\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 \(A^2/2\).
\subsection*Problem 8: Z-Transform of a Causal Sequence Find the Z-transform and ROC of \(x[n] = (0.5)^n u[n] + (2)^n u[-n-1]\). signals and systems problems and solutions pdf
\sectionLaplace Transform
\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*Solution The signal is periodic, so it has
\noindent\textbf11. \(x[n]=\delta[n]+\delta[n-1]+\delta[n-2]\), \(h[n]=\delta[n]+\delta[n-1]\). Convolution gives \(y[n]=\delta[n]+2\delta[n-1]+2\delta[n-2]+\delta[n-3]\). \subsection*Problem 8: Z-Transform of a Causal Sequence Find
\subsection*Problem 2: Even and Odd Decomposition Find the even and odd parts of \(x(t) = e^-atu(t)\), where \(u(t)\) is the unit step.
\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 \[ y(t) = \begincases 0, & t<0 \\ 1-e^-t, & 0\le t < 2 \\ e^-(t-2) - e^-t, & t \ge 2 \endcases \]