Filter IC sample rate vs. cutoff frequency

Hi Ole,

Think of it in terms of input and output to the black box (the LTC1064 - not really a black box but let's assume for now).

The input has to be a signal with no frequency components beyond half of the clock frequency (in practice this means a cut-off of less than half of the clock frequency). Your input filter (if you choose to have one) characteristic will be dependent on your chosen filter topology and component tolerance.

The output will be a signal with a (perhaps) sharper cut-off at some frequency (a small fraction basically, e.g. 1%, far below half of the clock frequency), but with a better defined characteristic and controllable, i.e. you can adjust it without swapping out Rs and Cs and Ls.

Are you sure you need this? If this is an audio solution, be aware this is not usually hi-fi (but been a while since I looked at switched cap filters). Good for speech processing or effects though, or sensor data processing.

I'm not sure I follow you. Most of the LTC1064 and friends have just one parameter you can adjust: the clock frequency. They don't depend on external resistors, capacitors, or inductors. By supplying a clock to the ICs, the ICs then determine both the cut-off frequency and the sampling frequency by dividing this supplied clock with two fixed ratios: one for the cut-off frequency and one for the sampling frequency. The way I read the datasheets, the sampling frequency of these ICs is fixed at twice that of the cut-off frequency, i.e., the lower Nyquist limit. For example, in the case of the 1164-6, the clock is divided by 100 to determine the cut-off frequency, and the sampling frequency is determined by dividing the clock by 50.

As you say, because of the Nyquist sampling theorem, this sampling frequency requires that the input signal have no frequency components above half the sampling frequency. But... since the filter IC's cut-off frequency is always half of the IC's sampling frequency, imposing this requirement on the input signal means there's no need to filter it at all! That's what I don't understand.

(I actually need it just for speech processing, so I'm looking at ICs like these instead of resorting to, say, biquads with op-amps and several components.)

Hi Ole,

Most of the LTC1064 and friends have just one parameter you can adjust: the clock frequency. They don't depend on external resistors, capacitors, or inductors.

That's my point - any external input filter will (typically) be composed of Rs, Ls or Cs. The black box doesn't need any of this, and has a defined characteristic - controlled by the clock as you say.

since the filter IC's cut-off frequency is always half of the IC's sampling frequency, imposing this requirement on the input signal means there's no need to filter it at all! That's what I don't understand.

No, the black box only functions if your input definitely has no components approaching the clock frequency (well, clock frequency minus your cut-off frequency, which in practice means a cut-off frequency below that). Hence the need for an input filter unless your input signal is already guaranteed to have no components that high. If you have components beyond this, then they will fold back with the possibility that your pass-band will now contain undesired content. If you're just doing speech processing as you mention then it is unlikely you'll have components close to the clock frequency, so you probably don't need an input filter. But that is just for your scenario. Other scenarios may need the low-pass filter on the input.

Bear in mind that it may still be advisable to have an input filter if you're chaining these devices, since there may be a bit of clock frequency in the output signal.

Thanks for your attempts to explain it to me. I'm still not sure I understand, though.

No, the black box only functions if your input definitely has no components approaching the clock frequency (well, clock frequency minus your cut-off frequency, which in practice means a cut-off frequency below that). Hence the need for an input filter unless your input signal is already guaranteed to have no components that high. If you have components beyond this, then they will fold back with the possibility that your pass-band will now contain undesired content.

If you're saying that the IC samples at (approximately) the supplied clock frequency, then certainly yes. This allows a trivial low-order low-pass filter to suppress virtually all frequency components this high above the cut-off frequency (and therefore aliasing will be practically eliminated).

But from how I read the datasheet, this isn't what the IC does; instead the IC seems to sample the input signal at a rate of twice the cut-off frequency. So unless I'm misunderstanding the datasheets: if I select a cut-off frequency of 3 kHz, the IC will select an internal sampling frequency of 6 kHz. But this requires the input signal to already have been low-pass filtered with a cut-off frequency of 3 kHz or less--and so the filter IC won't really be filtering anything.

So, basically I can't quite make sense of the datasheet. I'd understand if the datasheet said that if I select a particular cut-off frequency, then any subsequent sampling can be performed at twice that frequency, because that's the Nyquist limit. I can't make this of the datasheet, however.

If you're just doing speech processing as you mention then it is unlikely you'll have components close to the clock frequency, so you probably don't need an input filter. But that is just for your scenario. Other scenarios may need the low-pass filter on the input.

True. Although I'm only interested in the speech portion of the signal, the input signal contains higher audio frequency components than speech so this is in fact an "other scenario" where a low-pass filter is required.

Hi Ole,

if I select a cut-off frequency of 3 kHz, the IC will select an internal sampling frequency of 6 kHz. But this requires the input signal to already have been low-pass filtered with a cut-off frequency of 3 kHz or less--and so the filter IC won't really be filtering anything.

You're confusing filter frequency with clock frequency. As mentioned, your input signal needs to be low-pass filtered not to exceed clock frequency (well, less than that as mentioned).

The LT... device will filter the signal with a cut-off at a very small fraction (e.g. 1% as mentioned) of the clock frequency (this is how a switched cap filter works. It is nothing special about the LT.. device that you mention).

You can also see it in the datasheet, under "modes of operation". You can also see it in the application circuit, on page 13. It is visible from the graph there that this example application has a cut-off of around 100kHz, yet if you look at the circuit it says the clock frequency is 5MHz.

It's the datasheets that I don't understand, not analog or digital filter theory.

From what I can tell from the datasheet, one selects a certain clock frequency (say, 5 MHz). The IC divides this clock by a ratio that becomes the cut-off frequency, and by another ratio that becomes the sampling rate. My confusion is that according to the datasheets the ICs have been configured to divide the clock so that the sampling rate is invatiably twice that of the cut-off frequency (i.e., the lower Nyquist sample rate limit). If that is really the case, then this requires the input signal to already have been filtered to contain no frequency components above the cut-off frequency. But that wouldn't make much sense, and I hope this is not how the ICs really work.

However, if the IC samples at the supplied clock rate (that is, not at a rate that is derived by dividing the clock to twice the cut-off frequency as explained in the previous paragraph), then it makes perfect sense to me, and I suspect this is in fact how the LTC ICs work. I just can't make this from the datasheets.

So the question is: does the IC sample the input signal at the sample rate described in the datasheet (which happens to be derived as twice that of the cut-off frequency), or does the IC in fact sample the signal at a much higher rate?

Hi Ole,

So the question is: does the IC sample the input signal at the sample rate

It can't possibly be sampling at twice the rate of your desired cut-off frequency (this is obvious).

The switch caps are switching at the clock rate - hence your input signal is being sampled at the clock rate. This would apply to all switched cap filters. Nothing to do with this IC specifically.

The wikipedia page for switched capacitor shows this, and shows the formula too - you can see that the transfer function would be dependant on the clock frequency (switching frequency, represented by f on that wikipedia page) and a couple of Cs (the ratio of them). You arrange the clock freq (and the C ratio) to get your cut-off frequency.

Cut-off frequency != sample rate (nor half of it)

and

Cut-off frequency != clock frequency (nor half of it)

Shabaz,

Cut-off frequency != sample rate (nor half of it)

and

Cut-off frequency != clock frequency (nor half of it)

I know that perfectly, and I don't see why you think I've said otherwise. I've actually designed and built analog and digital filters before; it's just the LTC chips that I haven't played with before.

Reading the datasheets for some of the other ICs in the family actually makes me somewhat confident that I must have misunderstood the first datasheet that I read (unfortunately I don't recall which one that was) because they don't discuss any sample rate being twice the cut-off frequency at all.

However, the datasheet I first read did seem to suggest that if I selected the clock speed for a particular cut-off frequency (so that the clock speed is indeed many times higher than the cut-off frequency), somehow the IC would be sampling at a frequency of just twice the cut-off frequency. It was just te latter that didn't make sense to me.

In conclusion, I think I'll just believe the ICs are in fact sampling at a much higher rate than the cut-off frequency, as this is the only thing that makes sense to me.

With respect you yourself said:

But... since the filter IC's cut-off frequency is always half of the IC's sampling frequency

And I specified:

Cut-off frequency != sample rate (nor half of it)

This has nothing to do with the IC - it is a standard way all switched capacitor filters behave. This is not specific to the IC.

In conclusion, I think I'll just believe the ICs are in fact sampling at a much higher rate than the cut-off frequency

Right, exactly that -  in the very first response it was stated that the cut-off frequency would be a tiny fraction of the clock frequency.

The clock frequency is the switching frequency of the integrated circuit, and is the 'f' in the formula on the Wikipedia page, and the LT.. datasheet will

provide the exact divisor for you to determine the cut-off frequency.

Shabaz,

With respect you yourself said:

But... since the filter IC's cut-off frequency is always half of the IC's sampling frequency

That was of course referring to what the data sheet seemed to say about the cut-off frequency and its sampling frequency on this particular IC (hence the reference to the filter IC's behavior). I think I made it quite clear that this would render the IC rather ineffective.

Right, exactly that -  in the very first response it was stated that the cut-off frequency would be a tiny fraction of the clock frequency.

Yes, but that wasn't really my question. If the cut-off frequency is a small fraction of the sample rate, then everything is indeed fine. I was merely asking what the datasheet could possibly mean because it seemed to say otherwise.