# Convolution of exponential with variants of step.

Cuthbert Nyack
 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 with f2(t - t) shown opposite. f1(t) here corresponds to f1a(t) and f1a(t) is the same as f1a(t) with the variables changed.

 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 diagram opposite shows f1a(t) in red f2(t) in blue and fa(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 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 limits of the integration are 0 and 1 since the interval from 0 to 1 is the only one where both f1b(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) = 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.