Comparator: Noisy sine to square wave, how much phase noise?

Question

In a circuit a comparator is used to convert a sinusoidal signal to a square wave. The input signal however is not a clean sine wave, but has some noise added to it.

The comparator is supposed to be ideal and has a hysteresis which is much larger than the noise signal, thus there is no ringing at the zero crossings of the sine wave.

Yet due to the noise on the input signal, the comparator switches slightly earlier or later as it would for a clean sine wave, hence the produced square wave has some phase noise.

The plot below illustrates this behavior: the blue curve is the noisy input sine wave and the yellow curve is the square wave generated by the comparator. The red lines show the positive and negative hysteresis threshold values.

Given the spectral density of the noise on the input signal, how can I calculate the phase noise of the square wave?

I would like to do a proper analysis on this, but could not find any resources on the topic yet. Any help is much appreciated!

CLARIFICATION: I would like to analyze the phase noise produced by the given circuit and am NOT asking on how the reduce the noise!

Answer 1

The noise is sampled only once per zero crossing, or twice per cycle of the 1 MHz signal. Therefore, as long as the bandwidth of the noise is significantly wider than 1 MHz, its spectrum is folded many times into the 1 MHz bandwidth of the sampled signal, and you can treat the PSD of the phase noise as essentially flat within that bandwidth.

The amplitude of the output phase noise is related to the amplitude of the input signal noise by the slope of the sine wave (in V/µs) at the comparator threshold voltages. Analysis is simpler if the thresholds are symmetric around the mean voltage of the sinewave, giving the same slope for both. The amplitude of the phase noise (in µs) is simply the noise voltage divided by the slope, in whatever units you want to use, such as the RMS value of noise that has a Gaussian distribution. In other words, the PDF of the phase noise is the same as the PDF of the original voltage noise (after scaling).

Answer 2

Depending on how the spectral density is provided, it is essentially asin

Determine the phase error due to the hysteresis:

\$ \Theta_{low} = sin^{-1}(-0.3) \$

\$ \Theta_{high} = sin^{-1}(0.3) \$

This is the phase error purely due to the hysteresis if a pure sinewave was applied.

Assuming you have or have converted your spectral density into magnitude & equally assuming it is normally distributed. generate the MEAN and 1 standard deviation.

LOW:

\$ \Theta_{low_error\_mean} = sin^{-1}(-0.3) - sin^{-1}(-0.3 + mean) \$

\$ \Theta_{low\_error\_+\sigma} = sin^{-1}(-0.3) -sin^{-1}(-0.3 + \sigma) \$

HIGH:

\$ \Theta_{high\_error\_mean} = sin^{-1}(0.3) - sin^{-1}(0.3 + mean) \$

\$ \Theta_{high\_error\_+\sigma} = sin^{-1}(0.3) -sin^{-1}(0.3 + \sigma) \$

With the mean and the standard deviation "phase error" you can reconstruct a phase error distribution curve.

However... if the spectral density isn't normally distributed you will need to derive errors at a number of specific points to reconstruct a phase error curve specific to the information you have

Answer 3

For a random noise signal of Npp around 10% with a signal Vpp comparing peak-peak ratio it can be seen that if the signal is a triangle waveform that the amplitude noise is converted to phase noise in a linear equation where is S/N=1 each edge has T/2 jitter p-p.

However the amplitude of the sine fundamental component is 81% of a Vpp triangle waveform and thus it's slope is 1/81% or 1.23 steeper thus phase noise is reduced to 81% of the ratio with hysteresis set to just higher than the peak noise level.

Thus the jitter on each edge is 81% of the Vpp/Npp ratio. It could be shown that slope matches the triangle wave when the Npp reaches 75% of the Vpp or a Vpp/Npp ratio of 1.33.

Normally jitter errors are measure in RMS noise power and energy per bit and statistical probability of error, but this was shown from the perspective of the question for time jitter over any measurement time period.

This ignores any asymmetry error which may be caused by a DC offset or the comparator positive output feedback not biased properly. The phase shift and the edge jitter is also proportional to 81% of the % Npp/Vpp inverse SNR ratio for levels below the 20% range roughly.

e.g. Consider Noise is 10% in pp ratios then each edge will have jitter of 8.1% of T/2

Answer 4

This answer ----- The amplitude of the phase noise (in µs) is simply the noise voltage divided by the slope ----- is from Dave Tweed. Or $$TimeJitter = Vnoise / SlewRate$$

is the form I've used for over 2 decades.

I worked at a walkie-talkie company, who had converted from tiny 50_ohm RF modules to integrated circuits. Much less power demand, much longer battery life. But the close-in phase noise preventing shipping the product, because the transmitter would de-sensitize any nearby receiver; they needed a phasenoise level of -150dbc/rtHz and had no idea how to fix their problem. Line DOWN. No shipping. Using the above formula, and making assumptions about their frequency synthesizer's prescaler and the rbb' of the prescaler bipolar current-steering devices, we predicted the total Rnoise of the prescaler had to be less than 6,000 ohms. We were selectively burning power, only where the math/physics predicts power must be burned.

In ONNN Semi PECL, using Bandwidth of 10GegaHertz and Rnoise of 60 Ohm (1nV/rtHz), with Slewrate of 0.8v/40picoseconds, the TimeJitter is Vnoise = 1nV * sqrt(10^10) = 1nV * 10^5 = 100 microVolts RMS. SlewRate is 20 volts/nanosecond. The TimeJitter is 100uV RMS / (20v/nS) = 5 * 10^-6 * 10^-9 = 5 * 10^-15 seconds RMS.

What is the spectral density of the jitter? We simply scale down by the sqrt(BW) which is 10^5, yielding 5 * 10^-20 seconds/rtHz.

For your question: 1MHz, 1voltPeak, 20dB SNR and Tj = Vnoise/SR, we have Vnoise = 1V/10 = 0.1vRMS (ignoring any sin-peak-rms ratios) SlewRate = 6.3 Million volts/second, therefor TimeJitter = 0.1v/6.3Mega v/Sec = 0.1 * 0.16e-6 = 0.016e-6 = 16 nanoSeconds RMS.

EDIT/ENHANCE: converting a sin into a squarewave. One of the most risky of these is converting a CrystalOscillator sin into a rail-rail squarewave. Any casualness, or unawareness of the hidden trash generators, results in the typical jittery microcontroller clock. Unless the entire signal chain, from XTAL interface thru amplifiers and squarers and clock-distribution are provided private power rails, you end up with apparently random clock-timing upsets but not random at all, instead dependent on VDD collapses triggered by program-related energy demands. All of the circuits that touch, or bias any circuit that touches, the clock edge, should be analyzed using

$$Tjitter = Vnoise/SlewRate$$

The ESD structures are a problem. Why allow 3pF capacitors (the ESD diodes) to couple MCU-program-related energy-demand events into the clean sin from the CRYSTAL? Use private VDD/GND. And design the substrate and wells for charge control. To cross from XTAL domain into MCU domain, use differential current steering with a 3rd wire to pass along the expected trip points.

How serious is this? Consider typical MCU ringing to be 0.5 voltsPP. Running that into a 3pF ESD and then into a 27pF Cpi, we get a 10:1 reduction (ignoring any inductance), or 0.05 voltPP imposed atop the 2voltPP crystal sin. At 10MHz sin, the SlewRate --- d(1*sin(1e+7 * 2pi*t))/dt --- is 63MegaVolts / second. Our Vnoise is 0.05. The jitter right at that point in time is

Tj = Vn/SR = 0.05 volt / 63e+6 volt/sec == 0.05 / 0.063e+9 ~~ 1 nanosecond Tj.

What if you use a PLL to multiply that 10MHz up to 400MHz for MCU clock? Assume the divide-by-400 FlipFlops (8 of them) have 10Kohm Rnoise, with 50 picosecond edges over 2 volts. Assume the FFs have 1/(2*50pS) = 10GHz bandwidth.

Random noise density FF is 12nanoVolts/rtHz (4nv * sqrt(10Kohm/1Kohm)). Total integrated noise is sqrt(BW) * 12nV = sqrt(10^10Hz) * 12nV = 10^5 * 1.2e-9 == 1.2e-4 = 120 microVolts rms per FF. 8FF are sqrt(8) larger. We'll assume some gate noise, and make the factor sqrt(9): 120uV*3 == 360uVrms.

SlewRate is 25 picoseconds/volt or 40Billion Volts/second.

Tj = Vn/SR = 0.36milliVolts/40Billion volts/second = 0.36e-3/0.04e+12 = 9e-15 seconds Tj.

Seems rather clean, right? Except the FlipFlips have ZERO ability to reject VDD trash. And substrate trash is looking for a home.

Answer 5

As an advice, you could reduce the noise by adding a low-pass filter to your design before going into the comparator. This would cut-off the higher frequencies of your signal which is the noise in this case.

To calculate the frequency of the phase noise, you can use FFT or perform a spectrum analysis of the signal. A frequency spectrum would give you the frequency of your signal plus the frequency of the un-wanted noise.

The frequency spectrum of a time-domain signal is a representation of that signal in the frequency domain. The frequency spectrum can be generated via a Fourier transform of the signal, and the resulting values are usually presented as amplitude and phase, both plotted versus frequency.

Derive an equation for the signal you are getting, and perform a Fourier transform to get the amplitude and phase plotted against frequency.

Answer 6

Given the spectral density of the noise on the input signal, how can I calculate the phase noise of the square wave?

This is just a thought on how to possibly get to a value...

I think I'd be tempted to use a PLL (phase locked loop) to generate a squarewave from its VCO that tracks the basic fundamental signal. Your schmitt comparator is a good start and could feed a PLL nicely. The output from the PLL's phase comparator would need to be highly low-pass filtered so that the control voltage to the PLL's VCO would be very smooth and cause minimal jitter on the VCO.

The raw output from the phase comparator would be a very good measure of the phase noise. If there were no phase noise, that output would be very regular.

Anyway, it's just a thought.