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MTF

Participant

MTF

Any suggestions on how to get the average of the max and min values in the modulation image.

Thanks
Jason
125 REPLIES 125

MTF

MTF

Hi Val. Thanks.

This is great if I know the minimum values i.e. zero in your case, but thats the problem, I need to average the troughs and the peaks.

BTW, the only if syntax I know is different to yours.

Thanks
Jason

MTF

Here's a rather simple way to do it ... ie it won't work with noisy, more-general data.

- Guy

MTF

If what you want is the amplitude of the MTF there is a better approach than trying to find the averages of all the maxima and minima.

Richard

MTF

I should add that if you change the size of the bars a simple sinusoid may not adequately represent them. You can still get parameter estimates of the fundamental frequency using the FFT, but you'll have to change the fitting function for the second step.

Richard

MTF

On 5/13/2009 9:57:50 AM, rijackson wrote:
>If what you want is the
>amplitude of the MTF there is
>a better approach
>
>Richard

The way you did it isn't quite right - the MTF must be normalized to the zero-frequency response, so you cannot subtract off the mean value of the image to get an MTF.

Furthermore, the properly-normalized MTF of the image does not correspond to the MTF of the lens system. If your object has no feature at a spatial frequency f, then neither will your image. So your ability to measure MTF at a frequency f is affected by your object's spatial frequecny spectrum.

To obtain the MTF of your optics alone means dividing the spatial frequency spectrum of your image by the spatial frequency spectrum of your object - and the frequency spectrum at f of a sinusoidal object at f will be different from the f-component of a bar pattern at f.

- Guy

MTF

Richard & Guy, thankyou both very much for your help. I don't follow Guy's comments about object MTF.

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.

Thanks
Jason

MTF

Sorry - just read this after replying to a later message.

You ask about background subtraction for the contrast. The image enclosed is a raw image. I thought I could call the modulation as (Imax-Imin)/(Imax + Imin - 2*bg) where bg is a value of the image away from the bars.

Its great what you have done - I loove the plots of the points on the peaks and trough!

I think it was you who suggested using edges (or line pairs) as a means to measure image quality rather than beads.

Thankyou.
Jason

MTF

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.
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