As a simple example of evaluating the convolution, consider the functions f_{1a}(t) and f_{2}(t) shown in the diagrams opposite. 


The general expression for the convolution
f(t) of two functions f_{1}(t) and f_{2}(t) is 

Because of the discontinuity in
f_{1a}(t)
then the convolution
f_{a}(t)
of
f_{1a}(t)
and
f_{2}(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 f_{a}(t) above, however it is not zero for t > 1. 

For t > 1 the convolution of
f_{1b}(t)
and
f_{2}(t)
is given by f_{b}(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) = e^{1t } 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 f_{1b}(t) in red f_{2}(t) in blue and f(t) in purple. 

In the diagram opposite the function
f_{1b}(t)
is extended to
f_{1c}(t)
by adding a step for t ³ 2. For t < 2, the convolution is the same as for f_{1b}(t) but differs for t ³ 2. 

For t ³ 2, the convolution of
f_{1c}(t)
and
f_{2}(t)
is given by
f_{c}(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) = e^{1t } e^{t} for 1 £ t £ 2, and f(t) = e^{1t } e^{t} + 1  e^{2t} 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 f_{1c}(t) in red f_{2}(t) in blue and f(t) in purple. 

A modified version of f_{1b}(t) is shown in the diagram opposite as f_{1d}(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 f_{1d}(t) and f_{2}(t) is the same as f_{1b}(t) and f_{2}(t) but differs for larger values of t. 

In the interval 1 £ t £ 2 the convolution of f_{1d}(t) and f_{2}(t) is given by f_{d}(t) shown opposite. 

And for the interval t ³ 2 the convolution of f_{1d}(t) and f_{2}(t) bnnis given by f_{e}(t) shown opposite. 

The final result for the convolution
f(t) of
f_{1d}(t)
and
f_{2}(t)
is now given by: f(t) = 0 for t £ 0, f(t) = 1  e^{t} for 0 £ t £ 1, f(t) = 3  t  e^{1t } e^{t} for 1 £ t £ 2, and f(t) = e^{2t } e^{1t}  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 f_{1d}(t) in red f_{2}(t) in blue and f(t) in purple. 
