What is the difference when we use $e^{-j2\pi f}$ and $e^{-j\omega n}$ for Fourier transformation?
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1$\begingroup$ $\omega = 2\pi f$ $\endgroup$– endolithCommented Nov 10, 2016 at 17:44
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$\begingroup$ @endolith Thanks :) I know that, but I don't know how the meaning of Fourier transform and Fourier transform table pairs change if we use $f$ or $\omega$, Can you help me with that? $\endgroup$– user137927Commented Nov 10, 2016 at 17:46
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1$\begingroup$ It's exactly the same thing, just expressed in different units. You should really be comparing $e^{-j\omega}$ with $e^{-j\omega n}$, for instance. $\endgroup$– endolithCommented Nov 10, 2016 at 19:00
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$\begingroup$ An interesting side note: white noise has (two-sided) power spectral density $\frac{N_0}{2}$ regardless of whether you are denoting it $S_N(\omega)$ or as $S_n(f)$ !! In either case, the noise power in a bandwidth of $B$ Hz works out to be $N_0B$ watts or volts$^2$. $\endgroup$– Dilip SarwateCommented Nov 10, 2016 at 20:57
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$\begingroup$ Thank @endolith, What about normalization coefficient? As I know it is different in use of $f$ and $\omega$? Is there any pattern for this difference? $\endgroup$– user137927Commented Nov 11, 2016 at 6:30
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1 Answer
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You might be mixing two concepts, pertaining to (following Robert Bristow-Johnson)
- the "analog context" (more formally, the "continuous-time case"),
- the "digital context" (more formally the "discrete-time case").
The first concept corresponds to the continuous Fourier transform, for which you can use a form of normalized frequency cycles per second or Hertz ($e^{-j2\pi f}$), while the second is the angular frequency in radians per second ($e^{-j\omega}$).
The second one, with integer index $n$ relates first instance to discrete summations, e.g. the discrete-time Fourier transform (DTFT), where the Fourier kernel writes $e^{-j2\pi f n}$ with $f\in \mathbb{R}$ (for instance) or $e^{-j\omega n}$ with $\omega\in \mathbb{R}$ (for instance).
The first concept is in use in the inverse DTFT, since it involves integrals.
answered Nov 10, 2016 at 17:03
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2$\begingroup$ i would say, in the "analog case" (more formally, the "continuous-time case") it's $$ \Omega = 2\pi f$$ and in the "digital case" (more formally the "discrete-time case") it's $$ \omega = 2\pi \frac{f}{f_\text{s}} $$ in my book it's $$ s=j \Omega $$ and $$ z = e^{j \omega} $$ $\endgroup$ Commented Nov 10, 2016 at 23:57
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$\begingroup$ "in my book" is meant as a figure of speech. $\endgroup$ Commented Nov 11, 2016 at 0:02