5
$\begingroup$

I am using a superconducting $RLC$-circuit for detecting weak electrical signals. The center-frequency is around 650 kHz, and the $Q$-value approximately 50k. I am trying to better understand my signal background. Without the signal source, the electrical schematic be simplified to this:

parallel RLC circuit, amplifier on hot side

Note that there is no input to this circuit: We are looking at the Johnson-noise of the detection circuit in combination with the noise-floor of the amplifier's input stage.

Ideally, the power spectrum of the Johnson noise is given by $|Z_{RLC}|^2$, which can be shown to be a Lorentzian:

$$ \text{noise in dB} = 10 \log_{10}\left( 10^{A/10} \frac{\left(\frac{\nu_0}{2 Q}\right)^2}{\left(\nu-\nu_0\right)^2 + \left(\frac{\nu_0}{2 Q}\right)^2} + 10^{b/10}\right) \quad .$$

In this form, the amplitude $A$ and the background $b$ are given in dB. The center of the resonance is given by $\nu_0$, and the width given by the $Q$-value with $Q = \nu_0/\text{FWHM}$.

In reality, the resonance is almost a Lorentzian, but is lopsided. This becomes especially obvious when using a Lorentzian as a fit-function:

actual data showing lopsided Lorentzian

I checked a bit further away from the resonance to make sure that the transfer function of my amplifier is reasonably flat. Therefore, I don't think it can explain the lopsidedness. I suspect the lopsidedness is caused by an additional resonance, for example some by parasitic $R$, $L$ or $C$ somewhere in the system, like the series/parallel resonances in a Quartz oscillator. But I cannot find a reasonable model for that. What kind of parasitic component could cause this lopsidedness?

Possibly relevant: The amplifier is not directly connected to the "hot end" of the resonator, but to a 1:10 tap close to the "cold end" (ground).

$\endgroup$
  • $\begingroup$ I agree that your explanation (another resonance) is a good one to investigate - it was the thing that sprang to my mind on seeing the plot before I read your last paragraph. You've checked the amplifier presumably with a signal generator - but have you checked that the amplifier's input impedance is also flat? These are fairly low frequencies, so matching is unlikely to be too important. $\endgroup$ – WetSavannaAnimal Jun 14 '17 at 0:23
  • $\begingroup$ That looks a lot like another resonance. $\endgroup$ – DanielSank Jun 14 '17 at 2:48
  • 1
    $\begingroup$ This kind of lopside can come from a mismatch of impedance somewhere (typically the amplifier). empirically, adding a parameter of asymetrie in the numerator (v/2Q -> v/2Q + u in the numerator) should make the model more fitting $\endgroup$ – sailx Jun 14 '17 at 7:26
  • $\begingroup$ This is a good question, I hope the site gets more like this. $\endgroup$ – Emilio Pisanty Jun 14 '17 at 7:31
4
$\begingroup$

If you measured this with a frequency sweep excitation, There will be some lopsidedness if the sweep rate is too high relative to the Q of the circuit. The "undershoot and rebound" in amplitude above the resonance is what sweeping too fast usually looks like, assuming the sweep was increasing the frequency. If the sweep rate is much too fast, you can get several "bounces" in the response before it settles onto the true response of the system.

ISO-7626 gives standard values for sweep rate for mechanical vibration testing. The same principle applies to electronic circuits.

A typical order of magnitude for the maximum sweep rate to get acceptable results is $f^2/Q^2$ Hz/sec where $f$ is the resonance frequency in Hz. A simple check is to compare two frequency sweeps with the frequency increasing and decreasing through the resonance.

For mechanical vibration with low frequency resonances (e.g. 10 - 100 Hz) and high Q values, an accurate frequency sweep test can take several minutes, but with frequencies of 100 kHz - 1 MHz the test time will be too short to be an issue.

$\endgroup$
  • $\begingroup$ That's a good hypothesis, but the data was taken by mixing the signal down to 15kHz and sampling it with 40kHz for 4 seconds, then doing an FFT. The plot shows the (power) average of maybe 10 FFTs. Sampling, FFT and averaging is done by an SRS SR1 audio analyzer. $\endgroup$ – Martin J.H. Jun 14 '17 at 4:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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