Downsampling Theorem (Aliasing Theorem)

Theorem: For all $x\in{\bf C}^N$,

$$\zbox {\hbox{\sc Downsample}_L(x) \;\longleftrightarrow\;\frac{1}{L}\hbox{\sc Alias}_L(X).}$$

Proof: Let $k^\prime \in[0,M-1]$ denote the frequency index in the aliased spectrum, and let $Y(k^\prime )\isdef \hbox{\sc Alias}_{L,k^\prime }(X)$. Then $Y$ is length $M=N/L$, where $L$ is the downsampling factor. We have

$$\begin{eqnarray*} Y(k^\prime ) &\isdef & \hbox{\sc Alias}_{L,k^\prime }(X) \isd... ...n) e^{-j2\pi k^\prime n/N} \sum_{l=0}^{L-1}e^{-j2\pi l n M/N}. \end{eqnarray*}$$

Since $M/N=1/L$, the sum over $l$ becomes

$$\sum_{l=0}^{L-1}\left[e^{-j2\pi n/L}\right]^l = \frac{1-e^{-j2\p... ...right) \\ [5pt] 0, & n\neq 0 \left(\mbox{mod}\;L\right) \\ \end{array}\right.$$

using the closed form expression for a geometric series derived in §6.1. We see that the sum over $L$ effectively samples $x$ every $L$ samples. This can be expressed in the previous formula by defining $m\isdeftext n/L$ which ranges only over the nonzero samples:

$$\begin{eqnarray*} \hbox{\sc Alias}_{L,k^\prime }(X) &=& \sum_{n=0}^{N-1}x(n) e^{... ... & L\cdot \hbox{\sc DFT}_{k^\prime }(\hbox{\sc Downsample}_L(x)) \end{eqnarray*}$$

Since the above derivation also works in reverse, the theorem is proved.

An illustration of aliasing in the frequency domain is shown in Fig.7.12.