Consider a sinewave of frequency "19". It is routed to a light bulb through a synchronously reversing DPDT switch that is flipped at a frequency of "1".

Classic Fourier series would reveal that this waveform has a fundamental frequency of zero and a nineteenth harmonic of zero.

Yet the bulb lights.

Why?

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Many thanks,
Don Lancaster voice phone: (928)428-4073

The easiest way to see how it plays out is to remember that a multiplication of waveforms in the time domain is convolution in frequency domain and vice versa.

The squarewave discontinuity in the waveform means lots of extra frequencies in the modulated signal. It is easier to sketch the waveforms for a smaller number of cycles per DPDT switch throw than 19.

ITYM the amplitudes of the DC, fundamental and the 19th harmonic components are all zero.

But the amplitudes of lots of other frequency components are non-zero.

Worth noting here that power in the lamp goes as voltage squared.

Yes, not only that, but without applying pen to paper, we can know that the sum power of all the frequency components adds up to exactly the same value as if the switch hadn't been there.

Consider a sinewave of frequency "19". It is routed to a light bulb through a synchronously reversing DPDT switch that is flipped at a frequency of "1".

Classic Fourier series would reveal that this waveform has a fundamental frequency of zero and a nineteenth harmonic of zero.

Yet the bulb lights.

Why?

--
Many thanks,
Don Lancaster voice phone: (928)428-4073

Considering that a square-wave is a sum of weighted odd-multiples of its fundamental frequency (i.e. at f=3D1, 3, 5 ,...) and you are using it to do balanced amplitude modulation of a carrier at f=3D19, you should expect a result with components at sums and differences at f=3D18 & 20, 16 & 22, 14 & 24, ..., which are all even numbers. Therefore, as you have already observed, f=3D1 and 19 are missing.

The bulb acts like a full-wave rectifier plus resistor, since it ignores the sign of the voltage. Do a Fourier on the rectified sine, and you'll have a DC component, plus F=28, etc.

There's also a capacitor in there given the (often quite significant) thermal inertia of the filament.

But all the frequencies you predicted will still show up, just at reduced amplitudes.

One could use LEDs instead to effectively eliminate the capacitor in the model... although then you also need to be careful that the filtering effect of the human eye doesn't confound you...

I don't see much capacitive effect. The filament won't change resistance much in 1/120 second, and the "rectified" voltage will certainly cross through zero every cycle. The effect of filament temperature modulation will be a little distortion in the current waveform and a little phase lead.

What I meant was that, due to thermal inertia, if the output variable is "brightness" and the input variable is "voltage," the transfer function of the lightbulb includes a significant DC term. It's actually a pretty non-linear device with those definitions of input and output -- as you mentioned, it behaves like a rectifier as well.

Persistence because of the heating time of the filament makes it act like a "low-pass filter". I dunno - are any components other than DC visible in the output?

There's very little energy at f=3D28 (or 38) which correspond to the 9th (and 19th) harmonics of the square wave. Rather, most of the energy is at f=3D18 & 20. Do the math.

Your full wave rectifier is an unnecessary and unhelpful complication.

And for power you want FT{ [rectified sin]^2 }

Put more simply it acts like a square law detector if the load is held at constant temperature or like a V^1.6 ish detector for a real hot filament bulb leading to some harmonics.

The resulting light curve is not altered by flipping the phase synchronously with the zero crossings because (-1)*(-1)==1 I expect Xerox have got a spoiler patent on that identity too :(

The resulting light curve will be some average value with a sinusoidal variation at twice the fundamental frequency and some harmonics induced by the non-linearity of the resistance temperature curve for a lamp.

Into a pure resistive load held at constant temperature the power delivered would be exactly sinusoidal as V^2/R.

I have. You have taken the FT of the *voltage*. Power = V*I.

The power delivered into a pure resistive load is completely unchanged by randomly flipping the voltage by multiples of -1. (-1)*(-1) = 1

John is right about the main harmonic in the light output which is at f=38 with a tail of harmonics caused by the non-linearity of the filament resistance with voltage.

His full wave rectifier is an unnecessary complication.

Somewhere there is a stupidity in that analysis. In first second, 19 "in phase" sines are applied to load and obviously power it; the second second, 19 "out of phase" sines are applied to load and obviously power it. Repeat as desired, just do not wear out that switch. Frequencies present: 1, 19 main or fundamentals; switch = = square wave modulation --> odd harmonics of 1, 18 and 20 for first, 16 and 22 for third etc.

The phase-reversal method you describe is sometimes used to suppress a signal at X Hz when the signal is, in fact, operating at X Hz.

Parseval's theorem tells us that the energy in the time domain equals the energy in the frequency domain, so the energy that you're dumping into the light bulb has to appear in the spectrum _somewhere_. I ran an FFT of your problem and found that the energy is substantially (I didn't window the time domain data) at 37 Hz and 39 Hz.

Without doing the math, that does not sound right; clearly 1 and 19 exist and due to non-linearity the sum and differences: 1, 18, 19, 20; then the odd harmonics of those add in to a smaller degree - theoretically up to infinity if a perfect square wave transistion is assumed. Up to the 7th harmonic is good enough for almost everybody..

I took another look and realized that I had not scaled the FFT results correctly. The revised results are that significant terms appear at 18 and 20. There is nothing at 1 or 19.

This should be no surprise since, using sinusoids,

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