| As a simple example of evaluating the convolution, consider the functions f1a(t) and f2(t) shown in the diagrams opposite. |
|
|
|
The general expression for the convolution
f(t) of two functions f1(t) and f2(t) is  |
 |
Because of the discontinuity in
f1a(t)
then the convolution
fa(t)
of
f1a(t)
and
f2(t)
must be done in the 2 intervals of time
t < 0 and
t ³ 0. For t < 0, the convolution
is zero and for t ³ 0 it is given by
the expression below.
 |
 |
|
The calculation of the convolution can be extended by considering the
pulse opposite instead of a step. Here there are 2 discontinuities and the
convolution must be evaluated in 3 intervals. t < 0, 0 £ t £ 1, and t > 1. As above the convolution for t < 0 is zero, and for 0 £ t £ 1, it is fa(t) above, however it is not zero for t > 1. |
 |
For t > 1 the convolution of
f1b(t)
and
f2(t)
is given by fb(t)
in the equation below.
 |
|
|
The final result for the convolution
f(t)
is now given by:- f(t) = 0 for t £ 0 f(t) = 1 - e-t for 0 £ t £ 1, and f(t) = e1-t - e-t for t ³ 1. The convolution is a continuous function and the 3 "pieces" of f(t) must join up at the interval boundaries. The diagram opposite shows f1b(t) in red f2(t) in blue and f(t) in purple. |
 |
|
In the diagram opposite the function
f1b(t)
is extended to
f1c(t)
by adding a step for t ³ 2. For t < 2, the convolution is the same as for f1b(t) but differs for t ³ 2. |
 |
For t ³ 2, the convolution of
f1c(t)
and
f2(t)
is given by
fc(t)
in the expression below. The integral must be evaluated over the 2 intervals
0 to 1 and 2 to t for which both functions are nonzero.
 |
|
|
The final result for the convolution
f(t)
is now given by:- f(t) = 0 for t £ 0, f(t) = 1 - e-t for 0 £ t £ 1, f(t) = e1-t - e-t for 1 £ t £ 2, and f(t) = e1-t - e-t + 1 - e2-t for t ³ 2. The convolution is a continuous function and the 4 "pieces" of f(t) must join up at the interval boundaries. The diagram opposite shows f1c(t) in red f2(t) in blue and f(t) in purple. |
 |
| A modified version of f1b(t) is shown in the diagram opposite as f1d(t). Instead of the pulse dropping to zero at t = 1, it follows a ramp with slope -1 to reach 0 at t = 2. For t up to 1 the convolution of f1d(t) and f2(t) is the same as f1b(t) and f2(t) but differs for larger values of t. |
 |
| In the interval 1 £ t £ 2 the convolution of f1d(t) and f2(t) is given by fd(t) shown opposite. |
 |
| And for the interval t ³ 2 the convolution of f1d(t) and f2(t) bnnis given by fe(t) shown opposite. |
 |
|
The final result for the convolution
f(t) of
f1d(t)
and
f2(t)
is now given by:- f(t) = 0 for t £ 0, f(t) = 1 - e-t for 0 £ t £ 1, f(t) = 3 - t - e1-t - e-t for 1 £ t £ 2, and f(t) = e2-t - e1-t - e-t for t ³ 2. As in all the cases above, the convolution is a continuous function and the 4 "pieces" of f(t) must join up at the interval boundaries. The diagram opposite shows f1d(t) in red f2(t) in blue and f(t) in purple. |
 |