Let
denote any continuous-time signal having a Fourier
Transform (FT)
Let
denote the samples of
![$ x(t)$](http://www.dsprelated.com/josimages_new/mdft/img4.png)
at uniform intervals of
![$ T$](http://www.dsprelated.com/josimages_new/mdft/img73.png)
seconds,
and denote its
Discrete-Time Fourier Transform (
DTFT) by
Then the
spectrum ![$ X_d$](http://www.dsprelated.com/josimages_new/mdft/img1786.png)
of the sampled signal
![$ x_d$](http://www.dsprelated.com/josimages_new/mdft/img1787.png)
is related to the
spectrum ![$ X$](http://www.dsprelated.com/josimages_new/mdft/img55.png)
of the original continuous-time signal
![$ x$](http://www.dsprelated.com/josimages_new/mdft/img25.png)
by
The terms in the above sum for
![$ m\neq 0$](http://www.dsprelated.com/josimages_new/mdft/img1789.png)
are called
aliasing
terms. They are said to
alias into the
base band
![$ [-\pi/T,\pi/T]$](http://www.dsprelated.com/josimages_new/mdft/img1790.png)
. Note that the summation of a
spectrum with
aliasing components involves addition of
complex numbers; therefore,
aliasing components can be removed only if both their
amplitude
and phase are known.
Proof:
Writing
as an inverse FT gives
Writing
![$ x_d(n)$](http://www.dsprelated.com/josimages_new/mdft/img1792.png)
as an inverse DTFT gives
where
![$ \theta \isdef 2\pi \omega_d T$](http://www.dsprelated.com/josimages_new/mdft/img1794.png)
denotes the normalized discrete-time
frequency variable.
The inverse FT can be broken up into a sum of finite integrals, each of length
, as follows:
Let us now sample this representation for
at
to obtain
, and we have
since
and
are integers.
Normalizing frequency as
yields
Since this is formally the inverse DTFT of
![$ X_d(e^{j\theta^\prime })$](http://www.dsprelated.com/josimages_new/mdft/img1801.png)
written in terms of
![$ X(j\theta^\prime /T)$](http://www.dsprelated.com/josimages_new/mdft/img1802.png)
,
the result follows.
Next Section: Changing the BasePrevious Section: Reconstruction from Samples--The Math