Interpretation of a sampled signal in the frequency domain
Suppose we have a signal $y(t)$. For sampling we use a Dirac $\delta$ function. The sampled function is $\widetilde{y}(t)$. So $$\widetilde{y}(t) = y(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT).$$ The Fourier transformation is $$\widetilde{Y}(\omega)=\mathcal{F\{\widetilde{y}(t)\}} = \sum_{n=-\infty}^{\infty}y(nT)e^{-i2\pi\omega nT}$$ where $i$ is the imaginary unit and $\omega$ angular frequency. $y(nT)$ are the sampled points but I am having problems in understanding what $e^{-i2\pi\omega nT}$ of $\widetilde{Y}(\omega)$ means? I could write it down with $\cos$ and $\sin$ but this does not make it clearer. I would appreciate it if someone could explain it to me.
frequency-spectrum sampling frequency-domain
Laurent Duval
tumblertumbler
$\begingroup$ It's a complex sinusoid. I guess you have a textbook that introduces the Fourier transform – read that; this term is so fundamental that I'm almost 100% certain it's covered in there. $\endgroup$ – Marcus Müller Feb 11 '17 at 12:35
First, $i$ is not the imaginary part, but the imaginary unit. This tells you that the signal, under suitable conditions, admits a discrete time Fourier transform. Its is a linear representation of the discrete signal. It provides you with a spectral interpretation, ie a view of the content of the signal as a linear combination of waves, pure frequencies indexed by a continuous index $\omega$. Pure frequencies are cisoids or complex exponentials/sinusoids, by definition
$$e^{-i2\pi\omega nT} = \cos(-2\pi\omega nT)+i\sin(-2\pi\omega nT)\,.$$
The relative importance (weight) and "timing" of these waves is given by $\tilde{Y}(\omega)$, by its magnitude and phase, respectively. Another interpretation of the sum of products is a scalar product between the signal and the different waves. If the scalar product is high, the signal looks like the pure wave a lot, and the corresponding frequency is very "present" in the signal. If is is low, this component is not really present in the signal. The whole spectrum, made of all the $\tilde{Y}(\omega)$, tells you about the content in "each" frequency in your data.
The two links above drive you to the excellent site of J. O. Smith, and the mathematics of the discrete Fourier transform, an perfect starting point to understand these concepts.
Laurent DuvalLaurent Duval
The given expression for the Fourier transform of the sampled signal $y(t)$ is useful in the sense that it shows that the spectrum of the sampled signal equals the discrete-time Fourier transform (DTFT) of the discrete-time signal
$$y_d[n]=y(nT)$$
$$Y_d(\omega)=\sum_{n=-\infty}^{\infty}y_d[n]e^{-jn\omega}=\sum_{n=-\infty}^{\infty}y(nT)e^{-j\Omega nT}=\widetilde{Y}(\Omega)\tag{1}$$
where $\omega$ is defined as $\omega=\Omega T$, with the angular frequency $\Omega$ of the continuous-time signal, and the sampling interval $T$. Equation $(1)$ shows that $Y_d(\omega)=\widetilde{Y}(\Omega)$ is a periodic function with Fourier series coefficients $y(nT)$.
However, the expression $(1)$ does not give much insight into the relation between $\widetilde{Y}(\Omega)$ (or $Y_d(\omega)$) and the spectrum $Y(\Omega)$ of the original continuous-time signal $y(t)$. (And that's the way I interpret your question). This relation becomes explicit when you derive $\widetilde{Y}(\Omega)$ as follows:
$$\begin{align}\widetilde{Y}(\Omega)&=\mathcal{F}\left\{y(t)\cdot\sum_n\delta(t-nT)\right\}\\&=\frac{1}{2\pi}\mathcal{F}\{y(t)\}\star\mathcal{F}\left\{\sum_n\delta(t-nT)\right\}\\&=\frac{1}{2\pi}Y(\Omega)\star\frac{2\pi}{T}\sum_n\delta\left(\Omega-\frac{2\pi n}{T}\right)\\&=\frac{1}{T}\sum_nY\left(\Omega-\frac{2\pi n}{T}\right)\tag{2}\end{align}$$
where $\star$ denotes convolution. Note that $(1)$ and $(2)$ are equivalent, even if that's not immediately obvious. The expression $(2)$ very clearly shows that sampling results in periodization of the spectrum, and this will result in aliasing if the spectrum $Y(\Omega)$ is not properly band-limited.
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