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Laying out a mesh

In theoretical work and in programs, the unit delay operator definition $Z=e^{i\omega \Delta t}$ is often simplified to $\Delta t = 1$, leaving us with $Z=e^{i\omega}$. How do we know whether $\omega$ is given in radians per second or radians per sample? We may not invoke a cosine or an exponential unless the argument has no physical dimensions. So where we see $\omega$ without $\Delta t$, we know it is in units of radians per sample.

In practical work, frequency is typically given in cycles/sec or Hertz, $f$, rather than radians, $\omega$ (where $\omega = 2\pi f$). Here we will now switch to $f$. We will design a computer mesh on a physical object (such as a waveform or a function of space). We often take the mesh to begin at $t=0$, and continue till the end $t_{\rm max}$ of the object, so the time range $t_{\rm range} = t_{\rm max}$. Then we decide how many points we want to use. This will be the $N$ used in the discrete Fourier-transform program. Dividing the range by the number gives a mesh interval $\Delta t$.

Now let us see what this choice implies in the frequency domain. We customarily take the maximum frequency to be the Nyquist, either $f_{\rm max} = .5 /\Delta t$ Hz or $\omega_{\rm max} = \pi /\Delta t$ radians/sec. The frequency range $f_{\rm range}$ goes from $-.5/\Delta t$ to $.5/\Delta t$. In summary:

In principle, we can always increase $N$ to refine the calculation. Notice that increasing $N$ sharpens the time resolution (makes $\Delta t$ smaller) but does not sharpen the frequency resolution $\Delta f$, which remains fixed. Increasing $N$ increases the frequency range, but not the frequency resolution.

What if we want to increase the frequency resolution? Then we need to choose $t_{\rm range}$ larger than required to cover our object of interest. Thus we either record data over a larger range, or we assert that such measurements would be zero. Three equations summarize the facts:

$\displaystyle \Delta t \ f_{\rm range}$ $\textstyle =$ $\displaystyle 1$ (12)
$\displaystyle \Delta f \ t_{\rm range}$ $\textstyle =$ $\displaystyle 1$ (13)
$\displaystyle \Delta f \ \Delta t$ $\textstyle =$ $\displaystyle {1 \over N}$ (14)

Increasing range in the time domain increases resolution in the frequency domain and vice versa. Increasing resolution in one domain does not increase resolution in the other.


next up previous [pdf]

Next: INVERTIBLE SLOW FT PROGRAM Up: FOURIER TRANSFORM Previous: The Nyquist frequency

2013-01-06