Final answer:
The output of an LTI system with transfer function H(jω) = 2+jω when the input is x(t) = e-t u(t) can be found by taking the Laplace transform of the input, applying the transfer function in the s-domain, and lastly taking the inverse Laplace transform to return to the time domain.
Step-by-step explanation:
The subject of this question is the output of an LTI (Linear Time-Invariant) system when the input is a decaying exponential function. Given the transfer function of the LTI system H(jω) = 2+jω and the input x(t) = e-t u(t), we need to find the system's output. To do this, we can take the Laplace transform of the input signal, multiply it by the transfer function, and then take the inverse Laplace transform to obtain the output in the time domain.
To proceed with this process, observe that the Laplace transform of the input signal x(t) is X(s) = 1/(s+1), for s > -1. Now, we can apply the transfer function, which in the s-domain is simply H(s) = 2+s since the Laplace transform of jω is s. The output in the s-domain is therefore Y(s) = H(s)X(s) = (2+s)/(s+1). Finally, taking the inverse Laplace transform of Y(s) gives us the output in the time domain, y(t).