I have built a 555 oscillator and connected it to a speaker.

Using an oscilloscope I adjusted the 555 to generate a 2.5kHz square wave.

I then held a microphone up to the speaker and fed the input into a spectrum analyser.

What I expected to see was a single peak at 2.5kHz. However, what I actually got was this:

spectrum analyser screenshot showing harmonics every 2.5kHz

My question is, where have these harmonics come from if the 555 is only generating a 2.5kHz signal?

I know that a square can be constructed from sine waves:

diagram showing the sum of sine waves forming a square wave

However, the 555 does not generate sine waves or multiple frequencies, it generates a single square pulse. So where have these harmonic frequencies come from?

  • 16
    \$\begingroup\$ Odd harmonics are fundamental to a square wave as your second illustration shows. (It would be better explained as : a square wave can be decomposed into an infinite series of sine waves).. The even harmonics tell me the mark-space ratio is not 50%. \$\endgroup\$
    – user16324
    Oct 20, 2019 at 11:58
  • 63
    \$\begingroup\$ It's not that a square wave can be constructed from sine waves, as some optional thing or wacky view of it. A square is composed of sine waves. \$\endgroup\$
    – TonyM
    Oct 20, 2019 at 12:02
  • 12
    \$\begingroup\$ The FFT (actually a dft) is looking for sinusoidal content by definition. These are the glasses your spectrum analyzer is looking through. \$\endgroup\$ Oct 20, 2019 at 12:09
  • 10
    \$\begingroup\$ With no disrespect, it sounds like your level of understanding doesn't yet extend to Fourier analysis of electrical waves. Don't be looking for a series of sine wave generators and a mixer circuit here. Your square wave is inherently just that. But you'll find plenty of explanatory text on this on the interweb. \$\endgroup\$
    – TonyM
    Oct 20, 2019 at 12:29
  • 29
    \$\begingroup\$ The square wave is. The 555 created it by switching the output. By feeding it into a spectrum analyser, you're asking the question 'what sine waves make up this square wave?'. If you'd fed it into a power meter, you'd be asking the question 'what's the power in this square wave?' BTW, the 555 generates an approximation to a mathematical square wave, because the output voltage can't switch infinitely fast. It's pretty fast compared to 2.5kHz, but not fast compared to 100MHz. \$\endgroup\$
    – Neil_UK
    Oct 20, 2019 at 13:21

6 Answers 6


When you are holding a hammer, the world looks like a nail.

Roughly speaking, a spectrum analyser captures a time record and represents the resulting capture as a unique linear combination of sinusoids.

It does not mean that whatever generated the signal generated separate sinusoids, only that the resulting signal can be represented in this (very useful) way.

As other answers have pointed out, a square wave can be represented by the sum of sinusoids at odd harmonics, hence the harmonics on your analyser.

There are other systems of representation (cf. Walsh functions) that represent signals in terms of square waves, however these representations are not practical from current perspective. However, if one had a mythical Walsh spectrum analyser and you looked at a sinusoid, your question might then be asking where do all the square waves come from.

  • 2
    \$\begingroup\$ "Yeah, yeah, but your scientists were so preoccupied with whether or not they could" ...read a square wave with a spectrum analyzer... "that they didn't stop to think if they should.". :) \$\endgroup\$
    – longneck
    Oct 22, 2019 at 15:13
  • 5
    \$\begingroup\$ They felt it was their duty to cycle through the possibilities... \$\endgroup\$
    – copper.hat
    Oct 22, 2019 at 19:45
  • 1
    \$\begingroup\$ Older spectrum analyzers did not capture a time record. They (HP 8553B, in HP 141 display) merely use selectable crystal-filters (for big buck, you get down to 10Hz resolution bandwidth). \$\endgroup\$ Oct 23, 2019 at 17:36
  • 1
    \$\begingroup\$ I know, I was born before the FFT was 'discovered' :-). Sweeping with a narrow tunable band pass is basically computing the Fourier coefficients directly. \$\endgroup\$
    – copper.hat
    Oct 23, 2019 at 17:54

I know that a square can be constructed from sine waves. However, the 555 does not generate sine waves or multiple frequencies, it generates a single square pulse. So where have these harmonic frequencies come from?

Congratulations on your explanation of what you are seeing and your experimentation.

The key issue is that not only CAN a square wave be constructed from sine waves, it fundamentally IS a collection of sine waves.
You can generate a square wave by summing appropriate sine waves, but, however you do it, what you arrive at IS a waveform that can be represented by a collection of sine waves.

In ideal circumstances you would not expect to see on the spectrum analyser quite what you show, but impedance matching and a 555 and .... can easily combine to produce a non ideal result.

A square wave = a summation of \$ f + \frac{3f}{3} + \frac{5f}{5} + \frac{7f}{7} + ...\$ (if my brain has correctly retrieved the relevant long ago stored facts). So you would expect to see every second harmonic, and amplitudes should decrease.

  • 2
    \$\begingroup\$ Is the answer here really then that a "square wave" isn't really a wave at all? I suppose that a true square wave can't exist within the laws of physics, it would require a speaker cone to teleport between two positions and electrons to teleport within a wire. Therefore because both electrons and speakers are constrained to moving in sinusoidal patterns, the only way to construct a square "wave" is by combining a bunch of sine waves together? \$\endgroup\$ Oct 20, 2019 at 12:31
  • 6
    \$\begingroup\$ @JShorthouse Electrons and speakers are not constrained to move in sinusoidal patterns. That constraint only arises for certain particular systems (e.g. harmonic oscillators). It so happens that sinusoids happen to be a convenient basis for analyzing systems that are linear and time-invariant, but you're reading too far into things. \$\endgroup\$
    – nanofarad
    Oct 20, 2019 at 20:38
  • 3
    \$\begingroup\$ @JShorthouse You can't have it both ways :-). You wrote "Using an oscilloscope I adjusted the 555 to generate a 2.5kHz square wave." You know you didn't, of course. A 555 makes a "squarish wave". Other devices more-squarish (or less). But just as the true square wave needs an infinite sum of declining magnitude sinewaves (and so extra terms are unimportant at somewhere around the noise level) so too viewing, analysing , ... is going to run into non-ideal or non-infinite aspects. A speaker is not constrained to follow a single sinusoid - but any path folows is describeable by a set of sinusoids \$\endgroup\$
    – Russell McMahon
    Oct 20, 2019 at 23:34
  • 11
    \$\begingroup\$ @JShorthouse It's not that square waves can't be fundamental, but the job of a spectrum analyzer is to analyze things in terms of sine waves, so that's what it did. When you analyze a wave in this way, of course you will see that only sine waves are "fundamental" according to this kind of analysis. \$\endgroup\$
    – user253751
    Oct 21, 2019 at 15:23
  • 3
    \$\begingroup\$ at 2.5kHz, complaining the square wave can't be square is a bit like complaining your car's wheels can't be round because they're made of atoms. \$\endgroup\$
    – user253751
    Oct 22, 2019 at 9:01

A square wave can be viewed as a sum of the odd harmonics of a single frequency.

A square wave can be generated by summing a bunch of sine waves.

A square wave can also be generated by simply toggling the power on and off at the primary frequency of the square wave.

In either case, the spectrum will look the same.

You cannot tell how a square wave was generated by looking at the spectrum.

The simple act of turning the power on and off generates the primary frequency, but it also generates the harmonics.

Your spectrum shows even as well as odd harmonics.

The even harmonics are an artifact of distortion coming from your microphone or the microphone amplifier. Too much gain or the microphone too close to the speaker. Alternatively, the signal from the 555 caused distortion in the speaker.

In any case, you should only see odd harmonics (2.5kHz, 7.5kHz, 12.5kHz, etc.) for a 2.5kHz square wave. The even harmonics (5kHz, 10kHz, etc.) are not part of the square wave.

Connect the 555 output to the line in of your PC. You may need to use a voltage divider to reduce the level.

That should be cleaner, and closer to an undistorted square wave.

Baudline (the spectrum analyser you are using) has an oscilloscope view. Use it to check if your square wave is distorted. Check the signal from the speaker and microphone setup as well as the direct connection to the 555.

  • 3
    \$\begingroup\$ "Baudline (the spectrum analyse you are using) has an oscilloscope view. Use it to check if your square wave is distorted." thanks very much for this tip, the wave is indeed rather distorted. \$\endgroup\$ Oct 20, 2019 at 12:25
  • 3
    \$\begingroup\$ "The even harmonics (5kHz, 10kHz, etc.) are not part of the square wave." - This assumes the mark/space ratio is exactly 50%. That's rarely the case in a typical 555 circuit. In this case it looks like the ratio was very asymmetrical. \$\endgroup\$ Oct 21, 2019 at 3:33
  • 2
    \$\begingroup\$ @BruceAbbott: Absolutely correct. I was thinking only of nice, regular square waves with a 50% duty cycle. I should have mentioned that. I'd rather not change it just now, though. The question has landed on HNQ. If I go monkeying with it, it'll just attract more attention to a middling answer that's already gotten way more upvotes than is reasonable. \$\endgroup\$
    – JRE
    Oct 21, 2019 at 9:25

What I expected to see was a single peak at 2.5kHz.

I don't know why. You need to reset your expectations.

Think of this this way: If you just had a single peak, then the input would by definition be a sine wave. But you're feeding it a square wave, so how do you account for the difference?

I know that a square can be constructed from sine waves.

Change that to: A square wave is equivalent to an infinite series of sine waves. That's what the math of Fourier analysis is all about.

the 555 does not generate sine waves or multiple frequencies, it generates a single square pulse.

They are exactly equivalent. So it's actually doing both.

So where have these harmonic frequencies come from?

You can think of them as "coming from" the fast edges on the square waves. You can see in your own graphs that as you consider higher harmonics, the edges of the sum get steeper. In the limit (infinitely many harmonics), the edges become vertical.


our ears are correlators. The FFT is a correlator. The analog spectrum analyzers of Hewlett Packard are correlators: they use narrow-band analog filters.

Square waves and rectangular waves, and many other (non-pure-sin) waveforms will strongly correlate with ( Positive_Integer * Fundamental) sin basis functions.

Square waves are not composed of sinusoids. The 555, and any FlipFlop, do not build the rail-rail outputs using a big bucket of handy sinusoids.

You ask a fine question.

We model, and we measure, using sinusoidal basis functions, harmonically related.

Examine the integral of sin(1,000 * time) multiplied by sin(3,000 * time). Do this for 1 cycle, for 1.5 cycles, for 1.6 cycles, for 1.9 cycles, for 2 cycles, for 200 cycles.

Harmonics do not exist. Its the behavior of the correlators that confuse us.

  • \$\begingroup\$ so the down-voters merely vote against this heresy, remaining unable to craft a lucid rebuttal? \$\endgroup\$ Oct 21, 2019 at 10:13
  • 1
    \$\begingroup\$ "Harmonics do not exist." Then how do you explain the difference between a sine wave and a square wave at the same frequency? How do you explain the operation of frequency multiplier circuits? If the harmonics do not exist, then how do we extract real power from them? \$\endgroup\$
    – Dave Tweed
    Oct 21, 2019 at 11:10
  • 1
    \$\begingroup\$ frequency multipliers first run the sin thru a distorter; the distortion then will correlate (be filterable) with the desired multiple of the fundamental. Next question, please. \$\endgroup\$ Oct 21, 2019 at 11:18
  • 1
    \$\begingroup\$ This answer makes the most sense to me, I don't know why it's being downvoted so heavily, can anyone give a proper rebuttal? From quickly searching around it seems widely accepted that the human ear performs the equivalent of a Fourier transform (which has blown my mind somewhat). But this would clearly explain why you can hear harmonics in a signal that was generated by merely switching an output from rail to rail, and why a spectrum analyser sees the same thing. Like another commenter said, a 5cm stick can be said to be made of five 1cm segments, but that doesn't mean it was made that way. \$\endgroup\$ Oct 21, 2019 at 18:12
  • 1
    \$\begingroup\$ @JShorthouse our ears don't perform the equivalent of a Fourier transform, they do something much more simple and basic. Responding to sine waves is something that many (in some sense, almost all) physical systems do. Harmonics do exist, not just as a figment of the math, but as part of the behavior of real things. Digital circuits are an idealization; all real circuits are analog. The circuit is full of electrons that are very willing to do sine-wavey things, and its imperfections are well described by how it responds to some of those higher-frequency sines. \$\endgroup\$
    – hobbs
    Oct 22, 2019 at 2:28

I would suggest that it's due to semantics, and those semantics falsely colour our perspective of a square wave. The following is the internal architecture within a 555 chip:-


You can clearly see that it's a digital circuit (rise/fall times excepted). It does not output a series of sine waves, exactly as you suspect. The output toggles between high and low voltage levels. So you're correct.

But mathematically (and taking from Wikipedia), a theoretical square wave can be represented as an infinite summation of odd sine harmonics, thus:-


You can see the \$sin\$ operator in there. It's just that your spectrum analyser can't tell the difference. After all, you might be feeding it an analogue summation of a few sine wave oscillators all running at odd harmonic frequencies. It would be indistinguishable from a square wave.

Also don't forget the speaker, microphone and recording equipment, which are inherently analogue and have physical mass i.e. smoothing. Some peaks will therefore come from a unintentional filtering effects of your audio equipment.

  • 1
    \$\begingroup\$ A digital circuit is merely 'convenient analogue'. It's a circuit composed of analogue components that uses a simplified mode of operation. It's inputs and outputs still have analogue characteristics, as this question is showing. Meanwhile, a square wave being composed of sine waves isn't semantics, it's physics. \$\endgroup\$
    – TonyM
    Oct 20, 2019 at 13:15
  • 3
    \$\begingroup\$ @TonyM Hmm, I'm not sure. I think the interpretation is on behalf of the spectrum analyser. Otherwise, how is the inverter on pin 3 producing a 7.5 kHz sine wave? Where is that 7.5 kHz oscillator circuit? That's what you're suggesting it's doing. It's the same kind of mathematical trick/FFT/interpretation that proves that you can never reach where you're going by iteratively halving the remaining distance. \$\endgroup\$
    – Paul Uszak
    Oct 20, 2019 at 13:54
  • \$\begingroup\$ If you split white light, you realise it's composed by a 'mixture' of a spectrum of colours. That doesn't mean there's two 'white light's in existence: the one you see when you split it, the one you knew when you didn't. It means what you thought was an element is actually a compound, roughly speaking. Ivm afraid it reads like you should be reading the other answers, not writing one, downvoting accordingly. \$\endgroup\$
    – TonyM
    Oct 20, 2019 at 14:33
  • 1
    \$\begingroup\$ @TonyM Your's is a great example of what I mean. We're not making the white light in the first place are we? For light read square wave. There is no 7.5 kHz oscillator circuit to make the first harmonic in the 555, is there? Plus you'll find that this answer is entirely consistent with the others. Semantics :-) \$\endgroup\$
    – Paul Uszak
    Oct 20, 2019 at 17:06
  • 1
    \$\begingroup\$ @TonyM Does that mean a sine wave is fundamentally composed of Walsh functions? (like another commentator said) \$\endgroup\$
    – user253751
    Oct 22, 2019 at 9:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.