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I need a mathematical description of what happens to the spectrum of a signal when that signal is processed to divide its frequency.

As an example, construct an input signal as the sum of two sine waves of different frequencies (as commonly used in a "two tone" test). Processing this signal with a buffer that changes state at the rising zero crossings of the input signal produces a squared wave whose frequency can be divided by a flip-flop: enter image description here

The spectrum of the the input signal comprises tones at 102kHz and 105kHz: enter image description here "Squaring" the input preserves the two equal tones, albeit with added side tones (spaced at the difference between the two original tones) that result from removal of the amplitude envelope and added harmonics from the squaring: enter image description here

Passing the squared signal through a flip-flop produces a second square wave which in which the two-tone nature is no longer present: enter image description here

The single dominant tone is at 1/2 the frequency of the higher of the original two tones, while the spacing between the peaks continues to be the spacing between the two original tones.

Passing the squared result through an LC low-pass filter produces a sine wave with cycles that are half as frequent as the input: enter image description here

with the same general spectral distribution as the squared output: enter image description here

Can someone, please, explain what happened to the other tone?

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A sum of two sine waves can be expressed equivalently as the product of two sine waves. This is a phenomenon commonly referred to in musical circles as a beat frequency, which can be used to tune instruments by ear. For your example, your input can be expressed as a 103.5kHz frequency multiplied by a 1.5Hz frequency.

I wouldn’t use the term “squaring” for what you are doing as much as “clipping the ever living hell out of”. This would probably be expressed best using concepts from intermodulation distortion. However, in your application, the square nature of the output of the waveform can be analyzed more simply anecdotally. I would expect to see a square wave at the 103.5kHz with glitches at the 1.5Hz zero crossings. But notably, it would look a lot like a 103.5kHz waveform, with some side band nonsense caused by glitches.

I do not believe that the flip-flop can be expressed using any common signal processing techniques, but again it can be analyzed anecdotally. As you’ve hypothesized, I would expect the time between edges to double, because you’re skipping every other edge. However, this is a grosse generalization. It would be exactly true for a sinusoid, less true for the input you’ve proposed, and untrue for a generic input. So for you’re example, mostly energy at 51.75kHz.

At this point, you still have a square wave, roughly at 51.75kHz, with some side band garbage cause by the glitches. One could apply an arbititary amount of filtering to extract the sine wave at this frequency, resulting in the output waveform you described.

Something else to keep in mind. A frequency transform has practical limitations. A proper transform requires all time domain points of the input signal. Using less than infinite samples is equivalent to windowing, which causes side lobe patterns about dominant frequencies, which is a topic unto itself. Additionally, a signal sampled at discrete time intervals (like on a computer) has a limited bandwidth. When you “clip the ever loving hell” out of a signal, you are widening the bandwidth of the input signal significantly. If care isn’t taken to ensure that the discrete time sample rate is high enough, a significant amount of aliasing distortion will occur, hindering proper analysis.

Sorry for being a bit of a rant, but your question touches on a lot of points. Cheers.

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  • $\begingroup$ Thanks for those contributions, Dan. You prompted me to review lists of trig identities, leading me to some that may be useful. $\endgroup$ – Brian K1LI Mar 12 at 20:40
  • $\begingroup$ It has re-occurred to me that the average of the two tone frequencies becomes the "carrier" frequency of the SSB signal, while the difference of the tone frequencies is the frequency of the envelope "beat note." The biggest think I'd forgotten is that the phase of the signal flips at the zero point of each envelope "beat." Perhaps this contributes to the "disappearance" of the "other" tone. $\endgroup$ – Brian K1LI Mar 12 at 21:27

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