Energy and Power Signals MCQ Quiz in मराठी - Objective Question with Answer for Energy and Power Signals - मोफत PDF डाउनलोड करा
Last updated on Mar 19, 2025
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Energy and Power Signals Question 1:
The power of the signal
\(\rm{x\left( t \right) = {e^{ - 2t}}u\left( t \right)}\)
Answer (Detailed Solution Below)
\(\rm{0 \ W}\)
Energy and Power Signals Question 1 Detailed Solution
\(\rm{\begin{array}{l} P = \begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{2T}}\mathop \smallint \limits_{ - T}^T |x\left( t \right){|^2}dt = \begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{2T}}\mathop \smallint \limits_0^T |{e^{ - 2t}}{|^2}dt = \begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{2T}}\mathop \smallint \limits_0^T {e^{ - 4t}}dt\\ \Rightarrow P = \left. {\begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{2T}}.\frac{{{e^{ - 4t}}}}{{\left( { - 4} \right)}}} \right|_o^T = \begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{2T}}.\frac{1}{4}\left[ {1 - {e^{ - 4T}}} \right]\\ \Rightarrow P = \begin{array}{*{20}{c}} {lim}\\ {T \to \infty } \end{array}\frac{1}{{8T}}\left[ {1 - {e^{ - 4T}}} \right] = 0 \end{array}}\)