__link__ — Signals And Systems Problems And Solutions Pdf

\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: \(0.5 < |z| < 2\).

\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. signals and systems problems and solutions pdf

\section*Solutions to Selected Additional Problems \subsection*Solution First term: \(\frac11-0

\subsection*Solution The signal is periodic, so it has infinite energy but finite average power. \[ P = \lim_T\to\infty \frac1T \int_-T/2^T/2 |x(t)|^2 dt = \frac1T_0 \int_0^T_0 A^2 \cos^2(2\pi f_0 t + \theta) dt \] Using \(\cos^2(\cdot) = \frac1+\cos(2\cdot)2\), the integral of the cosine term over one period is zero: \[ P = \fracA^2T_0 \int_0^T_0 \frac12 dt = \fracA^22. \] Hence \(x(t)\) is a power signal with power \(A^2/2\). \] Hence \(x(t)\) is a power signal with power \(A^2/2\)

\sectionLinear Time-Invariant (LTI) Systems and Convolution

\begindocument