Moulton Lectures
On
Electro-Acoustics
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Lecture
14 Stability and
the Impulse Response
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Presented
by:
Dave
L Moulton
Location: Thales Acoustics Harrow UK
Date: 19-February-2003
Content
We now move onto the interesting subject of stability. In this lecture I want to define what we mean by stability and how it applies to closed loop systems, in particular the now familiar ANR closed loop system. In discussing stability I want to introduce a very important driving function, namely the impulse (also known as the Dirac Delta Function). I will define the delta function and show why it is important.
On with the lecture
Stability Definition
There are lots of ways of
defining the stability of a system, different technical authors have there own way
of describing a criteria that basically amounts to the same thing. I will define stability in the way that I
have come to understand it. My definition will be based around a linear system g(t).
Non Linear systems are a different ball game and will not be considered during
this lecture.
Firstly consider the function diagram shown in figure
14.1.
Figure 14.1
Here we have a system that is represented in the time
domain. By applying an input r(t)
to the system it will produce a corresponding output c(t) which is a
processed version of r(t). The
processing may take the form of combinations of Integration, differentiation
and scaling.
If we remove the input r(t) from the system the
output c(t) should either go to zero or remain in a defined state, for
example the system could have been designed to give a permanent offset between
the input and the output, so with no input the system output settles at a fixed
level greater or less than zero. In either case the offset is fixed and there
are no oscillations.
Figure 14.2 shows an example of a stable system. Depending on the design of the system g(t)
may or may not have an offset.
Figure 14.2
If we apply an input to the system it will produce a corresponding output. For example lets assume that if we apply a sine wave to the input we get a modified amplitude and phase shifted sine wave at the output.
Figure 14.3
Now consider the affect of applying an impulse to a stable system. Depending on the damping coefficient and natural frequency for the system we can expect the output to decay with or without oscillation to either the zero state or the offset state.
Figure 14.4
So a stable system can be defined in the following way:
A continuous system (discrete-
time domain) is stable if its impulse response Approaches zero as time
approaches infinity.
An unstable system is one that does not obey the above definition. In which case the impulse response will result in an output that continuously grows as time approaches infinity, or settles into a state of stable oscillation (constant amplitude oscillation).
Figure 14.5
The Delta Function
So far I have talked about stability in terms of the system impulse response. I will now give a more detailed explanation of the definition for the impulse function, commonly known as the Dirac Delta Function.
The delta Function is given the mathematical symbol d(t), and is defined such that:
This integral implies that d(t) has an area equal to unity. That’s is to say, the width of d(t) multiplied by its magnitude is equal to 1.
Now for the ‘not so obvious’ bit.
By definition d(t) is a delta function at time t=0.
Figure 14.6
A delta function at time t=t0 is defined as d(t – t0). This may look a little odd, however it must be remembered that a delta function is a discrete function and not a continuous time function. The definition indicates that a Delta function occurring at any time other than t=0 is basically the solution in the brackets of the function, thus t – t0 = 0 means that t = t0. This makes sense when you consider that d(t ) is the same as writing d(t - 0 ) thus t=0.
Getting to grips with this definition will take some time, but it will be worth the effort, especially in helping to get a good insight into Digital Sampling, Adaptive Signal Processing and Time Domain Convolution.
Figure 14.7
The integral in equation [14.1] and our definition of d(t ) (only occurring at time t=0), suggest that the Delta Function is infinitely thin and infinitely tall.
Thus we can further define d(t ) as:
Figure 14.8
The Delta Function has some very useful properties. For example, in the world of digital sampling if f(t) is a continuous function of t then we can use the Delta Function to sample the value of f(t) at some time t1.
Thus
Also
Now for a couple of questions:
How can we realize a Delta Function in the real world? and how can we establish the Laplace Transform of a Delta Function?
By definition the Delta Function is an ideal mathematical function with a height that tends to infinity and a width that tends to zero. In the real world a Delta Function is unachievable, however we can make an approximation for one by considering the addition of realizable step functions.
Creating a Delta Function From Step Functions
In the same way that a Delta function at t=0 is defined as d(t ), we can define a
step function at t=0 as U(t )
Figure 14.9
Likewise we can define a step function that starts at time
t=t0 as U(t - t0)
Which looks like:
Figure 14.10
We can create the general shape of a Delta Function by subtracting U(t – t0) from U( t ). Thus:
Figure 14.11
We now need to define the height and width of our derived function so that we have a resulting area under the curve = 1.
We can set the width of our function to be Dt and the height can be 1/Dt . Clearly if we now multiply the width and height together the area under the curve will always be =1.
If we now also let the limit Dt tend towards zero then we have found another way of defining the Delta Function in terms of Step Functions and Scaling.
Thus:
Figure 14.12
To express d(t) in a more mathematical way we would write:
Now to find the Laplace Transform of d(t).
By Definition:
We can replace d(t) in [14.5] with our definition in [14.4] and then look at the Laplace transform of the individual step functions.
Although this may look awkward, it really is quite simply a case of working out the Laplace transforms for the two step functions and then evaluating the function as Dt tend towards zero.
Here we go:
First I will remove the limit indicator and the 1/Dt out of the integral.
Evaluating the Laplace transform of the two step functions:
Thus:
Expanding out the series for the exponential term:
Reducing down to
Thus the Laplace transform of the Delta Function is =1.
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