What Type of Signal is x(t) in Parseval's Theorem?

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SUMMARY

In the context of Parseval's Theorem, the signal \( x(t) \) can be either a voltage signal or a power signal, as long as it is a complex-valued function for which the continuous Fourier transform \( X(f) \) exists. The theorem is typically interpreted in signal analysis as a means to compute the total energy of a signal, which is relevant for voltage or current signals. The discussion highlights a common misconception regarding the interpretation of power signals in relation to energy, emphasizing that the units of power and energy differ, which complicates the application of Parseval's Theorem in that context.

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Homework Statement


In parseval's theorem, what is type of x(t)?? I mean.. is this voltage wave? or power wave?

Homework Equations


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The Attempt at a Solution

 

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Mathematically, ##x(t)## can be any function ##x:\mathbb{R}\rightarrow\mathbb{C}## that is any complex valued function over the real line (or as a sub case any real valued function over the real line ##x:\mathbb{R}\rightarrow\mathbb{R}##) for which its continuous Fourier transform ##X(f)## exists . So it can be a function representing the (complex) voltage between two nodes of a circuit or the power between two nodes of a circuit.

However the usual interpretation of this theorem in signal analysis is that the two sides of the equation are just two different ways of computing the total energy of a signal ##x(t)## (a voltage or a current signal).
 
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you mentioned it doesn't matter that signal x(t) represents voltage or power. But I can't understand if x(t) represent power,
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this equation doesn't make sense as I think. In my opinion, How can power^2 be just power?? Unit of dimension is not same as I know.
 

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Yes I said that ##x(t)## can be any function for which the Fourier transform exists. However I also said that the usual interpretation is that of energy (the integrals in both sides represent energy) in the case ##x(t)## is a voltage or current signal. If ##x(t)## is a power signal then we can't give that usual interpretation of energy to this theorem.
 
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