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Proof of Aliasing Theorem

To show:

$\displaystyle \zbox {\hbox{\sc Downsample}_N(x) \;\longleftrightarrow\;\frac{1}{N} \hbox{\sc Alias}_N(X)}
$

or
\fbox{$x(nN) \;\longleftrightarrow\;\frac{1}{N} \displaystyle\sum_{m=0}^{N-1} X\left(e^{j2\pi m/N} z^{1/N}\right)$}

From the DFT case [248], we know this is true when $ x$ and $ X$ are each complex sequences of length $ N_s$, in which case $ y$ and $ Y$ are length $ N_s/N$. Thus,

$\displaystyle x(nN) \;\longleftrightarrow\;
Y(\omega_k N) = \frac{1}{N} \sum_{m=0}^{N-1} X\left(\omega_k + \frac{2\pi}{N} m \right), \; k\in [0,N_s/N)
$

where we have chosen to keep frequency samples $ \omega_k$ in terms of the original frequency axis prior to downsampling, i.e., $ \omega_k =
2\pi k/ N_s$ for both $ X$ and $ Y$. This choice allows us to easily take the limit as $ N_s\to\infty$ by simply replacing $ \omega_k$ by $ \omega$:

$\displaystyle x(nN) \;\longleftrightarrow\;
Y(\omega N) = \frac{1}{N} \sum_{m=0}^{N-1} X\left(\omega + \frac{2\pi}{N} m \right), \; \omega\in[0,2\pi/N)
$

Replacing $ \omega$ by $ \omega^\prime =\omega N$ and converting to $ z$-transform notation $ X(z)$ instead of Fourier transform notation $ X(\omega)$, with $ z=e^{j\omega^\prime }$, yields the final result.


Previous: Downsampling and Aliasing
Next: Differentiation Theorem Dual

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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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