Review of concepts from sampling theory - Signal Processing and Optimization for Transceiver Systems - page 807

because the sum is over all k. This proves Eq. (G.10) for any ω 0 .

Reflections of alias-free( T ) regions

Let A 1 and A 2 be two alias-free(2 π/ω s ) regions. Given any frequency ω 1 in A 1 ,

there is precisely one integer k such that

= ω 1 + kω s

ω 2

is in A 2 . If we have a signal X A 1 ( jω ) which is bandlimited to A 1 we can therefore

define a unique signal X A 2 ( jω ) bandlimited to A 2 such that

X A 2 ( jω 2 )= X A 1 ( jω 1 ) .

We say that X A 2 ( jω )isthe reflection of X A 1 ( jω )onto A 2 . Similarly, X A 1 ( jω )

is the reflection of X A 2 ( jω )onto A 1 .

Lemma G.2. With X A 1 ( jω )and X A 2 ( jω ) defined as above, the sampled
versions of X A 1 ( jω )and X A 2 ( jω ) (with sampling rate ω s ) are identical, that is,

♠

x A 1 ( nT )= x A 2 ( nT ) ,

for all n , or equivalently

X A 1 ( j ( ω + kω s )) =
k

X A 2 ( j ( ω + kω s )) , (G . 11)

k

for all ω. ♦

Proof. Given any frequency ω 0 there is at most one nonzero term on the
left-hand side of Eq. (G.11). Thus

X A 1 ( j ( ω 0 + kω s )) = X A 1 ( j ( ω 0 + k 1 ω s )) (for some k 1 )

k

= X A 2 ( j ( ω 0 + k 2 ω s )) (for some k 2 )

=
k

X A 2 ( j ( ω 0 + kω s ))

because the right-hand side is zero for k = k 2 anyway.