第 4 讲：卷积
本讲内容：信号的冲激表示；离散时间线性时不变（LTI）系统的卷积和表示，以及连续时间线性时不变系统的卷积积分表示；卷积的交换律、结合律和分配律。
授课教师：艾伦·V. 奥本海姆教授
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[MUSIC PLAYING]
PROFESSOR: In the last lecture, we discussed a number
of general properties for systems, which, as you recall,
applied both to continuous-time and to
discrete-time systems.
These properties were the properties associated with a
system having memory.
The issue of whether a system is or isn't invertible, we
talked about causality and stability, and finally we
talked about when linearity and time invariance.
In today's lecture, what I'd like to do is focus
specifically on linearity and time invariance, and show how
for systems that have those properties, we can exploit
them to generate a general representation.
Let me begin by just reviewing the two
properties again quickly.
Time invariance, as you recall, is a property that
applied both to continuous-time and
discrete-time systems, and in essence stated that for any
given input and output relationship if we simply
shift the input, then the output
shifts by the same amount.
And of course, exactly the same kind of statement applied
in discrete time.
So time invariance was a property that said that the
system didn't care about what the time origin
of the signal is.
Linearity was a property related to the fact that if we
have a set of outputs associated with a
given set of inputs--
as I've indicated here with the inputs as phi_k and the
outputs psi_k, then the property of linearity states
that if we have an input which is a linear combination of
those inputs, then the output is a linear combination of the
associated outputs.
So that linear combination of inputs generates, for a linear
system, an output which is a linear combination of the
associated outputs.
Now the question is, how can we exploit the properties of
linearity and time invariance?
There's a basic strategy which will flow more or less through
most of this course.
The strategy is to attempt to decompose a signal, either
continuous-time or discrete-time, into a set of
basic signals.
I've indicated that here.
And the question then is, what basic signal should we pick?
Well, the answer, kind of, is we should pick a set of basic
signals that provide a certain degree of analytical
convenience.
So we choose a set of inputs for the decomposition that
provide outputs that we can easily generate.
Now, as we'll see, when we do this, there are two classes of
inputs that are particularly suited to that strategy.
One class is the set of delayed impulses, namely
decomposing a signal into a linear combination of these.
And as we'll see, that leads to a representation for linear
time-invariant systems, which is referred to as convolution.
The second is a decomposition of inputs into complex
exponentials--
a linear combination of complex exponentials--
and that leads to a representation of signals and
systems through what we'll refer to as Fourier analysis.
Now, Fourier analysis will be a topic for a
set of later lectures.
What I'd like to begin with is the representation in terms of
impulses and the associated description of linear
time-invariant systems using convolution.
So let's begin with a discussion of discrete-time
signals, and in particular the issue of how discrete-time
signals can be decomposed as a linear
combination of delayed impulses.
Well, in fact, it's relatively straightforward.
What I've shown here is a general sequence with values
which I've indicated at the top.
And more or less as we did when we talked about
representing a unit step in terms of impulses, we can
think of this general sequence as a sequence of impulses--
delayed, namely, occurring at the appropriate time instant,
and with the appropriate amplitude.
So we can think of this general sequence and an
impulse occurring at n = 0 and with a height of x[0], plus an
impulse of height x[1]
occurring at time n = 1, and so that's x[1]
delta[n-1], an impulse at -1 with an
amplitude of x[-1], etc.
So if we continued to generate a set of weighted, delayed
unit samples like that, and if we added all these together,
then that will generate the total sequence.
Algebraically, then, what that corresponds to is representing
the sequence as a sum of individual terms as I've
indicated here or in terms of a general sum, the sum of x[k]
delta[n-k].
So that's our strategy--
the strategy is to decompose an arbitrary sequence into a
linear combination of weighted, delayed impulses.
And here again is the representation, which we just
finished generating.
Now, why is this representation useful?
It's useful because we now have a decomposition of the
sequence as a linear combination of basic
sequences, namely the delayed impulses.
And if we are talking about a linear system, the response to
that linear combination is a linear
combination of the responses.
So if we denote the response to a delayed impulse as
h_k[n], then the response to this general input is what
I've indicated here, where y[n], of course, is the output
due to the general input x[n].
h_k[n]
in is the output due to the delayed impulse, and these are
simply the coefficients in the weighting.
So for a linear system, we have this representation.
And if now, in addition, the system is time-invariant, we
can, in fact, relate the outputs due to these
individual delayed impulses.
Specifically, if the system is time-invariant, then the
response to an impulse at time k is exactly the same as the
response to an impulse at time 0, shifted over to time k.
Said another way, h_k[n]
is simply h_0[n-k], where h_0 is the response of the system
to an impulse at n = 0.
And it's generally useful to, rather than carrying around
h_0[n], just simply define h_0[n]
as h[n], which is the unit sample or unit impulse
response of the system.
And so the consequence, then, is for a linear time-invariant
system, the output can be expressed as
this sum where h[n-k]
is the response to an impulse occurring at time n = k.
And this is referred to as the convolution sum.
Now, we can--
just to emphasize how we've gone about this, let me show
it from another perspective.
We of course have taken the sequence x[n], we have
decomposed it as a linear combination of these weighted,
delayed impulses.
When these are added together, those correspond to the
original sequence x[n].
If this impulse, for example, generates a
response which is x[0]
h[n], where h[n]
is the response to a unit impulse at n = 0, and the
second one generates a delayed weighted response, and the
third one similarly, and we generate these individual
responses, these are all added together, and it's that linear
combination that forms the final output.
So that's really kind of the way we're thinking about it.
We have a general sequence, we're thinking of each
individual sample individually, each one of
those pops the system, and because of linearity, the
response is the sum of those individual responses.
That's what happens in discrete time, and pretty much
the same strategy works in continuous time.
In particular, we can begin in continuous time with the
notion of decomposing a continuous-time signal into a
succession of arbitrarily narrow rectangles.
And as the width of the rectangles goes to 0, the
approximation gets better.
Essentially what's going to happen is that each of those
individual rectangles, as they get narrower and narrower,
correspond more and more to an impulse.
Let me show you what I mean.
Here we have a continuous-time signal, and I've approximated
it by a staircase.
So in essence I can think of this as individual rectangles
of heights associated with the height of the continuous
curve, and so I've indicated that down below.
Here, for example, is the impulse corresponding to the
rectangle between t = -2 Delta and t = -Delta.
Here's the one from -Delta to 0, and as we continue on down,
we get impulses, or rather rectangles, from successive
parts of the wave form.
Now let's look specifically at the rectangle, for example,
starting at 0 and ending at Delta, and the amplitude of it
is x(Delta).
So what we have--
actually, this should be x(0), and so let me just
correct that here.
That's x(0).
And so we have a rectangle height x(0), and recall the
function that I defined last time as delta_Delta(t), which
had height 1 / delta and width delta.
So multiplying finally by this last little Delta, then, this
is a representation for the rectangle
that I've shown there.
Now there's a little bit of algebra there to kind of track
through, but what we're really doing is just simply
representing this in terms of rectangles.
What I want to do is describe each rectangle as in terms of
that function delta_Delta(t), which in the limit, then,
becomes an impulse.
So let's track that through a little further.
When we have that linear combination, then, we're
saying that x(t) can be represented by a sum as I
indicate here, which I can then write more generally in
this form, just indicating that this is an infinite sum.
We now want to take the limit as Delta goes to 0, and as
Delta goes to 0, notice that this term
becomes arbitrarily narrow.
This goes to our impulse function, and this, of course,
goes to x of tau.
And in fact, in the limit, a sum of this form is exactly
the way an integral is defined.
So we have an expression for y(t) in terms
of an impulse function.
There, I have to admit, is a little bit of detail to kind
of focus on at your leisure, but this is the general flow
of the strategy.
So what we have now is an integral that tells us that
tells us how x(t) can be described as a sum or linear
combination involving impulses.
This bottom equation, by the way, is often referred to as
the sifting integral.
In essence, what it says is that if I take a time function
x(t) and put it through that integral, the impulse as it
zips by generates x(t) all over again.
Now, at first glance, what it could look like is that we've
taken a time function x(t) and proceeded to represent it in a
very complicated way, in terms of itself, and one could ask,
why bother doing that?
And the reason, going back to what our strategy was, is that
what we want to do is exploit the property of linearity.
So by describing a time function as a linear
combination of weighted, delayed impulses, as in effect
we've done through this summation that corresponds to
a decomposition in terms of impulses, we can now exploit
linearity, specifically recognizing that the output of
a linear system is the sum of the responses to these
individual inputs.
So with h_kDelta(t) corresponding to the response
to delta_Delta(t-kDelta) and the rest of this stuff, the
x(kDelta) and this little Delta are
basically scale factors--
for a linear system, then, if the input is expressed in this
form, the output is expressed in this form, and again taking
the limit as Delta goes to 0, by definition, this
corresponds to an integral.
It's the integral that I indicate here with h_tau(t)
corresponding to the impulse response due to an impulse
occurring at time tau.
Now, again, we can do the same thing.
In particular, if the system is time-invariant, then the
response to each of these delayed impulses is simply a
delayed version of the impulse response, and so we can relate
these individual terms.
And in particular, then, the response to an impulse
occurring at time t = tau is simply the response to an
impulse occurring at time 0 shifted over to the time
origin tau.
Again, as we did before, we'll drop this subscript h_0, so
h_0(t) we'll simply define as h(t).
What we're left with when we do that is the description of
a linear time-invariant system through this integral, which
tells us how the output is related to the input and to
the impulse response.
Again, let's just quickly look at this from another
perspective as we did in discrete time.
Recall that what we've done is to take the continuous
function, decompose it in terms of rectangles, and then
each of these rectangles generates its individual
response, and then these individual
responses are added together.
And as we go through that process, of course, there's a
process whereby we let the approximation go to the
representation of a smooth curve.
Now, again, I stress if there is certainly a fair amount to
kind of examine carefully there, but it's important to
also reflect on what we've done, which is really pretty
significant.
What we've managed to accomplish is to exploit the
properties of linearity and time invariance, so that the
system could be represented in terms only of its response to
an impulse at time 0.
So for a linear time-invariant system--
quite amazingly, actually--
if you know its response to an impulse at t = 0 or n = 0,
depending on discrete or continuous time, then in fact,
through the convolution sum in discrete time or the
convolution integral in continuous time, you can
generate the response to an arbitrary input.
Let me just introduce a small amount of notation.
Again, reminding you of the convolution sum in the
discrete-time case, which looks as I've indicated here--
the sum of x[k]
h[n-k]
will have the requirement of making such frequent reference
to convolution that it's convenient to notationally
represent it as I have here with an asterisk.
So x[n]
* h[n]
means or denotes the convolution of x[n]
with h[n].
And correspondingly in the continuous-time case, we have
the convolution integral, which here is the sifting
integral
as we talked about, representing x(t) in terms of
itself as a linear combination of delayed impulses.
Here we have the convolution integral, and again we'll use
the asterisk to denote convolution.
Now, there's a lot about convolution that we'll want to
talk about.
There are properties of convolution which tell us
about properties of linear time-invariant systems.
Also, it's important to focus on the mechanics of
implementing a convolution--
in other words, understanding and generating some fluency
and insight into what these particular sum and integral
expressions mean.
So let's first look at discrete-time convolution and
examine more specifically what in essence the sum tells us to
do in terms of manipulating the sequences.
So returning to the expression for the convolution sum, as I
show here, the sum of x[k]
h[n-k]--
let's focus in on an example where we choose x[n]
as a unit step and h[n]
as a real exponential times the a unit step.
So the sequence x[n]
is as I indicate here, and the sequence h[n]
is an exponential for positive time and 0 for negative time.
So we have x[n]
and h[n], but now let's look back at the equation and let
me stress that what we want is not x[n]
and h[n]--
we want x[k], because we're going to sum
over k, and not h[k]
but h[n-k].
So we have then from x[n], it's straightforward to
generate x[k].
It's simply changing the index of summation.
And what's h[n-k]?
Well what's h[-k]?
h[-k]
is h[k]
flipped over.
So if this is what h[k]
looks like, then this is what h[n-k]
looks like.
In essence, what the operation of convolution or the
mechanics of convolution tells us to do is to take the
sequence h[n-k], which is h[-k]
positioned with its origin at k = n, and multiply this
sequence by this sequence and sum the product from k =
-infinity to +infinity.
So if we were to compute, for example, the output at n = 0--
as I positioned this sequence here, this is at n = 0--
I would take this and multiply it by this and sum from
-infinity to +infinity.
Or, for n = 1, I would position it here, for n = 2, I
would position it here.
Well, you can kind of see what the idea is.
Let's look at this a little more dynamically and see, in
fact, how one sequence slides past the other, and how the
output y[n]
builds up to the correct answer.
So the input that we're considering is a step input,
which I show here, and the impulse response that we will
convolve this with is a decaying exponential.
Now, to form the convolution, we want the product of x[k]--
not with h[k], but with h[n-k], corresponding to
taking h[k]
and reflecting it about the origin and then shifting it
appropriately.
So here we see h[n-k]
for n = 0, namely h[-k], and now h[1-k], which we'll show
next, is this shifted to the right by one point.
Here we have h[1-k]--
shifting to the right by one more point is h[2-k], and
shifting again to the right we'll have h[3-k].
Now let's shift back to the left until n is negative, and
then we'll begin the convolution.
So here's n = 0, n = -1, n = -2, and n = -3.
Now, to form the convolution, we want the product of x[k]
with h[n-k]
summed from -infinity to +infinity.
For n negative, that product is 0, and therefore the result
of the convolution is 0.
As we shift to the right, we'll build up the
convolution, and the result of the convolution will be shown
on the bottom trace.
So we begin the process with n negative, and
here we have n = -1.
At n = 0, we get our first non-zero contribution.
Now as we shift further to the right corresponding to
increasing n, we will accumulate more and more terms
in the sum, and the convolution will build up.
In particular for this example, the result of the
convolution increases monotonically, asymptotically
approaching a constant, and that constant, in fact, is
just simply the accumulation of the values under the
exponential.
Now let's carry out the convolution this time with an
input which is a rectangular pulse instead of a step input.
Again, the same impulse response, namely a decaying
exponential, and so we want to begin with h[n-k]
and again with n negative shown here.
Again, with n negative, there are no non-zero terms in the
product, and so the convolution for n negative
will be 0 as it was in the previous case.
Again, on the bottom trace we'll show the result of the
convolution as the impulse response slides along.
At n = 0, we get our first non-zero term.
As n increases past 0, we will begin to generate an output,
basically the same as the output that we generated with
a step input, until the impulse response reaches a
point where as we slide further, we slide outside the
interval where the rectangle is non-zero.
So when we slide one point further from what's shown
here, the output will now decay, corresponding to the
fact that the impulse response is sliding outside the
interval in which the input is non-zero.
So, on the bottom trace we now see the result of the
convolution.
OK.
So what you've seen, then, is an example of discrete-time
convolution.
Let's now look at an example of continuous-time
convolution.
As you might expect, continuous-time convolution
operates in exactly the same way.
Continuous-time convolution--
we have the expression again y(t) is an integral with now
x(tau) and h(t-tau).
It has exactly the same kind of form as we had previously
for discrete-time convolution--
and in fact, the mechanics of the continuous-time
convolution are identical.
So here is our example with x(t) equal to a unit step and
h(t) now a real exponential times a unit step.
I show here x(t), which is the unit step function.
Here we have h(t), which is an exponential for positive time
and 0 for negative time.
Again, looking back at the expression for convolution,
it's not x(t) that we want, it x(tau) that we want.
And it's not h(t) or h(tau) that we want, it's h(t-tau).
We plan to multiply these together and integrate over
the variable tau, and that gives us the output at any
given time.
If we want it at another time, we change the value of t as an
argument inside this integral.
So here we have x(t), and here we have h(t), which isn't
quite what we wanted.
Here we have x(tau), and that's fine-- it's just x(t)
with t relabeled as tau.
Now, what is h(t-tau)?
Well, here's h(tau), and if we simply turn that
over, here is h(t-tau).
And h(t-tau) is positioned, then, at tau equal to t.
As we change the value of t that change the position of
this signal, now we multiply these two together and
integrate from -infinity to +infinity with h(t-tau)
positioned at the appropriate value of t.
Again, it's best really to see this example and get the
notion of the signal being flipped and the two signals
sliding past each other, multiplying and integrating by
looking at it dynamically and observing how the
answer builds up.
Again, the input that we consider is a step input.
And again, we use an impulse response which is a decaying
exponential.
To form the convolution, we want the product of x(tau)--
not with h(tau), but with h(t-tau).
So we want h(t) time-reversed, and then shifted appropriately
depending on the value of t.
Let's first just look at h(t-tau) for t positive
corresponding to shifting h(-tau) out to the right, and
here we have t increasing.
Here is t decreasing, and we'll want to begin the
convolution with t negative, corresponding to shifting
h(-tau) to the left.
Now to form the convolution, we want the
product of these two.
For t negative, there are no non-zero contributions to the
integral, and so the convolution will be 0 for t
less than 0.
On the bottom trace, we show the result of the convolution,
here for t negative, and for t less than 0, we will continue
to have 0 in the convolution.
Now as t increases past 0, we begin to get some non-zero
contribution in the product, indicated by the fact that the
convolution--
the result of the convolution--
starts to build up.
As t increases further, we will get more and more
non-zero contribution in the integrand.
So, the result of the convolution will be a
monotonically increasing function for this particular
example, which asymptotically approaches a constant.
That constant will be proportional to the area under
the impulse response, because of the fact that we're
convolving with a step input.
Now let's carry out the convolution with an input
which is a rectangular pulse--
again, an impulse response which is an exponential.
So to form the convolution, we want x(tau) with h(t-tau)--
h(t-tau) shown here for t negative.
To form the convolution, we take the integral of the
product of these two, which again will be 0
for t less than 0.
The bottom trace shows the result of the convolution
here, shown as 0, and it will continue to be 0 until t
becomes positive, at which point we build up some
non-zero term in the integrand.
Now as we slide further, until the impulse response shifts
outside the interval in which the pulse is non-zero, the
output will build up.
But here we've now begun to leave that interval, and so
the output will start to decay exponentially.
As the impulse response slides further and further
corresponding to increasing t, then the output will decay
exponentially, representing the fact that there is less
and less area in the product of x(tau) and h(t-tau).
Asymptotically, this output will then approach 0.
OK.
So you've seen convolution, you've seen the derivation of
convolution, and kind of the graphical representation of
convolution.
Finally, let's work again with these two examples, and let's
go through those two examples analytically so that we
finally see how, analytically, the result develops for those
same examples.
Well, we have first the discrete-time case, and let's
take our discrete-time example.
In general, the convolution sum is as I've indicated here.
This is just our expression from before, which is the
convolution sum.
If we take our two examples--
the example of an input which is a unit step, and an impulse
response, which is a real exponential multiplied by a
unit step--
we have then replacing x[k]
by what we know the input to be, and h[n-k]
by what we know the impulse response to be, the output is
the expression that we have here.
Now, in this expression--
and you'll see this very generally and with some more
complicated examples when you look at the text--
as you go to evaluate these expressions, generally what
happens is that the signals have different analytical
forms in different regions.
That's, in fact, what we have here.
In particular, let's look at the sum, and what we observe
first of all is that the limits on this sum are going
to be modified, depending on where this unit
step is 0 and non-zero.
In particular, if we first consider what will turn out to
be the simple case--
namely, n less than 0--
for n less than 0, this unit step is 0 for k
greater than n.
With n less than 0, that means that this unit step never is
non-zero for k positive.
On the other hand, this unit step is never non-zero or
always 0 for k negative.
Let me just stress that by looking at the particular
graphs, here is the unit step u[k].
Here is the unit step u[n-k], and for n less than 0, so that
this point comes before this point, the product of these
two is equal to 0.
That means there is no overlap between these two terms, and
so it says that y[n], the output, is 0
for n less than 0.
Well, that was an easy one.
For n greater than 0, it's not quite as straightforward as
coming out with the answer 0.
So now let's look at what happens when the two unit
steps overlap, and this would correspond to what I've
labeled here as interval 2, namely for n greater than 0.
If we just look back at the summation that we had, the
summation now corresponds to this unit step and this unit
step, having some overlap.
So for interval 2, corresponding to n greater
than 0, we have u[k], the unit step.
We have u[n-k], which is a unit step going backward in
time, but which extends for positive values of n.
If we think about multiplying these two together, we will
get in the product unity for what values of k?
Well, for k starting at 0 corresponding to one of the
unit steps and ending at n corresponding to
the other unit step.
So we have an overlap between these for k equal to 0, et
cetera, up through the value n.
Now, that means that in terms of the original sum, we can
get rid of the unit steps involved by simply changing
the limits on the sum.
The limits now are from 0 to n, of the term alpha^(n-k).
We had before a u[k]
and u[n-k], and that disappeared because we dealt
with that simply by modifying the limits.
We now pull out the term alpha^n, because the summation
is on k, not on n, so we can simply pull
that term of the sum.
We now have alpha^(-k), which we can rewrite as
(alpha^(-1))^k.
The upshot of all of this is that y[n]
now we can reexpress as alpha^n, the sum from 0 to n
of (alpha^(-1)^k.
The question is, how do we evaluate that?
It essentially corresponds to a finite number of terms in a
geometric series.
That, by the way, is a summation that will recur over
and over and over and over again, and it's one that you
should write down, write on your back pocket, write on the
palm of your hand, or whatever it takes to remember it.
What you'll see is it that will recur more or less
throughout the course, and so it's one worth remembering.
In particular, what the sum of a geometric series is, is what
I've indicated here.
We have the sum from 0 to r, of beta^k.
It's 1 - beta^(r+1)--
this is one more than the upper limit on the summation--
and in the denominator is 1 - beta.
So, this equation is important.
There's no point in attempting to derive it.
However you get to it, it's important to retain it.
We can now use that summation in the expression
that we just developed.
So let's proceed to evaluate that sum in closed form.
We now go back to the expression that we just
worked out-- y[n]
is alpha^n, the sum from 0 to n, (alpha^(-1))^k.
This plays the role of beta in the term that I just--
in the expression then I just presented.
So, using that result, we can rewrite this summation as I
indicate here.
The final result that we end up with after a certain amount
of algebra is y[n]
equal to (1 - alpha^(n+1)) / (1 - alpha).
Let me just kind of indicate with a few dots here that
there is a certain amount of algebra required in going from
this step to this step, and I'd like to leave you with the
fun and opportunity of doing that at your leisure.
The expression we have now for y[n], is y[n]
= (1 - alpha^(n+1)) / (1 - alpha).
That's for n greater than 0.
We had found out previously there it was 0
for n less than 0.
Finally, if we were to plot this, what we would get is the
graph that I indicate here.
The first non-zero value occurs at n = 0, and it has a
height of 1, and then the next non-zero value at 1, and this
has a height of 1 + alpha, and this is 1 + alpha + alpha^2.
The sequence continues on like that and asymptotically
approaches, as n goes to infinity, asymptotically
approaches 1 / (1 - alpha), which is consistent with the
algebraic expression that we have, that we developed, and
obviously of course is also consistent with the movie.
That's our discrete-time example, which we kind of went
through graphically with the transparencies, and we went
through graphically with the movie, and now we've gone
through analytically.
Now let's look analytically at the continuous-time example,
which pretty much flows in the same way as
we've just gone through.
Again, we have the convolution integral, which is the
integral indicated at the top.
Our example, you recall, was with x(t) as a unit step, and
h(t) as an exponential times a unit step.
So when we substitute those in, this then corresponds to
x(t) and this corresponds to h(t-tau).
Again, we have the same issue more or less, which is that
inside that integral, there are two steps, one of them
going forward in time and one of them going backward in
time, and we need to examine when they overlap and when
they don't.
When they don't overlap, the product, of course, is 0, and
there's no point doing any integration because the
integrand is 0.
So if we track it through, we have again Interval 1, which
is t less than 0.
For t less than 0, this unit step, which only begins at tau
= 0, and this unit step which is 0, by the time tau gets up
to t and beyond.
For t less than 0, there's no overlap between the unit step
going forward in time and the unit step
going backward in time.
Consequently, the integrand is equal to 0, and consequently,
the output is equal to 0.
We can likewise look at the interval where the two unit
steps do overlap.
In that case what happens again is that the overlap, in
essence of the unit step, tells us, gives us a range on
the integration--
in particular, the two steps overlap when t is greater than
0 from tau = 0 to tau = t.
For Interval 2, for t greater than 0--
again, of course, we have this expression.
This product of this unit step and this unit step is equal to
1 in this range, and so that allows us, then, to change the
limits on the integral--
instead of from -infinity to +infinity, we know that the
integrand is non-zero only over this range.
Looking at this integral, we can now pull out the term
which corresponds to e^-at, just as we pulled out a term
in the discrete-time case.
We notice in the integral that we have e^(-a*-tau), so that
gives us the integral from 0 to t of e^(a*tau).
If we perform that integration, we end up with
this expression.
Finally, multiplying this by e^-at, we have for y(t), for t
greater than 0, the algebraic expression that I've indicated
on the bottom.
So we had worked out t less than 0, and we
come out with 0.
We work out t greater than 0, and we come out with this
algebraic expression.
If we plot this algebraic expression as a function of
time, we find that what it corresponds to is an
exponential behavior starting at zero and exponentially
heading asymptotically toward the value 1 / a.
We've gone through these examples several ways, and one
is analytically.
In order to develop a feel and fluency for convolution, it's
absolutely essential to work through a variety of examples,
both understanding them graphically and understanding
them as we did here analytically.
You'll have an opportunity to do that through the problems
that I've suggested in the video course manual.
In the next lecture, what we'll turn to are some general
properties of convolution, and show how this rather amazing
representation of linear time-invariant systems in fact
leads to a variety of properties of linear
time-invariant systems.
We'll find that convolution is fairly rich in its properties,
and what this leads to are some very nice and desirable
and exploitable properties of linear time-invariant systems.
Thank you.