On 5/13/2009 3:04:38 PM, Jbryant61 wrote:
>... I have a range of targets that
>have different frequencies
>(from 0.5um lines upto 5um
>lines) and wanted to plot the
>moudlation (Imax - Imin) /
>(Imax + Imin - 2*background))
>as a function of line pairs.
>So the image I included would
>just represent one point.
Yes, but technically the contrast you measure with bar patterns is not the same as the MTF. The MTF expresses the contrast you would see if your object had a sinusoidal variation of intensity - smoothly varying from dark to light in a sinusoidal way.
Bar patterns (step functions of intensity) are far easier to fabricate, however, so they are a common imaging tool. Strictly speaking, the contrast value you measure with bar patterns is not the value of the MTF.
This is easily explained. Because the peaks are brighter, on average, and the valleys darker, it is easier to see bar patterns at the edge of a system's response than sine patterns. Therefore, you'd expect to measure a higher contrast for bar patterns than you would for sine patterns at the same fundamental frequency.
How much higher? If the image of your bar pattern is degraded enough that it really looks like a sine wave[*], then the contrast you measure will be higher than the true MTF by a factor of 4/pi.
How do we arrive at that factor? The intensity of a sine pattern varies via 0.5*(1+sin(x)), and its normalized MTF at the fundamental frequency has an amplitude of 0.5. Blurry imaging of a square wave, however, reduces the image to just the fundamental Fourier component of a square wave, which has a normalized value of 2/pi. (See the Fourier peaks of the two waveforms in my last sheet in this thread.) So, contrast measurements of a blurry square wave are bigger than those of a sine wave by a factor of (2/pi) / (1/2) = 4/pi.
- Guy
[*] If the image is _not_ washed out, however, and still looks like a step function, then the contrast you measure should be much closer to the true value of the MTF ... of course it also means you're measuring a value close to 1 anyway.