Energy and Power Signals MCQ Quiz in मल्याळम - Objective Question with Answer for Energy and Power Signals - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

\(\rm{\begin{array}{l} P = \begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{2T}}\mathop \smallint \limits_{ - T}^T |x\left( t \right){|^2}dt = \begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{2T}}\mathop \smallint \limits_0^T |{e^{ - 2t}}{|^2}dt = \begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{2T}}\mathop \smallint \limits_0^T {e^{ - 4t}}dt\\ \Rightarrow P = \left. {\begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{2T}}.\frac{{{e^{ - 4t}}}}{{\left( { - 4} \right)}}} \right|_o^T = \begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{2T}}.\frac{1}{4}\left[ {1 - {e^{ - 4T}}} \right]\\ \Rightarrow P = \begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{8T}}\left[ {1 - {e^{ - 4T}}} \right] = 0 \end{array}}\)