Using direct evaluation find the convolution

Question

Using direct evaluaton the convolution of f(t)=t and g(t)= e^-2t

Answer 1

Assume that the convolution of f and g, or f(t) * g(t), is defined as thus from z = 0 to z = t:

f(t) * g(t) = ∫f(z)g(t - z) dz.

Here, z is a dummy variable.

Here, f(t)=t and g(t)= e^-2t, so:

f(t) * g(t) = ∫ (z)e^(-2(t - z)) dz, integrated from z = 0 to z = t.

Simplify:

∫ (z)e^(-2t)e^(2z) dz, integrated from z = 0 to z = t

The t-terms can be treated as constants.

e^(-2t) ∫ (z)e^(2z) dz, integrated from z = 0 to z = t

Use integration by parts, letting u = z and dv = e^(2z) dz. This results in:

e^(-2t)[(1/4)(2z-1)e^(2z)], evaluated from z = 0 to z = t

Evaluate over the bounds of integration and simplify:

e^(-2t)[(1/4)(2t-1)e^(2t) - (1/4)(-1)]

You should have this as a final answer:

(1/2)t - (1/4) + (1/4)e^(-2t)

Note that you can also get this answer by evaluating this integral, which is equivalent to f(t) * g(t):

g(t) * f(t) is defined as thus from z = 0 to z = t:

g(t) * f(t) = ∫f(t - z)g(t) dz

Here, g(t) * f(t) = ∫ (t - z)e^(-2t) dz, integrated from z = 0 to z = t.