# Quick question on Fourier transform

semc
Hi all, as a physics student, I seldom use Fourier transform but from my understanding, given a periodic function you can decompose the function into sine function with different frequencies. Also, to get a ultra short pulse in time domain, this would require mixing many frequencies. I would therefore like to ask what happens if I pass my monochromatic light through a optical chopper. Is Fourier transform not applicable in this case since ultra short pulse generation does not require many frequencies?

dauto
I would therefore like to ask what happens if I pass my monochromatic light through a optical chopper. Is Fourier transform not applicable in this case since ultra short pulse generation does not require many frequencies?

Ultra short pulses DO require many frequencies. Fourier transform theory applies

Gold Member
Of course when analyzing pulses, it may be advantageous to look at the wavelet transform instead depending on what you actually hope to get out of the analysis.

Gold Member
Fouier theory says that the spectrum Z of the product of two signals z(t) = x(t) y(t) is the convolution of their spectra, Z = X * Y. Your signal consists of a monochromatic wave times a periodic square wave, so its spectrum is the convolution of their spectra. You can easily figure it out from there.

cosmik debris
A single frequency like a pure sine wave is only a single frequency if it starts at the beginning of time and goes on forever. As soon as you start or stop it you introduce a step which is composed of higher frequencies. The sharper the step the higher are the included frequencies. So chopping introduces harmonics.

semc
Ultra short pulses DO require many frequencies. Fourier transform theory applies

I am aware that you need many frequencies to get short pulse. That's what Fourier transform tells us. In my example, I specifically said I only have single wavelength as source.

A single frequency like a pure sine wave is only a single frequency if it starts at the beginning of time and goes on forever.

So if I turn on and off my laser I introduce higher harmonics as well?

Fouier theory says that the spectrum Z of the product of two signals z(t) = x(t) y(t) is the convolution of their spectra, Z = X * Y. Your signal consists of a monochromatic wave times a periodic square wave, so its spectrum is the convolution of their spectra. You can easily figure it out from there.

Ain't this just the math? I am confused how higher frequencies are generated.

Gold Member
The rising and falling edges of the "chopper" are very short, and therefore contain high frequency components. An ideal square wave modulation has infinite slopes, thus having a spectrum that is non-zero to infinite frequencies.

semc
The rising and falling edges of the "chopper" are very short, and therefore contain high frequency components. An ideal square wave modulation has infinite slopes, thus having a spectrum that is non-zero to infinite frequencies.

dauto
I am confused how higher frequencies are generated.

That's an application of the uncertainty relation. When you chop down the beam you are making a measurement of the photon location. By the uncertainty principle the photon momentum becomes uncertain creating a spread on the wavelength. In other words, the uncertainty principle is the physical principle that expresses the spread in wavelength required by the mathematics of waves which can be studied and understood through Fourier transforms.

Gold Member
I second what dauto said. There is an uncertainty relation for Fourier transforms which is the analog of Heisenberg's Uncertainty Principle, and that applies to common classical signals. You can find some information on Wikipedia, and more in any text on Fourier transforms. I like Bracewell's text, which has an excellent chapter discussing relationships between the "two domains" (time and frequency).

"Quote by dauto
Ultra short pulses DO require many frequencies. Fourier transform theory applies

I am aware that you need many frequencies to get short pulse. That's what Fourier transform tells us. In my example, I specifically said I only have single wavelength as source."

Once you modulate the carrier (pure sine) with an envelope, you no longer have a pure sine of single wavelength. You have a time-varying envelope that may have a rich spectrum, as I indicated earlier.

• 1 person
DrBwts
A square wave (& therefore loosly speaking a square pulse) is the sum of all the odd harmonics.

Fourier decomposition of a (periodic) signal will give you a long string of harmonics along with their amplitudes that when added together will give you your original signal back. If the wave is square then you will only get odd harmonics. In practice you may well get low amplitude even harmonics as well.

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