Using the Continuous-time Fourier series, Parseval’s Theorem for Periodic signal describes that [...]: \(\displaystyle \frac{1}{T_{0}}\int^{T_{0}/2}_{-T_{0}/2}|x(t)|^{2}dt=\sum\limits^{\infty}_{k=-\infty}|C_{k}|^{2}\)
Answer
the average power of a periodic signal is the sum of the average powers of its Fourier components
Question
Using the Continuous-time Fourier series, Parseval’s Theorem for Periodic signal describes that [...]: \(\displaystyle \frac{1}{T_{0}}\int^{T_{0}/2}_{-T_{0}/2}|x(t)|^{2}dt=\sum\limits^{\infty}_{k=-\infty}|C_{k}|^{2}\)
Answer
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Question
Using the Continuous-time Fourier series, Parseval’s Theorem for Periodic signal describes that [...]: \(\displaystyle \frac{1}{T_{0}}\int^{T_{0}/2}_{-T_{0}/2}|x(t)|^{2}dt=\sum\limits^{\infty}_{k=-\infty}|C_{k}|^{2}\)
Answer
the average power of a periodic signal is the sum of the average powers of its Fourier components
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Parseval’s Theorem for Periodic signal Using the Continuous-time Fourier series, Parseval’s Theorem for Periodic signal describes that the average power of a periodic signal is the sum of the average powers of its Fourier components: \(\displaystyle \frac{1}{T_{0}}\int^{T_{0}/2}_{-T_{0}/2}|x(t)|^{2}dt=\sum\limits^{\infty}_{k=-\infty}|C_{k}|^{2}\)
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