Solving the Convolution Integral Using Unit Step Functions

Convolution integrals are commonly encountered in various fields, such as signal processing, control theory, and systems analysis. This article provides a detailed step-by-step solution to the convolution integral:

It integral-∞∞ e-2τ u(t - τ) e-(t - τ) u(τ) dτ

Role of Unit Step Functions

First, let's clarify the role of the unit step functions used in this problem. The functions u(t - τ) and u(τ) are defined as follows:

u(t - τ) is 0 for τ ≤ 0 and 1 for τ 0. u(τ) is 0 for τ ≤ 0 and 1 for τ 0.

Therefore, the integration will only take place over the interval where both step functions are 1. From the definitions of these functions, this interval is given by:

0 ≤ τ ≤ t

Rewriting the Integral

With the interval determined, we can now rewrite the convolution integral as follows:

It integral0t e-2τ e-(t - τ) dτ

Simplifying the Integrand

To simplify the integrand, we multiply the exponential functions:

e-2τ e-(t - τ) e-2τ e-t eτ e-t e-τ

Substituting this into the integral, we obtain:

It e-t integral0t e-τ dτ

Evaluating the Integral

Next, we evaluate the integral integral0t e-τ dτ:

integral0t e-τ dτ -e-τ |0t -e-t - (-e0) 1 - e-t

Substituting this back into the expression for It, we get:

It e-t (1 - e-t) e-t - e-2t

Thus, the solution to the convolution integral is:

Final Solution

boxed{e-t - e-2t}

Implications and Visualization

This result can be further interpreted visually. The convolution integral effectively considers the overlap of the two step functions u(t - τ) and u(τ). For a fixed value of t, the product of the two step functions is non-zero only when both are 1, which occurs for 0 ≤ τ ≤ t.

Consider a different representation of the convolution integral:

It e-t integral-∞∞ e-τ u(t - τ) u(τ) dτ

Using the properties of unit step functions, we can simplify the integral as follows:

u(t - τ) u(τ) 0 for τ ≤ 0 and τ ≥ t, and 1 for 0 ≤ τ ≤ t

This results in the integral:

It e-t integral0t e-τ dτ e-t (1 - e-t) e-t - e-2t