How can I prove the equation $(3)$? I can't understand why there is a $2/N$ in $(3)$. Why he just get the second term in DFT?

Consider a sinusoidal input signal of frequency $\omega$ given by

$$ x(t) = \sqrt 2 X \sin\left(\omega t + \phi\right)\tag{1} $$

This signal is conventionally represented by a phasor (a complex number) $\bar X$

$$ \bar X = Xe^{j\phi}=X\cos\phi+jX\sin\phi\tag{2} $$

Assuming that $x(t)$ is sampled $N$ times per cycle of the $60\textrm{ Hz}$ waveform to produce the sampled set $\left\{x_k\right\}$

$$ x_k = \sqrt 2 X\sin\left(\frac{2\pi}{N}k + \phi\right)\tag{3} $$

The Discrete Fourier Transform of $\left\{x_k\right\}$ contains a fundamental frequency component given by

$$ \color{red}{\boxed{\color{black}{\bar X_1= \frac 2N \sum_{k=0}^N x_k e^{-j\frac{2\pi}{N}k}}}}\tag{4}\\ $$

  • $\begingroup$ where did this text come from?? 1980 or before? IBM Selectric Typewriters with exchangeable balls? $\endgroup$ – robert bristow-johnson May 24 '16 at 5:13
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    $\begingroup$ and, in my Electrical Engineering training and experience, Eqs (1) and (2) are incompatible. if $$ x(t) = \sqrt{2} X \cos(\omega t + \phi) $$ (note that it's "$\cos()$", not "$\sin()$") then the phasor is $$ \bar{X} = X \ e^{j \phi} = X \cos(\phi) + j X \sin(\phi) $$ and $X \ge 0$ is the r.m.s. magnitude of the sinusoid. $\endgroup$ – robert bristow-johnson May 24 '16 at 5:20
  • $\begingroup$ to answer your question (once the premises are corrected), you first express what $\omega$ is (given that it's a "60 Hz waveform"). then compute what the sampling instances are (values of $t$) given that one cycle of your 60 Hz waveform is sampled with $N$ samples. since this is one cycle going into the DFT, the complex amplitude of the fundamental is in $X_1$ and $X_{N-1}$ of the DFT output (using the crappy subscript notation from pre-1980). and the "2" does not belong. $\endgroup$ – robert bristow-johnson May 24 '16 at 5:26
  • $\begingroup$ In that time there wasnt a clear phasor representation. $\endgroup$ – Felipe May 24 '16 at 14:03
  • $\begingroup$ I think there may be a way to get this 2 on equation (4), there is a lot of material using this. Like this one: nptel.ac.in/courses/108101039/download/Lecture-35.pdf $\endgroup$ – Felipe May 24 '16 at 14:04

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