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An integral which appears in almost all forms of engineering
is the convolution of two functions defined as
We will begin
with two examples as motivation, but these are for the students
who are interested in understanding why we are doing the convolution.
You may skip to the class material.
Motivation: A Leaky Mug
Consider the leaky mug shown in Figure 1.
Figure 1. A leaky mug.
If the mug contains of fluid at time
and the mug leaks at a rate of
, then the amount of fluid in the
mug may be described by the function in Figure 2.
Figure 2. The amount of liquid in a leaky mug starting with
If while the mug is emptying, another
is put into the mug at time
, the function describing the amount of liquid
becomes what is seen in Figure 3.
Figure 3. The amount of liquid in a mug when
are added at times
What happens, however, if the the mug is being filled from a tap, as
is shown in Figure 4.
Figure 4. A leaky mug with a faucet.
Suppose that we start turning on the faucet at time and
after two seconds, we reached a flow of
and we start shutting
off the faucet after 15 seconds until the faucet stops
running after 17 seconds. The plot of the rate of flow of
the faucet is shown in Figure 5.
Figure 5. The rate of flow of the faucet in units of centilitres per
What function describes the total amount of fluid in the mug?
In this case, the problem is actually very difficult to describe
because the differential equation which describes this problem
is non-linear! The rate of change of volume involves
the unit step function
and this is without a faucet.
With significant effort, is is possible to determine that
the amount of fluid in the mug is as is shown in Figure 6, the maximum
begin achieved at time . The
actual function is
Figure 6. The total amount of fluid in the mug while being filled
according to the function in Figure 5.
It is exceptionally fortunate that circuits made out of
resistors, inductors, and capacitors is linear, and we will use this
in the next example to justify the convolution integral.
Motivation: Discharging Capacitors
Consider the circuit in Figure 7. The differential equation of
this circuit is
is the charge on the capacitor.
Figure 7. A circuit with a capacitor, three resistors, and a voltage source.
voltage source is zero (the system is homogeneous) and the capacitor
has a charge of coulombs at time
, the capacitor would discharge
according to the function .
If the charge on the capacitor is initially zero and a charge of
size is only placed on the capacitor at time
(we shocked the system with the
charge ), the function describing the charge is
is the unit step function. If we shock
the system at time , the charge on the
capacitor then falls according to the function
For clarity, we will now define
Using the Impulse Function
From your ECE 140 course, you are able to determine
the response to a system when the voltage is constant. If you consider
the functions we previously used to define the impulse function,
it is possible to determine the response: initially, the charge is zero;
during the phase of duration , the
voltage is constant and the capacitor is being charged; and when
the voltage is turned off, the capacitor discharges again. Figure 8
shows the charge of the capacitor for
runs from 1 down to 0.01 in steps of 0.01.
Figure 8. The charge on the capacitor when the voltage is defined by
the function for
running from 1 down to 0.01 in steps of 0.01.
To emphasize the impulse, Figure 9 includes the impulse function.
You will note that the area under the pulse is always 1, but as it
becomes more narrow, its voltage increases.
Figure 9. The addition of the impulse to the function shown in
It seems plausible, therefore, to believe that the solution to the
where the capacitor is
initially discharged is
if the impulse is
, the response is
Suppose now we shock the system twice: first at time
and then again at time
. After 1 second, the charge
on the capacitor has already decayed to ,
but then we added another
and so now the charge begins to
decay starting from
, as is shown in Figure 10.
Figure 10. The response of a capacitor shocked with 2 and 3
coulombs at times 0 and 1, respectively.
We may already represent these two shocks using the
the response is . What
do we do, however, if the voltage is defined by a function
, as opposed to
Consider, for example, the function
shown in Figure 11.
Figure 11. The function .
Recall also that while electrical power is alternating, it is possible
to approximate a function using a rectified and modulated current, as
can be seen in Figure 12 where the rectified sinusoids are appropriately
scaled so that the total area remains the same.
Figure 12. An approximation of the function in Figure 11 using appropriately
scaled rectified sinusoids.
Using this idea, could we not not approximate a sub-interval by an
impulse function multiplied by the area of the function on that
sub-interval? In this case, we could approximate the function by a
sequence of impulses. For example, if we were to divide the interval
equally spaced sub-intervals, denoted the
width of each interval by
and chose a representative point
in each sub-interval,
it follows that an approximation of the area on each interval is
we may represent the function as
For example, consider the image in Figure 13.
Figure 13. An approximation of the function in Figure 11 using appropriately
Note: this approximation is only valid with respect to the operation of
The beauty of this method now comes from the observation that we
already know the response of the system to each of these impulse functions:
and therefore the total response is the sum of these responses:
This response is shown in Figure 14 where you will note that initially
the charge on the capacitor is 0 and then with each impulse, the charge
on the capacitor is correspondingly increased.
Figure 14. The response to a sequence of appropriately weighted impulse
Now, rewrite the sum as
and you will note that it looks like the Riemann approximation of
an integral, and as , it must
converge to the integral.
But this is the convolution integral defined at the very start of this
topic. Therefore, it follows that we can use this convolution to find
the response of a system:
You will note that after time
, the sinusoids cancel and the function is
just a decaying exponential. The plot of this function superimposed on Figure 14
is shown in Figure 15.
Figure 15. The convolution response plotted over the approximation shown
in Figure 14.
The Convolution and its Uses
The convolution function is so often used in engineering that not
only does it have a special name, but it also has a special
With a little effort, you can show that the operation of the
convolution is symmetric:
The Convolution and Causal Systems
is causal if the response to an input signal at time depends
only on the values of the function for (e.g., you do not expect your stereo to
respond until you actually push a button). One consequence of causality of a linear
system is that the response to the impulse
function is 0 for .
In many cases, we are also working with signals which start at time , and consequently, the
input will also be 0 for .
If both functions are zero for time , the convolution simplifies to
which is the form given in AMEM.
The Convolution and the Laplace Transform
Suppose we are working in the frequency domain by using the Laplace
transform. What is the Laplace transform of the convolution?
From the definition, we have that
Because the function is causal,
we may extend the inner integral to infinity:
We can now perform a change of variables:
. Hence, the transform
Switching the integrals, we have:
The conclusion is that
This is the Convolution Theorem.
An alternate proof is given in AMEM, p.196.
The convolution theorem may be used to find the inverse Laplace
transform; for example,
Therefore, by the convolution theorem,
You may now use the convolution:
Another Example: Integral Equations
Integral equations are equations where the unknown function appears
within an integrand. If the integral is in the form of a convolution,
the Laplace transform method may be used. For example, consider
where is called the kernel and is given
and is the forcing function.
Suppose, for example, that the kernel is
and the forcing function
is ; thus, we have:
and by taking the Laplace transform of each side,
therefore we have
Solving for , we have:
Using the Laplace transform, we have that