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Sweeping filters shows huge ripple

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neazoi

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Hi, I am sweeping across different IF filters with this circuit and all of my multi-pole crystal, ceramic, or mechanical filters show a huge passband ripple. I tested several of them:
  1. two 455KHz crystal filters, gave huge passband ripple.
  2. two 455KHz ceramic filters multi-pole, gave huge passband ripple.
  3. two 455KHz mechanical filters multi-pole, gave huge passband ripple.
  4. a single 455KHz ceramic resonator, gave no ripple.
  5. a single 3.6MHz crystal gave no ripple
  6. a single 7.1MHz ceramic resonator gave no ripple.

The ripple is always there, no matter if I reduce the power applied to the filter or if I load the detector input to 50 ohms.
Why is that happening on some filters and on some other not?
 

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Hi, I am sweeping across different IF filters with this circuit and all of my multi-pole crystal, ceramic, or mechanical filters show a huge passband ripple. I tested several of them:
  1. two 455KHz crystal filters, gave huge passband ripple.
  2. two 455KHz ceramic filters multi-pole, gave huge passband ripple.
  3. two 455KHz mechanical filters multi-pole, gave huge passband ripple.
  4. a single 455KHz ceramic resonator, gave no ripple.
  5. a single 3.6MHz crystal gave no ripple
  6. a single 7.1MHz ceramic resonator gave no ripple.

The ripple is always there, no matter if I reduce the power applied to the filter or if I load the detector input to 50 ohms.
Why is that happening on some filters and on some other not?
you show a screen capture of something with no scale on it. What are to make of that? Further, your schematic, with that inexplicable diode, is also a filter.
 

Consider that filters need to be operated with specified source and load impedance, otherwise the transfer curve might be quite different.
 

    neazoi

    Points: 2
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Consider that filters need to be operated with specified source and load impedance, otherwise the transfer curve might be quite different.
I suspected that as a problem too.
It might be that I am driving the input of the filter from a 50R generator and I get the output from a 100K or so impedance of the diode detector.
To verify it, I also loaded the detector input (filter output) with a 50R shunt resistor. Things did not change as far as concern the huge ripple in the passband, only response amplitude was reduced when loaded to 50R, as expected.

It is weird that the LC filters, ceramic resonators and single crystals I tried, do not have this behaviour.
But all the multi-pole crystal or ceramic filters I tried, did show a huge ripple in the passband.

To answer to barry, this is a storage CRT, with the curve and flat-line markers drawn manually by the sweeper. The top flat line is at 0dbm and the rest at -3, -6, -10dbm.
 

According to your acreenshot, the ripple is about 1 dB. That's not huge, designed passband ripple of the filter may be in this range.

The designed impedance of a 455 kHz filter is most likely higher than 50 ohm. You should have a datasheet or specification.

It is weird that the LC filters, ceramic resonators and single crystals I tried, do not have this behaviour.
Just expectable. A single LC or mechanical resonator has one transfer function maximum and thus no ripple.
 

What spectrum analyzer plugin are you using ? What speed is it running at ?

Ceramic filters more typically K area for impedance.


Passband ripple a couple of db.


Regards, Dana.
I am not using a spectrum analyzer.
I am using a storage CRT vacuum tube oscilloscope and a sweeper. I sweep very slowly through the filter and after that filter, I detect the signal with a peak envelope detector (the circuit at post #1).

So mystery solved. It is indeed an impedance mismatch on both input and output of the filter.
The attached image shows the same filter (different sweeper level signals applied to compensate for the loss and show the curves near the same levels) when more properly terminated and when non-properly terminated.

For the termination I used a series input 10k pot and a shunt output 10k pot. I applied these at the DUT ports.

I believe that the shunt pot at the output worked, because it brought down the ~120k input impedance of the detector to lower levels needed by the filter.
I believe that the series pot at the input worked because it increased the 50R generator impedance to higher levels needed by the filter.

I wanted to have a way to draw the horizontal reference lines on the screen (DUT bypassed, and vary the sweeper level). But when switching the DUT out of the circuit, the impedance will change and so these lines won't be at the right levels on the screen.
Any ideas how to cope with this problem?
--- Updated ---

Maybe this circuit with variable input and output impedance will be able to be used to set these reference lines when the DUT is bypassed?
The idea is that once the pots are set with the DUT (filter) in place, then the DUT is bypassed but the detector still sees the same impedance, and the generator impedance is the same as before.
So impedance is roughly kept the same with the DUT in place and with it bypassed.
Does it make sense or is this nonsense?
 

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Switch in equivalent Z's for Zin and Zout. Or since you have a storage scope
just add a sweep with a cal level to add the trace to the scope.

Can you get a spice model for the filter(s) ? So that you could approximate
the complex Zin, Zout and use that to create the dummy Z's.

Lastly dont forget Nanovna come pretty cheap .....




Regards, Dana.
 

    neazoi

    Points: 2
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Switch in equivalent Z's for Zin and Zout. Or since you have a storage scope
just add a sweep with a cal level to add the trace to the scope.

Can you get a spice model for the filter(s) ? So that you could approximate
the complex Zin, Zout and use that to create the dummy Z's.

Lastly dont forget Nanovna come pretty cheap .....




Regards, Dana.
Thanks Dana,
It is not a particular filter, it is a more general unknown specs filter (and amplifier blocks) response curve setup.

"Switch in equivalent Z's for Zin and Zout. Or since you have a storage scope
just add a sweep with a cal level to add the trace to the scope."

Exactly these two are what I am trying to do. Add a sweep with the filter bypassed and with a cal level to create the horizontal reference level line to the CRT.
BUT, to do this, you need to make sure that the signal level previously applied to the input of the filter, is the same as that applied to the detector when the filter is bypassed.
ALSO, you need to make sure that the detector is loaded to the same impedance when the filter is in place and when it is bypassed.

So I was thinking of this circuit, which keeps the input and output potentiometers at the same setting when the filter is in place or when it is bypassed. This is what you mentioned "Switch in equivalent Z's for Zin and Zout", which also affects the input output levels, apart from impedance (since it uses resistors in my case).

Does this circuit seem more or less right to you, or my thought is completely wrong?

Again, I am trying to make meaningful rough measurements and avoid huge measurement setup mistakes. I am not opting for extreme precision. And of course, I ignore reflections back to the generator, due to impedance mismatch. I case more of what is happening after that first potentiometer.
 

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Ideally you want DUT driven by a V source, and measurement circuit load
to not affect design output circuit load and Z.

Your circuit changes load effects on input pot hence drive side.

Of course you could always buffer the drive with a fast OpAmp follower, terminate
filter with typical load, buffer output, then Zdrive unaffected & output loading
of measurement circuit would not matter.

You can do the measurements with typical in circuit loads and Zin presentation,
or choose to evaluate with V driven input (read Zin of filter does not affect drive)
and unloaded output.

Because of freqs involved per your original post I would be inclined to use Manhattan
style prototype, single sided PCB (so no ground plane C) :

1696937876966.png


1696937975902.png







Regards, Dana.
 
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The high resonant peak filters are all undamped because you put the damping load after the diode rather than the filter output before the diode.

For BPF design,
 
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