefactions.  And so the cycle continues…..

 

Mathematical Representation of a Moving Wave

We now have some visual idea as to what is happening to the particles in a medium when a wave passes through it. In order to get a better if and more clear understanding we need to look deeper into the physics of the wave and its dynamic affect on the air particles.

 

Let us start off at a very simple level indeed.  If we consider the wave to be generated by a simple sinusoidal source, such as our piston, and that the whole process remains linear. Then we can describe the particle displacement as a very simple sine function:

 

 

Where: 

u is the particle displacement at any point within the cycle.

u₀ is the maximum particle displacement.

q represents the angular position within the cycle and can range from 0 and ± ¥

 

Graphically this function is shown in figure 20.3 below:

 

 

Figure 20.3

 

 

We now need to modify our sine wave into a function that describes a moving wave. Before we get too carried away Let us first introduce the element of time into equation [20.1].  Now we can describe angular velocity  w as the rate of change of angle q with time t. Thus:

 

Thus:

Resulting in:

 

We can now re describe our sinusoid as:

 

 

So for a sinusoidal wave with a fixed angular frequency we can clearly see that the particle displacement u is a function of time t and our wave diagram gets modified to show time on the axis.

 

 

Figure 20.4

 

If we look carefully at the expression in [20.4] we see that the function has values for both positive and negative time. It can be seen from figure 20.4 that the function exists on both the positive and negative wt axis. 

This means that from a mathematical view point the expression [20.4] describes the particle displacement in three time frames:  The Present time (t = 0), the Past time (t > 0) and the Future time ( t< 0).  Figure 20.5 shows the time frames added to the diagram.

 

 

Figure 20.5

 

Sound waves are theoretically created in the Future by means of a very predictable sound source such as a sinusoidal piston or diaphragm, although we cannot hear into the future we can predict that the waves will occur. Our perception of sound will always only happen in the Present and will very quickly become a memory as we cannot hear into the Past.

 

So without getting too philosophical from a mathematical perspective we can assume that sound waves travel from the future, through the present and into the past.

 

We now have to add that element of travel into our equation because at the moment expression [20.4] does not give any indication that the wave is  moving. In order for the wave to travel we need to give it a direction and an axis to move along. To keep things simple we will consider movement of the wave along a one dimensional x-axis from left to right.

 

NOTE:  By definition as the wave moves from the present into the past it travels in a positive direction along the x-axis. So although it is going forwards in direction it is in fact traveling backwards in time (Into the past).

 

Now Let us see how we can introduce forwards movement into equation [20.4]. Firstly we can look at figure 20.6 to get a better idea as to what we can do to modify the expression. Consider the top graph in figure 20.6. Let us assume that we can predict the future and that by deduction we know that the point ‘A’ on a Future part of the wave will be observed in the Present in t’ seconds time. We can also see that the point ‘B’ on the curve that we observe in the Present will become a memory as it travels for t’ seconds into the past

 

 

Figure 20.6

 

Our mathematical description can be made to reflect what we see in the present simply by applying a phase shift from the Future to the Present. Thus the wave will be described as moving from left to right on the wt axis.

 

Thus:                  

 

We know that the wave travels at the speed of sound C₀, so in a time t′ it will have traveled a certain distance in the positive x axis direction.

 

Thus:                   

 

 

Leading to:        

We can now see that our expression for the wave particle velocity has time t, wave displacement x as variables and the wave speed C₀ built into the expression. Thus u is now expressed as a function of t and x. u(x,t).

 

The Wave Number k

It is generally recognized that the ratio of the angular wave velocity ω to the wave speed c₀ is given the symbol k. This can also be used to relate the number of wavelengths per 2π radian cycle.

Thus

 

Equation [20.9] can be considered as a general solution for describing the displacement affect on a particle when a sinusoidal wave moves through the region of the  particle in the +ve x direction.  This is a very general solution and must come from a more advanced differential equation that describes the rates of change of particle displacement with time and wave displacement, independently. We will now establish what that relationship is.

 

Particle velocity ‘v’

This is simply the rate of change of particle displacement with time:

 

Particle acceleration ‘a’

This is the rate of change of particle velocity with time:

 

Thus

You should recognize this as the equation for simple harmonic motion, which is exactly what we are dealing with by using a simple continuous sine function.

 

 Strain ε_x

This is the rate of change of the instantaneous particle displacement u with respect to the wave displacement x. It is related to the Adiabatic Bulk modulus B of the medium and the localized pressure change dp. I will show this relationship in a later lecture,

 

Differential  of ε_x

It is not too easy to describe what this is in words, except that it has a very significant mathematical relevance in forming the wave equation.

 

 

Thus:

 

We can now combine equations [20.12] and [20.15] to form the wave equation:

 

Resulting in the One Dimensional Wave equation in its standard form

 

 

Where c is the wave velocity in the medium. In our case it is the speed of sound in air.

 

Equation [20.17] is the most simplest form of the wave equation but is none the less a very significant and important equation in wave acoustics. We have already seen that this equation has the sinusoidal equation in [20.9] as its solution. There is however another very important representation of the solution to the wave equation which is the exponential solution.   Let us take another look at equation [20.9] and rewrite it in exponential form:

 

Now both of the exponential terms are solutions to the wave equation but we are only really interested in one of them. The term with the –j(ωt-kx) represents the Present into the Future, whereas the term j(ωt-kx) represents the Present into the Past. We can there for simplify our wave equation solution to be.

Where A_f represents the amplitude of the particle velocity.  This may seem a little confusing considering we seem have just declared that A_f is in fact half the amplitude of u₀

Don’t panic, this purely a way of representing an amplitude. What is most significant is that the wave is propagating and is a solution to the wave equation in [20.17].  For example I could decide to describe a wave as:

 

 

Now Let us convince ourselves that this is a solution to the wave equation:

 

Thus:

 

Which gets us back to the wave equation. Note the wave equation is not dependant on the absolute value of the peak particle amplitude. So we can define the amplitude however we want without contradicting the wave equation.

 

NOTE: a wave traveling from the Present into the Past but in the negative x-axis direction (i.e a forward traveling wave is reflected off a wall) can be easily described in exponential form as.

 

 

I will leave it to the Audience to prove this very simple fact.

 

 

End Of Lecture