0
$\begingroup$

$$X(e^{j\omega}) = \sum_{n=-\infty}^\infty x[n] e^{-j\omega n} $$

The frequency term $\omega$ in DTFT is normalized as $\omega = \frac{\Omega}{f_\mathrm{s}}$

$\Omega= 2 \pi f$ is the angular frequency continuous-time Fourier Transform of $x(t)$.

A continuous-time signal that can be properly sampled will have non-zero energy components in its spectrum having frequency $|f| \lt \frac{f_\mathrm{s}}{2}$

Hence the range of $|\omega|$ should lie between $0$ through $\pi$. Given that the spectrum of a discrete time signal repeats every $2\pi$ radians, we should see this spectrum replicated from 0 to $\pi$, $2\pi$ to $3\pi$, $4\pi$ to $5\pi$ and so on.

I can extrapolate this to negative factors of $2\pi$ as well. So the spectra can additionally replicate from $-2\pi$ to $=3\pi$, $-4\pi$ to $-5\pi$ and so on.

I have however read that the range of $\omega$ is from $-\pi$ to $+\pi$. Where does this come from?

Improve this question
$\endgroup$
1
  • $\begingroup$ Brilliant editing by Robert Bristow-Johnson, especially the third paragraph which puts to rest my dilemma. Non-zero energy components are on both sides of fs/2. $\endgroup$ – Raj Apr 4 '19 at 6:17
2
$\begingroup$

The DTFT is independent of Nyquist sampling. The DTFT is applied to a sequence of numbers regardless of any analog waveform that may be sampled. Economic sequences are often inherently discrete time sequences. The DTFT applies to aliased sequences as well.

The summation is in general over $n$ from $-\infty$ to $\infty$. The summation limits can be truncated to the nonzero range of $x[n]$.

The nominal convention as presented in Oppenheim and Schaefer, which they introduce prior to their chapter on Nyquist sampling, is that the DTFT is a continuous valued, periodic function defined uniquely on $-\pi$ to $\pi$ which has a length of $2\pi$.

For real valued sequences, the interval $-\pi$ to $0$ is conjugate mirrored of the spectrum over $0$ to $\pi$.

In general for complex valued sequences, the mirrored conjugate symmetry can’t be assumed.

$0$ to $\pi$ is only one half of a length $2\pi$ period.

edit

$-\pi$ to 0 has an image at $\pi$ to $2\pi$ and $3\pi$ to $4\pi$ and ......

0 to $\pi$ has an image from $2\pi$ to $3\pi$ and $4\pi$ to $5\pi$ and ....

similarly for the negative frequencies.

Improve this answer
$\endgroup$
9
  • $\begingroup$ May we then conclude that although the bandwidth is defined as 2.pi (Either of "-pi to +pi" or "0 to 2.pi", useful information is restricted to "0 to pi" ? $\endgroup$ – Raj Apr 3 '19 at 17:48
  • $\begingroup$ no, complex valued sequences do not have symmetry $\endgroup$ – user28715 Apr 3 '19 at 17:50
  • $\begingroup$ even for reals, the segment-pi to 0 is identical to pi to 2pi, 0 to pi, is identical to 2pi to 3pi. it is a conjugate mirror reflected around 0. $\endgroup$ – user28715 Apr 3 '19 at 18:08
  • $\begingroup$ Stan, it doesn't matter whether $x(t)$ is complex or not. $X(e^{j\omega})$ will always be periodic with period $2\pi$. so i think your "no" answer is incorrect. $\endgroup$ – robert bristow-johnson Apr 3 '19 at 20:00
  • $\begingroup$ @robertbristow-johnson , the useful information is not limited to the interval 0 to pi. i stand by no. $\endgroup$ – user28715 Apr 3 '19 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.