[sdiy] Passive filters and impedances

[Tom Wiltshire]

No, obviously I can just simulate in LTSpice if I want particular values for a particular situation. But I don't learn much by so doing.

The intention here is to learn something about the how transfer functions can be derived from the actual parts in the circuit by trying it out on a few examples where I already know what the transfer function and results are supposed to look like. I could have chosen something else, but then I'd have no clue what the answer was!
So, yeah - it's an EE student exercise. I'm in EE student mode.

I've used the "x10 values" trick for actual passive filters I've built, and that's all good. It's been interesting to see from what I have got working (the single pole case) the way that changing the impedance at the output does affect the filter cutoff and level.

I agree that for most practical filtering applications, you want an op-amp and a "proper" filter. That said, I can think of several applications for cheap, lossy filters for slow control signals. Using the PWM output on a 5V uP to control the 3.3V ADC inputs on the Spin FV-1 is one I saw discussed on a forum recently (DIYStompboxes.com, I think). In that case, the lossy nature of the passive filter becomes a benefit.

Thanks,
Tom


On 31 Oct 2015, at 17:07, "Richie Burnett" <rburnett at richieburnett.co.uk> wrote:

> Do you actually *need* to know the transfer functions Tom, or can you just simulate in Spice?  They are complicated and messy to work out because of the series & parallel combinations of complex impedances, although you're doing it right starting at the output side and combining the impedances until you get back to the input.  The final equation reduces down to the potential divider equation with R1 as the top impedance in the divider, and the series/parallel combination of all the other complex impedances in the network as the bottom impedance in the potential divider.  This is the sort of problem that EE students are made to solve because it tests their maths skills but never ever use in practice!!!
> 
> In general we don't usually build filters like this because the lack of buffering causes loading effects that move the filter poles.  However, if you just want to make a quick and dirty 2nd or 3rd order filter without op-amp buffers, a good way to reduce the un-wanted interactions is to make the impedance of each RC filter stage about 10 times greater than the impedance of the preceding stage.  It doesn't stop the interaction, but it reduces it considerably.  The downside is that the output impedance of the filter will probably be quite high, so you'll likely need a buffer to follow it at least.
> 
> It might be okay at something like a cheap PWM reconstruction filter for generating a slowly varying control signal.  But for most other practical filtering applications I can think of you'd probably want something that's a bit more optimal.
> 
> -Richie,
> 
> 
> -----Original Message----- From: Tom Wiltshire
> Sent: Saturday, October 31, 2015 4:24 PM
> To: synthdiy diy
> Subject: [sdiy] Passive filters and impedances
> 
> Hi All,
> 
> I'm back to learning about impedances again. I've been looking at the following situations:
> 
> http://electricdruid.net/images/PassiveFilters.png
> 
> This shows a first order, second order, and third order passive filter, each feeding a destination which has its own impedance Rimp. Rimp might be infinity, in which case we could ignore it, or it might not.
> 
> For the first order case, I've been able to derive the transfer function by thinking of it as a potential divider with impedances. Ignoring Rimp for a moment, we have:
> 
> Vout / Vin = (1 / sC) / (R1 + (1/sC))
> 
> This works fine. Including Rimp doesn't make it much more complicated, except that we replace  1/sC with (1/sC || Rimp)
> 
> However, I can't make the second order (or by extension third order) case work the same way. It still looks like a potential divider, with the output across C2, but the total impedance is a bit more involved. I think we have to work backwards in from the end, so we have the output across C2. We have R2 in series with C2, which is then in parallel with C1, and the result of that is in series with R1. So the total impedance is something like:
> 
> Ztotal = R1+ (1/sC1 || (R2 + 1/sC2)) (ignoring Rimp again)
> 
> So my transfer function is:
> 
> Vout / Vin = 1/sC2  /  (R1+ (1/sC1 || (R2 + 1/sC2)))
> 
> Is this correct? Am I on the right lines with this "potential divider" understanding of the situation?
> 
> Thanks,
> Tom
> 
> 
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