ON MTF, Shift Invariance and Aliasing

[JimKasson]

In measurement we fulfil the first requirement by using a good sinusoidal target like a knife edge

?? Step function.

Which can be written as a sum of an infinite number of sinusoids. I think that's what Jack was getting at.

Of course, but if that's how you reckon it, any shape is sinusoidal. He said he was concerned with terminology.


A sinusoidal function?

Fair enough. I withdraw the comment.

--

the last word the last word - Photography meets digital computer technology. Photography wins -- most of the time.

Photography meets digital computer technology. Photography wins -- most of the time.

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[Jack Hogan]

In measurement we fulfil the first requirement by using a good sinusoidal target like a knife edge

?? Step function.

Which can be written as a sum of an infinite number of sinusoids. I think that's what Jack was getting at.

Yes.

Of course, but if that's how you reckon it, any shape is sinusoidal. He said he was concerned with terminology.


A sinusoidal function?

Hard to use that as a spatial resolution target, Thrilla, though one could use dead leaves:


Image of Dead Leaves low contrast target designed to have controlled scale and direction invariant features with a power law Power Spectrum.

https://www.strollswithmydog.com/introduction-to-texture-mtf/

I was instead referring to the fact that with a single edge one gets to measure the relative modulation of sinusoids of all different frequencies at once.

Alternatively one could do it very expensively by printing/etching a very large number of sinusoidal targets, each of a single spatial frequency, and then measure the modulation of each individually.

Happy to improve the semantics though. How else could I have phrased it?

Jack

 

[J A C S]

Basically, yes.

For example, the MTF frequently extends beyond Nyquist, and even beyond the sampling rate.

That cannot be represented after sampling.

I understand what you are saying.

So, if one wants to estimate the MTF of Voyager-1 camera, there is no way to do it because we only can get sampled data from it, right? And there is no way to bring it back to measure its lens separately.

What you say is true only if you care about the MTF beyond the Nyquist frequency. There are plenty of point sources visible where the spacecraft is located. Many samples should provide data that can be used to estimate the MTF up to 0.5 cy/px.

Well, John Vickers and JACS say that there is no MTF in sampled domain. It can only be measured in continuous domain.

I think that may be the difference between a mathematician and an engineer.

You ask a mathematician, “What’s 2 + 2?”
They say, “4.”

You ask a physicist,
They say, “It’s 4, give or take a small measurement error.”

You ask an engineer,
They say, “It’s 4, within tolerances.”

You ask an accountant,
They say, “What do you want it to be?”

You ask a lawyer,
They say, “Let’s negotiate.”

You ask a statistician,
They say, “The median is 4, but I’ll need a larger sample size to be sure.”

BTW, is the problem limited only to spatial sampling? For example, if the continues light intensity is digitized to, say, 12b, does it present a problem too?

This reminds me of another joke, copied and pasted from here:

Epicycle by P. J. Plauger. It was published in Analog, November 1973.

Are all odd numbers prime?

... "No, seriously," I hurried on. "If you were to ask a mathematician to test it he might say: 'Let me see, now. One is prime, three is prime, five is prime, seven is prime. Nine? Nine's not prime. Clearly the theorem is false.'

"But a physicist is more pragmatic. She, I mean he," the slip was calculated, and had the usual effect on a male listener, "might say: 'Let me see, now. One is prime, three is prime, five is prime, seven is prime. Nine? That may be an experimental error—let's go on.' "

Jenkins smiled.

"Eleven is prime, thirteen is prime, fifteen is… Well, that's a lot of data points. The theorem is probably true.'"

He laughed outright.

"But if you ask an engineer to test the theorem, he might say: 'Let me see, now. One is prime, three is prime, five is prime, seven is prime, nine is prime, eleven is…'"

 

[J A C S]

Basically, yes.

For example, the MTF frequently extends beyond Nyquist, and even beyond the sampling rate.

That cannot be represented after sampling.

I understand what you are saying.

So, if one wants to estimate the MTF of Voyager-1 camera, there is no way to do it because we only can get sampled data from it, right? And there is no way to bring it back to measure its lens separately.

What you say is true only if you care about the MTF beyond the Nyquist frequency. There are plenty of point sources visible where the spacecraft is located. Many samples should provide data that can be used to estimate the MTF up to 0.5 cy/px.

Well, John Vickers and JACS say that there is no MTF in sampled domain. It can only be measured in continuous domain.

"Measured" is a loaded term. The slanted method, for example, computes it, it does not measure it directly. Since all the information is in the samples, it can be computed from there.

 

[ThrillaMozilla]

In measurement we fulfil the first requirement by using a good sinusoidal target like a knife edge

?? Step function.

Which can be written as a sum of an infinite number of sinusoids. I think that's what Jack was getting at.

Yes.

Of course, but if that's how you reckon it, any shape is sinusoidal. He said he was concerned with terminology.


A sinusoidal function?

Hard to use that as a spatial resolution target, Thrilla, though one could use dead leaves:


Image of Dead Leaves low contrast target designed to have controlled scale and direction invariant features with a power law Power Spectrum.

https://www.strollswithmydog.com/introduction-to-texture-mtf/

I was instead referring to the fact that with a single edge one gets to measure the relative modulation of sinusoids of all different frequencies at once.

Alternatively one could do it very expensively by printing/etching a very large number of sinusoidal targets, each of a single spatial frequency, and then measure the modulation of each individually.

Happy to improve the semantics though. How else could I have phrased it?

I understand your point, and I'm familiar with the knife edge method. It's funny that you found the mushrooms inadequate compared to a knife edge or a dead leaves target. Perhaps if I found some mushrooms with a smoother surface. (This too is intended to be humorous.)

My point is purely about terminology. Although mushrooms or a knife edge can be represented by a Fourier transform or Fourier series (i.e., a sum of sinusoidal functions), that doesn't mean that they are sinusoidal. A mushroom doesn't look very sinusoidal to me, and neither does a knife edge. Maybe it would be better just to say that a knife edge can be represented by a sum of sinusoidal functions, rather than calling it a sinusoidal target.

 

[JimKasson]

Happy to improve the semantics though. How else could I have phrased it?

Instead of this:

In measurement we fulfil the first requirement by using a good sinusoidal target like a knife edge;

You might have said: In measurement we fulfil the first requirement by using a target that in the ideal case has frequency-domain components at all frequencies, like a knife edge;

Is that clearer?

 

[OpticsEngineer]

"You might have said: In measurement we fulfil the first requirement by using a target that in the ideal case has frequency-domain components at all frequencies, like a knife edge;

Is that clearer?"

The way you phrased it is much more in line with conventional terminology. So clearer just because it is consistent with the broader literature.

 

[Jack Hogan]

In measurement we fulfil the first requirement by using a good sinusoidal target like a knife edge

?? Step function.

Which can be written as a sum of an infinite number of sinusoids. I think that's what Jack was getting at.

Yes.

Of course, but if that's how you reckon it, any shape is sinusoidal. He said he was concerned with terminology.


A sinusoidal function?

Hard to use that as a spatial resolution target, Thrilla, though one could use dead leaves:


Image of Dead Leaves low contrast target designed to have controlled scale and direction invariant features with a power law Power Spectrum.

https://www.strollswithmydog.com/introduction-to-texture-mtf/

I was instead referring to the fact that with a single edge one gets to measure the relative modulation of sinusoids of all different frequencies at once.

Alternatively one could do it very expensively by printing/etching a very large number of sinusoidal targets, each of a single spatial frequency, and then measure the modulation of each individually.

Happy to improve the semantics though. How else could I have phrased it?

I understand your point, and I'm familiar with the knife edge method. It's funny that you found the mushrooms inadequate compared to a knife edge or a dead leaves target. Perhaps if I found some mushrooms with a smoother surface. (This too is intended to be humorous.)

My point is purely about terminology. Although mushrooms or a knife edge can be represented by a Fourier transform or Fourier series (i.e., a sum of sinusoidal functions), that doesn't mean that they are sinusoidal. A mushroom doesn't look very sinusoidal to me, and neither does a knife edge. Maybe it would be better just to say that a knife edge can be represented by a sum of sinusoidal functions, rather than calling it a sinusoidal target.

I like it, Thrilla!

 

[Jack Hogan]

Happy to improve the semantics though. How else could I have phrased it?

Instead of this:

In measurement we fulfil the first requirement by using a good sinusoidal target like a knife edge;

You might have said: In measurement we fulfil the first requirement by using a target that in the ideal case has frequency-domain components at all frequencies, like a knife edge;

Is that clearer?

Excellent, sold Jim!

 

[J A C S]

Happy to improve the semantics though. How else could I have phrased it?

Instead of this:

In measurement we fulfil the first requirement by using a good sinusoidal target like a knife edge;

You might have said: In measurement we fulfil the first requirement by using a target that in the ideal case has frequency-domain components at all frequencies, like a knife edge;

Is that clearer?

Excellent, sold Jim!

Many functions contain all frequencies, it is in fact a generic property in some sense. A Gaussian contains all frequencies, for example. The good thing about the step function is that it is bounded, as we expect our images to be, and its high-frequencies do not decay "too fast," in fact they decay like 1/frequency, which is more or less the slowest you can have. If the high frequencies (well, of the order of the Nyquist), have very small amplitudes, they would be hard to detect.

Also, its derivative is the Dirac delta, which would be the perfect target to recover the PSF.

 

[Jack Hogan]

Happy to improve the semantics though. How else could I have phrased it?

Instead of this:

In measurement we fulfil the first requirement by using a good sinusoidal target like a knife edge;

You might have said: In measurement we fulfil the first requirement by using a target that in the ideal case has frequency-domain components at all frequencies, like a knife edge;

Is that clearer?

Excellent, sold Jim!

Many functions contain all frequencies, it is in fact a generic property in some sense. A Gaussian contains all frequencies, for example. The good thing about the step function is that it is bounded, as we expect our images to be, and its high-frequencies do not decay "too fast," in fact they decay like 1/frequency, which is more or less the slowest you can have. If the high frequencies (well, of the order of the Nyquist), have very small amplitudes, they would be hard to detect.

Also, its derivative is the Dirac delta, which would be the perfect target to recover the PSF.

Yes, better to make explicit the uniform starting point for all frequencies. How about:

'In measurement we fulfil the first requirement by using a target that in the ideal case has practically uniform frequency-domain components at all frequencies, like a knife edge;'

uniform = equal energy.

But I am curious about the 1/f decay: in our context the step function always has limited extent (in the limit it is the length of the edge in the capture), so ideally its Fourier Transform is bounded and should be that of a pulse = a sinc function, no?

We do see the 'length of edge' sinc in the System MTF, that's one of the reasons why we like to crop the ESF at least +/- 16 pixels from the center of the transition if possible with current enthusiast kit: the undulations and decay in the frequencies of interest, the first couple of c/p, can then typically be ignored.

Jack

 

[Jack Hogan]

MTF is the magnitude of the Optical Transfer Function, which is the Fourier Transform of the Point Spread Function.

The PSF is itself the square magnitude of the Fourier Transform of the complex pupil function. With incoherent light, virtually always true in photography, we don't see phase in the PSF because the waves coming from the lens keep smashing into the sensing plane during exposure time and so their phase relationship is lost. Therefore the PSF is always made up of only real, as opposed to complex, intensity as we all know. (Incidentally, this implies Hermitian Symmetry: negative frequencies of MTFs are a mirror image of positive ones).

The FT of the PSF, however, is typically complex, giving rise to phase components in the Spectrum. In fact we can express the OTF in complex notation as a magnitude (MTF) multiplied by a phase component referred to as the Phase Transfer Function: OTF = MTF * PTF. We can estimate both quantities in the System OTF using the slanted edge method.

When the PTF is locally zero (e.g. unaberrated lens with circular aperture), clearly OTF = MTF.

When it is not it can be either linear or non-linear. A linear PTF simply moves the image around (e.g. Tilt, geometric distortion). Non-linear phase instead blurs the image (e.g. Spherical, Coma, Astigmatism, Field Curvature, etc.).

Using the Transfer Function framework, System OTF is the product of the OTF of the individual components of the system (e.g. Lens * Pixel aperture). The OTF will generally be real with no PTF if both components are real.

The OTF of a component of an imaging system is only real if the relative PSF is a centrally symmetric even function (i.e. PSF(x,y) = PSF(-x,-y). Examples of even functions are an Airy function and a square pixel aperture.

I hope this clarifies why comments about MTF are often qualified by the addition of 'ignoring phase'. Where sharpness is concerned it might as well be 'ignoring non-linear phase components'.

Thoughts, additions, deletions?

 

[JimKasson]

MTF is the magnitude of the Optical Transfer Function, which is the Fourier Transform of the Point Spread Function.

The PSF is itself the square magnitude of the Fourier Transform of the complex pupil function. With incoherent light, virtually always true in photography, we don't see phase in the PSF because the waves coming from the lens keep smashing into the sensing plane during exposure time and so their phase relationship is lost. Therefore the PSF is always made up of only real, as opposed to complex, intensity as we all know. (Incidentally, this implies Hermitian Symmetry: negative frequencies of MTFs are a mirror image of positive ones).

The FT of the PSF, however, is typically complex, giving rise to phase components in the Spectrum. In fact we can express the OTF in complex notation as a magnitude (MTF) multiplied by a phase component referred to as the Phase Transfer Function: OTF = MTF * PTF. We can estimate both quantities in the System OTF using the slanted edge method.

When the PTF is locally zero (e.g. unaberrated lens with circular aperture), clearly OTF = MTF.

When it is not it can be either linear or non-linear. A linear PTF simply moves the image around (e.g. Tilt, geometric distortion). Non-linear phase instead blurs the image (e.g. Spherical, Coma, Astigmatism, Field Curvature, etc.).

Using the Transfer Function framework, System OTF is the product of the OTF of the individual components of the system (e.g. Lens * Pixel aperture). The OTF will generally be real with no PTF if both components are real.

The OTF of a component of an imaging system is only real if the relative PSF is a centrally symmetric even function (i.e. PSF(x,y) = PSF(-x,-y). Examples of even functions are an Airy function and a square pixel aperture.

I hope this clarifies why comments about MTF are often qualified by the addition of 'ignoring phase'. Where sharpness is concerned it might as well be 'ignoring non-linear phase components'.

Thoughts, additions, deletions?

Well said, Jack.

 

[J A C S]

Happy to improve the semantics though. How else could I have phrased it?

Instead of this:

In measurement we fulfil the first requirement by using a good sinusoidal target like a knife edge;

You might have said: In measurement we fulfil the first requirement by using a target that in the ideal case has frequency-domain components at all frequencies, like a knife edge;

Is that clearer?

Excellent, sold Jim!

Many functions contain all frequencies, it is in fact a generic property in some sense. A Gaussian contains all frequencies, for example. The good thing about the step function is that it is bounded, as we expect our images to be, and its high-frequencies do not decay "too fast," in fact they decay like 1/frequency, which is more or less the slowest you can have. If the high frequencies (well, of the order of the Nyquist), have very small amplitudes, they would be hard to detect.

Also, its derivative is the Dirac delta, which would be the perfect target to recover the PSF.

Yes, better to make explicit the uniform starting point for all frequencies. How about:

'In measurement we fulfil the first requirement by using a target that in the ideal case has practically uniform frequency-domain components at all frequencies, like a knife edge;']

Well, that would be the Dirac delta. Its FT is one.

uniform = equal energy.

But I am curious about the 1/f decay:

See the FT section here, and ignore the delta term:

Heaviside step function - Wikipedia

in our context the step function always has limited extent (in the limit it is the length of the edge in the capture), so ideally its Fourier Transform is bounded and should be that of a pulse = a sinc function, no?

We do see the 'length of edge' sinc in the System MTF, that's one of the reasons why we like to crop the ESF at least +/- 16 pixels from the center of the transition if possible with current enthusiast kit: the undulations and decay in the frequencies of interest, the first couple of c/p, can then typically be ignored.

Are you talking about a finite extent along the edge or across it? If the latter, you do not want to create a new edge parallel to the main one. It is better to apply a gradual cutoff away from the edge first, if computing the FT after that. Even better, you can try to differentiate numerically first, and then apply such a cutoff. Whatever you do, the low frequencies would be computed not so accurately.

 

[Jack Hogan]

Many functions contain all frequencies, it is in fact a generic property in some sense. A Gaussian contains all frequencies, for example. The good thing about the step function is that it is bounded, as we expect our images to be, and its high-frequencies do not decay "too fast," in fact they decay like 1/frequency, which is more or less the slowest you can have. If the high frequencies (well, of the order of the Nyquist), have very small amplitudes, they would be hard to detect.

Also, its derivative is the Dirac delta, which would be the perfect target to recover the PSF.

Yes, better to make explicit the uniform starting point for all frequencies. How about:

'In measurement we fulfil the first requirement by using a target that in the ideal case has practically uniform frequency-domain components at all frequencies, like a knife edge;']

Well, that would be the Dirac delta. Its FT is one.

uniform = equal energy.

But I am curious about the 1/f decay:

See the FT section here, and ignore the delta term:

Heaviside step function - Wikipedia

Ah, I see what you mean. The slanted edge method takes the FT of the derivative of the estimated edge profile. Then there is no 1/f decay as you suggest, and that's what we see in the results.

...We do see the 'length of edge' sinc in the System MTF, that's one of the reasons why we like to crop the ESF at least +/- 16 pixels from the center of the transition if possible...

Are you talking about a finite extent along the edge or across it?...

I was referring to the length of the edge profile derivative (i.e. across the image of the edge), the FT of which is taken. This sinc we do see in the resulting spectrum: we can choose to normalize it out, though we typically don't since its effect is immaterial for the reasons mentioned above.

'In measurement we fulfil the first requirement by using a target that in the ideal case can result in practically uniform frequency-domain components at all frequencies, like a knife edge processed via the slanted edge method;'

Jack