Laying out a mesh

In theoretical work and in programs, the 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 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:

• $\Delta t \eq t_{\rm range} / N \quad$ is time resolution.
• $f_{\rm range} \eq 1/\Delta t \eq N /t_{\rm range} \quad$ is frequency range.
• $\Delta f \eq f_{\rm range}/N \eq 1/t_{\rm range} \quad$ is frequency resolution.

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:

$$\begin{eqnarray} \Delta t \ f_{\rm range} &=& 1 \\ \Delta f \ t_{\rm range} &=& 1 \\ \Delta f \ \Delta t &=& {1 \over N}\end{eqnarray}$$ (7) (8) (9)

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.