Electronics > Projects, Designs, and Technical Stuff
Square wave harmonic content
<< < (4/6) > >>
tggzzz:

--- Quote from: taydin on September 28, 2019, 04:47:13 pm ---The question isn't "what is Fourier transform and its mathematical foundation". The question is, if I switch a resistor with a MOSFET, how do the sinusoidals form? At the electron level, at the level of the magnetic fields or electric fields.

If I have three signal generators that generate 1 MHz, 2 MHz, and 3 MHz, and then I add these signals together, and then feed the results into a spectrum analyzer, I fully expect to see the three peaks in the spectrum, because those sinusoidals are there.

But if I switch a resistor using a MOSFET, there are NO SINUSOIDALS anywhere. So my points is, they aren't really there.

--- End quote ---

Yes, they are there, and (to use your terminolgy) have already been added together.

Band limit that square wave and put it into a spectrum analyser (just like you did for your added sinewaves), and you will see the three peaks (and more if you don't band limit the signal).


--- Quote ---What you are looking at is the effect of the square wave rise time. If the rise time is high, you will see much more harmonics spanning to higher frequencies. If the rise time is low, you will only see a limited number of harmonics.

--- End quote ---

Your language makes it appear you think the risetime causes sine waves or harmonics. It doesn't.

Square waves can be decomposed into sets of sinewaves.
Sets of sine waves can be composed into squarewaves.
The time domain and frequency domain are equivalent, as defined by the Fourier transform.
RoGeorge:

--- Quote from: taydin on September 28, 2019, 04:47:13 pm ---If I have three signal generators that generate 1 MHz, 2 MHz, and 3 MHz, and then I add these signals together, and then feed the results into a spectrum analyzer, I fully expect to see the three peaks in the spectrum, because those sinusoidals are there.

--- End quote ---

First, if you want to resemble something like a square wave, then you can have only odd harmonics, syncronized in a certain way and with a certain amplitude relative to each other, but let's say your generators are 1, 3 and 5 MHz, all with just the right amplitude and phase.  Once you add them together, and look at the resulting waveform, you won't see any sinusoid, you'll see almost a square wave.

So where are your original 1, 3 and 5 sinusoidal waveforms?  Did they just disappeared?  No, they are all there, in the (almost) square waveform of all 3 added together.  You don't believe your original 1, 3 and 5 MHz are still there?  Go ahead, make 3 filters at 1, 3 and 5 MHz respectively, and you will see your original signals.


--- Quote from: taydin on September 28, 2019, 04:47:13 pm ---But if I switch a resistor using a MOSFET, there are NO SINUSOIDALS anywhere. So my points is, they aren't really there. What you are looking at is the effect of the square wave rise time. If the rise time is high, you will see much more harmonics spanning to higher frequencies. If the rise time is low, you will only see a limited number of harmonics.

--- End quote ---

Yes, there are A LOT of sinusoidals in the square wave.  You CREATED them, all at once, when you turned your switch on through your resistor.  They ARE really there, they are real.  You can look at each of those components with an adjustable bandpass filter.

If you don't want to consider the composing parts of a more complicated object as real, then what is REAL?

Your body is made out of a bunch of various body organs.  When you look in a mirror you can "see" only the whole human body (analogy with the square wave), you are not seeing internal organs, but if you look with a tomography machine, you'll observe a bunch of internal organs (the 1, 3, 5 MHz sinus) so when you look in a mirror, are those internal organs REAL, or not?

Same with your square waveform.  It depends of what type of instruments you use to "look" at your signal.  With an oscilloscope, you'll see a square wave, with a spectrum analyzer/VNA you'll see 3 spectral components at 1, 3 and 5 MHz.

Why the difference between the 2 instruments?  Which instrument is lying to you, the oscilloscope or the spectrum analyzer?

The answer is that both instruments are correct.

What we call "reality" is just a preferred subset of what is really out there, a subset we get after we applied some sort of filter while we were observing (or measuring) the real world around us.

After all, what do you mean by "REAL", and by "are they (the sinusoids) really there"?



Many, many moons later (2021 edit):
Changed my mind.  Now I think those sinusoids are not physically there.
2N3055:

--- Quote from: taydin on September 28, 2019, 04:58:45 pm ---Or if I put it in another way, the rising edge, or the falling edge (or in fact, any dV/dt change that occurs in a signal) has energy change at a rate that can only be caused by a specific sine wave frequency. The fourier transform lets you know what those frequencies are. It's a mathematical tool.

The fact that you can represent a square wave by an infinite number of sine waves isn't proof that those sine waves are there. If they are there, there must be a physical mechanism (or quantum mechanism) causing them.

--- End quote ---

I cannot depict whether you are making fun of us or you are serious...

Lets take the picture and put it upside down..

If you take squarewave (as perfect as you can make it) at say 1 kHz. Make it very nice, clean and symmetric, both in duty cycle (50%) and amplitude wise.

And feed that into a nice analog filter, that will be made to filter out 3 kHz, 5 kHz, 7 kHz. A nice sharp filter with cutoff at 1 kHz.. Make filter at least 24dB/oct.  And feed output of that into scope.

What you will see? A nice clean 1 kHz sine wave. Where that one came from. It obviously was in there.... We didn't add it mathematically. We subtracted other sine waves from original signal, one by one, and what was left was fundamental tone of 1kHz. That thing WAS in there all the time, you just didn't see it from other crap.

And as Tim nicely said, actually there is no squarewave or triangle wave or any other wave that is 1 kHz. Only 1 kHz waveform would be insanely clean 1 kHz sine wave. Add phase noise, change waveform, you get a handful of frequencies (ideal sinewaves) combined into a waveform shape that repeats at 1 kcycles per second.


fourfathom:
The sine waves are there.  Try this:  Generate a square wave, say at 1MHz, or 100 KHz.  Connect this to a single-sideband radio receiver and tune across the spectrum.  You will hear a pure beat-note tone at all the odd harmonics of that square wave.  Those are your sine waves.
taydin:

--- Quote from: RoGeorge on September 28, 2019, 05:24:51 pm ---First, if you want to resemble something like a square wave, than you can have only odd harmonics, syncronized in a certain way and with a certain amplitude relative to each other, but let's say your generators are 1, 3 and 5 MHz, all with just the right amplitude and phase.  Once you add them together, and look at the resulting waveform, you won't see any sinusoid, you'll see almost a square wave.

So where are your original 1, 3 and 5 sinusoidal waveforms?  Did they just disappeared?  No, they are all there, in the (almost) square waveform of all 3 added together.  You don't believe your original 1, 3 and 5 MHz are still there?  Go ahead, make 3 filters at 1, 3 and 5 MHz respectively, and you will see your original signals.

--- End quote ---

You misunderstood. What I'm saying is, I have 3 sine wave sources, added together (NOT to form a squarewave, as you seem to think). Then I feed the sum to a SA, and I fully expect to see 3 peaks, because those sinewaves were created by the source.


--- Quote from: RoGeorge on September 28, 2019, 05:24:51 pm ---Yes, there are A LOT of sinusoidals in the square wave.  You CREATED them, all at once, when you turned your switch on through your resistor.  They ARE really there, they are real.  You can look at each of those components with an adjustable bandpass filter.

--- End quote ---

So can we say that the rising edge and the falling edge of the square wave implement a change of energy which resembles one half of multiple sine waves (the first 90 degrees). The bandpass filter will just act on these partial sine waves and produce the necessary output?
Navigation
Message Index
Next page
Previous page
There was an error while thanking
Thanking...

Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod