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MIP Maps


Filtering is easier to visualize in one dimension. Sound recordings are examples of one-dimensional signals.

Below is a signal that is a sum of low and high frequency waves.

Signal formula

Plot of signal

Fig 1. Signal that is a sum of strong low-frequency (2 Hz) and weaker high frequency (50 Hz) sine waves. f1 = 2, f2 = 50, A = 1, B = 0.25

In frequency space, this signal has two spikes at frequencies 2 and 50. Mathematically speaking, decomposing a signal into its frequency components is done by means of a Fourier transform. A Fourier transform of a signal is also called its spectrum.

Frequency distribution of signal

Fig 2. Fourier transform of the signal from Fig 1.

Filtering is done by averaging out multiple samples to get each new sample. The simplest averaging assigns the same weight to each sample within a small range. This corresponds to a box filter, a filter whose all components up to a fixed count have equal weight.

Box filter

Box filter Boxk = 1/5, k = 0..4. It averages out five samples at a time, giving each sample equal weight, 1/5.

This is the mathematical formula that describes the filtering process.

Convolution of box filter with the signal

The j'th sample of the result, yj, is a sum of five consecutive samples of the source signal, starting with xj, with weights given by the box filter. This formula is called a convolution of two sequences (Box and x). Filtering a signal means convoluting it with the filter.

The result of filtering is a new signal that has the low-frequency component intact, but with the high-frequency component gone. In every sum of five consecutive samples, the high-frequency component contributed an (almost) equal number of positive and negative terms. They averaged out to zero.

Filtered signal

Fig 3. The high frequency part of the filtered signal is averaged out.

Low-pass Filter

It so happens that a convolution of two sequences corresponds to multiplication of their Fourier transforms. So instead of looking at the input signal and the filter, and convoluting them, we can look at their Fourier transforms and multiply them.

The goal of low pass filtering is getting rid of high frequencies (or fine detail, in the case of pictures). If frequency space, this would correspond to multiplying the spectrum of the signal by a function that is one for lower frequencies and zero for higher frequencies. We want the Fourier transform of our filter to look like a step function.

Lowpass filter

Given its Fourier transform, we would like to see what the filter looks like, so that we can do the averaging. We have to apply the inverse Fourier transform (which is virtually the same as straight FT) to our step function. The result is a sinc function:

sinck = sin (k)/k


Fig 4. In blue, the ideal low-pass filter whose FT is a step function. For comparison, the purple plot is the box filter.

Unlike a box filter, the sinc has both positive and negative values. They make sense when averaging a sound signal with positive and negative values. They don't make sense when filtering a bitmap, which has only positive values of RGB (we might get, for instance, a negative value for the red component of some pixel!). In addition, the sync function extends to infinity (although its value quickly decreases like 1/x). It means that for each sample of the output we theoretically have to average out an infinite number of input samples.

In practice, one either has to window a sinc (multiply it by a step function), or use other, simpler filters, like the box filter.

Other popular types of filters are triangular and Gaussian (bell-shaped). The interesting thing about a Gaussian filter (exp (-ax2)is that its FT is also a Gaussian.

Texture Filtering

Imagine that a pixel is projected back to texture space. The four corners of the pixel will form a quadrilateral in texture space (it will usually be distorted due to perspective). The whole area covered by the back-projected pixel should somehow contribute to the color of the pixel. The brute-force thing would be to average out the colors of the texels in this area. This is not easy, so simpler prescriptions were invented.
  • Find a rectangle enclosing this area in texture space using min/max calculations. Average out the texels inside this bounding rectangle. (This algorithm may be further improved by pre-calculating summed area tables.)

    Pixel projected on texture

    Fig 5. A square pixel is projected back on the texture. Bounding rectangle is shown in yellow.

  • Find a square whose area is the same as the area of the quadrilateral and average out texels in this square. This is equivalent to using a 2-dimensional box filter. (This algorithm may be further improved by using MIP maps, see below).
  • Assume that a pixel is a circle. A circle projected back to texture space will always produce an ellipse (may be slanted). Use an elliptically shaped filter (Gaussian or windowed sinc). This is obviously the most expensive algorithm, but it gives most realistic results.

In addition, one should also take into account the fact that not the whole pixel might be covered by a polygon. So to be exact, one should clip the polygon to the pixel and back-project such a fragment to texture space. Such calculations might in fact be done when using the A-buffer for anti-aliasing.

Notice that all these algorithms perform more calculations the more distant the polygon is and the smaller its image on the screen. This doesn't seem like a good allocation of computing resources.

Summed Area Tables

There is a technique that can dramatically speed up the averaging over rectangular areas. Instead of storing the texture as a bitmap, one can create a summed area table. The elements of such a table are sums of the RGB components of texels over whole areas. An element indexed by (u, v) contains the sum of all texels located to the left and above the texel (u, v)--the upper-left rectangle. To calculate the sum of texels within a rectangle defined by (u1, v1, u2, v2), we add the elements

Sum [u2][v2] - Sum [u1][v2] - Sum [u2][v1] + Sum [u1][v1]

In Fig 5. we can see this construction geometrically. The small grey rectangle is obtained by subtracting the two red rectangles from the large blue rectangle and then adding back the smaller blue rectangle (to compensate for double coverage under two red rectangles).

Summed area table

Fig 5. Summed area table.

NextNext: Using mip-maps to speed up texture mapping.