8.10 Effect of finite  8. Spectral Analysis  8.8.5 The autoregressive process

8.9 Aliasing

If a continuous function f(t) is sampled at N finite intervals $\Delta T$, the observations $\{f_k\}$ contain no information on the behavior of f(t) at intermediate points, so it is necessary to consider what intermediate behavior is consistent with the data. When we interpret the data using a Fourier representation that matches the points exactly, as in the preceding parts of this chapter, we omit other possible contributions that might have higher frequency components or components intermediate to those included in the series. In effect, we assume that those components are zero by omitting them from the Fourier representation.

Consider the extreme case where the function g(t) assumes the observed values at each sampled point but is zero everywhere else. This is a possible time series that would give the observed values, and so must be consistent with the representation in Fourier components that is obtained from those observed values. The observed values can be written as the product of any time series having the observed values at the discrete observation times and a sum of delta functions that select the observed values:

$$g(t) = f(t)I(t) \ \ {\rm where}\ I(t)=\sum_m \delta(t-m\Delta t) \ .$$ (8.51)

 

The Fourier transform of a product of two functions is given by the "convolution integral" as

$$\tilde{g}(\nu) = \int_{-\infty}^\infty\tilde f(\nu-\nu^\prime)\tilde I(\nu^\prime)d\nu^\prime$$ (8.52)

 

where

$$\tilde I(\nu) = \int I(t)e^{i2\pi\nu t}dt = {{1}\over{\Delta T}} \sum_m\delta(\nu-{{m}\over{\Delta T}}) \ .$$ (8.53)

 

The Fourier transform of the observed time sequence is

$$\tilde g(\nu) = {{1}\over{\Delta T}}\sum_n\tilde f\left(\nu-{{n}\over{\Delta T}}\right)$$ (8.54)

 

so the observed Fourier transform includes the components

$$\tilde g(\nu) = \tilde f(\nu)+\tilde f(\nu+\nu_0)+\tilde f(\nu-\nu_0)+\tilde f(\nu+2\nu_0) + \cdots$$ (8.55)

 

where $\nu_0=N/T=1/\Delta T$ is the sampling frequency.

The Fourier transform determined from the discrete series thus will include contributions not only from the analyzed frequency $\nu$ but also from all other frequencies that differ from that frequency by an integer multiple of the sampling frequency. This mixing of contributions from different frequency components is called "aliasing." There is no way to separate the various components that contribute to $\tilde g(\nu)$ in (8.55), once they are mixed during sampling.

Figure 8.5 illustrates how this problem arises. Consider discrete samples from the continuous series as shown. There is no way to determine from the sampled series whether the underlying function is the solid or the dashed line, and if both components are present then both will contribute to the analyzed Fourier transform at either frequency.

 

Figure 8.5: Illustration of aliasing. At the times of sampling, indicated by vertical lines, the two signals shown both agree with the observations. Either of these signals could produce the indicated observations, shown by circled points, so the two signals are said to be "aliased."

For this reason, we can only analyze Fourier coefficients over a unique range of frequencies varying by $1/\Delta T$, because beyond this range the analysis will simply repeat. We can choose this frequency range to be $\pm 1/(2\Delta T)$, but then frequencies outside this range will still affect the measured frequency range as specified by (8.55). The maximum frequency $1/(2\Delta T)$is called the Nyquist frequency, $\nu_N$, and is half the sampling frequency. This limit can be thought of as arising from the need to sample at least fast enough to sample first the maximum and then the minimum in the sine wave corresponding to the maximum resolved frequency.
 

Exercise 8.2: Show that the frequencies that are aliased are separated by intervals of N/T where N is the number of samples and Tis the duration of the time series. Show that, when the spectrum is normalized over the frequency interval from zero to the Nyquist frequency, frequencies $\delta f$ above the Nyquist frequency are aliased with those $\delta f$ below the Nyquist frequency.

The effects of aliasing are illustrated well in the variance spectrum calculated from hourly observations of pressure (Fig. 8.3). The curve decreases steadily from about 0.01 to 0.1 h^-1, but for higher frequencies there is a departure from this trend. This increase in the spectrum near the Nyquist frequency is most likely caused by aliasing from above the Nyquist frequency.

To reduce aliasing, low-pass filters can reduce the high-frequency components of a signal before sampling. If this is not done, higher-frequency contributions from the variance spectrum will appear as false contributions to lower parts of the estimated spectrum, and they cannot be eliminated after sampling because information distinguishing the separate contributions from frequencies above and below the Nyquist frequency is lost. If the variance spectrum has a slope of -5/3, as is common, the values at the Nyquist frequency $\nu_N$will be approximately doubled by aliasing unless high-frequency components are suppressed by filtering before sampling. However, the effect at 0.1$\nu_N$ is an increase of less than 1% in that case, so the effects of aliasing on a spectrum having a -5/3 slope is mostly confined to the highest decade in frequency.
 

Exercise 8.3: Show that, for a variance spectrum that decreases with frequency as the -5/3 power of the frequency, the effect of aliasing is to approximately double the estimated variance at the Nyquist frequency unless frequencies above the Nyquist frequency are reduced by filtering before sampling.

   
 8.10 Effect of finite  8. Spectral Analysis  8.8.5 The autoregressive process