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I'm currently going through RF Microelectronics by Razavi. In chapter 3 he presents an example where the following signal is applied to a differential amplifier with a tail current source.

$$Acos(w_ct) + acos(w_c+w_m)t$$

He then decomposed the signal by the following:

$$Acos(w_ct) + acos(w_c+w_m)t=\frac{A}{2}cos(w_ct)+\frac{a}{2}cos(w_c+w_m)t+\frac{a}{2}cos(w_c-w_m)t+\frac{A}{2}cos(w_ct)+\frac{a}{2}cos(w_c+w_m)t-\frac{a}{2}cos(w_c-w_m)t$$

Where the first three terms of the equation represent the AM signal and the last three terms represent the FM signal.

His questions is to explain why the output of the differential contains the following and to assume that the differential pair experiences 'hard limiting' (A is large enough to steer Iss to each side):

$$\frac{A}{2}cos(w_ct)+\frac{a}{2}cos(w_c+w_m)t-\frac{a}{2}cos(w_c-w_m)t$$

Part of the solution he explains that "the one with sidebands of identical signs can be viewed as an AM waveform, which due to hard limiting is suppressed at the output. The spectrum with sidebands with opposite signs can be considered an FM waveform, which emerges at the output intact because hard limiting does not affect the zero crossings of the waveform."

My main confusion is why would hard limiting only suppress the AM signal and not the FM signal? They both have identical amplitudes and frequencies (besides one of the sidebands).

enter image description here

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    \$\begingroup\$ this bit from your question says it all ... FM waveform, which emerges at the output intact because hard limiting does not affect the zero crossings of the waveform. Don't get hung up that the spectra, the frequency domain representation, looks similar, because it only looks similar, it's ignoring phase. Concentrate on the time domain behaviour, that AM changes the amplitudes and not the zero crossings, FM changes the zero crossings but not the amplitudes. \$\endgroup\$ – Neil_UK Sep 18 '17 at 16:33
  • \$\begingroup\$ Yeah, I think that's my main problem. I'm too focused on the frequency domain. Thank you. \$\endgroup\$ – user367640 Sep 18 '17 at 16:43
  • \$\begingroup\$ Quite honestly, I am confused as well. And I am used to analyse distortions. If nothing else helps, you can always do the time domain analysis using Taylor approximation. This should give you the right terms for the frequency domain and you should see any cancellation/blocking in the terms. \$\endgroup\$ – Attila Kinali Sep 18 '17 at 18:40
  • \$\begingroup\$ hmmm, I've never tried doing a Taylor approximation. That sounds difficult and I'm assuming convolution would be involved? \$\endgroup\$ – user367640 Sep 19 '17 at 1:22
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Not an answer, but too big for a comment.

My only confusion is why did he change \$\omega_c \text{ to }\omega_0\$

I would consider only the phase modulation of the peak shift and not the zero crossing. This is not a limiter but some soft limiting is occurring if the A is say a hard peak limiting value. So the peaks are shifted early and late by PM not FM. So there are alias sum and difference frequencies but still linear gain and linear phase modulation above A but in opposite directions +/-cos for upper and lower alias or difference frequency.

Due to limiting A cannot be increased but the linear amp now multiplies phase for peak excursions of "a" in opposite directions. So the its power is now split to 1/2 and in opposite phase to the difference frequency as upper and lower sidebands due to peak phase shifting modulation.

Perhaps I am assuming wrong that the limiter action is a hard limiter at inputs above A.

The transfer function of a full limiter is to induce shifting in zero crossings and here the output still has some linear output of say unity gain.

But a hard limiter above some threshold on the output at an equivalent input amplitude A only shifts the peaks. He has rightly ignored harmonics at 2f, 3f etc , but some energy is lost there, so I would expect sidebands of 1/2a amplitude but not main carrier dropping to 1/2.

Also why is gm and Rd not factoring the ouput levels of A now reduced to A/2 . If it were a hard limiter we can get relative sidebands but not absolute A/2 , a/2 levels out deter ined by input. Perhaps he meant this as a relative level with some gain constant k on all spikes out, then zero-crossing mixer analysis would be done.

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  • \$\begingroup\$ I'm confused on why he is putting emphasis on hard limiting. I thought the whole point of the tail current source was to drive both transistors periodically. Where on one cycle the ID1 = ISS (M2 off) and when Vin1=Vin2 ID1=ID2 = ISS/2, etc. \$\endgroup\$ – user367640 Sep 19 '17 at 1:27
  • \$\begingroup\$ The challenge in tail current is to increase ICMR and CMRR in deep sub-micron, sub-threshold gates with constant gm. Some researchers seems to have solved this at V=+/-0.55 V with 160dB CMRR. In deep submicron ~1V output impedance of tail current source is low , causing tail current to vary as the common mode signal varies resulting in a poor CMRR. A Low Power Low Voltage Rail to Rail Constant gm Differential Amplifier with 150 dB CMRR and Enhanced Frequency Performance. Available from: ResearchGate \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 19 '17 at 1:46
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Examining the sidebands, the phasing is important. AM sidebands must generate amplitude variations only , thus two counter-rotating vectors are the descriptor. The counter-rotation ensures the quadrature energy always cancels.

For PM (phase modulation), the amplitude must be constant and the angle varies, thus 2 vectors rotating in the same direction suffice.

Granted frequency is the derivative of phase (degrees becomes degrees/second), the FM is well modeled as PM for purposes of sideband visualization.

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An AM signal is one in which the amplitude is varying with time. After passing through a hard limiter, the amplitude is constant, not time-varying. Thus, the limiter suppresses AM. This is all the author is saying.

An FM signal is one in which frequency is varying with time. The hard limiter has no effect on instantaneous frequency, so the FM modulation passes through the limiter.

I hope this makes sense. I don't think the author is trying to say anything more complicated than this.

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  • \$\begingroup\$ I don't know... I don't see how it would be suppressed. It's just a differential amplifier with a tail current source. I thought I had the at least the fundamentals of that circuit understood. I guess not. \$\endgroup\$ – user367640 Sep 20 '17 at 2:36
  • \$\begingroup\$ I am only going by the text. No math. A limiter clips the amplitude of the input signal. This clipping operation strips amplitude modulation from the signal for the simple reason that the amplitude is uniform after the clipping operation. Prior to the limiter, the amplitude is a function of time. After, it is a constant. \$\endgroup\$ – mkeith Sep 20 '17 at 2:45

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