Convolution of exponentials.

As a simple example of evaluating the convolution, consider the functions f_1a(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

with f₂(t - t) shown opposite. f₁(t) here corresponds to f_1a(t) and f_1a(t) is the same as f_1a(t) with the variables changed.

Because of the discontinuity in f_1a(t) then the convolution f_a(t) of f_1a(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 diagram opposite shows f_1a(t) in red f₂(t) in blue and f_a(t) in purple.

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 f₂(t) is given by f_b(t) in the equation below.

The limits of the integration are 0 and 1 since the interval from 0 to 1 is the only one where both f_1b(t) and f₂(t - t) 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, and
f(t) = e^1-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 f_1b(t) in red f₂(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 f₂(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^1-t - e^-t for 1 £ t £ 2, and
f(t) = e^1-t - e^-t + 1 - e^2-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 f_1c(t) in red f₂(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 f₂(t) is the same as f_1b(t) and f₂(t) but differs for larger values of t.

In the interval 1 £ t £ 2 the convolution of f_1d(t) and f₂(t) is given by f_d(t) shown opposite.

And for the interval t ³ 2 the convolution of f_1d(t) and f₂(t)  given by f_e(t) shown opposite.

The final result for the convolution f(t) of f_1d(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 - e^1-t - e^-t for 1 £ t £ 2, and
f(t) = e^2-t - e^1-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 f_1d(t) in red f₂(t) in blue and f(t) in purple.