Question

By direct integration [eq. (4.1)], find the Laplace transforms and the region of convergence of the following function: (a) u(t)-u(t-1)?

Answer 1

The Laplace transform of the function u(t) - u(t-1) is computed by integrating from 0 to 1, resulting in
`(1/s) [1-e^(-s)]` with a region of convergence for Re(s) > 0.

Step-by-step explanation:

The question pertains to the calculation of the Laplace transform of the function u(t) - u(t-1), which represents a rectangular pulse of unit height starting at t=0 and ending at t=1. The Laplace transform is defined as
`L[f(t)] = \int_0^\infty e^(-st) f(t) dt` , where s is a complex number and the region of integration is from 0 to infinity.

To find the Laplace transform of u(t) - u(t-1), we break down the integral piecewise corresponding to the defining intervals of the step functions u(t) and u(t-1). We get two integrals, one from 0 to 1 and the other from 1 to infinity. However, the function is 0 from 1 to infinity, so the integral to infinity doesn't contribute to the Laplace transform.

The resulting Laplace transform calculation is as follows:

• 
  `\( \int_0^1 e^(-st) dt = (1)/(s) [1-e^(-s)] \)`

The region of convergence (ROC) for the Laplace transform is the set of values of s for which the Laplace transform converges. In this case, the ROC will be Re(s) > 0, as the exponential function
`e^(-st)`requires positive real part of s to ensure convergence.