\subsection*Solution \(y(t) = \int_-\infty^\infty e^-\tauu(\tau) \cdot [u(t-\tau) - u(t-\tau-2)] d\tau\). For \(t < 0\): \(y(t)=0\). For \(0 \le t < 2\): \(y(t) = \int_0^t e^-\tau d\tau = 1 - e^-t\). For \(t \ge 2\): \(y(t) = \int_t-2^t e^-\tau d\tau = e^-(t-2) - e^-t\). Thus \[ y(t) = \begincases 0, & t<0 \\ 1-e^-t, & 0\le t < 2 \\ e^-(t-2) - e^-t, & t \ge 2 \endcases \]
\subsection*Solution First term: \(\frac11-0.5z^-1\), \(|z| > 0.5\). \\ Second term: \(-\frac11-2z^-1\), \(|z| < 2\). \\ Thus \(X(z) = \frac11-0.5z^-1 - \frac11-2z^-1 = \frac-1.5z^-1(1-0.5z^-1)(1-2z^-1)\), ROC: \(0.5 < |z| < 2\). signals and systems problems and solutions pdf
\sectionContinuous-Time Signals
\noindent\textbf13. Use Euler formulas and compare with exponential FS: \(x(t)=\sum a_k e^jk\omega_0 t\) with \(\omega_0=\pi\) (fundamental). For \(t \ge 2\): \(y(t) = \int_t-2^t e^-\tau
\noindent\textbf14. Use differentiation in Z-domain: \(n a^n u[n] \leftrightarrow \fracaz^-1(1-az^-1)^2\). \\ Thus \(X(z) = \frac11-0