第 6 讲：用微分方程表示的系统
本讲内容：一阶微分方程与差分方程；将解写为特解与齐次解之和；附加条件及其与系统线性、因果性、时不变性的关系；利用加法器、系数乘法器，以及离散时间中的延时单元或连续时间中的积分器，构建由差分方程或微分方程描述的线性时不变系统框图。
授课教师：艾伦·V. 奥本海姆教授
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PROFESSOR: Over the past several lectures, we've
developed a representation for linear time-invariant systems.
A particularly important set of systems, which are linear
and time-invariant, are those that are represented by linear
constant-coefficient differential equations in
continuous time or linear constant-coefficient
difference equations in discrete time.
For example, electrical circuits that are built, let's
say, out of resistors, inductors, and capacitors,
perhaps with op-amps, correspond to systems
described by differential equations.
Mechanical systems with springs and dashpots,
likewise, are described by differential equations.
And in the discrete-time case, things such as moving average
filters, digital filters, and most simple kinds of data
smoothing are all linear constant-coefficient
difference equations.
Now, presumably in a previous course, you've had some
exposure to differential equations for continuous time,
and their solution using notions like particular
solution, and homogeneous solution, initial
conditions, et cetera.
Later on in the course, when we've developed the concept of
the Fourier transform after that, the Laplace transform,
we'll see some very efficient and useful ways of generating
solutions, both for differential
and difference equations.
At this point, however, I'd like to just introduce linear
constant-coefficient differential equations and
their discrete-time counterpart.
And address, among other things, the issue of when they
do and don't correspond to linear time-invariant systems.
Well, let's first consider what I refer to as an
nth-order linear constant-coefficient
differential equation, as I've indicated here.
And what it consists of is a linear combination of
derivatives of the system output, y(t), equal to a
linear combination of derivatives of the system
input x(t).
And it's referred to as a constant-coefficient equation,
of course, because the coefficients are constant.
In other words, not assumed to be time-varying.
And it's referred to as linear because it corresponds to a
linear combination of these derivatives, not because it
corresponds to a linear system.
And, in fact, as we'll see, or as I'll indicate, this
equation may or may not, in fact,
correspond to a linear system.
In the discrete-time case, the corresponding equation is a
linear constant-coefficient difference equation.
And that corresponds to, again, a linear combination of
delayed versions of the output equal to a linear combination
of delayed versions of the input.
This equation is referred to an
nth-order difference equation.
The n referring to the number of delays of the output
involved, just as an Nth-order differential equation, the n
or the order of the equation refers to the number of
derivatives of the output.
Now, let's first begin with linear constant-coefficient
differential equations.
And the basic point of the solution for the differential
equations is the fact that if we've generated some solution,
which I refer to here as y_p(t), some solution to the
equation for a given input, then, in fact, we can add to
that solution any other solution which satisfies
what's referred to as the homogeneous equation.
So in fact, this differential equation by itself is not a
unique specification of the system.
If I have any solution, then I can add to that solution any
other solution which satisfies the homogeneous equation, and
the sum of those two will likewise be a solution.
And that's very straightforward to verify.
By simply substituting into the differential equation, the
sum of a particular and the homogeneous solution, and what
you'll see is that the homogeneous contribution, in
fact, goes to zero by definition of what we mean by
the homogeneous equation.
Now, the homogeneous solution for a linear
constant-coefficient differential equation is of
the form that I indicate at the bottom.
And it typically consists of a sum of N complex exponentials.
And the constants are undetermined by
the equation itself.
And this form for the homogeneous solution, in
essence, drops out of examining the homogeneous
equation, where if we assume that the form of the
homogeneous solution is a complex exponential with some
unspecified amplitude and unspecified exponent.
If we substitute this into our homogeneous equation, we end
up with the equation that I've indicated here.
The factor A and the e^(st) can in fact be canceled out,
and we find that that equation is satisfied
for N values of s.
And that's true no matter what choice is made for these
coefficients.
And the essential consequence of all of that is that the
homogeneous solution is of the form that I indicated
previously.
Namely it consists of a sum of N complex exponentials, where
the coefficients, the N coefficients, attach to each
of those complex exponential is undetermined or
unspecified.
So what this says is that in order to obtain the solution
for a linear constant-coefficient
differential equation, we need some kind of auxiliary
information that tells us is how to obtain these N
undetermined constants.
And there are variety of ways of specifying this auxiliary
information or auxiliary conditions.
For example, in addition to the differential equation,
what I can tell you is the value of the output and N - 1
of its derivatives, at some specified time, t_0.
And so the differential equation together with the
auxiliary information, the initial conditions, then lets
you determine the total solution, which namely lets
you determine these previously-unspecified
coefficients in the homogeneous solution.
Now, depending on how the auxiliary information is
stated or what auxiliary information is available, the
system may or may not correspond to a linear system,
and may or may not correspond to a linear
time-invariant system.
One essential condition for it to correspond to a linear
system is that the initial conditions must be 0.
And one can see the reason for that, if we refer back to the
previous lecture in which we saw that for a linear system,
if we put 0 in, we get 0 out.
So if x(t), the input, is 0, the output must be 0.
And so that, in essence, tells us that, at least for the
system to be linear, these initial conditions must be 0.
Now, beyond that, if we want the system to be causal and
linear and time-invariant, then what's required on the
initial conditions is that they be consistent with what's
referred to as initial rest.
Initial rest says that the output must be 0 up until the
time that the input becomes non-zero.
And we can see, of course, that that's consistent with
the notion of causality, as we talked about in
the previous lecture.
And it's relatively straightforward to see that if
the system is causal and linear and time-invariant,
that will require initial rest.
It's somewhat more difficult to see that if we specify
initial rest, then that, in fact, is sufficient to
determine that the system is both causal and linear and
time invariant.
But the essential point then is that it requires initial
rest for both linearity and causality.
OK, well, let's look at an example, and let's take the
example of a first-order differential equation, as I've
indicated here.
So we have a first-order differential equation, dy(t) /
dt + ay(t) is the input x(t).
And let's first look at what the homogeneous solution of
this equation is.
And so, we consider the homogeneous equation, namely
the equation that specifies solutions, which would
correspond to 0 input.
We, in essence, guess, or impose a solution of the form.
The homogeneous solution is an amplitude factor times a
complex exponential.
Substituting this into the homogeneous equation, we then
get the equation that I've indicated here.
What you can see is that in this equation, I can cancel
out the amplitude factor and this complex exponential.
So let's just cancel those out on both sides of the equation.
And what we're left with is an equation that specifies what
the complex exponent must be.
In particular, for the homogeneous solution, s must
be equal to -a, and so finally, our homogeneous
solution is as I've indicated here.
Now, let's look at the solution for a specific input.
Let's consider, for example, an input which is
a scaled unit step.
And although I won't work out the solution in detail and
perhaps using what you've worked on previously, you know
how to carry out the solution for that.
A solution with a step input is what I've indicated here: a
scalar, 1 minus an exponential,
times a unit step.
And you can verify that simply by substituting into the
differential equation.
Now, we know that there's a family of solutions.
In other words, any solution with a homogeneous solution
added to it, is, again, a solution.
And so if we consider the solution that I just
indicated, we generate the entire family of solutions by
adding a homogeneous solution to it, and so this then
corresponds to the entire family of solutions, where the
constant A is unspecified so far, and needs to be specified
through some type of auxiliary conditions.
Now, a class of auxiliary conditions is the condition of
initial rest, which as I indicated before, is
equivalent to the statement that the system is causal and
linear and time-invariant.
And in that case, for the initial rest condition, we
would then require in this equation above that this
constant be equal to 0.
And so finally, the response to a scaled step--
if the system is to correspond to a causal linear
time-invariant system, is then just this term, namely, a
constant times 1 minus an exponential, times the step.
Now, if the system is a linear time-invariant system, it can,
as we know, be described through its impulse response.
And as you've worked out previously in the video course
manual, for a linear time-invariant system, the
impulse response is the
derivative of the step response.
And just to quickly remind you of where that
result comes from.
In essence, we can consider two linear time-invariant
systems in cascade, one a differentiator, the other the
system that we're talking about described by the
differential equation.
And a step in here then generates an impulse into our
system, and out comes the impulse response.
Well, just using the fact that these are both linear
time-invariant systems, and they can be cascaded in either
order, then means that if we have the step response to our
system, and that goes through the differentiator, what must
come out, again, is the impulse response.
So differentiating the step response, we
get the impulse response.
Here again, is the step response as we just worked it
out, this time for a unit step.
If we differentiate, we have then, since the step response
is the product of two terms, the derivative of a product is
the sum of the derivatives.
And carrying that algebra through, and using the fact
that the derivative of the step is an impulse, finally we
come down to this statement for the impulse response.
And then recognizing that this is a time
function times an impulse.
And we know that if a time function times an impulse
takes on the value at the time that the impulse occurs, then
this term is simply 0.
And the impulse response then finally is an
exponential of this form.
And this is the decaying exponential for a-positive,
it's a growing exponential for a-negative.
And recall that, as we talked about previously, a linear
time-invariant system is stable if its impulse response
is absolutely integrable.
For this particular case, this impulse response is absolutely
integrable provided that the exponential factor a is
greater than 0.
Okay, so what we've seen then is the impulse response for a
system described by a linear constant-coefficient
differential equation, where in addition, we would impose
causality, linearity, and time-invariance, essentially,
through the initial conditions of initial rest.
Now, pretty much the same kinds of things happen with
difference equations as we've gone through with
differential equations.
In particular, again, let me remind you of the form of an
Nth order linear constant
coefficient difference equation.
It's as I indicate here.
And, again, a linear combination, this time of
delayed versions of the output equal to a linear combination
of delayed versions of the input.
Once again, difference equation is not a complete
specification of the system because we can add to the
response any homogeneous solution.
In other words, any solution that satisfies the homogeneous
equation and the sum of those will also satisfy the original
difference equation.
So if we have a particular response that satisfies the
difference equation, then adding to that any response
that is a solution to the homogeneous equation will also
be a solution to the total equation.
The homogeneous solution, again, is of the form of a
linear combination of exponentials.
Here, we have the homogeneous equation.
As with differential equations, we can guess or
impose solutions of the form A times an exponential.
When we substitute this into the homogeneous equation, we
then end up with the equation that I've indicated here.
We recognize again that A, the amplitude, and z^n, this
exponential factor, cancel out.
And so this equation is satisfied for any values of z
that satisfy this equation.
And there are N roots, z_1 through z_N.
And so finally, the form for the homogeneous solution is a
linear combination of capital N exponentials, where capital
N is the order of the equation.
With each of those exponentials, the amplitude
factor is undetermined and needs to be determined in some
way through the imposition of appropriate initial conditions
or boundary conditions.
So the general form then for the solution to the difference
equation is a sum of exponentials plus any
particular solution.
It's through auxiliary conditions that we determine
these coefficients.
We have N undetermined coefficients and, so we
require N auxiliary conditions.
For example, some set of values of the output at N
distinct instance of time.
Now this was the same as with differential equations.
In the case of differential equations, we talked about
specifying the value of the output and its derivatives.
And there, we indicated that for
linearity, what we required--
for linearity, what we required is that the auxiliary
conditions be 0.
And the same thing applies here for the same reason.
Namely, if the system is to be linear, then the response, if
there's no input, must be equal to 0.
In addition, what we may want to impose on the system is
that it be causal, and in addition to linear
time-invariant, and what that requires, again, is that the
auxiliary conditions be consistent with initial rest,
namely, that if the input is 0 prior to some time, then the
output is 0 prior to the same time.
So we've seen a very direct parallel so far between
differential equations and difference equations.
In fact, one difference between them that, in some
sense, makes difference equations easier to deal with
in some situations, is that in contrast to a differential
equation, a difference equation, if we assume
causality, in fact is an explicit input-output
relationship for the system.
Now, let me show you what I mean.
Let's consider the nth-order difference equation, as I've
indicated here.
And let's assume that we're imposing causality so that the
output can only depend on prior values of the input, and
therefore, aren't prior values of the output.
Well, we can simply rearrange this equation solving for y[n]
the leading term, with k = 0.
Taking all of the other terms over to the right side of the
equation, and we then have a recursive equation, namely, an
equation that expresses the output in terms of prior
values of the input, which is this term, and prior values of
the output.
And so if, in fact, we have this equation running, then
once it started, we know how to compute the output for the
next time instant.
Well, how do we get it started?
The way we get it started, of course, is through the
appropriate set of initial conditions or boundary
conditions.
And, if for example, we assume initial rest corresponding to
a causal linear time-invariant system, then if the input is 0
up until some time, the output must be 0 up until that time.
And that, in essence, helps us get the equation started.
Well, let's look at this specifically in the context of
a first-order difference equation.
So let's take a first-order difference equation, as I've
indicated here.
And so, we have an equation that tells us that y[n]
- ay[n-1]
= x[n].
Now, if we want this to correspond to a causal linear
time-invariant system, we impose initial rest on it.
We can rewrite the first-order difference equation by taking
the term involving y[n-1]
over two the right hand side of the equation.
Now, this gives us a recursive equation that expresses the
output in terms of the input and past values of the output.
And since we've imposed causality, and if we're
talking about a linear time-invariant system, we can
now inquire as to what the impulse response is.
And we know, of course, the impulse response tells us
everything that we need to know about the system.
So let's choose an input, which is an impulse.
So the impulse response is delta[n]
corresponding to the x[n]
up here, plus a delta of--
this should be h[n-1].
And let me just correct that.
This is h[n-1].
So we have the impulse response as delta[n]
plus a times h[n-1].
Now, from initial rest, we know that since the input,
namely an impulse, is 0 for n less than 0, the impulse
response is likewise 0 for n less than 0.
And now, let's work out what h[0]
is.
Well, h of 0, with n = 0, is delta[0]
plus a times h[n-1].
h[n-1]
is 0.
And so, h[0]
is equal to 1.
Now that we have h[0], we can figure out h[1]
by running this recursive equation.
So h[1]
is delta[1], which is 0, plus a times h[0], which we just
figured out is 1.
So, h[1]
is equal to a.
And if we carry this through, we'll have h[2]
equal to a^2, and this will continue on.
And in fact, what we can recognize by looking at this
and how we would expect these terms to build, we would see
that the impulse response, h[n], in fact, is of the form
a^n times u[n].
And we also can recognize then that this corresponds to a
stable system, if and only if the impulse response, which is
what we just figured out, if and only if the impulse
response is absolutely summable.
And what that will require is that the magnitude of a be
less than 1.
Now, we imposed causality, linearity, and
time-invariance, and generated a solution recursively.
And now, of course, if we want to generate the more general
set of solutions to this difference equation, we can do
that by adding all of the homogeneous solutions, namely,
all the solutions that satisfy the homogeneous equation.
So here, we have the causal linear
time-invariant impulse response.
In fact, with an impulse input, all of the possible
solutions are that impulse response plus
the homogeneous solutions.
The homogeneous solution is the solution that satisfies
the homogeneous equation.
That will, in general, be of the form an amplitude factor
times an exponential factor, and if we substitute this into
this equation, then we see that the homogeneous equation
is satisfied for any values of a and any values of z that
satisfy this equation.
Again, as we did with differential equations, the
factor A cancels out.
And also, in fact, I can cancel out a z^n
there and a z^n here.
And so what we're left with is a statement that tells us then
that the homogeneous solution is of the form a(z^n), for any
value of A and any value of z that satisfies this equation.
And that value of z, in particular, is z equal to a.
So the homogeneous solution then is any exponential of
this form with any amplitude factor.
And so, the family of solutions, with an impulse
input is the solution corresponding to the system
being causal, linear, and time-invariant, plus the
homogeneous term.
If we impose causality, linearity, and time-invariance
on the system, then of course, that additional exponential
factor will be 0.
In other words, A is equal to 0.
Now, we've seen the differential equations and
difference equations in terms of the fact that there are
families of solutions, and in order to get causality,
linearity, and time-invariance requires imposing a particular
set of initial conditions, namely, imposing initial rest
on the system.
Let's now look at the difference equation and then
later, the differential equation, interpreted in block
diagram terms.
Now, the difference equation, as I just simply repeated
here, is y[n]
= x[n]
+ ay[n-1], where I've taken the delayed term over to the
right hand side of the equation.
So, in effect, what I'm imposing on
this system is causality.
I'm assuming that if I know the past history of the input
and the output, I can determine the next value of
the output.
Well, we can, in fact, draw a block diagram that represents
that equation.
The equations says, we take x[n]
for any given value of n, say n0, whatever value we're
computing the output for.
We take the input at that time, and add to it the factor
a times the output value that we calculated last time.
So if we have x[n], which is our input, and if we have
y[n], which is our output, we, in fact, can get y of n by
taking the last value of y[n], indicated here by putting y[n]
through a delay, multiplying that by the factor a, and then
adding that result to the input.
And the result of doing that is y[n].
So the way this block diagram might be interpreted, for
example, as an algorithm is to say that we take x[n], add to
it a times the previous value the output.
That sum gives us the current value of the output, which we
then put out of the system, and also put into a delay
element, or basically, into a storage register, to use on
the next iteration or recursion.
Now, how do we get this started?
Well, we know that the difference equation requires
initial conditions.
And, in fact, the initial conditions correspond to what
we store in the delay register when this block diagram or
equation initially starts up.
OK Now, let's look at this for the case of difference
equations more generally.
So, what we've said is that we can calculate the output by
having previous values of the input, previous values of the
output, and forming the appropriate linear
combination.
So let's just build up the more general block diagram
that would correspond to this.
And what it says is that we want to have a mechanism for
storing past values of the input, and a mechanism for
storing past values of the output.
And I've indicated that on this figure, so far, by a
chain of delay elements, indicating that what the
output of each delay is, is the input delayed by one time
instant or interval.
And so what we see down this chain of delays are delayed
replications of the input.
And what we see on the other chain is delayed replications
of the output.
Now, the difference equation says that we want to take
these, and multiply them by the appropriate coefficients,
the coefficients in the difference equation, and so,
we can do that as I've indicated here.
So now, we have these delay elements, each multiplied by
the appropriate coefficients on the input, and by
appropriate coefficients on the output.
Those are then summed together, and so we now will
sum these and will sum these.
After we've summed these, we want to add those together.
And there's a factor of 1 / a_0 that comes in.
And so that then generates our output.
And so this, in fact, then represents a block diagram,
which is a general block diagram for implementing or
representing a linear constant-coefficient
difference equation.
Now, if you think about what it means in terms of, let's
say, a computer algorithm or a piece of hardware, in fact,
this block diagram is a recipe or algorithm for doing the
implementation.
But it's important to recognize, even at this point,
that it's only one of many possible algorithms or
implementations for this difference equation.
Just for example, I can consider that equation for
that block diagram.
And here, I've re-drawn it.
So here, are once again.
I have the same block diagram that we just saw.
And I can recognize, for example, that this, in
essence, corresponds to two linear time-invariant systems
in cascade.
Now, that assumes, of course, that my initial conditions are
such that the system is, in fact, linear.
And that, in turn, requires that we're assuming initial
rests, namely, before the input does
anything other than 0.
There are just 0 values stored in the registers.
But assuming that it corresponds to a linear
time-invariant system, this is a cascade of two linear
time-invariant invriant systems.
We know that two linear time-invariant systems can be
cascaded in either order.
So, in particular, I can consider breaking this cascade
here, and moving this block over to the other side.
And so let's just do that.
And when I do, I then have this combination of systems,
and, of course, you can ask what advantage there is to
doing that, and the advantage arises because of the fact
that in this form, exactly what is stored in these delays
is also stored in these delay registers.
In other words, it's this intermediate variable--
whatever it is--
down this chain of the delays and down this chain of delays,
and so, in fact, I can collapse those delays into a
single chain of delays.
And the network that I'm left with is the network that I
indicate on this view graph, where what I've done is to
simply collapse that double chain of delays into a single
change of delays.
Now, one can ask, well, what's the advantage to doing that?
And one advantage, simply stated, is that when you think
in terms of an implementation of a difference equation, a
delay corresponds to a storage register, a memory location,
and by simply using the fact that we can interchange the
order in which linear time-invariant systems are
cascaded, we can reduce the amount of memory
by a factor of 2.
Now, an essentially similar procedure can also be used for
differential equations, in terms of implementation using
block diagrams or the interpretation of
implementations using block diagrams.
And let me first do that--
rather than in general-- let me first do it in the context
of a specific example.
So let's consider a linear constant-coefficient
differential equation, as I've indicated here, and I have
terms on the left side and terms on the right side.
And with the differential equation, let's consider
taking all the terms over to the right side of the
equation, except for the highest
derivative in the output.
Next, we integrate both sides of the equation so that when
we're done, we end up with on the left side of the
equation with y(t).
On the right side of the equation with the appropriate
number of integrations.
And so the integral equation that we'll get for this
example y(t), the output, is x(t) plus b, the scale factor
times the integral of the input, and minus a, that scale
factor, times the integral of the output.
So to form the output in the block diagram terms, we form a
linear combination of the input, a scaled integral of
the input, and a scaled integral of the output, all of
that added together.
So we need, in addition to the input, we need the
integral of the input.
And so this box indicates an integrator.
In addition to the output, we need the
integral of the output.
And now, to form y(t), we multiply the integrated input
by the scale factor, b.
And add that to x(t), and we take the integrated output,
multiply it by -a, and add to that the result of the
previous addition, and according to the integral
equation, then that forms the output.
So just as we did with the difference equation, we've
converted the differential equation to an integral
equation, and we have a block diagram form very similar to
what we had in the case of the difference equation.
Now, the initial conditions, of course, are tied up in,
again, how these integrators are initialized.
Assuming that we impose initial rest on the system, we
can think of the overall system as a linear
time-invariant system, and it's a cascade of one linear
time-invariant system with a second.
So we can, in fact, break this, and consider
interchanging the order in which these
two systems are cascaded.
And so I've indicated that down below.
Here, I've simply taken the top block diagram,
interchanged the order in which the
two systems are cascaded.
And here, again, we can ask what the advantages to this,
as opposed to the previous one.
And what you can see, just as we saw with the difference
equation, is that now, the integrators--
both integrators--
are integrating the same thing.
In particular, the input to this integrator and the input
to this integrator are identical.
So in fact, rather than using this one, we can simply tap
off from here.
We can, in fact, remove this integrator, break this
connection, and tap in at this point.
And so what we've done then, by interchanging the order in
which the systems are cascaded, is reduced the
implementation to the implementation with a single
integrator.
Very much similar to what we talked about in the case of
the difference equation.
Now, let's just, again, with the integral equation or the
differential equation, look at this somewhat more generally.
Again, if we take the differential equation, the
general differential equation, integrate it a sufficient
number of times to convert it to an integral equation.
We would then have this cascade of systems.
And again, if we assume initial rest, so that these
are both linear time-invariant systems, we can interchange
the order in which they're cascaded.
Namely, take the second system, and move it to precede
the first system.
And then what we recognize is that the input to this chain
of integrators and this chain of integrators
is exactly the same.
And so, in fact we can collapse these together using
only one chain of integrators.
And the system that we're left with then is a system that
looks as I've indicated here.
So we have now just a single chains of integrators instead
of the two sets of integrators.
So we've seen that the situation is very similar here
as it was in the case of the difference equation.
Again, why do we want to cut the number of
integrators in half?
Well, one reason is because integrators, in effect,
represent hardware.
And if we have half as many integrators, then we're using
half as much hardware.
Well, let me just conclude by
summarizing a number of points.
I indicated at the beginning that linear
constant-coefficient differential equations and
difference equations will play an important role as linear
time-invariant systems throughout this course and
throughout this set of lectures.
I also stressed the fact that differential or difference
equations, by themselves, are not a complete specification
of the system because of the fact that we can add to any
solution a homogeneous solution.
How do we specify the appropriate initial conditions
to ensure--
how do we specify the appropriate initial conditions
to ensure that the system is linear and time-invariant?
Well, the auxiliary information, namely, the
initial conditions associated with the system being causal,
linear, and time-invariant are the
conditions of initial rest.
And, in fact, for most of the course, what we'll be
interested in are systems that are in fact, causal, linear,
and time-invariant.
And so we will, in fact, be assuming initial rest
conditions.
Now, as I also indicated, there are a variety of
efficient procedures for solving differential and
difference equations that we haven't yet addressed.
And beginning with the next set of lectures, we'll be
talking about the Fourier Transform and much later in
the course, what's referred to as the Laplace Transform for
continuous time and the Z-transform for discrete time.
And what we'll see is that with the Fourier Transform and
later with the Laplace and Z-transform, we'll have a
number of efficient and very useful ways of generating the
solution for differential and difference equations under the
assumption that the system is causal, linear, and
time-invariant.
Also, we'll see in addition to the block diagram
implementations of these systems that we've talked
about so far, we'll see a number of other useful
implementations that exploit a variety of properties
associated with Fourier and Laplace Transforms.
Thank you.