Square waves resolve better?

**Chaos' lil bro Order** · Mar15-06, 02:26 AM

Hi,

I have a question about using EM to resolve images. I know that resolving power is constrained by Rayleigh's Criterion.

Consider a setup where a normal incandescent light bulb is mounted 1ft. above a desktop. A penny is hung directly below the light bulb 10cm from its surface. We know that a penny looks perfectly circular to the naked eye, but we want to measure any variation that shows that it is not a perfect circle, due to manufacturing limitations or whatever.

Because the light bulb's light diffracts when it strikes the penny's surface, the image we see on the desktop has 'blurry' edges due to this diffraction. This blurriness (correct me if I'm wrong) comes from the fact that the light bulb's light is sinusoidal in addition to the limits implied from the emitted wavelength, as per Rayleigh's Criterion.

But what if we replaced the light bulb with an apparatus that shines light comprised of a square waveform instead of a sinusoidal waveform. Would the image of the penny on the desktop have less 'blurry', finer resolved edges?

Thanks.

**chroot** · Mar15-06, 02:36 AM

Electromagnetic square waves are not solutions to Maxwell's equations, and thus cannot exist.

- Warren

**nrqed** · Mar15-06, 03:07 AM

Originally Posted by chroot

Electromagnetic square waves are not solutions to Maxwell's equations, and thus cannot exist.

- Warren

??? Why not? One can build a square waveform from a superposition of sinusoidal waves...since a linear combination of solutions of Maxwell's equations is still a solution of Maxwell's equations, the square wave will be a solution...where's the problem?

Pat

**chroot** · Mar15-06, 03:11 AM

Okay, an approximation to a square wave may be made by Fourier series, but an actual square wave is not possible.

- Warren

**chroot** · Mar15-06, 03:15 AM

And, if you used such a Fourier series, the highest resolution would be obtained by the highest wavelength used in the series. The resolution would be no better than using that wavelength alone as a pure sinusoid.

A true square wave has frequency components out to infinity, which would give it infinite resolving power. Now, to set about building that infinite -frequency light source.....

- Warren

**ZapperZ** · Mar15-06, 04:16 AM

Not only that, Warren, but with that big of a range of frequency, the imaging object, which is the aim of this thread in the first place, would be horribly blurry. This is because each frequency will cause a different amount of diffraction.

This is why we try to make our "probing agent" as monochromatic or monoenergetic as possible, such as in x-ray diffraction measurement, SEM imaging, etc. Having too broad of a band is no good in this case.

Zz.

**nrqed** · Mar15-06, 12:50 PM

Originally Posted by chroot

Okay, an approximation to a square wave may be made by Fourier series, but an actual square wave is not possible.

- Warren

True... An actual square wave is impossible...as is an actual saw-tooth wave or even an actual sinusoidal wave... I agree.

I think that when most physicists talk about a square wave (whether it's an em wave or a sound wave or an em wave in a coaxial cable or any other wave) has in mind a Fourier series which can be made as close as one wants (in theory) to what we could call a "physicist's square wave".

I know that you will say it's only semantics and that my post is irrelevant but I think that a lot of people could have read your post and then look at books on wave motion or electromagnetism and see the use of square waves a bit everywhere and be confused since they would think that square waves are impossible (per your post).

I know that *you* know all that but I had in mind other people reading the thread....

**George Jones** · Mar15-06, 08:59 PM

Originally Posted by chroot

Electromagnetic square waves are not solutions to Maxwell's equations, and thus cannot exist.

Right - for example, square waves are not differentiable functions, and thus can't be solutions, as functions, to the (differential) wave equation.

But, square waves are perfectly valid weak solutions to the differential wave equation, and, as Patricks says, it is along these lines that physiscists often think.

Regards,
George

**Integral** · Mar15-06, 09:59 PM

But what if we replaced the light bulb with an apparatus that shines light comprised of a square waveform instead of a sinusoidal waveform. Would the image of the penny on the desktop have less 'blurry', finer resolved edges?

I am not sure I understand exactly you mean by "a square wave instead of a sinusoidal waveform"

As has been discussed in this thread, square waves must be generated by a superposition of sinusoidal waves. To generate a square wave you must sum ALL of the odd harmonics (of appropriate amplitude) of the desired fundamental frequency. Note the emphasis on ALL, this means you must sum to infinity, this is why you cannot generate a perfect square wave.

It has been pointed out the band of frequencies required to generate a square wave, result in a LOSS of resolution rather then an improvement. therefore your conjecture is incorrect.

**krab** · Mar15-06, 10:16 PM

Originally Posted by Chaos' lil bro Order

Consider a setup where a normal incandescent light bulb is mounted 1ft. above a desktop. A penny is hung directly below the light bulb 10cm from its surface. We know that a penny looks perfectly circular to the naked eye, but we want to measure any variation that shows that it is not a perfect circle, due to manufacturing limitations or whatever.

Because the light bulb's light diffracts when it strikes the penny's surface, the image we see on the desktop has 'blurry' edges due to this diffraction. This blurriness (correct me if I'm wrong) comes from the fact that the light bulb's light is sinusoidal in addition to the limits implied from the emitted wavelength, as per Rayleigh's Criterion.

OK, I'll correct you. The blurred edges you see on the desktop surface are not due to diffraction. We are talking a normal light bulb here, so it is an extended source, not a point source. It is this extension that causes blurriness.

**Chaos' lil bro Order** · Mar16-06, 04:34 AM

I was referring to a square waveform created from a fourier series of overlapping modes (the 'almost square' waveform, if you like). In math, f + 3f + 5f + 7f+...

The reply that a perfectly square waveform is impossible to create because you need infinte frequencies to do so, was a good insight.

So, getting back to the image on the desktop, would the diffraction pattern look different AT ALL if the light bulb used 'almost square' waveforms instead of sinusoidal waveforms?

**chroot** · Mar16-06, 03:07 PM

Haven't we already answered this in detail? The resolution would be no better than that of the highest single sinusoid you used in your Fourier series, and in fact would be worse due to all the other lower frequencies, which might as well be considered noise.

- Warren