One more possibility might be that: If Vin is the my incoming signal, then before reaching to the ideal differentiator, it gets attenuated by the resistive divider (Vin/2 is going into differentiator instead of Vin, which leads to 6 dB loss straightaway!). Means SNR degradation is taking place at the input itself, which is absolutely undesirable in RF system designs.
The gain of this circuit (from the sensor) can never become infinite even if every single element in this circuit was ideal. The transfer function (from the sensor) for the high-pass filter using the op-amp is not the same when the two R' resistors are added of course. The infinite noise issue you mentioned will not happen from the sensor. The highest gain this circuit can ever have (even with everything ideal) is -R/R'. Here I am making an important assumption that the noise is from the sensor only. If you assume the noise is from the resistors, well I guess you could say that part can get amplified towards infinity. If things are truly ideal, as soon as you turn the circuit on the world would end and so would the interview. ;)
Thank you so much for being here, we're big fans of your channel! :) Your answer is spot on! Once you mentioned the "transfer function (from the sensor) for the high-pass filter using the op-amp is not the same" we got a bit excited because most people ignore this during interviews. However, it's not only the gain that may be affected but the overall frequency response as we're introducing an extra pole through that path. We're you able to catch that?!
One possible alternative solution might be this... The left portion of the circuit can be converted into a series voltage source and a resistor using the Thevenin's theorem. Now, the circuit is basically a voltage source, a cap in series with a resistor (Rth = R/2 in this case) and the feedback resistor. Of course, the Op-Amp is still in its place. Now, the circuit essentially boils down to being a practical differentiator circuit (Note that the original circuit is not practically feasible since it contains one zero and no pole hence violating the causality principle) with a valid cutoff frequency. If the signal band around 1 GHz doesn't lie beyond the cutoff frequency of the practical differentiator circuit, then the circuit would introduce amplitude distortions in the signal i.e., for different frequency component of the signal, the output would be scaled differently in the frequency spectrum. Of course, this can be mitigated using an equalizer but it would only add to the cost of the design.
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I love your channel!! So nice to find a little community of IC engineers on TH-cam
I personally value more these interview question videos, pls put some priority on them, it is truly unique content :) Thank you for all of your work!
One more possibility might be that: If Vin is the my incoming signal, then before reaching to the ideal differentiator, it gets attenuated by the resistive divider (Vin/2 is going into differentiator instead of Vin, which leads to 6 dB loss straightaway!). Means SNR degradation is taking place at the input itself, which is absolutely undesirable in RF system designs.
The gain of this circuit (from the sensor) can never become infinite even if every single element in this circuit was ideal. The transfer function (from the sensor) for the high-pass filter using the op-amp is not the same when the two R' resistors are added of course. The infinite noise issue you mentioned will not happen from the sensor. The highest gain this circuit can ever have (even with everything ideal) is -R/R'. Here I am making an important assumption that the noise is from the sensor only. If you assume the noise is from the resistors, well I guess you could say that part can get amplified towards infinity. If things are truly ideal, as soon as you turn the circuit on the world would end and so would the interview. ;)
Thank you so much for being here, we're big fans of your channel! :) Your answer is spot on! Once you mentioned the "transfer function (from the sensor) for the high-pass filter using the op-amp is not the same" we got a bit excited because most people ignore this during interviews. However, it's not only the gain that may be affected but the overall frequency response as we're introducing an extra pole through that path. We're you able to catch that?!
One possible alternative solution might be this...
The left portion of the circuit can be converted into a series voltage source and a resistor using the Thevenin's theorem. Now, the circuit is basically a voltage source, a cap in series with a resistor (Rth = R/2 in this case) and the feedback resistor. Of course, the Op-Amp is still in its place.
Now, the circuit essentially boils down to being a practical differentiator circuit (Note that the original circuit is not practically feasible since it contains one zero and no pole hence violating the causality principle) with a valid cutoff frequency. If the signal band around 1 GHz doesn't lie beyond the cutoff frequency of the practical differentiator circuit, then the circuit would introduce amplitude distortions in the signal i.e., for different frequency component of the signal, the output would be scaled differently in the frequency spectrum. Of course, this can be mitigated using an equalizer but it would only add to the cost of the design.