QA402 96k and 192 k sampling

I bought the QA402 and QA471 recently (QA401 unavailable for new customer) to, in term, be able to measure and save impulse response files of loudspeaker.

I understand that the QA402 software is a brand new rewriting and that many features available in QA401 are not yet implemented.

For the time being, I recently played with chirp and freq response at both 96 & 192 k and I’m a bit surprised by the results. Below are response at both 96 and 192 with same settings. I can see that at the low end, the cut-off freq is higher at 192 k, and at the high end, we have the exact same response cut off at 96 & 192k (look at 40kHz for ex).

I’m really wondering why ?
Best regards from France

Hi @jcg, in loopback there are 4 factors at play:

  1. ADC filter (digital)
  2. ADC anti-alias filter (analog)
  3. DAC filter (digital)
  4. DAC anti-alias filter (analog)

In loopback, all three combine to give the composite response you see.

Now, you can use the FR right channel as reference mode. This will compare the difference between the left and right inputs and graph that, and that effectively take the ADC and DAC response out of the equation.

For example, tick the box below:

And then run L+ OUT through a BNC splitter and into L+ IN and into R+ IN. In this example, the L and R inputs are seeing the exact same signal (since we split it). And look at the response we get (it’s very flat at all sample rates):

What is left is the difference in analog antialias filters.

Now, switch to time domain. In the plot below, you can see a bit of “wobble” (for lack of better word) as the lowfreq sine wave gets pushed through the input caps at “A”. And you can see at higher frequencies, things roll off sharply at “B” (the very thing you noted seeing).

But if you click and drag a box over the time domain trace to really zoom in, you’ll see that while the left channel rolls off at B, so does the right channel by the same amount. And thus the response is shown as flat (that is since A and B are the same, A/B = 1 at every point)

So, if you really want to get a deep understanding of a circuit that will be operating in regions that might have impact from the ADC/DAC filters, then switch to right channel reference: Split L+ out. Run one path into your DUT, and the output of the DUT into L+ in. And then a split version of L+ out into R+ in and use “right channel reference” mode and you should be able to see true DUT response all the way up to Nyquist.

Hi Matt,
Thanks for the answer. First, let me say that English is by far not my native language nor am I a professional in electronics or electro acoustics. So maybe my understanding is all wrong but:

Of course, if you use the left generator and right ADC as a reference loopback, what remains only is the difference between L and R ADC digital and anti aliasing filters.

What I was trying to figure is why in the otherwise same conditions (chirp, level, FFT size…) do I have a higher high-pass cutoff freq at 192 ks (A region) and the SAME lowpass cutoff freq (B region) ?

It does look to me that in the B region there is a fixed low pass filter (at least at 96 & 192k) which does not depend on sampling rate.

Hi @jcg, yes there is a second order anti-alias filter that is common at all sample rates.

Hi Matt,

So ok for the fixed second order anti aliasing low pass filter common to all sample rates
Now for the A region, why is the cut-off freq is almost the double for 192 ks compared to 96 k ?

Hi @jcg, if you are using a fixed sweep time, the cutoffs will be very similar. By fixed sweep time, I mean that a 64K FFT at 48K has a duration of 1.37 seconds. A 256K FFT at 192K has a duration of 1.37 seconds. The lower-end 3dB point at 48K/64K looks to be 1.2 Hz, the lower-end 3 dB point at 192/256 looks to be 4 Hz. Pretty close.

But in general I’d not rely on chirps at band edges to give you a definitive answer. If you absolutely positively need to understand the response of something at the band edges, use stepped sines.

Hi Matt,
Hope that it will not take too long to have the functionalities of the previous QA401 software. To me is the ability to automate impulse response of drivers (via chirp) synchronized to my diy turntable via something like wifi stepper motor controller…

Hi @jcg,

The ability to play an exported chirp from a thumb drive on your external source (that doesn’t have analog inputs) should be going in to this week’s release (0.996). Do you plan to play the chirp from a record?? This project sounds interesting for sure

Hi Matt,

I didn’t express myself properly. My goal is to use the QA402 to get impulse response of drivers mainly at 96ks (so play a chirp, get the impulse response, save the file with a name like “driver_hor_0”, send a command to the turntable to move by 5 or 10°, get the new impulse response (“driver_hor_5°”) and so on.