And Systems Problems And Solutions Pdf | Signals
\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: \(e^-2tu(t) \leftrightarrow \frac1s+2\), \(\textRe(s) > -2\). \\ Second term: \(e^3tu(-t) \leftrightarrow -\frac1s-3\), \(\textRe(s) < 3\). \\ Thus \(X(s) = \frac1s+2 - \frac1s-3 = \frac-5(s+2)(s-3)\), ROC: \(-2 < \textRe(s) < 3\).
\noindent\textbf12. Using \(t^n e^-atu(t) \leftrightarrow \fracn!(s+a)^n+1\). signals and systems problems and solutions pdf
\subsection*Problem 5: Fourier Transform of a Rectangular Pulse Compute the Fourier transform of \(x(t) = \textrect(t/T) = 1\) for \(|t| < T/2\), 0 otherwise.
\sectionLinear Time-Invariant (LTI) Systems and Convolution For \(t \ge 2\): \(y(t) = \int_t-2^t e^-\tau
\section*Additional Problems (Brief Solutions)
\subsection*Problem 4: Trigonometric Fourier Series Find the Fourier series of the periodic square wave \(x(t)\) with period \(T=2\), defined on \((-1,1)\) as \(x(t)=1\) for \(|t|<0.5\) and \(x(t)=0\) otherwise. \\ Thus \(X(s) = \frac1s+2 - \frac1s-3 =
\subsection*Solution Modulation: \(x(t)\cos(\omega_0 t) \leftrightarrow \frac12[X(j(\omega-\omega_0)) + X(j(\omega+\omega_0))]\). \\ Thus \(\textrect(t/T)\cos(\omega_0 t) \leftrightarrow \fracT2\left[\textsinc\left(\frac(\omega-\omega_0)T2\pi\right) + \textsinc\left(\frac(\omega+\omega_0)T2\pi\right)\right]\).
